# Tag Info

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That's a nice video - good find :-) If you run a current through a coil; it generates an magnetic field inside the coil like this: (Image from the Hyperphysics site.) If the field lines are exactly parallel a bar magnet will feel no net force. However at the ends of the coil, where the field lines diverge, a bar magnet will be either pulled into the ...

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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 ...

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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 ...

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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 ... 13 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 ... 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 Let me try this more clearly than the other answers, which aren't wrong. You ask: So, can someone please elaborate what this EM field is with respect to \vec E and \vec B in the context of Helmholtz decomposition? There is no "EM field in the context of Helmholtz decomposition". Helmholtz just says that every vector field \vec V is decomposable ... 11 Classical electrodynamics is certainly studied in curved spacetimes to understand real phenomena. What better place for gravity and electromagnetism to work together than the ionized, magnetized plasma surrounding an accreting black hole? In particular, we observe quasars with extremely powerful relativistic jets. Quasars are the supermassive black holes at ... 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 ... 10 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$$in vacuum. In terms of ... 9 The fundamental particles we know today (of which the photon is one) are called fundamental exactly because they have no substructure, or indeed, spatial extent, we know of. They are point-like when localized. Note that these "particles" are quantum objects, not classical particles, so you should not imagine them as points whizzing about in space - they ... 9 The gradient of a scalar is again vector. 8 Retarded propagators are those with G(\dots, t,t')=0 for all t<t'. They're vanishing before t=t', the delta-function "stimulates" the field at t=t', and the Green's function for positive t-t' measures the response of the field. One may view this description as a construction of the Green's function which also proves that it's ... 8 Conservation of angular momentum does not predict that the disk stays motionless, because the field in this case has angular momentum. The charges produce an electric field, and the magnetic field is not parallel to it, so there is a Poynting vector going around in circles, and the field angular momentum is just converted to mechanical angular momentum when ... 8 The following passage has been extracted from Bohr's Nobel lecture: While in contradiction to the classical electromagnetic theory no radiation takes place from the atom in the stationary states themselves, a process of transition between stationary states can be accompanied by the emission of electromagnetic radiation, which will have the same ... 8 There is another 'infinity' (among others) lurking in classical electrodynamics which is evident when one calculates the electrostatic energy W of a uniform spherical charge distribution of radius a and total charge Q$$W = \frac{3}{5}\frac{Q^2}{4\pi \epsilon_0 a}$$Thus, by this result, a point (zero radius) particle of charge Q has 'infinite' ... 8 Formal accounts of EM do not need the Poincare lemma for the vector potential to exist: Formulating a theory like electromagnetism on any manifold is best done by viewing it as the \mathrm{U}(1) gauge theory on said manifold. Then, the "vector potential" is simply a connection on the circle bundle over the manifold, which exists regardless of whether ... 8 The necessary and sufficient condition for dB = 0 (more commonly written \nabla \cdot \mathbf B = 0) to imply B = dA is the vanishing of the second de Rham cohomology H^2 (M). This is guaranteed for a contractible manifold since cohomology is homotopy invariant. However this is for a 2-form B defined on all of M. Consider instead the restriction ... 7 In this answer, I'll start with a real expression for E, because I think the exposition is clearer. There is no loss of generality in doing that, because the real expression will always be equivalent to the real part of the complex version of E, for some appropriate choice of the origin. Thus, my starting point is$$E(z,t)=E_0\ sin(k z-\omega t)\ .$$... 7 I'm going to take a risk and try to answer this, even though my answer is different to Lubos's and he does have a reputation that is overwhelming right compared to mine. Static Magnetic fields don't don't do work, period (us), full stop (uk). So the work comes from the magnetic dipole itself whose internal energy is affected by the external force that ... 7 In recent years, it has become apparent that a class of materials called topological insulators can be described by an action where the term E\cdot B is added. The action is$$ S_{top} = S_{em} + \frac{\theta}{2\pi}\frac{e^2}{\hbar c·2\pi} \int d^3xdt\, E·B.$$For ordinary insulators, we have \theta=0 while for topological insulators, we ... 7 Note that since F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu, we have the equation$$ \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 ...

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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.

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Both sides of the equation given are vectors and so represent 3 equations which are, on a Cartesian basis: $E_x + \frac{\partial A_x}{\partial t} = -\frac{\partial V}{\partial x}$ $E_y + \frac{\partial A_y}{\partial t} = -\frac{\partial V}{\partial y}$ $E_z + \frac{\partial A_z}{\partial t} = -\frac{\partial V}{\partial z}$

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