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Is Newton's third law always correct? Newton's third law states that every force has an equal and opposite reaction. But this doesn't seem like the case in the following scenario:
For example, a person punches a wall and the wall breaks. The wall wasn't able to withstand the force, nor provide equal force in opposite direction to stop the punch.
If the force was indeed equal, wouldn't the punch not break the wall? I.e., like punching concrete, you'll just hurt your hand. Doesn't this mean Newton's third law is wrong in these cases?
| Newton's third law is not always correct, contrary to what you may have heard. It is correct in the context of newtonian mechanics, because we assume then that point particles are described only by their mass, and symmetry and conservation of momentum of the system imply that the third law must apply for the case of a closed system, which the universe is defined to be. It does not apply in the case of macroscopic bodies in general, because they deform themselves, as you described; energy is dissipated, hence the system is not closed. Also, it does not apply in the case of electromagnetism, because the field carries part of the momentum of the system.
| {
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What happens to atoms in extremely strong electromagnetic fields? I know that strong gravitational fields on the order of neutron stars (at the crust) atoms get compressed so tightly, the empty space between them is significantly reduced and it becomes denser. (http://www.newscientist.com/article/dn16948-star-crust-is-10-billion-times-stronger-than-steel.html#.U4aErvldWSo)
I know that in strong magnetic fields, the atoms polymerize, again becoming denser http://en.wikipedia.org/wiki/Neutron_star
So my question is what happens to atoms in super intense electromagnetic fields on the magnitude comparable to the g-fields and magnetic fields in neutron stars?
| Atoms are bound together by electrostatic forces. The energy is on the order of 10eV (e.g. 13.4eV for for an electron and proton in the ground state of a H atom), and the size is of the order of an Angstrom (Bohr radius is ~0.5 Å). Therefore when electric fields are greater than roughly
$$E\sim \frac{10\text{V}}{0.1\text{nm}}=10^{11}\,\text{V/m}$$
then the external field is greater than the binding force of the atom. When this happens you can treat the problem as independent particles in an external electric field (possibly with the electrostatic potential between particles as a perturbation).
If the the field is DC then you will just get constant acceleration of the particles (with the nucleus and electons accelerating in opposite directions). If the field is AC but below the plasmon resonance frequency the charges will oscillate at the field frequency. Above this frequency the field is oscillating too fast to couple to the particles (similar to x-rays passing through matter).
Strong pulses will ionize an atom as the positive nucleus and negative electrons will accelerate in opposite directions. However an AC field (that is linearly polarized) will cause the electrons and nuclei to seperate and then turn around and smash back into each other/recombine. When this happens the energy can be released as much higher energies and is called high harmonic generation which have also created attosecond pulses of light (which hold the record for shortest pulses of light created).
| {
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Why is the ratio of velocity to the speed of light squared in the Lorentz factor? Why is the ratio of velocity to the speed of light squared in the Lorentz factor?
$${\left( {{v \over c}} \right)^2}$$
My only guess is the value must be positive.
| I'll chime in with the hyperbolic geometry take:
The $\dfrac{v^2}{c^2}$ term in the Lorentz factor can be better understood if we look at the entire Lorentz factor:
$\gamma = \dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}$
Now we've got the term you asked about back into context. If we take a factor of $\frac{1}{c^2}$ out of the square root in the denominator we get
$\gamma = \dfrac{1}{\dfrac{1}{c}\sqrt{c^2-v^2}} = \dfrac{c}{\sqrt{c^2-v^2}}$
At this point, you might notice that we have something that looks an awful lot like a sine or a cosine definition. If only the two 'sides', v, and c in the square root were a sum instead of a difference, we'd be there. We're in luck though, because the definition of the hyperbolic cosine is exactly what we have: the adjacent side divided by the hyperbolic hypotenuse which is the square root of the difference, (not the sum) of the squares of the sides.
What's this tell us. Well first of all, $\gamma$ is known as the time dilation factor because it's the ratio of time experienced in the lab frame to time experienced by the particle, $\dfrac{dt}{d\tau}$. We can think of it as the speed of the moving particle through the time of the laboratory frame. If we write the sine function for the same triangle, we get
$\dfrac{v}{\sqrt{c^2 - v^2}}$, which is what we call the 'proper velocity', and denote by $\dfrac{dx}{d\tau}$. It indicates how quickly we move through the space of the laboratory frame with respect to the time of the moving frame.
So, to answer more concisely, the term you mentioned is squared because it's really the cloaked version of the square of a side of a velocity triangle in hyperbolic Minkowski space. Once we realize that, then we find out that the projection of our lab velocity $v$ onto the time velocity axis is our time dilation factor, and our projection onto the space velocity axis is our proper velocity. As a last note, proper velocity is often referred to as speedometer velocity, since it involves the moving frame measuring distance in the frame it started from, (think mile marker signs), and dividing by time in the moving frame, (the clock in the car). Proper velocity can exceed c. This is in with keeping wit special relativity, you just have to remember that your time's dilated, so you didn't exceed c in the laboratory frame.
| {
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What is the generating functional for a scalar theory with two different (interacting and real) fields? My question is specifically about how to use sources? For an interacting theory with one field, one puts a $J(x)\phi(x)$ term in the exponential in the path integral for $W[J]$. I now have two different fields ($\phi_1(x)$ and $\phi_2(x)$) and a number of interactions in the Lagrangian involving a combination of fields up to fourth order. To calculate the appropriate Green's functions and Feynman rules, should I use two different sources so that I have terms $J_1(x)\phi_1(x) + J_2(x)\phi_2(x)$?
Hope someone can help!
| Yes, what you want to do is to bring down $\psi_1(x)$ and/or $\psi_2(x)$ and you do that by multiplying them by independent currents and integrating over space-time, which then allows you to take the functional derivative w.r.t. either of the currents independently.
This is completely analogous to $\int J(x) \psi(x) d^4 x$, you now have a discrete index in addition to the continuous variable $x$.
| {
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Why Shock wave propagation is faster From The Blast Wave
A fraction of a second after a nuclear explosion, the heat from the fireball causes a high-pressure wave to develop and move outward producing the blast effect. The front of the blast wave, i.e., the shock front, travels rapidly away from the fireball, a moving wall of highly compressed air.
From Wikipedia
A shock wave is a type of propagating disturbance. Like an ordinary wave, it carries energy and can propagate through a medium (solid, liquid, gas or plasma) or in some cases in the absence of a material medium, through a field such as the electromagnetic field. Shock waves are characterized by an abrupt, nearly discontinuous change in the characteristics of the medium. Across a shock there is always an extremely rapid rise in pressure, temperature and density of the flow. A shock wave travels through most media at a higher speed than an ordinary wave.
How shock wave differs from ordinary wave, and how it can travel faster then ordinary wave in same medium. In first ex, In nuclear explosion why shock wave is traveling faster then fireball, when both are in same medium. Can someone please explain the phenomenon in detail.
| As per my answer here. The air on the inside of the explosion is also moving "faster then the speed of sound" so relative to that air, the shock wave is traveling at subsonic speeds. Thus the shock wave travels at a weighted average of the velocity that sound would travel in the mediums on either side of it.
| {
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Power fit to some experimental data I have to fit some data to a power law
$$ F=\alpha q^{\beta}$$
being $q$ and $F$ the experimental data points. What I usually do is taking logs so that
$$ \ln(F) = \beta \ln(q)+\ln(\alpha)$$
and apply least squares with uncertainties in both variables. How may I approach this in the case that some of the values of $q$ (and eventually $F$) are negative?. If it helps $\beta$ should be compatible with a value of $1$, but I want a general approach.
Edit:
The law to be fitted is Coulomb's law, $F$ is the force between to charged spheres. One of them has fixed charge, the other is a variable we call $q$. The proportionality constant $\alpha$ is related to $\varepsilon_0$ with a known expression in powers of $\beta=\frac{R}{d}$ where $d$ is the distance between the centers of the spheres and $R$ is their radius. So, I want to determine, from experiment, both $\alpha$ and $\beta$ and compare them with $\varepsilon_0$ and $1$.
The actual values are
$$
\begin{array}{|c|c|}
\hline
q / \mathrm{nC} & F / \mathrm{mN} \\ \hline
-26,8 \pm 0.8 & -1.5 \\ \hline
-18.8 \pm 0.5 & -1.1 \\ \hline
-9.9 \pm 0.2 & -0.6 \\ \hline
9.9 \pm 0.2 & 0.5 \\ \hline
18.8 \pm 0.5 & 1.0 \\ \hline
26.8 \pm 0.8 & 1.5 \\ \hline
35 \pm 2 & 2.1 \\ \hline
\end{array}
$$
with the uncertainty in the last decimal place for the forces.
| This sort of problem is reasonably straightforward to solve numerically using Gauss-Newton or similar algorithms and your favourite programming language. Matlab even has an entire curve fitting toolbox which does everything very nicely.
These non-linear least squares techniques should work for any (differentiable) function, so have quite a lot of applications.
| {
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Oscillations in forces other than the Weak As I understand it neutrino oscillations arise due to the neutrino mass eigenstates being distinct from the neutrino flavour eigenstates.
Flavour eigenstates are the states in which neutrinos interact via the weak force, and so are the eigenstates in which they are created and detected, and consist of a superposition of the the mass eigenstates. The mass eigenstates correspond to how the neutrino propagates through spacetime. The finite neutrino mass causes the relative phase of each mass eigenstate to change as the neutrino propagates, changing the probability that it will be detected in a given flavour eigenstate (hence an electron neutrino can oscillate into a muon neutrino and so on).
My question is, can other forces exhibit similar oscillations? For example, in principle is it possible for an electron to oscillate it's charge state as it travels, to be detected as a positron some distance later? Or are charge eigenstates not the similarly related to mass eigenstates?
| Only weak interactions can change the flavor of quarks and, therefore, produce mixing (oscillation) effects in mesons. Mixing always involves the weak interaction because it requires to change the flavor of the constituent quarks (strong and electromagnetic interactions can not change flavor).
Also, electric charge is conserved in all interactions and, therefore, mixing can only affect neutral particles.
| {
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Is it more efficient to stack two Peltier modules or to set them side by side? Is it more efficient to stack two Peltier modules or to set them side by side?
And why?
I have a small box that I want to cool down about 20 K below ambient -- cold, but not below freezing.
(I want to keep my camera cool, so I'm putting in this cool box.
The camera looks through a flat glass window on one side of the box).
The heatsink I have on hand is about twice as wide as the widest Peltier module I originally planned on using.
So there's room to put 2 Peltier modules side-by-side under the heatsink.
Or I could center a stack of 2 Peltier modules under the heatsink.
Which arrangement is more efficient?
I have to cut a bigger hole in the insulation for the side-by-side arrangement, so the unwanted heat that "back-flows" through the side-by-side arrangement is worse.
On the other hand, other effects are worse for the stacked arrangement.
(Is https://electronics.stackexchange.com/ a better place to post questions about Peltier coolers?)
| An efficiency of 10% does not mean you require 100W to produce 10W of cooling. This is complete nonsense. Efficiency is comparing the amount of heat shifted with the ideal Carnot cycle. In the ideal Carnot cycle 'coefficient of performance' (COP) values of over 5 are possible. That is why the efficiency of a TEM can be low but the COP actually quite reasonable, just no where as good as a vapour compression cycle. In most applications 100W of power supplied will shift about 50W of heat, give or take.
| {
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Quick question on sketching wavefunction in well
Usually for an infinite well, the sketch for n=3 level is this:
Now I think if one side of the potential barrier is higher, the particle will be more likely to spend time on the left side than the right side, so the wavefunction should have higher amplitudes on the left (skewed to the left):
| You are correct that for $n = 3$ there are $2$ non-boundary zero points.
Also, the modulus of $\psi(x)$ is lowest where $V(x)$ is lowest.
Where you are wrong is that $\psi(A)$ is not zero and $\psi(B)$ is not zero either, as you indicate in your schematic. For $x < A$ and $x > B$, $V(x)$ is not $\infty$ (your well is a finite, not infinite well) and in those regions the wave functions decay to $0$ exponentially ("tunnelling" effect).
See for example this simple case.
| {
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Where does the term "boost" come from for rotation-free transformations? I had never seen rotation free transformations called "boosts" (I think I have it right) before reading some questions here. I'm too old perhaps. I have not found the etymology after some searching, though it sounds like something V.I. Arnold would think up, or jargon from inertial navigation. Anyone know where/how it started or was popularized? (If it is in MTW or Ohanian (old edition) or Weinberg, I promise I'll facepalm)
| In Gravitation, Misner, Thorne and Wheeler first use the term boost in box 2.4 starting on page 67. They don't make a fuss of defining the term, so I assume it must have been in common use at the time of publication (1970).
| {
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Normal reaction - force without acceleration When a body lies on the surface of the Earth it is under the influence of gravity. The force on the body due to gravity causes it to exert a force on the ground and the normal reaction acts in the opposite direction causing the resultant force on the body to be zero.
However, how can the body exert a force on the ground when it does not have any acceleration? Since force equals mass times acceleration how does a body without acceleration experience a force?
|
When a body lies on the surface of the Earth it is under the influence of gravity. The force on the body due to gravity causes it to exert a force on the ground and the normal reaction acts in the opposite direction causing the resultant force on the body to be zero.
Correct
However, how can the body exert a force on the ground when it does not have any acceleration? Since force equals mass times acceleration how does a body without acceleration experience a force?
But you do have an acceleration, $g$, which you stated in the first paragraph (the force on the body due to gravity...). And this force is equal and opposite to the normal force:
$$
\mathbf F_{g} = -\mathbf F_N \\
mg\left(-\hat{z}\right) = -mg\hat{z}
$$
That the net force is zero only means that the object is not accelerating, not that there are no forces acting on it.
| {
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Dielectric boundary I am trying to determine why electric field may be confined to a certain region if there is a large difference in the permitivity for example if electric field flows through water and then reaches a water air boundary.
I have also been reading about EM waves, is it possible to model electric field as a wave because if so then the transmission T and reflection R coefficients given below in terms of n which is $\propto \epsilon_{r}^{1/2} $ and so if $\epsilon_{1}>>\epsilon_{2}$ then $n_{1}>>n_{2}$ ;
$\displaystyle R$ $\textstyle =$ $\displaystyle \left(\frac{n_1-n_2}{n_1+n_2}\right)^2, \rightarrow 1$
$\displaystyle T$ $\textstyle =$ $\displaystyle \frac{n_2}{n_1}\left(\frac{2 n_1}{n_1+n_2}\right)^2 \rightarrow 0.$
and so it is clear that the wave is reflected at the boundary, is this approach valid ?
http://farside.ph.utexas.edu/teaching/em/lectures/node103.html
| What you're saying is mostly correct, but your language is a bit imprecise, which makes me think your understanding is a bit incorrect.
To be clear: there are both static electric fields, as well as time varying electric fields, which occur together with magnetic fields electromagnetic (EM) waves. The equations you're using describe reflection and transmission of an EM wave at a boundary between two different media (e.g. air and water), but only at normal incidence; if the wave hits the interface at an angle, these formula fail. Based on these equations, some of the energy will transmit, while some will reflect. A larger different in index of refraction will result in greater reflection at the interface.
If you're trying to describe static fields, then these formula are completely irrelevant. You can't model static fields as waves.
| {
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What will happen after escaping earth's gravitational field? Suppose that I escaped the gravitational field of earth.
Then: am I going to be pulled by Sun's gravity?
| Yes, you will be pulled by the sun's gravity. However, that has nothing to do with whether you have "escaped" earth's gravitational field or not. That is a non-sensical concept because the gravitational fields of objects don't have a hard distance limit where they suddenly go to zero. Once you get far enough from a object, the gravitational pull from that object will fall off with the square of the distance.
Therefore, no matter where you are or how far you go, you will always be within the gravitational field of the earth, the sun, and any other piece of matter. The only issues is that if you go far enough, the forces from those objects will be so small that you can usefully approximate them as zero. However, that threshold varies by application, so again there is no hard distance where you can even pretend the earth's gravitational field is irrelevant.
| {
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A strange audio phenomenon, could there be a physical interpretation to it? https://mathoverflow.net/q/165038/14414
Motivation : Here is a motivation as to why this problem is so important.
Let $f(t)$ be an audio signal. We can safely asume it to be bandlimited to 0-20kHz as we cannot hear anything above that. Capture this signal in digital computer with appropriate sampling frequency and denote it as $f[n]$.
Now take Discrete Hilbert transform of $f[n]$ to get $f_h[n]$, (using the code $f_h$ = imag(hilbert(f)); in Matlab).
Compute the signal $f_{\theta}[n] = f[n]\cos\theta + f_h[n]\sin\theta$ for any value of $\theta$, then listen to the signal with different values for $\theta$.
They all sound exactly identical.
Similarly our $MI_{\omega_0,\omega_1}(t)$ is same for all $f_{\theta} = f\cos\theta + f_h\sin\theta$, for any value of $\theta$.
Question :
just try it. $<f,f_h> = 0$, they why do they produce same effect in the listner? Is it some quantum mechanical effect gone wrong?
Added :
Also see this metric space : metric space
I've recently filed a patent using this metric with a slight change, instead of arccos i used sqrt(2(1-cos(theta))), which makes it a Hilbertian metric. I had then embedded this metric space into an Hilbert space isometrically, to model using vectors.
MATLAB code :
[f,fs] = wavread('audio_file.wav');
fh = imag(hilbert(f));
theta = pi/4;
f_tht = fcos(theta) + fhsin(theta);
wavplay(f,fs);
wavplay(f_tht,fs);
| To your Question:
*
*There is no quantum mechanics involved. This is essentially a signal processing question, which is rooted in calculus.
*Why does it sound the same?
The ear works essentially as a power spectrum analyzer, i.e. what you hear of a signal $f(t)$ is mainly determined by the powerspectrum $|{F(\omega)}|^2$, where ${F(\omega)}$ is the Fourier-transform of $f(t)$.
In your case:
$F_{\theta}(\omega) = (\cos\theta) F(\omega) + (\sin\theta) F_h(\omega)$.
so that:
$|F_{\theta}(\omega)|^2 = (\cos\theta)^2 |F(\omega)|^2 + (\sin\theta)^2 |F_h(\omega)|^2 + K$.
Where $K \propto F(\omega)^* F_h(\omega) + F(\omega) F_h(\omega)^*$, and $*$ denotes the complex conjugtate.
Using the relation between $F$ and $F_h$ given by the Hilbert-Transform we find that $K=0$ and $|F(\omega)|^2 = |F_h(\omega)|^2 $.
We conclude that
$|F_{\theta}(\omega)|^2 = ((\cos\theta)^2 + (\sin\theta)^2) |F(\omega)|^2 = |F(\omega)|^2 $.
(Where in the last step there is a trigonometric identity)
Summary:
for all $\theta$ we find that $|F_{\theta}(\omega)|^2 = |F(\omega)|^2 $, so the ear (as a powerspectrum analyzer) hears the same.
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Why doesn't a block rotate due to friction? In a horizontal surface, a block (cube) is sliding due to a sudden push. When the block slides, there is frictional force which is acting on the block.
Frictional force will have a torque around the center of mass, so why does the block not rotate/roll around (a horizontal axis through) the center of mass instead of sliding?
| First consider the initial push. If the friction is high enough and the push is high enough on the block it will roll instead of slide. Generally static friction is greater than dynamic friction, so if it starts sliding it will continue.
If the friction gets great enough, the block can roll. Say you have a box sliding on ice that comes to the edge of the ice and starts sliding on dirt. The friction will increase considerably. You compare the torque the friction applies to the box against the torque required to lift the back edge against gravity. If it is greater, the box will start to roll.
| {
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Moment of inertia of a cylinder When I tried to calculate the moment of inertia ($I_C$) of a cylinder (mass M, height H, radius R) around the rotating axis going symmetrically through its middle, I came up with a different result than expected ($\frac{1}{2}MR^2$), but I do not spot my mistake, since my calculation makes perfect sense to me:
$$
I_C := \int_V{ρr^2dV} = \int_0^H{\int_0^{2π}{\int_0^R{ρr^2 dr dφ dh}}} = ρ \int_0^H{\int_0^{2π}{\frac{R^3}{3} dφ dh}} = ρ \cdot 2πH \frac{R^3}{3} = Vρ\frac{R^2}{3} = \frac{1}{3}MR^2
$$
Can anyone spot what's wrong?
| You made two mistakes:
*
*$dV = dr d\phi dh$ is wrong.
*
*$dV = r dr d\phi dh$.
*$V = 2πR H$ is wrong.
*
*$V = πR^2 H$.
$$
I_C := \int_V{ρr^2dV} = \int_0^H{\int_0^{2π}{\int_0^R{ρr^3 dr dφ dh}}} = ρ \int_0^H{\int_0^{2π}{\frac{R^4}{4} dφ dh}} = ρ \cdot 2πH \frac{R^4}{4} = Vρ\frac{R^2}{2} = \frac{1}{2}MR^2
$$
| {
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Why is the movement of electrons random in electric field? Suppose an uncharged conductor is present isolated and there is no net electric field applied. Then by assumption we have the motion of electrons random,i.e. isotropic.
But now suppose there is a closed conductor present. We apply a net electric field and the electrons starts moving. Suppose steady state current is achieved.
We notice that there is has been created a velocity of electrons on the opposite direction of the applied field.
I know that there is random movement of the electrons and they perform Brownian motion.
My question is that is it biased in one direction because of the field?
Is the motion of elections in a particular direction more probable or
they have more velocity in that direction?
|
Is the motion of elections in a particular direction more probable or they have more velocity in that direction?
Both are actually true in the Drude model. In presence of a field, the probability for the electron to move in the direction opposite to the field is enhanced. As a result, their Brownian motion is drifted, which results in an average velocity in the direction of the drift. However, the distribution of velocities will keep the same width because the Fokker-Planck equation is now
$$\frac{\partial P}{\partial t}=-\langle\vec v\rangle\cdot\vec\nabla P+D\nabla^2P \tag 1$$
where $\langle\vec v\rangle$ is the average velocity resulting from the force.
If the solution in absence of the field is $P(\vec r,\,t)$, then the solution of (1) is simply $P(\vec r-\langle\vec v\rangle t,\,t)$. This means that it corresponds to the zero-field solution in a reference frame performing a uniform translation with speed $\langle\vec v\rangle$.
| {
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Moon's pull causes tides on far side of Earth: why? I have always wondered and once I even got it, but then completely forgot. I understand that gravity causes high and low tides in oceans, but why does it occur on the other side of Earth?
| The Earth is free falling towards the Moon. Because gravity decays with distance, the side near the moon wants to fall faster than the center of the Earth, while the other side tends to fall slower. So observed on the Earth, the other side "lags behind" and therefore we have high tide there.
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Placing two similarly charged particles in space Now, I will make a hypothetical situation. Assume that we place two similarly charged particles (lets take electrons) in space. Imagine that there is no other force acting on the particles except the repulsive force and the gravitational force of the particles. In other words, only these two electrons are present in the universe. So they are free from any outside interference. Now by nature, these electrons will start moving away from each other due to the repulsive force. Since there is nothing to stop them (gravitational force will only slow them down and not stop them as it is of a lesser magnitude than the repulsive force) they will keep moving and never stop. Over here we exclude expansion of space also for no complications. Now since the particles will keep moving as there is a constant repulsive force acting on them, they will do infinite work because $Work = Force * displacement$ and the displacement over here will keep increasing. Please tell me what is the problem in my thought experiment because it violates conservation of energy.
| The problem is not so simple. In fact I think the final kinetic energy of the electrons is not equal to the initial potential energy. But this not means the energy conservation is violated. A moving charge that accelerate emits electromagnetic waves. This waves transport energy so the electron slows down while moving to infinity. The final total energy is:
$E_{final} = E_{waves} = \frac{e^2}{4 \pi \epsilon_0 x_0} = E_{initial}$
So there is no kinetic energy at infinity since, due to the self force (http://en.wikipedia.org/wiki/Abraham%E2%80%93Lorentz_force), the electron stops after some time.
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Irreversible heat engines strictly less efficient than reversible ones I understand how Carnot's theorem implies that irreversible heat engines must be no more efficient than reversible one's, but it is less clear why they need to be less efficient, as I have seen stated in some places.
If they could be equally efficient then an irreversible engine engine could be used to drive a reversible engine operating between the same heat reservoirs,
without any net energy transfer between the reservoirs. It would then be unclear what is irreversible about the irreversible engine. Does that constitute an actual contradiction though? If so can the argument be stated more tightly? It feels a little sloppy as is.
It could also be a question of how an irreversible engine is defined in Carnot's theorem. I understood it to mean one that cannot be run in reverse as a heat pump, which could presumably include a Carnot engine with a one way ratchet attached. If it really means an engine whose thermodynamic effects can not be undone, the implication would be trivial.
|
I understand how Carnot's theorem implies that irreversible heat engines must be no more efficient than reversible one's, but it is less clear why they need to be less efficient, as I have seen stated in some places.
The answer is that it is always possible to construct an incredibly inefficient engine, which is to say, it is always possible to obtain a lower efficiency than the reversible engine. This, coupled with the fact that it is very difficult to construct a truly reversible engine, places real engines at efficiencies below that of the Carnot limit.
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Entropy was created after inflation? I'm puzzeled by a statement in Big Bang Cosmology-review about the reheating phase subsequent to the exponential expansion during inflation:
In this reheating process, entropy has been created and the final value of $RT$ is greater than the initial value of $RT$.
(Taken from section 21.3.5. on page 17.)
How can I reconcile this with the first law of thermodynamics...
| There is nothing wrong with that statement (assuming that the meaning is: prior to inflation we have a total amount X of entropy, then after the inflation we have a lot more than X).
After the inflation, the scalar field (inflaton) is in the minimum of the potential well and is a super-cooled Bose-Einstein condensate whose constituents are very massive scalar bosons. Such system of very cold spin 0 bosons is unstable and is transformed by the decay process into energy of ultrarelativistic species, so that the universe undergoes a strong reheating phase.
The enormous increase in entropy is due to these decays. You can show that (the exact calculation depends on the model you choose) in a given comoving volume $ V $ the increase is something like:
\begin{equation}
S_{post \; Reh} \simeq e^{3 \mathcal{E}} S_{pre \; Inf}
\end{equation}
Usually $ \mathcal{E}\sim 60 $ and this means $ e^{180} \sim 10^{78} $. This result is certainly in agreement with the second law of thermodynamics! There are no contradiction with the first law (anyway you must pay attention to carefully define the quantity you want to use, the entire universe it's not a trivial thermodynamic system)
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Universal central charge in higher dimensional AdS/CFT? In the $AdS_3/CFT_2$ correspondence, the central charge of the dual CFT2 is universally given by
$$
c = \frac{3\ell}{2G}
$$
This is independent of the matter in the bulk of AdS3. Is it also universal in the higher dimensional analogues $AdS_d/CFT_{d-1}$ or does it depend on details of the matter in the bulk, etc.?
In particular, what is the central charge for $d=4$?
| The expression you've written derives from a proposal, due to Ryu and Takayanagi (See: http://arxiv.org/abs/hep-th/0603001), to calculate entanglement entropy of a CFT by making use of a gravity dual.
Define a CFT in $D$ space-time dimensions. Take a spatial slice, a ball of radius $R$ for example, and calculate the entanglement entropy associated with it. You'll get the following behaviours:
$S_{D=2}(R)=\frac{c}{3}\log \frac{R}{\epsilon}$
$S_{D=3}(R)=k_1 \frac{R}{\epsilon} -F$
$S_{D=4}(R)=k_2 \frac{R^2}{\epsilon^2} -a \log \frac{R}{\epsilon}$
where $c$, $F$, $a$ are universal terms. For the case of $D=2$ it's the central charge.
Now if this CFT admits a holographic dual then there's another expression for the entanglement entropy:
$S(R) = \frac{A_{\gamma(R)}}{4G_N}$
where $G_N$ is Newton's constant and $A_{\gamma(R)}$ is the area of a minimal surface $\gamma(R)$ in $AdS_{D+1}$ whose boundary is exactly the boundary of the spatial region you're computing $S(R)$ for. Like this picture:
The intuition for why this is true is that tracing out a region is like inducing a horizon beyond which you don't know what's going on. Then this holographic expression is like the Bekenstein-Hawking formula.
Matching expressions can give you a formula like you desire. An important thing to note though is that this $a$ for the case of $D=4$ isn't always the "$c$" central charge. (See my comment for a bit more on this) They differ in cases where your CFT is dual to a higher curvature gravity theory.(http://arxiv.org/abs/hep-th/9904179) This is related to its appearance in the trace anomaly.
This $a$ quantity does fulfill a role in generalizing the $c$-theorem of $D=2$ CFT to higher dimensions though; it's an RG monotone. This line of inquiry has been pursued by Rob Myers and Aninda Sinha. (See: http://arxiv.org/abs/1011.5819 and http://arxiv.org/abs/1006.1263)
EDIT:
As promised this will provide a formula for a 'central charge', or combination of them, in any number of dimensions which is just a function of constants and the AdS radius $R$.
For $D=4$ this yields: $a=\frac{\pi^{3/2}}{4 G^{(5)}_N} \frac{R^3}{\Gamma(\frac{3}{2})}$
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Energy of an EM Wave and its temperature and amplitude I'm trying to understand why classical physics fails to explain black body radiation.
I'm confused.
According to Boltzmann, energy calculation for em wave is based on temperature.
According to Maxwell, energy calculation for em wave is based on amplitude.
Are those different kinds of energies? How can we determine the energy of an em wave just taking temperature as a parameter, but not amplitude or frequency??
| M. Planck started to explain black-body radiation by the relation predicted with classical statistical mechanics (with continous energy equi-partition to every degree of freedom of the system). This was Wien's Law which indeed was accurate for high frequencies but divergent for low frequencies.
Then Planck decided to account for a quantized energy (or treat as a series of harmonic oscillators), which by the then statistical arguments was not justified, and managed to find a relation (Planck's Law) which correctly described the black-body radiation and was in accord to the original Wien Law in the area the latter was correct.
Now the whole point is why did the then statistical mechanics thought this was not a justifiable argument (in fact if one takes into account that a finitely bounded system can have such a propety the rest follows, since it is a physical argument of bounded energy of a bounded system). This was a start of what was to become Quantum Mechanics.
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Why can the entropy of an isolated system increase? From the second law of thermodynamics:
The second law of thermodynamics states that the entropy of an
isolated system never decreases, because isolated systems always
evolve toward thermodynamic equilibrium, a state with maximum entropy.
Now I understand why the entropy can't decrease, but I fail to understand why the entropy tends to increase as the system reach the thermodynamic equilibrium. Since an isolated system can't exchange work and heat with the external environment, and the entropy of a system is the difference of heat divided for the temperature, since the total heat of a system will always be the same for it doesn't receive heat from the external environment, it's natural for me to think that difference of entropy for an isolated system is always zero. Could someone explain me why I am wrong?
PS: There are many questions with a similar title, but they're not asking the same thing.
| We know that $ds_{\rm (universe)}$is equal to $ds_{\rm(system)} + ds_{\rm (surroundings)}$,and for an isolated system $ds_{\rm (surroundings)} = 0$ because $dq_{\rm (reversible)} = 0$; therefore, for an isolated system, $ds_{\rm (universe)}$ is equal to $ds_{\rm (system)}$.
Now, we know that the spontaneity criteria for any process is
$ds_{\rm (universe)} > 0$, or if not, at least should be $0$ for equilibrium.
Therefore, $ds_{\rm (system)} \geq 0$.
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Placing a lens in front of ultraviolet femtosecond laser with 10e38 J in pulse makes it a gun that fires microscopic black holes at speed of light? I calculated that in the focal spot of such a laser the critical energy density would be met.
Will these black holes really move at the speed of light, just as the the photons that made them? The conservation of momentum seems to give the speed of light here.
But if the black hole is a "matter-like" massive thing, how can it move at the speed of light?
| As Carl mentioned, such a contraption would be hard to build. However, I think it might work given a huge energy source and a HUGE lens.
The black hole would not move at light speed, though. Since you focused the light with a lens, the momenta of photons are not aligned. So the magnitude of the total momentum is smaller than the sum of magnitudes.
Also, the standard model and general relativity would probably break down at such energy densities. No one knows what would happen with a small black hole. It would probably decay instantly due to Hawking radiation.
A simpler example of light turning into matter is the Schwinger effect. Apparently, in electric fields larger than $E_{critical} \approx 10^{16} V/cm$ photons will collide with each other and produce an electron/positron pair and those do not move with the speed of light.
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What is the concept of cosmic strings? What is the concept of cosmic strings? Is it related to the strings in the string theory, and if it is, then how?
| In a paper by R. Gregory, Effective Action and Motion of a Cosmic String, the concept is explained well:
In high energy physics, a defect will generically occur during a symmetry breaking process where different parts of a medium choose different vacuum energy configurations, and the non-compatibility of these different vacua forces a sheet, line or point of energy where these non-compatible vacua meet... A defect may be topological, in that it is the topology of the vacuum that simultaneously allows formation, and prevents dissipation...
The fourth order effective action of the string is given by,
$$S=-\mu\int d^2\sigma \sqrt{-\gamma} \left[ 1-r_s^2 \frac{\alpha_1}{\mu}\mathcal{R}+r^4_s \left( \frac{\alpha_2}{\mu}\mathcal{R}^2 + \frac{\alpha_3}{\mu}K_{i\mu\nu}\mathcal{K}^{\mu\nu}_j \mathcal{K}_{i\lambda \rho} \mathcal{K}^{\lambda\rho}_j \right)\right]$$
where $\alpha_i$ are numerical coefficients, $\mu$ the tension of the string, and all geometric quantities evaluated with respect to the worldsheet of the string, which is the manifold traced out by the motion of the string. The worldsheet itself is within the target space or 'spacetime.' The $\mathcal{K}$ terms are the extrinsic curvature of the worldsheet. As R. Gregory states, a priori, it is difficult to determine whether these affect the rigidity of the string.
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Why is the constant velocity model used in a projectile motion derivation? I was re-studying university physics last week, I'm now in the chapter about kinematics in 2 dimensions and specifically the one treating projectile motion. In page 86 of his book (Serway - Physics for scientists and engineers) he derives the equation of the range of the projectile motion to be: $$R=\frac{{v_i}^2\sin2\theta_i}{g}$$
But I don't know why he used one of his assumptions
$\color{red}{\bf Question1:}$ Why $v_{xi}=x_{x\rlap\bigcirc B}$? Where $\rlap\bigcirc {\,\sf B}$ is the time when the projectile stops.
$\color{darkorange}{\bf Question2:}$ Why did he use the particle under constant velocity model to derive that formula, whereas here we deal with a projectile under constant acceleration?
Any responses are welcome, I'm disappointed a lot about those matters!
| "Why did he use the particle under constant velocity model to derive that formula, whereas here we deal with a projectile under constant acceleration?" You appear to have the classic confusion between the vertical component of the motion and the horizontal component. Vertically you have to include the acceleration due to gravity in your thoughts. Horizontally, strictly speaking, you should include a decelaration arising from atmospheric friction (at least where there is an atmosphere of some sort to consider). However it's much easier, and simplifies the problem enormously, to assume that you can ignore that deceleration. Then, all you have to deal with in the horizontal direction is speed = distance / time. I read however, that in the very practical application of these ideas to gunnery, the air friction is a significant modification to the simple theory. A shell will have a much shorter range in reality compared to its range computed using 'simple' projectile theory to predict where it will fall.
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Why is the speed of light arbitrarily the limit? I know Einstein was great and all. Why is it that exactly at the speed of light is where infinite energy is required to accelerate any object with mass? Is it simply because the math of relativity checks out and explains most of everything? Are there any physicists who disagree with Einstein's theory?
|
Why is it that exactly at the speed of light is where infinite energy
is required to accelerate any object with mass? Is it simply because
the math of relativity checks out and explains most of everything?
To say that the math checks out is the wrong way of putting what's going on because it separates the math and the explanation for the math. Einstein had explanations for why he picked the equations he picked. That is, he had an account of what was happening in reality to bring those outcomes about and without that explanation the math wouldn't have any relevance to physics. Einstein's explanation involved ideas about what kinds of measurements are allowed by the laws of physics, see:
http://www.fourmilab.ch/etexts/einstein/specrel/specrel.pdf
Are there any physicists who disagree with Einstein's theory?
General relativity is the best available explanation of gravity and the structure of spacetime. There is a problem with the theory in the sense that it is not a quantum theory of gravity. There is no reason to think that its replacement will involve going back to anything that resembles Newtonian mechanics.
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Why is charge not taken as a fundamental unit? According to the definition of electric current, it appears to be a derived quantity. Charge on the other hand seems more fundamental than electric current. Then why is current taken as fundamental quantity instead of charge?
Is it arbitrary choice? Is it because we can measure current more efficiently than charge or some other reason?
| I think that the question is why the SI system of units considers one ampere, the unit of current, to be the elementary one, rather than the unit of the electric charge.
Recall that one ampere is defined in SI as
"the constant current that will produce an attractive force of $2\times 10^{–7}$ newton per metre of length between two straight, parallel conductors of infinite length and negligible circular cross section placed one metre apart in a vacuum"
Note that this definition relies on magnetic forces; it is equivalent to saying that the vacuum permeability
$$\mu_0=4\pi\times 10^{-7} {\text{V s/(A m)}} $$
It's the magnetic force that has a "simple numerical value" in the SI system of units, and magnetic forces don't exist between static electric charges, just between currents.
If we tried to give a similar definition for the electric charge, using the electrostatic force, the numerical values would be very different.
Now, one may ask why the magnetic forces were chosen to have "simple values" in the SI system. It is a complete historical coincidence. The SI system was designed, up to the rationalized additions of $4\pi$ and different powers of ten, as the successor of CGSM, the magnetic variation of Gauss' centimeter-gram-second (CGS) system of units.
These days, both methods would be equally valid because we use units in which the speed of light in the vacuum is fixed to be a known constant, $299,792,458\,{\rm m/s}$, so both $\mu_0$ and $\epsilon_0=1/(\mu_0 c^2)$, the vacuum permittivity, are equal to known numerical constants, anyway.
At any rate, the unit of the electric charge is simply "coulomb" which is "ampere times second", so it is as accurately defined as one ampere.
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Why would we need to ground an AC source I'm new to this field hence this weird question. Why would we need to ground an AC source? Why wouldn't it be enough to have just one pole to get an AC current going? I understand why it wouldn't work in DC case where current is flowing in one direction. However, in case of AC source where the current is not flowing anywhere but rather just oscillating back and forth it's not that clear to me why connecting load to only one pole wouldn't work? Thanks.
| You need a return wire to complete the circuit. Otherwise the electrons would all try to pile up at the end and the current doesn't flow. In $60$ Hz circuits, the current flows in one direction for $1/60$ second. That doesn't mean you need a ground wire-AC works fine isolated from ground.
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Bragg diffraction and lattice planes Crystalline substances show, for certain sharply defined wavelength and incident directions, very sharp peaks of scattered X-ray radiation.
From the illustration below we see that we get constructive interference when the path-length difference is a multiple of the wavelength $\lambda$.
In real crystalline materials we have a large amount of closely packed lattice planes. This large amount accounts for the sharp peaks for certain $\theta$. I do not understand how this follows from the Bragg reflection formula $$ n\lambda = 2d \sin \theta , $$ since $d$ is not constant anymore. I understand the model for two lattice planes as in the illustration.
Is it true that $d$ can only take on values of the seperation of lattice planes, so $d$ is defined to be the seperation of points in the reciprocal lattice, or in others words, is $d$ constrained to be the absolute values of some reciprocal lattice vector?
How does the Bragg condition account for very sharp peaks when we let $d$ run through all such absolute values?
| The d in that formula is the distance between lattice planes, not points. Basically, one ray (the top one) bounces off the top "plane". The other ray (the bottom one) passes the first plane and "bounces" off the second plane. It then comes out and meets the first ray at some far away point.
Question: Since they travelled different distances, they have different phases. In other words, they will interfere. For this to happen, the "extra" distance the bottom ray traveled must be equal to a multiple of $ \lambda $ so that constructive interference occurs. What is the formula?
You can use some simple trig to show that the extra distance traveled by the bottom ray is $ 2dsin(\theta) $, and you want this to be a multiple of $ \lambda$, the wavelength of the light.
So: $ 2dsin(\theta) = n\lambda$ is the condition for constructive interference.
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Why doesn't time change in the non-relativistic limit of Lorentz transformations? A simple boost in the $x$ direction is given by:
$$ \Lambda = \begin{pmatrix}
\cosh(\rho) & \sinh(\rho) & 0 & 0 \\
\sinh(\rho) & \cosh(\rho) & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
\end{pmatrix} $$
Which get linearized to the following transformation:
$$
x^0 \mapsto x^0
, \quad
x^1 \mapsto x^1 + \frac vc x^0
$$
How come the zeroth component is not linearized to $x^0 \mapsto x^0 + \frac vc x^1$? Is that because there is another factor $c$ in the time components? Since $x^0 = ct$, that would mean the time is transformed like
$$ t \mapsto t + \frac v{c^2} x,$$
and $c^{-2}$ is just so small that is ignored?
Or is it just to fit the Galilei transformation?
| The Lorentz boost has two different low-velocity limits: the Galilean transformation appropriate for transforming ultra-timelike four-vectors, which is usually what we're interested in if we want to recover low-velocity kinematics, and the whimsically named "Carroll transformation" appropriate for transforming ultra-spacelike four-vectors. Your proposed time transformation is part of the latter.
Intuitively, the effect of the Lorentz transformation is a rotation along curves of constant spacetime intervals:
Image from here.
If we zoom in on the upper vertices of the hyperbola, the low-velocity Lorentz boosts of timelike vectors with $ct\gg|\mathbf{x}|$ are approximated by:
$$\begin{eqnarray*}ct' \approx ct\text{,}&\quad&\mathbf{x}' \approx \mathbf{x} - \mathbf{v}t\text{.}\end{eqnarray*}$$
Treating this low-velocity approximation as a transformation in its own right, we get the Galilean boost:
$$\begin{eqnarray*}ct' = ct\text{,}&\quad&\mathbf{x}' = \mathbf{x} - \mathbf{v}t\text{.}\end{eqnarray*}$$
This is sensible, because the tangent lines to the hyperbolas on ultra-timelike vectors are horizontal, so a low-velocity boost should not change the time coordinate by an appreciable amount.
If instead we zoom in on the right vertices, and repeat the above procedure on ultra-spacelike ($ct\ll|\mathbf{x}|$) vectors, we get the Carroll transformation:
$$\begin{eqnarray*}ct' = ct - (\mathbf{v}\cdot\mathbf{x})/c^2\text{,}&\quad&\mathbf{x}' = \mathbf{x}\text{.}\end{eqnarray*}$$
Note that this reflects the fact that a low-velocity boost of an ultra-spacelike vector does not change the spatial coordinates appreciably, as the tangent lines to the hyperbolas are vertical there.
On a purely formal level, they are equally valid low-velocity limits. The Galilean transformation is much more physically significant because material objects should follow timelike vectors, not spacelike ones, so we can interpret it as a possible transformation between inertial frames formed by ideal clocks and rulers (or some other means). In contrast, the Carroll transformation does not allow a sensible interpretation as a transformation between physical inertial frames.
“My dear, here we must run as fast as we can, just to stay in place. And if you wish to go anywhere you must run twice as fast as that.” ― Lewis Carroll, Alice in Wonderland
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How much electricity could be generated by cloths made of thermocouples? Lets say that we made a glove, shirt, pants and a hat out of the most effective thermocouple material available today. How much electricity would be generated by each, by a healthy person, on a cool day (98 degrees F vrs 79 degrees F?)
| As a starting point, figure out what the limit is due to the Carnot efficiency. The Carnot efficiency is the maximum possible work that can be extracted from heat flow. It is Tdiff / Thot.
You have Thot = 98°F = 310°K, and Tcold = 79°F = 299°K, so Tdiff = 11°K.
11°K / 310°K = 3.5%
That's the maximum theoretical portion of the heat flow that can be converted to work, given your parameters. In the case of a human giving off 30 W of heat, that leaves 1.1 W maximum theoretical that can be converted to electric from. Real heat engines, like thermopiles, of course have their own real inefficiencies. Due to the engineering difficulties, you'd be lucky to get a few 10s of mW.
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How is sweating a pipe an example of capillary action? I learned how to sweat a pipe today from my father. If you're not familiar with the process, this might help.
One thing that jumps out at me is this line (from the above link, as well as my father's explanation)
Solder, which melts at low temperatures, wicks into a joint by capillary action and bonds with copper at the molecular level.
This doesn't seem to be quite right to me. I always thought of capillary action as something like putting a piece of paper vertically into a dye and watching the dye rise up the paper. I also didn't think the solder bonded with the copper at the molecular level, just that it melted and filled in the gaps really well. Wikipedia seems to agree with my definition of capillary action
Capillary action ... is the ability of a liquid to flow in narrow spaces without the assistance of, and in opposition to, external forces like gravity.
When sweating a pipe you place the solder above the pipe, letting it drip down (ie with the assistance of gravity) which makes it not capillary action... right?
Is there something I'm just not understanding here?
| Molten solder has a low contact angle on (clean) copper. So if you looked at a cross section of the pipe joint as the solder was flowing in you'd see something like:
The solder is drawn into the joint in exactly the same way as water rises in a capillary tube. Both are correctly described as capillary action.
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"source": "stackexchange",
"question_score": "3",
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How exactly to show that s-matrix elements diverges because time-ordering is not well determined? Let's have s-matrix:
$$
S_{\alpha \beta} = \langle \alpha | \hat {S} | \beta \rangle ,
$$
$$\hat{S} = \hat{T}e^{-i\int \hat{L}(x)d^{4}x}, \quad \hat{T} \left( \hat{\Psi}(t) \hat{\Psi}(t') \right) = \theta (t - t')\hat{\Psi}(t)\hat{\Psi}(t') \pm \theta (t' - t)\hat{\Psi}(t')\hat{\Psi}(t).
$$
How to show that matrix elements diverges because time ordering is not well-defined operation when $t=t'$?
| Mathematically one way to see it is that the (combined) step functions become Dirac pulses i.e $\theta(t) \to \delta(t)$, which diverge. When $t \to t'$, the fields have nearly identical values, thus the time-ordered product involving step functions degenerates into dirac pulse (a dirac pulse is the derivative of the step function)
Physically another way to see this is: S-matrix describes interactions between states and particles, as such a matrix element for $t \to t'$, requires an interaction to take place instanteneously, thus it would require infinite energy at that point (also related to time-energy unceratinty).
| {
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Why $e$ in the formula for air density? I am reading a book that says that the density of air is approximately $D = 1.25 e^{(-0.0001h)}$, where h is the height in meters. Why is Euler's number $e$ used here? Was a differential equation used in deriving this formula?
| It's actually a surprisingly straightforward differential equation. If you assume that the acceleration due to gravity $g$ doesn't change with altitude (a good approximation if the atmosphere is thin compared to the radius of the earth), Bernoulli's relation tells you the change in the pressure $P$ with height $h$:
$$ \frac{dP}{dh} = -\rho g$$
Meanwhile the pressure and the density are also related by the ideal gas law
$$ PV = NRT $$
or
$$ P = \rho \frac{RT}{M} $$
where $M$ is the mass of one mole of the gas. If you're willing to neglect the changes in temperature $T$ and mean molar mass $M$, you can differentiate with respect to height and find
\begin{align}
\frac{dP}{dh} = \frac{d\rho}{dh} \frac{RT}M &= -\rho g \\
\frac{d\rho}{dh} &= -\rho \frac{gM}{RT} = -\frac{\rho}{h_0}
\end{align}
This is the classic differential equation for an exponential.
If I use nice round numbers $R=8\,\mathrm{\frac{J}{mol\cdot K}}$, $T=300\,\mathrm K$, $M=30\,\mathrm{g/mol}$, $g=10\,\mathrm{m/s^2}$, I get a scale height of 8000 meters, different from your textbook's approximation of $10^4$ meters by about 20%.
| {
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Why is the outside run of high presure line on a ductless mini-split airconditioner insulated? On every ductless mini-split air conditioner I've ever seen, both the high and low pressure lines are insulated between the compressor and the building.
It seems like the liquid refrigerant coming out of the condensing coil can never be cooler than the outdoor ambient air temperature because the outdoor air is what is cooling it in the condenser coil.
It can, however, be warmer than the outdoor ambient temp. Therefore, it seems like leaving that high pressure line coming out of the compressor uninsulated would at worst save costs for some insulation, and at best give the high pressure refrigerant some additional cooling before it gets back to the indoor unit.
What am I missing here?
NOTE: I can think of some reasons why you'd want to insulate that line on the inside of the building where the high pressure liquid refrigerant would be warmer than the ambient indoor air temp.
| Many AC units can be run in reverse as a heat pump. When that is the case, the high and low pressure sides are swapped.
| {
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Fermion as a mixture of particle and antiparticle The solution to the Dirac equation (in the Dirac basis) are 4 coupled fields. The first 2 of them represent a particle (spin up/down), the other 2 fields are the antiparticle (spin up/down). When the particle is observed from its rest reference frame, the antiparticle solutions are zero. However once the particle is moving, all the 4 fields become coupled.
Does it mean that a moving electron is a little bit of a positron at the same time (in the given reference frame)?
| Literally speaking the answer is negative. The charge of the state has to be always defined in view of the charge superselection rule. Thus for a particle described by Dirac equation, there are no things like coherent superpositions of electron states and positron states. A Dirac particle always stays in a quantum state which is proper of an electron or a positron, but never in a superposition of them.
Nevertheless the quantum field associated to these particle describes both particles and anti particles.
| {
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Is it ok to have two events $A$ and $B$ so that for one person $A$ occurs before $B$ but for another $B$ preceds $A$ Imagine two laser beams A and B are released at the same moment to bounce between two mirrors, A was moving and B was at rest, doing the calculations I found that for a person at rest B would reach the upper mirror before A because in his frame of reference A travels less distance. but for another person in the same reference frame of A, A would reach the upper mirror first.
Is that OK in relativity!
| If the two laser beams are emitted at the same moment in one frame, they will not be emitted at the same moment in another frame moving relative to the original frame. This is relativity of simultaneity. Since the light beams start at different times, it's not a problem for them to travel a different distances.
| {
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Moving wire relative to stationary charged particle "Now we turn our attention to what happens in $S'$, in which the particle is at rest and the wire is running past (toward the left in the figure) with the speed $v$. The positive charges moving with the wire will make some magnetic field $\ B'$ at the particle. But the particle is now at $rest$, so there is no magnetic force on it! If there is no magnetic force on the particle, it must come from an electric field. It must be that the moving wire has produced an electric field. But it can do that only if it appears $charged$-it must be that a neutral wire with a current appears to be charged when set in motion."
So i got this from Feynman's lectures in physics volume 2 pg 13-7 and 13-8. I find it hard to believe that the particle experiences no magnetic force due to the moving wire although it's stationary. Since this is classical and not quantum electrodynamics if we assume the charged particle to have inertia, then by Newton's second law, the particle will remain at rest until acted upon by an external force, which i assume is the $magnetic$ $wind$ created by the moving wire past the stationary charged particle. Could someone please clarify??
| The whole point of Feynman's paragraph is to show that what we might believe is not what must happen by physical law. The full electromagnetic force on a particle is the Lorentz force, which is
$$\vec{F} = q (\vec{E} + \vec{v} \times \vec{B})$$
Since the particle is stationary, the second summand is necessarily $0$ in the frame considered, and so, in this frame, there must be an electric field, since we know the total force is not zero.
I've just paraphrased Feynman above, this is exactly what he wants to tell you: As counterintuitive as it may seem, moving magnetic fields seem to be electric, and vice versa.
| {
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Why does the speed of the propellant limit the speed of a space ship in open space? Isn't speed a relative thing in space? If so, why would the speed of a propellant matter? Why can't a space ship accelerate infinitely?
| The maximum theoretical speed that a spaceship can reach isn't limited by anything (except the speed of light of course). However for a practical spaceship with a finite amount of fuel, the speed of the exhaust will set a practical maximum on the speed of the spaceship. This is because in order to accelerate to a higher speed, the spaceship would have to carry more fuel to begin with, but this additional fuel would increase the mass of the spaceship, making it even harder to accelerate. This relationship is exponential, which means for a reasonable rocket (one that you could actually build), the exhaust speed of the propellant sets a practical maximum on the final speed of the rocket.
If I recall correctly this practical limit is roughly twice the exhaust speed of the propellent. After this, the diminishing returns get too ridiculous.
| {
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How quickly should a fluid come to hydrostatic equilibrium? Let's say I'm holding a one-liter water bottle, full of water, which I then drop.
Before dropping the water bottle, the equilibrium is for there to be a pressure gradient in the water canceling the gravitational force on the water. While the bottle is in free fall, the new equilibrium is constant pressure everywhere. Should I expect the water to come to this new equilibrium in the few tenths of a second it takes the water bottle to fall?
I expect the answer is basically yes, because density changes (and therefore pressure changes) should propagate at around the speed of sound, and p-waves might bounce around a few times while exponentially dying away (depending on boundary conditions created by the material of the bottle?), at the end of which we have equilibrium. So for a 30-cm bottle with sound speed 1500 m/s, I might guess the time is a few times .02s, which is longer than the ~.5s it takes for the bottle to fall from my hand to the ground.
Does this sort of reasoning make any sense? How can I justify it in a less handwavy manner?
| My problem with the assumption is that sound is quite poorly absorbed in water. The $$30 cm = 1/4 \lambda$$ size means you'd look at waves of about 120 cm = 12 Hz. Absorption at those frequencies is measured in deciBels per kilometer. If we'd model the bottle as a cylinder, we might get a standing wave pattern that could persist for several kilometers (i.e. seconds).
Of course, there's likely going to be some non-axis aligned component, the bottle won't be a cylinder, so there's energy spilling over to other wave components, you get turbulence, and that does dissipate energy quickly. But quantifying that turbulence and its energy loss is a pain in the backside.
Also this is entirely ignoring cavitation.
| {
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Why are anti-de Sitter spaces so interesting when we believe the universe is expansionary? Perhaps this is a naive question, but in my recent (admittedly limited) readings about AdS spaces, I keep wondering why they seem to be such a hotbed for theoretical research (AdS/CFT correspondence, etc.). To my understanding, an AdS space has constant negative curvature in a vacuum, which should yield an attractive universe, not one with accelerating expansion. An AdS space can be thought of as having a negative cosmological constant, while a universe with accelerating expansion would imply that such a constant be positive. Since we observe that our universe's expansion is accelerating, it seems that if anything, we should be seeking to model it as a de Sitter space.
Am I mistaken? What aspects of our universe do AdS spaces attempt to model?
| An AdS universe can explain the cosmological observations. Our universe can be interpreted as an effective de Sitter brane in an Anti-de Sitter space. Therefore, you have to distinguish between a 5-dimensional cosmological constant from the bulk and the 4-dimensional constant from the brane (which is responsible for an accelerated expansion). Unfortunately, many physicists like to ignore this distinction.
Further reading
Standard Cosmology on the Anti-de Sitter boundary
Class. Quantum Grav., 2021
https://doi.org/10.1088/1361-6382/ac27ee
https://arxiv.org/pdf/2010.03391.pdf
| {
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Why are the backs of airplanes curved? I get the front part, but why is the back curved too? I do not see a problem with the back being flat.
| The reason why the aft ends of airplanes are streamlined is to preserve a smooth flow of air. Just as the fore ends of airplanes are streamlined to smoothly cleave the air, so too the aft ends are streamlined to smoothly reintegrate the flows. Turbulence is bad, regardless of where on the aircraft it occurs.
(source: answcdn.com)
Note that there is more turbulence behind the stumpy shape than the longer one, despite the fact that the front ends are shaped the same. That is why the back ends of aircraft are streamlined.
| {
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Uniformity in a solenoid I know the magnetic field strength increases as the number of turns in the solenoid increases.
However, I've learnt the field inside the solenoid is usually nearly uniform.
So, does the number of turns in the solenoid effect the uniformity of the field inside the solenoid? Does the field gets closer to uniform as the number of turns increases?
| The expression for the magnetic field due to a long solenoid can easily be derived using Ampere's Law. The expression is $$\vec{B}=\mu_onI\hat{z}$$
where $\hat{z}$ is a unit vector pointing along the axis of the solenoid, $n$ is the number of turns per unit length, and $I$ is the current running through the solenoid. This derivation assumes that the solenoid is infinitely long so that, by symmetry, the field only points in the $\hat{z}$ direction inside the solenoid. Note that the density of the windings affects the strength of the field, not the number of windings (which in this case is infinite).
Of course, no solenoid is infinite so the longer the solenoid is, the better the above expression approximates the field of the finite solenoid. So it's the length of the solenoid affects the uniformity of the field, not necessarily the number of turns.
| {
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Specific Internal Energy Decreases with Increasing Pressure? I was looking at "Properties of Compressed Liquid Water" (Table A-5) in Fundamentals of Engineering Thermodynamics by Moran. Table A-5 shows that the specific internal energy $u$ of compressed liquid water decreases as as the pressure increases at a fixed temperature. For instance, $u=418.24$ kJ/kg at $T = 100$ F and $P = 2.5$ MPa. While $u=412.08$ kJ/kg at $T = 100$ F and $P = 25$ MPa. Also, this same trend in specific internal energy is shown in Table 3. Compressed Water and Superheated Steam. This trend that the specific internal energy decreases with increases pressure goes completely against my intuition and the numerous examples of compressing a gas or two phase mixture in a cylinder with a piston. Any thoughts on how the internal energy is decreasing?
| If dealing with a gas, the internal energy must be reduced with increasing pressure, as the temperature will be constant and the gas will be more compact.
Also see the P-T diagram of water, when changing pressure keeping the temperature constant, the vapor phase will change to liquid phase or solid phase which have less energy than a vapor phase.
| {
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Black and white matters. But why and how? I know black conducts heat while white reflects it.
But they are colors after all.
If a metal is painted black, it conducts more heat or at a rapid speed than it would do before it was coated.
But, as far as I know, colors don't have any special "substance" in them, which might trigger the sudden absorption of heat or reflection of the same.
What is the physics behind this? Are colors by themselves, some catalyst kinda thing?
|
I know black conducts heat while white reflects it.
The correct term is "black absorbs light while white reflects it".
We have named colors of light we see in the visible spectrum .
White reflects most of the energy falling from the visible spectrum, black absorbs it. When the energy of light is absorbed it turns into heat . Any material painted black will absorb this heat further and its temperature will be raised but it will depend on the material how far the heat is transferred. If it is metal painted black, metal is a good conductor of heat and will distribute the energy fast on the whole body.
But they are colors after all.
They change the surface properties of materials on which they are painted thus changing the ability of absorption and emission of radiation.
The energy coming from the sun covers a much larger electromagnetic spectrum than the visible. The visible has about half of the energy coming from the sun on the surface, as seen in the link.
So a metal door in the sun will transfer the heat of the visible spectrum to the interior if painted black, will reflect it back and keep the interior cooler if painted white. It is a good reason for painting roofs and walls white in hot countries. A white car is also better in hot countries for this reason .
It is not always sure that the color properties ( absorption/reflection) are followed by the invisible part of the sun spectrum, infrared or ultraviolet. Each paint has to be studied as far as its response to the impinging radiation to be used efficiently for thermal protection.
| {
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How does AC current flow in an open circuit? In common house hold wiring we have the hot lead, neutral and ground.
If the hot lead in electrical wiring contacts earth ground (perhaps though a short circuit in the chassis of a device) then the current shorted to earth. This however doesn't appear to me to be a complete closed circuit. My intuition tells me current should flow, but my academic knowledge says that it shouldn't because we have an open circuit condition. So how does the current therefor flow?
I've heard the argument that because the neutral wire is earth grounded at the service panel (aka breaker box), that it is actually a complete circuit. So in my example, current travels though earth and back up into neutral at the service panel.
However IF at the service panel, the neutral was not connected to earth ground (so the neutral left floating), then we definitely don't have a closed circuit, yet my intuition tells me somehow current would still flow into earth in a short circuit condition.
I've looked up a similar question like this here on stackexchange. In that question, current flow is examined between earth and mars in an incomplete circuit. In that scenario the answer looked at the planets somewhat like two capacitor plates. That would not apply in my question however
| There is a LOT of capacitive coupling between the neutral wire and ground even if a DC current cannot flow. And we are talking about AC here.
| {
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Fork and Sheet Lightning I've noticed that during British lightning storms, I have only ever seen sheet lightning
However, on holiday in other countries, I frequently see Forked Lightning
Q) Is this just caused by cloud, or is it two distinct forms of lightning?
Q) Is there a reason Britain only seems to get the sheet kind?
| Sheet lightning is just lightning where you can't see the strike because there is cloud in the way - the phenomena is the same.
Lightning storms occur in Britain in regular rain storms, so along with lots of cloud which hide the strike. In places further inland the storms are associated with hot air rising over plains, so less clouds to hide the strike and so a more photogenic result.
| {
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Why ONLY Maxwell's equations are the basic equations of electromagnetism? In electromagnetism we say that all the electromagnetic interactions are governed by the 4 golden rules of Maxwell. But I want to know: is this(to assume that there is no requirement of any other rule)only an assumption, a practical observation, or is there a deeper theoretical point behind it? Could there be a deeper theory behind assuming that there is not requirement of rules other than Maxwell's equations?
| Depending on how "basic" you consider an equation to be to electromagnetism, you could consider other equations to be important enough to be thought of as basic, given the type of situation.
For instance, when dealing with electromagnetism in media (typically linear media), the Constitutive Relations also apply and are necessary:
$$\overrightarrow{D} = \varepsilon_{0} \overrightarrow{E} + \overrightarrow{P}$$
$$\overrightarrow{H} = \frac{\overrightarrow{B}}{\mu_{0}} - \overrightarrow{M}$$
Or, if you happen to be relating currents to charges, you may want to use the Continuity Equation (although you can derive this by taking the divergence of Ampere's Law with the Maxwell Correction):
$$\nabla \cdot \overrightarrow{J} + \frac{\partial \rho}{\partial t} = 0$$
Furthermore, if you are dealing with point charge with mass in addition to electromagnetic fields, the Lorentz Force Equation will be needed (although you can derive this from Newton's 2nd Law, Lagrangian Mechanics, or Hamiltonian Mechanics):
$$\overrightarrow{F} = q(\overrightarrow{E}+ \overrightarrow{v}\times \overrightarrow{B})$$
But if you are being strict, and want to have the most bare-bones version of Maxwell's Equations (in a vacuum), you can get away with only two equations, those of the vector and scalar potentials:
$$\overrightarrow{B} = \nabla \times \overrightarrow{A}$$
$$\overrightarrow{E} =- \nabla \Phi - \frac{ \partial \overrightarrow{A}}{ \partial t}$$
And by taking the divergence and curl of each of these equations, you can recover the four Maxwell's Equations.
The number of equations you need really boils down to what type of problem you are trying to solve.
| {
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Analytical problems with Green's function I have a question about the right definition of the Green's function in physics. Why do we introduce (or not) an infinitesimal, positive number $\eta$ to the following definition:
$$\left[ i\hbar\frac{\partial}{\partial t} - \hat{H}(\mathbf{r}) \pm i\eta\right]G(\mathbf{r},t;\mathbf{r'},t') = \delta(\mathbf{r} - \mathbf{r'})\delta(t-t')$$
| A Green's function is nothing but the (generally distributional) integral kernel of the inverse of a given operator. The point is that the operator $$A:= \left[ i\hbar\frac{\partial}{\partial t} - \hat{H}(\mathbf{r})\right]$$
does not admit a unique inverse. Conversely,
$$A_{\pm\eta}:=\left[ i\hbar\frac{\partial}{\partial t} - \hat{H}(\mathbf{r}) \pm i\eta\right]$$
admits a unique inverse (for every choice of the sign) and it is given by an operator whose integral kernel is $G_{\pm \eta}$.
It turns out that $\lim_{\eta \to 0^+}G_{\pm \eta} f$ exist and these limits select a pair of inverse operators (among the class of inverse operators) of $A$, whose physical meaning is relevant (advanced and retarded solutions).
In practice, the computation of the limits above can be performed in the complex plane using the residuum theory, after having written down $G_{\pm \eta}$ in terms of a Fourier expansion. Within this picture, the appearance of several inverses of $A$ is described in terms of the various ways to surround the singularity in the complex plane.
| {
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Electrostatics - Inserting a brass plate between two charges The question is: if I were to insert a brass plate between two charges, what will happen to the force between the charges? Would it increase, decrease or stay the same?
Does the brass plate increase the value of permittivity of the medium and therefore the force decreases?
The correct answer is that it will increase. But I do not understand how.
| When you take a brass plate of considerable thickness and place it in between two charges, say positive and negative, induction takes place in the brass plate since it is a conductor: the electrons shift to the end near the positive charge while the cations stay near the negative charge. Now, induction occurs in order to make the field outside a certain region zero. Here the regions in question are the sets of points between the plate and the charges. Thus the magnitude of charge on each side of the plate tries to be as close as possible but opposite in sign to the charge that it is facing. Hence, the (here) attractive force increases, due to lesser distance (the brass plate is closer to a charged particle than the other charged particle since it is kept between them) while the repulsive force is comparatively weaker since the brass plate is of sizeable thickness and increases the distance between like charges (like charges being the charged particle and the similar charge on the opposite face of the plate). A similar question can be found in the problem book by SS Krotov.
| {
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Why does intramolecular hydrogen bonding cause molecules to separate? Today I learned about intramolecular hydrogen bonding, which occurs in molecules such as ortho-nitro-phenol.
What I was told is that, in case of intramolecular bonding, the molecules separate from each other, opposite of what happens during intermolecular hydrogen bonding. I don't understand why this is. Why would attraction within a molecule cause separation of the molecules?
There is obviously some intermolecular hydrogen bonding present in this case, but I understand that this might be negligible. Still, why should the molecules spread away from each other?
| Hydrogen bonding arises when a chemical bond is polarised to one end of it has a slight positive charge and the other has a slight negative charge. In the case of o-nitrophenol it's mainly the OH bond that is polarised - the H atom has a slight positive charge and the O atom has a slight negative charge. The charge separation means the OH bond has an electric dipole.
Electric dipoles interact with each other, so the dipole on one o-nitrophenol molecule can attract the dipole on a neighbouring o-nitrophenol molecule. This creates the attractive force between the molecules known as a hydrogen bond. There is an extended description of this on this web site.
Just to complicate matters o-nitrophenol can also exhibit intra-molecular hydrogen bonding. This is because the OH and NO$_2$ groups are next door to each other, and the H on the OH group can interact with the O on the NO$_2$ group.
Response to comment:
Intramolecular hydrogen bonding will not cause molecules to repel each other, but it will reduce the attractive interactions between molecules. This is simply because in o-nitrophenol the OH group forms a hydrogen bond with the adjacent nitro group and that reduces its ability to form hydrogen bonds with other molecules. In p-nitrophenol the OH and nitro groups are on opposite sides of the molecule so there can't be any intramolecular hydrogen bonding and therefore the intermolecular hydrogen bonding is stronger. So o-nitrophenol has a melting point of 44ºC while p-nitro[henol has a melting point of 113ºC.
So intramolecular hydrogen bonding won't cause the molecules to separate, but it will make them bind together less strongly.
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The meaning of "heralded photon" I am not a native English speaker, and I have just started to study physics in English. However, I came across the term heralded photon while I was reading a review article about optical quantum memory. I don't understand what it means. A dictionary explains herald as, "to be a sign that something is going to happen," but it does not make any sense. Below is the excerpt from the article:
If a single-photon detector is placed in one of the emission channels of non-degenerate spontaneous parametric down-conversion, a detection event indicates emission of a photon pair, and thus the presence of a single photon in the other channel. Such a heralded photon is emitted at an arbitrary time, however, making it unsuitable for many application.
Would you help me understand this term?
| Spontaneous parametric down-conversion converts a single incoming photon to two outgoing photons. I think the article is saying that that if you measure one photon coming out there must be a second photon as well. The author is referring to the second as a heralded photon in the sense that measurement of the first photon is a sign that the second (heralded) photon is going to be emitted.
| {
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How could the multiverse theory be disproven? Theorists (physicists) suggest that there is the term/entity, the Multiverse that contains a huge number of universes not necessarily like our own.
I personally find this theory very elegant because its explains the probabilistic outcomes of the experiments with fixed conditions on a quantum level.
It's obviously very hard to prove the validity of the theory of multiverse, but
*
*How can we disprove the existence of multiverse?
| I agree with you that the mathematical multiverse theory of Max Tegmark is the triumph of Occam's razor in simplicity (Although I disagree in restricting it only to Godel computable, mathematical structures). Not only it gets rid of explaining the physical universe itself (why there exist a physical universe?) but has many other philosophical advantages that I will not discuss here.
Unfortunately for your answer, it is not a scientific but a metaphysical question, so you either believe it or not based on other benefits beyond it being provable.
Regardless of the efforts of many physicists includind Max Tegmark, I do not think it could ever be proved or disproved. It is an unfalsifiable proposal (yes, you can try some restrictions to make some "predictions", but in the end you can always change it to make it unfalsifiable).
| {
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Trying to combine red, green and blue to produce white I tried to mimic the mechanism of typical screens to produce white color out of red, green and blue.
What I did is displayed the attached image on the screen, and moved far away as to let the diffraction effects take place, so that the three colors appear as if they're coming from the same point.
Nonetheless, quite paradoxically, what I have seen was black instead of white.
I don't know if this question fits this place, so excuse me.
| What you are seeing at a distance is not black. It is a darkish shade of gray, RGB gray 85,85,85. The reason you aren't seeing "white" is because each of those three rectangles has an HSV value of only 33% and you are seeing that merged square against a white background.
That merged square will appear to be whitish if you make the background black rather than white and view the screen in a very dark setting.
| {
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Two air bubbles inside a liquid I know that if two air bubbles which are formed inside a liquid are somehow joined using something (say a small tube), then, as the bubble with the larger radius has less pressure and the one with the smaller radius has more air pressure, air will flow to the larger bubble from the smaller bubble as the excess pressure inside the smaller bubble is greater.
But will we be able to quantitatively calculate the time taken for this to happen?
If so then what is the relation connecting time and every other variable?
| It is possible to write a relationship between flow and pressure drop in a tube. If the flow is laminar, this will be given by the Hagen–Poiseuille equation (http://en.wikipedia.org/wiki/Hagen%E2%80%93Poiseuille_equation). For turbulent flow, phenomenological correlations can be used (http://en.wikipedia.org/wiki/Darcy%E2%80%93Weisbach_equation). With this, it is possible to calculate the flowrate between the two bubbles and how it varies with time.
| {
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Gradient is covariant or contravariant? I read somewhere people write gradient in covariant form because of their proposes.
I think gradient expanded in covariant basis $i$, $j$, $k$, so by invariance nature of vectors, component of gradient must be in contravariant form. However we know by transformation properties and chain rule we find it is a covariant vector. What is wrong with my reasoning?
My second question is: if gradient has been written in covariant form, what is the contravariant form of gradient?
| Allow me to try to provide the simplest explanation of why the gradient is a covariant vector.
By definitions, the components of a covariant vector transform obey the law :
$$ \overline A_i = \sum_{j=1}^n \frac {\partial x^j} {\partial \overline x^i} A_j \qquad \qquad (1) $$
and the the components of a contravariant vector transform obey the law :
$$ \overline A^i = \sum_{j=1}^n \frac {\partial \overline x^j} {\partial x^i} A^j \qquad \qquad (2)$$
If the components of gradient of a scalar field in coordinate system $ \Bbb {\mathit {x_j}} $ , namely $ \frac {∂f} {∂x_j} $ , are known, then we can find the components of the gradient in coordinate system $ \Bbb { \overline { \mathit {x_i}}}$, namely $ \frac {∂f} {∂ \overline x_j} $, by the chain rule:
$$ \frac {\partial f} {\partial \overline x_i} = \frac {\partial f} {\partial x_1} \frac {\partial x_1} {\partial \overline x_i} + \frac {\partial f} {\partial x_2} \frac {\partial x_2} {\partial \overline x_i} + \cdot\cdot\cdot+ \frac {\partial f} {\partial x_n} \frac {\partial x_n} {\partial \overline x_i} = \sum_{j=1}^n \frac {\partial x_j} {\partial \overline x_i}\frac {\partial f} {\partial x_j} $$
Obviously $ \quad \frac {\partial f} {\partial \overline x_i}=\overline A_i \quad $ , and $ \quad \frac {\partial f} {\partial x_j} = A_j \quad $, then we get the equation same as $ (1) \quad $
Therefore, the gradient is a covariant vector.
| {
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Why does a wine glass with less water resonate at a higher frequency? In this video https://www.youtube.com/watch?v=hWwM7F-zaHs, Professor Lewin showed that for the tube, the less water there is, the longer the effective length of the tube and therefore, the lower the frequency.
He then demonstrates an opposite effect for a wine glass. Namely, an empty wine glass resonates at a higher frequency than a filled one. Why is that so?
| The qualitative reason is in the case of the pipe, the walls can be assumed for practical purposes to be rigid (i.e. they don't vibrate), and the resonant frequency of the vibrations in the air inside is determined by the boundary conditions. In other words, the shorter the air column in the pipe (more water), the shorter the wavelengths of the acoustic modes, or the higher the frequency. Or less water, lower frequency.
Whereas for the wineglass, the walls of the glass are thin enough that they cannot be assumed rigid - indeed they vibrate, and it is the vibrations of the glass that determine its resonant frequency.
With no water in the wineglass, the walls are not mass loaded beyond the mass loading due to the air inside. But as you add water, the mass of the water mass loads the vibration of the walls. Just as with a simple harmonic oscillator, the larger the mass, the lower the frequency of oscillation.
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Is there a difference between "average acceleration" and centripetal acceleration? Question adapted from Examkrackers MCAT prep book:
A particle moves along a half circle (diameter=$10\text{ m}$) at a constant speed of $1\text{ m/s}$. What is the average acceleration of the particle as it moves from one side of the half circle to the other side?
A. $0$
B. $0.2/\pi$
C. $0.4/\pi$
D. $1$
The book says C is correct. Acceleration is change in velocity divided by time. Initial velocity is $1\text{ m/s}$ up; final velocity is $1\text{ m/s}$ down. The change in velocity is therefore $2\text{ m/s}$. The time is found from speed equals distance divided by time. Distance is $2\pi r/2$. Thus
$$a= \frac{2}{(2\pi(5)/2)/1} = \frac{2}{5\pi} = 0.4/\pi$$
I thought all of the answers were wrong because I thought they should've used the centripetal acceleration equation: $a= v^2/r$
SO my question like the title: Is there a difference between "average acceleration" and centripetal acceleration?
I searched for a couple hours and couldn't find this issue directly addressed.
| The problem with centripetal acceleration is that it is not a vector, and cannot possible have a negative sign. It should remain "constant" in this case, but its direction is changing. But actually, it's not remaining constant, just the magnitude is remaining constant while the direction is changing. Centripetal acceleration will just give you the magnitude of the acceleration a direction towards the center of the circle.
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Wrong calculation of work done on a spring, how is it wrong? So I would have thought that this would be how you derive the work on a spring: basically the same way you do with gravity and other contexts, use $$W=\vec{F}\cdot \vec{x}.$$ If you displace a spring by $x$, then it exerts a force $-k x$, so $F=-kx$, since the displacement is $x$.
So $$W=-kx^2.\qquad \leftarrow\text{ (however, apparently wrong!)}$$
I've seen the correct derivation of work in a spring (with an extra half) and don't doubt that it's correct, but also don't see where my logic fails in this alternate derivation.
| In school, generally you would do an experiment where you load a spring with weights one at a time. Then you use the weight times the change in height as the work done on the spring. You can use this because it is a closed system and the weight has lost potential energy of $mgh$. It doesn't matter that the force was also accelerating the weight, as this energy is then transferred to the spring. This energy would be equal to $kx^2/2$.
| {
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What determines whether a pool ball will bouce backwards after colliding with another pool ball? I'm no knowledgeable pool player, but I've noticed that sometimes when the cue ball hits another pool ball, they roll together; and sometimes the cue ball bounces back. And I have a very, very rough sense that a hard, sharp, and even strike of the cue ball tends to make it bounce back more while a slower or more angled strike will make it roll forward after collision. Can anyone give a more rigorous analysis of the phenomena, or point me to a resource for this? I've tried googling but haven't see anything that really seems to address this as far as I can tell.
[Edit: Upon more contemplation, I suppose a more general question is: In a collision, what determines how much of the combined momentum of the system is distributed to the parts? So in cars colliding, or pool balls, or a skater on ice throwing a baseball--what features of the system determine the amount of momentum imparted to each component?]
| The direction the ball will take depends on the angular momentum. The velocity with which the ball moves or bounces backwards but the chief determinant is the spinning effect of the incoming ball.
| {
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Ampere's law and Biot-Savart law gives different terms for magnetic field in middle of a current running in a loop
I would like if someone could clarify this issue for me:
When dealing with a current $I$ running in a loop with radius $R$ and looking for the magnetic field in the middle of the loop.
By using Ampere's law, I know that the current $I$ runs through a loop with the same radius $R$, we get that:
$$\oint_c\vec{B} \cdot d\vec{l} = \mu_0 I_{enc} $$
$$B= \frac{\mu_0 I}{2\pi R}$$
and when using Biot–Savart we get that
$$d\vec{l} \cdot \vec{r} = |d\vec{l}||\vec{r}|\sin(\frac{\pi}{2})$$ obtaining:
$$B = \frac{\mu_0 I}{2R}$$
Which is not the same result as with Ampere's law.
I obviously miss something, maybe I can't use Ampere's law?
Anyway, if someone could help me out here I would really appreciate it.
Thanks.
| You probably misapplied Ampere's law. This law is usually used to find magnetic field only in special cases when the contour integral can be found as a function of single field value based on symmetry.
Magnetic field of a circular current loop is not so simple and Ampere's law cannot be easily used to find it. In such cases, the method of choice is to use the Biot-Savart law (integrate the contributions to the field due to elements of the circuit) or find vector potential as a function of position and then derive magnetic field from it.
| {
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Why did nuclear testing not result in nuclear winter? According to Wikipedia over 2000 nuclear tests have been performed since the Manhattan Project. If nuclear war would bring about a nuclear winter, why didn't testing do? Were they too much spread out in time to cause any real climate damage?
| A nuclear winter would be a result of large amounts of smoke blocking light from the Sun. The smoke would be from the fires started by nuclear bombs on cities, not directly from the bombs.
Most bomb tests have been underground, and the above ground tests were mainly done where there wasn't much to burn, for example in the Nevada desert so they didn't generate any significant amount of smoke.
Burning cities and volcanos affect the weather in much the same way but for slightly different reasons. Volcanos produce sulpher dioxide aerosols while burning cities produce carbon dispersions, but both reduce insolation. In both cases their effect on the weather is only marked if they manage to get a significant amount of fine material past the tropopause. Anything in the troposphere gets rained out too quickly to cause much change, but once into the stratosphere smoke/dust can persist for several years and reduce the light reaching the ground. Rock vaporised by a nuclear explosion will condense quickly and is unlikely to reach the stratosphere.
| {
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How would an X-ray scanner identify a mirror? A mirror is under normal circumstance used to reflect Electromagnetic radiation also known as photons (light) and in airport security or medical facilities, they use X-rays to detect anomalies inside objects or bodies to detect narcotics or injuries. However, I always wonder what if I add a mirror inside the luggage or put a mirror in front of me during scanning?
That in mind, how would an X-ray scanner see the mirror? Would it be invisible? I am sure I am not the first one to think of this, as a lot of security and criminals thought of this, however I never got an answer, so can someone tell me please?
If there are X-ray reflecting mirrors? Why don't Airport security ban these items and Mirrors all together? Would X-Ray mirror look like a normal mirror? Do they reflect visible light spectrum as well?
| The thing that makes a mirror a mirror is a that it has a high reflectivity (and is very smooth of course, but that doesn't enter into this issue), but all optical properties including reflectivity are functions of wavelength.
The mirror is not reflective in the x-ray band, so it looks like a layer of glass (moderately dense) and a very thin layer of heavy metal (rather denser). It can be seen in the image but is in no way remarkable.
It will look just like any other thin layer of moderately dense material.
| {
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What is the advantage of AdS/CFT in studying strong coupled system comparing with the lattice method I often heard AdS/CFT correspondence provides a powerful framework to study strong coupled system, which perturbation is not applicable. However, lattice method still works in non-perturbative domain. My question is, what is the advantage of AdS/CFT? Is there any example impoosible to access by lattice method (I don't mind lattice get numerical than analytic results)?
| In general, lattice calculations are quite cumbersome and require advanced numerical techniques and computational power. In the AdS/CFT correspondence, the involved concepts surpass those of lattice QCD greatly in complexity, as they involve string theory and general relativity in geometric backgrounds that are far from trivial. On the computational side, however, many (but definitely not all) calculations can be done analytically or by solving relatively simple (ordinary) numerical problems. Determining particle spectra for example can be as easy as solving a simple eigenvalue problem in the context of an ordinary differential equation.
There are also specific aspects of strong coupling physics (or to be more precise, QCD) where AdS/CFT is quite useful while lattice techniques are still struggling. One are would be the calculation of decay rates, where the correspondence is far ahead of what lattice is capable of. For example, in a specific model of holographic QCD, the Witten-Sakai-Sugimoto (WSS) model, decay rates of various mesons can be determined with relative ease and considerable success in reproducing experimental data.
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Why do some hand dryers blow hot air? I am not sure why some hand dryers are blowing hot air and not just air at room temperature.
To me, hair dryers are just a way to dry one's hands using the same principle as when we shake our hands in the air, or when we blow some air over a hot drink.
Given that the blown air temperature's is not enough (is it ?) to vaporize water, why is hot air used?
| To take an extreme case, suppose your hand dryer is located in a room without climate control, such as in an isolated restroom in the middle of a park. Occasionally the temperature in the room will fall to the dew point. In that case a room-temperature hand dryer would blow saturated air and have zero drying effect. However, warming the air brings it above the dew point and allows it to hold moisture from your hands.
There's probably also a comfort issue. Evaporation absorbs heat. Blowing unwarmed air on wet hands might dry them, but would make them feel cool. By blowing warmed air, the heat of vaporization can come from the air rather than from your hands.
| {
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Radiation emission and absorption Any object can emit and absorb radiation and the power of emission can be represented by the Stefan-Boltzmann law:
$$P=A\epsilon\sigma T^4$$
In many texts the net power radiated is the difference between the power emitted and the power absorbed:
$$P_{net}=A\epsilon\sigma (T^4-T_s^4)$$
where $$T_{s}$$ is the temperature of the surroundings.
Why can the surrounding and the object share the same $\epsilon$ ?
If we try to find out the radiation emitted from the surrounding it should be $P_s=A\epsilon_s\sigma T_s^4$, and if $\epsilon_s<\epsilon$, we will get a strange result that energy radiated from the surrounding is less than the radiation absorbed by the body from the surrounding. What am I missing?
| I think the key to the paradox is that you can't ignore the reflection coefficent. Lets say you have a lump of coal inside a shphere of polished steel. Yes, the coal emits much more heat than the steel; but that doesn't mean there is a net transfer of heat from the coal to the steel. Because the steel reflects back the excess heat.
There is a fixed relationship between the absorption, emission and reflection coefficients.
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Proof of Liouville's theorem: Relation between phase space volume and probability distribution function I understand the proof of Liouville's theorem to the point where we conclude that Hamiltonian flow in phase-space is volume preserving as we flow in the phase space. Meaning the total derivative of any initial volume element is 0.
From here, how do we say that probability distribution function is constant as we flow in the phase-space?
What's the relation between phase space volume and the density function, which instantaneously tells us the probability of finding the system in a neighborhood in phase-space?
| To obtain the result $\frac{\text d \rho }{\text d t}=0$ you need two facts: the first is that the hamiltonian flow preserves the volume of phase space. The second fact is the conservation of probability, that is, the probability that the system is found in a volume $U$ at time $t=0$ equals the probability of finding it within $\Phi _t U$ at time $t$, where $\Phi _t$ denotes the hamiltonian flow. This is a direct consequence of the deterministic nature of classical mechanics: the two propositions “$(p(0),q(0))\in U$” and “$(p(t),q(t))\in \Phi _t U$” are equivalent.
Using conservation of probability, for an arbitrary volume $U$ we can write an equation: $$\int _U \rho(p,q,0) \text d p \text d q=\int _{\Phi _t U} \rho(p,q,t)\text dp \text d q .$$
By Jacobi's theorem: $$\int _{\Phi_t U} \rho (p,q,t)\text d p \text d q=\int _U\rho (\Phi _t (p,q),t)\text J_{\Phi _t}d p \text d q.$$
The Jacobian $J_{\Phi _t}=1$, because the flow preserves volumes. It follows that: $$\int _U \rho (p,q,0)\text d p \text d q =\int _U \rho (\Phi _t (p,q),t)\text d p \text d q,$$
and, since the volume $U$ was arbitrary, $\rho (p,q,0)=\rho (\Phi _t (p,q),t)$, or $\text d\rho /\text d t=0$.
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Why is the pressure inside a soap bubble higher than outside? Apparently, the air inside a soap bubble is under higher pressure than the surrounding air. This is for instance apparent in the sound bubbles make when they burst. Why is the pressure inside the bubble higher in the first place?
| It is like a balloon. The pressures of the inner and outer air tend to equilibrate, creating a force over the balloon surface from the higher pressure to the lower one, trying to make them equal (the force goes from inside to outside, when you inflate it, from outside to inside when you deflate it). That's why it changes its size, because the gas pressures tend to be the same. At the other hand, the elastic material of the balloon when it is inflated tries to return to its resting state (deflated), so it creates an inward force compressing the inner air. The balloon reaches its balance (it gets inflated at a certain level) when these two forces are equal. That means that the inner air will always be at a higher pressure than the outer air, being this difference related to the balloon material elasticity (the more the material resistance to inflate/deform is, the more the pressure difference will be).
In a bubble this resistance to inflate is due to the surface tension (a contractive force that always tries to leave a liquid surface at its minimum; that's why bubbles and droplets are spherical).
(I'm not a native english speaker, please be comprehensive with expression)
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Pool in a submarine A common theme in aquatic science fiction is the submarine pool/access to the ocean. That terrible TV show Seaquest had it, The Deep & Deep Blue Sea (Samuel L Jackson is standing in front of it when the shark chomps him). My question is how this could possibly work? From what little knowledge I have, I'd say the cabin where the pool resides would have to be pressurized to the water at that depth. The implications are that you'd have to pass through an airlock to get to the room, and that it would only work to a certain depth.
Is this correct, or it too far to the "fiction" side of the science fiction axes?
| You can do it without raising the internal pressure. First you have your pool, at 1 atm or ambient inside the vessel. Then you have a wet chamber which can be sealed at both ends. Enter the pool, then only the inner door of the wet lock opens, enter this area, and close the door to the moon pool behind you. When the internal door is shut you can open the external door. External water pressure never reaches the internal pool. The mass / volume of water inside the wet lock won't change so it won't push back into the atmosphere of the habitat. You couldn't use this as an unprotected diver since you would explosively decompress every time you re-entered the vessel interior (because it's at 1 atm and you have just entered from a significantly higher pressure zone) But it would serve to protect the atmosphere inside the vessel and can be used with 1 atm suits or subs.
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Which ball touches the ground first? This is a very well known problem, but I can't find an answer in the specific case I'm looking for.
Let's consider two balls :
*
*Ball 1 weighs 10 kg
*Ball 2 weighs 1 kg
*Balls have identical volumes (so Ball 1 is much more dense)
*Balls have identical shapes (perfect spheres)
Let's drop them from a rather important height, on earth, WITH air. (That's the important thing, because all the proofs that I browse take place in a vaccum).
I am arguing with a colleague. He thinks that ball 1 will fall faster in air, and that the two balls will fall at the same speed in a vacuum. I think that the identical shapes and volumes make air friction identical too and that the vaccum has no importance here. Could someone tell who's right and provide a small proof?
| Since air creates a force that is approximately proportional to the square of the velocity, the acceleration for each sphere is $a_r = kv^2/m (where \text{ } k = \frac{1}{2} C_x\rho\ S) $ The net acceleration on each sphere is $ a_n = g - a_r$. As the velocity increases, the $a_r $ increases until the net acceleration $a_n $ becomes zero $(a_r = g)$, and thus each sphere reaches its terminal velocity. $$Given: m_1 = 10kgr, \text{ } \text{ } m_2 =1kgr, \text{ } k = 0.01, \text{ }g = 9.8m/s$$ $$ For \text{ } m_2, ( v_2 = m_2g/k)^{1/2} = (1x9.8/.01)^{1/2} = 31.3 m/s$$ $$For \text{ } m_1, (v_1 = m_1g/k)^{1/2} = (10x9.8/.01)^{1/2} = 98.99m/s$$
After using an iterative method, I determined that the 1kgr mass $(m_2)$ reaches the terminal velocity in about 10 seconds and the 10kgr mass $m_1$ in about 33 seconds. Although the spheres reach their terminal velocity at different times, the larger mass reaches a higher velocity because the lighter mass reaches its terminal velocity sooner and does not increase after that. The heavier mass, takes longer to reach its terminal velocity, and thus it becomes larger. So, the heavier mass will reach the ground sooner.
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Calculate water flow rate through orifice I'm not very good with fluid physics, and need some help. Imagine the following setup with water contained in-front of a wall with an opening on the bottom:
How do I calculate the water flow $Q$?. I have made some re-search and found I need to (partially) calculate the pressure across the opening (orifice). But I don't know the pressure on the back side of the orifice. Can this be solved in any way?
Note: I'm not saying "please give me the solution, I'm lazy". I want to figure it out myself. But since, in this case, I only found formulas involving calculating pressure drop, I canno't use them to solve the problem. Therefore I'm turning my face to you, to see if there's another way to solve this problem.
Update: The "tank" holding the water is actually a big lake, and the opening is how much the water gate have opened. I need to very precisely calculate how much water flows through the opening.
| From a Wolfram article we get the simplified Bernoulli equation:
$$Q = a c \sqrt{2 g h}$$
Where
*
*$Q$: the flow rate ($\mathrm{m^3/s}$)
*$a$: the area of the hole ($\mathrm{m^2}$)
*$c$: flow coefficient (dimensionless)
*$g$: the gravity acceleration ($\mathrm{m/s^2}$)
*$h$: the depth of the hole ($\mathrm{m}$)
That is valid for a small enough hole, but since your hole can be big, we have to use integral calculus. Moreover, I think that the flow coefficient can be set as 1 for a big hole. And the area of the hole can be calculated as the width of the hole times the height (assuming a square hole).
So $$\begin{align}\renewcommand{\intd}{\,\mathrm{d}}
Q &= \int_{h-d}^h \sqrt{2 g y}\,w \intd y \\
&= \int_{h-d}^h w \sqrt{2 g} \sqrt{\vphantom{2}y} \intd y \\
&= w \sqrt{2 g} \int_{h-d}^h \sqrt{\vphantom{2}y} \intd y \\
&= w \sqrt{2 g} \left[\frac{2}{3} \sqrt{y^3}\right]_{h-d}^h \\
&= \frac{2}{3} w \sqrt{2 g} \left[\sqrt{y^3}\right]_{h-d}^h \\
&= \frac{2}{3} w \sqrt{2 g} \left(\sqrt{h^3} - \sqrt{(h-d)^3}\right)
\end{align}$$
| {
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AM Receiver and sideband frequencies How sorry for "dummy" question. I have difficulty to understand how the AM receiver.
Being a child I was taught that AM receiver resonates only with the radio frequencies it is tuned to, and "passes" only this frequencies to "other" circuits.
Recently I have read in articles that somehow the AM frequencies have the component of carrier frequency and component of side band frequencies. I even can more or less understand the formula derivation of it, which is quite simple. What I have difficulty to understand is: If AM receiver tuned to receive only one frequency, e.g the carrier frequencies, how all this side band frequencies are not getting filtered out by the receiver?
Please be patient all you radio gurus.
| There is no realizable filter that blocks all frequencies except one.
The fact is that frequencies above and below the tuned frequency are passed but with greater and greater attenuation the farther away from the carrier frequency.
More importantly, the actual information is in the side bands, not the carrier. If the filter only passed the carrier, there would be only the constant amplitude carrier frequency at the output of the filter.
Mathematically, to produce an amplitude modulated carrier requires frequencies above and below the carrier frequency to 'beat' with the carrier.
When the three constant amplitude frequencies, on the left are combined (summed), the result is the amplitude modulated signal on the right. Thus, we must pass not only the carrier but also the side band frequencies to the amplifiers and detector of the receiver.
(Note: There are several types of amplitude modulation but that is beyond the scope of this answer.)
| {
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How would an X-ray mirror work? I was wondering if light can be reflected how can someone reflect X-ray of what material does it need to be made of and is its design completely different to that of our original mirrors? Does this mean during long-space voyages in which radiation is an problem why can scientists not develop large panels of X-ray mirrors and Gamma-Mirrors and simply reflect the radiation off rather than worry about that?
| Yes it is possible but as BarsMonster points out it isn't like an optical mirror. X-ray reflectors are used in the construction of nuclear weapons and are critical to increasing the yield. How they work is the initial fusion reaction releases high energy radiation, this is then reflected back into the reaction mass increasing the energy levels of the reaction mass causing a sort of radiation implosion.
The reflector is typically a cylinder made of a material such as
uranium. The primary is located at one end of the cylinder and the
secondary is located at the other end. The interior of the cylinder is
commonly filled with a foam which is mostly transparent to x-rays,
such as polystyrene.
The term reflector is misleading, since it gives the reader an idea
that the device works like a mirror. Some of the x-rays are diffused
or scattered, but the majority of the energy transport happens by a
two-step process: the x-ray reflector is heated to a high temperature
by the flux from the primary, and then it emits x-rays which travel to
the secondary. Various classified methods are used to improve the
performance of the reflection process.
Since these type of reflectors are typically used in weapons research there is not much public data on them. However here is a rough illustration of what an x-ray reflector looks like in the second stage (reference).
| {
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Critical size and Radioactive Nuclei Nuclear fission requires the mass of the fissile material above the critical mass. So that the explosion takes place at least in the case of a nuclear bomb. But once a single nucleus got involved in the reaction, the reaction can't be stopped easily if I'm right.
A single nucleus how do know about the critical mass of the entire fissile material?
| There's spontaneous fission, a rare decay mode for some superheavy nuclei like uranium and plutonium. There's also induced fission, where some interaction with the environment (typically neutron capture) causes a nucleus to split. Both fissions typically produce a couple of free neutrons, which may be captured on other fissionable nuclei, or may be captured on some nearby neutron absorber (a "control rod" or a "neutron poison"), or may escape from the reaction volume to infinity.
The "critical mass" is the amount of fissionable material needed so that each fission is likely to trigger, on average, one other fission. It's not a property of a single nucleus, but of the fuel assembly, its geometry, and its environment.
| {
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Who proposed the bulk-edge correspondence principle? Who proposed the bulk-edge correspondence principle?
The principle is often quoted in counting the number of zero energy states localized on the interface between two insulators with distinct band topology. However, I could not retrieve who was the first to say that.
| I believe it was Xiao-Gang Wen in 1989, see also this 1994 paper by him and his collaborators
http://dao.mit.edu/~wen/pub/ednab.pdf
He's at MIT. I was once hosting a seminar by him, he is one of the most creative and playful folks in this segment of condensed matter physics. The paper above contains some other relevant references, including a paper by Wen and Tony Zee.
| {
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Where does wave frequency come from? I am trying to wrap my head around where do oscillations in electromagnetic waves come from. As an example if I would take a string of guitar and ring it, it would produce a certain sound based on the amount of vibrations per second. That amount of vibrations would be the sum of moves of string per amount of time, e.g there is one oscillation happening many times until string runs out of energy.
When I see the visible light it must be same thing something is vibrating and all the oscillations must the the sum of 'something' of one.
What is producing that one oscillation?
| For low-frequency radiation, it's quite simple: there's some electronic circuit that works (simple case) analogous to a tuning fork, but instead of building up mechanical tension it charges a capacitor and instead of the inertia in the fork's arms it has a magnetic field in a solenoid. You can measure the voltage against time, count the oscillations in one second, and know your frequency in Hertz.
For visible light, this explanation doesn't quite work anymore, but still it's kind of sort of vibrations – on an atomic scale! These systems must be described in quantum states, and there's this thing that if a state has energy $E$ then you can assign it a frequency $\nu = E/h$, where $h$ is the Planck constant. This frequency can't be observed directly, but what you can observe is, for a quantum superposition of two states with different energy $E_1, E_2$, that the system kind of "wiggles" with a frequency $\Delta\nu = \tfrac{E_1 - E_2}{h}$. And that wiggling frequency is the frequency of light emmited by a transistion from state 1 to state 2.
(Of course this explanation does not quite reflect how quantum mechanics works, just a very rough picture.)
| {
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How to calculate the drag coefficient using terminal velocity? I was wondering if it were possible to calculate the drag coefficient by allowing an object to reach terminal velocity. Can you rearrange the terminal velocity formula to give the drag coefficient?
| Yes you could. Since the force on an object from drag is given by
$$F_D = \frac{1}{2}\rho v^2 A C_D$$
where $C_D$ is the drag coefficient, then all you would need to know are your velocity ($v$), your fluid density $\rho$, your cross sectional area ($A$) and the force of gravity on the body, which would then be equivalent to $F_D$ since the body would have no net force on it. Thus, you could isolate for $C_D$, getting
$$C_D = 2mg/\rho v^2 A$$
| {
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Does Free Will Theorem imply that quantum mechanics plays crucial role in our brain’s functioning (consciousness)?
*
*Does Free Will Theorem imply that quantum mechanics plays crucial role in our brain’s functioning (consciousness)?
*Is opposite statement of Free Will Theorem right: If elementary particles have a certain amount of free will, then so must we?
Because to me elementary particles does have a bit of free will – quantum mechanics guarantees that nobody can predict what one is going to do, say in double slit experiment.
*So Penrose was right and origins of our consciousness lie in the laws of quantum mechanics?
*Is the only way our free will can come from is that of quantum mechanics?
| Penrose is saying things much more subtle in his book "The Emperor's New Mind". I wouldn't attempt to describe it.
Another view is by Minsky in "Society of Mind".
Again, I think he does a far better job of explaining it than I could, though in that video he's talking more about consciousness. Free will is just another one of those descriptions we make of ourselves when we don't really have any idea how the whole thing works.
| {
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Airplane on a treadmill I've heard conflicting answers, and would like to see the record set straight:
An jet/propeller airplane is traveling on a giant treadmill at takeoff speed. Will the plane takeoff, or will it remain on the runway, and why?
|
An jet/propeller airplane is traveling on a giant treadmill at takeoff
speed.
With respect to what does the plane have takeoff speed?
I believe the following two statements are uncontroversial:
(1) if the plane has takeoff airspeed (or greater), the plane can takeoff.
(2) if the plane does not have takeoff airspeed (or greater), the plane cannot takeoff.
Now, the plane's airspeed and the speed of the plane with respect to the treadmill belt are, within practical bounds, unrelated.
Thus, if you only specify the speed of the plane with respect to the treadmill belt, we don't know what the plane's airspeed is so we can only refer you to (1) and (2) above.
To be clear, it is possible for the plane to have both takeoff airspeed and, say, twice takeoff speed with respect to the treadmill belt.
For concreteness, assume the takeoff airspeed is 100 mph. Then, assuming low enough friction and strong enough wheels/tires, there is nothing to prevent the plane on the treadmill from having an airspeed of 100 mph and a speed relative to the treadmill belt of 200 mph.
The plane's propeller pulls the plane through the air as normal and the wheels rotate at twice the speed compared to an identical plane alongside on the runway next to the treadmill.
| {
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Determine the acceleration in a double Atwood machine Trying to solve the following double Atwood machine:
Suppose there is a mass of 12kg hanging on an ideal rope that wraps around an ideal pulley, and that the other end of the rope is attached to another ideal pulley's axis. This second pulley has a rope around it which on one side holds a mass of 4kg and on the other a mass of 8kg. What is the acceleration of the first mass?
So here's what I've done so far: First I've determined that the tension in the first rope is twice the tension in the rope for the "subsystem", and that the acceleration of the first mass is the average of the accelerations of the two other masses but in the opposite direction. That is to say
$$T' = T/2$$
$$a_{1}=-\frac{a_{2}+a_{3}}{2}$$
From free-body diagrams we can determine that
$$-12a = T-12g$$
$$4a_{2} = T/2 - 4g$$
$$-8a_{3} = T/2 - 8g$$
However, when I solve this system I obtain $T=32a$ which seems impossible that the tension would be greater than the greatest force of gravity of any mass.
Part of me wonders if this has anything to do with it: How do I know which direction the third mass is traveling? The first mass is causing the subsystem to accelerate upwards but the third mass is accelerating downwards relative to the subsystem. So how do I know the direction of acceleration relative to the ground?
| Let
\begin{align}
m_1 = 12\,\mathrm{kg}, \qquad m_2 = 4\,\mathrm{kg}, \qquad m_3 = 8\,\mathrm{kg}
\end{align}
If you solve this problem symbolically, then you'll find that the tension $T$ applied to mass $m_1$ satisfies
\begin{align}
T = \left(\frac{8m_1m_2m_3}{m_1m_2+m_1m_3+4m_2m_3}\right)g.
\end{align}
If you plug in the values given for the various masses, then you obtain
\begin{align}
T=\frac{192}{17}g \approx (11.30\,\mathrm{kg})g <\text{weight of mass $m_1$},
\end{align}
so it seems that your claim
the tension would be greater than the greatest force of gravity of any mass
is false. For reference, here are the equations you obtain using Newton's Second Law:
\begin{align}
T-m_1g&=m_1a_1\\
\frac{T}{2}-m_2g &= m_2a_2\\
\frac{T}{2}-m_3g&=m_3a_3,
\end{align}
and the constraint you wrote down is correct;
\begin{align}
a_1 = -\frac{1}{2}(a_2+a_3).
\end{align}
| {
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Is this "cloaking in time" serious and what is really meant? I just found this news article: http://www.nature.com/nature/journal/v481/n7379/full/nature10695.html
What did those researchers actually do? The article itself doesn't sound to me like it can be taken seriously, so... what was the experiment and what was observed?
| I do not understand all the nitty details, but the gist goes as follows:
You have a constant probe beam, which is cast on a sample and analysed later.
Suppose the sample has a very predictable behaviour, with an event every $24\mu \text{s}$. Then the analysis of the beam will show exactly that. This is the blue line in Fig. 4.
Now, suppose you want to hide these events, but you don't want to divert the beam around the sample (for whatever reason). They do it in the following way:
*
*Increase/decrease the wavelength of that part of the probe beam, which would encounter the event. This only serves as "marking" the part of the wave you want to manipulate.
*Speed up waves with lower frequency and slow down waves with higher frequency. This creates a beam gap in the middle. That means, the sample will see the probe beam "turned off" during the event, and the probe beam is not affected.
*After the sample, reverse step 2, i.e. slow down waves with lower frequency and speed up waves with higher frequency.
*Reverse step 1. Now your probe beam looks roughly as if it was not messed with at all and probed the sample at all times.
Note, that this is a layman explanation, and the actual speeding up/slowing down is a rather complicated feat.
| {
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Kinetic Energy of a Particle Consider a particle of mass $m$ in $6$ dimension. Its coordinate w.r.t. origin $\left(0,0,0,0,0,0\right)$ is given as $\left(x,y,z,\dot{x},\dot{y},\dot{z}\right)$. If we denote $r = \sqrt{x^2+y^2+z^2}$, then which of the following two is the kinetic energy of this particle:
*
*$T = \frac{1}{2}m\left(\dot{x}^2+\dot{y}^2+\dot{z}^2\right)~?$
*$T = \frac{1}{2}m\left(\frac{dr}{dt}\right)^2 = \frac{1}{2}m\frac{\left(x\dot{x}+y\dot{y}+z\dot{z}\right)^2}{r^2}~?$
| Hard for me to say what you're asking. If you have a particle of mass $M$ in three dimensions such that its positions is described by coordinates $\vec x(t)=(x(t),y(t),z(t))$, then the velocity vector $\vec v=\frac{d\vec x}{dt}=(\dot x(t),\dot y(t),\dot z(t))$
The kinetic energy is then defined as,
$$T=\frac{1}{2}M\vec v\cdot \vec v=\frac{1}{2}M\bigg(\dot x(t)^2+\dot y(t)^2+\dot z(t)^2\bigg)$$
If you're asking about a six dimensional space that has coordinates, $$\vec x(t)=(x_1(t),x_2(t),x_3(t),x_4(t),x_5(t),x_6(t))$$
$$\vec v = (\dot x_1(t),\dot x_2(t),\dot x_3(t),\dot x_4(t),\dot x_5(t),\dot x_6(t))$$
Then the kinetic energy is similar to the first one (assuming a 6-dim Euclidean metric)
$$T=\frac{1}{2}M\vec v\cdot \vec v=\frac{1}{2}M\bigg(\dot x_1(t)^2+\dot x_2(t)^2+\dot x_3(t)^2+\dot x_4(t)^2+\dot x_5(t)^2+\dot x_6(t)^2\bigg)$$
If the question is "why is that the form of the kinetic energy?" then you could say its a low velocity approximation of the energy of a particle. There are other neat motivations i've seen for that question as well.
Hope that helps.
| {
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Diffusion and Drift currents in a PN junction In a forward-biased PN junction, the potential barrier decreases, allowing more majority carriers from one side to diffuse to the other side where they are minority carriers. After they cross the potential barrier, they form a diffusion current, the drift current of minority carriers is insignificant, then they recombine with majority carriers and form a drift current under the effect of the applied electric field.
Why do minority carriers form a diffusion current not a drift current after they cross the potential barrier? It is counter-intuitive that the main current is diffusion when there is an applied electric field.
This is according to all the microelectronics book I'm currently reading. There is one which says this can be proved but without providing anything. Can someone please provide a proof for this.
| The unbiased PN junction is in an equilibrium where the diffusion current is cancelled by the drift current.
The applied electric field in a forward biased PN junction "cancels" the electric field in the depletion region, eliminating the potential barrier and allowing diffusion current to flow.
| {
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What if an asteroid the size of the moon hits earth? Would we all perish due to excessive heat? Or would that be limited to the area near the impact while the people on the rest of the earth would die from other phenomena such as mega earthquakes, volcanic activities, tsunamis etc.? Does it matter where the impact is - if it landed in Antarctica, would we have massive floods, but if it landed in the middle of the Eurasian continent, would the effect be different?
| The University of Arizona had a neat website that addressed precisely such questions. It's moved here now.
You can choose all kinds of parameters - kind of projectile (ice / rock), impact angle, velocity, landing site and get a prognostication, expected damage etc. Really cool stuff.
| {
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Is it possible to 3D print a mirror to create a high quality telescope? Is it possible to 3D print a mirror with todays available materials?
If so, would there be a reduction in image quality?
| The answer is "Yes" but not the way you might expect. It is possible to construct a telescope mirror from rotating liquid metal.Mercury used to be used but something like Gallium is safer and better.
So print a cradle for it, put in the Gallium, raise the equipment past the melting point (about 30 degC), spin gently to get a parabolic surface, and then cool.
| {
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Differences between probability density and expectation value of position The expression $\int | \Psi\left(x\right)|^2dx$ gives the probability of finding a particle at a given position.
If wave function gives the probabilities of positions, why do we calculate "expectation value of position"?
I don't understand the conceptual difference, we already have a wave function of a position. Expectation value is related to probabilities.
So what is the differences between them? And why do we calculate expectation value for position, although we have a function for probability of finding a particle at a given position?
| Expectation value is a different concept from probability. In fact, you can have an expectation value of energy, angular momentum, etc., not just for position.
An expectation value of an observable for a given state $\Psi$ is the average value of a large number of measurements of that observable, assuming each measurement is made on the same state $\Psi$. For example, if you have probability 0.5 to measure energy $E_o$ and probability 0.5 to measure $-E_o$, the expectation value is $0.5\times E_o + 0.5\times(-E_o) = 0$. As this example shows, the expectation value doesn't have to be one of the "allowed" measurements. This also shows that knowing the probabilities is a different concept than knowing the average value of a large number of measurements.
The same goes for position. You might know what the probability density is at a particular position, but you would need to do extra computations to figure out what the average value of many position measurements would be.
| {
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Why do liquids boil when their vapor pressure equals the ambient pressure? Given that the boiling point of a liquid is the temperature at which the vapor pressure is equal to the ambient (surrounding) pressure, what significance does a liquid's vapor pressure have in the formation of bubbles that happens at and above the boiling point?
The definition of boiling point seems to imply that the pressure inside of the bubbles must be at least as great as the liquid's vapor pressure in order to balance the outside pressure, but is there any particular reason why the pressure inside of the bubbles is related to the vapor pressure?
The vapor pressure seems to be a measurement describing the tendency of the molecules to escape from the surface of the liquid, but I don't see how that relates to bubble formation within the liquid.
This question has bothered me for a while, so any help would be much appreciated.
| Once boiling starts, there is an equilibrium between the liquid and vapor phase. The pressure, temperature and partial molar Gibbs energy are equal for each phase so that water molecules have no preference for one phase or the other. That's for intensive variables. However, the total enthalpy of the liquid and vapor is not fixed : it keeps increasing as more energy is brought in. The proportion of vapor to liquid is fixed through this energy balance.
If the pressure is maintained externally, say by a piston, there is no possibility for the water bubbles to form at any other pressure than the vapor pressure. However when you boil water at the bottom of a pot, two phenomena alter the situation slightly. The first is the hydrostatic height of the water column, which increases the pressure at the bottom (where bubbles form) and raises the equilibrium temperature. The second is the surface tension of water, which increases the pressure required to form a bubble. If the bottom surface is perfectly smooth, nucleation of bubbles is difficult, and the onset of boiling can be delayed to higher temperatures as water remains in a metastable liquid state.
Note that when boiling water you will often see bubbles form at an early stage and then disappear as the temperature increases : those bubbles are not water vapor but dissolved gases.
| {
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What is $c + (-c)$? If object A is moving at velocity $v$ (normalized so that $c=1$) relative to a ground observer emits object B at velocity $w$ relative to A, the velocity of B relative to the ground observer is
$$ v \oplus w = \frac{v+w}{1+vw} $$
As expected, $v \oplus 1 = 1$, as "nothing can go faster than light".
Similarly, $v \oplus -1 = -1$. (same thing in the other direction)
But what if object A is moving at the speed of light and emits object B at the speed of light in the exact opposite direction? In other words, what is the value of $$1 \oplus -1?$$
Putting the values in the formula yields the indeterminate form $\frac{0}{0}$. This invites making sense of things by taking a limit, but $$ \lim_{(v,w)\to (1,-1)} \frac{v+w}{1+vw}$$ is not well-defined, because the limit depends on the path taken.
So what would the ground observer see? Is this even a meaningful question?
Edit: I understand $1 \oplus -1$ doesn't make sense mathematically (thought I made it clear above!), I'm asking what would happen physically. I'm getting the sense that my fears were correct, it's physically a nonsensical situation.
| Recalling that the relativistic velocity addition formula in 1+1 dimensions in terms of rapidity
$$\beta~:=~\tanh^{-1}\frac{v}{c}$$
is just ordinary addition of rapidities
$$\beta_1+ \beta_2,$$
then it becomes clear that OP is essentially asking
What is $\infty+(-\infty)$ ?
which is not defined mathematically.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to approximate acceleration from a trajectory's coordinates? If I only know $x$- and $y$- coordinates of every point on a trajectory without knowledge of time information, is there any way to approximate Cartesian acceleration angle at each point? Time interval between every two points is very small, ~0.03 second.
| In your case, lets $\Delta t = 0.03s $
By the method alemi explained,
$$a_{x}(t)=\frac{x(t-\Delta t)-2x(t)+x(t+\Delta t)}{(\Delta t)^{2}}$$
and
$$a_{y}(t)=\frac{y(t-\Delta t)-2y(t)+y(t+\Delta t)}{(\Delta t)^{2}}$$
| {
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Conserved current for a constant translation of a free massless scalar field In Zinn-Justin's Quantum Field Theory and Critical Phenomena they start with an action for a free massless scalar field:
$$S(\varphi) = \frac{1}{2}\int d^{2}x\left[\partial_{\mu}\varphi(x)\right]^{2}$$
And say that the action is invariant under constant translations of the field
$$\varphi(x) = \varphi^{\prime}(x) + \theta$$
This is all fine, the problem I'm having is with the next part. They say a conserved current $J^{V}_{\mu}$ corresponds to this symmetry:
$$J^{V}_{\mu} = \partial_{\mu}\varphi(x)$$
I just can't see how to get that out with $\theta$ being a constant - unless I'm missing something in the notation.
| For each continuous symmetry, infinitesimal transformations may be expressed, by a bracket involving the conserved charge operator associated to the symmetry :
$$\delta_\epsilon \phi(x) = i\epsilon [Q, \phi(x)] \tag{1}$$
In our case, we must have :
$$\epsilon \theta = i\epsilon [Q_\theta, \phi(x)] \tag{2}$$
A solution is then :
$$Q_\theta = \theta \int d^3 x' \,\partial_0\phi(x',t) \tag{3}$$
This is because the quantization of fields is expressed as :
$$[ \phi(x,t),\partial_0\phi(x',t)]= i \delta^3(x-x') \tag{4}$$
From $(3)$, we see that the zero component $J_0$ of the current associated to the conserved quantity $Q_\theta$, is $\theta \, \partial_0\phi$, so the current $J_\mu$ is simply $\theta \,\partial_\mu \phi$, and the fact that $Q_\theta$ is a conserved quantity, implies that the associated current is conserved : $\partial^\mu J_\mu=0$, that is in fact $ \theta \,\partial^\mu \partial_\mu \phi=0$, which is nothing, dividing by a non-zero $\theta$, that the equations of movement for the field $\phi$, obtained from the Lagrange-Euler equations.
| {
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The Gluon - Does It Exhibit Wave Properties? Do Gluons have frequencies and wavelengths? I assume that they do, but have been unable to find anything on point in SE or Wikipedia. Just beginning to study university-level physics here.
| They do, just as all quantum objects do. They have momenta, and since they are massless, their frequency/wavelength/energy/momentum relations are the same as for photons.
But since you will never detect a free gluon, as they are color-charged and thus confined, this is not a sensible thing to say. Quantum objects are not waves (just as they are not classical particles), and if you cannot examine a free gluon, you cannot do something like the double-slit with it, and the "wavelength" you might want to associate with it is not really useful.
| {
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How to determine the sign of the s-wave scattering length? I guess it is relatively easy to determine the magnitude of the scattering length $a$.
We just need to measure the scattering cross section. In this way, we can determine the value of $a^2$.
But how to determine its sign?
| I think the appropriate method will depend on what system you're studying. I don't know about a general method, but since you also asked about experiments which determined the sign of $a$, I can offer a specific example:
In ultracold atomic physics experiments there is the case of Feshbach resonance, where you have a large positive scattering length due to a weakly bound state, and then as you adjust the bound state energy across the resonance (by varying the applied magnetic field) the scattering length changes sign. As was done in this experiment, one can measure the binding energy of the molecules (at various magnetic field strengths) using rf spectroscopy, and then use this information to determine the magnitude and sign of the scattering length.
| {
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Can an electron make a transition between sub energy states of the same energy level? Electrons make transition between different energy levels - say, $n =3$ to $n=2$ or $n =1$, as per the applicable selection rules.
My question is: can an electron make transitions between sub energy states of the same energy level?
For example: if an electron is in the $n=3$ energy level, can it make a transition between different sub energy states of $n=3$ following the selection rules?
| It depends on whether you're considering only hydrogen, or whether you're considering multi-electron atoms.
In hydrogen the states with different $\ell$ are (very nearly) degenerate in energy, so a transition like $2s^1 \rightarrow 2p^1$ is in principle allowed but in practice unobservable. However in Helium the $2s$ and $2p$ levels are not degenerate and indeed the transition line for $1s^12s^1 \rightarrow 1s^12p^1$ can be observed and has a wavelength of about a micron.
| {
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How to write QM operator if I know all of it's eigenfunctions? Suppose I have selected enough orthogonal functions in representation of operator A and I want to derive operator B which has these functions as it's eigenfunctions.
How to do that?
| It's difficult to answer your question because a collection of eigenvectors $\{ v_i\}$ does not uniquely specify an operator. For example, any two operators that are simultaneously diagonalizable are, by definition, operators that share the same set of eigenvectors. Moreover, an operator (and its matrix representation) can be defined by its action on basis vectors. You say that your eigenfunctions are orthogonal, so they automatically form a basis for your vector space. Thus, you can define an operator $T$ that acts on each function by $T(f_i) = c_if_i$, where $c_i \in F$ is a constant from the field over which your vector space is defined.
| {
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Complex semi-definite programming I'm doing some calculations and I want to simulate them in python or matlab (or whatever). However I use hermitian matrices and I don't really manage to find a library which enables me to calculate primal problems in complex form.
Do you know of any obvious extension to the real problem to calculate the problem
using real numbers?
The extension of taking a matrix $A = A_{R} + i A_{I}$ and then construct
$$\tilde A = \begin{pmatrix} A_R & -A_I \\ -A_I & A_R \end{pmatrix} $$
will NOT work since in general (and unfortunately in my case too) $\tilde A \neq \tilde{A}^T$, which is needed for real semi definite programs.
So the questions are:
*
*Does anyone knows of some software allowing complex semidefinite programming?
*Does anyone knows about some algorithm to implement in python to solve primal problems of semidefinite programs?
Let me know if the questions are unclear.
| You may use PICOS for Python: "PICOS is a user friendly interface to several conic and integer programming solvers, very much like YALMIP under MATLAB."
Since the version 1.0.1, it is possible to do complex semidefinite programming with Picos:
http://picos.zib.de/v101dev/complex.html
| {
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Which way does the scale tip? I found the problem described in the attached picture on the internet. In the comment sections there were two opposing solutions. So it made me wonder which of those would be the actual solution.
So basically the question would be the following. Assume we would have two identical beakers, filled with the same amount of the same liquid, lets say water. In the left beaker a ping pong ball would be attached to the bottom of the beaker with a string and above the right beaker a steel ball of the same size (volume) as the ping pong ball would be hung by a string, submerging the steel ball in the water as shown in the picture. If both beakers would be put on to a scale, what side would tip?
According to the internet either of the following answers was believed to be the solution.
*
*The left side would tip down, because the ping pong ball and the cord add mass to the left side, since they are actually connected to the system.
*The right side would tip down, because of buoyancy of the water on the steel ball pushing the steel ball up and the scale down.
Now what would the solution be according to physics?
| The weight on the left bowl would be the weight of the water plus vase plus ping-pong ball (plus thread, ignored).
The weight on the right bowl would be the weight of the water plus vase plus the buoyancy of the steel ball (plus the buoyancy of the submerged thread, ignored). That buoyancy is the weight of an equivalent volume of water.
Since the ping-pong ball is lighter than water, the scale will tip to the right.
Why is that the weight on the right bowl? Look at it this way: the ball is in equilibrium, so the sum of all forces on it will be 0. These forces are weight, tension on the thread and buoyancy. So the tension on the thread is $tension = ball - buoyancy$ (obvious?). And the weight on the right plate is the sum of all weights minus the tension on the thread. That is $water + vase + ball - tension$, which is the same as $water + vase + buoyancy$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/130688",
"timestamp": "2023-03-29T00:00:00",
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Is our universe an emulation? I was watching one of Neil Degrasse Tyson talks and there was a scientist (can't recall his name sorry) who was talking about a recent discovery:
"Doubly-even self-dual linear binary error-correcting block code" has been discovered embedded within the equations of superstring theory.
Is this for real? Does it imply that our universe just a sophisticated emulation running on some supercomputer?
EDIT:
Would it be possible to set up an experiment that would be able to test this hypothesis?
I am thinking about World of Warcraft, for example, how would an elf in WoW test if his world is an emulation or not? Is it even possible?
| As Count Iblis pointed out, The Church–Turing–Deutsch principle makes this impossible to decide using the structure of the laws of physics as it will always be compatible with the universe being simulated by a quantum computer.
Nevertheless, in this well-known paper the author argues that if we accept some very reasonable assumptions, then is is almost certain that we live in a simulation.
Therefore, if for any reason you are uncomfortable believing that we will live in a simulation, then you will have to either:
i)Reject the Church–Turing–Deutsch principle, (This will allow you to refute the claim by directly proposing an experiment that proves otherwise,)
ii)reject one or more or the very reasonable assumptions in the paper or
iii)find some fault in the reasoning in the paper.
In any case, this whole story provides plenty of food for thought...
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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How does a fixed amount of transmitted radio energy supply an unknown number of destinations? I did some maths and physics up to the age of 18, and hold an amateur radio licence. This thing has puzzled me for a while - does reception of an electromagnetic wave imply an interaction with the transmitter? Does it drain some of the transmitter's energy?
| A wavefront (your signal) has a fixed amount of energy given to it by the transmitter. Whatever happens to the wave once it leaves the transmitter is independent of the transmitter, thus receiving a signal does not drain any additional energy from the transmitter (though it can drain energy from the wavefront itself).
EDIT: As pointed out by @Alfred Centauri in the comments, the transmitter would be affected if the receiver was in the near field. For amateur radio purposes, the near field ceases to exist well within 200 meters of the transmitter (for the vast majority of cases), thus it is unlikely that anyone "tuning in" to your broadcast would be directly affecting your transmitter.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/130877",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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