Q
stringlengths 18
13.7k
| A
stringlengths 1
16.1k
| meta
dict |
|---|---|---|
Is true black possible? Black is the absence of light because it absorbs light, but when we create black paint or black objects, light is always reflected, either in all directions in matte or smoothly in shiny black objects, making it never a true black. Would it be possible to use polarization to create an object that does not reflect any light, creating a truly black substance, without any shadows or reflection of light?
| The problem with the suggestion of using polarization is that you now have the reflections off the polarizers to contend with.
I think the short answer is "it depends on how 'black' you want it to be". "Truly black" = reflectance of 0. I am quite sure that is impossible - there will always be some probability of light scattering off a surface. All you can do is make that probability "quite small".
The world record for "blackness" appears to be held currently by Ventablack, a material with a special surface structure (nanotubes) that traps incident photons, and reflects less than 0.04 % of incident light. That is indeed very nearly black (but nowhere near "perfect"). Just look at this picture to get a sense of just how black that is. Of course if most cameras have 12 bit sensors, then one LSB is 1 part in 4000 - and 0.04% is 1 part in 2500. So indeed, this is almost invisibly black for a typical camera. Uncanny.
(Image from the above linked source).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/183473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 1
} |
How to calculate the period of the movement from a potential? I have an assignment, where I have an object moving in 1-D with a given mass and energy, and the potential V(x), and I'm supposed to calculate the period of the movement as a function of the energy
$$
V(x)=\begin{cases}\infty &x < -a \\
0 &-a < x < 0\\
\alpha x^2 & x>0
\end{cases}
$$
Should I find 3 Lagrangians for the 3 separate parts of the potential? And then how would I come to the period of the movement? Thanks in advance!!
| It seems, based on the comments above, that you have figured it out. Just for closure, I am writing the steps out.
If you had just a parabolic potential well, $V(x) = \alpha x^2$, you could get the period quite easily - for a given mass $m$, the frequency would be
$$\omega = \sqrt{\frac{2\alpha}{m}}\\
T = \frac{2\pi}{\omega} = \pi \sqrt{\frac{2m}{\alpha}}$$
For half the parabola, the time from the bottom to the side and back will be half this:
$$T_1 = \frac12 T = \pi \sqrt{\frac{m}{2\alpha}}$$
The mass will slide along the bottom of the well with a constant velocity given by the energy (which is all kinetic at this point):
$$v = \sqrt{\frac{2E}{m}}$$
So the time taken to cover the distance $a$ (there and back) is
$$T_2 = \frac{2a}{v} = a\sqrt{\frac{2m}{E}}$$
When the mass gets to the infinite potential wall it will just turn around.
Summing these two times give you the period of the oscillation:
$$T = \pi \sqrt{\frac{m}{2\alpha}} + a\sqrt{\frac{2m}{E}}$$
As expected, one term depends on how steep the potential well is on one side, and the other term decreases as the energy goes up. And when the width of the well increases, the period increases as well.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/183645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Polarisation by Reflection - oscillation direction I'm currently studying polarisation by reflection, and have come across two pieces of information from the same source, which I can't seem to understand on how they differ.
*
*The oscillation direction of the field polarised in the plane of incidence is parallel to the direction of the reflected ray.
*As EM waves are transverse waves, this polarisation cannot propagate along the reflected ray.
Could someone please explain, the direction of oscillation of the reflected ray and what each piece of information means.
| I have a site which also explains this phenomenon clearly here.
In short here you would find that when an unpolarised light hit at the interface of the two medium the reflected light will be generated such that only that component of electric field is oscillating, which is perpendicular to the direction from point of incidence to the point of observation which will also be the path for propagation.
And one would see that this component is parallel to the plane of incidence and and a component of the unpolarised light.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/183739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Notation of vectors It's very common to see $\text{F} = 30 \text{ N}$ when the problem is unidimensional. Yet, force is a vector. Shouldn't I write $|\overrightarrow{F}| = 30 \text{ N}$? Because if I write $\overrightarrow{F} = 30 \text{ N}$ I'm saying that the vector is equal to an scalar. On the other hand, I rarely see $\overrightarrow{F} = (30, 0, 0)$.
| Force is indeed a vector. Technically you should write $|\overrightarrow{F}| = 30N$, however there is usually context given that let you omit this.
If you are working in one dimension, then the vector-like direction is all encapsulated in the sign once you've defined your coordinate system (e.g. -30N is 30N downwards.) Beyond that, it is typically just a shorthand to write $F = 30N$ as the magnitude of a vector $\overrightarrow{F}$ and the equation/problem will typically give you context into what is meant.
Sometimes you will see vectors that are written as $\overrightarrow{F} = (30N, 0, 0)$ (don't forget the units.) It's probably less common than something like $\overrightarrow{F} = 30N\hat{x}$, but if you are just learning this then expect all types to be thrown at you.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/183856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Is there a major difference between neutron-neutron collisions compared to proton-proton collisions? For the sake of argument, assume the LHC was able to accelerate / focus / collide neutron beams, with the same energy levels it runs at for p-p collision.
Would the collisions produce any major differences vs the p-p collisions?
| Major differences no, because what really collides are constituents of the protons (called partons): the valence quarks, the quarks and anti-quark due to quantum fluctuation and the gluons. The same thing happen with neutron.
However, since in proton $u$ quarks carry more momentum than $d$ quarks (because proton contains 2 $u$ valence-quarks and 1 $d$ valence-squark), processes involving u quark happen more likely. For example $u \bar{d} \to W^+$ happens more often than $d \bar{u} \to W^-$. If neutrons collide instead of protons, the situation will be the opposite. It would be easier to get $d$ quark with large momentum, the neutron containing $ddu$ as valence quarks. Hence, more $W^-$ would be produced.
So in conclusion, no major differences, but generally speaking, all charge asymmetries would be opposite.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/184011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Explain relationship between angular diameter distance and luminosity distance, Etherington Theorem I have a question relating to the Etherington Theorem.
The luminosity distance is defined by the equation for flux, i.e.
$F=\frac{L}{4\pi D_L^2}$
where flux is in units energy per unit time (luminosity) per unit area.
The angular diameter distance is defined by
$D_A=\theta/R$, where $\theta$ is the observed angular size measured by a telescope, and $R$ denotes the proper size of an object.
These two quantities are related by $D_L=(1+z)^2D_A$
I have never read a clear explanation for this relationship, nor have I come across a derivation. Could anyone explain to me where the redshift dependence $(1+z)^2$ comes from?
| See D. Hogg's Distance measures in cosmology, 2000
http://arxiv.org/abs/astro-ph/9905116
Section 7, Luminosity Distance, p. 6
$D_L=(1+z)^2 D_A$
follows because the surface brightness of a receding object is reduced by a factor $(1+z)^{−4}$, and the angular area goes down as $D^{-2}_A$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/184115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
How electrons move so fast in a electric circuit? Whenever we switch on a bulb......it takes almost no time to glow up.....But we know that the atoms of a solid are tightly packed and there is a very little space between them.
So how the electrons travel through them irrespective of so much blockages in the conductor???
| In fact, electron's speed is not so fast that light bulb glows up immediately. It is the electromagnetic field which travels in the circuit at near the speed of light that is resposible for it.
After turn on the light, electron only acquires a little speed in addition its thermal speed. The thermal speed of electron can be estimated by $mv^2/2\approx k_BT/2$, where $k_B$ is the Boltzmann constant and $T$ is the absolute temperature and nearly 300 K. So the thermal speed is about 67.4 km/s.
In the electric field inside the circuit, the distance the electron travels freely before hitting the nucleus is estimated to about 300 nm in copper. If the voltage is 110 V, the speed increases by 86 m/s. In fact, the increase is the upper bound, because I assume the voltage change in 300 nm is 110 V, which is not true. Voltage changes over a much longer distance.
So you see, the increase in speed of electron is only about 86 m/s. The draft speed of electron is not really hight (compared to thermal speed).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/184317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 1
} |
Entanglement of Tripartite States Is there any simple algorithm to determine the entanglement of a tri-partite state?
In particular, what is the proof for entanglement of $ |GHZ\rangle $ and $ |W\rangle $ states?
$ |GHZ\rangle =\frac{1}{\sqrt{2}}(|000\rangle+|111\rangle) $
$ |W\rangle =\frac{1}{\sqrt{3}}(|001\rangle+|010\rangle+|100\rangle) $
| You can calculate the density matrix of the state $$\rho = |\varphi\rangle\langle\varphi|$$ And then, the reduced density matrix for one of the particles taking the partial trace $$ \rho_A = \textrm{tr}_{B,C}(\rho)$$ The state is entangled if and only if the reduced matrix is a mixed state. This can be checked, for example, calculating the von Neumann entropy $$ S = -\textrm{tr} (\rho_A \log \rho_A)$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/184426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Why is the electric field inside a charged conductor zero in the electrostatic case? I am trying to understand the idea (or the fact) that most books introduce which is about the electric field inside a charged solid conductor.
Books tell that the field has to be zero everywhere inside solid conductor, otherwise charges will move around. Using this idea and Gauss's law, the charges inside the solid conductor is zero.
Now let us take for example four extra positive charges (each =1.2x10^-10 coulomb) inside a solid conductor of radius 1.
According to the idea of charge at the surface and due to the symmetry, the charges will distribute as follows:
I have plotted the electric potential (V=Q/(4πε0r)) and electric field (E=-∇V) using principle of superposition and the plot is:
Clearly the electric potential inside the conductor is not constant and the electric field is not zero.
How can this issue be explained?
| You have ignored the mobile charges in the conductor. In your plot the field lines are not perpendicular to the surface, particularly near the charges. That will cause the conduction electrons to move. The positive charges will attract electrons until the field inside the conductor is zero. This means that the whole conductor, including the inner surface, is an equipotential. After that, Gauss' law says the field inside is zero.
You figure is a fine one if the four charges are in empty space.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/184606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Simple generalization of the Feynman rules for QFT to thermal QFT? Assuming that one knows Feynman rules for QFT, what is the simplest way to generalize them for $T \neq 0$ case? What is the main difference? Can we just read them off from Lagrangian the same way as in QFT?
Since the partition function $Z$ in thermal quantum field theory is just $Z_{QFT}$ with substitution $t \rightarrow i\tau$, with $\tau$ being integrated from $0$ to $\beta$, as opposed to time $t$ being integrated from $-\infty$ to $\infty$, and with different boundary conditions for the fields (periodic/anti-periodic for bosons/fermions), can we use the same analogy for the generalization of the Feynman rules?
| Finite temperature Feynman rules are simply zero temperature Feynman rules for Euclidean ($t\to i\tau$) QFT in periodic imaginary time. So instead of continuous values for the momenta, you will have a discrete spectrum for the timelike moments (such as in the infinite potential well in basic quantum mechanics). It's called the Matsubara formalism, if you want to google for more.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/184722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Calculating voltage in piezoelectric material The piezoelectric constitutive law is defined by the two equations:
$$S=s T+dE\\
D=d T+\epsilon E$$
$S$: Strain. $T$: Stress. $E$: Electric field. $D$: electric charge-density displacement. $\epsilon$: Permittivity (evaluated at constant stress). $d$: piezoelectric strain constant. $s$: Elastic compliance (evaluated at constant electric field).
Apart from that we have the expression:
$$g=\frac{d}{\epsilon}$$
$g$ is piezoelectric voltage constant (or open-circuit constant).
For a piezoelectric material we can look up $g$, $d$, $\epsilon$ and $s$ in data tables, and let's assume that all other material data can be found likewise if needed. I also know the physical lenght/height $L$ of my material bulk. If I calculate the electric field generated at a certain applied stress $T$, then I can use that $E$ to find the voltage over the material piece:
$$V=EL$$
BUT I am stuck trying to calculate the $E$. I feel that I do not know $D$ and $S$ in the above equations. I am even unsure of exactly what the $E$ in those equations is - isn't it the generated electric field?
Thank you in advance and good day.
| If the material is not connected to a circuit and doesn't have any external charges added then $D =0$ so you can use
$$
D= dT + \epsilon E = 0
$$
along with the known stress $T$ to obtain $E$. Then use $V=EL$ to obtain the voltage $V$.
Here's a screenshot from Learnpiezo.org video series Lecture 3 Part F which covers the situation you described.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/184806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Does General Relativity imply loops in space? Everyone who has been interested in modern science has heard explanations (certainly simplifications) of general relativity, mostly that space is curved. The analogy with a rubber sheet is popular. In such an analogy, orbiting planets are said to be naturally following "a straight line in a curved space".
Assuming that is not an oversimplification, would it mean that orbits are loops generated in space by massive objects ?
Also, if we consider spacetime as a curved structure, thus lines are not necessarily straight, what would be the meaning of momentum in such a frame?
| "Straight lines" in curved space are geodesics. But the geodesics that define particle paths are in the pseudo-Riemannian manifold of space-time, they are not geodesics in space!
As a planets path is not closed in space-time (but some kind of spiral), massive bodies do not induce loops in the topology of the space-time. (Actually the loop is not even closed in space, due to the perihelion shift).
The momentum is defined in terms of the energy-stress tensor, which fulfils the conservation law
$$T^{\mu\nu}_{;\nu} = 0.$$
As this reduces to the classical momentum and energy conservation locally upon taking the limit that leads to Newtonian mechanics.
Furthermore, if the space has a Killing vector field $\xi^\mu$ (that is a continuous symmetry), the quantity $\xi^\mu x_\mu$ is conserved along geodesics, and can be interpreted as the momentum component of the particle along the direction of the Killing vector field in space-time.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/184898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 3
} |
When does light reach a shell observer in Schwarzschild metric? I am trying to simulate the trajectory of light in the Schwarzschild metric (as seen by a far away observer) with fixed $\theta = \pi/2$. According to my source (Chapter 18, section 18.5) the trajectory is then governed by:
$$\frac{dr}{dt} = \dot{r}$$
$$\frac{d\phi}{dt} = \dot{\phi}$$
$$\frac{d\dot{r}}{dt} = \frac{-4M^2+2Mr+(r-5M)r^3\dot{\phi}^2}{r^3}$$
$$\frac{d\dot{\phi}}{dt} = \frac{2(-3M+r)\dot{r}\dot{\phi}}{(2M-r)r}$$
I have a situtation where shell observer sits at $(r_T, \phi_T)$ and I know that $r(0) = r_0$, $\phi(0) = \phi_0$, $r(T) = r_T$, $\phi(T) = \phi_T$ where $r_0$, $r_T$,$\phi_0$ and $\phi_T$ are known, but $T$ is unknown. It seems to me that I need an additional constraint to figure out $T$ since I have 4 equations (the ones above), but 5 unknowns ($r(t), \dot{r}(t), \phi(t), \dot{\phi}(t), T$).
Do I need an additional constraint to figure out $T$ and what would that constraint be?
| The way the equations are presented seems unnecessarily obscure, as there are only two equations that matter:
$$ \frac{d^2r}{dt^2} = \frac{-4M^2+2Mr+(r-5M)r^3}{r^3}\,\dot{\phi}^2 $$
$$ \frac{d^2\phi}{dt^2} = \frac{2(-3M+r)}{(2M-r)r} \, \dot{r}\dot{\phi} $$
These come from the geodesic equation expressed using coordinate time.
So you start at some convenient $(r, \phi)$ with initial coordinate velocity $(\dot{r}, \dot{\phi})$ and integrate forward in time to calculate $r$, $\phi$, $\dot{r}$ and $\dot{\phi}$ as a function of the coordinate time $t$.
You get to pick whatever initial values for $r$, $\phi$, $\dot{r}$ and $\dot{\phi}$ you want, but obviously $\dot{r}$ and $\dot{\phi}$ are related because you're describing a light beam. The relationship comes from the Schwarzschild metric. For a light ray $ds = 0$, and we can take $\theta = \pi/2$ and $d\theta = 0$, so we get:
$$ 0 = -\left(1-\frac{2M}{r}\right)dt^2 + \frac{dr^2}{\left(1-\frac{2M}{r}\right)} + r^2d\phi^2 $$
or:
$$ \left(1-\frac{2M}{r}\right) = \frac{\dot{r}^2}{\left(1-\frac{2M}{r}\right)} + r^2\dot{\phi}^2 $$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/184985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Where does the majority of the mass of the usual matter come from? I apologize in advance to experts for the naivety of the question. It should be a duplicate but I didn't find any satifying question or answer about that.
The proton is composed by two up quarks and one down quark.
mass(proton) $\sim 940 \ MeV/c^2 $
mass(up) $\sim 2.3 \ MeV/c^2 $
mass(down) $\sim 4.8 \ MeV/c^2 $
It follows that: $2$mass(up) $+$ mass(down) $\sim 9.4 \sim \frac{1}{100}$ mass(proton).
Question: Where does $99\%$ of the remaining mass of the proton come from?
| The three quarks you talk about are usually called the valence quarks of the proton, and their contribution to the mass of the proton is not it. In particle accelerators, when we hit protons with high energy beams, we discover that protons are made of a cluster of smaller constituents (like quarks and gluons, which constantly are created and destroyed in particle-antiparticle pairs.) At lower energies, the proton appears to be made of three quarks, but for higher energy collisions, we find that the proton is actually made up of loads of such particles. These particles make up the 'missing' mass of the proton.
Edit: Looking at the proton like it is a particle would be wrong, because it is actually made up of quantum fields. (like everything else.) These fields 'act out' differently depending on how much energy you 'supply' to observe them. For lower energies, the proton behaves like three particles, but you can observe that it is made up of a much denser mix at higher energies.
(Sorry if I used weird words like act out and supply, QM and words doesn't go too well for me.)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/185149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Crane Balancing, Center of mass I am working on the ICPC 2014 Problem C "Crane Balancing"
The initial idea was to calculate the center of mass of the polygon, which I did via this equation:
Where the Area A:
Now, the solution is to binary search over the mass and look for th e maximum mass M where the crane is still balanced.
But I have a problem figuring out how the mass added affects the position of the polygon center of mass,
Thanks in advance.
| You have to find mass center of the whole system polygon - added mass and ensure that this center is located above polygon bottom base (without going over the extreme points)
CommonCenter.X=(PolyCenter.X*Poly.Mass + AddedMass.X*AddedMass.Mass)/(Poly.Mass+AddedMass.Mass)
CommonCenter.Y=(PolyCenter.Y*Poly.Mass + AddedMass.Y*AddedMass.Mass)/(Poly.Mass+AddedMass.Mass)
(look at A system of particles and Barycentric coordinates parts here)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/185232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
What is the lowest level things have temperatures? Something can be cold. If you split it in half, it will still be cold, keep doing that and it will still be cold. My question is, what is the lowest level things can have temperature? Do atoms have temperatures? Do quarks?
| One definition of temperature is that it is the parameter which determines the distribution of velocities of an ensemble of particles. Note that I refer to an ensemble (a group) of particles. If you truly continue breaking down into smaller and smaller pieces, you no longer have the concept of a distribution and things become tricky.
You can actually define the temperature of only a few particles by looking at only a handful of fermions, and asking how long do they typically occupy their single/doubly occupied states. Quantum dots, only a few nanometres in size typically, can have a well defined temperature assigned to them. See Wikipedia for a bit more detail.
All this being said, temperature is a very weird concept, and I'm not sure that it is well understood in the extremes.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/185293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Why did the electric potential energy fall in this situation? Suppose you have two capacitors configured as shown in the image below. One of them has charge $q$ and the other one is uncharged. Then, you close the switch, allowing charges to flow. After everything stabilizes, we now have another charge distribution, as shown in the image.
We know that
$$\frac{q_1}{C_1} = \frac{q_2}{C_2}$$
$$q_1 + q_2 = q$$
Let's calculate the electric potential energy stored before closing the switch (let's call that $U_0$) and after closing the switch (let's call that $U_f$).
I calculated those and found that
$$U_0 = \frac{q^2}{2C_1}$$
$$U_f = \frac{q^2}{2(C_1+C_2)}$$
This means that the electric potential energy was lowered. What happened to the rest of the energy?
At first, I thought this difference of energy became kinetic energy for the charges to move around. But, in the end of everything, those charges stop. The total change of kinetic energy is zero!! What happened? What am I missing?
| The rest of the energy is basically emitted as heat energy.
Why?
You have two capacitors in the circuit, and the connecting wires offer negligible resistance.
Hence, when electrons flow from the charged capacitor to the uncharged one, the electrons basically face no resistance, and they collide with high speed with the uncharged capacitor. This collision leads to loss of energy in the form of heat energy.
After the first few electrons reach the capacitor, they start repelling the other electrons coming with high speeds, and hence, the loss of energy reduces as the capacitor is being charged.
Try this practically. You'll find the uncharged capacitor has become hot, due to this.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/185391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Confusion about two forms of connection coefficients I am new to GR. In one book I found that the connection coefficient expression is given by
$$ \Gamma^\mu_{\nu\lambda} = -\frac{1}{2} g^{\mu\rho} (\partial_\nu g_{\lambda\rho} + \partial_\lambda g_{\rho\nu} - \partial_\rho g_{\nu\lambda}). $$
In another book I found
$$ g_{\rho\mu} \Gamma^\mu_{\nu\lambda} = -\frac{1}{2} (\partial_\nu g_{\lambda\rho} + \partial_\lambda g_{\rho\nu} - \partial_\rho g_{\nu\lambda}). $$
Is one of them wrong; if not, how can I go from one to the other? Multiplying both sides by $g_{\rho\mu}$ seems to give the trace, which is 4 rather than 1.
| Since $\rho$ is already a summed-over dummy index in the first equation, we can't introduce it again. Instead, multiply both sides of the first equation by $g_{\sigma\mu}$:
\begin{align}
g_{\sigma\mu} \Gamma^\mu_{\nu\lambda} & = -\frac{1}{2} g_{\sigma\mu} g^{\mu\rho} (\partial_\nu g_{\lambda\rho} + \partial_\lambda g_{\rho\nu} - \partial_\rho g_{\nu\lambda}) \\
& = -\frac{1}{2} \delta_\sigma^\rho (\partial_\nu g_{\lambda\rho} + \partial_\lambda g_{\rho\nu} - \partial_\rho g_{\nu\lambda}) \\
& = -\frac{1}{2} (\partial_\nu g_{\lambda\sigma} + \partial_\lambda g_{\sigma\nu} - \partial_\sigma g_{\nu\lambda}).
\end{align}
This is your second equation, with $\sigma$ replacing $\rho$ as one of the free indices.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/185497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is an event horizon absolute to all observers? Recently I had discussion whether the event horizon of a black hole is absolute or relative to different (outside) observers. Does someone just 1m above the horizon (disregarding effects of tidal forces, stability of orbits etc.) perceive it at the same depth as someone at infinity?
I'm not able to prove it theoretically. My only justification is that the relation "information from point $A$ can reach point $B$" is transitive, so all photons that can reach 1m above the horizon can also reach distant observers.
| The event horizon is not absolute. It all depends on the observer. THAT is relativity. Space/time is relative to the observer. An event horizon is the consequence of space and time curving to a degree that light cannot escape relative to an observer at a certain point in space/time. An observer close enough to the black hole to be affected by its gravitational “force” would observe time differently, therefore the event horizon would be closer to the singularity. The closer to the singularity an observer, the more the event horizon would change from their POV. If able to observe the black hole from the singularity everything would become infinite. The observer would basically witness the end of the universe.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/185692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 4
} |
Is the Planck length the smallest length that exists in the universe or is it the smallest length that can be observed? I have heard both that Planck length is the smallest length that there is in the universe (whatever this means) and that it is the smallest thing that can be observed because if we wanted to observe something smaller, it would require so much energy that would create a black hole (or our physics break down). So what is it, if there is a difference at all.
| None of the above. Though there are many speculations about the significance of the Planck length, none is proven in any currently accepted theory.
It is expected, though, that quantum gravity effects become definitely non-neglegible at the energy/distance scale set by the Planck length, so it provides a heuristic scale at which we should not expect our current theories to make accurate prediction.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/185939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "83",
"answer_count": 4,
"answer_id": 0
} |
Meniscus attached to an inclined plate To be more specific, suppose a hydrophilic infinite plate is stuck into a semi-infinite region of water, above the water is a semi-infinite region of air, when the plate is stuck into the water vertically, the contact angle is $\alpha$, as shown in the figure below:
Needless to say, the menisci on both sides are symmetric, but what will happen when the plate is inclined for an angle $\beta$? Will the contact angle $\alpha$ remain unchanged? The meniscus on which side will be higher?
| To answer the question in your comment:
Yes the contact angle remains constant. The contact angle is determined by surface energies of the three materials and microscopic surface roughness, neither of which depend on the direction of gravity so tipping the plane will not effect that.
The height of the meniscus (distance from the asymptotically flat horizontal surface to the contact point) is given by:
$$z=\sqrt{\frac{2\gamma}{\rho g}(1-\cos(\alpha))}$$
derived here
Where $\alpha$ is the angle of the water at the contact point relative to horizontal
The extended domain version that properly predicts both positive and negative heights is:
$$z=2\sqrt{\frac{\gamma}{\rho g}}\sin\left(\frac{\alpha}2\right)$$
For a contact angle $\theta$ and a plane tipping angle $\phi$ off vertical the heights of the two contact points would be:
$$z_1=2\sqrt{\frac{\gamma}{\rho g}}\sin\left(\frac{\pi-2\,\theta+2\,\phi}4\right)$$
$$z_2=2\sqrt{\frac{\gamma}{\rho g}}\sin\left(\frac{\pi-2\,\theta-2\,\phi}4\right)$$
Here's a plot from horizontal to horizontal with an assumed contact angle of 45 degrees.
And here's a gif of the surface:
However, this model assumes a singular contact angle whereas real materials have contact angle hysteresis. So if we assumed ideal contact angle hysteresis, then if we start from vertical and slowly tip we'd get something more akin to this:
Where near the beginning there is an upper and lower bound where the height would be somewhere in-between them. This figure assumes that the rotation happens about the mean contact point. If the rotation origin were below the lower contact point then both sides would tend towards the higher limit while if the rotation origin were above both contact points then they would both tend towards the lower limits.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/186034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
If the speed of light is constant in all reference frames, why does the mirror clock experiment show light travelling on an angle? I was recently looking for answers as to why time slows down the faster you travel and regularly came across the mirror clock experiment. This experiment has a beam of light bouncing between two mirrors as the mirrors travel relative to a stationary observer. The experiment says that from the point of view of the stationary observer the light has to travel a longer distance, down on an angle (due to the sources movement) then after hitting the mirror, backup on an angle to hit the top mirror - ad infinitum. However I thought that the speed of light is the same in all reference frames and not affected by the source. So if correct (I assume I'm wrong here) wouldn't the light still go straight down relative to the stationary observer, and essentially the moving clock would move away from the light?
| The point of that experiment is not that the light goes slower but that the light has a longer distance. This means that a single bounce of the light off the mirror takes longer for the observer's point of view. This is due to time dilation. So it is not because the light is slower, but it is because the light has to travel longer
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/186189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Causality and how it fits in with relativity I was talking to my teacher the other day about Einstein's spacetime and there's one thing he couldn't explain about the nature of Cause. I may be being stupid or just unable to comprehend, thanks for any replies.
According to Einstein and relativity, two observers will agree on what things happened but not necessarily on the chronological order in which they happen. I understand how this radically alters our view of time into something that isn't the same or experienced equally by all. What I don't understand is how this fits in (or doesn't) with cause. If event A causes Event B, but we're saying that two people could experience them in a different order, how can event B happen before event A which caused it.
I'm intrigued?
| It's a bit more complicated than that. Given any two events, there is a quantity, called the interval (also 'spacetime interval' or 'invariant interval'), denoted $\Delta s^2$, and which equals $\Delta s^2=c^2\Delta t^2-\Delta \mathbf r^2$, which determines how the two events can relate to each other causally.
*
*If $\Delta s^2>0$, then we say $A$ and $B$ are "timelike separated" (or lightlike separated if $\Delta s^2=0$). In this case all observers will agree that (say) $A$ happened before $B$, and $A$ can causally influence $B$.
*If $\Delta s^2<0$, then we say $A$ and $B$ are "spacelike separated". In this case $A$ and $B$ are causally disconnected, and neither can influence the other. Different observers will disagree on their temporal order, and in fact you can always find observers for whom $A$ happened before $B$, $A$ happened after $B$, and $A$ happened at the same time as $B$.
*Finally, is $\Delta s^2=0$, then we say that $A$ and $B$ are "lightlike separated", or that the interval between them is "null". This is identical to timelike separations: all observers will agree that (say) $A$ happened before $B$, and $A$ can causally influence $B$; moreover, a light ray emitted at $A$ in the direction of $B$ will reach that position at the exact instant that $B$ is happening, and it will do so in all frames of reference.
The set of all events $B$ which are at lightlike separations from $A$ is called the light cone of $A$, and it separates space in three regions: the interior, with timelike separations, itself split into the causal future and the causal past of $A$, and the exterior, with spacelike separations, which contains all events that are causally disconnected from $A$, and which are simultaneous with it in some frame of reference.
Thus, as you succinctly put it,
if $A$ and $B$ are linked (one causes the other), then they have to be timelike [or lightlike] separated and all observers will agree on their temporal order.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/186296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 1,
"answer_id": 0
} |
Are spherical coordinates distances or angles? I've become confused about spherical coordinates when dealing with electric fields.
The way I always understood spherical coordinates is something like the below picture. To define a vector, you give it a distance outwards (r), and two angles to get a final position. Below, the $\theta$ and $\phi$ components are measured in radians.
(Courtesy Wikipedia.org)
However, you can also have, say, an electric field in spherical coordinates. In this case, the unit vectors $\theta$ and $\phi$ don't define angles but rather values of the vector fields. So, in the case of electric fields, we might have $E_\theta = 10\text{ Vm}$. That is, at every point there will be this electric field component in the theta direction.
So, it seems there are two different ways of dealing with spherical coordinates. One, where the $\theta$ and $\phi$ components represent angles, and one where they represent values of the components in those directions.
This would then give you two different measures of lengths of the vectors. In the first case, the length of the vector is always given by the r component. In the second case, you take $|\vec{E}|=\sqrt{E_r^2+E_\theta^2+E_\phi^2}$.
What am I mixing up here?
| You can't take the root mean square of the spherical coordinate parameters because they aren't all the same units (one is a length measurement while the other two are angle values). Well you can, but the output is meaningless.
To convert spherical parameters to Cartesian coordinates, you use simple trig:
\begin{align}E_z & = r \cos(\theta) \\
E_x & = r \sin(\theta)\cos(\phi) \\
E_y &= r \sin(\theta)\sin(\phi) \end{align}
Now if you take the root mean square of $E_x$, $E_y$ and $E_z$ you will get the correct $r$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/186388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Obtaining a copy of Hawking's Ph.D thesis - Properties of Expanding Universes Due to its popularity, I am interested to know the 4 chapter titles and topics covered in S.W. Hawking Ph.D, Properties of Expanding Universes. I also ask this because that thesis is hardly available.
| It seems it went online at last! It went global and public, and it collapsed the wave-function! Here https://www.repository.cam.ac.uk/handle/1810/251038
From this, the chapter titles are...
*
*Title: Properties of expanding universes.
*Abstract.
*Introduction.
*Acknowledgements.
*Chapter 1. The Hoyle-Narlikar theory of gravitation.
*Chapter 2. Perturbations.
*Chapter 3. Gravitational radiation in an expanding universe.
*Chapter 4. Singularities.
*References.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/186539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
What about a surface determines its color? Light falls on a surface. Some wavelengths get absorbed. The other are reflected. The reflected ones are the colors that we perceive to be of the surface.
What is the property that determines, what wavelengths are reflected and what are absorbed? Is it electronic configuration of the object on which the light falls?
If yes, then if we know the electronic configuration of a surface can we make a model, which will predict the color it will show?
| A material displays a color when light is reflected off a surface.
When a certain wavelength reaches a surface, if the energy $E=\frac{hc}{\lambda}$ of the photon correspond to the difference between two electronic states then it has a certain probability to be absorbed. The probability of being absorbed depends on the density of electronic states of course.
If a photon isn't absorbed it'll go through the material. This material will then be said to be transparent to its wavelength $\lambda$.
If the photon is absorbed, it will be re-emitted in no particular direction. But when emitted toward the inside of the material it'll be absorbed again, therefore the direction toward the exterior of the material will be favored (because in this direction the photons won't be absorbed). Those wavelength are being reflected.
When looking at the material, you'll see a spectrum made of all the reflected wavelengths, and that will make its color.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/186642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 3
} |
Is there any tension in a massless spring that connects two free falling bodies in different horizontal planes? Two bodies A and B of same mass $m$ are attached with a massless spring and are hanging from a ceiling with a massless rope. They are in same vertical plane but not in same horizontal plane.
Now the string that connected A with the ceiling is cut and the system is experiencing free fall.
1. Is there any tension in the spring?
My attempt:
Now the whole system should descend with the acceleration $g$ and the body B (and also A) experiences a gravitational pull $mg$.
Let the tension in the spring be T.
Therefore, from the free body diagram of B, $mg - T = mg$,ie. $T=0$.
*But A also moves downwards, so puts a force on B, how to take account of that? Will there be an relative acceleration between A and B? I am confused about the free body diagrams of A and B.
*Will the tension change if the mass of A and B are different?
| There is tension in the spring. It it extended and hence there is tension! It is the centre of mass that falls with acceleration $g$ rather then each individual mass. So the equation $$mg-T=mg$$ is invalid. As the two masses fall they will oscillate (getting closer and further away) and the tension will cycle.
Let us call the distance fallen by mass $A$, $x_A$ and that fallen by mass B $x_B$ the equation of motion for each mass is given by:
$$m \ddot x_A=mg+T$$
$$m \ddot x_B=mg-T$$
$T$ is a function of $x_A$ and $x_B$, ($T=k(x_B-x_A-L)$ where $k$ is the spring constant, and $L$ is the natural length) and we cannot assume that $\ddot x_A=g$ or $\ddot x_B=g$. These sorts of equations are called coupled differential equations and can be solved a number of ways.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/186736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
The Fermi Surface of the Free Electron Model for T>0K For the Free electron model, we can easily describe the Fermi surface at T=0 due to the uniqueness of the Fermi-Dirac Distribution at T=0; below the Fermi-level, a state is definitely filled, above, the state is not filled. So due to the this, the Energy contours are just regular spheres right?:
$k_{F}$ fixed depending on how many electrons are present to fill the states. Hence: $E = \frac{\hbar^2}{2m}k_{F}^2$, which is a sphere in reciprocal space.
However, when the temperature is greater than T=0, then what happens? The states are now not in a filled or not filled position but follow instead a probability distribution: $n = \large \frac{1}{1 + e^{(E-E_{f})/kT} }$. So now what happens to the Fermi-sphere?
Could it be that now the temperature is non-zero, Bragg diffraction has an effect on the sphere at the edges of the Brillouin Zone where Bragg Diffraction is most prominent. If this were the case, in what way does this change the Fermi sphere anyway?
| The Fermi surface is a $T = 0$ property. But as the Fermi energies in metals are on the order of a few electron volts (that is a few $10^4\,\mathrm{K}$), it is a good concept for approximately discussing matters at $T > 0$ (where the sharp surface will not be present, but there will only be excitations states close to the Fermi surface).
Furthermore for interacting electrons there will not even at $T = 0$ be a sharp Fermi surface in the sense, that the occupation drops to zero above the Fermi energy. Rather the quasi-particle density will be smeared out around the Fermi surface (which survives as a jump in the occupation density).
The sphere is just an approximation to the form of the Fermi surface. To handle this correctly you have to consider the dispersion relation of the electrons in the lattice, which is determined by the lattice periodic potential, and will certainly be anisotropic. So the Fermi surface will be a deformed sphere anyway. Often the Fermi surface in real metals even intersects the Brillouion zone boundary! (While I heard an explanation for some qualitative features in terms of Bragg diffraction, I don't think it is a very fruitful concept to understand this.)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/186986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Calculating Dipole Magnetic Moment Given Magnetic Field Strength I am trying to figure out how to solve the following:
The Earth's magnetic field can be represented, in a first approximation, by a magnetic dipole placed in the Earth's center, at least up to
distances of a few Earth radii ($R_E$)·
Using the fact that, at one of the magnetic poles, the field has a magnitude of approximately 0.5 gauss near the surface, calculate the dipole magnetic moment, $\mu$.
How does one go about calculating the dipole magnetic moment given the magnetic field strength at one the Earth's poles?
It appears as though applying the equation
$
\vec{B} = \frac{\mu_0}{4\pi r^3}(3(\vec{\mu}\cdot \hat{r})\hat{r}-\vec{\mu})
$
will be useful but I am not sure if that is the correct approximation to make here.
| Indeed, we can approximate the Earth's magnetic field as a dipole and apply:
$
\vec{B} = \frac{\mu_0}{4\pi r^3}(3(\vec{\mu}\cdot \hat{r})\hat{r}-\vec{\mu})
$
Where we know via the problem statement what the magnetic field strength at the North pole is, $|B_{N_{pole}}|$, and that the radius at the pole will be one earth radii, $R_{Earth}$, since it is assumed in the problem that the dipole approximation should be made from the Earth's center. Further, $\vec{\mu}$ points from South to North, and thus at the North pole we have
$\vec{\mu}\cdot \hat{r} = |\vec{\mu}||\hat{r}|\cos(0^\circ)= |\vec{\mu}|$
, so applying the magnetic field of a dipole approximation at the North pole with $|\vec{B}_{N_{pole}}|$ we have:
$\begin{align}
|\vec{B}_{N_{pole}}| &= \Big|\frac{\mu_0}{4\pi R_{Earth}^3}(3(\vec{\mu}\cdot \hat{r})\hat{r}-\vec{\mu})\Big|\\
\therefore \frac{4\pi R_{Earth}^3|\vec{B}_{N_{pole}}|}{\mu_0} &= \Big|3|\vec{\mu}| \hat{r}-\vec{\mu}\Big| = |\vec{\mu}|\Big|3 \hat{r}-\hat{r}\Big| = 2|\vec{\mu}|\\
|\vec{\mu}| &= \frac{4\pi R_{Earth}^3|\vec{B}_{N_{pole}}|}{2\mu_0}\\
|\vec{\mu}| &= \frac{2\pi R_{Earth}^3|\vec{B}_{N_{pole}}|}{\mu_0}
= \frac{2\cdot \pi \cdot (6.38 \times 10^6 m)^3\cdot (0.5 \times 10^{-4} T)}{(1.256 \times 10^{-6} NA^{-2})}\\
\therefore |\vec{\mu}| &\approx 6.48 \times 10^{22} Am^2
\end{align}$
Some insights were gleaned from this source: http://geophysics.ou.edu/solid_earth/notes/mag_earth/earth.htm
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/187088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
On energy levels and emission of photons This is a very basic question but I cannot seem to find the answer anywhere.
Say we have an atom in ground state. Its first energy level is 2 eV. An incoming photon of energy 2.5 eV hits an electron in the atom (with the lowest energy level) which is excited and moves up one enery level. However it does not have enough energy to reach the next energy level. When it loses energy it emits a photon of energy 2 eV. What has happened to the remaining 0.5 eV? Is it some other type of energy in the atom such as kinetic energy?
|
An incoming photon of energy 2.5 eV hits an electron in the atom (with the lowest energy level) which is excited and moves up one enery level.
In most cases, this won't happen. The incoming energy must match the transition that is being excited.
It's possible for a 2.5 eV photon to excite a 2 eV transition, but only if there is some other particle available to carry away the other 0.5 eV. For example there could be a phonon or there could be a new 0.5 eV photon generated. However these processes tend to be much less likely to occur, because they are three-particle interactions rather than two particles.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/187196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How can the D'Alembertian of a field be interpreted intuitively? The D'Alembertian operator is defined as
$$
\Box = g^{\nu\mu}\nabla_\nu\nabla_\mu
$$
For the Minkowski metric in Cartesian coordinates that is
$$
\Box=\frac{1}{c^2}\frac{\partial^2}{\partial t^2} - \frac{\partial^2}{\partial x^2} - \frac{\partial^2}{\partial y^2} - \frac{\partial^2}{\partial z^2}
$$
Can it be intuitively described, just as a gradient or divergence, curl or Laplacian may be?
I'm looking for something similar to the interpretation of a Laplacian given in this question and answer.
| The operator is just $\partial_t^2-\nabla^2$. So it is the difference between a "temporal laplacian" and a "spatial laplacian". Since laplacian measures curvature, this is basically telling you the difference in curvature between the spatial and temporal variation of the field.
One reason this comes up in physics is in describing elastic sheets under tension. In an elastic sheet, if there is (spatial) curvature at a point, the tension in the sheet will pull the point in order to flatten out the curvature. Thus the point feels a force in the same direction as the curvature. So by newtons second law, the point on the sheet will accelerate, that is, have a second time derivative, in the direction of the curvature. This is why you would expect the difference in the "temporal laplacian" and "spatial laplacian" to be zero.
If this operator is non-zero, then it means the temporal and spatial variations are inconsistent with each other, and it looks like there is an external force acting on the point in your elastic sheet.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/187284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Are Fluted Barrels More Rigid Than Standard Ones? This seems to be quite a debated topic in the shooting community and is something I'm not sure about.
A fluted barrel is a barrel that has grooves milled into it to reduce weight, help it cool faster and supposedly to make it less flexible (see picture below)
I don't see how removing material like this would make it more rigid, because surely by removing mass the barrel would have less inertia to help keep it rigid when under the stress of firing, unless it had the same effect as corrugated sheet metal. For example if you had a flat strip of material compared to a corrugated piece the corrugated piece would have a lot less flex.
So if you had two barrels made of the same material being the same length and width with one being fluted and the other being a smooth cylinder which would be more rigid under the stress of firing?
(Sorry if this is not how you should post on here, I normally only use StackOverFlow.)
| If you have the same mass, then the fluted beam will be more rigid because the second moment of area is larger - in the same way that an $\mathrm{I}$ beam is more rigid than a circular rod of the same mass.
In the case of a gun barrel, lighter weight helps in a number of ways - portability, ability to hold the gun still as you aim... but making it lighter will make it less rigid. So people do a few different things.
First - the barrel is thicker near the stock: the internal pressure is higher there (so it needs to be stronger), but so is the bending moment (induced by the recoil). A little bit of bending there will cause a big deviation of the muzzle and you want to avoid that.
Second - they can add a bit of mass at the tip of the barrel (near the muzzle). This increases the moment of inertia; when done right ("tuning"), it will actually make the gun more accurate as well (see for example this earlier answer I wrote on that topic).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/187370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Is entanglement a consequence of the uncertainty principle? I am an aspiring physicist and once, I asked my professor on what triggers quantum entanglement and he graciously remarked "The great uncertainty principle!" - I was slightly confused and didn't say anything but pondered about it quite heavily on my way to home. Can somebody explain this?
| As far as I understand, it is not.
Entanglement is a consquence of the Hilbert space structure of composite systems in quantum mechanics.
When you postulate that if you have system A (with Hilbert space $H_A$) and system B (with Hilbert sapce $H_B$) then the Hilbert space of both systems together (i.e., when they interact) is $H_A \bigotimes H_B$, then entanglement comes naturally.
I think that you can be a little far-fetched to give meaning to the claim of your professor, but, IMHO, this is a more natural way to see things.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/187487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Turning points of particle
A particle of mass $m$ and energy $E<0$ moves in a one-dimensional Morse potential:
$$V(x)=V_0(e^{-2ax}-2e^{-ax}),\qquad V_0,a>0,\qquad E>-V_0.$$
Determine the turning points of the movement and the period of the oscillation of the particle.
I have started learning for my exam and this is one of the exercises in my textbook.
Never dealt with this type of question, so these are my thoughts so far:
To get the turning points I was thinking of solving the equation $E=V_0(e^{-2ax}-2e^{-ax})$. I was doing the arithmetics with the absolute value of $E$. But still I couldn't seem to find the values for $x$. At the end I used Wolfram Alpha to find the values but it gave me results with complex values. Is there a simple way to solve this type of equations for $x$?
Anyway, about the period, I assume it's the time it takes for the particle to get from $x_1$ to $x_2$. But how am I supposed to approach this? How would I get a time value just out of the equation for the potential?
I hope someone can help me out here.
| Energy conservation dictates
$$ E = \frac{1}{2}m\dot{x}^2 + V(x) = \text{const}$$
With some arithmetic it follows
$$ \dot{x} = \frac{dx}{dt} = \sqrt{2m^{-1}(E-V(x))}$$
This ODE can be solved via separation of variables,
yielding
$$ \int_{t_1}^{t_2}dt = \int_{x_1}^{x_2} \frac{dx}{\sqrt{2m^{-1}(E-V(x))}}$$
The integral on the left hand side can be evaluated immediately, where $t_1$ and $t_2$ are understood as the times when the particle is at $x_1$ or $x_2$ respectively. So it is simply half the period.
Observe that the integral on the right diverges when $x$ approaches the turning point $E=V(x)$.
This method of solving Newtons equations in a 1d potential should be treated in any textbook on mechanics.
For a general potential it is in general hard or impossible to find the turning points in closed form. Here however, you can substitute $y=\exp(-ax)$ and solve the corresponding quadratic equation. I'll leave the explicit calculation to you.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/187622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Force of an ideal spring Suppose you have an ideal spring (constant of the spring $k$) attached to a uniform disc of radius $R$ as in the picture below:
The force $F$ in red is from the spring.
My question is the following:
How should I decompose the force $F$ into its $x$ and $y$ components??
My intuition would tell me to multiply the module of the force by the sine and cosine of the angle between $F$ and the $x$ axis or something similar, however I noted in some exercises and exams, the solution simply states that $F_x = -kS(u_x)$ where $S$ is the distance from the y axis, $u_x$ the unit vector of the $x$ axis and $F_x$, of course, the $x$ component of $F$. What is the correct approach? I could not find anything useful on the internet so far...
| If $S$ is de distance from $C$ to de $y$ axis, then $S=|r|\cos(\theta)$ where $r$ is the position of $C$ with respect to the origin and $\theta$ is the angle between the spring and the $x$ axis. However if you want to write $F_y$ then you would have $F_y=-k|r|\sin(\theta)u_y$. So you were correct in thinking in the decomposition of $F$ as a function of $\theta$ observer however that $S$ is not $|r|$ which is a crucial difference.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/187726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
conservation of angular momentum/conservation of energy? Perhaps a dumb question to ask, but
I was given the following problem to solve:
A 10 g bullet traveling at 400 m/s strikes a 10 kg, 1.0-m-wide door at the edge opposite the hinge. The bullet embeds itself in the door, causing the door to swing open. What is the angular velocity of the door just after impact?
When I saw this problem my first attempt to solve this was to use conservation of energy:
$m_b$=mass of bullet
$m_d$=mass of door
$L$=length of the door
$I=I_{bullet}+I_{door}= \frac 13m_dL^2+ m_bL^2$
$\frac 12m_bv^2=\frac 12Iw^2$
$w=\sqrt{\frac {m_bv^2}I}$
which is far off from the actual answer. Now, I know that the correct method of solving this type of problem is to use conservation of angular momentum, which will give:
$m_bvL=Iw$
$w=\frac {m_bvL}{I}$
Why is the first method wrong?
Thanks in advance.
| In any situation, momentum (linear or rotational) is conserved. So if there are no external forces, you can use that.
But energy is only conserved in certain situations (such as elastic collisions). The deformation of the door by the bullet is inelastic. You can consider it similar to energy losses from friction. In this case, the (mechanical) energy in the final state is less than the initial state because of losses.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/187780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why do our ears pop? Have you ever been on a train going through a tunnel or plane and your ears pop?I was wondering why this happens and I know it relates to pressure but don't know exactly the reason
| Part of your question is about human physiology; that's true. And that part is simple. Air pressure outside your eardrums needs to equalize with air pressure inside your eardrums, and this happens when the Eustachian tubes open up a little to let some air into the cavity behind your eardrums.
The physics parts of your question are about trains and airplanes. And there are two parts to that. First, the airplane goes up really high, where there is less air between you and outer space, so there's less air pushing down, so the air pressure is lower. If your ears were adjusted to the pressure at ground level, they'll need to adjust when the plane moves to higher altitude -- and vice versa.
The second (and more interesting) phenomenon is due to the speed of either the airplane or the train through the air. Bernoulli's principle says that an increase in the speed of the air around the train or plane implies a decrease in the pressure of that air.
Now, when the train enters a tunnel, suddenly all that air in the tunnel has to start moving out of the way -- or just gets pushed ahead of the train. Either way, the pressure is doing funny things. Ultimately, this changes the pressure outside of your ear, which requires your Eustachian tubes to adjust that pressure, which it does with a little pop.
Bernoulli's principle is also used to measure the speed of a plane relative to the air in something called a pitot tube, which uses the difference between the pressure at the front of a forward-facing tube and the pressure on the side of the plane (at a static port) to figure out how fast the air is moving.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/187905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Can change in temperature cause a change in mass of an object? If a gold bar is heated to say 200 degree Celsius then will it have the same mass at say 10 degree Celsius. Does energy has mass? If so then does this increased 'heat energy' cause an increase in the mass of an object
| In the modern way of viewing things, no, (rest) mass is invariant. What happens is that the energy content of the body changes and some people still interpret this as a change in mass (which is an old point of view that, unfortunately, is fairly common).
A nice discussion about this can be found here:
http://profmattstrassler.com/articles-and-posts/particle-physics-basics/mass-energy-matter-etc/more-on-mass/the-two-definitions-of-mass-and-why-i-use-only-one/
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/187975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How can there be entropy change in this system? How can there be an entropy change in this system?
Suppose if I have a system consisting of liquid water, $1\, \mathrm{kg}$ at $290\,\mathrm{K}$, I stir it, and do say, $10\, \mathrm{J}$ of work on it, I can calculate the temperature change of the system given that:
$$U = cT \quad\mbox{ and }\quad S = c \ln \Omega$$
for $c$ constant.
From the fundamental equation of thermodynamics:
$$dU = dQ + dW = 0 + dW = 0 + 10 = 10\,\mathrm{J}$$
Hence:
$$dT = \frac{dU}{cM} = \frac{1}{410}\,\mathrm{K}$$
But how can there be a change of entropy in the universe when $dQ = 0$. I understand that we can calculate it using the formula for $S$ given, but I don't understanding how the fundamental equation allows this?
$$dQ = 0$$
and
$$dS = T^{-1}\,dQ$$
Hence, it may be concluded that:
$$dS = 0$$
Can someone tell me where my understanding is lacking, because obviously the entropy change is not zero in this case?
| The formula
$$
dS = dQ/T
$$
only applies to thermodynamic processes that can be described by a path in thermodynamic state space (representing a quasi-static process, where the system is in thermodynamic equilibrium at all stages). Only for such processes it is meaningful to talk about continuous change of thermodynamic entropy.
Stirring a fluid is not such a simple process. If any work is to be done on the fluid, the fluid needs to get into state of flow. In such a state of flow, how soever slow, supplying work via stirring is an irreversible process. It is similar thing as when we're pushing a book from one side of a table to another; work is being done, but irreversibly. Both dry friction and fluid viscosity are forces that change their direction once the direction of displacement is changed, so the process is not reversible and consequently, the relation $dS=dQ/T$ cannot be used.
The whole process of stirring is a process whose intermediate stages are outside of the domain of classical thermodynamics. The only thing one can say about it based on thermodynamics is that if the process begins in an equilibrium state 1 and ends in equilibrium state 2, the change of entropy after the state 2 is attained obeys the Clausius inequality
$$
\Delta S(1\rightarrow 2) \geq \int_1^2 \frac{dQ}{T_\textrm{res}}
$$
where $Q$ is total energy increase of the system due to heat transfer since state 1 and $T_\textrm{res}$ is the temperature of the reservoir that enables the heat transfer (provided it can be ascribed temperature). If no energy transfer except due to stirring occurs, $dQ=0$ and whatever $T_\textrm{res}$, the right-hand side vanishes:
$$
\Delta S(1\rightarrow 2) \geq 0.
$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/188153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Explanation of Michelson Interferometer Fringe Shift I have been working on an experiment where 2 glass microscope slides are pinched together at one end (so that there is a "wedge" of air between them) and placed in the path of a laser in one leg of a Michelson interferometer. When I move the glass slides (fractions of a mm at a time) so that the path of the laser is closer or further from the place where the slides are pinched, a fringe shift occurs. I cannot seem to explain why this is happening! Any help with explaining this phenomenon would be greatly appreciated! If any more specifics about the setup or dimensions of the slides are needed, please let me know.Also, a full "light to dark" fringe shift occured roughly every 4mm of moving the slides.
| This is speculating - but if your slides are of non-uniform thickness, or they are bent as a result of the pinching, they will present a different path length in one leg of the interferometer (and therefore give rise to a shift in the fringe pattern). This may become clear by looking at this diagram:
In the diagram on the left, the total path length is independent of the position of the ray - in all cases the light bends by the same amount as it interacts with the different surfaces. In the diagram on the right, the rays closer to the "pinch point" will traverse less glass than the ones that are further away (which intersect the glass at a greater angle). This means that the path length will change as you move the slides left to right.
It is not clear whether there is a spacer as part of your "wedge" (I imagine there must be one, but I can't see it in your photo). If there is, then the big clip you use will surely bend the slides; and a Michelson interferometer is very, very sensitive to path length differences...
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/188247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
What was the largest object/particle tunneling observed? What is a current record? Reference to that would be nice. and what can be expected in near future? what are the theoretical limits?
| In some sense, there is some kind of classical tunneling. For example, a high jumper, if (s)he bends her/his body over the bar, can pass over the bar in such a way that its center of gravity is always below the bar, and his kinetic + potential energy is always less than mgh, where m is her/his mass and h is the height of the bar. See, e.g., A. Cohn, M. Rabinowitz, Classical Tunneling, Int'l Journ. Theor. Phys., v. 29, #3, 1990, p. 215.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/188447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Is this definition of orthohelium and parahelium incorrect?
"One electron is presumed to be in the ground state, the 1s state. An electron in an upper state can have spin antiparallel to the ground state electron ($S=0$, singlet state, parahelium) or parallel to the ground state electron ($S=1$, triplet state, orthohelium)."
From HyperPhysics
When they say "parallel to the ground state electron" then is it assuming that they are both spin up, or both spin down? If so, isn't it then ignoring the $S=1$ state with spin up and spin down:
$$|1\rangle |0\rangle = \frac{|+\rangle|-\rangle + |-\rangle|+\rangle}{\sqrt 2}$$
Therefore, if one electron is presumed to be in the ground state, 1s, state, if the spins can be opposite, a second electron can also occupy the ground state in $S=1$ orthohelium. Is this correct?
| The spins are not antiparallel in this state as @my2cts answer correctly states (as demonstrated by $\vec S_1 \cdot \vec S_2$ having a positive eigenvalue).
I was confused by this at first too, especially the argument, that the measurements of $S_{1z}$ and $S_{2z}$ will point to opposite directions due to entanglement seemed convincing to me.
But it just looks that way because the $S_z$ basis is the wrong basis to look at. They have to point in opposite directions in this basis, because the total $m_z = 0$. So you are just asking a spin state in the $x$-$y$-plane for its $z$-component. This way you can't find out whether the spins are parallel or antiparallel in that plane.
To make the parallel orientation of the spins manifest in the expression for the state, you'll have to rewrite the state in a basis in the plane.
We rewrite our state terms of the eigenstates $\lvert \pm \rangle$ of $S_x$: $$ S_x \lvert \pm \rangle = \pm \frac \hbar 2 \lvert \pm \rangle$$
In terms of those states we have
$$ \lvert \uparrow / \downarrow \rangle = \frac{1}{\sqrt{2}} \big( \lvert + \rangle \pm \lvert - \rangle \big)$$
in terms of these states the two product states become
$$ \lvert \uparrow \downarrow \rangle = \frac 1 2 \big( \lvert + \rangle + \lvert - \rangle \big) \otimes \big( \lvert + \rangle - \lvert - \rangle \big) = \frac 1 2 \big( \lvert++\rangle - \lvert+-\rangle + \lvert -+ \rangle - \lvert -- \rangle \big) $$
$$ \lvert \downarrow \uparrow \rangle = \frac 1 2 \big( \lvert++\rangle - \lvert-+\rangle + \lvert +- \rangle - \lvert -- \rangle \big) $$
So the triplet state with $m_z = 0$ is given by the following in this basis:
$$ \frac{\lvert \uparrow \downarrow \rangle + \lvert \downarrow \uparrow \rangle}{\sqrt{2}} = \frac{\lvert ++ \rangle - \lvert -- \rangle}{\sqrt{2}}$$
And in this form we can see manifestly that the spins are in parallel. (Similarly, this could be done for the eigenstates of $S_y$.)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/188543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Different signatures I was working out the christoffel symbols, once where the metric that I am using has (+---) signature and another time where it has (-+++) signature because two books had different signatures and I had to check for any inconsistencies. I had the christoffel symbols the same for both but the curvatures $(R_{\mu\nu})$ has opposite signs. Why is that?
| The signature is one convention (both in special relativity and in general relativity).
But in general relativity there are many different conventions besides just the signature. The front inside cover of Misner Thorne and Wheeler lists conventions for signature, for the Riemann Tensor, for the Einstein Tensor, and for the use of Greek and Latin indices and lists 34 texts and what conventions they use. And then spells out on the facing side where the signs go.
So it's just another convention and the Ricci tensor inherits the convention from the Riemann tensor but unfortunately also depends on a convention for the Einstein tensor. So even if you know the convention for one of those tensors you still don't know the sign of the Ricci tensor.
So you will have to also pay attention to the convention for the Einstein tensor as well as the convention for the Riemann tensor.
And the Einstein equation itself can look different depending on the convention for the Einstein tensor.
Finally, that book was published in the 1970s so maybe newer books have decided to have even more conventions to break. I can't say I've read absolutely every new book.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/188663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Probability density for momentum in Quantum Mechanics In a Quantum mechanics book, I found the following equations:
$$ \Phi(k)=\frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty \Psi(x,0)e^{-ikx}dx $$
and
$$ \Psi(x,t)=\frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty \Phi(k)e^{ikx-\frac{\hbar k^2}{2m}t}dk $$
So, with $Ψ(x,t)$ I can find $Φ(k,t)$ because the following theorem exists (Fourier transformation):
$$f(x)=\frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty F(k)e^{ikx}dk~~\Leftrightarrow~~ F(k)=\frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty f(k)e^{-ikx}dx $$
So, I suppose that $\Phi(k)$ is the probability density of the momentum. Is this true?
If so, why don't I see in books the use of the integral relation that gives us $\Phi(k)$, in order to, say, find the probability of measuring a range of values of the momentum?
Lastly, I think that this only holds if the allowable values of momentum are very close to each other (like the case of a free particle), so as to be able to make the sum of the superimposed states an integral. Am I right?
|
So, I suppose that $Φ(k)$ is the probability density of the momentum. Is this true?
Almost. $\Phi(k)$ is the probability amplitude for the momentum of the particle. The probability density is obtained as usual by squaring the amplitude, giving $|\Phi(k)|^2$.
For a free particle, all values of momentum are always allowed, which enables the superposition to be expressed as an integral. The only times when this breaks down is when you have a particle confined to a finite interval or when you impose periodic boundary conditions; this does restrict the allowed momentum values to a discrete set and turns the integral into a Fourier series.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/188780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How is the Full Standard Model group representation displayed? I have often seen, on YouTube lectures and textbooks, the direct product gauge group representation listed below and it is often accompanied with a statement to the effect that "this is how we sum up the standard model":
$$SU(3) × SU(2) × U(1)$$
My question is: As the above only deals with the gauge boson side of things, (SEE ANSWER BELOW) then is presumably (much) larger group representation that covers all of the elements of each of the 3 generations, including quarks and leptons (and the Higgs), how they interact, how their properties are related, etc, displayed?
In short, is there a group that represents/combines everything that we know about both fermions and bosons, that we can mathematically express in group representation form.
My usual disclaimer, if you can point me to a website X or textbook Y, that is as good as an answer and hopefully I can come back with better understanding and a better question.
Also, if I have, not surprisingly, got something badly mixed up, either in concepts, mangled terminology, or how groups actually work in practice, please just tell me that and I will delete this question and come back to it later with more background covered.
|
As the above only deals with the gauge boson side of things
This is a wrong assumption.
The format represents the total knowledge from innumerable data of particle physics that have been fitted with SU(3)xSU(2)xU(1) . The particles are slotted into representations of the groups and there are rules of how the interactions happen within the structure of the groups. There exists a standard model lagrangian and the groups are a symmetry of this lagrangian. For an example how the known particles are within a representation see this.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/188862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How fast does heat travel via conduction? I have read this question which seems to ask an identical question, but I'm not sure - it had far too many words I don't understand, let alone the equations. Perhaps someone can answer with a heat-for-dummies answer.
I understand that given thermal conductivity, and thermal mass, it takes a while for heat applied to one end of a material to make the other side rise in temperature. Like for a heat sink for electronics, with the component dissipating a constant power, and the ambient temperature staying constant, a thermal gradient will develop until it reaches an equilibrium, where the power dissipated by the electronic component equals the power dissipated by the heat sink to the ambient environment.
But my question is: say the heat sink is exactly at ambient temperature. And say the electronic component instantly starts dissipating to the heatsink a given power. How long would it take for the heatsink to start dissipating even the smallest amount of power to the environment? My guess is it would be equal to the speed of the molecules' vibrations that is otherwise known as heat.
Or perhaps at the speed of light, since thermal radiation would penetrate the material, even if it is very, very little.
I hope this makes sense?
| At near the speed of light. The same maxwells equations that describe the conduction of electric discribe the conduction of thermal energy.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/188943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Why does a Yo-Yo sleep, and then awaken? What are the mathematics / mechanics principles behind a sleeping Yo-Yo,
and in particular, what changes with a wrist-snap flick that
causes it to "awaken" and return to your hand?
(Image from www.wikihow.com/Make-a-Yoyo-Sleep.)
There must be some stability principle that keeps it asleep, and
which is disturbed by the tug.
| "Sleeping" is when the YoYo rotates in place because the string is freely rotating around the bar between the halves. If something is done to prevent the string from freely spinning then the energy that was driving the YoYo to rotate in place is now used to reel in the string. Flicking your wrist or otherwise jolting it can cause the string to double up on itself in the narrow gap between the halves, giving it enough friction to no longer freely spin around the bar and to instead begin spooling up. Note that this friction is usually not sufficient to prevent the YoYo from unraveling the doubled up string on its next trip down. The "sleeping" process can begin again when the string is again unraveled into the state where the loop can freely rotate around the bar in the middle, if the jolt of the YoYo reaching the end of its string isn't enough to cause the string to bunch up again as described above.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/189367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 3,
"answer_id": 1
} |
Why is probability of finding the electron at a certain point when one of the slits is closed $|\Psi|^2 $ & not $|\Psi|^2 dx$? Let in a given physical condition, the wave-function to a particle be assigned as $|\Psi (x_i,0,0,t)|^2 dx$.
Now, at the double-slit experiment , the probability of finding the particle at any $x$ when hole 2 is closed is given as $P_1 = |\Psi|^2$.
My problem is here. At the first, I learnt that $P_i = |\Psi_i|^2 dx$. But in the double-slit experiment, I found only $P_1 = |\Psi_1|^2$ devoid of $dx$. Why is it so?
| The square of the wavefunction, $|\Psi(x,t)|^2$, is the probability density function for finding the particle. This means that the probability of finding the particle in an interval of (infinitesimal) width $\mathrm dx$ at position $x$ equals $|\Psi(x,t)|^2\mathrm dx$. On occasion, however, authors will drop the $\mathrm dx$ if it is convenient and does not affect the results they are demonstrating. This is an example of the balance one must always strike, particularly in textbook material, between clarity of exposition and thoroughness and rigour; this particular instance is probably justified as long as there is little chance of confusion (though if you want further examination of this particular usage you should state which book you are reading).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/189468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to interpret vector operators in quantum mechanics? To the point:
How should I think about the equation $$\hat{\mathbf{x}}\mid\mathbf{x'}\rangle = \mathbf{x'}\mid\mathbf{x'}\rangle~?$$
Is it a triple of equations
$\hat{x}\mid x'\rangle = x'\mid x'\rangle$
$\hat{y}\mid y'\rangle = y'\mid y'\rangle$
$\hat{z}\mid z'\rangle = z'\mid z'\rangle$
or can the 1-D operators act directly on the (vector) state $\mid\mathbf{x'}\rangle$ such as
$\hat{x}\mid\mathbf{x'}\rangle = x'\mid\mathbf{x'}\rangle$?
In the latter case can we interpret $\mid\mathbf{x'}\rangle$ as a simultaneous eigenket of the operators $x, y$ and $z$, given they all commute? ( i.e $\mid\mathbf{x'}\rangle = \mid x',y',z' \rangle$ )
I have tried thinking about this problem in terms of discrete states as sometimes I have an easier time with matrix notation but in this case I am again stuck. As above should I interpret the vector equations as a triple of matrix equations (or a matrix of matrices)?
Finally, how do I interpret operator functions of $x, y$ and $z$ such as $F(r)\mid\mathbf{x'}\rangle$, where $r = \sqrt{x^2+y^2+z^2}$. See, for example Q1.27 Sakurai, Modern Quantum Mechanics.
| Your second option is correct.
It is a simultaneous eigenket of all three commuting operators. And that is how it should be interpreted. To be explicit, the equation $$\hat{\mathbf{x}}\mid\mathbf{x'}\rangle = \mathbf{x'}\mid\mathbf{x'}\rangle~?$$
Is a triple of equations
$\hat{x}\mid \mathbf x'\rangle = x'\mid \mathbf x'\rangle$
$\hat{y}\mid \mathbf x'\rangle = y'\mid \mathbf x'\rangle$
$\hat{z}\mid \mathbf x'\rangle = z'\mid \mathbf x'\rangle$
And you can use a product to have all three components as part of one object.
As for operators inside functions, if you consider a simultaneous eigenket then it isn't strange to define the operator to act by scalar multiplication of the scalar value determined by putting the eigenvalues into the function.
You can work out the entire functional calculus but that is not what most everyday physicists do and probably not what is expected in a first course. For instance in the position representation pointwise multiplication of a function of the coordinates does exactly this and usually a fuss is not made.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/189585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How to find tangential/radial/angular velocity for motion in any curve? Is the radial velocity responsible only for changing distance between objects and the component perpendicular to it only for change in direction? If so why?
Please try to give a different explanation than saying that the radial velocity points in the line of sight can only increase the distance, and radial velocity is not affected by the component perpendicular to it, because I find this difficult to understand as velocity can be decomposed into two vectors that are not perpendicular, by using non-perpendicular coordinate axes.
What is the proof of the relation between tangential and angular velocity along any curve?
The formula for angular velocity (I am referring to proof of this relation) is given by
$v \cos(\beta)/R$ (where $v$ is the speed and $R$ distance from the origin or observer). $v \sin(\beta)$ is the radial velocity. Is there a specific name for the $v \cos(\beta)$ component?
| You would be pretty accurate in saying: "the radial velocity responsible only for changing distance between objects and the component perpendicular to it only for change in direction"
Why?
Consider each case individually:
1) An object has only "radial velocity" that is, it points directly away (or towards) the observer: $\mathbf v = v_0 \hat {\mathbf r}$.
Well, it moving directly towards or away from you, as 'directly' as possible.
2) An object has only "angular velocity" which we'll take to be in the $\hat \theta$ direction (imagine a 2D world), so $\mathbf v = v_0 \hat {\mathbf \theta}$.
This is the case for something whose radius from you never changes, as it's always moving in the 'theta' direction from the observer. This is circular motion. Note, this changes with position! (and for this reason is unlike Cartesian coordinates, that are time/orientation independent).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/189683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
What are the symmetry criteria for continuous phase transitions in Landau theory? My understanding is that within Landau theory, a continuous phase transition is only possible if certain symmetry rules are satisfied. (These rules represent necessary but not sufficient conditions for a continuous phase transition.) One criterion is that the symmetry group of one phase must be a subgroup of the other phase's symmetry group (cf. p. 782 of Rep. Prog. Phys. 50, 783). What are the other criteria, and where can I find a derivation of them?
| This "criteria" is neither sufficient nor necessary.
There are many symmetry-breaking transitions which are not continuous. For example, it is well-known that $n$-state Potts model has a thermal symmetry-breaking phase transition, and when $n>4$ it is first order. For a more realistic one, I think melting transition is first order. Basically there is no way to tell whether a transition is continuous or not just from the symmetry. It really depends on the details a lot.
There are also "exotic" continuous phase transitions between states with the same symmetry, or states with very different symmetry breaking. A famous example of the latter type is the so-called deconfined quantum criticality, which happens in lattice spin models where one phase spontaneously breaks translation symmetry and the other phase spontaneously breaks spin rotation symmetry. Neither is a subgroup of the other. The critical theory is a gauge theory, going beyond the scheme of Landau theory.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/189807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
If Coulomb and Esu have different dimensions, why can we convert simply by 1C= 3*10e9 esu? I was always told in school we cannot convert an apple into a potato, so we have to have the same dimensions on both sides.
When converting coulomb to esu, we do not seem to bother with dimensions. Why is it still possible?
| An important point here is that you don't just convert from Coulombs to ESU. You also change the units of various constants that appear in the expressions of electromagnetism.
The most obvious example is, Coulomb's Law
$$\begin{align*}
F_{SI} &= \left (8,99 \times 10^9 \,\mathrm{N \, m^2/C^2}\right)
\frac{q_{1, C} \, q_{2, C}}{r^2} \\
F_{guass} &= \left (1 \,\mathrm{dimensionless} \right)
\frac{q_{1,ESU} \, q_{2,ESU}}{r^2} \,.
\end{align*}$$
Similarly in Gauss's Law the factor of $1/\epsilon_0$ (which is dimensional) that appears in SI units must also be converted to a $4\pi$ (dimensionless) in Gaussian units.
The upshot is that the statement
$$ 1\,\mathrm{C} = 2997924580 \,\mathrm{ESU} \, \tag{!}$$
isn't really an equality, but is one step in a recipe for converting between different systems of units.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/189942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Friction in Rolling Motion My understanding of rolling motion:
When a body rolls, at every instant, there is just one point of contact between the body and the plane and this point has no motion relative to the plane.
Now if there is no motion relative to the plane, then how does the rolling friction came into being?
I researched a bit about the origin of rolling motion but could not understand what does the following paragraph means:
During rolling, the surfaces in contact get momentarily deformed a little (Why?), and this results in a finite area of the body being in contact with the surface.
| Your first quote is correct for an idealised model. There is no rolling friction then. Both wheel and surface are considered completely rigid.
Ideal model - no rolling friction
Non-ideal/more realistic model - rolling friction comes into the picture
These pictures are from this link that gives a very good graphic view on this. Going away from an ideal model introduces rolling friction since the wheel touches more than just one point, and not all points press back directly through the center - forces from such points cause counteracting torques, which is perceived as the rolling friction.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/190057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Radius of curvature I have come across a question that asked me to find the radius of curvature of a projectile.
As far as I know, the path of a projectile is a parabola and I have found mention of the radius of curvature referring to lenses and mirrors. But in optics, the lens and mirrors were assumed to be part of a circle.
My questions are:
*
*How can a parabola have a center from which a radius is to be measured?
*Does the radius of curvature change with the position of body (in projectile motion)?
*In mechanics and mathematics, what is the radius of curvature and how does one calculate it (in the case of a parabola)?
| We can draw a tangent (Call it 't') at any point (call it 'p')on a curve. We can also draw a circle to which 't' is a tangent. The radius of this circle is the radius of curvature to the given curve at the point 'p'.
An analogy from motion of a body along a curved path may help easier understanding.
When a body moves along a curved path, its velocity keeps changing. However, we can talk of the instantaneous velocity of the body at each and every point along the curve.
On similar lines, for a given curve the radius of curvature keeps changing along the curve. However, we can talk of radius of curvature at each and every point along the curve. It refers to the radius of the circle which has a common tangent with the given curve at the point under consideration.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/190262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 3,
"answer_id": 2
} |
What is "forward peaking"? In "Research and Development for a Gadolinium Doped Water Cherenkov Detector" the phrase "forward peaking" is used to describe a signal. This comes up in lots of other contexts too, but I still can't work out what's meant by it. The sentence in question is;
"But due to a strong forward peaking of the neutrino-electron
elastic scattering cross section, a solar neutrino signal can be extracted from the data using the directionality of the
electrons."
Can you define "forward peaking" please?
| It means that when the neutrinos hit electrons, the electrons are moving preferentially in the same directions that the neutrinos were moving. So when we are building a water Cherenkov detector for solar neutrinos, the Cherenkov signal will be coming from the direction of the sun. This is very advantageous to suppress background and because of the daily and annual movement of the sun we can alo suppress other (nondirectional) detector bias effects that could distort a neutrino signal from the sun very effectively.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/190376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
"Center of a black hole is a time" $\newcommand{\d}[1]{\mathrm{d} #1}$In one lecture (around 1:33:15) of the series of lectures "Theoretical Minimum" of Prof. Susskind he talks about black holes and the Schwarzschild metric:
$$\d \tau^2=\left(1-\frac{r_s}{r}\right)\d t^2 - \left( \cfrac{1}{1-\frac{r_s}{r}} \right) \d r^2 - r^2 \d \Omega^2$$
where $r_s$ is the Schhwarzschild radius.
He says that there is nothing funny going around for $r<r_s$ because both $\d t^2$ and $\d r^2$ term changes sign. I'm perfectly find with this but then he says whilst discussing about black holes that "$\;r=0$ is not a place but more like a time and that's why you can't avoid falling into black hole. You can't avoid the future."
I really don't understand the above statement. I think he is referring to the fact that $\d t$ becomes a space component because of the minus sign and $\d r$ becomes a time component because of the plus sign but I don't even know what that means intuitively. It would be really nice if someone can explain the quote I gave above.
| The Schwarzschild metric as you've written it is only one particular coordinate system and the fact that $r$ and $t$ switch roles at the event horizon is an artifact of that coordinate system. There are other coordinate systems which make certain properties of the metric more intuitive. The ones that might be most useful for you are ones which can be drawn as a Penrose diagram. In a Penrose diagram time is always up and radius is always side to side and light always travels at 45 degrees. On the diagram of a black hole you can see that the event horizon is just the point where all light (and therefore everything) must hit the singularity because the singularity encompasses all of its possible future positions.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/190451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
Can we find actual rest mass of things on Earth Earth moves around the Sun and the Sun moves around the galaxy and the galaxy moves with unknown speed and direction. We have speed so the mass of us all altered.
Can we know the real rest mass? If so, can we deduce our speed in the universe?
| The rest mass of an object, by definition is the total energy of an object as measured by an inertial observer who is at rest relative to the object. If the object is not moving uniformly, then you can measure its rest mass from a momentarily co-moving freefall frame. This rest mass is also the constant in Newton's law, i.e. the inertial mass as well.
So you need to be at rest relative to the object, and you need to be confident that so-called tidal effects are not significant over the object's spatial and temporal extent (i.e. that special relativity applies to the object and the comoving laboratory you fit it into, or, from a Newtonian perspective, that the gravitational field over this laboratory can be treated as uniform).
Once you fulfill these mild conditions, then you can measure the object's rest mass, most easily by giving it a whack of known impulse and measuring the change in velocity that follows. So, yes, the mass we measure in our laboratories of things here on Earth is going to be extremely near to the "true" rest mass, with only fantastically small tidal effects tainting this measurement. But, as in Acid Jazz'z Answer, we don't have theories that give us rest masses of objects.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/190558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 5,
"answer_id": 0
} |
Phase on Aharonov-Bohm effect doubts How I show that $$\Lambda(\textbf{x}')=\frac{q}{\hbar}\int \mathbf{A} \cdot d\mathbf{x'}$$ on $$
\tilde{\psi}(\textbf{x}',t)=e^{[\frac{iq\Lambda(\textbf{x}')}{\hbar c}]}\psi(\textbf{x}',t)$$ for Aharonov-Bohm effect? I don't know about path integral formulation, that was suggested on Sakurai,J.J. book... There is other way to show that? Using the gauge transform for $$\tilde{\mathbf{A}}=\mathbf{A}+\nabla\mathbf{\Lambda}~ ?$$
| You should be able to show by direct substitution that your proposed wave function $\tilde{\psi}(\mathbf{x},t)$ solves the Schroedinger equation, where the vector potential is included via the minimal coupling prescription
$$ \mathbf{p} \to \mathbf{p} - \mathrm{i} q \mathbf{A}(\mathbf{x})$$
(up to a sign convention for $\mathbf{A}$ and some dimensionful constants which I may have got wrong), so long as $\psi(\mathbf{x},t)$ is the solution when $\mathbf{A} = 0$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/190628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Galaxy bias and baryon acoustic oscillations (BAO) I have a doubt with the concept of galaxy bias and how it affects baryon acoustic oscillations, it is supposed to mean that by measuring the distribution of galaxies we are not measuring the actual distribution of dark matter (I imagine this is due to processes of galaxy formation?). But since baryon acoustic oscillations affect the baryons, why do we need to know the dark matter distribution?
| You are correct about the bias factor - because the dark matter distribution is not measured directly, but via tracers (galaxies), there may be some bias in the tracers. This is put into the analysis as an unknown bias parameter $b$ that needs to be fit.
Now the second part of your question - why do we care about the dark matter? Briefly, it is because the dark matter and baryons mutually gravitate. Due to hydrodynamics, BAOs result in an overdensity of baryons in a shell at the sound horizon scale around an initial (random) baryon overdensity. But these overdense baryon regions are massive, and their gravity pulls in a bit of extra dark matter than the average, so there is also an overdensity in the dark matter distribution tracing the overdensity in the baryons.
Then later, galaxies form. We think every galaxy forms inside a dark matter "halo", and because of BAOs there should be more galaxies than average forming at separations equal to the sound horizon scale because (1) there is some extra dark matter there to help draw in baryons for galaxy formation, (2) there are some extra baryons to fuel galaxy formation (and their self-gravity will also help along gravitational collapse).
The baryons and dark matter are gravitationally coupled "fluids" (dark matter is typically modelled as a collisionless fluid), so they need to be considered together.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/190747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is it possible for man to break Earth into 2 parts? Many countries have extreme devastating nuclear weapons. Also they have weapons in very large numbers.
I want to ask that
Is it possible for man to break earth into 2 parts with the help powerful weapons like nuclear weapons or any other technology?
|
break earth into 2 parts in which each part is nearly equal to half of volume of the earth
Yes and no.
No, as in you can't neatly split the planet in half. Most of the earth is liquid and will re-form once the cutting device has passed through it. Much like cutting pudding.
Yes, if you want 2 half-earth balls when you are done and don't care about the intermediate states. You would need to apply enough energy to overcome the gravitational binding forces plus just the right amount to accelerate the bits to lunar orbit. Then just arrange for half to reform here and half to reform there. After a few millenia you should end up with two half-earth size balls orbiting each other.
Does Man have the energy to do this? Not by many orders of magnitude. Every nuke ever made, piled in one place and detonated all at once, will make a most awesome explosion and crater. The shockwave will (literally) be heard around the world, and there will be a rather large glowing crater left behind. It's possible the environmental and radioactive effects will extinguish all life, but there will be zero measurable effect on the planet or it's orbit. We've been hit by pretty big rocks in the past, some of them made bigger bangs.
You will need either a Death Star or an Illudium Q-36 Explosive Space Modulator to pull off the Big Split. The latter is preferred as it is apparently pocket-size.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/190844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Why is the introduction of a quantization volume necessary for quantization of the EM field I have been working through the quantization of the electromagnetic field, and every source I find introduces a quantization volume with periodic boundary conditions in the process, in which we fit the general solution of $A(\boldsymbol{x},t)$. Why is this necessary? I understand that this allows us to consider a countably infinite sum over wave vectors, rather than an uncountable one, as the wave vectors are made to satisfy the periodic boundary conditions.
I have the vague impression that it has something to do with the orthogonality of the wave functions (solutions to the wave equation of the field before quantization), so that integration as follows yields a delta function
$$\sum\limits_{kk'} \int dx \, e^{i(k-k')x} = \delta(k-k')$$
but then I think this should work equally in the continuous case
$$\int\int {dk \,dk'} \int dx \, e^{i(k-k')x} = \delta(k-k')$$
What am I not understanding? Thanks in advance for any help!
| Quantizing in a finite volume is not specific to the electromagnetic field, and it is not a necessity, neither for the electromagnetic field nor for any other.
It is generally more well-behaved to quantize in a finite volume because no infrared-like divergences appear from allowing arbitrarily low momenta (since no arbitrarily long wavelengths fit into the finite volume), and because extensive quantities like energy will not turn out to be infinite, while, naturally, e.g. for non-zero (vacuum) energy density and infinite volume the (vacuum) energy will also be infinite.
But you may also quantize in infinite volume and have Fourier integrals over momenta instead of Fourier sums, there is nothing prohibiting it, and "usually" (i.e. in my limited experience) QFT is indeed done in infinite volume.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/190948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
What is the meaning of coherent decay rate and population decay rate? In quantum optics, especially the part when talking about atom-light interaction, there are two different kinds of decay rate, coherent decay rate $\gamma_{ij}$ where $i$ and $j$ are atomic energy states, and population decay rate $\Gamma_{ij}$.
I don't clearly understand these two. Could you explain in term of both physical and mathematical meaning?
Thank you!
| I am not a specialist of quantum optics but I am familiar with laser and classical optics.
I would say that coherent decay is stimulated emission and population decay rate is simply spontaneous emission.
When atoms/electrons are interacting with light they have a certain probability of absorbing photons to get into higher energy level states. Once they are in the higher energy states they have 2 different possibilities to relax back to original state :
They can loose their energy by spontaneous radiation which only depends on the nature of the transition. The spontaneous emission probability can be assessed from quantum vacuum fluctuation / virtual photon. The most important is that this radiation does only depend on the transition and is random in phase and direction, hence referred to as incoherent .
The other way to radiate back into original state is through stimulated emission. It actually consists in a radiation that is excited by another photon. The newly created photon is this time in phase with the exciting photon and has the same direction hence said as being coherent . The likelihood of stimulated emission depends on the transition and the amount of exciting photons.
Mathematically :
The coherent decay is given by :
$\gamma_{ij}=B_{ij}u(\nu)$
The population/incoherent decay is given by :
$\Gamma_{ij}=A_{ij}=\frac{1}{\tau_{ij}}$
$A_{ij}$ and $B_{ij}$ are the Einstein coefficients and their ratio can be established with thermodynamic considerations(photons being bosons whereas atoms follows Boltzmann statistics) :
$\frac{A_{ij}}{B_{ij}}=\frac{8\pi h\nu^{3}}{c^{3}}$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/191020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Kinematics of Euler angles relative to a rotating frame I have a rotating body $B$ and a rotating frame $F$ whose orientations are described by the quaternions $q_B$ and $q_F$ respectively. I also have the angular velocity vectors $\omega_B$ and $\omega_F$.
I'm then interested in the Euler angles (extrinsic x-y-z) of B relative to F. That is, I convert $q_F^* q_B$ to Euler angles $\phi, \theta, \psi$.
My question is how to calculate $\dot\phi, \dot\theta, \dot\psi$ from the angular velocities. I'm currently using
$$M(\phi,\theta) = \begin{pmatrix}1&\sin\phi\tan\theta&\cos\phi\tan\theta\\
0&\cos\phi&-\sin\phi\\
0&\sin\phi\sec\theta&\cos\phi\sec\theta\end{pmatrix}$$
$$\begin{pmatrix}\dot\phi\\\dot\theta\\\dot\psi\end{pmatrix}=M(\phi,\theta)\omega_B$$
But clearly this isn't correct when F is rotating.
| First Thought (probably not the fastest)
Let us assume you have a vector space in $R^{3}$ with a quaternion defined as:
$$
\mathbf{q} = q_{F}^{*} \ q_{B} \\
= a + b \hat{\mathbf{x}} + c \hat{\mathbf{y}} + d \hat{\mathbf{z}}
$$
where $(a, b, c, d)$ are the Euler parameters and $(\hat{\mathbf{x}}, \hat{\mathbf{y}}, \hat{\mathbf{z}})$ defines the reference unit basis set.
If we define the axis of rotation as $\mathbf{n}$ and the angle through which we rotate as $\zeta$, then the Euler parameters are defined as:
$$
a = \cos{\left( \frac{\zeta}{2} \right)} \\
b = n_{x} \ \sin{\left( \frac{\zeta}{2} \right)} \\
c = n_{y} \ \sin{\left( \frac{\zeta}{2} \right)} \\
d = n_{z} \ \sin{\left( \frac{\zeta}{2} \right)} \\
$$
Thus, if you know $\mathbf{q}$, or rather $(a, b, c, d)$, you can find $\mathbf{n}$ and $\zeta$. Once you know the axis of rotation and the angle of rotation, you can determine the Euler angles. First we define the cross product matrix as:
$$
\left[ \mathbf{n} \right]_{x} = \left[
\begin{array}{ c c c }
0 & - n_{z} & n_{y} \\
n_{z} & 0 & - n_{x} \\
- n_{y} & n_{x} & 0
\end{array} \right]
$$
and the outer product of $\mathbf{n}$ with itself given by:
$$
\left[ \mathbf{n} \otimes \mathbf{n} \right] = \left[
\begin{array}{ c c c }
n_{x} \ n_{x} & n_{x} \ n_{y} & n_{x} \ n_{z} \\
n_{y} \ n_{x} & n_{y} \ n_{y} & n_{y} \ n_{z} \\
n_{z} \ n_{x} & n_{z} \ n_{y} & n_{z} \ n_{z}
\end{array} \right]
$$
Then we can define the rotation matrix as:
$$
\overleftrightarrow{\mathbf{R}} = \cos{\zeta} \ \overleftrightarrow{\mathbf{I}} + \sin{\zeta} \ \left[ \mathbf{n} \right]_{x} + \left( 1 - \cos{\zeta} \right) \ \left[ \mathbf{n} \otimes \mathbf{n} \right]
$$
where $\overleftrightarrow{\mathbf{I}}$ is the unit or identity matrix.
Second Thought (probably faster/easier)
An easier method is to follow the procedure given here. Following that procedure, we define:
$$
\alpha = \frac{ 2 \left( a \ b + c \ d \right) }{ 1 - 2 \left( b^{2} + c^{2} \right) } \\
\beta = 2 \left( a \ c - d \ b \right) \\
\gamma = \frac{ 2 \left( a \ d + b \ c \right) }{ 1 - 2 \left( c^{2} + d^{2} \right) }
$$
which gives us the Euler angles:
$$
\phi = \tan^{-1}{ \alpha } \\
\theta = \sin^{-1}{ \beta } \\
\psi = \tan^{-1}{ \gamma }
$$
Since you already have $\mathbf{q}$ and you can numerically/analytically determine $(\phi, \theta, \psi)$, then I would just take the time derivative of each of these angles to find $(\dot{\phi}, \dot{\theta}, \dot{\psi})$ rather than using the angular velocities.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/191088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Why do elements on the Binding Energy per Nuclear Molecule after Iron (most stable) even form? So I was reading about the stability of elements based on Nuclear Binding Energy, and I saw that the 'Iron group' of elements were most tightly bound and hence most stable, and that is why the graph peaks there. Why do elements that come after Iron, which are less stable even exist? And if they do, why do they not constantly strive to achieve Iron-like stability?
| There are a couple of related questions:
*
*What elements can be created in the fusion process of different types of stars?
*What is the heaviest element possible produced in a supernova?
though surprisingly I can't find an exact duplicate (which probably just means I didn't look hard enough).
Iron is the most stable nucleus so in principle all other nuclei should fuse or fission to form iron, but the reaction is extremely slow because large kinetic barriers exist. If heavy nuclei are formed faster than they can decay then we end up with a signficiant concentration of the heavy nuclei.
In supernovae and stars heavy nuclei can be formed by the r-process and the s-process respectively. In normal stars the temperature is not high enough for the heavy nuclei to decay to iron at any significant rate (though they can be destroyed in other nuclear reactions). The temperature in supernovae may be high enough, but the high temperature lasts for too short a time. In either case the end result is a significant concentration of the heavy nuclei.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/191195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Cylindrical capacitor in an electric circuit I've come across a tricky question and would appreciate some hints or explanations as to why the given solution is the way it is. The question reads as follows:
A coaxial cable consists of a wire with radius $a$ (the core of the cable), which is wrapped with insulating material with dielectric constant $\varepsilon$, until radius $b$ (called the insulator). Around the cable there is a layer of conducting material (radius $c$ from the center of the cable and is called the wrapper).
The wire's length is $d$ such that $d \gg a,b,c$. At one side of the cable, a voltage source $V_0$ with inner resistance $R_0$ is connected to both the wire and the wrapper, and at the other side, a resistor $R$ is connected instead of a voltage source.
It asks to find the magnetic and electric fields $B(r)$, $E(r)$, where $r$ is the distance from the center of the cable (from $z$-axis in the picture), when $t\rightarrow+\infty $.
In the solution, they said that when $t\rightarrow +\infty$, no current will pass through the cylindrical capacitor so: $I=\frac{V_0}{R_0+R}$ therefore $V\left(\text{final}\right)=V_0 \frac{R}{R_0+R}$.
I do not get this, how can one imagine how this circuit works? Is there an equivalent and more simple circuit? According to what they said, after infinite time, no current passes through the capacitor, but the wires are connected to the wrapper so how can there be current at all in the circuit? All I know is when an uncharged capacitor is charged, it will act as an open switch in the circuit after a long time.
Possible equivalent Circuit?:
| There is only one connection to the wrapper, at the power supply end. One can read the problem to imply that the resistor at the other end is connected to the wrapper, but you shouldn't. If there were a connection there, the resistor would be shorted out. As the wrapper has only one connection, at long $t$ it will have come to the proper equilibrium potential and there will be no current in the wrapper. Your basic circuit is then an ideal battery in series with two resistors, which is where the final current calculation comes from.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/191424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Field expansion in Peskin & Schroeder Peskin and Schroeder state something which I'm not fully understanding. More specificially I think it's just phrased in a way I'm not understanding.
In the Schrodinger picture we can expand the real scalar field $\phi(x)$ which satisfies the Klein-Gordon equation as
$$\phi(\textbf{x})=\displaystyle\int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_p}}(a_pe^{i\textbf{p.x}}+a_p^\dagger e^{-i\textbf{p.x}}).$$
Then of course we find $\phi(x)=\phi(\textbf{x},t)$ by switching to the Heisenberg picture.
Now, on page 83 they say
At any fixed time $t_0$ we can of course expand $\phi$ in terms of ladder operators
$$\phi(\textbf{x},t_0)=\displaystyle\int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_p}}(a_pe^{i\textbf{p.x}}+a_p^\dagger e^{-i\textbf{p.x}}).$$
Then to obtain $\phi(\textbf{x},t)$ for $t\neq t_0$ we just switch to the Heisenberg picture
$$\phi(\textbf{x},t)=e^{iH(t-t_0)}\phi(\textbf{x},t_0)e^{-iH(t-t_0)}.$$
The first problem is that they say we switch to the Heisenberg picture, implying we were in the Schrodinger picture to begin with. But then how can the $\phi$ be time-dependent i.e. why is it depending on $t_0$, even though $t_0$ appears nowhere in the expansion?
Are they just saying somewhat awkwardly that $\phi$ is (obviously) not time-independent in the Schrodinger picture, we pick a certain time slice (where our states are now time fixed) and then time evolve from there? It shouldn't matter since I'd imagine we should have $\phi(t_0)=\phi(t')$ for some other time $t'$
| The upshot is that we need one condition to specify how the operators in the Schrödinger and Heisenberg picture are connected. This is usually done by declaring that the two pictures agree at some fixed instant $t_0$.
To summarize: The Schrödinger operator $\phi(\textbf{x},t_0)$ does not depend on time $t$, while the Heisenberg operator $\phi(\textbf{x},t)$ does depend on time $t$. For kets and bras it is the other way around.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/191625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Is magnetic reconnection reconcilable with magnetic field lines neither starting nor ending? According to Maxwell's equations, magnetic fields are divergence-free: $\nabla \cdot \mathbf{B} = 0$. If I understand this correctly, this means that magnetic field lines do not start or end. How can we reconcile this with magnetic reconnection?
| $\nabla\cdot\mathbf B=0$ does indicate that there are no magnetic monopoles, so there isn't a "starting" or "ending" point for field lines is mostly correct. So this must mean that magnetic field lines either
*
*form closed loops
*extend to infinity
*intersect the domain boundary (wall, stellar surface, etc)
So the "starting & ending points" issue is nuanced beyond what you've stated. With reconnection, we can usually assume the middle option: field lines extend to infinity (though invoking that they intersect the boundary is just as valid).
For those unawares, magnetic reconnection is the when magnetic field lines pointing in opposite directions pinch together (reconnect) and form new lines:
(source)
To model this, one needs to modify the ideal MHD equations (this is because if we assume $\mathbf B\parallel\delta\mathbf x$ where $\delta\mathbf x$ is some displacement of field lines, it will remain so for all time $t$). For typical plasmas, one uses Faraday's law in conjunction with the Lorentz force to model the magnetic evolution,
$$
\frac{\partial \mathbf B}{\partial t}=-\nabla\times\mathbf E=\nabla\times\mathbf u\times\mathbf B\tag{1}
$$
But when considering magnetic reconnection, the conductivity isn't assumed to be infinite, so we have to use Ohm's law and add the current density, $\mathbf J\sim\nabla\times\mathbf B$:
$$
\frac{\partial \mathbf B}{\partial t}=\nabla\times\left(\mathbf u\times\mathbf B+\eta\nabla\times\mathbf B\right)\tag{2}
$$
where $\eta$ is the magnetic diffusivity. So now the magnetic field can diffuse, rather than simply moving along the flow; this is what allows for reconnection to occur in the plasma.
However, because the divergence of the curl of any vector is identically zero, $\nabla\cdot\nabla\times\mathbf A=0$, both (1) and (2) satisfy the divergence-free condition.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/191942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Clarifying some notation, the square of a vector derivative I'm reading a text which asserts that, if $\vec{F}(\vec{x})=-\nabla V(\vec{x})$ then we define
$$E = \frac{m}{2} \left( \frac{d\vec{x}}{dt}\right)^2-V(\vec{x}) \, .$$
However, I don't understand how you can square $d\vec{x}/dt$ since this is a vector function. Does this mean that the vector is dotted with itself? Or are we not taking the coordinate-wise derivative of $\vec{x}$?
| The kinetic energy of a particle whose motion is described by $\textbf{r}(t) = \left(x(t),\,y(t),\, z(t)\right)$ is, at the point $(x,y,z)\in\mathbb{R}^3$, defined as
$$
T(x,y,z) = \frac{1}{2}m\left(\dot{x}^2 + \dot{y}^2 + \dot{z}^2\right) =
\frac{1}{2}m\,\textbf{v}\cdot\textbf{v}
$$
and yes, the square of a vector means (by abuse of notation) its scalar product with itself.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/192028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Raychauduri equation for Milne universe I am trying to solve this kind of problem:
Consider the Milne model, i.e., the empty $ \kappa = −1 $ Friedmann model. Verify by a direct calculation that the expansion $ \Theta $ of the unit normal to the $ \tau = \mathrm{const}$ hypersurfaces satisfies
$$
\dot{\Theta} + \Theta^2/3 = 0.
$$
I started from Raychauduri equation
$$
\dot{Θ} + Θ^2/3 - \dot{u}^a_{;a} + 2(σ_{ab}σ^{ab} − ω_{ab}ω^{ab}) + R_{ab}u^au^b = 0
$$
Where for empty space $ R_{ab}=0 $ and geodesic flow $\dot{u}=0$.
Then I found that if the congruences are hypersurface orthogonal, then $\omega_{ab}=0$. But I don't really understand what that means(or why is it true), can someone please explain it to me please?
| It can be shown that $\omega_{ab} = 0 \Leftrightarrow \omega^a \equiv \epsilon^{abcd}u_b \nabla_c u_d = 0$. The latter quantity is known as the twist (or vorticity). In a local inertial frame it is easy to see that $\vec{\omega} \sim \vec{\nabla}\times \vec{v}$ where $\vec{v}$ is the 3-velocity field of the flow.
This lends to the following interpretation of $\omega^a$ (and hence of $\omega_{ab}$). If an observer $\mathcal{O}$ in the congruence carries with them a local Lorentz frame $e_{\hat{\alpha}}$ such that the spatial axes $e_i$ are Fermi-Walker transported, so that these spatial axes constitute a set of mutually orthogonal torque-free gyroscopes, then the average rotation, relative to the gyroscopes, of a sphere of observers in the congruence centered on $\mathcal{O}$ will vanish.
Now, a geodesic flow in flat space-time is just a congruence of inertial observers. Furthermore, and more importantly, the expansion, just as in the usual FRW metric, is isotropic and homogenous meaning relative to a fiducial observer $\mathcal{O}$, the neighboring observers only have a relative radial velocity in the Milne coordinates. Thus they clearly do not rotate relative to $\mathcal{O}$, which is why $\omega_{ab} = 0$, basically by construction.
As for why hypersurface orthogonality implies $\omega_{ab} = 0$ on an intuitive level, it is simply because if $\omega_{ab} \neq 0$ then the worldlines of the congruence would be twisting around one another in which case it is impossible to find a non-intersecting family of surfaces that is everywhere orthogonal to the worldlines.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/192118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Is UV catastrophe same as IR catastrophe? I am currently studying quantum physics from Serwey-jewet. Where in the topic of Planck's law, infrared catastrophe is alternatively used for UV catastrophe while explaining how Plancks constant explains Rayleigh jeans law and UV catastrophe. Is it the same thing or it is an error,because as far as i know infrared catastrophe is an different phenomenon that occurs in photons.
| It has to be an error. Rayleigh-Jeans theory fails at high frequency, but it's a good approximation for low frequency.
Infrared catastrophe has nothing to do with the black body radiation problem. Only at high frequencies it was evident that Rayleigh-Jeans model failed.
In the following plot you can see how at low frequencies (where infrared belongs), the agreement between Planck's model and Rayleigh-Jeans model is good
Image borrowed from here, you can read it too.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/192207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Help recognizing partial differential equation I would be very grateful if someone could tell me something about the following partial differential equation:
$$
\frac{\partial U}{\partial t} = K * (\frac{\partial^2 U}{\partial r^2} + (1/r)\frac{\partial U}{\partial r}).
$$
A friend told me that the equation models the heat equation, but I don't think he is right.
Any help?
| That is the heat equation in polar coordinates with axial symmetry. The (isotropic) heat equation without sources or sinks is
$$
\frac{\partial U}{\partial t} - K\nabla^2U =0.
$$
If you look up the Laplacian operator in cylindrical coordinates, you will find that your expression matches this exactly.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/192341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Simple Harmonic Motion in Special Relativity I was trying to see what results I would get if I were to incorporate relativistic corrections into the case of a harmonic oscillator in one dimension. I thought that if the maximum velocity of the oscillating body were to approach relativistic velocities, the measurement of the time period of motion and related measurements should change. I assumed a lab frame $(t,x)$.
The equation of motion should be
\begin{equation}\frac{dp}{dt}+kx=0
\end{equation}
where $t$ is the time measured in the lab frame. $\frac{dp}{dt}$ is calculated to be $\gamma^3m_0\ddot{x}$(since only $a_{\mid\mid}$ exists for this motion). Naively, I took this as my solution in the following way:
\begin{equation}\gamma^3m_0\ddot{x}=-kx
\end{equation}
\begin{equation}\implies\ddot{x}=-\frac{k}{\gamma^3m_0}x
\end{equation}
\begin{equation}\implies\ddot{x}=-\omega^2x
\end{equation}
So I took the angular frequency $\omega$ simply like that. However, I forgot that $\gamma$ is a function of $\dot{x}$ and this obviously complicates things... for instance the period becomes a function of velocity. I'm not sure if my approach is correct and I don't have the tools to solve the differential equation I would get if i were to open up $\gamma$, so my questions are the following:
*
*Am I doing this right?
*Am I missing something conceptually?
*Is there a better way to approach the problem?
*And if, by some miracle, my last expression for $\ddot{x}$ is correct, could I have
some help solving the differential equation?
EDIT: If you are posting the relativistic expression, please post the corresponding evaluation of the Classical limit of your expression.
| You can get an exact solution for $t(p)$, although it involves a rather nasty integral that I'm not sure can be written in closed form. Here's how:
The equations of motion are
$$
\frac{dp}{dt} = -kx \qquad \frac{dx}{dt} = \frac{1}{m} \frac{p}{\sqrt{1 + p^2/m^2 c^2}}.
$$
This second equation can be obtained by taking the equation $p = m v /\sqrt{1 - v^2/c^2}$ and solving for $v$. Differentiating the first equation and plugging in the second then yields
$$
\ddot{p} = - \omega^2 \frac{p}{\sqrt{1 + p^2/m^2 c^2}}.
$$
where $\omega^2 = k/m$ as usual.
This is a second-order ODE of the form $\ddot{p} = f(p)$, and so can be solved (at least in principle) via the method of quadratures:
\begin{align}
\dot{p} \ddot{p} &= - \omega^2 \dot{p} \frac{p}{\sqrt{1 + p^2/m^2 c^2}} \\
\frac{d}{dt} \left( \frac{1}{2} \dot{p}^2 \right) &= \frac{d}{dt} \left( - \omega^2 m^2 c^2 \sqrt{ 1 + p^2/m^2 c^2 }\right) \\
\frac{1}{2} \dot{p}^2 &= - \omega^2 m^2 c^2 \sqrt{ 1 + p^2/m^2 c^2 } + C,
\end{align}
where $C$ is a constant determined by the initial conditions. In particular, if we take the case where the particle is released from rest a distance $A$ from the origin, then we have $p_0 = 0$ and $\dot{p}_0 = - kA$, implying that
$$
C = \frac{1}{2} k^2 A^2 + \omega^2 m^2 c^2;
$$
and so we have
$$
\frac{dp}{dt} = \pm \sqrt{ k^2 A^2 + 2 \omega^2 m^2 c^2 - 2 \omega^2 m^2 c^2 \sqrt{ 1 + p^2/m^2 c^2 } }.
$$
This is a separable equation, so in principle, then, we have an implicit solution for $t(p)$:
$$
t(p) = \int_0^p dp \left[ 2 C - 2 \omega^2 m^2 c^2 \sqrt{ 1 + p^2/m^2 c^2 } \right]^{-1/2}.
$$
Making everything as dimensionless as possible, we redefine $\tilde{p} = p/mc$ and $\tilde{C} = C/\omega^2 m^2 c^2$; the integral then becomes
$$
t(\tilde{p}) = \frac{1}{\omega} \int_0^\tilde{p} d\tilde{p} \left[ 2 \tilde{C} - 2 \sqrt{ 1 + \tilde{p}^2 } \right]^{-1/2}.
$$
I don't know how to solve this integral, personally. Mathematica tells me that there's a solution in terms of incomplete elliptic integrals, but it's not terribly illuminating; and it seems pretty hopeless to try to invert this to get $p(t)$ and thereby get $x(t)$.
However, this expression can still be used to find the period of oscillation. If we think about the oscillation, it will take one-quarter of the object's period to go from $p = 0$ to a maximum of $p$ (where $\dot{p} = 0$.) This will be precisely where quantity in square brackets in the above integral vanishes. In other words, the period $\tau$ will be given by
$$
\tau = \frac{4}{\omega} \int_0^{\sqrt{\tilde{C}^2 - 1}} d\tilde{p} \left[ 2 \tilde{C} - 2 \sqrt{ 1 + \tilde{p}^2 } \right]^{-1/2}
$$
Note that the dependence on amplitude of the period is encoded in the constant $\tilde{C}$. That said, I'm still not sure how to solve this integral, and Mathematica doesn't much care for it either. If nothing else, it's amenable to numerical integration now.
EDIT: In the Newtonian limit, we will have $\tilde{p} \ll 1$ at all times, and
$$
\tilde{C} = 1 + \frac{1}{2} \frac{k^2 A^2}{\omega^2 m^2 c^2} = 1 + \frac{\omega^2 A^2}{2 c^2} \quad \Rightarrow \quad \sqrt{\tilde{C}^2 - 1} \approx \frac{\omega A}{c}.
$$
So the period in this limit becomes
\begin{align*}
\tau &\approx \frac{4}{\omega} \int_0^{\omega A/c} d\tilde{p} \left[ 2 + \frac{\omega^2 A^2}{c^2} - 2 - \tilde{p}^2 \right]^{-1/2} \\
&= \frac{4}{\omega} \int_0^{\omega A/c} d\tilde{p} \left[ \frac{\omega^2 A^2}{c^2} - \tilde{p}^2 \right]^{-1/2} \\
&= \frac{4}{\omega} \left[ \arcsin \left( \frac{c \tilde{p}}{\omega A} \right) \right]_0^{\omega A/c} \\
&= \frac{4}{\omega} \left[\frac{\pi}{2} \right]= \frac{2\pi}{\omega}.
\end{align*}
So the Newtonian result for the period is recovered. One could do a similar expansion to solve for $p(t)$ in this case, since we would get something like $\omega t = \arcsin (c \tilde{p}/\omega A).$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/192426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Will a bullet travel the same after going through 2 substances (in different orders)? This may seem like a trivial question, however I am a little curious as to whether 'order matters' in the case of this system, or whether you would be able to treat any object as a 'black box' with given input and output.
My scenario is this: If you have a block of ballistics gel with a steel plate at the end, and you fire a bullet through it (Such that the bullet has enough energy to easily pierce both gel and steel) will the amount of kinetic energy (or speed) of the bullet vary depending on if it went through the gel first and than steel, or through the steel first and than the gel?
Assume that the materials are held in place such that when shot they won't move vertically or out of place.
Would the solution change if the bullet didn't get deformed?
| No - hitting the steel plate first will deform the bullet resulting in greater drag in the gel and lower penetration
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/192818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
A misconception in an application of Newton's laws When a body is kept in ground, and is at rest, the downward gravitational pull is balanced exactly by the Normal Reaction if we assume the earth to be an inertial frame. But this would mean that any external force provided to that body would lead to its motion. if this was the case, moving things both heavy and light would be same. what is the external force that abstains it from happening?
|
When a body is kept in ground, and is at rest, the downward gravitational pull is balanced exactly by the Normal Reaction if we assume the earth to be an inertial frame.
This is not correct. The normal force is a constraint force that acts in one direction (upward). Suppose you exert an upward force of five newtons on an object that weighs eight newtons and another object that weighs four newtons. In the case of the eight newton object, the upward normal force exerted by the ground will decrease from eight newtons to three newtons as you apply the force. In the case of the four newton object, the object accelerates upwards. The normal force vanishes the moment the object leaves the surface. The normal force is not glue; it does not switch from an upward force that keeps objects from sinking into the Earth to a downward force. It acts in one direction, normal to and away from the surface.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/192963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Free fall into circular motion If I'm on a roller coaster free falling from height $h$ and then suddenly start going into horizontal motion with a radius $r$ of turn what is the $g$-force I experience?
I worked out the equation like this but am not sure if it is correct:
*
*(1) instant velocity of free-fall $v=\sqrt{2 g h}$
*(2) uniform circular motion acceleration $a = \frac{v^2}{r}$
*(3) $g$-force $ gf = \frac{a}{g} = \frac{v^2}{g r}$
My doubts are:
*
*I don't know if I can use uniform circular motion equation since $v$ is not constant
*Where is the g-force directed towards? The center of the turn?
| There are two accelerations involved: The gravitational acceleration $g$ that points down, and the centripetal acceleration $a_r = \frac{v^2}{r}$ that points along the radius vector of the curve. The component of the gravitational acceleration that is tangential to the curve does not contribute to the g-force as it accelerates the cart and us in this direction. We do feel the component of the gravitational acceleration that points along the radius vector given by $a_g = g \text{sin}\phi$ where $\phi \in [0,\frac{\pi}{2}]$ is the angle that starts with zero when you enter the curve and ends with $\frac{\pi}{2}$ when you enter the horizontal part. We can add the two accelerations as they are parallel:
$$a_{tot} = a_g +a_r$$
We now observe: $ v = \sqrt{2g(h+r\text{sin}\phi)}$ and so
$$a_r = \frac{v^2}{r} = \frac{2g(h+r\text{sin}\phi)}{r}$$
thus:
$$a_{tot} = g \text{sin}\phi + \frac{2g(h+r\text{sin}\phi)}{r} = 3g \text{sin}\phi + \frac{2gh}{r}$$
The g-force is $ a = \frac{a_{tot}}{g}$ thus
$$a = 3 \text{sin}\phi + \frac{2h}{r}$$
So the g-force is no constant but depends on where you are in the curve (on $\phi$). And as sin is monoton increasing over $[0,\frac{\pi}{2}]$ so is the g-force. it reaches it's maximum as it enters the horizontal part.
What you might also wanto observe is that if you build a rollercoaster like this people would run screaming, as the g-force is not continues. There is a jump as you enter the curve and a big jump as you go onto the horizontal. And basically at these discontinuities the g-force is infinite :)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/193042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
Must we test whether e.g. $A=B$ and $A=C$ implies $B=C$ by experiment? Chaper 10, conservation of momentum in "The Feynman Lectures on Physics" in the chapter entitled, the authors write that
Suppose we know from the foregoing experiment that two pieces of
matter, $A$ and $B$ (of copper and aluminum), have equal masses, and we
compare a third body, say a piece of gold, with the copper in the same
manner as above, making sure that its mass is equal to the mass of the
copper. If we now make the experiment between the aluminum and the
gold, there is nothing in logic that says these masses must be equal;
however, the experiment shows that they actually are. So now, by
experiment, we have found a new law. A statement of this law might be:
If two masses are each equal to a third mass (as determined by equal
velocities in this experiment), then they are equal to each other.
(This statement does not follow at all from a similar statement used
as a postulate regarding mathematical quantities.)
I assume that by "a similar statement used as a postulate regarding mathematical quantities" the authors refer to the transitive axiom of algebra, that is, if $A=B$ and $A=C$, then $B=C$.
It seems to me that, using Feynman's definition of mass and the transitive axiom of algebra, one must conclude that it is the case that $B=C$ even without making an experiment. Why do Feynman et al claim that
there is nothing in logic that says these masses must be equal; however, the experiment shows that they actually are
| We know how real numbers (in the mathematical sense) behave. There is no a priori reason however to assume that masses of objects behave as real numbers. The proposed experiment can be seen as a check to see if masses actually can be modeled by (a subset of the) real numbers.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/193266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Why are bandstructures plotted only along certain symmetry points? Why is it that bandstructures are usually represented along certain symmetry points ? What determines these symmetry points ?
| Here's an example bandstructure
(image from here). Each point on the horizontal axis is a point in k-space. Then the vertical axis shows the bands.
k-space is three dimensional. In a perfect world we would make a four-dimensional graph with k-space on three axes versus band energy on the fourth axis. Alas, we live in a 3D universe and can most easily look at 2D graphs. So the best we can do is to pick a path through k-space and use that as the horizontal axis. What path? There is no right or wrong path, all that matters is that the plot you make illustrates the thing that you're trying to explain.
Typically people choose a path that goes on straight segments between "high symmetry points" (center of the BZ and corners of the BZ and side-centers of the BZ), simply because people are often discussing things that happen at those points (like band minima) so we want to illustrate them. Also this path is typically (qualitatively) representative of the whole 3D k-space - for example, if there is a bandgap along this path then there is typically a bandgap in all of 3D k-space (this is not required, but it is usually the case).
Note that the straight-line path is still kinda weirdly meandering, for example it revisits the $\Gamma$ point twice in this example.
But if you wanted to chart a different course through k-space -- some random windy path -- you are welcome to make such a bandstructure plot. There's nothing wrong with that. But it would usually be a slightly less useful plot than the traditional type that follows straight line-segments between high-symmetry points.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/193347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
With a machete, why is a diagonal cut more effective than a right angle one? When cutting back some thick growth in the garden a question that always nagged me. Why is cutting diagonally seemingly more effective than cutting at right angles? Part of the answer is obviously to do with the ease of cutting vertically as opposed to horizontally (with vertical stems), but this also seems to be true at most other growth angles as well.
| The flexural rigidity of a vertical beam will be greater with a diagonal cut. This allows more of the force to be channeled into cutting the target as opposed to bending the target.
The determinants of flexural rigidity affected by a diagonal as opposed to a horizontal cut are the cross section of the beam and distance from the supported end of the beam. The elasticity of the material is measured by Young's modulus, but that's not at issue here as it applies to the material itself, not to its positioning or structure.
A diagonal cut effectively increases the beam's cross section, which increases rigidity.
A diagonal cut transmits more force into the ground (as Jerry Shirmer and CoilKid noted in their comments), which effectively shortens the distance to the supported end of the beam.
For a beam in tension or compression (as one would expect of a vertical beam), the axial stiffness is:
k = A * E / L
where:
A is cross section area,
E is Young's modulus or the modulus of elasticity (Bending stiffness is also a function of the modulus of elasticity, cross section, and length: https://en.wikipedia.org/wiki/Bending_stiffness),
L is length (effectively length to the ground).
The diagonal cut effectively increases A and decreases L, resulting in greater k and more force put into cutting, less into deforming the target.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/193455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
drift velocity of gas needed to get into a pressurised vessel Would a gas entering a narrow tubular opening of an initially empty vessel at a given velocity, fill up the vessel to a definite pressure?
If so, how is the velocity of the gas related to the final pressure of the gas in the vessel?
If now, the gas is let out of the opening, what would be the initial velocity of the gas stream?
| Preliminary:
Pressure is the driving (or source) term in Naiver-Stokes equation that governs fluid flows in continuum region, if we know pressure at inlet and outlet we can find the velocity for simple low speed flows.
This process is isentopic if there is no heat flow. That problem can be solved by Bernoulli equation if the flow is irrotational, and the equation is (this can be used for this flow). Here i'm neglecting surface tension because I'm not sure about the length to diameter of the tube you want to use.
$$ p_1 +\rho \frac{v_1^2}{2} =p_2 +\rho \frac{v_2^2}{2}$$
If there is any diverting in that passage you can consider continuity equation
$$ A_1v_1=A_2v_2$$
Here
$p_1$, $v_1$ are inlet velocity and pressure respectively. For big vessel or chamber $v_1$ =0 (approx.)
$A_1$ and $A_2$ are cross sectional area normal to the flow at inlet and exist respectively.
I'm not aware of any analytical method to solve transient equation. For that only way is solving NS equation using numerical methods or some Computational Fluid Dynamics packages.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/193527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
A circuit that is net charged What differences would you measure if a circuit were significantly charged negatively? Would the resistance change? To be clear, I mean that excess electrons are added to the system. The circuit can be of any kind you can imagine.
| To do this experiment, you will need:
1> a grounded, sheilded cable, (to prevent magnetic interference)
2> an ability to apply a voltage from the cable to ground, (say a battery, or better a variable DC source, with the postive terminal grounded and the negative one attached to the cable) and
3> a sensitive ohm-meter.
Then compare the ohms through the cable with voltage applied or not to the cable, to generate more electrons. Keep a log of your results. It might be wise to compare various materials for the cable, to see if some materials have more "hole effect" than others. I would also advise not going above two car batteries in voltage unless you are a trained electrician. (24 volts)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/193618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Why are angles dimensionless and quantities such as length not? So my friend asked me why angles are dimensionless, to which I replied that it's because they can be expressed as the ratio of two quantities -- lengths.
Ok so far, so good.
Then came the question: "In that sense even length is a ratio. Of length of given thing by length of 1 metre. So are lengths dimensionless?".
This confused me a bit, I didn't really have a good answer to give to that. His argument certainly seems to be valid, although I'm pretty sure I'm missing something crucial here.
| "In that sense even length is a ratio. Of length of given thing by length of 1 metre. So are lengths dimensionless?"
No, if they where then where did the 1 meter come from, why wasn't it 1 feet, or 1 mile?
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/193684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "55",
"answer_count": 12,
"answer_id": 6
} |
Are Cosmic Rays net neutral in charge? As I understand them, Cosmic rays mainly consist of high energy charged particles. I began to wonder if they would eventually net charge the Earth and then assumed that they must come in roughly equal amounts of charge. If they don't I suppose the charged Earth would deflect like charge particles and attract the opposite until it is neutral once more. Any thoughts on the matter would be interesting to hear.
| The net charge of earth due to incoming CR particles may not be static and may vary over periods of time. Earth may have a semi-stable ferromagnetic core and transient peaks of positive or negative absolute-earth-charge may inductively contribute to earth’s magnetic pole reversal. Assuming CR particles with directed energy, if earth is bathed in positively charged CR particles the earth should take on a positive charge or a negative charge for the opposite scenario. Significant static-attraction or static-repulsion of CR particles due to earth’s accumulated charge may only occur at the extremes of Earth’s transient potentials (magnitude of potential extremes would then depend on the typical velocity and mass-to-charge ratio of the particles). Periods of the transient extremes may be similar to the periods of Earth magnetic pole reversals.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/193771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Why do the waters of the different oceans of the world have different ratios of their H and O isotopes? I was reading about the Kelvin, then water, which defines it, so I ended up reading about the VSMOW. It's based on "average" ocean water.
From https://en.wikipedia.org/wiki/Vienna_Standard_Mean_Ocean_Water :
"VSMOW is a recalibration of the original SMOW definition and was created in 1967 by Harmon Craig and other researchers from Scripps Institution of Oceanography at the University of California, San Diego who mixed distilled ocean waters collected from different spots around the globe. VSMOW remains one of the major isotopic water benchmarks in use today."
Hence the question in the title.
For further clarification more from the above link:
"The isotopic ratios of VSMOW water are defined as follows:
*
*${}^2H/{}^1H$ = $1455.76 ±0.1$ ppm (a ratio of 1 part per approximately $6420$ parts)
*${}^3H/{}^1H$ = $1.85 ±0.36 × 10^{−11}$ ppm (a ratio of 1 part per approximately $5.41 × 10^{16}$ parts, ignored for physical properties-related work)
*${}^{18}O/{}^{16}O$ = $2005.20 ±0.43$ ppm (a ratio of 1 part per approximately $498.7$ parts)
*${}^{17}O/{}^{16}O$ = $379.9 ±1.6$ ppm (a ratio of 1 part per approximately $2632$ parts)"
So these are averages gained from the mix, because, apparently, the components vary in their individual ratios. So that's what the question is. Why do they vary?
| The water in the Oceans is partly
1) the original water from their formation and this will have come into equilibrium with the underlying strata of earth. These strongly depend on location for the chemicals composing them and this will apply to the isotopes in these chemicals
2) Rain water which brings down whatever dust exists in the atmosphere, and again there are different chemicals in the dust. You should see the rust rain from the winds coming from the Sahara desert to the Mediterranean.
3) drainage of rain water by the rivers from the continents surrounding the ocean
All this has been happening from the creation of the oceans and the continents , which define the oceans, are largely barriers to homogenizing currents.
Now the quantity you are asking about is for distilled water. This would reflect the isotopic composition of the ocean for chemicals that can be distilled, which according to 1-3 will be different from ocean to ocean.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/193950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 3,
"answer_id": 2
} |
Ozone elimination I have a potential problem with airflow through a high voltage capacitively coupled RF discharge (in a tube) producing ozone. How can I remove ozone from the airflow? The use of liquids is not possible. Flow rate is quite low, in the mL per minute.
| The absolute cheapest thing could be to wire a variable resistor in series to try to get the discharge voltage down: it's possible that you could reach a regime where you're only ionizing some of the gases in the air, without melting the resistor.
Since ozone's "badness" comes from being hyper-reactive you might be able to remove it chemically by putting some combustion process e.g. a candle in the tube. You might also be able to filter it out.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/194171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
If two objects have all the same conditions except different masses. Will their terminal velocity be different? I can't seem to find a straight forward answer to this. I really just want to know if changing mass of an object affects the terminal velocity. If two objects of the same dimensions except one had twice the mass, fell from a plane, would the one with higher mass reach a faster terminal velocity, therefore making it hit the ground before the one with less mass? I know all objects have the same gravitational pull which makes a marble and a bowling ball hit the ground at the same time if you drop them. But if they were both dropped from a plane would the marble max out at terminal velocity slower than the bowling ball, making the bowling ball hit first?
| Imagine 3 objects. One is a flat piece of paper. The second is an identical piece of paper rolled into a ball. The third object is the exact same shape and size as the rolled up piece of paper but it's made of iron.
If you drop the flat paper and the rolled up paper at the same time, the rolled up paper hits first because it has less air resistance due to having less surface area against incoming wind. (I'm sure you are of aware of that). Now drop the rolled up paper and the piece of iron at the same time. The iron will hit first. Air resistance depends on the velocity of the object and its surface area, which are the same in this case. So what gives? Well just because the wind resistance is the same doesn't mean they fall at the same rate. The iron is more massive so a wind resistance force slows it down less than it does the rolled up piece of paper.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/194476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Is double-slit experiment dependent on rate at which electrons are fired at slit? I am a mathematician and I am studying string theory. For this purpose I studied quantum theory. After reading Feynman's book in which he described the double-slit experiment (Young's experiment) I was wondering if I send one electron per day or per month (even more), could I see the interference pattern?
| Imagine that the sended electron interacts with the surface electrons from the slits edges. Together they form a quantized electric field. This field is not static in the sence that the position of the incoming electron is slightly different and the surface electrons are not standing still. The incoming electrons get deflected from the surface electrons (or hits the wall or goes trough without influence if it is fare away from the edge) and this deflection is quantized.
The intensity distribution on the detector's screen shows this quantized field. So it does not matter you shot a electron per day or per month. About the somehow similar interaction between the surface electrons and photons see here.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/194570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 2
} |
How does a magnet cause magnetic force and its magnitude/direction? Magnetic fields are created due to electron spin in the magnet. But how exactly does an electron "spinning" create forces around the magnet? Also, the magnetic force on a charge moving in the magnetic field is $qvB\sin\theta$, and its direction is perpendicular to the field and velocity of the charge. Why are these two facts true?
I read the explanation for how magnetic force arises next to an electric current (How do moving charges produce magnetic fields?), but does the special relativity argument apply to magnets?
| Magnets are made from metals. Metals have crystalline structures and this means that the atoms somehow are fixed. Under the influence of an external magnetic field the magnetic dipole moments of the atoms get somehow aligned and sometimes could stay aligned after removal of the external field. The magnetic dipole moment in atoms arises from the magnetic dipole moments of the electrons, the protons and neutrons too.
Now about moving particles. In this case the mentioned from you intrinsic spin comes into game. The intrinsic spin and the magnetic dipole moment of an electron are linked unambiguously. Say, the magnetic dipole moment is parallel to the spinning axis for electrons, then it is antiparallel for positrons. When move electrons in a bended wire due to gyroscopic effect the intrinsic spins get aligned and by this the magnetic dipole moments get aligned too. A coil under current flow will be a magnet.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/194628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Time evolution of scalar field Consider the quantized real scalar field acting on the vacuum state $\vert 0 \rangle $. We can interpret the state $\phi(\textbf{x})\vert 0 \rangle $ (defined in the Schrodinger picture at $t=0$) as a particle created at $(t=0,\textbf{x})$.
Peskin and Schroeder say that the state $\phi(x)\vert 0 \rangle $ is particle prepared at spacetime point $x$. I see how this works in the Heisenberg picture. But I want to work in the Schrodinger picture to convince myself of whats going on. So I time evolve the state $\phi(\textbf{x})\vert 0 \rangle $ by acting with the time evolution operator $U=e^{-iHt}$. However this time evolution gives an incorrect expression, in particular instead of having $e^{iEt}$ which is what I need, I have a $e^{-iEt}$ term which doesn't coincide with the Heisenberg picture result (or the result you get from just solving the Klein Gordon equation).
What's the problem?
Edit: A more in depth explanation;
We have $\phi(\textbf{x})\vert 0 \rangle = \displaystyle\int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_\textbf{p}}}e^{-i\textbf{p}.\textbf{x}}a^\dagger_\textbf{p} \vert 0 \rangle$. In the Heisenberg picture we can show that at time $t$ we have
$\phi(x)\vert 0 \rangle = e^{iHt}\phi(\textbf{x})e^{-iHt}\vert 0 \rangle$.
Expanding $e^{-iHt}$ as a Taylor series and since the state $\vert 0 \rangle$ is such that $H\vert 0 \rangle=0$ we have $e^{-iHt}\vert 0 \rangle=\vert 0 \rangle$. This gives us
$\phi(x)\vert 0 \rangle = e^{iHt}\phi(\textbf{x})\vert 0 \rangle$. Using the commutation relations between the ladder operators and $H$ we obtain the correct result.
The problem is that if I take the state $\phi(\textbf{x})\vert 0 \rangle$ and time evolve with the time evolution operator $U=e^{-iHt}$ we obtain $\phi(x)\vert 0 \rangle = e^{-iHt}\phi(\textbf{x})\vert 0 \rangle$. Comparing our expressions we have $e^{-iHt}=e^{iHt}$ implying $t=0$ or $H=0$, the former coinciding with the fact that the pictures agree at $t=0$.
Why am I not obtaining the same result in the Schrodinger picture?
| This problem arises because you cannot compare states in the Heisenberg picture and states in the Schrödinger picture (except for $t=0$) because they are physically different objects.
The only thing that coincides and can be compared are matrix elements or expectation values. If you have a look at expectation values, you will see (quite trivially) that there is no difference.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/194702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Coulomb's law with an $r^3$, not $r^2$, in the denominator I am reading an older physics book that my professor gave me. It is going over Coulomb's law and Gauss' theorem. However, the book gives both equations with an $r^3$, not $r^2$, in the denominator. Can somebody please explain why it is given as r^3? An image is attached for reference.
Also for equation 1-24, can somebody please explain how the middle side is equal to the right side with the del operator?
| It does give "Coulomb's law" with $\frac{1}{r^3}$, it gives it in its proper vectorial form
$$ \vec E \propto \frac{\vec r}{r^3}$$
which, when taking the absolute values, yields the form you are probably more familiar with
$$ E \propto \frac{1}{r^2}$$
since $\lvert \vec r \rvert = r$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/194894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Observations of erratic rotation of asteroids An asteroid generally has an irregular shape, therefore, one would expect its rotation is quite erratic in some sense. Are there any observational examples?
| Rigid bodies with three distinct moments of inertia have two stable rotation axes, the axes with the greatest and least moments of inertia (typically the shortest and longest axes). Non-rigid bodies have but one stable rotation axis, the axis with the greatest moment of inertia. The axis with the least moment of inertia becomes unstable thanks to entropy.
Assuming no external torques act on the body, the body's angular momentum will be conserved. It's rotational energy, however, is not conserved. The tumbling of a non-rigid body causes the object to heat up a bit. That heat gets radiated into space. The object loses energy to the universe as a whole (thereby maximizing entropy). The axis with the smallest moment of inertia represents the largest rotational energy configuration for a fixed angular momentum. That axis gets locked out as the body radiates away energy. The axis with the largest moment of inertia represents the smallest rotational energy configuration for a fixed angular momentum. This is the configuration toward which non-rigid rotators eventually migrate.
As a consequence, most asteroids rotate about their shortest axis (the axis with the greatest moment of inertia). There are a few that don't. Toutatis exhibits a complex rotational motion.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/195022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
The Makeup of the Pentaquark Why is it that when they have the artist's rendition of the Pentaquark it shows two downs, two ups, and one anti-strange quark? Is this or is this just for show? Follow up to this question: if this configuration is just for show, what is the Pentaquark truly made of?
| I suggest you to read my answer here for a quick look into exotic hadrons. These particles mostly(at least so far) are very unstable appears for a short time and decay into another particles, hadrons and mesons. Hadrons made of 3 quarks and mesons one quark and its anti-matter. These unstable particles(resonances) have a quark content based on initial and final state particles.
$$
\Lambda_b^0\rightarrow Resonance \rightarrow J/\psi p K^-
$$
$J/\psi$ has a charmonium quark content $c \bar{c}$ where proton $uud$. And our penta-quark has five-leafed clover.
But this not as easy as seen. There are lots of possibilities about quark content and you need to eliminate the others by technical analyzes. I strongly recommend the LCHb article for further details.
Today, researchers are confident about the exotic state, pentaquark and its form of quarks. What they should do further is studying the "internal mechanism" of the particle. I mean quark interactions of the particle. Further studies will be about whether this particle is tightly bound with strong gluonic interactions or softly bound via pion exchange as I showed in my previous answer.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/195310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Engine fuel consumption vs power Something that has bothered me for a long time is why a 600 hp engine uses more fuel per kilometer than a 80 hp engine. Let pretend I have two equal cars (same shape same weight etc) except for the engine, one is 600 hp and other is 100 hp. If I manage the throttle to accelerate both with the same acceleration, my knowledge says that will require the same energy to get to a certain speed, so why will the 600 hp will use more fuel/energy?
Also if I am right is it possible that in the future cars will be so efficient that a big engine will only consume more if pushed to the limit?
| To generate more power, an engine needs to be able to burn more fuel; laws of physics, chemistry and thermodynamics dictate this requires a larger displacement (bigger volume in which you burn the fuel).
Larger volume = larger area over which you generate friction, more more importantly, more air being pulled through per stroke. Some engines with a lot of cylinders have actually been engineered to "shut off" a certain number of cylinders when running in low power mode. This is done by closing the valves for the entire stroke. Rather than sucking in air, compressing it, and letting it go through the exhaust (which takes energy, even if you didn't burn any fuel), these cylinders keep the air inside - it loses a bit of energy due to friction, but much less than is wasted by moving air.
See for example variable displacement technology and this article on the Mercedes Benz engine which claims up to 30% improved efficiency because of this technology. That demonstrates that the air flow is indeed a major component of the efficiency of small displacement engines.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/195479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
How is energy dissipated in a travelling em wave How is energy dissipated in a travelling em wave. Will there be any dissipation if it were to travel trough vaccum ?
| Individual photons will not lose any energy as long as they do not interact with any other particle. If you are referring to the intensity of the EM emission, that depends of the angle incidence from their source.
So basically, if you imagine a laser that could emit just a single photon in the vacuum of space, that photon would maintain its frequency, and therefore its energy, up until the moment it interacts with another particle.
$$E=hf$$
Where $E$ is the energy of the photon, $h$ is Plank's constant, and $f$ (or alternatively $\nu$ is the frequency of the photon.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/195616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proof for Resistance is directly proportional to length and inversely proportional to the area of cross-section? I have heard that Resistance is directly proportional to length and inversely proportional to area of cross-section. Can someone give me a practical explanation for this?
| Well, I guess you could "derive" it in the Drude model (see my post here), where the proportionality of current density $\vec j$ and electric field $\vec E$ is the conductivity $\sigma$ or inverse resistivity $\rho$:
$~~~~~~\vec j = \cfrac{q^2}{m} \tau n \vec E = \sigma \vec E = \cfrac{1}{\rho} \vec E$
Using the current density $|\vec j| = I / A$ as quotient of current $I$ to cross section of your conductor $A$ and the electric field $|\vec E| = U/d$ as quotient of the voltage drop $V$ over a piece of wire of length $d$ one obtains
$~~~~~~\cfrac{I}{A} = \cfrac{1}{\rho} \cfrac{U}{d}$.
The definition of resistance $R = U/I$ than provides the sought relation
$~~~~~R = \rho \cfrac{d}{A}$.
From an engeneering standpoint you could see it as discrete portions of resistance, as a electrical circuit component. Put two in series, you double $d$, and you double the resistance. Put two parallel, well you figure it out youself ;).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/195725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
What happens to a radioactive element or isotope's electrons when it undergoes alpha decay? It seems to make sense that when an atom loses two protons, it would lose two electrons as well, but I don't actually know what happens.
| It is complicated and we ignore it, but your intuition is right. When the nucleus loses an alpha particle its charge decreases by two. The atomic physicists now claim their job is done and don't care. The solid state physicists don't consider radioactivity, so they don't care either. If it is an atom floating freely in space, two electrons will move off in some direction, but who cares? If it is an atom in a solid crystal, you should ask how much the recoil moves the atom and whether it dislocates the crystal.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/195831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Why do semiconductors remain neutral outside the depletion region? Why there is a sharp cut off of the charged region outside the depletion region, like on this image?
For example why don't electrons on the conduction band in the n-type side rush towards the positively charged area making the whole piece positively charged somewhat, not just at the area near the depletion region?
The source of the confusion is that I know if you charge up a regular conductors the internal currents will uniformly distribute the charge along the whole piece, while insulators are only locally charged up, since they cannot carry current.
Semiconductors here seem to act like insulators, but diodes do carry current when used. How?
| Look up the depletion region approximation. Which by the way is actually pretty good, especially when the junction is revered biased.
When it is forward biased, or one side is highly doped sometimes it is not as good and it would be better to work out the solution numerically. Then you could find that the edges are not sharp.
Part of the reason the edges are sharp when you make the approximation is that the carrier concentrations are pretty large and the doping is supposed to be uniform. Then assumption is that there are no free carriers since they get swept out by the electric field.
I think the other thing to consider in answering your question is that you end up balancing drift and diffusion currents to maintain charge neutrality.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/195928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
How covalent bonding lower the energy of potential energy? Considering the potential energy of interacting particles, how does covalent bonding lower the energy of the system?
| They would not bond if it was not lower energy. Like I said look at it from the other direction - it takes energy to pull them apart. An O2 molecule is actually smaller than an oxygen atom. It is a tight orbitals that fill the orbitals. Like hudling (spooning) from the cold. S likes to fill (2) and then P likes to fill up (6 I think). Actually 3 P orbital with 2 electrons per orbital as I recall. If has been a few years since I took chemistry.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/196026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why is it difficult to hear when speaking in the presence of strong wind? We say that sound waves require medium for their propagation, but during heavy storms or strong wind we cannot hear sound properly. Even if wind is flowing towards specific direction and we speak in that direction we cannot hear it well. Why is it so?
| Sound waves in air are a series of high and low pressure areas moving through the elastic medium of the air. If the medium is being compressed and rarefied by other areas of high and low pressure, the integrity of the sound wave may be broken and may become unrecognizable through the background noise of the medium's turbulence. Here is an explanation of the high and low pressure nature of sound waves in air: http://www.physicsclassroom.com/class/sound/Lesson-1/Sound-is-a-Pressure-Wave. You may be able to visualize what happens to pressure wave fronts if they become involved with other pressure waves, particularly those carried by strong wind, which generally is not a smooth laminar flow.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/196307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Is there some no-go theorem for $D=9$ Kaluza Klein QCD+EM? While QCD is a typical product of AdS/CFT and some other research trends in extra dimensions, I have never found in the literature an example producing the non-chiral part of the standard model, colour plus electromagnetism, or even color alone, from D=9 Kaluza Klein.
In principle such theory could be obtained
*
*by compactification on the 5-sphere $S^5$ and adding some Higgs to break $SO(6)$, or
*by compactification on the product manifold $CP^2 \times S^1$, producing directly the gauge group $SU(3) \times U(1)$.
Besides, QCD alone could be got from a $D=8$ theory on $CP^2$.
The traditional argument about the absence of chiral fermions do not apply here as both QCD and EM are defined with Dirac fermions. So if there is a non-go theorem at work, forbidding the scheme, must be a different one. Here the question: Is there one? Or, as a counter-proof of my question, is the example actually done in the literature and it only happens that I have not searched deep enough?
| While it is certainly possible to get an $SU(3) \times U(1)$ gauge group from the metric alone, if one started with a 9d theory, there are several issues with using Graviphotons as gauge bosons in a 4d theory.
Most prominently, in addition to vector bosons you will always create scalars in the adjoint from internal components of the metric, which we do not observe.
Furthermore, while the effective gauge group active at everyday energies is indeed a vector-like $SU(3)_\mathrm{c} \times U(1)_\mathrm{em}$, we do know for certain that there are effects that are due to a chiral $SU(2)_L \times U(1)_Y$ which is broken to $U(1)_\mathrm{em}$ via the Higgs mechanism. I'm not aware of someone reproducing spontaneous symmetry breaking on off-diagonal blocks of a metric (I think Witten tried some time in the 80s, but I don't have there reference at hand right now).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/196610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Is it true that spring has more force acting on it at its positive maximum amplitude than than at the negative one? Am I missing something?
It seems obvious to me that at $+A$ and $-A$, the spring has restorative forces equal in magnitude but opposite in direction.
But since gravity is always pulling it down, the spring at position $-A$ must have less net force acting on it. But my book says that at both positions, the spring has its maximum $\sum F = ma$. How does this make sense?
| The answer is no. The net force at the maximum elongation points has the same magnitude.
This is because the rest point of the spring is modified by the gravitational weight of the mass. The mass oscillates around this new rest point, and at the points of maximum amplitude the net force is the same. One can tell that the gravitational force
can be "integrated out".
To see this in more details, consider the net force on the mass, which in this case is the sum of the elastic force and the gravitational force (which is constant):
$$
F=mg- k (x-x_0)
$$
where $x_0$ is the rest position of the spring (without the mass).
Now you can calculate the rest position $x_0'$ of the spring including the effect of gravitation, which is defined as the point where $F=0$. So you obtain that at the position $x=x_0'$ you have
$$
0=mg- k (x_0'-x_0)\Rightarrow x_0'=x_0+mg/k
$$
Now, what happens if one changes coordinates?
$$
F=mg- k (x-x_0)=mg- k [x-(x_0'-mg/k)]=mg- k [x-x_0']-mg=-k(x-x_0')
$$
which means that, if one considers the displacement with respect to the new rest point $x_0'$, the force is simply given by
$$
F=-k(x-x_0')
$$
Now it is easy to see that at the two points of maximum elongation the net force is the same in magnitude, but opposite in sign.
Edit:
Consider the maximum elongation $A$, which is measured with respect to the new rest point $x_0'$. One can reverse the transformation I'v done and obtain that the net force at $x-x'_0=\pm A$ (i.e., $x=x'_0\pm A$) is
$$
F=mg- k (x'_0\pm A-x_0)=\mp k(x-x_0')
$$
Note that the net force is the same in magnitude, but the elongation of the spring with respect to the original rest point $x_0$ (without considering the mass $m$) is not the same but it is $(x'_0-x_0\pm A)$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/196718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.