# Tag Info

## Hot answers tagged classical-electrodynamics

29

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 ...

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 ...

13

J. D. Jackson in the introductory remarks of his chapter on 'Radiation Damping, Classical Models of Charged Particles' (3rd edition), says that the problem of radiation reaction on motion of charged particles is not yet solved. He says that we know how to find motion of charged particles in given configuration of EM fields and also how to calculate EM fields ...

12

Assume for simplicity that the speed of light $c=1$. The existence of the gauge $4$-potential $A^{\mu}=(\phi, \vec{A})$ alone implies that the source-free Maxwell equations $$\vec{\nabla} \cdot \vec{B} ~=~ 0 \qquad \text{no magnetic monopole"}$$ $$\vec{\nabla} \times \vec{E} + \frac{\partial \vec{B}}{\partial t} ~=~ \vec{0}\qquad \text{Faraday's ... 10 Your TA is right that energy density alone does not trigger black hole formation. Consider a ball that's sitting still. Now speed up and look at the ball again. It will have gained (kinetic) energy. Relativistically, you can make the ball's energy density arbitrarily large by moving sufficiently near the speed of light. But the ball hasn't done anything in ... 10 Yes, of course that if a field - magnetic field - is able to make a bar magnet rotate or move, it is doing work. The statement that magnetic fields don't do any work only applies to point-like pure electric charges. Magnetic moments may be visualized as objects with a forced motion of charges (solenoids have the same magnetic field as bar magnets), and if ... 8 I) Let us just for fun generalize OP's question to n spacetime dimensions, and check how the counting of eqs. and degrees of freedom (d.o.f.) work out in this general setting. We shall use Lubos Motl's answer as a template for this part. Also we shall use a special relativistic (-,+,\ldots,+) notation with c=1, where \mu,\nu\in\{0,\ldots,n-1\} denote ... 7 In recent years, it has become apparent that a class of materials called topological insulators can be described by an action where the term E\cdot B is added. The action is$$ S_{top} = S_{em} + \frac{\theta}{2\pi}\frac{e^2}{\hbar c·2\pi} \int d^3xdt\, E·B.$$For ordinary insulators, we have \theta=0 while for topological insulators, we ... 6 Electromagnetism is parity-symmetric. Because all other terms in the action - such as mv^2-V(x) for particles - are parity-even, the electromagnetic contribution has to be parity-even, too. Otherwise the different terms would transform differently and the combined theory would violate parity. "Parity-even" simply means that the Lagrangian density is a ... 6 Both sides of the equation given are vectors and so represent 3 equations which are, on a Cartesian basis: E_x + \frac{\partial A_x}{\partial t} = -\frac{\partial V}{\partial x}  E_y + \frac{\partial A_y}{\partial t} = -\frac{\partial V}{\partial y}  E_z + \frac{\partial A_z}{\partial t} = -\frac{\partial V}{\partial z}  6 Note that since F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu, we have the equation$$ \partial_\mu F_{\nu\alpha} + \partial_\alpha F_{\mu\nu} + \partial_\nu F_{\alpha\mu} = 0 $$This equation is called the Bianchi Identity. This equation is separate from the equations of motion one obtains from varying the action. It can be shown that the Bianchi ... 6 Magnetic fields never do work directly. This is because the magnetic force on any charged particle,$$\mathbf{F}=q\mathbf{v}\times \mathbf{B,}$$is always orthogonal to the velocity, and therefore the power transferred, \mathbf{F}\cdot\mathbf{v}, is zero. On the other hand, this seems to contradict much of our intuition about how magnets behave. If you ... 5 Retarded propagators are those with G(\dots, t,t')=0 for all t<t'. They're vanishing before t=t', the delta-function "stimulates" the field at t=t', and the Green's function for positive t-t' measures the response of the field. One may view this description as a construction of the Green's function which also proves that it's ... 5 classical electrodynamics mainly deals with two kinds of proplems: a) The action of a field on a charged particle and b) the fields arising from the motion of such a field. Of course, this can only be approximative but it turns out that a lot of phenomena can be described in this way. However, you are right, an entire treatment would include a) and b) ... 5 The superscript * is a common notation for complex conjugate. Going back to check, (3.53) in the blue English edition states$$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),$$making is clear that it has to be complex conjugation. 5 There are still some very important open problems in the classical electromagnetism of relativistic charges, and there is indeed no satisfactory resolution of the reaction force and self-field problems for a relativistic point charge. One good resource for this is Conservation laws and open questions of classical electrodynamics. Marijan Ribarič and Luka ... 5 First, \vec{r}^\prime is a vector that goes from the origin to the source of charge. If the source is a volumetric distribution, one must sum all contributions of charge, that's why one integrates over all the volume, say \mathcal{V}; the (correct) expression for the potential should be$$V( \vec{r}) = \frac{1}{4 \pi \epsilon _{0}} \int_\mathcal{V} ...

5

From a geometric perspective, the last two equations are a consequence of: (1)F = dA (Faraday tensor is the exterior derivative of the four-potential) (2) dd = 0 (the exterior derivative of the exterior derivative vanishes) Thus (3) dF = 0 which gives the last two equations in your question.

5

Note. As indicated by user23660, the EM wave must be transverse which means the $x$'s in your phases should instead be $z$'s. At a given time $t$ and spatial point $\mathbf x = (x,y,z)$, the electromagnetic wave you consider is a combination of both of the fields; \begin{align} \mathbf E(t,\mathbf x) &= E_0\sin(kz-\omega t) \hat{\mathbf x} \\ ...

5

Yes. There is a standard way to generalize to field theory. Let a theory of $n\geq 1$ fields $\phi^i$ with a Lagrangian density $\mathcal L = \mathcal L(\phi^i, \partial_\mu\phi^i)$ be given. Here we use that standard abuse of notation in which $\phi^i$ denotes the vector whose components are the fields; $\phi^i = (\phi^1, \dots, \phi^n)$. To obtain the ...

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

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

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 ...

4

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 $$\tag{1} {\cal L} ~=~{\cal L}_{\rm Maxwell} + {\cal L}_{\rm matter} ,$$ $$\tag{2} {\cal L}_{\rm Maxwell}~=~ -\frac{1}{4}F_{\mu\nu}F^{\mu\nu},$$ $$\tag{3} {\cal L}_{\rm matter}~=~{\cal L}_{\rm ... 4 In general physics course we assume Maxwell's equations as the result of many experiments. After that, in field theory we build the Lagrangian which satisfies the Maxwell equations. We also can build non-linear field theories of EM interactions, but there is a requirement to getting of the Maxwell equations in the limit of weak fields. So these methods ... 4 Well F_{\mu\nu} is gauge invariant, so you can't gauge it away. The other reason is to remember that the photon has two polarizations (this is really one and the same, as you can show two gauge redundancies), so you can't set more than two components of A_\mu = 0. In short, if you start with the Lorentz gauge, \partial_\mu A^\mu = 0, you can further ... 4 I) For a general Lagrangian L(q,v,t), the Legendre transformation may be singular, i.e. the velocities v^i in the momentum relations$$\tag{1} p_i~:=~\frac{\partial L(q,v,t)}{\partial v^i}$$cannot be isolated. How to perform a singular Legendre transformation to achieve the corresponding Hamiltonian formulation goes under the name Dirac-Bergmann ... 4 This is not correct: "Since the relativistic Lagrangian is transitionally invariant, there is no dependence on coordinates so the last term is zero." In fact,$$ \frac{\partial L}{\partial {\bf r}} = q\nabla\left[ \frac{1}{c} \left({\bf u} \cdot {\bf A}\right) - \phi \right] = q\left\{\frac{1}{c} \left[\left({\bf u} \cdot \nabla \right){\bf A} + {\bf u} ...

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