Q stringlengths 18 13.7k | A stringlengths 1 16.1k | meta dict |
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Why is the surface integral ignored when deriving the current equation from Ginzburg-Landau theory? The Ginzburg-Landau free energy is defined as
$$F = \int dV \left \{\alpha |\psi|^2 + \frac{\beta}{2}|\psi|^4 + \frac{1}{2m^*} \left| \left(\frac{\hbar}{i}\nabla - \frac{e^*}{c}\vec{A}\right)\psi \right|^2 + \frac{h^2}{8\pi}\right \},$$
where $\vec{h}=\nabla\times \vec{A}$. The current equation is derived by taking the variation of the vector field $\vec{A}$. My concern is mainly about the variation of the last term in the free energy:
$$\delta \int dV\,\left(\nabla\times \vec{A}\right)^2 = 2\oint_S \left(\delta \vec{A} \times \vec{h}\right)\cdot d\vec{S} + 2\int dV \,\delta\vec{A}\cdot \left(\nabla\times \vec{h}\right).$$
However, in arriving at the final current equation, usually the surface integral is ignored. Why is that so?
In the original paper of Ginzburg (above equation 9), he said:
... by varying this expression with respect to $\psi^*$ and $A$, we
obtain the equations defining $\psi$ and $A$ (it must be assumed that
${\rm div} A = 0$)....
Is he implying that the surface term is ignored because $\nabla\cdot \vec{A}=0$? If so, why?
I can also think of other two possibilities:
*
*We manually require $\delta\vec{A}$ to vanish at surface boundary. If so, why is a similar requirement not applied to $\delta \psi^*$?
*We can also ignore the surface integral by an appropriate choice of gauge, which is complicated but exist.
What is your opinion?
| Short answer: the key is that $\delta \vec{A}$ is arbitrary, so you can set it to whatever you need.
In the equation
$$\delta \int dV\,\left(\nabla\times \vec{A}\right)^2 = 2\oint_S \left(\delta \vec{A} \times \vec{h}\right)\cdot d\vec{S} + 2\int dV \,\delta\vec{A}\cdot \left(\nabla\times \vec{h}\right),$$
you get rid of the surface term in the same breath as you get rid of $\int dV$ to pass from an integral expression to an expression valid point by point:
$\delta \vec{A}$ is arbitrary, so you set it equal to zero everywhere apart from an infinitesimal volume $dV$ around an arbitrary point $x$.
| {
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Law of refraction In my textbook it is given that "The refractive index of a substance does not depend on the angle of incidence"
But
Refreactive index =sine of angle of incidence/sine of angle of refraction
It is clear from above relation that refractive index depends on angle of incidence.Where am I wrong?
| Both the equation and the statement are right. You just misunderstood the meaning. Actually, the refractive index of a medium does not depend on the angle of incidence. It remains constant. That is, when the angle of incidence changes the angle of refraction also changes and adjusts. Thus the refractive index remains same whatever the angle of incidence be.
| {
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How was the Kelvin Scale created? How was it found that if we go $\rm 273°C$ below $0$, we would reach absolute zero. What experiment gave this result.
Charles' law states that :
Volume is directly proportional to the absolute temperature.
Was the Kelvin perhaps created as scale in which temperature and pressure of a gas are proportional?
I tried to find absolute zero in this manner with experimental data from this video.
This is my work:
$v_1$ $= \rm 112\ mL$
$t_1$ $= \rm 23°C$
$v_2$ $= \rm 138 \ mL$
$t_2$ $= \rm 100°C$
Let's define $x$ as the quantity to add to °C measurement's to satisfy Charles' law. In other words $\rm -x°C$ would be absolute zero.
\begin{equation}
\frac{v_1}{v_2} = \frac{x + t_1}{x + t_2} \\
\frac{112}{138} = \frac{x + 23}{x + 100} \\
26x = 112×100 - 138×23 \\
x = 308.69
\end{equation}
I also tried other values but I am not getting the desired answer. Is there a flaw in my reasoning or does this have nothing to do with the Kelvin scale at all? If so, then how was the Kelvin scale formulated.
| You reasoning and method is totally correct. Your final answer is almost correct. The problem is that you are looking at high temperatures and trying to find a number which is far away. This means that a little rounding in your numbers will have a huge impact on your answer. For example, if you try these numbers,
v1 = 111.1mL,
t1 = 23.9°C,
v2 = 138.9mL,
t2 = 99.1°C,
You will end up with 276.629 which is much closer to absolute zero.
| {
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Am I misunderstanding something, or are these Wikipedia statements about quantum tunneling wrong? Badly stated? From https://en.wikipedia.org/wiki/Quantum_tunnelling#Introduction_to_the_concept :
The reason for this difference comes from treating matter as having properties of waves and particles. One interpretation of this duality involves the Heisenberg uncertainty principle, which defines a limit on how precisely the position and the momentum of a particle can be simultaneously known.[7] This implies that no solutions have a probability of exactly zero (or one), though it may approach infinity. If, for example, the calculation for its position was taken as a probability of 1, its speed, would have to be infinity (an impossibility). Hence, the probability of a given particle's existence on the opposite side of an intervening barrier is non-zero, and such particles will appear on the 'other' (a semantically difficult word in this instance) side in proportion to this probability.
(Emphasis added.)
The bolded part doesn't make sense to me. I don't see how a probability can meaningfully be said to approach anything above one, let alone infinity. And regarding the latter bolded sentence, Heisenberg's uncertainty principle describes knowledge of position and momentum (not velocity) as complementary, so before that (seemingly misplaced) last bolded comma, should that say "momentum", not "speed" (with infinite momentum implying an impossible speed of c)?
| EDIT: this was a sloppy answer, well deserving of a downvote. I had the right idea, but I should've included the words "in any non-zero size region" - because of course the wavefunction can easily be zero at discrete points. I'll leave this bad answer as is, because I wrote a better answer (to redeem myself :-)).
They were trying to say that the wavefunction is nowhere exactly 0, though it may approach 0 at infinity.
Of course, fixing that sentence will still leave the passage not very well written.
| {
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Difference in the equlibrium of the buckets Let there be a bucket of mass on earth resting on floor with water reaching the brim. And now consider another identical bucket with water till brim in outer space with no gravitational influence, and also a 3rd identical bucket with water till brim falling freely near Earth's surface. Here, as we know according to newton's laws, 1'st and 2'nd bucket are in equilibrium as they have no net force acting on them. And the 3rd bucket is accelerating at $g$ in $m/s^2$.
But we tend to say 2'nd and 3'rd buckets are equvalently in identical conditions (saying this because pressure difference pattern in 2'nd and 3'rd buckets is identical).
What is this supposed to be?
| Not all equilibria are equivalent.
In the first case, while the net force on the system is $0$, this does not mean there are no forces at all. For the first bucket on Earth's surface the fluid pressure changes as you go deeper into the bucket due to the weight of the fluid above the point in question.
In the second and third cases there is no fluid pressure gradient as you travel down the bucket. This is because fluid higher up is not pushing down on fluid below it. And this is how the second and third cases are equivalent.
| {
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Why do Hopping Hamiltonians have physical significance? When studying quantum systems, we often use "hopping" Hamiltonians to represent the system energy. For example, we might represent a 1D ring of N sites with a single particle as:
$H = -t\left[\sum_{i}^{N-1}(\sigma^+_i\sigma^-_{i+1}+\sigma^-_i\sigma^+_{i+1}+)+\sigma^+_L\sigma^-_1+\sigma^-_L\sigma^+_1\right]$
Or a graphene lattice as:
$H=-t\sum_{i, j}\left(a^\dagger_ib_j+b_j^\dagger a_i\right)$
I'm struggling to understand how a series of hopping operators can produce the energy of a system. Any insights will be greatly appreciated!
| The inspiration for the hopping Hamiltonians are the tight-binding Hamiltonians in condensed matter physics (hint - this is where one should look for derivations). One typically solves such a Hamiltonian in terms of plane waves, expands the dispersion near the band bottom (effective mass approximation) and thus obtains the usual kinetic term $\hbar^2k^2/(2m)$ (although one may prefer expanding the dispersion in other points - typically near the Fermi level - e.g., in graphene or one treating Luttinger liquids.)
The Hamiltonian is usually called hopping rather than tight-binding in the contexts where the underlying lattice is fictitious rather than real, i.e., where the form of the Hamiltonian is a matter of computational convenience rather than an approximation to real lattice.
Where the real difficulty lies is that such Hamiltonians cannot be rigorously obtained from the expression for the true energy/potential. One typically considers sites as isolated atomic orbitals and postulates ad-hoc coupling between them. The coupling parameters are then estimated experimentally or using more sophisticated band structure calculation techniques.
Another term used for this type of Hamiltonians is transfer Hamiltonian - typically in a context where discrete states are coupled via tunneling to continuous energy bands.
| {
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Formula for centripetal acceleration: simple proof that does not use calculus? I teach physics to 16-year-old students who do not know calculus and the derivates. When I explain the formula for centripetal acceleration in circular uniform motion, I use this picture:
Here,
$$\vec{a}_{\text{av}}=\frac{\Delta \vec{v}}{\Delta t}=\frac{\vec{v}_2-\vec{v}_1}{\Delta t}$$
and
$$\vec{v}_1=(v\cos\phi){\bf \hat x}+(v\sin\phi){\bf \hat y}, \quad \vec{v}_2=(v\cos\phi){\bf \hat x}+(-v\sin\phi){\bf \hat y}.$$
Combining these equations gives
$$\vec{a}_{\text{av}}=\frac{\Delta \vec{v}}{\Delta t}=\frac{-2v\sin\phi}{\Delta t}{\bf \hat y}, \tag 1$$
which shows that the average acceleration is towards the center of the circle.
Using $\Delta t=d/v=2r\phi/v$, where $d$ is the distance along the curve between points $1$ and $2$, gives
$$\vec{a}_{\text{av}}=-\frac{v^2}{r}\left(\frac{\sin \phi}{\phi}\right){\bf \hat y}.$$
As $\phi\to 0$, $\sin \phi/\phi\to 1$, so
$$\vec{a}_{\text{cp}}=-\frac{v^2}{r}{\bf \hat y}, \tag 2$$
which shows that the centripetal acceleration is towards the center of the circle.
Does there exist another simple proof of Equation $(2)$, in particular, that the centripetal acceleration is towards the center of the circle?
|
I think this is easy to understand:
with $~\Delta v=v\,\Delta\varphi$
$$a=\frac{\Delta v}{\Delta t}=v\,\frac{\Delta\varphi}{\Delta t}=v\,\omega$$
and
$$ v=\frac{\Delta s}{\Delta t}=
\frac{R\,\Delta\varphi}{\Delta t}=R\,\omega$$
where $~\frac{\Delta\varphi}{\Delta t}=\omega=~$constant and $\Delta s~$ is the line element on the circle.
thus :
$$a=R\,\omega^2=\frac{v^2}{R}$$
| {
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How to calculate radiation length of muon from radiation of electron $X_o=37 \rm gm/cm^2$? I want to calculate the radiation length of muon. I understand I have to use formula of bremsstrahlung because it is a particle but I do not understand how
| There is a section about Passage of Particles Through Matter in the Particle Data Group review, where you may find something (even though it is very technical). But according to this quote from there, it may be irrelevant to define the radiation length of muons:
Radiative effects dominate the energy loss of energetic muons found in
cosmic rays or produced at the newest accelerators. These processes
are characterized by small cross sections, hard spectra, large energy
fluctuations, and the associated generation of electromagnetic and
(in the case of photonuclear interactions) hadronic showers.
As a consequence, at these energies the treatment of energy loss as a
uniform and continuous process is for many purposes inadequate.
| {
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Why is it impossible to measure position and momentum at the same time with arbitrary precision? I'm aware of the uncertainty principle that doesn't allow $\Delta x$ and $\Delta p$ to be both arbitrarily close to zero. I understand this by looking at the wave function and seeing that if one is sharply peeked its fourier transform will be wide.
But how does this stop one from measuring both position and momentum at the same time?
I've googled this question, but all I found were explantions using the 'Observer effect'. I'm not sure, but I think this effect is very different from the intrinsic uncertainty principle.
So what stops us from measuring both position and momentum with arbitrairy precision?
Does a quantum system always have to change when observerd? Or does it have to do with the uncertainty principle?
Thank you in advance.
| To measure speed, you measure time between two positions. Once you have a speed determined, which of the two positions would you associate with it simultaneously? You can't rightly do it with either position. To associate it with the average of the two positions would require that you assume constant velocity but you only measured the average velocity between the two positions and have no way to know if it was constant between them.
| {
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Why is it easier to handle a cup upside down on the finger tip? If I try to handle a tumbler or cup on my fingertip (as shown in fig), it is quite hard to do so (and the cup falls most often).
And when I did the same experiment but this time the cup is upside down (as shown in fig), it was quite stable and I could handle it easily.
In both the cases, the normal force as well as the weight of that cup is the same but in first case it falls down and in the other it is stable.
I guess that it is falling because of some torque but why is there no torque when it is upside down.
What is the reason behind this?
| Take a look at this picture of a cup slightly out-of-balance :
In case (A), generated torque is directed out of your reference axis and in case (B) - towards your reference axis. So in case A), you need to compensate out of balance movement with your finger contra-movement. But in case B),
torque assists you and makes balancing for yourself, so that you need minuscule additional efforts.
| {
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How do I prove that an electron beam has a plane wave function? I have been told that an electron beam has a wave function equivalent to a plane wave $\psi(x) = Ae^{ikx}$, however I would like to know why? Also, if an electron beam can be shown to have a wave function $\psi(x) = Ae^{ikx}$ how do we reconcile this with the fact that this function can not be normalized?
I would like to stress this is not a homework problem, I just genuinely would like to understand why this is the case.
| The plane wave is indeed an idealized approximation. As it was correctly addressed by Thomas Fritsch, in many experiments the electrons in the beam should have a well defined momentum. He also answered the question on normalizablity. The only thing I want to add is a response to the first question on why a plane wave: Even if you start with a relatively localized electron wave packets, as they accelerate through the electric field in a cathode ray tube, the wave packets spread very quickly so that within a very short time you will have a widely dispersed wave packet, which can be approximated as a plane wave.
| {
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Please explain the meaning of below statement Newtons second law is a local law.
(In the book,it says that it means that it applies to a particle at a particular instant without taking into consideration any history of the particle or its motion.)
Um, I couldn't understand what do they mean by " taking into consideration any history of the particle or its motion ".
If possible ,please explain it with an example.
| The second law states that the rate of change of momentum of a body is directly proportional to the force applied, and this change in momentum takes place in the direction of the applied force.
$$\mathbf{F}=\frac{d\mathbf{p}}{dt}=m\frac{d\mathbf{v}}{dt}=m\frac{d^2\mathbf{r}}{dt^2}$$
If we consider mass to be constant.
As you can see it's an ordinary differential equation. Now differential equation are local is the following sense
Suppose one has the following differential equation with some initial condition
$$y'=y, \ \ \ \ \ \ \ \ \ y(0)=1$$
It's local in the sense that if you know the value of $y$ at time $t=0$, that's $1$ in this case, You can immediately what's the value of $y$ at time $t=dt$ as
$$dy=ydt$$
$$y(0+dt)=y(0)+dy=y(0)+y(0)dt$$
That's what meant to be local. Differential equation required only one(for first order) initial condition to determine the complete trajectory.
| {
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Why is acceleration directed inward when an object rotates in a circle? Somebody (in a video about physics) said that acceleration goes in if you would rotate a ball on a rope around yourself.
The other man (ex Navy SEAL, on YouTube too) said that obviously it goes out, because if you release the ball, it's going to fly in outward direction.
Then somebody said that the second man doesn't know physics; acceleration goes in.
| Let's consider an everyday example: Driving a car or a bike. If you drive on a straight line at constant speed you do not experience any force. That's boring (not part of your question), so let's drive in a circle. If we drive in a circle in the counter-clock-wise direction, we are constantly turning to the left.
However, in order to move to the left we must experience a force, which is pushing/pulling us to the left. Hence, taking this perspective it becomes clear that the force we are experiencing must be directed inwards, to the center of the circle.
The situation in reversed if we take the perspective of being the inwards pulling force. So if we have a mass on a string and we rotate it in a circle, the mass becomes the car/bike of the former story and we take the role of the inwards pulling force. Since the mass experience an inwards pulling force, and since any force must be balanced (see Newtons law), we must experience an outwards pushing force.
Hence, whether we experience a force with is inwards or outwards directed depends on the role we play. Hope this helps.
| {
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Can Compton scattering generate entangled photons and electrons? Since the Compton scattering process is where a high energy photon scatters off an electron, would it be a plausible result for the energies of the scattered photon and the scattered electron to be entangled?
| The photon and electron will definitely be entangled after scattering.
This is easy to see conceptually in two ways. First way: we can view this process as measuring the photon's position, with the electron playing the role of the apparatus. As usual, after the measurement, the photon and the apparatus will be entangled.
Second way: consider running a million experiments where one billiard ball scatters off another. The final momenta and energies of the balls will be distributed differently depending on the exact place where the balls hit. So there will be a classical correlation between the two balls.
Now replace this classical ensemble with a quantum superposition, and the classical correlation becomes the entanglement. What do I mean by that? Take those million initial states from the previous paragraph and consider a quantum superposition of all of them. That's the analog of the initial state in Compton scattering, since both the photon and electron are in a superposition of different positions. Now, of course time evolution in quantum mechanics is linear, so the final state of the billiard balls will be a superposition of all the million final states, i.e. an entangled state. (Of course you don't really need a million to get entanglement, even just two would suffice.)
| {
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Is there an accepted Lagrangian for the transport equation? Perhaps because it is so simple, I have not seen a lagrangian form of the transport equation
$$(\partial_t + a \partial_x)q = 0.$$
This equation is first order, which makes obtaining it from the Euler-Lagrange equation a bit tricky.
It would appear that the lagrangian $\frac{m}{2}q_{t}^2 + \frac{T}{2}q_{xt}$ yields $q_{tt} + \frac{T}{m}q_{xt}=0$, which in turn yields the transport equation with constant forcing
$$q_t + \frac{T}{m}q_x + c =0$$
after integrating with respect to $t$ (the constant c comes from here).
Alternatively we might consider the lagrangian $q^2(q_{tt} + q_{xx}) + q(q_t^2 + q_x^2)$, which via the Euler-Lagrange equations yields
$$-q_t^2 - q_x^2 = 0 $$
which factors into the two transport equations describing motion in opposite directions.
Is there some classic approach or equation that I am missing?
| I don't know of any simple way to get the equation $(\partial_t+c\partial_t)\phi=0$ from an action principle.
I belive that the best one can do is to use the action integral for a chiral boson field, which is usually taken to be
$$
S= \frac 12 \int dx dt\left( \pm \partial_x \phi \partial_t \phi - c(\partial_x \phi)^2\right).
$$ From this we get
$$
\delta S= \int \delta \phi(x,t) \left( \mp \partial^2_{xt} \phi + c\partial^2_{xx} \phi\right)dxdt,
$$
which gives
$$
\partial_x(\partial_t\pm c\partial x)\phi=0.
$$
The latter equation has solution $\phi(x,t) = f(x\mp ct)+h(t)$ where $f$ and $h$ are arbitrary. The function $h(t)$ is a kind of gauge ambiguity, and in the theory of chiral bosons physical variables are required to be things like $\partial_x\phi $ which are insensitive to $h(t)$.
| {
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Variable Exchange $F(t) \to F(x)$? I want the value of the work done by drag I got a nonlinear equation, which describes the magnitude of a force as a function of time, but I don't know how to calculate the work done by the force. Given:
$$F(t) = kv(t)$$
with
$$v(t)= \left(1-\frac{k}{m}\right)^t v_{0}$$
and I want to derive:
$$W = \int F(t) dx$$
and also want to convert $F(t)$ to $F(x)$
| $$ \begin{align}F(t)&=\space kv(t) \\
v&= \frac{dx}{dt}=(1-\large{\tfrac{k}{m}})^{\small{t}}\, v_0\\
{\rm d}x&=(1-{\tfrac{k}{m}})^{\small{t}} v_0 {\rm d}t\\
\int_0^x {\rm d}x&= \int_0^t (1-{\tfrac{k}{m}})^{\small{t}}\, v_0 {\rm d}t\\
x&=\frac{(1-{\frac{k}{m}})^{\small{t}}-1}{\ln(1-{\frac{k}{m}})} v_0\\
(1-{\tfrac{k}{m}})^{\small{t}} v_0&= v(t)= v_0 + x \space \ln(1-{\tfrac{k}{m}})\\ \end{align}$$
Substituting the values in the expression for $F(t)$ $$ \begin{aligned}
F(x)=k \left[v_0 + x \space \ln(1-{\tfrac{k}{m}})\right]
\end{aligned}$$
I hope you can calculate forward.Regards.
| {
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Flow through parallel pipes with filters present I have a system where water flow is split into two pipes, A and B. Each of A and B have a filter in the pipe. After the filter, the pipes recombine to outlet. The filters clog at different rates.
I'm very confused as to how the Bernoulli equation will relate the pressure build up at one filter as it clogs with a reduction in relative flow.
Any help would be massively appreciated.
| Pick three points on the systems. point A before the pipe splits. point B in the with weak filter and point C in the strong filter.
$$B_i = P_i + \rho g h_i + \frac{1}{2} \rho v_i^2 - P_i(v)$$
where $P_i(v)$ represents the effect of the filter. It may or may not be a function of velocity. Now we have,
$$B_A = B_B + B_C$$
and the continuity equation,
$$Q_A = Q_B + Q_C$$
If you know the effect of the filter, you should be able to solve this problem now.
| {
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Euler angles - geometry
Hello, I don't quite see why should the angle between $\hat{\dot{\theta}}$ and the projection of $\hat{\dot{\phi}}$ onto the $x_0$, $y_0$ plane be a right angle. Does it have something to do with pure geometry or physics? Of course it gives the right answer for the angular velocity as a function of the Euler angles, but I don't see why it is necessarily a right angle.
| They don't have to be, but convention has it that consecutive rotations in Euler angle schemes be orthogonal to each other.
The orientation is set by a sequence of three rotations, each being perpendicular to the previous rotation such that the influence of one angle to all the other rotations is zero.
And since the sequence $\mathrm{R} = \mathrm{R}_z \mathrm{R}_x \mathrm{R}_y$ consists of three mutually orthogonal rotations, the resulting rotational speed is a sequence of mutually orthogonal relative rotations
$$ \vec{\omega} = \hat{z}_0 \dot{\psi} + \mathrm{R}_z \left( \hat{x}_1 \dot{\theta} + \mathrm{R}_x (\hat{y}_2 \dot{\phi}) \right) $$
In this case, $\hat{z}_0$ is perpendicular to $\mathrm{R}_z \hat{x}_1$, since $\hat{z}_0$ is perpendicular to $\hat{x}_0$ and the rotation $\mathrm{R}_z$ does not change that. Also, $\hat{x}_1$ is perpendicular to $\mathrm{R}_x \hat{y}_2$, since $\hat{x}_1$ is perpendicular to $\hat{y}_1$ and the rotation $\mathrm{R}_x$ does not change that also.
| {
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How does the difference in energy levels decrease as you increase $n$ in 1D particle in a box (correspondence principle)? I found from multiple sources that the difference between energy levels must decrease as you increase quantum numbers in order to follow the correspondence principle. The energy of a particle in the box increases proportionally to $n$-squared. This means that the energy spacings increase at high values of $n$. How does the correspondence principle is applicable with respect to energy level differences?
| The idea that the difference between energy levels must decrease as you increase the quantum number in order to recover the correspondence principle must be wrong: if it were true, then the quantum harmonic oscillator could not exist in the physical world. What we may require is the fact that the relative energy difference must decrease as a function of $n$. Relative to what? Well, relative to the macroscopic energy of the system $E(n)$. We therefore postulate:
$$
\lim_{n\rightarrow\infty} \Delta_r(n)=\lim_{n\rightarrow\infty} \frac{E(n+1)-E(n)}{E(n)}=0.
$$
If this is true, then for high $n$ (classical limit) increasing the energy quantum number does not produce an appreciable energy difference, at least if compared to the macroscopic energy $E(n)$ we experience. In other words, we lose track of the discreteness of the energy levels of $E(n)$.
Note that for the case of the particle in a 1D box we have $E(n)=\frac{n^2\pi^2\hbar^2}{2mL^2}$, where $m$ is the particle mass and $L$ the box length. Hence, $\Delta_r(n)=(2n+1)/n^2$, which correctly vanishes when $n$ goes to infinity. In contrast, a physical system with energy dependence $E(n)= E_0 e^n$ would display quantum effects in the macroscopic world, since $\Delta_r(n)=e-1$. As far as I know, however, we still haven't found such a quantum system and the classical limit is always recovered.
| {
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Magnetic dipole moment of volume current derivation The multipole expansion of the magnetic potential yields the dipole term :
$$\mathbf A_{dipole}(\mathbf{r}) =\frac{\mu_0}{4\pi r^{2}}\int (\hat {\mathbf r} \cdot {\mathbf r}') \mathbf J {dV}'$$
How do I get from this expression to the final expression :
$$\mathbf A_{dipole}(\mathbf{r}) =\frac{\mu_0}{4\pi r^{2}}\left (\frac{1}{2}\int ( {\mathbf r}'\times \mathbf J) \ {dV}'\right) \times \hat{\mathbf r} $$
Thanks in advance.
| Consider the following identity, that can be obtained with the BAC-CAB-rule:
$$
(\mathbf{r'} \times \mathbf{J}) \times \mathbf{\hat r} = -\mathbf{\hat r} \times (\mathbf{r'} \times \mathbf{J})=\mathbf{J} (\mathbf{r'} \cdot {\mathbf{\hat r}})
-\mathbf{r'} (\mathbf{\hat r} \cdot \mathbf{J}).
$$
We will show, that the two expressions on the right yield the same integral:
$$
\int dV' (\mathbf{r'} \times \mathbf{J}) \times \mathbf{\hat r} =
\int dV' [\mathbf{J} (\mathbf{r'} \cdot \mathbf{\hat r})
-\mathbf{r'} (\mathbf{\hat r} \cdot \mathbf{J})]=2\int dV' \mathbf{J} (\mathbf{r'} \cdot \mathbf{\hat r})
$$
(which would give you your second equation)
Proof: First show that the volume integral of the product of the current density with any gradient $\mathbf{\nabla'} f$ is $0$. Note that the multipole expansion assumes that the current density $\mathbf{J}(\mathbf{r'})$ is localized $(r'<<r)$ and considers a magnetostatic szenario (continuity equation yields: $0=\partial_t \rho=-\mathbf{\nabla'}\mathbf{J}$ (1)), for a magnetic field far away from the origin (where $\mathbf{J}$ is localized and the only charges $\rho$ are). Since $\mathbf{J}$ is localized at the origin we can choose a Volume $V$ big enough, such that the integration over the surface $\partial V$ of $\mathbf{J}$ yields $0$ (2):
$$
\int dV' (\mathbf{J}\cdot(\mathbf{\nabla'} f))=\int dV' (\mathbf{\nabla'}\cdot (f\mathbf{J}))-\int dV' f \, (\mathbf{\nabla'}\cdot\mathbf{J})\stackrel{(1)}{=}\int dV' (\mathbf{\nabla'}\cdot(\mathbf{J}f)) \\
\stackrel{Gauß}{=} \int_{\partial V} \mathbf{dA'}\cdot\mathbf{J}f\stackrel{(2)}{=}0
$$
Now choose $f:=r'_i r'_k \Rightarrow \mathbf{\nabla'} f= r'_i \mathbf{\hat r'_k}+r'_k \mathbf{\hat r'_i}$. We get:
$$
0=\int dV' (\mathbf{J}\cdot(r'_i \mathbf{\hat r'_k}+r'_k \mathbf{\hat r'_i}))= \int dV' (r'_i J_k+r'_k J_i). (3)
$$
With this we finally get:
$$
\int dV' (-\mathbf{r'}(\mathbf{\hat r}\cdot\mathbf{J}))_i=\sum_k \int dV' (-r'_i \hat r_k J_k)\stackrel{(3)}{=} \sum_k \int dV' (r'_k \hat r_k J_i)=\int dV' (\mathbf{J}(\mathbf{\hat r}\cdot\mathbf{r'}))_i,
$$
which concludes the proof. $\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \square$
| {
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Is High-Fidelity Quantum-Entanglement Data-Transfer Real? I was discussing quantum entanglement with a friend and explaining that faster than the speed of light data transfer isn't a reality with our current understanding of physics. He brought up quantum entanglement and I explained that while they are entangled over a distance and in a sense are "exchanging data" (eh) faster than the speed of light, you can't really exchange custom data faster than light because trying to impose a state change on the particles would break the entanglement.
He later sent me this journal entry in PRX Quantum which boasts faster than light data transfer with 90% fidelity via quantum entanglement. Is my understanding of how entanglement works way off? Is this journal being interpreted out of context? Or is there something else entirely that's gone awry?
| The paper is about quantum teleportation, which is a process enabled by entanglement. However, it does not provide faster-than-light communication. The complete process of teleportation always needs a conventional communication link without which the process cannot be completed. In the theoretical description of entanglement, the outcome of the joint measurement needs to be communicated to the person receiving the final state to know which unitary operation to perform on that state to produce the original state. In practical implementations of teleportation, such a conventional communication link is used to register coincidences in detections.
So, one can emphatically state that faster-than-light communication is not possible, not even with entangled states. There is a theorem, called the no-signaling theorem, that shows that entangled states do not allow faster-than-light communication.
| {
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Measuring the correct mass or weight of an object in the presence of atmospheric pressure? I know due to atmospheric pressure the weight of an object increases, but when we take measurements we do not omit the weight of the column of air above it. So, doesn't it affect the accuracy of our calculations?
|
I know due to atmospheric pressure the weight of
an object increases, but when we take measurements
we do not omit the weight of the column of air above it.
No, it is actually the other way round.
The weight of an object decreases due to the
buoyant force from the displaced air.
According to Archimedes' principle the buoyant force
is equal to the weight of the displaced air.
The buoyant force points upward because the
air pressure is larger at the bottom of the object
than at the top of the object.
The density of air is around $1.2$ gram/liter.
So the difference is actually not difficult to measure,
for example by weighing a solid container filled with air,
and comparing it to the same container emptied by a vacuum pump.
| {
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How does a capacitor get charged instantly in AC whereas it takes infinite time in DC? I have seen when a capacitor is connected to a dc source it takes infinte time to charge, but when connected to ac it takes the potential of the source instantly,
probably the approach in the books is not adequate, please clarify,
Here is a link that mentions the time constant for DC https://www.electronics-tutorials.ws/accircuits/ac-capacitance.html
And here is one that describes the AC https://physicscatalyst.com/elecmagnetism/growth-and-delay-charge-R-C-circuit.php
| Regarding the transient currents (transient regime actually), it is there when you turn on an AC circuit that was off. I mean when you switch on the power. What you see in most description is the solution after the transients have "died" but the complete solution includes both. In most introductory physics we don't see this mentioned but electrical engineering books should have it. If you had AC circuit with very large inductance you could actually see the delay until the non-transient regime takes over.
| {
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How can information be communicated with entanglement? Forgive my overly simplistic view of this. But I have been wondering this for years. Any insight would be greatly appreciated.
So imagine you have two entangled particles, and they are separated and put in box A and box B. Box A is transported across the universe and Box B is left where it is. When someone opens Box B (measures Box B) and say for argument, the particle is measured as spin up, they immediately know that the particle in Box A is spin down.
At a glance this seems like they instantaneously "know" something about the other box, and hence seems like they have communicated instantaneously. But the state of the particle in Box B would have a 50/50 chance of being spin up or spin down.
So to me it seems as if there is nothing deterministic about this system. Every few days I read quasi-bogus pop sci articles indicating that scientists have successfully "teleported" particles or communicated faster than the speed of light.
It seems to me there is no way for information to be actually communicated (deterministically) faster than the speed of light. So how can we engineer anything with entanglement?
| You are right. Bob can only determine the state of his particle (without measuring it) once he receives information from Alice about the outcome of her measurement - and this information can be transmitted no faster than the speed of light.
Similarly, quantum “teleportation” requires Bob to receive information from Alice, transmitted no faster than the speed of light, which he can then use to reproduce the pre-measurement state of Alice’s particle. And this does not contradict the no-cloning theorem because Alice must make a measurement on her particle in order to provide this information, which changes the state of her particle.
| {
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Lorentz Transformations for Polar coordinates or Inertial Frame in Polar Coordinates Do polar coordinates define an inertial frame or not?
Everywhere in GR, the authors of all the books talk about bring the metric to diag(-1, 1,1,1) which would show that a Local Inertial Frame exists at each point on the manifold. And the coordinate in this local frame would be $(t, x, y, z)$.
But can't Polar coordinates define an inertial frame. Can't the metric be brought to \begin{equation} diag(-1, 1, r^2, r^2 \sin^2(\theta))\end{equation} at every point. And the frame attached to the point is still inertial but just the coordinates used will be spherical instead of cartesian. Is this wrong.
In fact books begin constricting inertial frames using perpendicular rods and this cartesian coordinates. Can't we construct inertial frame using polar coordinates.
Edit- after an answer
Lorentz Transformations are transformations between different frames. Will a transformation from cartesian coordinate to polar coordinates be called a Lorentz transformation or called just a coordinate transformation. Now I have read that Lorentz transformations are linear. Transformation from cartesian to polar or from $(r, \theta, \phi) -> (r', \theta', \phi') $ would be Non Linear.
So will they be Lorentz Transformation.
| Polar coordinates are not inertial coordinates. Only those coordinate systems that can be reached from an inertial coordinate system by the action of the Poincaré group are inertial coordinates (it should be noted that the Lorentz transformations are a special case of the Poincaré transformations that preserve the origin).
If you want to work in polar coordinates you should be careful as objects that are Lorentz covariant are not necessarily diffeomorphism covariant which is the appropriate notion of covariance in general relativity. That being said, there is no problem working in polar coordinates in Minkowski space, you must just be careful since you are not in an inertial coordiante system and so some assumptions may not hold.
| {
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Is inertia and gravity determined by relativistic mass or invariant mass? As far as I know, mass fundamentally determines inertia and the gravitational force. But since there are two types of mass, which mass determines which? From what I have read so far, and correct me if I'm wrong, the relativistic mass determines the inertia, but not the gravitational force. Then why does one determine inertia and another determine gravity? Also since relativistic mass represents the total mass-energy of an object taking into account the kinetic energy, does that imply that the gravity is not determined by the total energy content of an object, but only by its invariant mass which doesn't take into account its kinetic energy?
| Neither inertia nor gravity is determined by either the mass $m$ or by the so-called relativistic mass $m\gamma$.
If you write Newton's second law in terms of the acceleration three-vector and the force three-vector, it looks like $F=m\gamma a_\perp+m\gamma^3 a_\parallel$. Although it's true that you can write the second law in terms of four vectors as $F=ma$, the four-vector force is not the force that any observer actually measures, and it doesn't behave the way newtonian forces behave for purposes of computing work (its inner product with the velocity is always zero).
The source of gravity is the stress-energy tensor, not a scalar such as $m$ or a single real number such as $m\gamma$.
BTW, relativistic mass is becoming deservedly extinct. It's no longer used in writing by professional physicists or in textbooks. It's only used these days in popularizations. See Oas, "On the Abuse and Use of Relativistic Mass," 2005, http://arxiv.org/abs/physics/0504110 .
| {
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Does a non-rotating satellite always show the same side towards the main body? If an initially non-rotating (zero angular momentum) satellite starts moving around a main body, such as the earth,
*
*will then the satellite continue to have zero rotation or
*will it show the same face towards the main body, thereby having angular momentum relative to the stars.
One could argue for the latter by considering the orbit around earth as a ”straight ahead” in curved space.
| Well, the fuel inside the booster tank would slosh around, creating a torque for which the satellite would slowly have to rotate to provide an anti-torque. Of course, other factors like thermal expansion, bombardment from space dust and all are not included.
| {
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Implications of massless pulleys in Atwood machines Consider the double Atwood machine below:
I understand the classic approach to a solution for the acceleration of each mass involves setting the tension in the upper string equal to twice that of the lower string to achieve equilibrium in the lower pulley. The common reasoning is that "since the pulley is massless, it must be in equilibrium otherwise it would have infinite acceleration".
While I understand this reasoning must be correct, surely if the lower pulley is in equilibrium there would be no acceleration of m1 at all, which can't be correct? Therefore, under what conditions are the subsequent accelerations derived physically useful - are they just in the limit as the mass of real massive pulleys tend to zero?
| A way to avoid confusion in analyzing motions of pulleys is to write the dynamical equations of motion sans the assumptions of low inertia, which may apply for some of the pulleys, and then neglect the masses of the pulleys. On doing so, it is clear that terms such as $m^{pulley}_i a^{pulley}_i$ and $m^{pulley}_i g$ which arise in the dynamical equations can be ignored if $m^{pulley}_i \rightarrow 0$. On noticing that applying this assumption required no additional assumption on the magnitude of $a^{pulley}_i$, it is evident that the pulley denoted by $i$ need not be in equilibrium in the analysis which includes the assumption of negligible mass. In general, such confusions are a result of conflating concepts of static analysis when conducting dynamical analysis and are readily avoided by doing the latter as a matter of policy towards as a safety measure (although it might make the analysis more time consuming than required when the system is indeed completely static, i.e. has no accelerating bodies).
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What's considered an interaction in Quantum Mechanics? Clearly I'm not an expert of QM but recently I came across the fact that particles do not have intrinsically a specific location until they interact (very ambigous term to me) with something else and so they wave function collapses into a specific location. My question is: what's considered an interaction? Doesn't particles interact with each other trough gravity and electromagnetism all the time? The effect of gravity is an interaction mediated by the deformation of space-time and so may be excluded but to me EM should make the wave function collapse every instant.
| to add to what Quantum-Collapse said:
interaction between particles (or quantum systems) is manifested in the Hamiltonian operator, and taken into account through the wave function.
for example, coloumb interaction between 2 electrons is represented in the Hamiltonian equation as a potential term, in addition to the kinetic terms of both particles:
$$H=\frac{|\textbf{p}_1|^2}{2m_e} +\frac{|\textbf{p}_2|^2}{2m_e}+\frac{Ke^2}{|\textbf{r}_1-\textbf{r}_2|}$$
with the standard definitions of momentum:
$$\hat{p}_j=-i\hbar\nabla_{(j)}$$
in such case you have to solve the Schrodinger equation for a wave function of the 2 positions (or momenta) $\psi(\textbf{r}_1,\textbf{r}_2)$
| {
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Will this spaceship collide with the star? (time dilation) I thought of the above thought experiment and arrive on 2 conflicting conclusions. I can't seem to identify the flaw in my reasoning.
Suppose there is a star 4 light years from earth that has will explode and turn into a white dwarf in 3 years (as measured in the earth frame). A spaceship travels to the star at 86% the speed of light.
According to earth's frame of reference, the journey to the star will take 4.5 years so the star will have turned into a white dwarf.
according to the ship's frame of reference, however, the journey will only take 2.25 years. Moreover, since the star is travelling relative to the ship in its own frame, the event of the star exploding will actually take 6 years. So the spaceship will collide with a white dwarf instead of a star.
I thought this had something to do with simultaneity, but I know that the events must be same in all frames of reference. My collusions imply that collisions occur between different bodies depending on the frame of reference, which can't possibly be true, can it? Where am I going wrong?
| The spaceship leaves earth.
B: The star becomes a white dwarf.
C: The spaceship reaches the star.
*In the earth frame, these have the following coordinates:
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Why don't radio waves sent by electronic devices intefere with each other? You know how phones, computers and other electronic devices that use wireless communication use radio waves to communicate?...well since almost everyone has gotten a smart phone wouldn't radio waves from different smart phones interfere with each other...well...they don't (other wise we wouldn't be able to call people)... But I want to know why radio waves from different electronic devices or from two different sources don't interfere with each other... if there are two explanations then pls tell me both of them (specially if one of them is a quantum mechanical theory)..help will be appreciated...thanx in advance
| You are basically asking, "How does all of telecommunication multiplexing work", which is a question too large to answer comprehensively, so a vignette or two will have to do.
There is frequency multiplexing on radio and television, in which various signals are transmitted a different frequencies, commonly referred to as "stations" in the former and "channels" in the latter.
Your stereo (if those still are a thing) uses "space division" multiplexing, in which the left and right signals are carried on physically separate cables from source to speaker.
Civilian GPS uses code division multiplexing. All the GPS satellites transmit at the same frequency (and bandwidth), but each satellite has a pseudo random binary code which XORs the phase of their signal. The received signal can then be correlated with pseudo random code to recover each of the satellites data simultaneously in parallel, albeit at a lower bandwidth.
Another common method is Time Division Multiplexing, whose most familiar applications are CPU sharing on "the mainframe", or worse, time-share properties, in which multiple people (signals) share a vacation house (channel).
Many other more complex and/or subtle forms of multiplexing exists, but those are the basics.
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Is the brick-concrete connection the 'weakest link' in brick made buildings? When an earthquake strikes,do buildings made of bricks and concrete break mostly on concrete-brick connection spots?If Yes does the brick on the picture prevents this from occuring?
| Brick structures fail in tension at the brick-to-mortar interface. The older the brick building is, the weaker becomes that interface, and 100 year old brick buildings are mostly held together by gravity. Earthquakes cause old brick buildings to easily collapse into low mounds of loose bricks, crushing the contents of the building.
The tongue-and-groove feature you propose makes the strength of the joint dependent on the tensile strength of the brick, which will fail readily when subjected to a suddenly-applied shock load.
The most common way to reinforce brick buildings is to use hollow bricks through which steel reinforcing rods have been threaded. The hollow spaces are then backfilled with cement to trap the steel in place. The steel takes the tension loads and prevents the structure from being shaken into rubble, and keeps the building from collapsing completely under a shock load.
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When we push a box, the box applies an opposite force on my hand, but why does my hand move with the box as I push the box if the net force of is 0? When we push a box, the box applies an opposite force on my hand, but why does my hand move with the box as I push the box if the net force of my hand is 0? What is happening?
|
Why does my hand move with the box as I push the box if the net force of my hand is $0$?
No , the net force on your hand is not $0$. It is true that the box is exerting a reaction force equal in magnitude to the applied one on your hand but while considering the motion of your hand you have to consider all the forces on it which will be evident when you draw the free body diagram exclusively of your hand.
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Correlation Functions: How can I prove this simple equation The correlation functions of the Transverse Ising Model is beautifully explained in "Quantum Ising Phases and Transitions in Transverse Ising Models" Quantum Ising Phases and Transitions in Transverse Ising Models book.
Second Edition
"https://link.springer.com/book/10.1007/978-3-642-33039-1"
But I cannot deduce equation (2.A.30), page 42, even though it seems straightforward, where:
$$
{\left\langle {\psi_0 } \right|A_iA_j \left| {\psi_0 } \right\rangle }={\left\langle {\psi_0 } \right|(\delta_{ij}-c_j^{\dagger}c_i+c_i^{\dagger}c_j)\left| {\psi_0 } \right\rangle }=\delta_{ij},
$$
where $A_l=c_l^{\dagger}+c_l$ and $c_i^{\dagger}/c$ are the fermion creation/annihilation operators.
Does anyone know where this equation came from?
| It's quite straightforward. You need the anti-commutator $[c_i,c_j^\dagger]_+ = \delta_{ij}$ (anti because they're fermionic) and then
$$A_iA_j = (c_i^\dagger + c_i)(c_j^\dagger + c_j) = c_i^\dagger c_j^\dagger + c_i^\dagger c_j + c_i c_j^\dagger + c_i c_j$$
We want to calculate the matrix element with $|\psi_0\rangle$. We have the relation that the annihilation operator annihilates the ground state ket while the creation operator annihilates the ground state bra, that is $c_i |\psi_0\rangle = 0$ and $\langle\psi_0| c_i^\dagger = 0$. Therefore, we drop the terms that have two creation and annihilation operators. What remains is
$$\langle \psi_0 | A_i A_j | \psi_0 \rangle
= \langle \psi_0 | c_i^\dagger c_j + c_i c_j^\dagger | \psi_0 \rangle
= \langle \psi_0 | c_i^\dagger c_j - c_j^\dagger c_i + [c_i,c_j^\dagger]_+ | \psi_0 \rangle.$$
We can insert the anti-commutator and realize that the other terms act with $c_i$ or $c_j$ on the ground state ket, so they vanish. That is
$$\langle \psi_0 | A_i A_j | \psi_0 \rangle
= \langle \psi_0 | \delta_{ij} + c_i^\dagger c_j - c_j^\dagger c_i | \psi_0 \rangle
= \delta_{ij}\langle \psi_0 | \psi_0 \rangle
= \delta_{ij}.$$
We used the identity
$$AB = AB + BA - BA = [A,B]_+ - BA$$
along the line.
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Quantum expressions involving Dirac delta function
I want to find the following quantum expression:
$$ \langle x|PX|x'\rangle.$$
A. If I use $X|x'\rangle = x'|x'\rangle $, I will get:
$$ \langle x|PX|x'\rangle = \langle x|Px'|x'\rangle = x'\langle x|P|x'\rangle = x'\frac{\hbar}{i}\frac{\partial}{\partial x} \langle x|x'\rangle = x'\frac{\hbar}{i} \delta'(x-x').$$
B. But If I start from the other side I will find:
$$ \langle x|PX|x'\rangle = \frac{\hbar}{i}\frac{\partial}{\partial x} \langle x|X|x'\rangle = \frac{\hbar}{i}\frac{\partial}{\partial x} (x\langle x|x'\rangle) = \frac{\hbar}{i}(\delta(x-x')+x\delta'(x-x'))$$
Also using $XP-PX = i\hbar$:
$$ \langle x|PX|x'\rangle = \langle x|XP-i\hbar|x'\rangle = \langle x|XP|x'\rangle -i\hbar\langle x|x'\rangle = x\langle x|P|x'\rangle -i\hbar\langle x|x'\rangle = \frac{\hbar}{i}(\delta(x-x')+x\delta'(x-x')).$$
I don't know which answer is correct.
| The equivalence of OP's two methods A & B follows from the following identity
$$ \{f(y)-f(x)\}~\partial_x \delta (x-y)~=~ \delta (x-y) ~f^{\prime}(x)\tag{1} $$
for the Dirac delta distribution, where $f\in C^{\infty}(\mathbb{R})$. Eq. (1) in turn follows from differentiation of the identity
$$ \{f(y)-f(x)\}~ \delta (x-y)~=~ 0.\tag{2} $$
| {
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How to compute gauge potential $A$ from the field strength $F$? Let $F=dA+A \wedge A$ be the field strength that solves vaccum Yang-Mills equation.
The question is: how to recover the gauge potential $A$? Is there any standard way? or any theorem stating the solvability? Suppose the metric is $g=g_{\mu \nu}dx^{\mu}dx^{\nu}=\eta_{ab}e^ae^b$, $e^a$ is tetrad basis.
| Even if you only consider the gauge potentials up to gauge equivalence, you still can't recover a potential from a field in the non-abelian case. Not even locally! This is called the "field copy problem" in the literature, and goes back to Wu and Yang.
There are some special cases where the potential can be recovered from the field, e.g. if the holonomy group is as big as possible on every open subset.
| {
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How is the no-cloning theorem equivalent to the fact that two non-orthogonal states cannot be distinguished between in one measurement? I have seen a few isolated proofs for each of the above claims, but do not really see their 'if and only if' interdependence.
| First, assume you are given one of two states. If you can clone, you can use unambiguous state discrimination (i.e. a measurement with 3 outcomes: (i) must be state 1, (ii) must be state 2, (iii) not sure) to make copies and measure until you get one of the two unambiguous outcomes (i) or (ii), thereby distinguishing the two states.
Conversely, if you want to clone an unknown state which is in one of two non-orthogonal states, you can use a measurement which allows to distinguish them to identify which state you have, and then make copies of it.
Whether the converse also allows to derive a general cloner for a completely unkown quantum state from the ability to just distinguish two copies: Probably not in such a straightforward fashion. On a deeper level, they both lead to violations of linearity, and then pretty much anything will be possible.
| {
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How can $1/V$ be equal to $0$ in Boyle's Law? In relation to ideal gases, Boyle's Law states that pressure is inversely proportional to volume under constant temperature. In other words,
$$P \propto 1/V$$
Below is a graph that plots pressure, $P$, against inverse volume, $1/V$.
How can $1/V$ ever equal zero? How is this possible?
| We meet this issue in many contexts in Mathematics and in Physics. I don't think you'll go far wrong by thinking of it like this...
We can consider larger and larger container volumes, that is values of $1/V$ that become smaller and smaller. So the points on the line on your graph can be plotted closer and closer to the origin. In fact as close as you care to demand, so that you can't see, even with huge magnification, that there isn't actually a point at the origin itself. The case of $1/V=0$ is the so-called 'limiting case'.
| {
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Cases where renormalisation isn't needed In quantum field theory I've came across splitting up a Lagrangian (which has an interaction part) into renormalized parts with counterterms. Are there any lagrangians for which renormalisation is not required? When do you know you need to introduce counterterms?
| Insertion of counterterms is needed, when you have certain non-vanishing diverging diagrams. For instance, the self-energy diagram in QED (Quantum Electrodynamics) for electrons leading to the mass renormalization, or polarization loop diagram for photon, which is logarithmically divergent, and renormalizes the charge.
One doesn't need counteterms definitely for the case of free theories, there is nothing to renormalize. It is a trivial example.
Less trivial case are the supersymmetric theories, where non-renormalization follows from the fact, that bosonic and fermionic contributions annihilate each other. One of the famous examples in non-renormalization of F-terms. Terms of form:
$$
\int d^4 \theta \ W (\Phi)
$$
Where $W(\Phi)$ is some superpotential of the superfield $\Phi$ do no receive perturbative corrections. This can be proven via supergraphs, but the most beautiful argument due to N.Seiberg uses the power of holomorphy - https://arxiv.org/abs/hep-th/9408013v1.
There is a nice review by S.Weinberg - https://arxiv.org/abs/hep-th/9803099v1
| {
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Where does Coulomb's constant, $k_e$, come from? I'm familiar with Coulomb's law, $\vec{F} = k_e\frac{q_1q_2}{r^2}$, and know how to apply it (having used it extensively in practice, on exams, etc). I find that the $\frac{q_1q_2}{r^2}$ "part" of Coulomb's law is at least somewhat intuitive, however I've never quite understood precisely where Coulomb's constant, $k_e$, comes from.
I know that it can be measured experimentally, of course, but where do we get that $k_e = \frac{1}{4\pi \epsilon_0}$ and, further, how to we get $\epsilon_0 = \frac{1}{\mu_oc^2}$?
I'm looking for both a "rigorous" derivation, and intuition (where possible).
| Note that Coulomb’s Law is experimentally determined, though because Maxwell’s equations describe all electromagnetic phenomena you can consider doing it this way (so you can see how the constant $k$ comes about):
Consider this equation which is one of Maxwell’s equations,
$$\oint_S \vec E \cdot \vec{dS} = \frac{Q}{\epsilon_0}$$
where this surface integral $S$ is taken over a closed sphere surrounding the charge $Q$ then
$$E (4\pi r^2) = \frac{Q}{\epsilon_0}$$
and given that the electric field $E$ is defined as the force per unit charge ie.,
$$E = \frac{F}{q}$$
then we can write
$$F = \frac{Qq}{4\pi \epsilon_0 r^2}$$
which is Coulomb’s Law.
The vacuum permitivity constant is a measure of how much a vacuum permits an electric field. Most problems we assume that there is nothing else in the space between the charges we are studying, which is why we use $\epsilon_0$. In cases where we do, we use the dielectric constant $\epsilon$ which is a measure of how much the medium permits an electric field.
When we manipulate Maxwell’s equations, we end up with the electromagnetic wave equation and as you can see in the link
$$v = \frac{1}{\sqrt{\mu \epsilon }}$$
where $v$ is the speed of an electromagnetic wave in a region with permeability $\mu$ and permitivity $\epsilon$. Of course if we are in a vacuum this becomes
$$c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$$
where $c$ is the speed of light in a vacuum.
| {
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How can there be current without charge? This might be a stupid question, but I actually think that it is not so obvious:
When solving Maxwell equations, depending on the problem usually a charge and current density are assumed.
However, how can there be any current without charge density?
I understand that in most problems where this question arises there is no net charge. So for example in a conductor where a voltage is applied across it there is no net charge, however nevertheless a current flows.
That is quite clear to me, but formally should one not have to assume the existence of charges even if the net charge vanishes.
Is this problem solved in quantum mechanics where the charge and current are manifestations of a function satisfying the Dirac equation?
|
From the example I understand that the net charge vanished but that there can be a current. But I was wondering whether this is just a good approximation and that actually one could take into account that there exist charges.
Nobody is saying that charges don't exist. If $\nabla \cdot \mathbf E = 0$, then that means that in every volume of space, there are an equal number of positive and negative charges (which could be zero or non-zero). If you consider a classical model of electrons and protons where they are simply tiny charged spheres, then this is an approximation which would break down on atomic length scales, but this would be very quickly averaged out on larger scales.
| {
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Free Body Diagram for a Body on a Smooth Inclined Plane Consider a body of mass $m$ on a smooth inclined plane with the normal force and weight labelled as shown:
Now, usually when the forces are broken into its components, we get an FBD like this:
From this we can obtain the equation:
$$ \Sigma F_\perp = 0\quad \implies\quad N = mg\cos(\theta) \tag1$$
However, I tried resolving the forces along the x and y axes and obtained a diagram like this:
But this time I got a different equation:
$$ N\cos(\theta) = mg \tag2$$
Clearly, equation (1) and (2) aren't the same. So what am I missing here?
| Nothing wrong with you analysis so far but you now need to consider what you are going to do with the horizontal $N\sin \theta$ force noting that the acceleration is going to be parallel to the slope.
| {
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Where does the law of conservation of momentum apply? Take the scenario of a snowball hitting a tree and stopping. Initially, the snowball had momentum but now neither the snowball nor tree have momentum, so momentum is lost (thus the law of conservation of momentum is violated?). Or since the tree has such a large mass, is the velocity of the tree is so small that it's hardly noticeable?
If the explanation is the latter, this wouldn't hold for a fixed object of smaller mass. So in that case, how would the law of conservation of momentum hold?
| The tree is attached to the earth. The momentum from the snowball is transferred to the tree and then distributed troughout the whole earth. Because a tree is a solid object this transfer happens almost instantaneously. The earth is so large that this tiny amount of momentum won't be noticable at all.
| {
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Can thermodynamics be considered logical? One of the laws says that heat won't flow from cool to warm and at the same time this same theory claims that there is a finite (albeit tiny) chance that it will, because there is always such a microstate.
We can also have a situation where all air molecules in the room can be found in the left side of the room and none in the right side, because it is one of the microstates therefore it can happen and the entropy will drop. So how can we say that the entropy always increases when it can decrease too sometimes?
| Physical laws behave differently for different scales. Even if we know the microscopic law of nature is Quantum Mechanics, we do not use them to build a bridge. The collective quantum phenomenon somehow results in a classical phenomenon when we have many atoms together.
It is the same with thermodynamics. A few molecules can have wild fluctuations and it is not very useful to define a temperature and describe how they will fill the box. Keep increasing the number of molecules and the situation will become entirely different. When you have $10^{23}$ molecules it is very useful to state a law that is supported by overwhelming probability. The probability is so overwhelmingly in favor of the law that no one has ever seen it fail for $10^{23}$ molecules. For a theory this is as good as it can get.
| {
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What makes a standing wave a wave? Well, this is my physics professor's question and it really made me think a lot about standing waves, realising I don't actually understand it. What makes a standing wave a wave? How could I explain it? What actually makes wave a wave?
| I like the general question without being specific.
In this sense, the simplest explanation to me is that a wave is a "thing" that transports energy. In a wave you can measure energy everywhere, but not at any time.
Waves then can be added to one another. A standing wave can then be seen like Solomon Slow explained it.
| {
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What experiment confirms $\mathbf{J}^2 = \hbar^2 j(j+1)$? I learned that if we measure the spin angular momentum of an electron in
one direction $J_z$, we get $\pm \frac{1}{2} \hbar$. But if we measure
the magnitude of the angular momentum $\mathbf{J}^2$, we should get
$\frac{3}{4} \hbar^2$. What experiment gives the latter result?
As @user1585635 notes, measuring $J_x$, $J_y$, and $J_z$ separately and
summing their squares gives $\frac{3}{4} \hbar^2$. This is not what I'm
looking for. First, if I measure $J_z$, measure angular momenta in
three directions separately, and measure $J_z$ again, the two
measurements of $J_z$ aren't guaranteed to be the same, since $J_z$,
$J_x$, and $J_y$ don't commute. But $\mathbf{J}^2$ commutes with $J_z$.
Second, when the spin is not $\frac{1}{2}$, say it's $j$, summing the
squares of components of $\mathbf{J}$ gives 3ℏ²², where ² should
be ℏ²(+1) doesn't always give $\hbar^2 j(j+1)$. (Thanks to
@MichaelSeifert for pointing that out)
This question is not a duplicate of
Why is orbital angular momentum quantized according to $I= \hbar \sqrt{\ell(\ell+1)}$?.
That question is about how $\mathbf{J}^2$ is derived mathematically.
Mine is about how it is confirmed experimentally.
| While it's not in the context of an atom, molecular rotations provide a particularly "clean" demonstration of the fact that $\mathbf{J}^2 = \hbar^2 j (j+1)$. In particular, it's not hard to show that the rotational KE of two point masses connected by a rigid rod is
$$
H = \frac{J^2}{2 I} \quad \Rightarrow \quad E_j= \frac{\hbar^2 j(j+1)}{2 I},
$$
where $I$ is the moment of inertia about the center of mass. In particular, this means that transition energies between adjacent rotational energy states are
$$
\Delta E = E_{j} - E_{j-1} = \frac{\hbar^2}{2I} j [j + 1 - (j-1)] = \frac{\hbar^2 j}{I},
$$
i.e., they are evenly spaced by energies of $\hbar^2/I$.
This even spacing has been confirmed by many experiments, and the values of $I$ inferred from these spectra are consistent with other measures of the molecular dimensions. Below is shown the absorption spectrum of carbon monoxide; the evenly spaced absorption lines are evident. While the large absorption lines of the molecules composed of carbon-12 and oxygen-16 (the most common isotopes of carbon and oxygen) are the most obvious features, we can also distinguish smaller absorption lines from molecules composed of carbon-12 and oxygen-18. Molecules composed of different isotopes will have the same bond lengths but different moments of inertia due to the the different masses of their constituent atoms; this results in a different spacing of the rotational lines.
From Hollas, Modern Spectrosopy, 1992; taken from these lecture notes.
| {
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Does a planar object balance on a unique point? Consider a horizontal planar convex 2D object (say lying on x-y plane) with uniform density. Under constant gravitational force (say in -z direction), does it always balance on a unique point lying on the object (i.e. sum of the torques vanishes with respect to a unique point)?
I guess the answer is yes and if so then I want to conclude that the point must be the object's center of mass (that is the object will balance in any orientation w.r.t. that point), assuming the uniqueness of center of mass.
Possibly the question is trivial but I have been confused over this for some time. Any comments or answer will be appreciated.
EDIT: Added the assumption of convexity, as otherwise the point of balance may not lie on the object.
Let me add that, one may assume existance of the point. I am more interested in showing that there can not be two or more points of balance.
| Put origin at the centre of mass
We have then
$\vec{r}_{\text {cm }}= \frac{1}{M} \sum_{i}^{N} m_{i} \vec{r}_{i}=0$
Assuming uniform $\vec{g}$ and denoting $\vec{w} _{i} =m_i g$ as the weight of ith particle
Then the torque will be
$\begin{aligned} \vec{\tau} &=\sum_{i} \vec{r}_{i} \times \vec{w}_{i} \\ &=\sum_{i} m_{i} \vec{r}_{i} \times \vec{g} \end{aligned}$
=$\left(\sum m_{i} \vec{r}_{i}\right) \times \vec{g}$
=$M\left(\frac{1}{M} \sum m_{i} r_{i}\right) \times \vec{g}=0$
Hence proving that the total torque about the centre of mass is zero, and hence the body if pivoted there will stay there .
| {
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Why have Majorana fermions not been detected? There are two types of fermions - Dirac's and Majorana's. Majorana's fermions are their own antiparticles and they have not been detected yet. Sometimes, it is conjectured that e.g. neutrinos could be Majorana fermions.
However, it seems to me that there should be no Majorana fermion in current Universe. When particle and its antiparticle meet, they anihilate each other. Since there is no distinction between particles and antiparticle in case of Majorana fermions, a bunch of Majorana fermions should completely anihilate (maybe one particle can surive in case there were odd number of the particles in the bunch). This should happen shortly after Big Bang when all type of particle were created and the Universe was still small. Perhaps we could create Majorana fermions in accelerators, however, I suppose we should not be able to detect them in the Nature.
Does my reasoning make sense? If not, what am I missing?
| Dirac equation is not all we have - there are also "boundary conditions" or some experimental input. Only that together determines uniquely the solutions. No Majorana input data, no Majorana solutions in practice.
| {
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Force analysis in a compound pulley system At 4:56 of this video on compound pulley systems with the following schematic set up:
The professor does a force analysis of the forces on the pulley A , and strangely (for me) , he finds that there is tension acting vertically at two diametrically opposite ends of the pulley.. but why does it specifically act there? Why does tension not push the pulley down for the region where the rope curves around the pulley ?
| There are no frictions between the rope and the wheel of pulley. Therefore, the wheel will not rotate, and the rope slips over the wheel frictionlessly.
Consider a finite segment of the rope between $\theta_1$ and $\theta_2$, the total force acting on this segment of the rope (see the Figure):
$$
\vec{F}_{total}(\theta_1 \to \theta_2) = \vec{T}_1 + \vec{T}_2 + \int_{\theta_1}^{\theta_2} \vec{N}(\theta) d \theta = 0;
$$
where the vectors $\vec{T}_1$ and $\vec{T}_2$ are the draging forces from th rest of the rope ( in the tangential direction), and $\vec{N}(\theta)$ is the normal force per angle exerted from the wheel to this arc of rope. The normal force may be a function of angle. Since the mass of rope is neglected, the total foce has to be zero.
Then, we focus on the infinitesmal segment of rope betweem angle $\theta_2$ and $\theta_2 + d \theta$:
$$
d\vec{F}({\theta_2}) = \vec{F}_{total} (\theta_1 \to \theta_2 + d\theta) - \vec{F}_{total}(\theta_1 \to \theta_2) = \frac{\partial \vec{F}_{total}(\theta_1 \to \theta_2) }{\partial \theta_2} d \theta = 0.
$$
The force on the infinitesmal segment should also vanish, as detailed balance of the force.
The force $\vec{T}_1$ depends only one $\theta_1$, thus vanishes as derived by $\theta_2$ :
$$
\frac{d \vec{T}_2 (\theta_2) }{d \theta_2} + \vec{N}(\theta_2) = 0.
$$
Since $ \vec{T}_2$ is in the tangential direction, $ \vec{T}_2 = T \hat{\theta}_2$. The derivative of a unit tangential vector renders a unit vector in negative $r$-direction, $-\hat{r}$. This render the normal force at angle $\theta_2$ to be:
$$
\vec{N}(\theta_2) = - T \frac{d \hat{\theta_2}}{d\theta_2} = T \hat{r}.
$$
Surprisingly, the magnitude of the normal force per angle is a constant independent of the angle, equals to tension of the rope.
| {
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Can the thermal state be associated with a single pure state? I'm trying to understand better the quantum thermal state defined by
\begin{equation}
\rho_{0}=\frac{e^{-\hbar\omega_{\mu}}\left|n_{\mu}\right\rangle \left\langle n_{\mu}\right|}{\sum_{n_{\mu}}e^{-\hbar\omega_{\mu}}}
\end{equation}
More specifically, I'm interested whether or not we could associated to the above density matrix a state ket defined through by $\rho_{0} =\left|\psi_{0}\right\rangle \left\langle \psi_{0}\right|$ with perhaps
\begin{equation}
\left|\psi_{0}\right\rangle =\sum_{n_{\mu}}\frac{e^{-\frac{\hbar\omega_{\mu}}{2}}}{\sqrt{\sum_{n_{\mu}}e^{-\hbar\omega_{\mu}}}}\left|n_{\mu}\right\rangle
\end{equation}
I believe this is not the correct answer since if I use this formula it will give rise to terms like $\left|n_{\mu}\right\rangle \left\langle n_{\mu}+l\right|$. Any thoughts on that?
Thanks
| No. The state $\rho_0$ is not a pure state, i.e. it cannot be written in the form $\rho_0=|\psi_0\rangle\langle\psi_0|$.
This can be seen by noting that $\mathrm{trace}(\rho_0^2)<1$, while for a pure state, the trace would have to be $1$.
$\rho_0$ can, however, be seen as one half of the "thermofield double" state
\begin{equation}
\left|\psi_{0}\right\rangle =\sum_{n_{\mu}}\frac{e^{-\frac{\hbar\omega_{\mu}}{2}}}{\sqrt{\sum_{n_{\mu}}e^{-\hbar\omega_{\mu}}}}\left|n_{\mu}\right\rangle \otimes \left|n_{\mu}\right\rangle \ .
\end{equation}
| {
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Why does falling water become less and less focused as it falls down? So, I was observing my water purifier fill up my water bottle and I noticed that the beam of water coming through it became less and less focused as it falls down more.
Now, I do understand the concept of surface tension but I don't get it why this happens. Any answers will be helpful.
|
I noticed that the beam of water coming through it became less and less focused as it falls down more.
You are probably referring to the waterflow becoming turbulent. The flow rate of water is constant especially when it is from a water purifier. But this phenomenon is not due to the fluctuations in the flow rate of water.
Since the flow rate of water is constant, its velocity too is constant. Therefore the flow of water is streamlined. But as it falls down, its velocity increases due to gravity and hence the linear dimension too. These increase in velocity($u$) and linear dimension($L$) causes the increase in the value of Reynolds number($Re$). Reynolds number is used to predict the type of flow of a fluid. Higher is the value of Reynolds number, higher is the probability that the fluid will be under turbulent flow.
The Reynolds number can be calculated by the following formula:
$$Re=\frac{ρuL}{μ}$$
Here,
$ρ$ is the density of the fluid
$u$ is the velocity of the fluid
$L$ is the linear dimension
$μ$ is the dynamic viscosity of the fluid.
So you see, how the change in velocity and linear dimension affects the flow of water:)
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Does Schwarzschild metric in Kruskal-Szekeres coordinates admit asymptotic ($r \to +\infty$) timelike observers? I thank in advance whoever will answer my question.
Schwarzschild metric in Schwarzschild coordinates in $\mathbb{R}^{1,3}$ is [1]:
$$ds^2=-\bigg(1-\displaystyle\frac{2M}{r}\bigg)dt^2+\bigg(1-\frac{2M}{r}\bigg)^{-1}dr^2+r^2d\Omega^2.$$
When $r\to+\infty$ then $g_{\mu\nu} \to \eta_{\mu\nu}$, with $\eta_{\mu\nu}$ Minkowski metric (which obviously admits timelike observers e.g. $(1,0,0,0)$). By operating the change of coordinate of Kruskal-Szekeres we have [1]:
$$ ds^2=\frac{32M^3}{r}e^{-r/2M}(-dT^2+dX^2)+r^2d\Omega^2$$
with $r=2M(1+W_0[(X^2-T^2)/e])$ and $W_0$ Lambert function. In this case, when $r \to +\infty$ we get $ds^2 \to r^2d\Omega^2>0$ i.e. any vector becomes spacelike there. How can this discrepancy be explained?
Thanks a lot.
References
[1] http://bascom.brynmawr.edu/physics/courses/325/Problem_Sets/Review_PS_3.pdf
| You're getting tripped up in two different ways here.
(1) When you reason about the sizes of the terms in the K-S metric, you're ignoring the fact that the expression you're using is written using a mixture of variables from the two coordinate systems. This point is easier to understand with a simpler example. Suppose that $ds^2=e^{-r}dX^2$, where $r=\ln X$. (As pointed out in a comment by A.V.S., the log is the approximate behavior of the W function.) Implicit differentiation then gives $dX=e^r dr$, so that the line element can be rewritten as $ds^2=e^{r}dr^2$, and the coefficient clearly doesn't approach zero exponentially.
(Of course every point has a complete light-cone, including the timelike portions, so we knew in advance that there had to be a mistake in your conclusion that all vectors are spacelike in K-S coordinates at large $r$.)
(2) If we want a curve to approach spacelike infinity, it's not enough that it have $r\rightarrow\infty$. For example, a null geodesic can have $r\rightarrow\infty$, but it's going to approach null infinity, not spacelike infinity. The difference between the different idealized boundaries has to do with how fast $r$ approaches infinity compared to how fast $t$ approaches infinity.
If you look at a Penrose diagram for the Schwarzschild spacetime, it's clear that you can't have a timelike world-line that approaches spacelike infinity.
| {
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Polarity in a magnetized Möbius strip When a flat iron or Alnico washer is magnetized one of the faces develops a north polarity and the other, south. The geometric shape here is simple.
However, when a standard Möbius strip (or one of given thickness, radii of curvature and torsion of edges) is magnetized, which regions develop a north polarity and which regions south and according to which geometrical or other mass distribution criterion/law?
I am curious to know because such a Möbius strip (of rectangular section) has only one surface and only one edge.
Is magnetic polarity and strength distribution after magnetization influenced by changed geometry ( by homeomorphism ) ?
It may be easy to make a flattened thin Möbius strip looking like a recycling symbol to apply a magnetizing current. Thanks in advance for references.
| There is an electrical component called a Möbius resistor that takes advantage of this weird geometry.
The current flows in through the wire marked (+) and out through the wire marked (-). Because current flows in both directions around the ring, only a negligible magnetic field is generated. This can be important in high-power, high-frequency electrical circuits such as radar systems.
More mathematically, a Möbius strip is an example of a non-orientable surface. Magnetic field can be represented by arrows (vectors). So, you can place an arrow on a point of the surface pointing away from the surface (perpendicularly) to represent the magnetic field coming out of that surface. If you continue putting arrows along the surface, you will eventually loop back around and put an arrow on the surface facing the opposite way. There is no consistent way to assign a magnetic field to the surface of a Möbius strip, hence the advantageous properties--no stray magnetic fields, hence no inductance--in the electrical component above.
You can take a Möbius strip and magnetize it using an external magnetic field, but the resulting magnetic field will not keep a consistent angle with the surface. In fact, there has to be at least one part of the strip where the magnetic field is parallel to the surface.
| {
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Problem regarding components of forces in conical pendulum
For the motion of conical pendulum we can write equations as
$$T_{F}\cos\theta =mg$$
$$T_{F}\sin\theta=\frac{mv^2}{R}$$
$T_{F}$ represents tension in the string.
$v$ is the velocity of bob at this instant.
$R$ is the radius of circle.
But if we split $mg$ into its components,we can write
$$T_{F}=mg\cos\theta$$(because length of the string is constant)
What does then $mg \sin\theta$ will do? Also $mg \sin\theta$ is not in plane of circle. So what causes centripetal acceleration if I split components like this?
| I don't think you can write $T_F = mgcos \theta$, since in the first line, you've clearly (and correctly) have written $T_F cos \theta = mg$. To be more intuitive, imagine a vector in direction of $T_F$, which is a component of $mg$. For the rope's length to remain constant, this vector's magnitude should be equal to $T_F$, only in the opposite direction of $T_F$. From the image of the question, you can see that this vector's length is clearly longer than $mg$, so this must be equal to $mg/ cos \theta$ and not $mg cos \theta$.
| {
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$\beta$-function in free massive scalar field It is well-known that interacting QFTs with conformal symmetry preserved at the quantum level have a vanishing $\beta$-function. Another common statement is that mass terms break conformal invariance.
The free massless scalar field is conformal, and adding a mass term breaks the conformal invariance. Although it is still commonly called "free", I think you could view the mass term $m^2 \phi \phi$ as an interaction term between two scalar fields. After all, what is the difference between $m^2 \phi^2$ and $\lambda \phi^4$, apart from the number of fields being coupled in the vertex?
So is there some sort of a $\beta$-function related to the mass? If yes, how would it look like? Or does the statement above only make sense when there is a coupling constant for $n \geq 3$ fields for some reason?
| The concrete answer is actually in this post: Beta-function non-zero at classical level?
In short, the $\beta$-function for the mass is just:
$$\beta(m^2) = -2 m^2. \tag{1}$$
I am sure there are many ways to see that, e.g. the one described in the post. Here is another (trivial) way to check it. The renormalization group equation reads:
$$\left\lbrace p \frac{\partial}{\partial p} - \beta(m^2) \frac{\partial}{\partial m^2} + 2 - 2 \gamma_m \right\rbrace G^{(2)} (p;m^2) = 0\,, \tag{2}$$
where $g^{(2)} (p;m^2)$ is the exact propagator, i.e. (in Euclidean space):
$$G^{(2)}(p;m^2) = \frac{1}{p^2 + m^2}\,. \tag{3}$$
In a free theory the mass does not have an anomalous dimension, so $\gamma_m = 0$. The RGE can now be solved for $\beta(m^2)$, and we obtain the result mentioned above.
| {
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What are the applications of calculus of variations, if any, to the subject of thermodynamics? If we apply calculus of variations to Newtonian mechanics, we can discuss of functionals such as the lagrangian and how optimizing it leads to the equations of motion. However, does there exist application of the subject in thermodynamics?
I'm not sure how to phrase it precisely but I think there is a similarity in the notions of the concept of path dependence and functionals. In the sense that, in both we are talking function of functions, for example the work:
$$ W= \int_{V_1}^{V_2}P dV$$
Is a function dependent on the kind of path you take in the PV curve, this is extremely similar to the notion of a functional!
| The change in entropy of a system between an initial thermodynamic equilibrium state and a final thermodynamic equilibrium state is an extremum. It is the maximum value of the integral of $dq/T_{boundary}$ over all possible process paths between the initial state and the final state, where $T_{boundary}$ is the temperature at the boundary (interface) of the system through which the heat dq flows.
| {
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If a jet engine is bolted to the equator, does the Earth speed up? If a jet engine is bolted to the equator near ground level and run with the exhaust pointing west, does the earth speed up, albeit imperceptibly? Or does the Earth's atmosphere absorb the energy of the exhaust, and transfer it back to the ground, canceling any effect?
| Total angular momentum with respect to the center of mass, in practice the center of Earth, is conserved. The plane will contribute angular momentum to the atmosphere and to Earth, totaling to zero. After a short while the air will give back this angular momentum to Earth and the original value is restored. In the mean time Earth has gained a tiny advance on its rotation, while the atmosphere has gained a tiny lag.
| {
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Meeting point of magnets Suppose, we leave two magnets in space close to each other with no other force acting on them apart from the attractive force between them. One is stronger than the other but, they have the same mass. Would they meet at the middle or closer to the stronger?
It’s important to clarify that although both magnets have the same mass we have to assume that a percentage of the material of the weaker one is not magnetizable, otherwise we could assume that the stronger magnet would induce a field in the weaker one making the strength of both fields similar.
| As you probably know, there's no magnetic monopoles thus it would be difficult to compare the magnitude of the force between two magnets unlike in the case of two electrically charged particles where its pretty straightforward.
The simplest interaction would be magnetic dipole--dipole interaction.
If you leave two opposite charged particles like electron and proton and if you compare the gravitational and electromagnetic force between them, you can get the following ratio
$$\frac{F_g}{F_e} \approx 10^{-40} $$
If you take the electron and proton example into consideration, and if you consider everything classically you can say they will meet somewhere closer to the proton. But this picture is flawed.
| {
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Normalization Constant in Time Evolution of Density Matrix Given the Hamiltonian:
$$%H = \omega \left(|0\rangle \langle1| + |1\rangle \langle0| \right) = \begin{bmatrix}
0 & \omega \\
\omega & 0
\end{bmatrix}$$, I want to find the final state $\rho(t_f)$of the given density operator:
$$\rho(0) =|0\rangle \langle0| = \begin{bmatrix}
1 & 0 \\
0 & 0
\end{bmatrix} $$
To do so I started by stating:
\begin{equation}
\rho(t_f) = U\rho(0)U^\dagger\\
U=e^{-i\frac{t_f}{\hbar}H} \approx 1-i\frac{t_f}{\hbar}H\; \; \Rightarrow \;\;U^\dagger \approx 1+i\frac{t_f}{\hbar}H
\end{equation}
Although once I compute $\rho(t_f)$ using the above formula I obtain a non normalized state:
$$Tr(\rho(t_f))\neq 1 \; \; \forall \omega \neq 0$$
Of course this problem could be solved if out of nowhere I multiplied my $\rho(t_f)$ with a normalization constant N:
$$N = \frac{1}{Tr(\rho(t_f))}$$
My question is: is there something wrong with my thought process or calculations? Or do I really just have to introduce a new normalization constant? I would not mind an explanation in the option that the latter was the case(even if just as a reference).
I worked with it for a bit, and this is what I got:
P.S.
As suggested, I fully expand the U operator:
$$\%mathbf{U}=e^{-i\frac{t_f}{\hbar}\mathbf{H}} = \sum_n^\infty \left(\frac{c^n}{n!}\mathbf{H}^n \right)$$
Where for simplification I defined $c =i\frac{t_f}{\hbar}$.
By introducing a new operator denoted as $\mathbf{H}'$ ($\mathbf{H'} = \frac{1}{\omega}\mathbf{H}$), I notice the property:
$$\mathbf{H}^n=\left\{\begin{matrix}\omega^{n} \mathbf{I},& if \;\; n = even \\
\omega^n \mathbf{H}',& \; \; \; \; if \;\; n = odd
\end{matrix}\right.$$
Hence, the problem to solve becomes:
$$\mathbf{\rho}(t_f) = -\left(\sum_n^\infty \frac{c^n}{n!}\mathbf{H}^n \right)\rho(0) \left(\sum_n^\infty \frac{c^n}{n!}\mathbf{H}^n \right)$$
$$=%-\left( \sum_n^\infty\frac{c^{2n}}{2n!}\omega^{2n}\mathbf{I} + \frac{c^{2n+1}}{(2n+1)!}\omega^{2n+1}\mathbf{H'}\right)\rho(0)\left( \sum_n^\infty\frac{c^{2n}}{2n!}\omega^{2n}\mathbf{I} + \frac{c^{2n+1}}{(2n+1)!}\omega^{2n+1}\mathbf{H'}\right)$$
$$=-\left[ \sum_n^\infty \omega^{2n} \left( \frac{c^{2n}}{2n!}\mathbf{I} + \frac{c^{2n+1}}{(2n+1)!}\omega\mathbf{H'}\right )\right]\rho(0)\left[ \sum_n^\infty \omega^{2n} \left( \frac{c^{2n}}{2n!}\mathbf{I} + \frac{c^{2n+1}}{(2n+1)!}\omega\mathbf{H'}\right )\right]$$
$$=-\left( \sum_n^\infty \omega^{2n} \begin{bmatrix}
\frac{c^{2n}}{2n!} & \omega\frac{c^{2n+1}}{(2n+1)!}\\
\omega\frac{c^{2n+1}}{(2n+1)!}&\frac{c^{2n}}{2n!}
\end{bmatrix} \right)\left( \sum_n^\infty \omega^{2n} \begin{bmatrix}
\frac{c^{2n}}{2n!} & \omega\frac{c^{2n+1}}{(2n+1)!}\\
0&0
\end{bmatrix} \right)$$
$$%= \sum_n^\infty \omega^{4n}\begin{pmatrix}\frac{c^{4n}}{\left(2n\right)!\left(2n\right)!}&\omega\frac{c^{4n+1}}{\left(2n\right)!\left(2n+1\right)!}\\ \omega \frac{c^{4n+1}}{\left(2n\right)!\left(2n+1\right)!}& \omega^2\frac{c^{4n+2}}{\left(2n+1\right)!\left(2n+1\right)!}\end{pmatrix} = \mathbf{\rho}(t_f)$$
| If you expand $U$ to linear order in $t$, your density matrix will also only have trace one to linear order $t$, so $\mathrm{tr}(\rho(t))=1+O(t^2)$. As long as you get this, you did everything fine. Of course, your results will only be correct as long as the terms of order $t^2$ and higher will be small compared to the rest.
| {
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Electric field near surface of conductor derivation In page-50 of I.E irodov Basic Laws of Electromagnetism, he derives the relation for the force on surface of a conductor by every other charge distributed across the rest of the surface area.
The proof goes as follows: Let $E_{\sigma}$ be the intensity of field created by the charge on surface area, let $ E_0$ be the field created by rest of the charges.
Inside the conductor, for electrostatic equilibrium:
$$ E_0 - E_{\sigma} = 0 \tag{1}$$
Near the surface:
$$ E = E_0 + E_{\sigma} = 2E_o \tag{2}$$
After some more algebra, we can write:
$$F= \frac{\epsilon_o E^2}{2} \hat{n}$$
My doubt is how we can we say that the fields caused by each surface area portion under consideration for inside portion and outside portion are equal in magnitude?
As in, how do we know:
$$ E_{\sigma inside} = E_{\sigma outside}$$
| Start considering a small charged planar disc of surface $\Delta S$, which remains the same if you mirror it. The magnitude of the field is then the same below and above the disc by symmetry. See also these lecture notes, example 2.1.3 for the calculation of the field of a disc.
When you then shrink the area to the point that its size becomes negligible with respect to the distance at which you calculate the field, the shape becomes then irrelevant because you can consider it a point charge. And this has a symmetric field too.
| {
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In QFT why does the degree of the interaction terms in Lagrangian start from 3? I'm new to QFT so it's not obvious to me why there is no quadratic interaction terms in Lagrangians.
For example, the Lagrangian for a real scalar field is
$$L=\frac{1}{2}\partial_\mu \phi \partial^\mu \phi-\frac{1}{2}m^2\phi^2-\sum_{n\geq 3}\frac{\lambda_n}{n!}\phi^n.$$
What's the reason that we can't add terms like $g\phi^2$ to the free field Lagrangian?
| Such a term can be absorbed into a redefinition of the propagator.
| {
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Are all atoms spherically symmetric? If so, why are atoms with half-filled/filled sub-shells often quoted as 'especially' spherically symmetric? In my atomic physics notes they say
In general, filled sub-shells are spherically symmetric and set up, to a good approximation, a central field.
However sources such as here say that, even for the case of a single electron in Hydrogen excited to the 2p sub-shell, the electron is really in a spherically symmetric superposition $\frac{1}{\sqrt3}[2p_x+2p_y+2p_z]$ (which I thought made sense since there should be no preferred direction in space).
My question there now is, why is the central field approximation only an approximation if all atoms are really perfectly spherically symmetric, and why are filled/half-filled sub-shells 'especially spherically symmetric?
| No coherent superposition of 2p orbitals is spherically symmetric. Your example $\frac{1}{\sqrt3}[2p_x+2p_y+2p_z]$ is a 2p orbital pointing in the 111 direction and is not spherical. The proper description is by a diagonal density matrix, which states that the atom is in an incoherent superposition of the three states.
| {
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Einstein's light speed postulate I have seen two statements of Einstein's 1905 light speed postulate; for instance, in Andrew Steane's Relativity Made Relatively Easy:
*
*There is a finite maximum speed for signals.
*There is an inertial reference frame in which the speed of light in vacuum is independent of the motion of the source.
Does anyone have a proof that these statements of the postulate are equivalent? Can their equivalence be shown without resorting to the relativity postulate?
| They are not equivalent:
Just from the fact (or assumption) that there is a maximum signal speed, you cannot conclude that light travels by that speed. It could very well be that light travels at a smaller velocity than $c$, even depending on its frequency, due to some finite photon rest mass.
You can still formulate RT all the same using $c$, it would just be the case that $c$ is not the speed of light but the speed of causality.
| {
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How does rubbing soap on wet skin produce foam, and does it really enhance cleaning? We tend to rub soap after applying it to the skin. I found it interesting that the mere act of sliding our hands on the wet skin surface produces millions of air bubbles in the liquid, that later becomes foam. I wonder how exactly we manage to do that?
[Image source] This is the kind of foam I am talking about (foam/lather/froth... I find these words confusing).
Talking of foam, I have an unexplainable feeling that the effectiveness of a wash/bath is directly proportional to the amount of foam produced. Coming to think about it, it seems like the opposite should be true.
Soap without foam has a lesser amount of soap solution protruding out as bubbles; most of it is in contact with the skin surface, where actual cleaning takes place. I suspect that this is a misconception that got imprinted to our minds because soap does not clean or foam well in hard water (but that has an entirely different reason).
So to sum up,
*
*How exactly does rubbing soap on the skin produce foam?
*Is there any plausible reason why a soap with foam can do better cleaning than the same soap without any foam?
Simple and straightforward answers are welcome.
| Molecules of soap are composed by one hydrophobic and one hydrophile end. They clean because the hydrophobic end sticks to dirt stuff that is normally greasy, while the hydrophile end allows the product (soap + dirt) be washed up with water.
When a thin layer of water has 2 layers of soap molecules, (one at each side), it is possible for all the hydrophile ends be in contact with water, and the hydrophobic end be as far as possible of it.
When we rub, or agitate a solution of soap, we facilitate that configuration by adding air into the solution.
The spherical form minimizes the surface energy. That is the same reason for the growth of the bubbles with time.
About the cleaning efficiency, what can be said is that bubbles are an indication of the presence of soap molecules. And soap cleans.
| {
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Photon - Neutrino interaction Suppose we have a laser source, that means a coherent laser beam formed by 'in phase' photons. Is it possible to calculate how many photons can change their wavelength in neutrino - photon interaction if any? If there is no such wavelength change, can some photon loose the phase coherence in the interaction with neutrino? Obviously I suppose there could be a difference between muonic, eletronic and tau neutrino but how take into account this? Thanks
|
Is it possible to calculate how many photons can change their wavelength
Photons do not have wavelength , just energy = to h*nu, where nu is the frequency of the classical electromagnetic light, the laser light. Photons can scatter off various particles and if the scattering is inelastic, a new photon goes off.
in neutrino - photon interaction if any?
Neutrinos can couple to electromagnetic fields only in higher orders in perturbation theory, and the coupling constants are so small that any scatter will be unmeasurable.
If there is no such wavelength change, can some photon loose the phase coherence in the interaction with neutrino?
The phase coherence can only exist with the laser light, as individual photons just scatter, it would be the emergent beam that could display a classical phase difference, as laser light on a medium does. As explained above neutrinos are not real targets to laser light.
Obviously I suppose there could be a difference between muonic, eletronic and tau neutrino but how take into account this
See above.
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What is the difference between semi-relativistic and non-relativistic limit of an equation? In special relativity, we often take the semi-relativistic and non-relativistic limit of an equation. When the relative velocity is much smaller than the velocity of light, i.e., ${v/c}\ll1 $, we consider the motion to be non-relativistic and the Lorentz factor $\gamma=\dfrac{1}{\sqrt{1-(v/c)^2}}$ becomes unity. But I couldn't clearly understand the notion of semi-relativistic limit.
| I don't think it makes sense to talk about a "semi-relativistic limit."
In the non-relativistic limit, as you say, we have $v/c \ll 1$, and we can use binomial expansion on the Lorentz factor to say things like
$$
E = \gamma mc^2 = mc^2 \times \left(
1 + \frac12\frac{v^2}{c^2} + \cdots
\right)
\approx mc^2 + \frac12 mv^2
\tag{slow}
$$
In the ultra-relativistic limit, we have $\gamma \gg 1$, and we can say things like
$$
E^2 = (pc)^2 + (mc^2)^2 \approx (pc)^2
\tag{fast}
$$
But if you're not working in either of these limits (if, for instance, you're in the regime where $\gamma\simeq2$ or $\gamma\simeq 10$), the neither of these approximations is correct or useful, and you have to keep track of all the $\gamma$s rather than using one of the above tricks to make them go away. I have probably referred to this region as "semi-relativistic" in order to help my collider-physics colleagues remember that electrons do have mass. But it doesn't make sense to refer to this intermediate range as a "limit." It's just relativity.
If you're a three-significant-figures person, you might think "semi-relativistic" rather than the fast or slow limit for the interval $1.01 < \gamma < 100$. Sometimes it's nice to think about $\beta\gamma = \gamma v/c$ over this intermediate range, since the behavior of that product doesn't run up against an asymptote like $\beta$ or $\gamma$ do.
| {
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Why is the slope of pressure and volume almost zero below critical point for liquefaction of gas?
Image shows the Andrews isotherms regarding liquefaction of gases.
Why is the $PV$ curve almost horizontal from point $B$ to $C$?
I want to understand why does a small increment in pressure result in a large change in volume in the above mentioned zone ? What is the atomic reason that leads to this?
Thank you in advance.
| Inside the gray zone you have liquid in equilibrium with vapor. At any given temperature, this equilibrium exists at only one pressure. If you hold temperature constant and change the volume (move along the x-axis) you will cause some liquid to evaporate or some vapor to condense, but provided you change the volume slowly enough that equilibrium can more or less be maintained, the pressure stays essentially the same.
| {
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Calculation of time for electronic transitions How do we determine the time for electronic transitions in atoms or in semiconductor devices, from one energy level to another?
| In some cases one can measure the time, for example with pump-probe experiments with a variable delay between the laser pulses.
In cases where the line width is limited by the lifetime, one can use Heisenberg's indeterminacy principle for energy and time: $\Delta E \ \Delta t > \hbar/2.$
I understand this as the same relation as in the spectral bandwidth of sound pulses: the fewer oscillations there are in the pulse the greater is the uncertainty in frequency.
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What 'sets' the curvature in the FRW metric? In the FRW metric, the way I understand it, there are 4 key quantities, $P$ (pressure), $\rho$ (matter density), $\Lambda$ (the cosmological constant) and $K$ (the curvature constant).
$P$ and $\rho$ are fairly self explanatory, while $\Lambda$ we attribute to 'dark energy', whatever that turns out to be.
My question is, what is it that 'sets' the value of $K$? Is it the matter density that sets the global curvature, or does the universe simply have a global curvature as an intrinsic property?
| $k$ is the curvature of spatial slices, not the full spacetime, and it varies with $\dot a$ because it's essentially defined by the local motion of matter.
Take a bunch of parallel lines in Euclidean space, draw small local neighborhoods of the plane perpendicular to each one, and stitch them together. You'll get a plane. Now do the same with lines that are not quite parallel. You'll get a curved surface. The curvature will increase linearly with the degree to which the lines deviate from being parallel. If they're radiating out from a common point then you'll get a portion of a sphere whose radius is the distance to the origin point. If they are actually curves and don't converge at that point, you'll get a sphere whose radius is a linearly extrapolated apparent origin point.
In general relativity the same thing happens, with planes of simultaneity in local frames at rest with the Hubble flow standing in for the perpendicular planes in Euclidean space. There are two differences. First, "spheres" with timelike radius have negative curvature, which is why $\dot a^2+k$ instead of $\dot a^2-k$ appears in the first Friedmann equation (note that $k$ is the square of what I called the curvature above). Second, the spacetime you're slicing can be curved as well, which is why the right hand side of the equation is not zero.
| {
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Why does a body in SHM have more probability (of being observed) at the extreme positions and not at other positions? There is a question on the number of times a body comes to a place in a simple harmonic motion. Have a look:
I thought that the answer was B because in each oscillation a body is at the extremes only once, while for other positions, it comes two times. But the correct answer is C which I don't know why.
How come this is true?
| This is basically because the particle has a slow speed at the ends $( \ at \ \pm x_0)$ and hence a greater chance of being captured there.
Let us look at the equation of motion of the particle:
$x=x_0 sin(\omega t) $
The $X$ axis shows time and the $Y$ axis shows displacement of particle $(x)$.
I have marked points on the curve with a fixed time interval between them. Each of these dots represent a snapshot of the oscillator. We can see from the figure that at the turning points: $ x = x_0$ and $x=-x_0$, points cluster together, giving a greater chance of seeing the particle there.
This can be mathematically shown using density of states, but I feel for multiple choice questions, intuition is what is necessary.
| {
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Translational and Rotational Kinetic Energy Does an object that only spins have translational kinetic energy?
Since $v=\omega r$ ($\omega$ is angular velocity), if substituting $v$ with $\omega r$, won't that allow an object that only spins to have translational kinetic energy?
Thinking about the other way, won't an object that only slides would have rotational kinetic energy?
Therefore, for an object that only spins, the total energy would be...
$$E = \frac{1}{2}m(\omega r)^2 + \frac{1}{2}I\omega^2$$
And for an object that only slides, the total energy would be...
$$E = \frac{1}{2}mv^2 + \frac{1}{2}m(v/r)^2$$
But, the second one doesn't make sense, because $E = \frac{1}{2}mv^2$ for object that only slides
I'm confused. Can someone explain about this please?
| An object spinning about a point not the center of mass will have translational kinetic energy since it has momentum.
To find the proportion of kinetic energy due to translation and the proportion due to rotation you must express the kinetic energy at the center of mass, and then the $\tfrac{1}{2} m v_{\rm CM}^2$ part is the translating energy and the $\tfrac{1}{2} I_{\rm CM} \omega^2$ is the rotational energy.
Consider a rod pinned on one end. It has at the center of mass $I_{\rm CM} = \tfrac{m}{12} \ell^2$ and $v_{\rm CM} = \omega \tfrac{\ell}{2}$
$$ \begin{aligned}
K & = \tfrac{1}{2} m v_{\rm CM}^2 + \tfrac{1}{2} I_{\rm CM} \omega \\
& = \underbrace{ \tfrac{1}{8} m \ell^2 \omega^2 }_\text{trans} + \underbrace{ \tfrac{1}{24} m \ell^2 \omega^2 }_\text{rot} \\
& = \underbrace{ \tfrac{1}{6} m \ell^2 \omega^2 }_\text{total}
\end{aligned}$$
But take the same situation and calculate the kinetic energy at the pivot A with $I_{\rm A} = \tfrac{m}{3} \ell^2$ and $v_{\rm A} = 0$
$$ \begin{aligned}
K & = \tfrac{1}{2} m v_{\rm A}^2 + \tfrac{1}{2} I_{\rm A} \omega \\
& = 0 + \tfrac{1}{2} ( \tfrac{1}{3} m \ell^2) \omega^2 \\
& = \underbrace{ \tfrac{1}{6} m \ell^2 \omega^2 }_\text{total}
\end{aligned}$$
Same total kinetic energy, but the proportion of translating and rotating isn't evident at all.
In the case where the body is purely translating with $v_{\rm CM} \neq 0$ and $\omega =0$ then the kinetic energy calculation is
$$ \require{cancel} K = \frac{1}{2} m v_{\rm CM} + \cancel{ \frac{1}{2} I_{\rm CM} \omega} $$
| {
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Why is the tension equal to the spring force here?
Here the block is oscillating and to solve this question I took the tention in the string to be equal to the spring force
But if that's the case a particle in the junction of the spring and the rope will expirence a net force of zero yet it still it goes up and down every oscillation
So the tention should be higher than the spring force at some intervals
But most questions like these take them to be the same value
| We do not solve homework problems on this site but the following should help.
Considering m accelerating, consider the forces on m alone, then consider the forces on m and the string at the connection to the spring.
Assuming a massless pulley (no inertia) and massless string the tension in the string is equal on both sides of the pulley.
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What information is sufficient for describing a thermodynamic system? For a single-component system, why are the energy, volume, and number of particles sufficient for describing the thermodynamics of the system? Why just three variables and those three variables in particular?
In the book that I am using (Callen, $\textit{Thermodynamics and an Introduction to Thermostatistics}$) he postulates that the macroscopic equilibrium state is characterized by the energy, volume, and particle numbers of its components, but what is the reason for this?
| Typically, the state variables used to identify a macrostate are N, V, E for an isolated system, or you can replace E with T if the system is in contact with an heat bath. Other choices are also possible (the different formulations are related by the Legendre transform).
Note that other substances may require an enlarged set of state variables (for example, magnets or superfluids, see https://arxiv.org/abs/cond-mat/0405111 ).
Generally, apart from V (that sets the "size") and the internal energy E (that is related to the microscopic Hamiltonian), the other variables should be "conserved quantities" under the (possibly dissipative) dynamics of the system: for a simple system N is a conserved quantity (it is a "Noether charge"), other systems (like superfluids) may have extra conserved quantities.
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Converting energy from units of joules into units of wavenumber If we have an expression for energy $E$ in Joules why is it that if we wish to convert the energy into wavenumber we divide the expression by $hc$?
I know that $$E = \frac{hc}{\lambda}$$
So surely this implies that $$\frac{E}{hc} = \frac{1}{\lambda}$$
and since $$ k = \frac{2\pi}{\lambda}$$
Would it not make sense that in order to convert energy in joules to wavenumber we multiply $E$ by $\frac{2\pi}{hc} = \frac{1}{\hbar c}$.
So my question is why in converting energy from the units of joules to the units of wavenumber do we divide $E$ by $ hc $ rather than by $\hbar c$
| You're using the 'angular' wavenumber $k = \frac{2\pi}{\lambda}$ (i.e. the number of radians accumulated per unit length), but the wavenumber used by spectroscopists is the "spectroscopic" wavenumber,
$$
\tilde\nu=\frac1\lambda,
$$
which measures the number of wavelengths accumulated per unit length. If you want to get $k$ then indeed you should divide by $\hbar c$, and for $\tilde\nu$ you should divide by $hc$.
Unfortunately, the notation for these choices is very muddy and there are very few universal and objective standards, so the same word ends up getting used for distinct concepts. This is highly suboptimal, but it's mostly an unavoidable consequence of the fact that there isn't that much linguistic room around this notation to begin with. (Or maybe it is just that we are too lazy to say "angular wavenumber" like we do to distinguish angular frequencies from frequencies.) It isn't too terrible (you just need to be careful to check what factors of $2\pi$ need to go where), but it does require some attention (you need to be careful with the factors of $2\pi$).
As a general rule, though, if it's reported in $\rm cm^{-1}$, then it is almost certainly $\tilde\nu$ instead of $k$.
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Will liquid nitrogen evaporate if left in an unopened container? SOS! I left work today and got a horrible feeling that I forgot to put the lid back on a large container of liquid nitrogen which contains many racks of frozen cells in it. If this did happen, how long would it take liquid nitrogen to evaporate? Does it start to evaporate as soon as it is exposed to oxygen? Will all the liquid nitrogen be gone from the container when I go back tomorrow?
|
Does it start to evaporate as soon as it is exposed to oxygen?
Exposure to oxygen has nothing to do with it.
Actually the liquid will evaporate whether the lid is off or on. But, the rate of evaporation depends on how well the container is insulated. Evaporation is what keeps the liquid cold.
Leaving the lid off allows heat from the room to enter the container at a greater rate, and the liquid will evaporate faster in proportion to that rate. The open lid allows thermal radiation from above to "shine" down into the container, and it allows the cold gas near the surface of the liquid to mix with the warm room air. The rate of that mixing will depend on the height of the liquid in the container, and on how much the building's ventilation system stirs the air.
P.S., I presume that the room has some kind of forced air ventilation. You wouldn't want to walk into the room in the morning if the "air" was all "evaporated" nitrogen, and no oxygen. At least, not unless you were wearing SCBA gear.
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Different values for R134a property Tables I need internal energy and volume value for R134a refrigerant, to get the values went through steam tables/property tables for R134a
For a particular steam table
At temperature 30 degree celsius, Specific internal energy = 246.14 kJ/kg, specific Volume of gas = 0.026622 m^3/kg,
For one more steam table from another resource
At temperature 30 degree celsius, Specific internal energy = 394.48 kJ/kg, specific Volume of gas = 0.02671 m^3/kg,
Which table values must be considered and why there is such a difference?
| As @Chet Miller pointed out in his comment, the reason could be different zero reference states for internal energy for the two tables. Notice the last sentence of the notes at the bottom of your first table. It reads "The enthalpy and entropy values of saturated liquid are set to zero at -40 C".
But heres the important point: For all practical purposes it doesn't matter where the zero reference is set and what the specific absolute values are. That's because when we use these tables we are almost always only interested in changes in the values of internal energy, enthalpy, and entropy, not the absolute values. For that purpose either table would be ok since changes in these properties should be approximately the same.
To illustrate, consider the increase in the specific internal energy of the fluid going from 30 C to 40 C for the two tables.
From the first table:
$\Delta u=250.97-246.14=4.83$
From the second table:
$\Delta u=399.46-394.48=4.98$
So you see the change in specific internal energy of the saturated vapor between the same two temperatures is about the same for both tables.
By the way, your first table appears to align with the following table:
https://www.cambridge.org/us/files/2313/6697/5548/Appendix_C.pdf
Note that in this table the zero reference state for the specific enthalpy and specific entropy of saturated liquid is -40 C per the notes in your first table. The zero reference state for the specific internal energy of the saturated liquid falls between -35 C and -40 C.
Hope this helps.
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Is it even theoretically possible for a perfect clock to exist? I have heard that even atomic clocks lose a second every billion years or so. That raises the question, is it even theoretically possible for a perfect clock to exist, one that never gains or loses time?
| The words "perfect" and "exist" are not compatible. The only way we can imagine perfect things is in our imagination, not in anything that exists. For things that exist, we can ask how close they are to being perfect, but even that question will not have a perfect answer. The best we can do is to estimate how close they are to being perfect. Losing 1 second in a billion years is pretty darn close to being perfect (since it is 1 part in 31,557,600,000,000,000) but it is not perfect and nothing physical can ever be perfect.
| {
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Who predicted the existence of the muon neutrino? The Wikipedia article on the muon neutrino says:
The muon neutrino is a lepton, an elementary subatomic particle which has the symbol $\nu_\mu$ and no net electric charge. Together with the muon it forms the second generation of leptons, hence the name muon neutrino. It was first hypothesized in the early 1940s by several people, and was discovered in 1962 by Leon Lederman, Melvin Schwartz and Jack Steinberger.
But it doesn't say who hypothesized it, and I haven't been able to find that out.
So: who predicted the existence of the muon neutrino, and in what papers did they do it?
| I may have found it. I'm quoting Wiki's article on Schoichi Sakata:
Sakata and Inoue proposed their two-meson theory in 1942.[3] At the time, a charged particle discovered in the hard component cosmic rays was misidentified as the Yukawa’s meson ($\pi^\pm$, nuclear force career particle). The misinterpretation led to puzzles in the discovered cosmic ray particle. Sakata and Inoue solved these puzzles by identifying the cosmic ray particle as a daughter charged fermion produced in the $\pi^\pm$ decay. A new neutral fermion was also introduced to allow $\pi^\pm$ decay into fermions.
We now know that these charged and neutral fermions correspond to the second generation leptons $\mu$ and $\nu_\mu$ in the modern language. They then discussed the decay of the Yukawa particle,
\begin{equation}
\pi^+\rightarrow \mu^+ + \nu_\mu
\end{equation}
Sakata and Inoue predicted correct spin assignment for the muon, and they also introduced the second neutrino. They treated it as a distinct particle from the beta decay neutrino, and anticipated correctly the three body decay of the muon. The English printing of Sakata-Inoue’s two-meson theory paper was delayed until 1946,[4] one year before the experimental discovery of $\pi\rightarrow\mu\nu$ decay.
And there's the link to said paper:
https://academic.oup.com/ptp/article/1/4/143/1846220
I might be wrong but it's my best shot.
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Where is this magnetic rail gun getting its energy? Please watch this youtube video of a magnetic rail gun moving a marble.
So as you already know, Conservation of Energy states that "energy can be neither created nor destroyed, but can only change form". Where is this rail gun getting the energy to move the marble? It looks like it's being created out of thin air. Is it not possible to create a perpetual motion machine by modifying this rail gun? For example, what if a funnel and tube was added that can catch the moving marble, which feeds the marble back to the rail gun to continue the loop? I'm not 100% sure, but I can't imagine any kind of friction will cause this loop to stop.
Thanks!
| A permanent magnet will attract a ferromagnetic object. But when you apply force to remove the object from the magnetic field you will find that there is no net energy gain.
| {
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Normalisation of wavefunction given by the form $Ae^{i(kx-wt)}$ Question 1.
Let's say that the wavefunction is given in the form
$$\Psi(x, t) = Ae^{i(kx-wt)}$$
Then because of the normalisation condition, the following should hold.
$$\int \Psi^*\Psi dx = A^2 \int_{-\infty}^{\infty} e^{-i(kx-wt)}\times e^{i(kx-wt)} \ dx = 1$$
Because $e^{-i(kx-wt)}\times e^{i(kx-wt)} = e^{-i(kx-wt) + i(kx-wt)} = 1$, the condition demands that
$$A^2 \int_{-\infty}^{\infty} dx = 1$$
As the integral value diverges to $+\infty$, we reach the conclusion that $A$ should converge to zero.
What's wrong here?
Question 2.
This is another question that should be classified and asked separately but as it is a short one I will just put this one into here. When expressing the wavefunction as a linear combination of basis functions, especially in discrete cases, is it that the index varies from $-\infty$ to $\infty$? That means, is it that
$$\Psi(x) = \sum_{-\infty}^{\infty} c_i \psi_i \ ?$$
Apologies in advance if the questions are trivial. I am a newcomer to quantum mechanics.
| (1) Nothing wrong there. Plane waves are states of infinitely precise momentum and cannot be properly normalized in position space due to having infinite spread from Heisenberg uncertainty. In practice they still help e.g. in the scattering matrix formalism to get an amplitude for reflection and an amplitude for transmission, and to settle e.g. the basic physics of an Aharonov-Bohm ring where the actual lengths one cares about are finite.
(2) You always can and you never have to. There is a bijection between $\mathbb Z$ and $\mathbb N$ so however you number things is up to you. There is a slight reason to prefer $\mathbb N$ which is that a large class of these basis states are eigenfunctions of a Hamiltonian which is bounded from below, and thus these eigenvalues go on infinitely in one direction but not the other.
| {
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Intrinsic carrier concentration in doped semiconductors For an intrinsic semiconductor, due to thermal energy we get some charge carriers whose concentration is known as intrinsic carrier concentration.
Now if we dope the material we'll have carriers both due to donation and due to thermal energy generation.
In this case is the intrinsic carrier concentration still defined to be equal to the thermally generated carriers?
And in what way might this doping affect the number of intrinsic carrier concentration?
| In heavily-doped semiconductors, acceptor and donor energy levels (which are close to the valence and conduction band edges, respectively) form energy bands, effectively moving the valence and conduction band edges toward the middle of the band gap, and hence effectively reducing the band gap. This is called band gap narrowing. Since the intrinsic carrier concentration is related to band gap, i.e.
$$ n_i = \sqrt{N_CN_V}\exp\left(\frac{-E_g}{2k_BT}\right),$$
heavy doping can increase the intrinsic carrier concentration, in the sense that in a semiconductor compensated by large and equal concentrations of acceptors and donors, the electron and hole densities will be greater than $n_i$ of a pure sample.
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How can we say load is static in case of simple tension test if it is changing with time? In simple tension test the load acting is static load(which doesn't change with time), but in this test the load varies with time i.e., the load increases, so how can we say it is a static load test?
| By static, it means slowly changing with time. It is the case of a standard uniaxial tensile test.
One example of the opposite is the Charpy test, where a notched sample is broken by a weight in a pendular movement. If the same weight were just placed slowly on the same test specimen, it is possible that it didn't break.
| {
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Will the magnitude of electric/magnetic force reduce if an non-magnetic or neutrally charged body is kept in between two charges magnets? If we keep two magnets or two charges in a space and keep a glass block or anything which is neutrally charged (should be big enough so that apparently it would look like its obstructing)in betweeen those,will the magnitude of force reduce in this case compared to the force that should act when tha neutral thing is absent?
| No. Physical objects cannot block block the electric field created by the first charge at the position of the 2nd charge (the electric field in the region where the object is placed DOES change but the force on charge 2 depends on the electric field present at its own position)
The (net) electric force between 2 charges does decrease however if the medium is changed from vacuum to air or water for example. This is because when a charge creates an external electric field in a medium, the electron clouds of the atoms/ molecules of the medium shift slightly in the opposite direction of the external electric field under the influence the electric force. This is called polarization and results in an internal electric field of the medium in the opposite direction of the external electric field, thereby weakening the net electric field and this electric force exerted at every point.
The strength of the induced electric field depends on polarizibilty of the material and Is measured by "electrical permittivity" ε of the material.
This decrease in net electric force due to induced electric fields is directly incorporated in the coulomb constant 1/4πε
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Why is Minkowski metric diagonal? Why is the Minkowski metric a diagonal in a 4x4 matrix? What does the diagonal do?
| For a given spacetime the metric tensor may be written in both (globally) diagonal and non-diagonal forms depending on what coordinates we choose. For example for flat spacetime one diagonal form (not the only diagonal form) is the Minkowski metric:
$$ ds^2 = -dt^2 + dx^2 + dy^2 + dz^2 $$
But we could choose to write the metric using rotating cylindrical coordinates, and we would get:
$$ ds^2 = -\Big(1 - \frac{r^2\Omega^2}{c^2}\Big)dt^2 + r^2d\theta^2 + 2r^2\frac{\Omega}{c} d\theta dt + dr^2 + dz^2 $$
This is not diagonal because we have the term $2r^2\frac{\Omega}{c} d\theta dt$, but it's the same spacetime, and the metric is only non-diagonal because we have made a poor choice of coordinates.
So the interesting questions are:
*
*what determines whether for a particular geometry there is a choice of coordinates for which the metric is (globally) diagonal?
*why do we want the metric to be diagonal anyway?
(1) is hard. Technically we require: In d dimensions, you can write the metric in diagonal form if and only if your manifold can be foliated by d mutually orthogonal families of surfaces. But good luck with that. There is no simple way to take some arbitrary geometry and find coordinates in which the metric is diagonal. Not all metrics can be written in a diagonal form - for example the Kerr metric cannot, regardless of how inventive we are in choosing coordinates.
As for (2): having the metric in diagonal form is computationally convenient. It makes its determinant easy to calculate so it is easy to invert. It also reduces the amount of work needed to calculate the Christoffel symbols.
As an addendum to this, during a discussion of this issue in the chat room ACuriousMind pointed out that for specific cases the "hypersurface orthogonality" condition is equivalent to the "Frobenius condition" (also just called "integrable" by mathematicians) $\xi \wedge \mathrm{d}\xi = 0$ for the one-form $\xi$ dual to the Killing vector. For more on this see this article on the Schwarzschild metric (the link is a PDF).
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Is strain during free thermal expansion zero?
If a steel rod of length 1 m rests on a smooth horizontal base is heated from $T_1$ to $T_2$, what is the longitudinal strain developed?
I think that due to thermal expansion/contraction, the length of the rod should change, so strain should be developed. However, after googling my doubt it turns out the strain developed is zero.
Can I get an explanation as to why it is zero?
Here's a link to one of the pdfs I found on the Internet about this topic:
https://ocw.mit.edu/courses/aeronautics-and-astronautics/16-20-structural-mechanics-fall-2002/lecture-notes/unit9.pdf
See page no. 5, figure 9.1.
| Thermal strain is strain that develops when a material is heated or cooled. So the strain should not be zero.
In addition to the site you provided, thermal strain for a non-constrained object is not zero as long as the coefficient of thermal expansion is not zero, per the following:
http://emweb.unl.edu/NEGAHBAN/Em325/05-Thermal-strain/Thermal%20strain.htm
http://www.ce.memphis.edu/3322/Pdfs/PaulsPDFs/Thermal%20Strain.pdf
https://www.engineeringtoolbox.com/stress-restricting-thermal-expansion-d_1756.html
https://en.wikipedia.org/wiki/Thermal_expansion#Effects_on_strain
https://www.britannica.com/science/thermal-strain
Thermal stress (though a misnomer) is the internal stress that occurs if the object being heated or cooled is constrained. That would be the case if the steel rod was wedged between two rigid walls prior to heating.
When the heated rod is constrained from increasing in length, its thermal strain is cancelled by the compressive mechanical stress of the walls. If it was fixed to the rigid walls (i.e., unable to separate from them) and cooled so that it is constrained from decreasing in length, its thermal strain is cancelled by the mechanical tensile stress of the walls.
Hope this helps.
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What does the "true" visible light spectrum look like? When I google "visible light spectrum", I get essentially the same image. However, in each of them the "width" of any given color is different.
What does the "true" visible light spectrum look like, then? It can't be that each and every image search result is correct.
I could not find any information about this on the web, so I turn to the experts.
| When you look at a spectrum, you're using some kind of optical effect to spread out the different frequencies of light. Most commonly this is done by a triangular prism, where the difference between the speed of light in air and the speed of light in the prism material ('refractive index') causes different frequencies of light to bend by different amounts. Generally, lower-frequency red bends least and higher-frequency violet bends most, with the other colors spreading to varying amounts between the two.
Depending on the refractive index difference, the amount of bending changes. One kind of prism (say, clear plastic) might be rather weak so that the difference in angle between red and blue are very small, producing a very narrow rainbow, while another prism can use a better material (such as a diamond) and spread red and blue onto very different angles, giving you a wide rainbow. Both of those are real and valid spectra. Neither is more "correct" or "real" than the other; it's just different materials spreading out the light to different amount.
And then of course the actual size of the spectrum produced depends on how far the target wall is from the prism. A prism that produces a very wide spectrum with a wall that's only two inches away will have a narrower resulting image than a less powerful prism projecting on a wall ten feet away.
This question seems bit like looking at an 18" television and a 50" television and asking "But which one is showing the real TV signal?" The question doesn't really make sense.
| {
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How does Inertia opposes change in state? The first line of Wikipedia states that Inertia opposes change in velocity. My teachers also told that Inertia opposes change in state of rest or motion.
But I would like to know how?
Suppose a body is at rest as seen from an inertial frame and a net external force is applied then according to the above definition, shouldn't every body oppose the acceleration caused from that force because that force will change its state from rest to motion?
If yes, then why don't we count inertia force when calculating the net acceleration of a body ? i.e why isn't:
$$F_{(net\; external)}= inertial\; force+ ma$$
|
The first line of Wikipedia states that Inertia opposes change in velocity. My teachers also told that Inertia opposes change in state of rest or motion...shouldn't every body oppose the acceleration caused from that force because that force will change its state from rest to motion?
Changing from rest to motion is a change in velocity (the velocity changes from $0$ to not $0$), so there is no contradiction, although the Wikipedia version is more general and includes the other.
However, saying inertia "opposes" something is very qualitative. "Opposes" here doesn't mean "stops". We are essentially just saying "more inertia $=$ more 'resistance' ". Once you start learning more physics you can abandon/bolster your qualitative descriptions with more accurate, quantitative ones. "Inertia opposes change in motion" becomes simply $F=ma$.
| {
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Are there fields (of any kind) inside a black hole? It is said that nothing escapes from black holes, not even light. All particles are now thought to be excitation of different fields (electric field, electromagnetic field, photon field, etc).
Does it follow that there are no fields (of whatever kind) inside the event horizon of a black hole?
IFFY (if and only if) there are fields, are particles created inside the black hole?
| We don't know for certain what is inside the event horizon of a black hole, but we expect there to be fields inside the event horizon that are just extensions of the fields outside of the event horizon. The only restriction is that events that happen inside the event horizon cannot be causally linked to effects outside the event horizon.
However, a complete description of the event horizon at the very smallest scales requires a theory of quantum gravity, which we do not yet have.
| {
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What are the necessary and sufficient conditions for a wavefunction to be physically possible? Often times it is stated in books that a quantum state is physically realizable only if it is square integrable. For example, in Griffiths (2018 edition) page 14 he stated
Physically realizable states correspond to the
square-integrable solutions to Schrödinger’s equation.
But when we have an operator with continuous eigenvalues like the position operator $\hat{X}$ our eigenstates are not square-integrable. Like the eigenfunction for the eigenvalue $x'$ of $\hat{X}$ is $\delta(x-x')$ which is not square-integrable and we also know -
$$ \langle x|x'\rangle=\delta(x-x')\Rightarrow \langle x|x\rangle=\infty$$
But clearly $|x\rangle$ is a physical state as wavefunction collapses to that after measurement. So definitely the definition by Griffiths is not completely correct. I came across about Rigged Hilbert space while searching in Stack but I am still not completely sure whether all physical states lie in Rigged Hilbert space or all states in Rigged Hilbert space are physical. So, my question is what are the necessary and sufficient conditions for a wavefunction to be physically possible?
| If you want to use the theory of probability, a necessary condition for a wavefunction to be physically meanigful is $$\psi \in L^2(\mathbb{R}^3,d^3x)\:.$$ That is because, as a basic postulate of QM, we have that:
$\qquad\qquad\qquad\qquad$ $|\psi(x)|^2$ is the probability density to find the particle at $x$,
and the total probability must be $1$:
$$\int_{\mathbb{R}^3} |\psi(x)|^2 d^3x =1 <+\infty$$
(Values different from $1$ can be next obtained by considering non-normalized wavefunctions).
That this condition is also sufficient is a much more delicate issue which largely depends on the physical hypotheses you assume on realizable pure states.
In principle all vectors
$\psi \in L^2(\mathbb{R}^3,d^3x)$ are admitted.
No continuity or differentiability requirements make sense, even if it is sometimes erroneously stated. That is because all observables, as another basic postulate of QM, are self-adjoint operators and no differential operator is selfadjoint: in fact, one should deal with selfadjoint extensions of these differential operators, whose domains are made of non-differentiable functions, generally speaking.
Rigged Hilbert spaces, i.e., the rigorous version of Dirac (fantastic!) formalism elaborated by Gelfand and coworkers, has to be considered a mere formal/mathematical tool.
In particular distributions as $\delta(x-x')$ are not physically meaningful as they do not satisfy the condition to be elements of $L^2$.
| {
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Why does the bullet have greater KE than the rifle? A rifle is fired, and the bullet moves much faster than the rifle, and momentum is conserved.
My question is whether the kinetic energy of the bullet is greater than the kinetic energy of the rifle because the mass of it is smaller, therefore the force acting on it (for the same time it acts on the rifle) results in greater acceleration, and thus greater final velocity (basically, the change in momentum needs to be equal and in opposite direction to the rifle, which has greater mass). On the other hand, kinetic energy depends on work done, which is force times displacement. Therefore, we can say that the force acted on the bullet for a greater distance traveled.
Are these explanations stating the same thing, or are they causally different? (From the expression of the momentum formula as m. dx)
| Basic principal of ballistic physics, Once the round leaves the barrel it begins to lose energy. Kinetic energy is energy PUT to work in motion.
The bullet may have less mass than the rifle but remember the cartridge stores propellant, that's Chemical energy (Potential)
Also most rifles this day have some sort of recoil management to mitigate the energy opposing delivered.
The rifle weighs (depending on model) from 5-10 lbs? the energy recoil is mitigated through the body of the rifle.
| {
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Why do fans spin backwards slightly after they (should) stop? Today, I've decided to observe my PC fan as I shut the computer down. The fan slowly lost angular momentum over time. What I've found really interesting is the fact that the momentum vector change did not stop at the zero vector, but instead flipped its orientation and "went to the negatives", albeit very small in the absolute value compared to the powered spin; this caused the fan's angle to deviate by a few degrees (opposite to the powered spin rotation) compared to the observed angle when momentum was equal to the zero vector.
If I let $\overrightarrow{L}$ be the momentum vector, $\overrightarrow{L}_0$ be the momentum vector at $t_0$ (= poweroff time), and $\overrightarrow{L}(t) = y(t) * \overrightarrow{L}_0$ (with $y_0 = y(t_0) = 1$), then these are the plots of $y$ through the course of time.
Expected fan poweroff behavior:
Observed fan poweroff behavior:
Can anyone explain why may this happen?
| Your PC is a box with limited vents for air to enter/exit. Especially if these clog up with dust, the fan could create a noticeable pressure differential between inside and outside of the box. After the fan turns off, that equalizes, forcing some air backward through the fan, and causing it to rotate backward. Maybe.
Another possibly, as pointed out in comments, and since you report much less than a full rotation backward, is that the DC motor running the fan likes to be in certain orientations and not others when it is unpowered, due to alignment of the permanent magnet with other motor parts. As the fan slows down, at some point it passes one of these preferred orientations but lacks the momentum to make it over the hump to the next one, and rotates back.
| {
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Speed of light in the Schwarzschild metric or Gullstrand metric In the Schwarzschild metric, the speed of light will be given by:
$$\frac{\mathrm{d}r}{\mathrm{d}t}= \Big(1-\frac{2M}{r}\Big)$$
and in the Gullstrand metric, it will be given by:
$$\frac{\mathrm{d}r}{\mathrm{d}t}= 1- \sqrt{\frac{2M}{r}}$$
As is evident they are clearly different from $c$ in suitable regions.
But SR tells us that the speed of light is an invariant and a constant of the universe.
So what is happening here? Why is it not an issue that we are getting that the speed of light is different from $c$ in different coordinates?
|
But SR tells us that the speed of light is an invariant and a constant of the universe.
So what is happening here? Why getting the speed of light different from c in different coordinates not an issue?
SR only tells us that the speed of light in an inertial frame is equal to c. Neither the Schwarzschild nor the Gullstrand coordinates are inertial everywhere. Even in flat spacetime it is easy to get speed of light violations in non-inertial coordinates, and in curved spacetime there are no global inertial frames so it is essentially unavoidable in global coordinates.
| {
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Why does a layer of rubbing alcohol increase the transparency of frosted glass? While cleaning a piece of glass with 70% ethyl alcohol, I made the observation that after swabbing a frosted portion of the glass, the glass appears to lose the "frosted" property temporarily and become almost as transparent as the surrounding glass. What about the act of applying rubbing alcohol produces this effect?
My initial hunch is that since the glass gets the frost effect from creating impurities on the surface by some means such as sandblasting, the alcohol when applied creates a temporarily smooth surface by filling the imperfections to a relatively similar effective height. However, I attempted to recreate the effect with water in place of alcohol and it was significantly less effective, which is the opposite of what I expected given the overall stronger intermolecular forces.
What am I missing? Do I have it backwards and the properties of water like higher surface tension make the effect less pronounced?
| Alcohol has a smaller tension surface than water, so when applied it's less likely to form "little bubbles" that works as little lenses and tend to "blur" the light passing trough them (this is caused by the refraction of the light rays changing material). Your assumptions about "filling the holes" is correct. Infact glass and alcohol have pretty similar refractive index, so smoothing the surface gives a smaller "blur effect" because the surface in no more irregular, which when alcohol is not applied cause the scattering of light rays.
| {
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Is there a limit to the energy density of a battery? Better battery technology is very important today: improving the energy stored per volume or mass. This led me to wonder whether there is a theoretical limit. (I'm not expecting that we are at all close to it. Real life just inspired the question.)
One extreme battery would have a reservoir of anti-matter which it could combine with ordinary matter in a controlled fashion. Could anything beat that for stored energy per mass?
| If you put a maximum amount of energy in a volume you get a black hole. Note a black hole can be used as a battery because the energy will be released as hawking radiation which can be used to do work. The limit to the maximally dense battery then is the Schwarzschild limit $r=2GM/c^2$ and $E=Mc^2 $ so $r=2GE/c^4$.
| {
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Question about the non-renormalization of conserved currents It is asserted in many places, e.g, sec. 12.4 in Peskin&Schroeder, that the current
$J^\mu=\bar\psi\gamma^\mu\psi$ needs no renormalization.
The following diagram
contributes a divergence of $Z_1$-like, which is equal to a $Z_2$-like divergence by Ward identity. Then this divengence can be removed by the renormalization factor $\sqrt{Z_2}$ of $\psi$ in $J^\mu$. So there's no need for a further renormalization of $J^\mu$ as asserted.
However, the following diagram
seems to contribute a $Z_3$-like divergence. $J^\mu$ has to mix with $A^\mu$ to remove this divergence, contradicting that $J^\mu$ needs no renormalization. Is there any problems?
On the other hand, the equation of motion reads
$$ Z_3[\partial^2 A^\mu - \partial^\mu(\partial\cdot A)] = - Z_1 e J^\mu \,.$$
While $A^\mu$ is already renormalized, $J^\mu$ also needs no renormalization as asserted. So we should have $Z_3=Z_1$ up to finite terms, which is not correct?
| It seems that the current does mix with $A^\mu$, or $\partial_\nu F^{\nu\mu}$ more precisely, as discussed in this paper written by J. C. Collins et al. (I fount it from another question in PSE). I quote from the abstract:
It is commonly asserted that the electromagnetic current is conserved and therefore is not renormalized. Within QED we show (a) that this statement is false
and
The current mixes with the four-divergence of the electromagnetic field-strength tensor.
| {
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Will a muon decay in an empty universe? Imagine, a deep empty universe consisting of only one particle muon. Will it decay?
As there isn't any change in its surroundings and thus time will lose its meaning. But if the muon will decay in that situation, doesn't it prove that time is something deeper than our experience or what a clock shows!
| A muon is a kind of clock in relativity theory. But in an empty universe there would be no other clock to compare it to and no relativity. . Without relativity the universe as we know it is not defined and cannot exist.The answer to your question is therefore no
| {
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Normalisation of QCD Lagrangian In QCD, and more generally in representations of $\mathfrak{su}(N)$, there is a freedom to choose the normalisation of the generators,
$$
\mathrm{Tr} \, \left[R(T^a) R(T^b)\right] = T_R \delta^{ab}.\tag{1}
$$
I am trying to work out the implications of this for the kinetic term for the gluon field in the QCD Lagrangian density. I've looked in various textbooks and notes, and every source I can find either glosses over this entirely, or gets the details trivially wrong (e.g. eqn. 33 in these or eqn. 4.66 in these).
Conventionally physicists choose $ T_F = \frac{1}{2}$ and write
$$
\mathcal{L}_\mathrm{kin} = - \frac{1}{4} F_{\mu\nu}^a F^{\mu\nu}{}^a,\tag{2}
$$
where the field strength tensor has been expanded into components
$$
\mathbf{F}_{\mu\nu}
=
\sum_a
F_{\mu\nu}^a T^a.\tag{3}
$$
It seems clear to me that
$$
\mathrm{Tr} \, \left[\mathbf{F}_{\mu\nu} \mathbf{F}^{\mu\nu} \right]
=
T_F \; F_{\mu\nu}^a F^{\mu\nu}{}^a,\tag{4}
$$
and so the correct, Lie-algebra-normalisation-convention-independent expression for the Lagrangian density should be
$$
\mathcal{L}_\mathrm{kin}
=
-
\frac{1}{2}
\mathrm{Tr} \, \left[\mathbf{F}_{\mu\nu} \mathbf{F}^{\mu\nu} \right]
=
-
\frac{1}{2}
T_F \; F_{\mu\nu}^a F^{\mu\nu}{}^a,\tag{5}
$$
which restores the usual result in the conventional case.
Is this correct, and is there a good source for it? (ie one that could be cited in an academic work).
Why is the convention-specific component expansion used preferentially in the literature?
| The standard normalization of the kinetic term in components is OP's eq. (2)
$$ -\frac{1}{4}F_{\mu\nu}^{ a}F^{\mu\nu a}~=~\underbrace{\frac{1}{2} \sum_{i=1}^3\dot{A}^a_i\dot{A}^a_i}_{\text{kinetic term}}+\ldots, $$
cf. e.g. this Phys.SE post. How this is written with a trace depends on the author's normalization of the trace.
| {
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Why does water cast a shadow even though it is considered 'transparent'? If you pour water from a container, the flowing water stream seems to cast a shadow. I am not sure you can call it a shadow, but it definitely is not letting all light through it. How is this possible and what uses can it have?
| I guess some light will get reflected at the interface air-water, which will lower the intensity of light through the stream. That and possibly lens-like phenomena which leads to a non-uniform distribution of light intensity on the other side.
| {
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Motion of a rigid body on application of constant force perpendicular to a point which is not the center of mass? Consider a rod which has a uniform mass distribution which is free to move in space. Now, it is known that if an impulse (perpendicular to rod) is applied which does not pass through the center of mass, it will undergo pure rotational motion (with translation).
Now consider a situation where the force is applied constantly, and it remains perpendicular to the point of application throughout the situation even as the body undergoes motion.
What would be the motion of the body in such a case? How could this motion be described mathematically?
| If the force is of constant magnitude, so is the torque applied about the center of mass. So the rotational motion is easy to calculate, as fixed rotational acceleration of $$ \alpha = \frac{F d}{I}$$ is applied.
The orientation is thus
$$ \theta(t) = \tfrac{1}{2} \alpha t^2 $$
But the direction of the force changes with time and so the acceleration of the center of mass
$$ \vec{a} = \frac{\vec{F}}{m} $$
As there is planar motion here, the force vector is
$$ \vec{F}(t) = \pmatrix{ F \cos \theta(t) \\ -F \sin \theta(t) } $$
Unfortunately the integral $$ \vec{v} = \tfrac{1}{m} \int \vec{F}(t)\,{\rm d}t $$ does not an analytical solution. But it is easily evaluated numerically, or experimentally.
| {
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What is the uncertainty in time if I am doing a video analysis of a falling object using my phone camera at 60 FPS. Is it just ±1/60s? As a physics experiment where I was trying to find the terminal velocity of an object, I decided to record the drop and analyse it on Loggerpro frame by frame, but did not use a stopwatch. I created a position vs. time graph, so I could use the gradient as the velocity. I recorded at 4k 60 FPS, but I am unsure of the uncertainty in my time.
| The best source for questions on estimating and reporting uncertainty is the NIST guide: https://emtoolbox.nist.gov/Publications/NISTTechnicalNote1297s.pdf
In section 4.6 it describes how to treat a measurement like this. Basically, at 60 FPS if you can identify the single frame where an event happened then that gives the time to within $\frac{1}{60}\text{ s}$.
The actual time of the event could be anytime in that window, with equal probability. So you would model it as a uniform distribution so, per section 4.6, the estimated standard uncertainty would be $$u=\frac{1}{2\sqrt{3}}\frac{1}{60}\text{ s}$$ The factor in front accounts for the shape of the distribution.
| {
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