Q
stringlengths 18
13.7k
| A
stringlengths 1
16.1k
| meta
dict |
|---|---|---|
Why can we see the cosmic microwave background radiation? This radiation (CMBR) is said to have its origin at the surface of last scattering that exposed itself when the big bang universe had expanded for less than a million years.
In order to see radiation from a source, one has to be on its future light cone. In a universe that is flat and open, which our Universe is asserted to be at the large scale, we are not on the future light cone of this radiation, but almost maximally remote from it. One can also say that the surface of last scattering is not on our own past light cone.
How is this visibility to be understood within standard big bang cosmology?
(This question is different from an earlier one with the same wording.)
| I get the feeling you think the CMBR is the flash from the explosion that was the Big Bang - like when Ripley blows up the Nostromo at the end of Alien.
If so, that's not really it. Rather, the universe was full of this radiation (heading in all directions) as it expanded. At our random position, we are now seeing photons that have been travelling for 13.7 billion years from a distant point in the universe. When they set out from that point, we were much closer (and they had a shorter wavelength), but as space expanded they had further to go to get to us.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/430581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
What happens to a radioactive material's atom when it disintegrates? Suppose you initial had radioactive $2^n$ atoms (where $n$ is an integer). Now after a number of halflives the number of left out atoms becomes 1. Now what will happen to it will it disintegrate and the leftover would be half an atom? Now if the reaction stops then the statement "The decaying radioactive atom would never end" then it'll be wrong.
| Radioactive decay is a stochastic process. This means that there is random chance involved, so the exponential model used to represent radioactive does not say exactly how many atoms of the original substance will be left at a given time, rather it tells you the expected value of atoms remaining. If you begin with n=1 atom, after some time the exponential model gives you n=0.5. This does not mean there are 0.5 atoms remaining, it rather means that there is a 0.5 chance that the atom has not decayed yet.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/430764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Work done by friction on a body which is rolling on an inclined plane Why is the work done by friction zero during translational motion but nonzero when the body is rolling on an inclined plane?
| During pure rolling, at any instant of time, the point of contact between the roller and the ground will act as an instantaneous centre(the entire roller appears to rotate about that point at that instant).There is no sliding between the roller and the ground against friction.So the work done by friction is zero during pure rolling.But during sliding, work done by friction is not zero.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/431037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Once a black hole is formed, is there anything other than Hawking radiation which shortens its life? Hawking radiation is supposed to very slowly evaporate a black hole (terms and conditions apply :] ).
Apart from Hawking radiation, is there any mechanism or effect that can make a black hole cease to exist? Or once they are formed are they expected to exist in this form "forever"?
| Just FYI, you probably want to wait longer before you just answer your own question like that...
Anyway, black holes can merge thereby forming a new black hole - the two from the beginning no longer exist! Without appealing to some hand wavy argument about how energy is extracted from a black hole, instead black hole binaries lose energy by emitting gravitational radiation. So far 6 black hole binaries have been observed by aLIGO emitting gravitaional radiation as they inspiral and merge, destroying each other to give birth to a new black hole.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/431175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
"Iron Core" in Inductive Charigng Inductive charging used for wireless charging often faces the hindrance of being too short ranged for many cases. There appear to be some workarounds such as using a capacitor to resonate them at the same resonant frequency. Please excuse the naiviety of the question, but when looking back at the humble solenoid, a simple iron core can drastically boost it's magnetic field strength. So, why not just stick an iron core into the middle of the inductive charging coils?
|
So, why not just stick an iron core into the middle of the inductive
charging coils?
From the energy and field strength perspective, at high frequencies, used in wireless charging, magnetic cores are not required and, at higher end frequencies, just would not work.
From the coupling perspective, cores would not be very effective, considering that wireless charging involves two coils residing in two different physical devices separated in space, i.e., they would not be able to share the same core.
For resonant charging, where high Q is essential, the losses associated with cores (if they could work at all at those frequencies) would be counterproductive.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/431282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Could you have sand pipes like water pipes? It's common knowledge that sand behaves like water when in small grains. So can you make a pipe that carries sand in the same way pipes carry water? If not, is there another way you could?
| Yes! If you go back to see how the old steam locomotives were built, one of the domes on the steam engine held sand that flowed down through a pipe were it ended near the tracks in front of the front drive wheel. The sand was added to provide traction between the steel wheel and the steel track. There was also a washer that washed the sand from the tracks just after of the rear drive wheel. For the sand to flow, it needed to be completely dry. The railroads had a house, not surprisingly called the sandhouse, where the sand was dried by heating it using a coal stove.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/431485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 0
} |
Angular velocity by velocities of 3 particles of the solid Velocities of 3 particles of the solid, which don't lie on a single straight line, $V_1, V_2, V_3$ are given (as vector-functions). Radius-vectors $r_1, r_2$ from third particle to first and second are given aswell. How could I find the angular velocity $w$ of the solid?
I tried to solve this problem using Euler's theorem : $V_2=V_3+[w \times r_2]$, $V_1=V_3+[w \times r_1]$.
After this step I tried to consider different cases: if $V_1 $ is not collinear to $V_2$ we could write $w = k*[(V_2-V_3) \times (V_1-V_3)]$. However, it doesn't really help. The second case is even more difficult to analyze.
Second attempt consisted in solving this system by multiplication (scalar product or vector work) equations by appropriate vectors. However, I didn't really succeed.
| The algebra is not especially nice, but it is just algebra. This is rigid body rotation, taking point 3 as the origin of coordinates,
so effectively
$$\mathbf{r}_1=\mathbf{R}_1-\mathbf{R}_3, \qquad
\mathbf{r}_2=\mathbf{R}_2-\mathbf{R}_3.
$$
We start as you suggested, and abbreviate
$$
\mathbf{v}_1=\mathbf{V}_1-\mathbf{V}_3, \qquad
\mathbf{v}_2=\mathbf{V}_2-\mathbf{V}_3,
$$
so that
$$
\mathbf{v}_1 = \boldsymbol{\omega}\times\mathbf{r}_1, \qquad
\mathbf{v}_2 = \boldsymbol{\omega}\times\mathbf{r}_2.
$$
Now since the three points are not collinear, we can let
$$
\boldsymbol{\omega} = a\,\mathbf{r}_1 + b\,\mathbf{r}_2
+ c\, \mathbf{r}_1\times\mathbf{r}_2
$$
but we must remember that $\mathbf{r}_1$ and $\mathbf{r}_2$ will not in general be orthogonal.
We can obtain $c$ directly,
from either of the two equivalent equations
\begin{align*}
\mathbf{r}_2\cdot\mathbf{v}_1
&= \mathbf{r}_2\cdot\boldsymbol{\omega}\times\mathbf{r}_1
= \boldsymbol{\omega}\cdot\mathbf{r}_1\times\mathbf{r}_2
= c |\mathbf{r}_1\times\mathbf{r}_2|^2
\\
\mathbf{r}_1\cdot\mathbf{v}_2
&= \mathbf{r}_1\cdot\boldsymbol{\omega}\times\mathbf{r}_2
= -\boldsymbol{\omega}\cdot\mathbf{r}_1\times\mathbf{r}_2
= -c |\mathbf{r}_1\times\mathbf{r}_2|^2
\\
\Rightarrow\quad c&=
\frac{\mathbf{r}_2\cdot\mathbf{v}_1}{|\mathbf{r}_1\times\mathbf{r}_2|^2}
=
-\frac{\mathbf{r}_1\cdot\mathbf{v}_2}{|\mathbf{r}_1\times\mathbf{r}_2|^2}
\end{align*}
where we took advantage of the properties of the
scalar triple product.
The other coefficients come from scalar products with $\mathbf{r}_1\times\mathbf{r}_2$.
We use the general identity
$$
(\mathbf{A}\times\mathbf{B})\cdot(\mathbf{C}\times\mathbf{D})
=
(\mathbf{A}\cdot\mathbf{C})\,(\mathbf{B}\cdot\mathbf{D}) -
(\mathbf{B}\cdot\mathbf{C})\,(\mathbf{A}\cdot\mathbf{D})
$$
and a special case of this, which we use, is
$|\mathbf{r}_1\times\mathbf{r}_2|^2=|\mathbf{r}_1|^2|\mathbf{r}_2|^2-(\mathbf{r}_1\cdot\mathbf{r}_2)^2$.
\begin{align*}
\mathbf{r}_1\times\mathbf{r}_2 \cdot \mathbf{v}_2
&=
(\mathbf{r}_1\times\mathbf{r}_2 ) \cdot (\boldsymbol{\omega}\times\mathbf{r}_2)
\\
&=
\left( a|\mathbf{r}_1|^2 + b(\mathbf{r}_1\cdot\mathbf{r}_2) \right)\,
|\mathbf{r}_2|^2-
\left( a(\mathbf{r}_1\cdot\mathbf{r}_2) + b|\mathbf{r}_2|^2 \right)\,
(\mathbf{r}_1\cdot\mathbf{r}_2)
\\
&= a |\mathbf{r}_1\times\mathbf{r}_2|^2
\\
\Rightarrow\quad
a&=\frac{\mathbf{r}_1\times\mathbf{r}_2 \cdot \mathbf{v}_2 }{|\mathbf{r}_1\times\mathbf{r}_2|^2}
\\
\mathbf{r}_1\times\mathbf{r}_2 \cdot \mathbf{v}_1
&=
(\mathbf{r}_1\times\mathbf{r}_2 ) \cdot (\boldsymbol{\omega}\times\mathbf{r}_1)
\\
&=
\left( a|\mathbf{r}_1|^2 + b(\mathbf{r}_1\cdot\mathbf{r}_2) \right)\,(\mathbf{r}_1\cdot\mathbf{r}_2) -
\left( a(\mathbf{r}_1\cdot\mathbf{r}_2) + b|\mathbf{r}_2|^2 \right) \,
|\mathbf{r}_1|^2
\\
&= -b |\mathbf{r}_1\times\mathbf{r}_2|^2
\\
\Rightarrow\quad
b &=-\frac{\mathbf{r}_1\times\mathbf{r}_2 \cdot \mathbf{v}_1}{|\mathbf{r}_1\times\mathbf{r}_2|^2}
\end{align*}
I hope I haven't made any slips, you should definitely check!
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/431674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What if... you had a bowl of electrons? My chemistry teacher used to tell us that if you had a soup bowl with only electrons in it, the explosion could make you fly to Pluto. Was he right? Could this happen?
| The answer would depend how densely the electrons are packed. Let's say we have 1 kg of electrons, meaning we would have about $N = 10^{30}$ of them. For simplicity, let's approximate by arranging all of these electrons arranged in a spherical shell of radius $r=0.1$ meters. By symmetry, the voltage at the location of each charge would be constant and would have the following value.
$$V = -\frac{1}{4\pi \epsilon_0}\frac{Ne}{r}$$
We can thus compute the potential energy $U = \frac{1}{2}Q V$ associated with this configuration.
$$U = \frac{1}{2} \cdot Ne \cdot \frac{1}{4\pi \epsilon_0}\frac{Ne}{r} = \frac{N^2 e^2}{8 \pi \epsilon_0 r} \approx \boxed{1.15 \times 10^{33} \text{ Joules}}$$
Ignoring air friction, the Earth's escape velocity is $v =11200$ meters / second, which for a $m=2\times 10^6$ kg space shuttle, would only require $E = \frac{1}{2} mv^2 = 1.25 \times 10^{14}$ Joules.
So, yeah, you'd have way more than plenty sufficient energy.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/431751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Instantaneous velocity So here’s a question I’ve been thinking of for a while. Suppose we say, “an object is having an instantaneous velocity along a particular direction ( say 10 m/s along the $x$-direction)” . Is it fair to conclude that it is traveling in a straight line along the $x$-axis? Well my opinion on this is,
For instance, a projectile ( on earth ) , the instantaneous velocity ( which is constant through out the journey ) is always in the $x$-direction while the body is executing a parabola in the $x$-$y$ plane? Please acknowledge me if I’m wrong.
| The equation $\overrightarrow{v}=\frac{\Delta_\overrightarrow{x}}{t}$, with $\overrightarrow{v}$ being the velocity vector, $\Delta_\overrightarrow{x}$ the change in the position vector, and $t$ being the time passed only applies to a constant velocity. The more general equation for velocity would be $\overrightarrow{v}=\overrightarrow{x}'(t)$ with $\overrightarrow{x}'(t)$ being the derivative of the position vector with respect to time. So the instantaneous velocity is the derivative of the position vector with respect to time at that instant.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/431908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Lateral momentum of Gaussian beam A beam of light carries momentum. What fraction of this is lateral rather along the propagation direction if we assume something like a Gaussian beam?
Wikipedia claims in the entry on Gaussian beams that the Poynting vector is entirely along the z-axis. But this expression is based on solving the paraxial approximation of the Helmholtz equation rather than Maxwell's equations, implicitly assuming that there is indeed no lateral energy or momentum flow.
(Simon, Sudarshan & Mukunda 1986) give an expression for the Poynting vector of an actual Maxwell gaussian beam as $$\mathbf{S}(x,y,z)=\frac{2}{\pi}\frac{1}{\sigma(z)^2}\left [\frac{x}{R(z)},\frac{y}{R(z)},1\right]$$ where $R(z)=z [1+(k\sigma_0^2/2z)^2]$ and $\sigma(z)=\sigma_0 \sqrt{1+(2z/k\sigma_0^2)^2}$. They find that the energy flow follows the geometrical optics rays that are normal to the phase curvature.
So by this account the fraction of energy going laterally would be $$E_{lat}/E_{total}=\frac{r/R(z)}{\sqrt{r^2/R(z)^2+1}},$$ which reasonably approaches 1 as $r\rightarrow \infty$. But there seems to be a Gaussian term missing from their expression since integrating the lateral energy flow across a plane diverges.
(Allen 2000) looks at Laguerre-Gaussian beams and gets a lateral Poynting component of $p_r=\epsilon_0 \frac{\omega k r z}{z_R^2+z^2}|u|^2$ and an axial component $p_z = \epsilon_0 \omega k |u|^2$ (assuming no azimuthal component). The fraction of lateral momentum is $\frac{rz}{(z_R^2+z^2)\sqrt{\frac{r^2z^2}{(z_R^2+z^2)^2}+1}}$. Integrating this times $|u|^2=(C/w(z)^2)\exp(-2r^2/w(z)^2)$ where $w(z)^2=(2/k)(z_R^2+z^2)/z_R$ gives $$P_r = \frac{Cz}{w(z)^2(z_R^2+z^2)}\int_0^\infty
\frac{r}{\sqrt{\frac{r^2z^2}{(z_R^2+z^2)^2}+1}}\exp(-2r^2/w(z)^2) dr$$
which at least converges, but doesn't look like it has a neat analytic solution.
I assume there is a much simpler argument for how much of the momentum ends up sideways when we project a beam through a finite radius aperture or have a finite Gaussian beam waist width.
(My application is a consideration of the ultimate limitations of photon rockets; lateral momentum is wasted from a propulsion standpoint. Sure, there are other problems with photon rockets too, but one thing at a time.)
| Firstly, Gaussian beam (and also Laguerre-Gauss beams) are solutions of the paraxial wave equation. Therefore, these beams are valid under the paraxial condition, which can be imposed by a requirement that the beam divergence angle is small:
$$ \theta_B = \frac{\lambda}{\pi w_0} \ll 1 , $$
where $\lambda$ is the wave length and $w_0$ is the radius of the beam at its waist.
For a paraxial optical beam, the Poynting vector is given directly by the gradient of the phase. As such, it is equal to the optical current. For an optical field $\psi(\mathbf{x})=|\psi(\mathbf{x})| \exp[ i\theta(\mathbf{x})]$, the Poynting vector or optical current is given by
$$ \mathbf{S} = |\psi(\mathbf{x})|^2 \nabla \theta(\mathbf{x}) . $$
To determine the "fraction of energy going laterally," one first needs to define exactly what is meant by that phrase. Energy is a scalar quantity. It cannot be divided into components like a vector. So the problem is one of definition.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/432282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Why does pair production produce an electron and positron with opposite spin? So I was trying to understand quantum entanglement and the example that was used to describe an entangled pair of particles was of an electron and positron after it is formed from a photon in pair production . So I was wondering why is it assumed that the particles produced from a photon in pair production have opposite spin?
If anyone is looking for the video that I saw https://www.youtube.com/watch?v=tafGL02EUOA Here is the link
| The video is wrong on this point, at 4:47 he talks of a "photon spontaneoulsy creating an electron positron pair". The speaker is using a wrong example because there is no way a single photon can "decay" spontaneously , as mentioned in comments, due to energy and momentum conservation at the center of mass of the pair. The photon has no center of mass frame as its mass is zero.
The photon needs to interact with a field, with a virtual photon, in order to create a pair, and then all spin possibilities are open as it is a three body effect.
A correct example of entanglement is the $π^0$ decay into two photons, where the photons have to have opposite spins because the $π^0$ has spin $0$. Also the decay of the $Ψ$ resonance to a pair would be correct, it is spin 1 and decays to a pair.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/432586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why does the bulk of a superconducting material expel magnetic field but not that of a perfect conductor? It is well-known that when a superconductor (SC) is cooled below the transition temperature $T_c$, the magnetic field passing through the bulk of the SC is completely expelled. In Zee's book on Quantum Field Theory in a Nutshell, he explains why a superconductor abhors magnetic field in the bulk along the following lines
A hallmark of superconductivity is the Meissner effect, in which an external magnetic field $\textbf{B}$ permeating the material is expelled from it as the temperature drops below $T_c$. This indicates that a constant magnetic field inside the material is not favored energetically. The effective laws of electromagnetism in the material must somehow change at $T_c$. Normally, a constant magnetic field would cost an
energy of the order$\sim\textbf{B}^2V$, where V is the volume of the material. Suppose that the
energy density is changed from the standard $\textbf{B}^2$ to $\textbf{A}^2$ (where as usual $\nabla\times\textbf{A}=\textbf{B}$).
For a constant magnetic field $\textbf{B}, \textbf{A}$ grows as the distance and hence the total energy would grow faster than V. After the material goes superconducting, we have to pay an unacceptably large amount of extra energy to maintain the constant magnetic field and so it is more favorable to expel the magnetic field.
Statement 1 "Suppose that the
energy density is changed from the standard $\textbf{B}^2$ to $\textbf{A}^2$..."
Question 1: What does this mean? Energy density depends on $\textbf{B}^2$ and not on $\textbf{A}^2$.
Statement 2 "After the material goes superconducting, we have to pay an unacceptably large amount of extra energy..."
Question 2 Why?
| I'll address the question in the title. I don't find the argumentation in the textbook convincing.
In a superconductor single electrons are not scattered at all. Therefore the electrons will alter their motion in the presence of a magnetic such the the flux is opposed and at sufficient depth into the bulk will be cancelled.
In a perfect conductor the electrons are still scattered at the boundary of the conductor. In the presence of a magnetic field there is a current that opposes the magnetic field but it is cancelled by the current resulting from the scattering at the boundary.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/432737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Debye Temperature for Copper I am trying to calculate the Debye temperature, $\theta_D$, of copper using the following:
$$
\theta_D = \frac{\hbar v_s}{k_B} \left( \frac{6\pi^2N}{V} \right)^{1/3}
$$
I have the following values: $\rho = 8900$ kgm$^{-3}$, $v_s = 3800$ms$^{-1}$, atomic mass $ M_a=63.5$gmol$^{-1}$. Now, the speed of sound, $v_s$ is not correct for copper according to online tables, and it seems to be closer to $4600$ms$^{-1}$.
However, I also know that the Debye temp for copper is about $343$K.
Using the fact that,
$$
{N\over V} = \frac{N_A\rho}{M_a} = 8.44\times10^{27}
$$
where $N_A$ is Avogadro's number, I get,
$$
\begin{align}
\theta_D &= \frac{(1.055\times10^{-34})(4600)}{(1.381\times10^{-23})}\cdot \left( 6\pi^2\cdot8.44\times10^{27} \right)^{1/3} \\\\ &=279K
\end{align}
$$
Which just isn't right. And using the value for $v_s$ provided in the question gives an even lower answer of $230$K...which isn't right either.
Am I missing something here?
| There are few things going on here. The first is that you seem to be mixing units for density and the molar mass, using kg in one case, and g in the other. If you fix that, you will correctly get a number density on the order of $10^{28}$. However, you still won't find good agreement with the $~345K$ value you expect. Why is this?
Well, there's a second and subtler thing going on, which is that you are using a single speed of sound. In reality, the speed of sound is different in the (one) longitudinal and (two) transverse directions. If you instead use a mean speed calculated through
$$\bar{v}_s = 3^\frac{1}{3}\left( \frac{1}{v^3_{\mathrm{transverse}}}+\frac{2}{v^3_{\mathrm{longitudinal}}}\right)^{-\frac{1}{3}} $$
you'll get a lot closer to the experimental value. You'll note that this isn't an ordinary average velocity, it's simply a constant defined in the derivation of the Debye temperature.
The third thing worth mentioning is that the speed of sound is going to depend on how your sample of copper was made. The value $v_s=3800$m/s is the longitudinal sound speed in thin copper rods, whereas $v_s=4600$m/s is a (rather low) value for the bulk material.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/432850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
How did the Moon get into orbit? It's the accepted theory that Moon was formed when an object collided with Earth, throwing off a lump of matter which became the Moon.
But also, an object cannot be launched into orbit from a planet without extra force being applied once it's in space, because otherwise, either it leaves at faster than escape velocity and never returns, or if it does return, its orbital path - being an ellipse - ends up where it began and so the object would just collide back into the planet.
So how did the Moon manage to achieve orbit?
| You're right in that, absent post-collision interactions, material ejected during the impact would either leave on a hyperbolic orbit or be placed in an elliptical orbit that would have it falling back to the proto-Earth within one orbit.
However, the hypothesized collision is an extremely messy environment, and you do have significant interactions between all components of the system, both mechanically and gravitationally, which are able to re-distribute both energy and angular momentum within the ejected material (and to exchange both quantities with the proto-Earth core). This is what enables enough material to be placed on an accretion disk with (i) enough energy to be in orbit, but not enough energy to reach escape velocity, and also (ii) enough angular momentum that the perigee is high enough to avoid falling back to the surface. This accretion disk will then re-coalesce into the Moon on relatively short timescales.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/433232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Classical Theory explanation of Compton Effect We all have studied in introduction to quantum mechanics about Compton Effect. In all the books I have read, it says that classical theory can't explain the shift in wavelength because the incident EM wave will oscillate the electron at the frequency of light, and the oscillating electron will emit radiation of the same frequency. This is all well and good, but while reading Quantum Physics[Berkeley Series],the author mentioned something among the lines that ( not exact sentence) , the em waves oscillate electrons and that in turn produces the light of same frequency, also some loosely bound electrons are ejected from the atom which radiate the light of the slightly different frequency.
Is this explanation correct, if so what was the problem in mathematically describing this theory? Also , if this is the case, how is the ejected electron different from the ones in photoelectric effect.
| Compton scattering, unlike the photoelectric effect, can occur for a free electron. For a free electron, the classical theory can't explain the shift in wavelength. A theory of Compton scattering has to explain all observations, not just some of them, so it needs to explain the case where the electron is free.
Of course there is no such thing as a perfectly free electron, since there is no such thing as an exactly vanishing electric field. However, the electrons in ordinary matter are an excellent approximation to free electrons, because the atomic binding energies, on the order of eV, are negligible compared to the MeV energies of the photons.
There is also a problem with classical explanations of the photoelectric effect and Compton scattering, because they can't explain the entanglement between electrons in different atoms. Without this entanglement, each atom's probability of being ionized is independent, and then conservation of energy and momentum hold only at the statistical level. This lack of correlation was a prediction of the BKS theory, which was disproved by the Nobel prize-winning 1925 Bothe-Geiger experiment.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/433336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How is momentum conserved in this example? Suppose a sticky substance is thrown at wall. The initial momentum of the wall and substance system is only due to velocity of the substance but the final momentum is 0. Why is momentum not conserved?
| Remember Newton’s 3rd law. The change in momentum is $F \: \Delta t$ (also known as impulse). So, since by Newton’s 3rd law the forces are equal and opposite then the change in momentum must also be equal and opposite.
Therefore, Newton’s laws guarantee conservation of momentum, and to see where the momentum goes all you have to do is look for the Newton’s 3rd law pair. So here momentum is transferred between the sticky substance and the wall, and the wall (being so massive) gains a little momentum which makes it move imperceptibly.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/433634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Why do we use the RMS but not the fourth root mean quad? Why do we use the power of $2$? What is the relation between this and having the same heat energy in both AC and DC?
| Average power $\left <\rm power \right > = \dfrac{\left <\rm \rm voltage^2 \right >}{R} = \left <\rm \rm current^2 \right > R$ so it is the mean of the values squared which you need to use.
In the example above you will see that the areas $A$ and $B$ above and below the $\left <\rm \rm i^2 \right > = \left <\rm \rm current^2 \right >$ line are equal so the extra energy delivered during the time period of $A$ is exactly compensated for by the reduced amount of energy delivered during the time period of $B$. (Think of it as levelling some ground which originally had a lot of bumps on it.)
Thus a steady current of magnitude $\rm i_{dc} = i_{\rm rms} = \sqrt{\left <\rm \rm current^2 \right >} $ will, over a cycle, dissipate the same amount of electrical energy as the alternating current does over a cycle.
If you wanted the mean of the power squared then you would raise the quantities to the power of four but note that in general $\left <\rm power^2 \right >\ne \left <\rm power \right >^2$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/433726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Kosterlitz-Thouless transition and correlation function I’m studying Kosterlitz transition on this book: https://tinymachines.weebly.com/uploads/5/1/8/8/51885267/kardar._statistical_physics_of_fields__2007_.pdf#page173 . At page 165 it says:” The gradient expansion applies to configuration that can be continuosly deformed to the uniformly ordered state”. My question is why? I know that at low temperature there are vortex antivortex pair (that can be transformed into ordered state) and an higher temperature vortex that cannot transformed into ordered state.
| When doing gradient expansion, we are implicitly assuming $\theta(x)$ is a single valued function of $x$ and there is no discontinuity. Otherwise the gradient of $\theta$ diverges.
As the textbook says, any single-valued function can be continuously deformed to a uniformly ordered state. Here is how. Define a uniformly ordered state by $\theta_{uni}(x) = 0$ for all $x$. Then we can define a continuous deformation from $\theta_0(x)$ by
$$
\theta(x,\lambda) = (1-\lambda)\theta_0(x) - \lambda\theta_{uni}(x) = (1-\lambda)\theta_0(x)
$$
where $\lambda = 0$ correspond to the initial state and $\lambda=1$ corresponds to the final state. As long as $\theta_0(x)$ is smooth and continuous, this transformation is also smooth and continuous.
Vortex configurations cannot be described as a single valued function. For any single valued function $f(x)$ we have
$$
\int_C \nabla f(x) dx = 0
$$
for any closed loop C. However, for a loop C containing a vortex we have
$$
\int_C \nabla \theta(x) dx = 2\pi n
$$
where n is defined to be the vorticity of the vortex (+1 for vortex, -1 for anti vortex and so on). Thus, $\theta(x)$ cannot be a single valued function, and we need to put extra care in order to take vortices into account.
That explains why gradient approximation fails. The more interesting questions is why gradient approximation succeeds in some cases. As you have pointed out, at low temperature there are very few vortices present, so we can get away with ignoring their presence and using gradient expansion.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/433844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Relationship between strain energy function and strain or stress How one can get the strain or stress from the strain energy function ?
And if one cannot do it, what is the use of that function ?
| You can determine the 3D stress-strain equation for a material from its strain energy function. To get a certain component of the stress, you take the partial derivative of the strain energy function with respect to the corresponding component of strain. And, to get a certain component of the strain, you take the partial derivative of the strain energy function with respect to the corresponding component of stress. The following reference gives the strain energy function for Hooke's law in Eqn. 8.2.19:
http://homepages.engineering.auckland.ac.nz/~pkel015/SolidMechanicsBooks/Part_I/BookSM_Part_I/08_Energy/08_Energy_02_Elastic_Strain_Energy.pdf
Try taking the partial derivative of this function with respect to any of the strain components and see what you get.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/434101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
(Fluid Dynamics) Euler's equation including gravity In fluid dynamics, we can write down the Euler's equation as
$\dfrac{\partial \mathbf{v}}{\partial t} + ( \mathbf{v} \cdot \mathbf{\text{grad}} ) \mathbf{v} = - \dfrac{\mathbf{\text{grad}} \; p}{\rho}$ .
If the fluid is in a gravitational field, we can add an extra term on the RHS, for example
$\dfrac{\partial \mathbf{v}}{\partial t} + ( \mathbf{v} \cdot \mathbf{\text{grad}} ) \mathbf{v} = - \dfrac{\mathbf{\text{grad}} \; p}{\rho} + \mathbf{g}$ .
My question is rather simple, in the above equation, does it imply that the direction of gravitational acceleration is in the same direction as the acceleration of the fluid? If so, what if the fluid is accelerating upwards due to a huge force $- \mathbf{\text{grad}} \; p$?
| $\mathbf{g}$ in that equation is a generalized vector and its direction is undefined. It could be in any direction, depending what vector you choose to use. If you are using a standard cartesian reference frame, for $\mathbf{g}$ you would use (0, 0, -9.81).
All of the terms in that equation have undefined directions. So if there is a huge pressure gradient pointing upwards, that will dominate the gravitational term, and the fluid will accelerate upwards.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/434261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Can the horizon of a black hole move? Because of time dilation we cannot observe a black hole forming in a finite amount of time. For the same reason I suppose we also cannot observe the horizon moving: everything happening on the horizon takes an eternity to witness from the outside perspective.
Therefore, would a moving black hole result in new horizons (almost) forming according to an outside observer? The initial horizon would remain frozen in time, followed by the horizons around the moving singularity. Meaning, would moving black holes leave a trail of blackness behind, everywhere it passed?
An important issue here is the moving reference frames. Could one really claim that there is a perspective where the observer moved, rather than the black hole? The observer isn't curving spacetime to extremes, while spacetime is a medium: it's a fabric, it seems more than something described by coordinate systems.
Edit:
I changed the question title, it used to be "Moving reference frame of a black hole" but the new title better suits my question. The issue with reference frames is more a follow-up question.
| Since I found an answer to my own question, which I first only wrote as a comment, I'll put it here to wrap things up:
Just like photons don't age but still move, black hole horizons don't age but still move.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/434342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Imprecision in experiments When dealing with breadboards, and electronic circuits in general, in my case finding the total (equivalent) resistance using an ohmmeter, what are the factors that make the experimental value not equal to the theoretical one?
Cables? The resistances themselves?
| The major factors are:
*
*Ohmmeter is not properly calibrated ("zeroed") to account for resistance of probe wires and clamp tips.
*Contact resistance between the probe tip and the circuit element due to the presence of solder resist residues, solder flux residues, or oxides/corrosion products.
*For old-style analog meters, low or defective batteries in the power supply or corrosion of battery contacts.
*Imprecision in the component itself (i.e., resistor labeled as 1K which measures out to 990 ohms) which is within the tolerance band for the component.
*For resistance measurements, the presence of electrolytic capacitors in the circuit measurement path which perturb the measurement by using the sense voltage injected by the meter to charge themselves.
*The presence of transistors in the measurement path, which introduce polarity-dependent current leakage during measurements.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/434727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why does increasing the volume in which a gas can move increase its entropy? Let's say we have a box with a non-permeable wall separating the box in half. There is gas on the other side of the wall. Now we remove the wall so that the gas can diffuse to the other half of the box.
It is said that the entropy of the gas increases because the molecules now have more room to move, and therefore there are more states that the gas can be in. I can understand this well.
But the change in entropy is also defined as follows:
$\displaystyle \Delta S = \frac{Q}{T}$
Where $T$ is the temperature of the gas and $Q$ is the change in heat of the system. But if we look at this definition, why did the entropy change for the gas inside the box? By just removing the wall, the kinetic energy of the molecules does not change, therefore the temperature does not either. We also didn't add any heat to the system, so $Q$ is zero as well. So why did the entropy change?
| This is a classic example of making sure you know what your equations actually mean. You are thinking along the lines of
$$Q=T\Delta S$$
Where your change in entropy determines the heat exchange. This is not the right way to view the equation.
The actual meaning of the equation is "if you have reversible heat flow $Q$ at temperature $T$ for a reversible process, then you have a change in entropy $\Delta S$". Since there is no heat flow, this equation does not tell you anything useful. All you can say is there is no entropy change due to a heat flow.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/435119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Is Quantum Mechanics Compatible with Conservation of Information? What is exactly the law of conservation of information? In quantum mechanics we have truly random outcomes in experiments, but doesn't this randomness mean that new information is produced and the law of conservation of information is violated?
| As Dominic stated, local conservation laws hold in quantum mechanics, i.e. the conservation equation for probability density.
But in a simpler sense, "information is conserved" in that the total probably to measure an observable is always 1. More technically, if $\Psi(x)$ is a function that represents a quantum state in the position basis, then we have that
$$ 1 = \int_{\infty}^{\infty} |\Psi(x)|^{2} dx$$
which is the well known normalization condition. In our context here, it means that we will always find our particle at some position if we measure it. So in this sense, all the information of the wave function is conserved as well.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/435255",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
What if we shine a laser to east and another to west, will they arrive at the same time? If we have two lasers based at the equator and we direct one of them east and the other west on to two screens each 10 km away, will light emitted synchronously at the lasers arrive at the same time as each other on thier respective screens?
My interest is that one laser moves with the rotation of the earth and the other one the opposite.
From what I know the light shouldn't be affected by the rotation of the earth so the laser on the west side will arrive faster than the east side since the screen on the west side will come to the laser reference point.
Correct me if I'm wrong. (with mathematics if you can)
| It depends on your frame of reference.
If you are positioned by the laser on the Earth's surface you would observe the two laser beams to have the same speed and your would observe the two targets to be stationary and you would observe the light to take the same amount of time to reach each target.
On the other hand if you were not at rest compared to the laser and screens, say for example you are observing from the moon, you would still observe the laser beams having the same speed but you would see the screen as moving at about 460 m/s. From this frame of reference you would see the light hit the screen to the west first.
What is simultaneous in one frame of reference is not necessarily simultaneous in a different frame of reference.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/435404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "28",
"answer_count": 2,
"answer_id": 0
} |
Why does current conservation involve an arbitrary function? In section 6.1 of Peskin's quantum field theory introduction, right after equation 6.3, the four current density $j^{\mu}$ is said to be conserved because for any function $f \left( x \right)$ that falls off at infinity, we have
$$
\int f \left( x \right) \partial _{\mu} j^{\mu} \left( x \right) \mathrm{d}^4 x = 0
$$
I am just so confused on the fact that there is a function $f \left( x \right)$ involved in this evaluation. Is current conservation not just $\int \partial _{\mu} j^{\mu} \left( x \right) \mathrm{d}^4 x = 0$?
Thanks in advance!
| Current conservation is $\partial_\mu j^\mu=0$, not integrated over ANYTHING (note the zeroth component of $j$ is the current density, so this is equivalent to the usual statement of current conservation $\dot\rho=-\vec\nabla\cdot\vec j$). So, Peskin wants to show $\partial_\mu j^\mu$ is zero. He does it by showing its integral over any test function $f(x)$ is zero. If $g(x)$ is a function such that $\int f(x)g(x)=0$ for all test functions $f(x)$, then $g(x)=0$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/435597",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Conservation of energy in rotational motion Suppose a boy is standing on a platform which is free to rotate about an axis passing through its center. The Kinetic energy of the boy and the platform is K.
If the boy stretches his hands so that the moment of inertia of the sytem(boy + platform) gets doubled. Then I have to find the new Kinetic energy of the system.
As there is no torque so angular momentum remains conserved. Since M.O.I gets doubled so angular velocity gets halved.
Therefore the new kinetic energy is:-
$ K'= \frac{1}{2}I'×(w')^2 $
$ K'= \frac{1}{2}×2I×(\frac{w}{2})^2 $
$ K'= \frac{1}{2}×\frac{1}{2}×I×w^2 $
$ K'= K/2 $
Thus the kinetic energy gets halved.
So is the law of conservation of energy being violated here? Is the total energy not conserved?
| The boy is doing work against his own spin by extending his arms outwards. Taken to the differential limit, we can see that the force extending his arms has a tangential component that decreases angular velocity. Work is therefore done against his own rotational kinetic energy.
A similar example is the commonly-asked textbook problem of a mass tied to a string, moving in circular motion on a frictionless table, around a hole through which the string runs. When the string is pulled inwards, work is done on the system as the act of pulling inwards adds to the tangential velocity, as user ja72 has excellently explained here.
So a radial outward force has a tangential component that decreases angular velocity (and kinetic energy), while a radial inward force has such a component that increases it. Energy is not conserved.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/435870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is partial trace the inverse operation of Kronecker product? Computer science student here, who is interested in quantum information theory.
Suppose I have these pure states:
\begin{bmatrix}1&0\\0&0\end{bmatrix} and
\begin{bmatrix}0&0\\0&1\end{bmatrix}
The Kronecker product of these is:
\begin{bmatrix}0&0&0&0\\0&1&0&0\\0&0&0&0\\0&0&0&0\end{bmatrix}
By using partial trace, I can extract information of the original states:
And if I were to tensor multiply them again, I'd get back the previous 4x4 matrix:
Below is a short python code, which works for the above example.
However it doesn't work for mixed states, as you can see on the right, in the console. If we were to tensor the two 2x2 density matrices extracted from the 4x4 matrix, we don't end up with the original 4x4 matrix.
My question is that partial trace can only restore pure states without information loss?
| It is in the sense that given any pair of states $\rho$ and $\sigma$, you have
\begin{align}
\operatorname{Tr}_2(\rho\otimes\sigma)&=\rho,\\
\operatorname{Tr}_1(\rho\otimes\sigma)&=\sigma.
\end{align}
This happens because $\rho\otimes\sigma$ represents a product states, in which the two parts of the system are independent from each other.
However, as soon as you have entanglement, this stops being true. In mathematical terms, it stops being true as soon as you take the partial trace of a sum of tensor/kronecker products.
For example, if $E_+$ and $E_-$ denote the two two-dimensional matrices you defined, then you can easily verify that taking the partial trace of
$$E_+\otimes E_++E_-\otimes E_-$$
gives you something completely different than both $E_+$ and $E_-$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/435988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Where is mass density information of a Black Hole? Have read that all the information of a black hole is contained on the Event Horizon. Does that include mass density of interior as a function of position? The shape of an event horizon depends on the mass-energy-charge inside the black hole and possible effects by nearby objects. In principle, the effects of nearby objects can be predicted and subtracted away from whatever behavior the Event horizon experiences.
If the information is on the surface then I'd think it sufficient "space telling matter how to move" to effect local fields without signals from nearby masses effecting the mass at the center of the black hole.
Alternative formulation: Is there a difference between information about mass being stored on the Event Horizon vs. mass being at the event horizon? As I recall, charge distributed on a a straight line has the same field far enough away as some other charge density located on one of its equipotential surfaces. Is that part of what's going on here?
| Are you asking about the viewpoint of the external observer, or somebody inside the black hole?
Because for somebody outside, yes everything about the inside of the black hole is frozen on the event horizon. For an external observer, everything is frozen on the event horizon, and so anything that entered the black hole, must be visible on the event horizon. All the mass density of the black hole is obtainable from looking at the event horizon.
For somebody inside the black hole, it is different. When you enter the black hole, nothing special happens (other then being spagettified, but that is a question of taste if you like spagetti). Other then that, time passes normally for you, and you seem to pass the event horizon like nothing happened, you don't even know you passed it.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/436148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Converting $\mathrm{ps/nm}$ to $\mathrm{ps}^2$ I have a dataset in the unit $\mathrm{ps/nm}$ for many different $\lambda$ which I want to convert to $\mathrm{ps}^2$.
I guess I can assume that I only deal with Gaussian bandwidths such that $1\ \mathrm{ps}$ corresponds to $1\ \mathrm{THz}$.
$$\Delta f = \frac{c\Delta \lambda}{\lambda^2} \approx 125\ \mathrm{GHz} \quad\text{for}\ \Delta \lambda = 1\ \mathrm{nm}\ \text{at}\ \lambda=1550\ \mathrm{nm}.$$
$1\ \mathrm{ps}$ corresponds to $\lambda_d \approx 8\ \mathrm{nm}$ at this wavelength.
Does that mean that I have to multiply by $\lambda_d \approx 8\ \mathrm{nm}$ if I want to convert from $\mathrm{ps/nm}$ to $\mathrm{ps}^2$?
| Since you are wanting to "convert" between two units with different physical meanings, there is not a standard way to do this. Multiplying a number with units of $\rm{ps/nm}$ by any number with units of $\rm{nm}$ will give you a number with units of just $\rm{ps}$. You would need to multiply by a number with units of $\rm{ps\cdot nm}$ to get what you desire. What value you actually use is up to you, and must be interpreted accordingly based on the context of your problem.
Typically values are chosen that are good characteristic scaling numbers for the problem at hand. For example, if you are dealing with time scales of $1\ \rm{ps}$ and wavelengths of $8\ \rm{nm}$, then perhaps you could use a "conversion factor" of $1\ \rm{ps}\cdot8\ \rm{nm}=8\ \rm{ps\cdot nm}$. Without knowing more about your specific dataset or the experiment you are looking at, I don't think I can give more insight into what this conversion physically means though.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/436239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Work done by a gas In the expression for work done by a gas,
$$W=\int P \,\mathrm{d}V,$$
aren't we supposed to use internal pressure?
Moreover work done by gas is the work done by the force exerted by the gas, but everywhere I find people using external pressure instead of internal pressure.
| The work done by an expanding gas is the energy transferred to its surroundings. In effect, as the gas expands it is compressing its surroundings so the work done is the force exerted on the surroundings (i.e. the pressure of the surroundings times the area) times the distance moved.
The extreme case of this is a Joule expansion where a gas expands into a vacuum i.e. the pressure of the surroundings is zero. In this case the expanding gas does no work regardless of the initial pressure of the gas.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/436339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 7,
"answer_id": 2
} |
Determining the acceleration of the Universe from a single star? It Occurs to me we might be able to find an entirely independent method of determining the Universe's acceleration using a single source.
If one was to watch a single high source consistently one should be able to simply simply watch for the change in it's redshift with time. I know it would be a small effect (perhaps choosing a high Z source would help).
Is such a task feasable?
Maybe something like finding a source that could line up with the recoiless resonant absorption (Mossbauer effect) in some crystal would have enough sensitivity? (ie a cosmological scale Pound Rebka experiment)??
Anyway, I hadn't heard of it, but maybe someone can tell me why its a bad/good idea. Thanks!
NOTE: I get that ultimately one would use several sources for a proper analysis.
| You just need to measure the recession velocity of any astronomical source.
The proper distance to a given source $d$ is related to the comoving distance $\chi$ through:
$$ d(t) = a(t) \chi$$
where $a(t)$ is the scale factor for expansion of the universe. Then the recession velocity can be written as:
$$\dot{d} = \dot{a} \chi = \frac{\dot{a}}{a} d$$
So the Hubble constant $H\equiv \dot{a}/a$ measures the rate of expansion of the universe in the above relation from the recession velocity of the source, and proper distance to it.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/436657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Are uncertainties higher than measured values realistic? Whenever I measure a positive quantity (e.g. a volume) there is some uncertainty related to the measurement. The uncertainty will usually be quite low, e.g. lower than 10%, depending on the equipment. However, I have recently seen uncertainties (due to extrapolation) larger than the measurements, which seems counter-intuitive since the quantity is positive.
So my questions are:
*
*Do uncertainties larger than the measurements make sense?
*Or would it be more sensible to "enforce" an uncertainty (cut-off) no higher than the measurement?
(The word "measurement" might be poorly chosen in this context if we are including extrapolation.)
| An uncertainty greater than the value for a known-positive value makes sense in the context of a non-gaussian credence.
If that first sentence was clear, you can stop reading. If not, let me back up. When we talk about "uncertainty", there isn't some sharp line involved. For any given range, we can express a probability that the value is within that range based on what we know. That probability is the integral over the range of our "credence", or "probability density function" ("pdf" for short). You can think of this function as the probability we assign every possible value, multiplied by infinity in such a way that its integral is one (somewhere, a mathematician flinched).
Very often, our credence is gaussian, in which case we can describe it with two parameters: mean and standard deviation. The mean is also the expected value and the maximum likelihood value, so it makes a great point estimate. We can then call the standard deviation "the uncertainty".
If the credence is not gaussian, which seems to be the case here, then we need to describe it a bit more verbosely.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/436747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
} |
Unpolarized Light: is it always linear or could it be elliptical? Unpolarized light is essentially polarized light in every direction i.e. there are so many waves radiating, that each wave oscillates in a different direction. Polarized light can either be linear or elliptical. Is light from the sun or a lamp all linearly polarized or could they/are they elliptical? Is most polarization in nature linear?
| It can either be linear, or elliptical... or any other! There are not just 2 possibilities.
The terms "linear" or "elliptical" refer to the basis that we have chosen. If we choose a linear description, we can be lucky and find a beam which has only one-axis component. That's linear light. However, most beams will have both components, because most light isn't purely "linear".
The same happens if we choose a circular basis. We can be lucky to find a clockwise polarized beam, but that's not very probable actually.
So, going to your final question, most light in nature is randomly polarized. This is often called "incoherent light", and it means that the electric field points in a random direction every single time. There's no way to predict where it will be pointing afterwards.
Since the light from the Sun is like this, this is usually called "natural polarisation". Most lamps also emit this way. You can find some screens where pixels make it linearly polarized, but most of it is random. Notice that most light bulbs are made by fluorescence.
To be precise, the atmosphere adds a small linear component. And there are many lightbubs that emit with a certain polarization. Of course, they will if their glass is a polariser.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/436977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Continuity equation in the Lagrangian flow picture approach In deriving continuity equations using Lagrangian.
We consider the element of fluid which occupied a rectangular parallelopiped having its centre at the point $(a,b,c)$ and its edges $\delta a$ , $\delta b$ ,$\delta c $ parallel to the axes . At the time $t$ the same element for an oblique parallelepiped . The centre now has for its co-ordinates $x$ , $y$ , $z \ ;$ and the projections of the edges on the co-ordinate axes are respectively
$$ \frac{\partial x}{\partial a} \delta a \ , \ \frac{\partial y}{\partial a} \delta a \ , \ \frac{\partial z}{\partial c} \delta a$$
$$\frac{\partial x}{\partial b} \delta b \ , \ \frac{\partial y}{\partial b} \delta b \ , \ \frac{\partial z}{\partial b} \delta b$$
$$\frac{\partial x}{\partial c} \delta c \ , \ \frac{\partial y}{\partial c} \delta c \ , \ \frac{\partial z}{\partial c} \delta c$$
How can i get these projections ?
The volume of the parallelepiped is therefore
$$\begin{vmatrix}
\frac{\partial x}{\partial a} & \frac{\partial y}{\partial a} & \frac{\partial z}{\partial a} \\
\frac{\partial x}{\partial b} & \frac{\partial y}{\partial b} & \frac{\partial z}{\partial b} \\
\frac{\partial x}{\partial c} & \frac{\partial y}{\partial c} & \frac{\partial z}{\partial c}
\end{vmatrix} \delta a \delta b \delta c$$
or as its often written $$\frac{D(x,y,z)}{D(a,b,c)} \delta a \delta b \delta c$$
since the fluid mass is unchanged and the fluid is incompressble we have
$$\frac{D(x,y,z)}{D(a,b,c)} =1$$
Is there a way to prove that $$\frac{D(a,b,c)}{D(x,y,z)}= 1$$
without expanding the determinant?
| *
*In the Lagrangian flow picture ${\bf a}\equiv(a,b,c)$ typically denote continuous labels of a fluid parcel distributed such that
$$d(\text{mass})~=~da~db~dc,\tag{2.1}$$
cf. e.g. Ref. 1.
*On the other hand ${\bf x}\equiv(x,y,z)$ typically denote the position coordinates of a fluid parcel. Therefore the mass density becomes
$$ \rho~=~ |\det\frac{\partial{\bf a}}{\partial{\bf x}} |.\tag{2.2}$$
*The flow velocity is defined as $${\bf u}~\equiv~ \frac{d{\bf x}}{dt}.\tag{2.4}$$ The mass continuum equation follows in the Lagrangian flow picture from
$$\begin{align} -\frac{d\ln\rho}{dt}~=~&\frac{d}{dt}\ln |\det \frac{\partial{\bf x}}{\partial{\bf a}}|\cr
~=~&{\rm tr}\left( \frac{\partial{\bf a}}{\partial{\bf x}}\frac{d}{dt}\frac{\partial{\bf x}}{\partial{\bf a}}\right)\cr
~=~&{\rm tr}\left( \frac{\partial}{\partial{\bf x}}\frac{d{\bf x}}{dt}\right)\cr
~=~&{\bf \nabla}\cdot {\bf u}.\end{align} \tag{2.3}$$
For an incompressible flow, the density $\rho$ is constant along the fluid flow.
References:
*
*R. Salmon, Hamiltonian Fluid Mechanics, Ann. Rev Fluid. Mech. (1988) 225. The pdf file can be downloaded from the author's webpage.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/437144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Can any body be uniform in the universe? If I take any body in the shape of a rod and stretch that, after it reaches breaking stress it breaks at one point.
Even though we apply the same the stress on each and every part of the rod it broke at one point. If it's uniform it should break at all points because breaking stress is same for all the parts of body as it's uniform.
| You should consider the implicances of your question, which is mostly a philosophical issue regarding our human subjectivity. You are implying an ideal uniformity of states which wouldn't allow quantum behavior.
That would mean that atoms should have all the exact same state before the breaking, that is, considering them as particles, they should be all at the same angle, have the same rotational speed, same energy, etc. In such case, the momentum of the expansion process would be exactly the same for all couples, and the object would dissipate perfectly as smoke.
A very beautiful and utopic idea. But reality is not like that. Things might break even without stressing them. The probability of a body to remain intact without external action is not 100%.
Taking such logic to the extreme, it means that even if the material is perfect, all its couples of structural interactions have different degrees of resistance to stress... That means that absolutely all couples have different values, even if they are infinitesimally small!
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/437495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 4
} |
Is this possible to focus common light (not laser) to a small point in the wall and control it with mirrors? I want to make something like a laser projector for an experiment but with common light.
My question is: is this possible to focus a small point of common light to the wall and control it with mirrors? What combination of lenses I'll need for it?
| "A common light" needs to be defined better. I assume you want to use an incoherent source such as an LED or an incandescent light. The size of the smallest spot to which a light beam can be focused depends on the size of the light source. Imagine, for example, trying to focus light from the Sun to a small spot. In essence, this means forming a small image of the Sun. If you study how a lens works (e.g., this figure from Wikipedia illustrating the geometry of image formation by a lens), you'll see that the angle from the lens to the two sides of the image is the same as the angle from the lens to the two sides of the Sun.
A smaller spot can be formed from sunlight only by imaging a smaller portion of the Sun. So, in order to form a small spot from an ordinary incoherent light source, with the spot at a distance from the lens, you would need a source that is extremely bright and extremely small. There really is nothing that easily fits the bill. A diode laser works fine, but is coherent and therefore not "ordinary".
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/437647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding the Eigenvalues and Eigenvectors of the Hamiltonian for three spin-1/2 particles coupled antiferromagnetically Problem
Given three spin-1/2 particles with the total spin operator $\vec{S}=\sum\limits_{i=1}^3 \vec{S}_i$ and its $z$ projection $S_z=\sum\limits_{i=1}^3 S_{z,i}$, and the Hamiltonian
$$H = J\sum\limits_{i=1}^3 \vec{S}_i \cdot \vec{S}_{i+1}
$$
(assuming for $i=3$ that $i+1=1$), calculate the eigenstates and the eigenenergies.
Hint: Rewrite $H$ as a function of $S^2$ and $S_i^2$.
Work
I've already calculated the basis for $\vec{S}^2$ and $S_z$
$$
\vert 3/2,3/2\rangle \equiv \vert\uparrow\uparrow\uparrow \rangle \\
\vert3/2,1/2\rangle \equiv \frac{1}{\sqrt{3}}\Big( \vert\downarrow\uparrow\uparrow \rangle + \vert\uparrow\uparrow\downarrow \rangle + \vert\uparrow\downarrow\uparrow \rangle \Big)\\
\vert3/2,-1/2\rangle \equiv \sqrt{\frac{2}{3}}\Big( \vert\downarrow\downarrow\uparrow \rangle + \vert\downarrow\uparrow\downarrow \rangle + \vert\uparrow\downarrow\downarrow \rangle \Big)\\
|3/2,-3/2\rangle \equiv \vert\downarrow\downarrow\downarrow \rangle \\
$$
with eigenvalues according to
$$S^2\vert s,m \rangle = \hbar^2s(s+1)\vert s,m \rangle = \frac{15\hbar^2}{4}\vert 3/2,m \rangle \\
S_z\vert s,m \rangle = \hbar m\vert 3/2,m \rangle \text{ for } m=-\frac{3}{2},-\frac{1}{2},\frac{1}{2}\frac{3}{2}.
$$
I'm now attempting to rewrite the Hamiltonian according to the hint, using
$$\vec{S}_i \cdot \vec{S}_{i+1} = \frac{1}{2}\Big[\Big(\vec{S}_i + \vec{S}_{i+1}\Big)^2 - \Big(\vec{S}_i^2 + \vec{S}_{i+1}^2\Big) \Big].$$
Issue
I'm not certain I'm applying the hint correctly. With the above,
$$H = \frac{J}{2}\Big[ \Big(\vec{S}_1 + \vec{S}_2 \Big)^2 + \Big(\vec{S}_1 + \vec{S}_3 \Big)^2 + \Big(\vec{S}_2 + \vec{S}_3 \Big)^2 -2\Big( \vec{S}_1^2 + \vec{S}_2^2 +\vec{S}_3^2\Big) \Big],$$
which once expanded and using the expansion of $\vec{S}^2 = \Big(\vec{S}_1 + \vec{S}_2 +\vec{S}_3\Big)^2$ gets me to
$$H = J\sum\limits_{i=1}^2 S_i^2,$$
which just seems wrong to me, since it's not written as a function of $S^2$ and $S_i^2$, as the hint suggests.
| Just for fun, there's a second way to approach this problem. A generic basis state is of the form $| a_1 a_2 a_3 \rangle$ where $a_i$ is an up- or down-arrow. Let $P$ be the operator that exchanges the different spins: $P| a_1 a_2 a_3 \rangle = | a_2 a_3 a_1 \rangle$, so it's like a discrete momentum. It's easy to write down all of the eigenstates of $P$ (there are three possible eigenvalues). Now notice that $H$ commutes with $P$. This means that $H$ only mixes states that can have the same eigenvalue of $P$, so it's a much "smaller" problem to diagonalize $H$ in this basis.
The above technique is really common in condensed matter physics, so it's cool to master (Google the term "magnon").
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/437898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Focal length vs working distance in an infinity corrected objective Consider the following microscope objective (infinity corrected):
Magnification $= 50 \times$
Glass Thickness $= 3.5$mm
NA $= 0.50$
WD $= 13.89$mm
Focal Length $= 4$mm
I'm confused about the difference between $f$ (focal length) and $WD$ (working distance).
Also, what does $M=50 \times$ mean in the context of an infinity corrected objective? The magnification should be determined by whatever tube lens I use right (i.e. $M=\frac{f_{tube}}{f_{objective}}$)?
Can someone explain this?
| You are correct that for a single lens the working distance would be the focal length. For compound lenses, like microscope objectives, you have to look at the entire optical system to figure out the working distance. The short answer to your question is that the focal length and working distance are not what you expect because the objective is a compound lens. The magnification for microscope objectives is confusing because to calculate the magnification you must know something about the tube lens that is used along with the objective.
I'm going to go through an example, for quantities I don't define explicitly here, please see Figures 9 and 12 in refrence 1.
Suppose you have two focusing lenses (f1, and f2) separated by a distance d were the first lens is a distance W from the object you are trying to image. Microscope objectives have to fit in a specific space so they have a required parfocal distance PD = d + W (I am assuming thin lenses).
The ray tracing optics for this lens system would be:
\begin{equation}
M=
\left(
\begin{array}{cc}
M_{11} & M_{12} \\
M_{21} & M_{22} \\
\end{array}
\right)
=
\left(
\begin{array}{cc}
1 & 0 \\
-\frac{1}{\text{f2}} & 1 \\
\end{array}
\right).\left(
\begin{array}{cc}
1 & PD-W \\
0 & 1 \\
\end{array}
\right).\left(
\begin{array}{cc}
1 & 0 \\
-\frac{1}{\text{f1}} & 1 \\
\end{array}
\right).\left(
\begin{array}{cc}
1 & W \\
0 & 1 \\
\end{array}
\right)
\end{equation}
Actual performing the math we get:
\begin{equation}
M=
\left(
\begin{array}{cc}
1-\frac{\text{PD}-W}{\text{f1}} & \text{PD}+\left(1-\frac{\text{PD}-W}{\text{f1}}\right) W-W \\
-\frac{1-\frac{\text{PD}-W}{\text{f2}}}{\text{f1}}-\frac{1}{\text{f2}} & -\frac{\text{PD}-W}{\text{f2}}+\left(-\frac{1-\frac{\text{PD}-W}{\text{f2}}}{\text{f1}}-\frac{1}{\text{f2}}\right) W+1 \\
\end{array}
\right)
\end{equation}
In order for this objective to image properly we set the requirement that M12 = 0. When we are happy that our objective correctly images we can figure out the effective focal length by calculating the M21 element.
Let's work an example. Suppose that PD = 45 mm, f1 = 11 mm, f2 = 16.5 mm. I mostly made these numbers up, but PD for microscope objectives is 45 - 60 mm. This is on purpose to make objectives interchangeable.
From the requirement that M12 = 0 we can calculate W:
\begin{equation}
M_{12} = W \left(1-\frac{\text{PD}-W}{\text{f1}}\right)+\text{PD}-W = 0 \\
\implies W = 19.15 mm,
\end{equation}
so the working distance is 19.15 mm.
Next we want to calculate the focal distance of this lens, which we can get from the M21 element:
\begin{equation}
M_{21} = -(1/f2) - (1 - (PD - W)/f2)/f1 = -0.0091 mm^{-1},
\end{equation}
but focal lengths are usually written like $M_{21} = -1/f_{o}$ so the focal length of our compound lens system is $f_{o}$ = 110 mm. So very clearly, compound lenses do not behave like single lenses.
What about the magnification? To answer that question we have to look at the M matrix with all the numbers plugged in:
\begin{equation}
M=
\left(
\begin{array}{cc}
-1.35037 & 0. \\
-0.00906831 & -0.740536 \\
\end{array}
\right).
\end{equation}
From inspection you'd expect the magnification of this lens to be 1.35 (the minus sign just means the image is upside down). This is where your confusion comes in: magnification for microscope objectives require knowledge of the tube lens. From this example of someone trying to sell you a microscope objective you can see that it clearly states that:
All stated magnifications are based on a tube lens focal length of 200mm.
So for our example above using a 200 mm tube lens the magnification would be $M=f_{tube}/f_{objective} = 200 / 110 = 1.82$. Scrolling down the page of reference 2 you can see that the FL=100 mm lens has a magnification of 2.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/438073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Do we need a small displacement to create a oscillatory motion on the spring? Do we need a small displacement to create a oscillatory motion on the spring with a mass attached to it? Whats the limit of the displacement that we can give initally to create a oscillatory motion? Is it has to be small or it can large?
| The displacement sets the initial amplitude of the oscillations, which will go down from that point. So, the greater the displacement, the greater the amplitude.
The only limit here is the sterchability of the spring.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/438337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Uncertain principle mistake? I believe I have a hole in my understanding of Heisenberg's Uncertainty Principle, but I'm not sure where it is.
1) Assume you have a source of monochromatic light, a laser, perhaps.
2) Assume it emits from a point isolated to a sphere of some small radius, e.g. 3cm.
3) Assume I can isolate when it was fired to within a few femptoseconds.
4) Monochromatic light implies no uncertainty in frequency and so no uncertainty in its energy.
5) Light travels at a finite speed.
6) I know where and when the associated photons were emitted to within some error.
7) So the emitted photons must be within a region of space having a radius of less than $1.1ct$ where $c$ is the speed of light, and $t$ is time since emission.
8) I have zero uncertainty in energy and therefore zero uncertainty in momentum, yet I have finite uncertainty in position.
9) Does this contradict the uncertainty principle?
Possible sources of error?
Is it possible to have a source of monochromatic light? Perhaps lasers or LEDs.
Does failing to measure in the final step mean there is no violation?
| If the duration of the light is finite--that is, the laser beam is not infinitely long--then there is a non-zero uncertainty in the energy of the light. This is true regardless of the construction of the light source. In fact, the shorter the pulse, the broader the energy spectrum. This fits with the uncertainty principle since a short pulse takes up a small amount of space, and so must have a large spread (uncertainty) in its monemtum (energy).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/438609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Orbital parity of simple bound states in atomic and particle physics The parity operator commutes with the Hydrogen atom Hamiltonian. The energy eigenfunctions are parity eigenstates with orbital parity $(-1)^\ell$ which follows from the fact that $Y_{\ell m}(\theta,\phi)$ is an eigenstate of parity with parity eigenvalue $(-1)^\ell$.
Question 1 What about the Helium (He) atom? The Hamiltonian of the He atom also commutes with parity but is not central. But I am not sure whether the energy eigenfuctions are still given by a product of a radial and an angular part with the angular part given by $Y_{\ell m}(\theta,\phi)$. So I cannot decide whether or not the He atom energy eigenstates also have the orbital parity $(-1)^\ell$?
Question 2 What about a meson $\bar{q}_1q_2$ i.e., a bound state of a quark and an antiquark? The strong interaction Hamiltonian commutes with parity. Mesons being strong interaction eigenstates (in absencce of weak interaction contamination) should also have definite parity. Can we say that mesons also carry an orbital parity $(-1)^\ell$? Apart from orbital parity there exists contribution from intrinsic parity which is not my concern here.
| In the central field approximation, the Hamiltonian of a many-electron atom is assumed to comprise of a large spherically symmetric part and a small spherically asymmetric component. Since the asymmetric part is small, it can be treated as a perturbation. The total angular momentum operator $\textbf{L}=\sum\limits_{i}\textbf{L}_i$ commutes with the spherically symmetric component $H_{\rm central}$ of the Hamiltonian. Same for the total spin $\textbf{L}=\sum\limits_{i}\textbf{S}_i$. Hence, it's possible to find simultaneous eigenfunctions of $$H_{\rm central},\textbf{L}^2,\textbf{S}^2,L_z,S_z$$ labelled as $$|\gamma LSM_LM_S\rangle$$ where $\gamma$ is additional quantum numbers and rest of them the usual meanings. Since $H_{\rm central}$ is the dominant component of the many-electron Hamiltonian, the total orbital angular momentum of any approximate eigenstate $|\gamma LSM_LM_S\rangle$ will have orbital parity $(-1)^L$ where $L$ can have values from $|\ell_1-\ell_2|$ to $(\ell_1+\ell_2)$ in steps of unity for the He atom.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/438663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
What is the relation between physical theory and physical law? Gravitational law was explained by Newtons theory of gravity. So a law was described by a theory. What is the theory for Newton's laws of motion?
| The laws are the theory. Newton did not explain why the gravitational force is $GMm/r^2$. He theorized this formula, and showed that it explained the solar system and apples falling from trees.
Similarly his three laws of motion, plus some assumptions about absolute space and time, are the theory of Newtonian mechanics.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/438756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Are there any general results about the nodes of energy eigenfunctions in higher dimensions? A well-known result of quantum mechanics is that for a single particle in one dimension in a bounding potential $V(x)$ that goes to $+\infty$ as $x \to \pm \infty$, the energy eigenfunctions are discrete and the $n$th eigenfunction has exactly $n-1$ nodes at which $\psi(x) = 0$. (Moreover, we can say more - for example, between any two consecutive nodes in the $n$th eigenfunction, there exists a node in the $(n+1)$th eigenfunction.)
Do any similar results apply for single particles in higher than one dimension, or for multiparticle systems (for which the wave function is defined on configuration space rather than real space)? If not, is there an explicit example of a higher-dimensional or multiparticle system whose ground state wavefunction has a node?
| I found the answer on Chemistry SE. Apparently, for systems of one particle in higher than one dimension, the only known general result is that the number of nodes in the $n$th eigenfuntion is $\leq n-1$, and only in 1D is the inequality always saturated. (See the answer for more info and precise term definitions.) Moreover, in higher dimensions the sequence of node numbers is not necessarily increasing; in fact, the question gives a simple explicit example of a one-particle system in 3D such that a higher excited state has fewer nodes than a lower excited state. For multiparticle systems, I don't know of any general results at all.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/438920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Why does gauge invariance in electrodynamics mean that there are redundant degrees of freedom? It is possible to choose different gauges in electrodynamics. I am familiar with two of them: Coulomb gauge and Lorenz gauge. Let us stick to the Coulomb gauge. It sets $$\nabla\cdot\vec{A}=0.$$ The wisdom is that with this choice the physical electric and magnetic fields $\vec{E},\vec{B}$ do not change. But there is more to it. It is also important for me to understand why this gauge condition implies that there are superfluous degrees of freedom.
What are these superfluous and non-superfluous degrees of freedom? With which mathematical quantities should we identify them?
First of all, at each spacetime point, we have four numbers $$\phi(\vec{x},t),A_1(\vec{x},t),A_2(\vec{x},t),A_3(\vec{x},t).$$ I understand these four numbers as the four degrees of freedom. Now, Coulomb gauge means that the latter three can be related, without any loss of generality, through the differential equation $$\partial_1A_1+\partial_1A_2+\partial_3A_3=0.$$ Given this, how to understand the rest?
| The scalar and vector potentials are gauge covariant which is to say $\Phi' \ne \Phi,\,\mathbf{A}' \ne \mathbf{A}$ where
$$\Phi'(\mathbf{x},t) = \Phi(\mathbf{x},t) - \frac{\partial}{\partial t}\chi(\mathbf{x},t)$$
$$\mathbf{A}'(\mathbf{x},t) = \mathbf{A}(\mathbf{x},t) + \nabla\chi(\mathbf{x},t)$$
The electric and magnetic fields, on the other hand, are gauge invariant since the gauge transformation above leaves the fields unchanged
$$\mathbf{E}' = -\nabla\Phi' - \frac{\partial}{\partial t}\mathbf{A}' = -\nabla\Phi + \frac{\partial}{\partial t}\nabla\chi - \frac{\partial}{\partial t}\mathbf{A} - \frac{\partial}{\partial t}\nabla\chi = -\nabla\Phi - \frac{\partial}{\partial t}\mathbf{A} = \mathbf{E}$$
$$\mathbf{B}' = \nabla\times\mathbf{A}' = \nabla\times\mathbf{A} +\nabla\times\nabla\chi = \nabla\times\mathbf{A} = \mathbf{B}$$
Thus, it must be that $\Phi$ and $\mathbf{A}$ have a non-physical degree of freedom which is essentially the freedom to choose the gauge function $\chi(\mathbf{x},t)$.
Choosing a gauge amounts to specifying (in full or in part) the gauge function $\chi$. For example, an arbitrary vector potential $\mathbf{A}$ can be made to satisfy the Coulomb gauge with a gauge function $\chi$ that satisfies
$$\nabla\cdot\nabla\chi = -\nabla\cdot\mathbf{A}$$
such that
$$\nabla\cdot\mathbf{A}' = \nabla\cdot(\mathbf{A} + \nabla\chi) = 0$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/439044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Where does the principle of equal a priori probabilities come from in statistical mechanics? I have studied that the principle of equal a priori probabilities yields maximum entropy principle and minimum free energy principle and we can define and calculate other thermodynamic variables. However, where does the principle of equal a priori probabilities come from?
| Principles ( and laws, and postulates) in physics are the equivalent of axioms in a mathematical theory. The mathematical format used to study physics is very broad . A subset of the possible solutions allowed by the mathematical axioms is picked up by the use of laws, as the conservation laws, and principles, as for example the least action principle, to formulate models which can fit existing data, and, very important, be predictive of future data.
Physics, in contrast to mathematics, does not end with "quod erat demonstrandum". The models have to be validated or falsified by data.
Principles can be formulated from observations. The simplest and most studied by all probability distribution is the throw of the dice. The equal probability for all the faces is important for the dice to be unbiased, and this is an observation based on the randomness of the throws and the weight uniformity of the dice.
For dice it is easy to "prove" that if the throw is random and the dice matter uniform the probability distribution will be flat. It is not so for many particle systems treated in statistical mechanics, discussed here . Assuming it as a principle allows for picking the mathematical formats used in modeling many particle systems with statistical mechanics.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/439193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Relativistic space rocks. Are they possible? I was thinking that the Universe is full of extreme events. Colliding galaxies, exploding stars, colliding planets (like in one of models of our Moon formation), colliding black holes...
Can it be possible, that such event would accelerate some rocks to let's say $0.1c$ ?
Do we observe such objects? If not, why?
| Suppose the total energy release of a typical supernova is on the order of 1044 J, for a baseline. The question then becomes, if 100% of that energy goes into hurtling a physical object, how massive would the object be if the ejection speed was 0.1c?
At 100% efficiency (which is entirely unrealistic), the mass of the ejecta reaching 0.1c would be ~2 x 1029 kg or ~0.11 solar masses (or ~116 Jovian masses). This is also unrealistically small for a remnant core of a supernova.
Let's suppose we had a hypernova at 1046 J, just for comparison. Again, even with the unrealistic assumption of 100% efficiency the mass of the ejecta reaching 0.1c would be ~2 x 1031 kg or ~11 solar masses. This is more reasonable in that most cores of supernova are larger than one solar mass but the issue is with the assumption of 100% efficiency. Most of the energy in a supernova goes into the neutrinos and only ~1% goes into the blast waves that would accelerate the ejecta. So at 1% efficiency we are back at the original estimate for a supernova.
Suppose we invert this and impose a net energy imparted to one solar mass and find the maximum speed of the ejecta. For 100% efficiency and 1046 J, the speed of the ejecta could reach ~0.32c. If we are more realistic and assume 1% efficiency, then the net speed would be ~0.033c or ~10,000 km/s, which is still extremely fast for a one solar mass object. For a regular supernova at 1% efficiency, the ejecta speed is even smaller at ~1000 km/s, which is much closer to observed ejecta speeds as noted in other answers and comments.
Update
I wrote a slightly more detailed comment to the following question: Can supernovas or other explosive events release relativistic bombs?
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/439315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Why don't all gasses have infinite entropy? Entropy of an ideal gas is defined as the logarithm of the number of possible states the gas can have multiplied by Boltzmann's constant:
$${\displaystyle S=k_{\mathrm {B} }\log \Omega .}$$
In deriving the Maxwell-Boltzmann distribution, we initially start by counting a finite number of states, so this definition of entropy makes sense. But in the end we say that the number of possible states is so high that we can acctually say the distribution is continuous. But if the distribution is continuous, the number of possible states is infinite. So why is entropy not always infinite when a continuous distribution is used?
| The distribution is continuous is exactly true only in the thermodynamic limit, in which case $E$, $V$, $N$, and $S$ are all infinite, while $E/V$, $N/V$, and $S/V$ are all finite (or, equivalently, $E/N$, $V/N$, and $S/N$ are finite). In a real-life situation in which the volume isn't infinite, the distribution is technically discrete. But usually the volume is macroscopic while the length scales in the problem are microscopic, in which case taking the distribution as continuous is an extraordinarily good approximation. Then the total entropy is enormous but finite.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/439522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Is the amplitude of an EM wave the combination of the electric AND magnetic fields added together? For instance, to get the TOTAL energy of an EM wave(s) or intensity you square the amplitude. But do you first add or combine the strengths of the e and m fields?
| Suppose you have two sources of Electric field, $E_1$, and $E_2$
Then $\vec{E_{tot}}=\vec{E_1}+\vec{E_2}.$
So the total intensity is
$$E^2_{tot}=(\vec{E_1}+\vec{E_2})^2=E^2_1+E_2^2+2\vec{E_1}\cdot\vec{E_2}$$
So the intensity of the sum of two electric fields is not the sum of their intensities individually.
The same thing applies to a magnetic field.
Now both the E field and the B field can carry energy. Not only is there the complexity of the above, summing the associated intensities has it's own complications.
Are you familiar with the Poyting Vector?
$$\vec{S}=\frac{1}{\mu_0}\vec{E}\times\vec{B}$$
It satisfies the energy conservation equation of electro magnetic waves.
$$\nabla\cdot \vec{S}+\frac{\partial u}{\partial t}=0$$
$$u=\frac{1}{2}\epsilon_0E^2+\frac{1}{2\mu_0}B^2$$
$u$ is the energy desnity of the fields. The Poyting Vector indicates the direction of the flow of energy of the EM field.
So the energy density is a linear combination of the sum of their intensities, but neither exactly the sum of their intensities nor the intensities of their sums.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/439761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is there a traditionally accepted threshold probability at which highly unlikely becomes impossible? Many events which by any practical definition are impossible have extremely low but nonzero probability of occurrence. For instance, the positions of oxygen molecules in a room are basically random and independent, but it is clearly impossible that they all end up in one half of the room at the same time. Is there a standard order of magnitude of probability where the transition from unlikely to impossible occurs? In terms of coin flips, I don’t think 50 consecutive heads would be considered impossible, but I guess that 500 consecutive would.
I was hoping for a general answer, application independent, but it seems the consensus is that there isn’t one.
| This idea that events with probability under some small value are to be treated as impossible is known as Borel's or Cournot's law, but is not actually an accepted law, as far as I know.
It is however understandable that in applications of probability calculus, the events with extremely low probability (such as Sun exploding nova or gas in a room collapsing in one of its corners) are routinely ignored (thus treated as if impossible).
However, the value of probability under which events should be so ignored depends on the question that we try to answer. Sometimes such simplification is not possible. As is often said, extremely improbable events happen all the time.
Borel's paper Sur les probabilites universellement negligeables can be read here:
https://gallica.bnf.fr/ark:/12148/bpt6k3143v/f539.image
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/439909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Formula for potential energy? Conservation of energy? How would we know what formula to use for potential energy?
In my class, $mgh$ was used, but when dealing with a spring, it's ${1\over2}kx^2$. Is that because that's the elastic potential energy formula?
Also, for elastic and inelastic collisions, momentum is conserved. But kinetic energy is conserved only in elastic collisions, what does this really mean?
| For conservative forces the Potential energy can be defined as $$U=-\int_{x_i}^{x_f} \vec{F} d\vec{x}$$ For example think about $$\vec{F}=mg(-\vec{i})$$
I put a minus sign because it means that I choosed upward direction as positive. If we put the equation we have
$$U = -\int_{x_i}^{x_f} mg(-\vec{i})dx\vec{i} = mg(x_f-x_i)$$ which it means that if you move the object upper the potential energy of the ball will increase since $x_f>x_i$
For the spring case its the same idea. Maybe even you can try it. We know that $$\vec{F}=-kx(\vec{i})$$ (Lets assume motion happens in $i$ direction)
If you put it into the Potential energy equation you'll see that its equal to $1/2kx^2$
For the second part of your question. In inelastic conditions we assume that some of the energy goes to heat and internal energy that merges the two particles.That is why energy is not conserved.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/439978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Geodesics of anti-de Sitter space It is said that (p. 9), given the anti-de Sitter space $\text{AdS}_2$, let's say in the static coordinates
$$ds^2 = -(1 + x^2) dt^2 + \frac{1}{(1+x^2)} dx^2$$
Every timelike geodesic will cross the same point after a time interval of $\pi$. That is, if $(x_0, t_0) \in \gamma$, then $(x_0, t_0 + \pi) \in \gamma$.
So I've been trying to find out how to show it. The non-zero Christoffel symbols are
$${\Gamma^x}_{xx} = - \frac{x}{1+x^2},\ {\Gamma^x}_{tt} = x + x^3, {\Gamma^t}_{xt} = {\Gamma^t}_{tx} = \frac{x}{1+x^2}$$
So that the geodesic equation is
\begin{eqnarray}
\ddot{x}(\tau) &=& \frac{x}{1+x^2} \dot{x}^2 - \dot{t}^2 (x + x^3)\\
\ddot{t}(\tau) &=& -2 \frac{x}{1+x^2} \dot{x} \dot{t}\\
\end{eqnarray}
We also have the two following equality : the timelike geodesic is such that $g(u,u) = -1$
$$\frac{1}{(1+x^2)} \dot{x}^2 -(1 + x^2) \dot{t}^2 = -1$$
and since the metric is static, there is a timelike Killing vector $\xi$ such that $g(\xi, u)$ is a constant.
$$(1 + x^2) \dot{t} = E$$
or
$$\dot{t} = \frac{E}{(1 + x^2)}$$
This gives us
$$\dot{x}^2 = -(1 + x^2) + E^2$$
And so
\begin{eqnarray}
\ddot{x}(\tau) + x &=& 0\\
\ddot{t}(\tau) &=& -2 x \dot{x} \frac{E}{(1 + x^2)^2}\\
\end{eqnarray}
Which gives us for a start that $x(\tau) = A \sin(\tau) + B \cos(\tau)$. Not quite periodic in $\pi$ (it should be $2\pi$ here), but more importantly this periodicity is in $\tau$ only and not in $t$, and it doesn't seem that $t = \tau$ in this scenario. Is there something wrong here or did I commit a mistake, either in interpreting the statement or the derivation here?
Given $x(\tau) = \sin(\tau)$, Wolfram Alpha gives out the following solution for $t(\tau)$, for instance :
$$t(\tau) = c_1 \tau + c_2 - \frac{1}{2\sqrt{2}} \arctan(2 \sqrt{2} \tan(\tau))$$
which doesn't seem to be particularly helpful here.
| First, the statement
will cross the same point after a time interval of $\pi$
is wrong. In the cited paper the actual statement
… each timelike geodesic which intersects the $t$ axis at the point $t=t_0$
intersects that axis again at $t=t_0+\pi$.
So the $\pi$ interval refers to passing through the $x=0$, the actual period for a massive particle moving along a geodesic (as in, not only position but also velocity of the particle is the same) is $2 \pi$.
To make the “focusing property” of AdS space intuitive let us recall the canonical embedding of AdS space into the ambient pseudo-Riemannian $\mathbb{R}^{2,1}$ space with two timelike and one spacelike coordinates: $ds^2=-dU^2-dV^2+dX^2$.
AdS2 is defined as a hyperboloid $-U^2-V^2+X^2=-1$.
Internal static coordinates $(t,x)$ are connected with coordinates of ambient space via:
$$ (U,V,X) = (\sqrt{1+x^2}\cos(t),\sqrt{1+x^2}\sin(t),x) .$$
It is easy to see that the points with static coordinates $(x_0,t_0)$ and $(x_0,t_0+2\pi)$ are actually the one and the same. If we “unroll” the $t$ variable by making them distinct we actually go from AdS space proper to universal covering space of AdS. Timelike geodesics on AdS are the sections of hyperboloid by a
timelike plane of an embedding space passing through the origin. To show that, one could start by showing that circle $X=0$, $U^2+V^2=1$ (or alternatively $U=\cos \tau$, $V=\sin\tau$, $\tau$ is proper time) is a geodesic and then use AdS isometries (which is a Lorentz group $SO(2,1)$ of an embedding space) to make this geodesic into all other timelike geodesics. Since these sections are closed curves (ellipses) (for the AdS space proper), or winding curves periodic in $t$ coordinate with a period $2\pi$ (for the covering space) we have proven the statement in question (with a correct period), without explicit calculations. Incidentally, the solution $x(\tau) = A \sin(\tau) + B \cos(\tau)$ becomes kind of obvious by way of embedding space, with $A$ and $B$ coming from Lorentzian transformations of $U$ and $V$.
The actual calculations in the OP's question for the geodesic equation are correct up until the last equation. One should remember, that the condition $g(u,u)=-1$ gives us dependence between $A$ and $B$ constant of the $x(\tau)$ and the energy constant $E$. Namely, $1+A^2+B^2=E^2$. As a result if we shift $\tau\to \tau+\delta$ to eliminate $A$, we could integrate $\dot{t}=f(\tau)$ to obtain
$$\tan(t-t_0)=\frac{\tan(\tau)}{\sqrt{1+B^2}}.$$
We see, that the phase difference between $t$ and $\tau$ is never large and becomes zero after every $\pi$. And so $x(t)$ would also be periodic with a period of $2\pi$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/440470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Wave group and phase velocities When talking about phase velocity and group velocity they tend to be both expressed in terms of another $v$. From my understanding this $v$ comes from the wave equation, but what does it represent? And regarding phase velocity my understanding was that it is the velocity of a specific frequency component, however within a guided wave it is possible for this velocity to tend to infinity, which does not agree so well with SR. Any help for where I'm getting this wrong are appreciated.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/440573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
|
How does an object in space travelling at constant velocity have a net force of zero acting upon it? If the definition of balanced forces is "two opposing forces that are equal" and an object with a net force of zero acting upon it means that the forces are balanced on the object, then it should follow that an object travelling through space at constant velocity should have a net force of zero acting upon it. But there was an initial force acting upon it when it was thrown into space. What is the force that is balancing that initial force to make the net force zero?
| I think I understand your point, and it does make sense.
The way I see it is the following : the object you're throwing in space is firstly acted upon a force (let's call it F1). You're suggesting that since the object is traveling with a constant velocity, then the net force on that object must be 0, which is correct. However, where you might get confused is what kind of force F2 might have cancelled F1 to get a net force of 0, right?
Technically, no force F2 was applied. In fact, you threw the object in space with a force of F1 (the object currently accelerates from a velocity of 0 to velocity V). The object keeps on accelerating as long as F1 is applied.At some point, you're going to stop applying F1, and leave the object alone. The object already has a velocity, so it will keep on moving. Since F1 is no longer applied, no acceleration takes place (F = ma, if F = 0 then a = 0). So then your net force is 0, which explains the constant velocity and the 0 acceleration.
Hope this helps!
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/440838",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Cylinder vs cylinder of double the radius roll down an incline plane, which one wins? A solid cylinder and another solid cylinder with the same mass but double the radius start at the same height on an incline plane with height h and roll without slipping. Consider the cylinders as disks with moment of inertias I=(1/2)mr^2. Which one reaches the bottom of the incline plane first?
According to this, the velocity of any body rolling down the plane is
v=(2 g h/1 + c) ^½
where c is the constant in moment of inertia (for example, c=2/5 for a solid sphere).
My thought process was that since the radius doubled, c=2. So, the velocity of the doubled cylinder would be less, therefore finishing later. Similarly, if it’s moment of inertia increases, it’s angular and linear acceleration decreases. However, my other peers and even my professor disagree, saying that radius and mass do not play a role in the velocity of the body, since both m and r will cancel in an actual calculation of the velocity.
Could anyone elaborate on whether I am right or wrong?
| Conservation of energy tells us potential energy becomes kinetic energy as the disks fall. If they roll without slipping, some energy goes into translational kinetic energy and some goes to rotational kinetic energy.
The rolling without slipping condition requires that the velocity of the disk be equal to the rotatoinal velocity times the radius $v=\omega R$.
Total kinetic energy = $\frac{1}{2}Mv^2+\frac{1}{2}I\omega^2$
So:
$$Mgh=\frac{1}{2}Mv_1^2+\frac{1}{2}I_1(\frac{v_1^2}{R^2})=\frac{1}{2}Mv_2^2+\frac{1}{2}I_2\frac{v_2^2}{4R^2}$$
$I_1=\frac{1}{2}MR^2$
$ I_2=\frac{1}{2}4MR^2=2MR^2$
We can take the ratio of the squared velocities:
$$\frac{1}{2}Mv_1^2+\frac{1}{2}(\frac{1}{2}MR^2)(\frac{v_1^2}{R^2})=\frac{1}{2}Mv_2^2+\frac{1}{2}(2MR^2)\frac{v_2^2}{4R^2}$$
$$\frac{1}{2}Mv_1^2+\frac{1}{4}(M)(v_1^2)=\frac{1}{2}Mv_2^2+\frac{1}{4}(M)v_2^2$$
$$\frac{v_1^2}{v_2^2}=\frac{1+\frac{I_2}{4MR^2}}{1+\frac{I_1}{MR^2}}=\frac{\frac{3}{2}}{\frac{3}{2}}=1$$
So I stand corrected. Using the proper radii consistently, the velocities end up the same.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/440946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Spatial part of Robertson-Walker metric The spatial part of the FRW metric can be written as $$d\Sigma^2=d\rho^2+f^2(\rho)(d\theta^2+{sin}^2\theta d\phi^2)$$ where $f(\rho)$ satisfies $$\frac{df}{d\rho}=\frac{f(2\rho)}{2f(\rho)}.$$ I am trying to derive the form of $f(\rho)$ by using a power series expansion $f(\rho)=\sum a_n \rho^n$ and show that $f(\rho)$ can be $\rho$, $R\sin(\rho/R)$ or $R\sinh(\rho/R)$. I am getting stuck.
What should be further steps?
| Should be:
$f(\rho) = \sum a_n \rho^n$
$df/d\rho = \sum n a_n \rho^{n-1}$
$f(2 \rho) = \sum a_n (2 \rho)^n = \sum 2^n a_n \rho^n$
You plug what above in the equation, you multiply L.H.S. and R.H.S. times the expression for $f(\rho)$ and move all the terms to the L.H.S. Then order the terms as per powers of $\rho$, e.g. $\rho^0$, $\rho^1$, $\rho^2$, etc. For each power you equate the coefficient to zero. You should get a recurring formula relating $a_n$ to $a_{n-1}$.
DERIVATION
$df/d\rho = f(2 \rho) / (2 f(\rho))$
$(df/d\rho) (2 f(\rho)) = f(2 \rho)$
$(\sum_n n a_n \rho^{n-1}) 2 (\sum_m a_m \rho^m) - \sum_l 2^l a_l \rho^l = 0$
$\sum_n \sum_m 2 n a_n a_m \rho^{n-1+m} - \sum_l 2^l a_l \rho^l = 0$
Equating the coefficients of power $l$ requires $n - 1 + m = l$, that is $m = l + 1 - n$, hence
$a_l = 2^{-(l - 1)} \sum_{n = 0}^{l + 1} n a_n a_{l + 1 - n}$
Solution I
$l = 0, n = 0, 1$
$a_0 = 2 a_1 a_0$
$a_0 (2 a_1 - 1) = 0$
Let us consider $a_0 = 0$ and $a_1$ undefined
$l = 1, n = 0, 1, 2$
$a_1 = a_1^2$
$a_1 (a_1 - 1) = 0$
$a_1 = 1$ defined
$l = 2, n = 0, 1, 2, 3$
$a_2 = 2^{-1} (a_2 + 2 a_2)$
$a_2 = 0$
$l = 3, n = 0, 1, 2, 3, 4$
$a_3 = 2^{-2} (a_3 + 3 a_3)$
$a_3$ undefined
If we choose $a_3 = 0$, then the equation for $a_l$ with $l \ge 4$ provides a zero value if the previous coefficients up to $a_{l - 1} $ are zero, except for $a_1$.
To summarize we got the following solution:
$a_1 = 1$
$a_l = 0$ for $l \ne 1$
That is $f (\rho) = \rho$
It is the zero curvature geometry of FRW metric, i.e. a flat universe.
Note: To look for the positive/negative curvature solutions you have to exploit the fact that $a_3$ is undefined. You assume it is non zero and then proceed. However it is more laborious.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/441052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is the normal force equal to weight if we take the rotation of Earth into account? In my physics class we were doing problems such that we set $N$ (normal force) $= mg$. I understand that by Newton's Third Law, if I exert a force on the ground, then the ground will exert an equal and opposite force on me. However, the part that I am slightly confused about is that when the Earth rotates, and thus I rotate too, I am accelerating with the centripetal force towards the center of the earth (assuming I am at the equator). How am I doing this if the normal force equals $mg$? If the normal force doesn't equal mg then why isn't the ground exerting an equal and opposite force?
| I'd like to finalize the conclusion from @AaronStevens's great answer. In the truer expression for normal force (on flat ground) that he arrives at,
$$N=mg-\frac{mv^2}{r}=m\left(g-\frac{v^2}{r}\right)\quad ,$$
Earth's rotation adds the term $\frac{v^2}{r}$ so it deviates from the expected $N=mg$. How much is the influence of $\frac{v^2}{r}$?
Earth's radius is around $r=6400\;\mathrm{km}$. In one day, which is $t=24\,\mathrm{hr}=86400\,\mathrm s$, we move through the entire circumference of Earth, which is $d=40200\,\mathrm{km}$. That gives us a constant speed of around $v=d/t=465\,\mathrm{m/s}$. I am aware that I have used rough numbers here, from the top of my head, mainly fitting from the Equator. You can try to redo the calculations with more accurate values.
If we plug in $r$ and $v$, we get something like:
$$\frac{v^2}{r}=0.0338\,\mathrm{m/s^2}$$
Compare this with $g=9.80\,\mathrm{m/s^2}$. The contribution of Earth's spin to the effective gravitational acceleration $(g-\frac{v^2}{r})$ is thus only something like 0.3 %. You can try to calculate a normal force for an object with and without this influence and see if there is a significant difference within significant figures.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/441245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
Lagrangian of EM field: Why the $B$-field term has a minus sign in front of it in the Lagrangian? I know that $L = T - U$ and that, in the non-relativistic case
$$L= \frac{1}2mv^2 - q\phi(r,t) + q\vec{v}\cdot\vec{A}(r,t).\tag{1} $$
My lecturer used the following form of the Lagrangian density to derive Maxwell's equations:
$$L = \vec{j}(r,t)\vec{A}(r,t) - \vec{\rho}(r,t)\vec{\phi}(r,t) + \frac{\epsilon}2 \vec{E}^2(r,t)-\frac{1}{2\mu}\vec{B}^2(r,t). \tag{2}$$
Comparing the two equations for $L$, I can see that the KE term in the first equation is substituted for the energy density of the EM field. What I do not understand is why the $B$-field term has a minus sign in front of it in the Lagrangian (2)?
Can someone please shed some light on this for me please?
P.S - I have checked the related posts and none of them address my issue.
| In the gauge $\phi=0$, the $E$ term is $\frac12\dot A^2$, which is kinetic energy, and the $B$ term is $(\nabla\times A)^2$, which is potential energy and therefore gets a minus sign.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/441516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Do centrifugal force and gravity differ in their effects on objects? If the type of object matters, consider the human body. If the situation matters, consider standing on the inside wall of an O'Neill cylinder compared to standing on the surface of Earth.
"Differ in their effects on objects" means: Would the object be able to tell the difference? That is, is there an instrument that could tell whether it is placed in an O'Neil cylinder or on the surface of a planet from the effects (acceleartion, I suppose) of centrifugal force and gravity alone?
| General Relativity is compliant with the Strong Equivalence Principle. According to this principle:
The outcome of any local experiment (gravitational or not) in a freely falling laboratory is independent of the velocity of the laboratory and its location in spacetime.
This implies that locally gravity is indistinguishable from acceleration. Thus the answer is that locally the effects of gravity and the centrifugal force are the same. Here "locally" means a region small enough where the force is uniform. For example, if the rotating cylinder is large and you are confined inside an elevator, you would have a very hard time telling gravity from acceleration. However, in a larger region, many different experiments and observations would easily reveal differences between the centrifugal force and gravity, as justly stated in the comments and the other answer.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/441606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Why does the warm air rises up? Warm air has more energy than cold air. This means that according to the Einstein equation $E = mc^2$ the warmer air has a greater mass than the cold one. Why is the warm air rising, if it has a greater mass, which means that the attraction of gravity between the Earth and the warm air is greater?
| Even though $E=mc^2$ is only for objects that are not moving and we really should be using $E^2=p^2c^2+m^2c^4$, you are conceptually correct in that objects with more energy have more mass. The issue is that c is a huge number so it takes a ton of energy to actually give something mass in a significant way. c is 299,792,458m/s and $c^2$ is 89,875,517,873,681,764$m^2/s^2$. The reason hot air rises is due to a combination of the ideal gas law and the buoyant force. According to the ideal gas law, an increase in temperature will also increase in volume (assuming pressure and # of air molecules are held constant of course). You can think of the air molecules all bouncing against each other. The hotter they are the faster they are moving and the more they push up against the air around it and expand. Now it occupies more space but is still the same mass(extremely close). The buoyant force is to the air around it also having weight. That air is "trying" to get underneath anything that takes up space. The more space it takes up the more surrounding air there is exerting a force on it. This force exceeds gravity when the air is less dense than the surrounding air causing it to rise. For more information consult the following sources:
http://physics.bu.edu/~duffy/sc527_notes01/buoyant.html
http://hyperphysics.phy-astr.gsu.edu/hbase/Kinetic/idegas.html
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/441954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Relationship between freefall velocity time dilation and gravitational time dilation in a Schwarzschild metric If you drop an object into a gravitational field, is its final velocity equal to what it would have to be in flat space in order to generate the same time dilation that you get at a given radius for an object that is stationary relative to the gravitational body (sitting on the surface in the case that it isn't a black hole)? I don't have enough GR background to do the calculation myself but this seems consistent with the effects on photons going into a gravitational well.
Here's what I've already figured out (mostly from http://jila.colorado.edu/~ajsh/bh/schwp.html)
*
*The distance toward the black hole is contracted/expanded by an amount $\dfrac{1}{\sqrt{1−r_s/r}}$ where $r$ is "circumferential radius" that you get from dividing the orbit length by $2\pi$ and $r_s=2GM/c^2$ is the Schwarzschild radius.
*Time dilatation relative to "Schwarzschild time" is $\sqrt{1−r_s/r}$.
| Yes, this is correct. From Wikipedia:
"Time dilation in a gravitational field is equal to time dilation in far space, due to a speed that is needed to escape that gravitational field. Here is the proof.
*
*Time dilation inside a gravitational field $g$ is $t_0 = t_f \sqrt{1 - \frac{2GM}{rc^2}}$
*Escape velocity from $g$ is $\sqrt{2GM/r}$
*Time dilation formula per special relativity is $t_0 = t_f \sqrt{1-v^2/c^2}$
*Substituting escape velocity for v in the above $t_0 = t_f \sqrt{1 - \frac{2GM}{rc^2}}$
Proved by comparing 1. and 4."
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/442032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Would a supersonic object without a combustion power source leave behind a contrail? Contrails, as far as I understand them, are caused by either a pressure change that forces the condensation of H2O(g) OR by the release of warm H2O from a combustion engine. Most plane contrails, I would assume, operate largely by this second mechanism as they burn jet fuel and release warm CO2 and H2O.
My question is, can the first mechanism alone be enough to create a contrail? Would a supersonic object passing through Earth's atmosphere leave behind a contrail? If so, what conditions are required for this - presumably high pressure (close to the surface of the Earth) and high speed?
For those curious, this started out as a question on Worldbuilding and has been asked again here per a wise comment.
| The space shuttle is an example of the phenomenon of a supersonic craft leaving behind a contrail.
The Fact that the space shuttle flies as a glider on re-entry is stated by NASA here.
The velocity of the space shuttle at different altitudes during descent is depicted here in the section "Repeat explanation of a Shuttle reentry and landing".
The space shuttle leaving behind a contrail is shown in the third space shuttle image on this page, and by an enthusiast here.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/442267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
If the molecular collisions are elastic will there be any dissipation in a fluid? Viscosity arises due to collisions of the molecules of one layer of a fluid with another in contact. But viscosity is a dissipative element leading to heating and dissipation. Where does it heat come from? Does it come from the molecular collisions being inelastic? If the collisions were elastic, would there be no viscosity or dissipation in a fluid?
| The viscosity arises because of the elastic collisions, not in spite of them.
When temperature in the fluid rises, it is because the molecules are moving faster, due to elastic collisions. The energy is called "heat" because the motion is disorganized (random).
Its entropy has increased.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/442363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
What is the shape of a gravitational wave form? What is the shape of a gravitational wave as it hits the Earth, particularly the time portion.
Does time start at normal speed, then slow slightly, and then return to normal speed?
Or does it start at a normal speed, slow down slightly, then speed up slightly, and then return to normal speed?
Those other questions only concerned whether time dilation exists. I'm more concerned with the shape of the wave form. So not the same questions at all.
| A simple monochromatic gravitational plane wave has two possible transverse polarizations. If the wave is traveling in the $+z$ direction, then the metric for one of the polarizations can be written in the simple form
$$ds^2 = -dt^2 + (1+h_+)\,dx^2 + (1-h_+)\,dy^2 + dz^2$$
where the small metric perturbation $h_+$ is wavelike:
$$h_+ = A_+ \cos{(k z-\omega t)}.$$
The metric for the other polarization is similar, but rotated by 45 degrees in the $x$-$y$ plane.
So, at least in the coordinate system where the metric takes this form, time does not speed up or slow down and distances along the direction of the wave do not grow or shrink. But distances perpendicular to the wave’s direction do grow and shrink, in a wavelike fashion.
For this polarization, when distances in the $x$-direction grow slightly, distances in the $y$-direction shrink slightly. So a circular ring in the $x$-$y$ plane would be distorted slightly into an ellipse elongated in the $x$ direction, then back to a circle, then to an ellipse in the $y$ direction, then back to a circle, etc.
When LIGO detects a gravitational wave, it is because the length of LIGO’s two “arms” oscillates. When one gets slightly longer, the other gets slightly shorter. But the effect is very small and hard to detect, because the amplitude $A$ of gravity waves that it is detecting is roughly $10^{-21}$. This tiny metric perturbation makes LIGO’s 4-km arms grow and shrink by less than one-thousandth of the size of a proton!
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/442588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How should I imagine a multi-particle state in a free QFT? It is reasonable to think of single-particle Focks states as of plane waves. Indeed, since $|p\rangle=a^\dagger_p|0\rangle$ and $\langle x|p\rangle\sim \operatorname{e}^{ipx}$, we conclude that the state $|p\rangle$ can be thought of as a plane wave in the position representation.
What about multi-particle states, such as $|p_1,p_2\rangle=a^\dagger_{p_1}a^\dagger_{p_2}|0\rangle$? Naturally, I would imagine those to be superpositions of plane waves (since in the free theory we are dealing with equations obeying the superposition principle). But is it really like that? Is the matrix element $\langle x|p_1,p_2\rangle$ really equal to the sum of plane waves?
| Take a compound system of many strongly interacting particles at $T=0$ and give it a shake. The system in an excited state can be approximately described as a set of free quasi-particles corresponding to the normal modes of the system.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/443083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Minimum separation from the spacetime interval I've been working through invariant spacetime interval questions recently, and I came across a question in my lecture notes where;
$$\Delta s^2=\Delta x^2 -(c\Delta t)^2 > 0 $$
Now it is clear to me that there is no frame where $\Delta x' = 0$ which I have already proven as the question required. Now, out of curiosity, I'm wondering if there is a way to determine the minimum value that $(\Delta x')^2$ can take?
I am assuming that the spacetime interval is the same in every frame, so
$$\Delta s'^2=\Delta x'^2 -(c\Delta t')^2 > 0$$
which would give
$$\Delta x'^2 > (c\Delta t')^2$$
But since $t'$ can be equal to 0, I'm not sure where to go from here. Is there anybody that can either show me how, or point me in the right direction? Any help is much appreciated.
|
But since $t'$ can be equal to 0, I'm not sure where to go from
here.
$\let\D=\Delta$
You meant $\D t'$, didn't you? You were on the right track, but went
lost. Your first equation says
$$\D s^2 = \D x^2 - (c\,\D t)^2 > 0\tag1$$
Then you wrote
I am assuming that the spacetime interval is the same in every frame
which is right, but wrote
$$\D s'^2 = \D x'^2 - (c\,\D t')^2 > 0$$
forgetting what you had just written: $\D s'=\D s$. So this equation should
have been
$$\D s^2 = \D x'^2 - (c\,\D t')^2 > 0$$
If $\D t' = 0$ we have
$$\D s^2 = \D x'^2.$$
Comparing with (1)
$$\D x'^2 = \D x^2 - (c\,\D t)^2$$
and this is the least $\D x'$, given $\D x$ and $\D t$. It obtains in a frame where $\D t'=0$.
There is just one point to be understood: are we sure a frame where
$\D t'=0$ actually exists? The answer is yes, but can you give a proof?
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/443198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Is there a simple way to calculate Clebsch-Gordan coefficients? I was reading angular momenta coupling when I came across these CG coefficients, there is a table in Griffith's but doesn't help much.
| It might depend on your definition of "simple".
For easy cases (low numbers, direct steps), yes, it is simple. However, it gets complicated too fast. What we do is: calculate only the easy ones, and let the rest for computers. I really encourage you to do this.
The trick is: at the top of the ladder, there is only one possibility, except for a global phase factor. For example:
$| j_1 \ j_2 ; m_1 m_2 \rangle = |1\ ½; \ 1 \ ½\rangle $ can only be one possibility: $|J \ M\rangle= |3/2\quad 3/2\rangle $
BEcause it is the maximum value of both $j_1$ and $j_2$. They can only yield the maximum $J,M$.
So we know one equivalence:
$|1\ ½; \ 1 \ ½\rangle \equiv|3/2\quad 3/2\rangle $
or, if you want,
$|1\ ½; \ 1 \ ½\rangle \equiv 1\cdot|3/2\quad 3/2\rangle $
I know, I know, except for a phase factor, but let's ignore that.
So, what do we do now? We apply $J_-$ at both sides of the equation.
On the one hand, $J_-=J_{1-}+J_{2-}$, so we know what it does:
$J_- |1\ ½; \ 1 \ ½\rangle = J_{1-}|1\ 1\rangle +J_{2-}|½\ ½\rangle $ (carry on yourself).
On the other hand, the result is also an angular momentum, so:
$J_- | 3/2 \quad 3/2\rangle = \hbar\sqrt{3/2\cdot(3/2+1)-3/2\cdot(3/2-1)} \ \ |3/2 \quad ½\rangle $
So, equating both sides, you can get
$|3/2 \quad ½\rangle $ as a function of $|1\ ½; \ 1 \ -½\rangle$ and $|1\ ½; \ 0 \ ½\rangle$.
You can get an orthogonal vector to that one, so you complete the sub-space.
Now, you can keep applying $J_{-}$ to both of them. Like that, you get all C-G coefficients.
Of course, you can also start by the bottom of the ladder and go upside with $J_+$.
This is the standard procedure. You do it once with the simplest case (for example, two ½ spins). Then, you make the computer program and breath. By the way, there are already many of them, even online.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/443496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Particle in a 2d box, using de broglie wavelength to find momentum of the particle Consider a proton in a 2-dimensional infinite potential box with widths, L and K.
a) Using de Broglie wavelength, find the momentum of the particle.
I find some online sources that finds the energy (not momentum) using the wavefunction and seperation of variables. But the question clearly states that we should find the momentum using the de Broglie wavelength
I am a bit confused actually. How can we find the momentum just using the wavelength ?
If the box was one dimensional, the momentum could be expressed as $p=hn/2L$
since we have 2 dimension we will have 2 momentum component ?
Can I express my solution as $p=p_x+p_y=hn/2K+hm/2L$ ?
I am thinking that this is wrong. But also thinking about what question asks, I get confused
Thanks.
| Momentum is a vector. The norm is given by $p=\sqrt{p_x^2+p_y^2}$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/443691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Could quantum fluctuations spawn real matter? Would it be plausible for fluctuations in the QED vacuum to spawn actual matter (such as quarks, electrons the constituents of a hydrgen atom) given enough time and space?
| Note: People seem to think that an empty universe is an eigenstate of the full QED Hamiltonian. This is not true. The Hamiltonian is
$$H= \sum_{e^+e^-} \omega_p (b^\dagger_p b_p +c^\dagger_p c_p)+ \sum_{\rm photons}\!k\,a^\dagger_k a_k\,+\, \sum_{k+p+p'=0}(a^\dagger_k + a_k)(b_p c_{p'} + c^\dagger_p b^\dagger_{p'})$$
If we define the vacuum as
$$a_k|0\rangle = b_p|0\rangle = c_p|0\rangle =0,$$
Then
$$H|0\rangle = \sum_{k+p+p'=0}a^\dagger_kc^\dagger_p b^\dagger_{p'}|0\rangle \neq 0$$
That means that the true vacuum
$H|\Omega\rangle =0$
has
$$a_kb_kc_k|\Omega\rangle \neq 0.$$
The number of particles is not conserved. Neither is energy with respect to the free Hamiltonian. Even for a single particle.
====================================
Normally, a process that violates energy conservation (of the free Hamiltonian!) by an amount $\Delta E$ can only survive for a time $\hbar/\Delta E$. Energy is only strictly conserved (in scattering for instance) for processes where the initial and final states are at $t=\pm \infty$. For example, you could create the state with two positrons and two electrons through the process (time going up):
but it would only last a time $\hbar/4m_e \sim 10^{-22}$ seconds. The reason in that in the interaction picture for scattering from $t=-T$ to $t=+T$, the scattering probability will look something like
$$P \sim \left|\int_{-T}^T\!dt\,e^{i(E_f-E_i)t}\,\langle f | V|i\rangle\right|^2\sim \frac{\sin^2(E_f-E_i)T}{(E_f-E_i)^2}\left|\langle f | V|i\rangle\right|^2$$
You can check that this is non-zero for times $T\lesssim 1/(E_f-E_i)$.
However, when gravity is included, time can be stretched, making short-lived states very long-lived. This happens when you have a horizon, like when there is a black hole or in an accelerating frame, in which case time can be infinitely stretched, producing matter. I think this also happens in inflation but there the particles get pulled away from each other. Not sure how to show this in terms of quantum mechanics.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/443802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 4,
"answer_id": 1
} |
Does it make sense to view all solutions to Einstein's field equations as $g_{\mu \nu}=\eta_{\mu \nu}+h_{\mu \nu}$, for a given $h_{\mu \nu}$? I have a question about General Relativity that so far I have never been able to solve.
Let's take Einstein's field equations:
$$ R_{\mu \nu}-\frac{1}{2}g_{\mu \nu}R=\frac{8 \pi G_{N}}{c^3}T_{\mu \nu}. \tag{1}$$
By imposing proper initial conditions and an energy-momentum tensor, one is able to find solutions for the metric tensor that include the Schwarzschild's metric, the Freedman-Robertson-Walker one and so on.
Now, let's define the following
$$g_{\mu \nu} \equiv \eta_{\mu \nu}+h_{\mu \nu}\tag{2}$$
and let's insert this expression into the previous Einstein's field equations.
My question is the following
"Given a proper $T_{\mu \nu}$, are all solutions in $g_{\mu \nu}$ and $h_{\mu \nu}$ equivalent?"
Is it possible that there exist certain solutions that cannot be expressed as a Minkowski background $\eta_{\mu \nu}$ plus fluctuations $h_{\mu \nu}$?
|
Is it possible that there exist certain solutions that cannot be expressed as a Minkowski background $\eta_{\mu \nu}$ plus fluctuations $h_{\mu \nu}$?
Yes. There are links between curvature and topology, and some topologies are not consistent with a Minkowski metric. For example, closed FLRW spacetimes have the spatial topology of a 3-sphere, which is not compatible with a Minkowski metric.
Even for spacetimes that topologically could admit a Minkowski metric, there would be the question of what coordinate system to impose on the spacetime in order to identify some coordinates with the Minkowski coordinates. In general this coordinatization is not unique.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/443866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
When to use sine or cosine when computing simple harmonic motion For simple harmonic motion (SHM), I am aware you can start of using either sine or cosine, but I am a bit confused as to when you would start off with sine rather than cosine. I know that a sine graph starts at $y=0$ and a cosine graph starts at $y=1$. So therefore, I would say you use sine for equilibrium positions?
However, I came across a question asking to write down the equations for position, velocity and acceleration of a particle starting from rest at time $t=0$, then undergoing SHM with maximum amplitude 0.2 m and period 5 sec.
I worked out the angular frequency to be $2\pi/5$ from the period formula. And then used the position formula to be of the form with sine and differentiated to get cosine velocity equation, etc. However, the answer says I should have started with cosine and I am now unsure when I should start with sine or cosine.
| "a particle starting from rest at time t=o" This is the key to why you start with cosine. So you know that instantaneous velocity is the derivative at that point. Look at a sine graph at t=0 there is a positive gradient thus the particle doesn't start at rest with sine. But with cosine at t = 0 the gradient is zero thus the particle is initially at rest.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/444247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Does a rock use up energy to maintain its shape? A rock sitting on land, the ocean floor, or floating in space maintains its shape somehow. Gravity isn't keeping it together because it is too small, so I'm assuming it is chemical or nuclear bonds keeping it together as a solid. If not it would simply crumble apart. So, what type of energy maintains the shape of a rock, where did this energy come from, and is it slowly dissipating?
As a corollary, if a large rock is placed on top of a small rock, is the energy required to maintain the shape of the small rock 'used' at a greater rate?
| The amount of work done is equal to the distance moved times the force in the direction of motion. As the rock is staying the same shape it does not need to exert energy.
You may be thinking that the rock needs to expend energy in order to hold up its heavy mass in the same way our muscles do if we hold up a heavy weight. But muscles need to contract to lift a heavy weight and this requires continuous activity at the cellular level as explained in the answer to this question.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/444307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 5,
"answer_id": 1
} |
Two People Pushing Off of Eachother, Newton's Third Law, and Unbalanced Force Different versions of this question have come up all over the internet. Usually it deals with tension in a rope or two people pushing on each other with the same force. I am trying to understand 2 people pushing each other with different forces.
Say two people of equal mass are standing on a frictionless surface, touching palm to palm, and they push off of each other at the same time with different amounts of force. Say the person on the left pushes with 100 N and the person on the right pushes with 70N. What happens to this system? It seems like there should be a net force of 30 N to the right, but at the same time, each person pushing should feel the same force pushing back as per Newton's 3rd Law.
If I push on you with 100 N then I also feel the 100 N from the force pair and we slide apart each having been accelerated by 100 N. If you push on me with 70 N then I feel the 70 N force and I exert 70 N back on you, we slide apart each having felt 70 N this time. But what happens when we both push with a 100 N and 70 N respectively?
Also, I have seen elsewhere on this site that if 2 people were to push on each other with the same force under conditions like those in the scenario above, they each feel the same force but fly apart at double the final velocity because the same force has been applied for twice the distance. Is this actually the case?
| The problem here stems from understanding closed systems.
The center of mass of a closed system (no external forces) does not move. This comes from the fact that Newton's third law states that any internal force applied will be canceled by another opposite and equal internal force.
The forces you and your friend are applying on each other are internal forces. The argument with 30 N that you made only applies if some external third person pushed your friend 100 N and then you 70 N. Those are forces outside the system where the equal and opposite force was applied to an object outside your system (the third person) causing a net change to your system. Because the forces in your description comes from within what you define as your system, the center of mass doesn't change i.e. no net force. Hope that helps.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/444396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Commutive property of the Bra-ket notation I'm struggling when it comes to understanding the commutive properties of the Bra-ket notation in quantum mechanics. I understand how to work with constants, bra and kets. However, the second I start introducing eigen-equations such as
$$\hat{x}|x'\rangle = x'|x'\rangle$$
to solve problems like
$$\langle\psi_p|\hat{x}|x'\rangle = x'\psi_p^*(x')$$
I instantly lose grip on the rules for the Bra-ket notation.
Am I allowed to simply pull the $x'$ out of the bracket like this: $\langle\psi_p|x'|x'\rangle=x'\langle\psi_p|x'\rangle$?
My book only state the rules for operators between a bra and a ket, $\langle a|\boldsymbol{A}|b\rangle$, but not how to handle problem including eigen-equations such as this one.
| When writing $\hat x\vert x'\rangle=x'\vert x'\rangle$, the $x'$ is actually a number (aka a scalar) and so can be moved about like a regular number, so that
$$
\langle \psi_p\vert\hat x\vert x'\rangle =
\langle \psi_p \vert x'\vert x'\rangle = x'\langle\psi_p\vert x'\rangle
$$
because $x'\in \mathbb{R}$, much in the way that $\langle b\vert 3\vert b\rangle=3\langle a\vert b\rangle$.
Indeed if $\hat A\vert b\rangle = \alpha \vert c\rangle +\beta \vert f\rangle$ then
$$
\langle a\vert \hat A\vert b\rangle=
\langle a\vert\left[\alpha \vert c\rangle +\beta \vert f\rangle\right]
=
\langle a\vert\alpha \vert c\rangle +\langle a\vert \beta \vert f\rangle
=\alpha \langle a\vert c\rangle +\beta \langle a\vert f\rangle.
$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/444515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Laplace transform on thermodynamics equation I'm trying to create a simple model of a single-flash geothermal plant which consist of 3 main parts (flash-separator, turbine, and condenser). Is there any way to create a transfer function using Laplace Transform (s-domain) of each parts based on the thermodynamics equation?
| The first thing you need to do is write down the time-dependent mass- and energy balance equations for the turbine (using the open system control volume version of the 1st law of thermodynamics). Then you express each the variables as the sum of a steady state part and a transient part. Then you solve for the steady state part. Then you linearize the time-dependent equations with respect to the perturbations about the steady state. Then you can use the Laplace transform for the linearized perturbations.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/444843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Lorentz' Derivation of Lorentz force I have one simple question: Can someone point me to the paper where H.A. Lorentz published Lorentz force? I was digging through the usual literature in electromagnetics (Jackson, Griffiths, Thide, Sommerfeld, Stratton) for the reference on the original Lorentz paper with no success. Any help is greatly appreciated
| What is nowadays known as the Lorentz force law was originally due to Maxwell, equation 77 in Part 2 of his 1861 paper On Physical Lines of Force (p. 482 of vol. 1 of his Scientific Papers), which in more modern vector notation looks like:
$$\mathbf{E}=\mu\mathbf{v}\times\mathbf{H}-\frac{\partial\mathbf{A}}{\partial t}-\nabla{\psi},$$
where $\mathbf{E}$ is EMF.
References:• Tombe, F. D. (2012a). Maxwell’s original equations. The General Science Journal. pp. 4-5.• See the references in p. 225 fn. 26:Assis, André Koch Torres; Chaib, J. P. M. C; Ampère, André-Marie (2015). Ampère's electrodynamics: analysis of the meaning and evolution of Ampère's force between current elements, together with a complete translation of his masterpiece: Theory of electrodynamic phenomena, uniquely deduced from experience (PDF). Montreal: Apeiron. ISBN 978-1-987980-03-5. p. 225:[It] was first obtained by J. C. Maxwell between 1861 and 1873, and by H. A. Lorentz in 1895.²⁶
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/444936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
What is the function of this complicated tensioning system? I saw this arrangement for tensioning overhead cables from my train window (schematic below). Why not just have one pulley wheel leading directly to the weights? What function do the additional pulleys serve? For that matter, what are the cables for? They're clearly not power lines.
| This is a so-called block and tackle arrangement which is often used for tensioning of overhead lines. Tensioning is required to keep a desired line geometry and, in case of contact wires, to avoid standing mechanical waves (waves in a tensioned line travel faster).
Apart from mechanical advantage, such systems often provide a safety braking mechanism which prevents the line from unravelling completely in case it breaks.
The location in your photo most probably corresponds to an overhead line section break: the power line on the left side has ended, was pulled outside of the tracks and terminated by a tensioning mechanism. About a hundred meters to the left, a new section must have started (probably with another tensioning mechanism) and pulled to the middle of the track at the point where the photo is taken. That's the wire you see just above the mechanism in question.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/445062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
"answer_count": 5,
"answer_id": 3
} |
Why are green screens green? How/why do green screens work? What's so special about the color green that lets us seamlessly replace the background with another image and keep the human intact?
Are there other colors that work similarly?
| It's partly about how human colour vision works, partly about avoiding colours you want to keep, such as those of the actors.
Colour cameras record concentrations of red, green and blue light to mimic human colour vision. Before digital techniques, blue screens were preferred because, of the three primary colours, that's the one rarest in human skintones.
When digital cameras were invented, they were given greater sensitivity to green light to mimic a bias in human vision. Green screen doesn't require as much illumination of the screen as blue screen does, which prevents the risk of chroma spill onto the foreground subject's edge, which can cause a special effects failure called a chroma halo.
In the pre-digital era, when the foreground-background distinction had to be much larger than is required today (because of the complicated optical process involved in achieving chroma key), it was almost impossible to get away with any colour beyond blue. Nowadays both colours are very common, with green almost the new default; but, unlike the blue-only era of the past, typically both colours are now on standby.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/445158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 2,
"answer_id": 1
} |
Movement of fluid in a container filled with that same fluid If a cylinder with the bottom end closed and the top end open was filled with water and then dropped in a pool of water. Would the water inside the cylinder stay in the cylinder?
| You could do this experiment at home. Take a glass of water, add some food colouring und then drop it into your bathtub (fill it first with enough water, otherwise your 'container' will burst...). My prediction:
The water in the glass gets accelerated, first, together with the glass. The glass will then hit the water and will feel a strong deceleration. The water inside it will feel the same, but as a non-rigid fluid the pressure on the vertical axis will get re-distributed to the other sides, particularly to the open top of the glass. This is especially true if your container rotated by 90° while falling. If not, and the top of the glass is still the highest point when landing, then it still holds that there is no counter pressure and the fluid will even out the local pressure terms to get into equilibrium. An additional effect might occur, when the glass sinks deep enough to be covered by water: hydrogen bonds might form and dilute your coloured water with the surrounding water.
Still, I'd like to see the experiment! :D
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/445228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Examples of central forces on the path of orbit? In solving a problem from Goldstein (3.13), I solved for multiple properties of a circular orbit with the attractive central force where the path of orbit crosses the point of the force (at origin).
The solutions were simple enough to find, but what's been in the back of my mind is what type of physical system does this represent? I am used to Kepler type problems where the central force is located within the orbit and not on it. What system would this be applied to? Or is it merely an exercise?
| Are you asking purely about an orbit system? Or a general system? One I can think of is simple harmonic motion. Take a mass on a spring. The force will always act towards the origin, and the mass will go through the origin. Although the definition of a central force is one that always acts towards a fixed point (in this case, the origin), I'm not 100% sure if simple harmonic motion counts as a central force. Perhaps someone can correct me here if necessary.
I'm struggling to think of an orbit example in space which fits your problem however.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/445323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Neutral buoyancy for cylinder/bladder in water A 25 liter flexible plastic bladder consists of three parts: an air chamber located on top of the bladder, a large middle water chamber, and a lower weighted area that serves as a counterbalance and ballast found below the bladder. The bladder is cylindrical in shape and is designed to float on the surface of a swimming pool. The air chamber (holds 2.5 liters of air) keeps the bag floating at the surface, while the water chamber (holds up to 25 liters), remains below the surface of the water, as in the picture below. The counterbalance stabilizes the bladder and keeps it in a neutral buoyant position, as shown in the image below.
If the air chamber is filled to maximum capacity (2.5 liters of air), the water chamber will hold only 22.5 liters, for a 25 liter maximum capacity for the bladder. In this example, the water to air ratio is 10:1. In its current state, the bladder floats at the surface of the water, but if additional weight is added, the bladder sinks.
What I want is to determine the exact amount of weight needed to suspend the bladder in a neutrally buoyant position at the surface and at 20 cm below the surface. Is there a formula to determine neutral buoyancy?
I have put together a short video showing the bladder in a neutrally buoyant position on my website at: www.romaquatics.com/videos
| Since your just looking for an estimated ratio to get you close from the jump, consider the volume of air in the bag, and the weight of volume of water it is displacing. Counterweight for the weight of the water displaced and fine tune it from there.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/445433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
In what context is enthalpy a convenient concept? Internal energy $U$ is clearly an important concept; the first law of thermodynamics states that for an isolated system internal energy is constant $(\Delta U=0)$ and that for a closed system the change in internal energy is the heat absorbed by the system $Q$ and work done on the system $W$ $(\Delta U=Q+W)$.
Enthalpy $H$ is the sum of the internal energy and the product of the pressure and volume of the system $(H=U+PV)$. I was taught that enthalpy is a preferred quantity to internal energy for constant pressure systems where $\Delta H=Q$, as opposed to constant volume systems where $\Delta U=Q$. But why would anyone care about what quantity is equal to $Q$ under certain conditions instead of simply reporting $Q$? Enthalpy seems redundant in this context.
Is enthalpy a convenient concept in other contexts, such as systems with varying pressure? Is it described by any fundamental laws as internal energy is by the first law?
| Enthalpy is a particularly convenient concept when analyzing components, such as turbines, compressors, pumps, condensers, evaporators, and expansion valve, in open steady flow processes. Here are some examples for the case of an ideal Rankine power cycle where the $h$ is enthalpy:
Reversible, adiabatic turbines: $\dot W_{out} =\dot m(h_{in}-h_{out})$
Reversible, constant pressure condensers (heat extraction): $\dot Q_{out}=\dot m(h_{in}-h_{out})$
Reversible, adiabatic compressors, pumps: $\dot W_{in}=\dot m(h_{in}-h_{out})$
Reversible, constant pressure, expansion (boiler heat addition): $\dot Q_{in}=\dot m(h_{out}-h_{in})$
Hope this helps
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/445649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Can the second law of thermodynamics be derived from Quantum randomness? The second law of thermodynamics says that the entropy of an isolated system continuously increases. Can we say that this is due to Quantum mechanics, which continuously increases the information contained in the system by producing random numbers? Is the entropy of a classical system without randomness always constant?
| The laws of thermodynamics can be derived (or at least motivated) from statistical physics.
To do so, we assume that there is an "underlying theory" describing the microscopic physics of the system.
However, we do not know have perfect information about the state of the system (we say that we do not know the "microstate") - we only know some macroscopic properties, the "macrostate".
This derivation is remarkably universal, it does not care much whether the underlying theory is classical or quantum.
Is the entropy of a classical system without randomness always constant?
No, it is not.
While the microscopic processes are deterministic, there is a randomness which comes from the fact that we don't have perfect information about the microstate.
The second law of Thermodynamics only says that in this situation of imperfect information, the system is extremely likely to assume the most typical macrostate.
If we consider the case where we have perfect knowledge of a classical system, that means that the entropy of the system is zero and it will remain zero.
(You could say that entropy measures the amount of missing information.)
Note that the same is true for a quantum system, if you have perfect information and don't measure it, its entropy will not change.
Only if you measure a quantum state (and don't look at the result of the measurement), you suddenly have less information about the system state than before:
its entropy increases.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/446092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
What is the length of null geodesic? There are many questions about this but none of them adresses my concrete question. If it is indeed true that for light we have $ds^2 = 0$ does that mean that in 4d spacetime total "distance" is zero for light? By distance I mean lenght of a geodesic that light moves on? Describes?
| Spacetime manifolds are observer's manifolds, and there is no spacetime manifold all observers agree upon. However, according to the second postulate of special relativity, lightlike phenomena are observed by all observers as moving at c, so there is always observed a traveled distance > zero.
In contrast, the spacetime interval cannot be observed by observers, it may only be calculated, in the same way as the proper time of a particle. For this calculation, due to the structure of spacetime, all observers agree that the corresponding result is zero.
Example: In the following diagram where c is set to 1, a straight worldline of an object is observed which is moving 8 light minutes within 8 minutes - a lightlike movement.
As you can see, the observer observes a 45° worldline which is not zero. The zero cannot be found anywhere in the diagram, it is not part of the observation. However, he can calculate the spacetime interval, and he will find zero.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/446404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Can we ever "measure" a quantum field at a given point? In quantum field theory, all particles are "excitations" of their corresponding fields. Is it possible to somehow "measure" the "value" of such quantum fields at any point in the space (like what is possible for an electrical field), or the only thing we can observe is the excitations of the fields (which are particles)?
| One must be careful not to confuse formalism for physics. What do we mean when we talk about a field? In quantum field theory (which is a formalism) a field is often represented by an operator. What does that operator represent in the physical world? The field operator is in fact used to define the observables (correlation functions and all that). These observables are then used to observe states (which are the things that we have in the physical world). In other words, the measurement would look like $\langle\psi| \hat{\phi}(x) \hat{\phi}(y) |\psi\rangle$, where $\hat{\phi}(x)$ is the field operator and $|\psi\rangle$ is the state.
So what would it then mean to "measure a field"? Another way to look at it is to consider the path integral formulation of quantum field theory where the field becomes an integration variable for the path integral. In that sense, the field is any solution of the equations of motion. In certain cases, one can have a state associated with a specific solution. In that case the solution becomes a parameter function for the state. When one makes measurements of that state, one can in fact observe (aspects of) its parameter function, which may then in some sense correspond to the idea that one "measures the field."
Does that make sense?
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/446492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 3
} |
Work-Energy Theorem for Non-Constant Mass "The net work done on an object is equal to its change in kinetic energy."
Let's say that a rocket is moving upwards while expelling gas and is thus losing mass. (Non-constant mass)
As the object loses mass, it gains even more kinetic energy.
Is the work-energy theorem still applicable?
| I believe the problem lies with how you are stating the work-energy theorem.
Every good reference to it I can find states something along the lines of:
"The work done on a particle is equal to it's change in kinetic energy"
(emphasis mine) The word particle is extremely important here (sometimes they also use "rigid body"). For a single particle or rigid body, it cannot really do much beyond move as a reaction to forces. When you have an object that is made up of multiple interacting particles (such as a rocket ship expelling fuel), you cannot treat it as a rigid body, and so the interactions between work and energy aren't as direct as "work done is the change in kinetic energy" on the macroscopic scale of the spacecraft itself.
Work done can also change potential energy, thermal energy, or electrical energy, depending on how it interacts with the system. In this case, there are changes in the potential energy and thermal energy as a result of the fuel combustion, and this is able to increase the kinetic energy of the parts of the spacecraft we want to increase the kinetic energy of (i.e. the payload). (a bit more elaboration can be found here)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/446628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Having trouble making sense of Einstein's thought experiment So I was reading about Einstein's thought experiment where he tries to show that simultaneous events in one frame may not be simultaneous in another frame.
So, in the given pic, light from B' reaches Mavis before light from A' and I get that because she is moving to the right. But what if Mavis had another way to check whether the lightning hits the two ends simultaneously or not? What if I have two clocks attached at the ends A' and B' and the two clocks are synchronized. Attached to the ends A' and B' are devices that record the time when the lightning hits them. So if the lightning hits both A' and B' simultaneously with respect to Stanley, then by looking at the recorded times from A' and B', can't Mavis also come to the same conclusion?
| The speed of light is constant for all observers, so if Stanley sees lightning reach A’ and B’ simultaneously, then Mavis must see it reach B’ first, as you said. The diagram shows that this occurs, from Stanley’s perspective, because Mavis is moving to the right and the light from B’ reaches her before the light at A’. Of course, she thinks that the speed of light from both sources relative to her is identical, so she must conclude that lightning struck B’ earlier since the light from this event reached her first. This thought experiment shows that Mavis must observe a time difference, so if Stanley records the strikes using synchronized clocks, Mavis will see them as non-synchronized. This is the only way to preserve her view of events. The thought experiment of Stanley using synchronized clocks actually demonstrates that, in relativity, spatially separated clocks which are synchronized in one frame of reference may not be synchronized in another. This can be seen in the Lorentz transformation for time: If two observers moving relative to one another, A and B, start at the same position with initially synchronized clocks, and observer A later records time $t$ at position $x$, observer B moving relative at velocity $v$ in the $x$-direction will record time $t'=\gamma\big(t-\frac{vx}{c^2}\big)$, where $\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$. The change in recorded time in this formula depends on position, which demonstrates how two distant clocks may appear synchronous in one frame of reference but not another.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/447051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Does a particle with infinite energy escape an infinite well? Currently, my modern physics class is going over particles in finite and infinite wells, general quantum formalism, and tunneling.
What happens to a particle as it gains an infinite amount of energy? Does it stay inside of the infinite well? Does it escape? Can it not be determined? Does it depend?
Are there any issues with this question? Is it valid? Is there anything I need to define or presume before I ask it? Do I need to define the rates at which the potential of the walls go to infinity, or the rate at which the particle's energy goes to infinity?
|
Are there any issues with this question? Is it valid? Is there anything I need to define or presume before I ask it? Do I need to define the rates at which the potential of the walls go to infinity, or the rate at which the particle's energy goes to infinity?
Your question is valid, but only if you take an infinite well as a purely hypothetical idea, unachievable in practice. It's a device to illustrate how we go about getting solutions to (equally hypothetical) one dimensional problems.
It's an infinite well, so it's use is to demonstrate a basic, but actually impossible to achieve, cutoff to the probability of tunnelling.
Unless an infinite potential is used, there is always the possibility of the solution to it being incorrect.
What happens to a particle as it gains an infinite amount of energy? Does it stay inside of the infinite well? Does it escape? Can it not be determined? Does it depend?
The particle can't gain an infinite amount of energy in real life.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/447199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Could someone explain what is a potential? In many part of physics, me talk about potential (electrical potential, gravitation potential, elastic potential...). All those definition looks very different, and I would like to know how all those quantity are related. The mathematical definition of a potential is : $F$ is a potential if $F=\nabla f$ for some scalar field $f$. But I don't understand how I can relate this to all potential that exist in physic. Also, many people relate potential and energy, what doe it mean exactly? Because if a potential $U$ is something that can be written as a gradient, every integrable function (at least in $\mathbb R$) could be a gradient. So what the thing with these potential?
| A scalar potential whose gradient is the force exists in many branches of physics, for example heat conduction, static magnetism, static electricity, etc. In all these cases the gradient (or negative gradient) of the potential field is the force that can "move" things from here to there.
For example, in static magnetism as described by Ampere's law (steady state case) $\textrm{curl}\textbf{H}=\textbf{J}$ in regions where the current is zero we also have $\textrm{curl}\textbf{H}=0$ hence a scalar $\psi$ such that at least in all singly connected domain we also have $\textbf{H}=\textrm{grad}\psi$. Note the restriction to singly connected domain, it would not be true for a domain containing a toroidal magnet but would still be true locally and when calculated locally the force field is gradient and this force can move things.
Another example is steady state temperature distribution calculated from Fourier's law. We have heat (internal energy and entropy) flow as being proportional to the spatial gradient of temperature. What moves is heat (internal energy + entropy) and the driving force behind that motion is the temperature variation between neighboring points.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/447313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
} |
Stack/Chimney Effect: a physical explanation on how height of chimney affects $\Delta P$ A fellow engineer told me that there are greenhouses which exploit the stack effect, in order to cover some or all of their electrical energy needs. This is achieved by installing small electrical generators with fans mounted on the rotor, on or near a chimney, which has to be large. Due to pressure difference between the greenhouse and the environment, there is a flow of cold air, from outside, which is capable of rotating the fans and therefore producing electrical energy.
By reading the wiki article on the effect, I understood why the chimney has to be large, since $\Delta P$ is proportional to the height h:
$\Delta P = C\alpha h(\frac{1}{T_0}-\frac{1}{T_i})$
where
*
*ΔP = available pressure difference [Pa]
*C = 0.0342, the temperature gradient [K/m]
*a = atmospheric pressure [Pa]
*h = height or distance [m]
*To = absolute outside temperature [K]
*Ti = absolute inside temperature [K]
But can someone explain to me why the height is proportional to ΔP in a physical/mechanical way? or recommend an article/book/paper that explains it?
EDIT:
Something similar (if not exactly the same) is the solar updraft tower. Again, tower has to be as tall as possible, in order to maximize the power generation. But how is this mechanically explained? Is it because of greater pressure, the speed of the air increased, so fans rotate faster?
| The chimney only has to be high enough to contain an extended column of warm/light air. Too short and the warm air might spill out into the ceiling of the greenhouse instead of being siphoned up the chimney. Every home fireplace relies on this effect to keep smoke from filling the room.
Pressure decreases with height because the pressure comes from the weight of the air above. The higher you go, the less air is above you. The same effect occurs in liquids.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/447463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
What is the "lowest energy"? In many textbooks I come across the term lowest energy. For example in atomic structures, electrons are placed in orbitals in order for the atom to have the lowest energy. But what is this energy? Potential- or kinetic energy or the sum of the two?
| The sum of the two.
An electronic state like an orbital is an exact or approximate solution of some time independent Schrödinger equation, i.e. an eigenstate of the hamiltonian made by a kinetic and a potential energy term. The corresponding eigenvalue is the expectation value of such hamiltonian, evaluated on the state described by the orbital.
The expectation value of the hamiltonian can always be written as the sum of the expectation value of kinetic and potential energy. The lowest energy, is the eigenvalue of the hamiltonian with the minimum value.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/448569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Can a battleship float in a tiny amount of water? Given a battleship, suppose we construct a tub with exactly the same shape as the hull of the battleship, but 3 cm larger. We fill the tub with just enough water to equal the volume of space between the hull and the tub. Now, we very carefully lower the battleship into the tub.
Does the battleship float in the tub?
I tried it with two large glass bowls, and the inner bowl seemed to float. But if the battleship floats, doesn't this contradict what we learned in school? Archimedes' principle says "Any floating object displaces its own weight of fluid." Surely the battleship weighs far more than the small amount of water that it would displace in the tub.
Note: I originally specified the tub to be just 1 mm larger in every direction, but I figured you would probably tell me when a layer of fluid becomes so thin, buoyancy is overtaken by surface tension, cohesion, adhesion, hydroplaning, or whatever. I wanted this to be a question about buoyancy alone.
| For the interested, here is a little illustration I made once, to defend someone called 'Marilyn vos Savant'.
I didn't believe what people wrote in there against her statement, about a battleship floating in a bucket of water.
I am very happy to see, that everybody in here knows what is right.
Here is the illustration:
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/448673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "39",
"answer_count": 8,
"answer_id": 4
} |
How is $\int \frac{d^{3}\mathbf{p}}{(2\pi)^3}\frac{1}{2\sqrt{|\mathbf{p}|^2+m^2}}$ manifestly Lorentz-Invariant? When writing integrals that look like
$$
\int \frac{d^{3}\mathbf{p}}{(2\pi)^3}\frac{1}{2\sqrt{|\mathbf{p}|^2+m^2}} \ = \int \frac{d^4p}{(2\pi)^4}\ 2\pi\ \delta(p^2+m^2)\Theta(p^0)
$$
it is often said that this is manifestly Lorentz invariant (where $\Theta$ is the Heaviside step function). Why is this true?
If I consider a Lorentz transformation $p \to q = \Lambda p$, then $|\det(\Lambda)| =1$ so the Jacobian is just 1, and:
$$
\int \frac{d^4p}{(2\pi)^4}\ 2\pi\ \delta(p^2+m^2)\Theta(p^0) = \int \frac{d^4 q}{(2\pi)^4}\ 2\pi\ \delta(q^2+m^2)\ \Theta\big([\Lambda^{-1}]^0_{\ \nu} q^\nu \big)
$$
After this transformation $p^0$ gets taken to $[\Lambda^{-1}]^0_{\ \nu} q^\nu$, and things do not look Lorentz invariant to me. If it were invariant wouldn't this get mapped to just $q^0$?
| The measure is invariant under proper orthochronous Lorentz transformations (those continuously connected to the identity, or equivalently those that do not represent a spatial reflection or time reversal) because $\Theta(\Lambda^0_\nu q^\nu) = \Theta(q^0)$ for such transformations as they preserve the sign of the temporal component of any time-like vector.
Proof of the claim that a proper orthochronous transformation preserves the sign of the temporal component: The "mass shell hyperboloid" cut out by $p^2 = -m^2$ has two connected components, one with positive, one with negative temporal component. A transformation continuously connected to the identity cannot change the connected component of any vector without removing it from the mass shell for an intermediate transformation along the path to the identity, but the mass shell condition is invariant under all Lorentz transformations and therefore this cannot happen.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/449122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
2D global conformal transformations and the $z= \frac1w$ argument
For instance in Blumenhagen's CFT, there is a standard argument which determines that globally defined conformal transformations on the Riemann sphere where
$$l_n = -z^{n+1} \partial_z$$
is an element of the Witt algebra. In this argument we note that $l_n$ is non-singular at $z=0$ only for $n\geq -1$. Also substituting $z=-\frac1w$, we find
$$l_n = -\left(-\frac1w\right)^{n-1}\partial_w$$
is non-singular at $w=0$ only for $n\leq +1$. Therefore the global transformations are generated by $\{l_{-1},l_0,l_1\}$.
Why is the substitution $z=-\frac1w$ special? For instance if I use $z = -\frac{1}{w^2}$ I could repeat the argument above and conclude $n \leq 1/2$.
| The minus sign is not essential, so let's remove it for simplicity. An atlas of the Riemann sphere, say $M$ ( which is a 1-dimensional complex manifold), is given by two coordinate charts
*
*$z:M\to\mathbb{C}$, which covers the whole $M$ except the "infinity" $i\in M$.
*$w:M\to\mathbb{C}$, which covers the whole $M$ except the "origin" $o\in M$.
There is a holomorphic transition map: $w=1/z$ in the overlap $M\backslash\{i,o\}$. In polar coordinates $z=re^{i\theta}$, we have $w=e^{-i\theta}/r$.
Therefore the monodromy $z\to e^{2\pi i}z$ leads to the correct corresponding opposite monodromy $w\to e^{-2\pi i}w$ on the other chart, i.e. if we circle around the origin $o$ one time in one chart, it corresponds to circle around the infinity $i$ one time in the opposite direction in the other chart.
In contrast, a transition map of the form $z =1/w^m$, where $m\in \mathbb{N}\backslash\{1\}$, as OP suggests, would lead to wrong monodromy properties $w\to e^{-2\pi i/m}w$, and would hence not correspond topologically to a Riemann sphere.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/449324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
How can two electrons repel if it's impossible for free electrons to absorb or emit energy? There is no acceptable/viable mechanism for a free electron to absorb or emit energy, without violating energy or momentum conservation. So its wavefunction cannot collapse into becoming a particle, right? How do 2 free electrons repel each other then?
| Your first statement is false: energy can indeed be added at will to electrons by accelerating them with electrostatic charge distributions, as for example in the case of rapidly varying radio frequency (electromagnetic) fields. Neither energy nor momentum conservation is violated in this case. Search on SLAC for more details about this.
Your other questions are unclear. I recommend you do the search, read a bit, and return here if you have further questions.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/449431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 2
} |
Rule of addition of velocities in water Is the speed of light still the same for all inertial observers in water?
If not, what are the rules of addition of velocities in water according to special relativity?
| No, the speed of light in water depends on the observer's speed. If you're at rest relative to the water, you'll see a wave with 4-velocity:
$$ u_{\mu} = \gamma_n(c, 0,0, c/n) = \gamma_n c (1, 0,0,\frac 1 n ) $$
where:
$$ \gamma_n \equiv \frac 1 {\sqrt{1-\frac 1 {n^2}}}$$
Now you can boost it any which way, and you'll find:
$$||v_i'|| = \frac{||u_i||}{\gamma_n} \ne c/n$$
For example, co-linear with the light (boosting with $\beta$ and $\gamma$):
$$u'_{\mu} = \gamma_n c (\gamma-\beta\gamma/n, 0, 0, -\beta\gamma+\gamma/n)$$
$$ u'_{\mu} = \gamma_n\gamma c(1-\beta/n, 0, 0, -\beta+1/n) $$
or
$$ u'_{\mu} = \gamma_n\gamma c[1-\frac{\beta}n](1, 0, 0, \frac{\frac 1 n-\beta}{1-\frac{\beta}n}) $$
which means the 3 velocity is:
$$ v' = \frac c{\gamma_n\gamma} \frac{\frac 1 n-\beta}{(1-\frac{\beta}n)^2} = c \frac {\sqrt{1-\frac 1 {n^2} }}{\gamma} \frac{\frac 1 n-\beta}{(1-\frac{\beta}n)^2} $$
Note what should have been obvious from the beginning: there are (collinear) boosts that can change the direction (and one that zeros it out), which means a priori we should have known the answer is "no".
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/449552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Violating Newtons First Law! Suppose you are inside a very large empty box in deep space , floating ( i.e not touching the box from anywhere initially).The box is at complete rest.
Now you push the box forward from inside.
Now you would go backwards but the box will move forward to conserve momentum.
However since you were inside the box your force is an internal force but the box would have moved forward.
So doesnt this violate newtons 1st law
as an internal force made a body move from state of rest?
| Your confusion stems from the misunderstanding of the term "internal force".
Internal force in this context is not a force which acts from the geometrical inside of some body. It is internal in the sense that it results from the interaction of the parts of the system, whose collective motion (the acceleration of its center of mass) you are considering.
So when analysing the motion of the whole system (the box and the person together), we see that its center of mass has indeed not accelerated. But if we consider only the box, we see that it had accelerated, but Newton has no need to worry, as an external force (your hand) had caused the acceleration.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/449699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
} |
How can we define energy other than the definition that it's a capability to do work? It is actually a property of energy that it can do some work not an actual mean to define it because we cannot define a thing on the basis of what it is doing or what it can do.
| From my perspective, it is just a number associated with a system or object. For instance, a 2 kg ball moving with velocity 5 m/s has kinetic energy of 25 J. This number is always conserved for a system of objects (for instance 2 balls colliding, or two charges acting on each other) provided that no external forces act on it.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/449793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Is it the current which create magnetic field, or vice versa, or both? Talking about stationary magnetic field, it is said that if a conductor rotates inside the field, a current is induced. Also, I read that current (moving charges) generate magnetic field, too. How are these connected, and what's the best approach to undertand both phenomenons?
| Very briefly, here is how to get started:
You are right- if a conductor experiences a changing magnetic field in its vicinity (as, for example, when it is moved past a magnet in such a way that it cuts across the magnet's "lines of force") then a current will be induced to flow in that conductor. This is called FARADAY'S LAW.
You are also right that anytime a current flows through a wire, it will set up a magnetic field surrounding the wire. This is called OERSTED'S LAW.
These laws are tied together in their most general form by MAXWELL'S LAWS, which show how these phenomena are mathematically related to one another on a deep level.
You will find ample material explaining Faraday's, Oersted's, and Maxwell's laws on-line, or in an introductory book on electromagnetism.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/449916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
What was wrong with the old definition of temperature scale in kelvin? Wikipedia's article on the recent change to the definition of the SI base units states, as the reason for changing the definition of the kelvin:
A report published in 2007 by the Consultative Committee for Thermometry (CCT) to the CIPM noted that their current definition of temperature has proved to be unsatisfactory for temperatures below 20 K (−253 °C; −424 °F) and for temperatures above 1,300 K (1,030 °C; 1,880 °F).
Sure, I understand that tying temperature to a physical artifact, even a highly-reproducible one like the triple point of water, is unsatisfying. But the way it's worded implies there's some more significant problem, as if temperature measurements outside that range are less accurate or less reliable. What is that problem?
| It is difficult to determine Boltzmann's constant from the temperature of the triple point. That is easier at extremely low temperatures (with statistical mechanics) or at very high temperatures (from radiation).
At ambient temperatures, the connection to Boltzmann's constant is by using gas thermometers. Very cumbersome, and not as precise as in the other ranges.
ITS-90 will remain in effect. For calibration of thermometers at ambient temperatures, the triple point of water will continue to be used.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/449975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.