Hot answers tagged classical-electrodynamics
18
The details of your analysis are not quite right - that's not what the electric field of a moving charge looks like, for example. This is probably because you haven't learned all the rules of electromagnetism yet. Still, the spirit of your question is hitting at an important point.
Charges do not conserve momentum and don't obey Newton's third law. You have ...
17
It isn't a problem because two of the eight equations are constraints and they're not quite independent from the remaining six.
The constraint equations are the scalar ones,
$$ {\rm div}\,\,\vec D = \rho, \qquad {\rm div}\,\,\vec B = 0$$
Imagine $\vec D=\epsilon_0\vec E$ and $\vec B=\mu_0\vec H$ everywhere for the sake of simplicity.
If these equations are ...
9
Your TA is right that energy density alone does not trigger black hole formation. Consider a ball that's sitting still. Now speed up and look at the ball again. It will have gained (kinetic) energy. Relativistically, you can make the ball's energy density arbitrarily large by moving sufficiently near the speed of light. But the ball hasn't done anything in ...
7
Yes, of course that if a field - magnetic field - is able to make a bar magnet rotate or move, it is doing work. The statement that magnetic fields don't do any work only applies to point-like pure electric charges.
Magnetic moments may be visualized as objects with a forced motion of charges (solenoids have the same magnetic field as bar magnets), and if ...
7
In recent years, it has become apparent that a class of materials called topological insulators can be described by an action where the term $E\cdot B$ is added.
The action is
$$ S_{top} = S_{em} + \frac{\theta}{2\pi}\frac{e^2}{\hbar c·2\pi} \int d^3xdt\, E·B.$$
For ordinary insulators, we have $\theta=0$ while for topological insulators, we ...
6
Electromagnetism is parity-symmetric. Because all other terms in the action - such as $mv^2-V(x)$ for particles - are parity-even, the electromagnetic contribution has to be parity-even, too. Otherwise the different terms would transform differently and the combined theory would violate parity. "Parity-even" simply means that the Lagrangian density is a ...
6
I) Let us just for fun generalize OP's question to $n$ spacetime dimensions, and check how the counting of eqs. and degrees of freedom (d.o.f.) work out in this general setting. We shall use Lubos Motl's answer as a template for this part. Also we shall use a special relativistic $(-,+,\ldots,+)$ notation with $c=1$, where $\mu,\nu\in\{0,\ldots,n-1\}$ denote ...
5
Retarded propagators are those with $G(\dots, t,t')=0$ for all $t<t'$. They're vanishing before $t=t'$, the delta-function "stimulates" the field at $t=t'$, and the Green's function for positive $t-t'$ measures the response of the field. One may view this description as a construction of the Green's function which also proves that it's ...
5
Is it correct to say that magnetic fields DO do work?
Yes! I show this quantitatively:
Each charged particle experiences action of magnetic force. This force is transmitted to a conductor in which the charges move. As a result, the magnetic field acts with a certain force on the current-carrying conductor. Let the volume charge density, (electrons in a ...
5
Conservation of angular momentum does not predict that the disk stays motionless, because the field in this case has angular momentum. The charges produce an electric field, and the magnetic field is not parallel to it, so there is a Poynting vector going around in circles, and the field angular momentum is just converted to mechanical angular momentum when ...
5
The superscript $*$ is a common notation for complex conjugate. Going back to check, (3.53) in the blue English edition states
$$Y_{l,m} = \sqrt{\frac{2l+1}{4\pi}\frac{(l-m)!}{(l+m)!}}P^m_l(\cos\theta)e^{im\phi}$$
which is followed by (3.54)
$$Y_{l,-m}(\theta,\phi) = (-1)^m Y^*_{l,m}(\theta,\phi),$$
making is clear that it has to be complex conjugation.
4
classical electrodynamics mainly deals with two kinds of proplems: a) The action of a field on a charged particle and b) the fields arising from the motion of such a field. Of course, this can only be approximative but it turns out that a lot of phenomena can be described in this way.
However, you are right, an entire treatment would include a) and b) ...
4
The claim that accelerated charges must radiate is simply false. There are very many simple situations in which they do, but in general things should be examined on a case-by-case basis; there is not simple thumb rule like "acceleration yields radiation."
The simplest way to see this is to consider a wire carrying a constant current. This situation is ...
4
The answer is: the Motion Mountain book is wrong, your TA and John Baez are correct
A couple answers here are discussing specific solutions in GR. I think there is a much easier and more general way to answer this.
In special relativity, we can choose to use a different coordinate system (we don't need to physically change our motion as some posters seem ...
4
Because there is a changing configuration of charges with time, a time-dependent dipole moment. The electron's field is coming from a different center at different times, so that the field is oscillating in magnitude with a period the orbital period of the electron. When you have an oscillation of electric fields, it sets up oscillations of the entire ...
3
The stress-energy tensor $T_{\mu\nu}$ on the right side of the Einstein equation characterizes all of the various forms of "stuff" in the spacetime. If there are electromagnetic fields in the spacetime, then the stress-energy tensor of the electromagnetic field is part of that $T_{\mu\nu}$, along with contributions from other forms of energy, mass, etc.
3
Whoever the PRL referee(s) was/were, they should have sent it back to the author to put the argument into a manifestly covariant formalism. The editors should have done the same before the paper got to a referee. As it is, everybody has to waste time unpicking the 3-d vector mess. 3-d vectors have a perfectly legitimate place in Physics, but not if one is ...
3
No, it is not possible, and the argument is simple--- there is no dimensional parameter with unit of length, so if there were a stable equilibrium at one radius, there would be many such equilibria obtained by rescaling the original solution to a one-parameter family of solutions.
In fact, it is easier to see that the stable solution is for the electron to ...
3
It may be a reference to the fact that you can reproduce the characteristics of the photoelectron production in a model which treats the incident light classically, but treats the matter in the target quantum mechanically. This is explained in Mandel and Wolf's book (chapter 9), which explains how a simple semiclassical calculation can be used to derive the ...
3
By the word classical we will mean $\hbar=0$, and we will use the conventions of Ref. 1.
The Lagrangian density for Maxwell theory with various matter content is
$$\tag{1} {\cal L} ~=~{\cal L}_{\rm Maxwell} + {\cal L}_{\rm matter} , $$
$$\tag{2} {\cal L}_{\rm Maxwell}~=~ -\frac{1}{4}F_{\mu\nu}F^{\mu\nu},$$
$$\tag{3} {\cal L}_{\rm matter}~=~{\cal L}_{\rm ...
2
Put speed of light $c=1$ and use sign convention $(-,+,+,+)$. To have that the canonical momentum
$$\vec{p}~=~\frac{\partial L}{\partial d_t \vec{x}}~=~ \gamma m_{0}d_t \vec{x} +q\vec{A}$$
is conserved $d_t \vec{p}=0$, and a $\vec{x}$-translational symmetry of the Lagrangian
$$L~=~T-U, \qquad T~=~-\frac{m_0}{\gamma}, \qquad U~=~ q (\phi -\vec{A}\cdot ...
2
We have a thermodynamic understanding of energy violation, in terms of perpetual motion, but momentum violation is somewhat less immediately paradoxical in a changing background, maybe because we have intuition that the field has momentum in this case. But the field momentum is irrelevant for this question. Mechanical momentum should be conserved in a ...
2
Solving both, Lorentz force and Maxwell's equation, does include radiation reaction. Radiation reaction emerges from the interaction of the charge with the field it has emitted itself, the self-field. If you do not want to solve Maxwell's equations and treat the fields as external, you can use Lorentz-Abraham-Dirac equation with some known inconsistencies ...
2
Here is an experimentalist's view of the question:
1) one photon hits the antenna and raises a molecular electron band to a higher energy level, and it will fall back to its lower one, with the characteristic electromagnetic transition time of the order of 10^-16sec, giving the energy to the antenna grid of molecules.
One photon will just disappear.
2)a ...
2
Generally speaking, the answer to both questions is linked to some number becoming increasingly large so that, for atoms you have a large density of higher excited states (think to Rydberg atoms as an example) or for electromagnetic field one has such a large number of photons that a coherent state is a good description of it and an average field can be ...
2
I'm not sure there is a generic answer to your questions other than the trivial "don't bother including the quantization when the accuracy of your result isn't compromised by making this approximation". I know that doesn't really help much, because you may not be able to verify this until you've done the calculation including the quantization anyway. You ...
2
I think i came to the origins of this equation. In all likelihood, this equation describes not a speed of an electrical impulse but a direct current power transmitted via a superconducting coaxial cable.
A proof:
Consider a simple transmission DC coaxial cable. To eliminate the energy losses due to Joule heating in the cable, the inner(of radius $r$) ...
2
There is a comment by Daniel A. T. Vanzella with a counter argument that essentially removes the paradox. He uses the natural covariant formulation of the problem. In this, you can see that the lorentz force has no spatial component in the charge/dipole rest frame, but the four force is not null. The dipole develops a time dependent angular momentum ...
2
Chapter 17 precedes chapter 27 which covers field momentum and so he's looking for a simple explanation involving mechanical angular momentum. The initial angular momentum of the system is carried by the initial current in the coil, and so there's no paradox.
Note also that the magnetic field can't collapse immediately, but has to disspiate the stored ...
2
The circuit likely was closed by his body, or by a grounding wire he was holding. At least that's one way the demonstration has been done. I assume the other end was in contact with the generator. Another way is to suspend the tube so it is not in electrical contact with anything, and swing it around, so that you observe a momentary discharge when the ...
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