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33

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 ...


14

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 ...


13

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 ...


13

Let's take a slightly more general case: Consider a wave with wave vector $\vec k=(k_x,k_y,k_z)$, with the electric field given by $$\vec E=\vec E_0\ e^{i(\vec k \cdot \vec r-\omega t)} $$ where $\vec r=(x,y,z)$. Now, we wan't to satisfy Maxwell's equations in the vacuum, including Gauss' law: $$\vec \nabla \cdot \vec E=0$$ The derivative is quite easily ...


12

You are missing nothing. The Bohr model of the atom is false, and nowadays we replace the idea of the semi-classical "orbit" of Bohr with the fully quantum mechanical notion of orbitals or electron clouds, which give a probability distribution for the position of the electron around the nucleus, but do emphatically not imply that the electron is moving in ...


12

D.J. Griffith's Introduction to Electrodynamics must be mentioned. To my knowledge this text is ubiquitous in junior-level E&M courses. The writing is extremely friendly and is excellent for self-study. The author frequently tells you what he is doing and provides motivation, unlike the ubiquitous graduate-level text by Jackson. Equations often use a ...


11

You are correct when you concluded that two classical point electrons could never touch each other. It would take infinite energy.


11

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 ...


11

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 ...


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 ...


9

The gradient of a scalar is again vector.


9

Purcell is a good non-Griffiths option. I would judge the completeness of the material between Griffiths and Jackson, but with an intuitive level of understanding close to Griffiths. I used it to study for graduate qual exams when Jackson was making me feel particularly obtuse. Some positives: Touches more ideas than Griffiths Uses some real-world ...


9

The Maxwell Lagrangian is given by, $$\mathcal{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$$ where $F_{\mu\nu}$ is the field-strength of the gauge field, or alternatively may be interpreted as the curvature of a $U(1)$ Lie algebra valued connection, $A_{\mu}$. By applying the variational principle we obtain, $$\partial_\mu F^{\mu\nu}=0$$ in vacuum. In terms of ...


8

There is another 'infinity' (among others) lurking in classical electrodynamics which is evident when one calculates the electrostatic energy $W$ of a uniform spherical charge distribution of radius $a$ and total charge $Q$ $$W = \frac{3}{5}\frac{Q^2}{4\pi \epsilon_0 a}$$ Thus, by this result, a point (zero radius) particle of charge Q has 'infinite' ...


8

The following passage has been extracted from Bohr's Nobel lecture: While in contradiction to the classical electromagnetic theory no radiation takes place from the atom in the stationary states themselves, a process of transition between stationary states can be accompanied by the emission of electromagnetic radiation, which will have the same ...


8

Formal accounts of EM do not need the Poincare lemma for the vector potential to exist: Formulating a theory like electromagnetism on any manifold is best done by viewing it as the $\mathrm{U}(1)$ gauge theory on said manifold. Then, the "vector potential" is simply a connection on the circle bundle over the manifold, which exists regardless of whether ...


8

The necessary and sufficient condition for $dB = 0$ (more commonly written $\nabla \cdot \mathbf B = 0$) to imply $B = dA$ is the vanishing of the second de Rham cohomology $H^2 (M)$. This is guaranteed for a contractible manifold since cohomology is homotopy invariant. However this is for a 2-form $B$ defined on all of $M$. Consider instead the restriction ...


8

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 ...


7

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 ...


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 ...


7

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.


7

In this answer, I'll start with a real expression for $E$, because I think the exposition is clearer. There is no loss of generality in doing that, because the real expression will always be equivalent to the real part of the complex version of $E$, for some appropriate choice of the origin. Thus, my starting point is $$E(z,t)=E_0\ sin(k z-\omega t)\ .$$ ...


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

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

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

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) ...


6

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 ...


6

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 ...


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 ...



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