# Tag Info

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You can only calculate electric fields or magnetic fields after fixing a reference frame, so no, you can't move P around in that formula. It is assumed in that formula that you are working in a specific frame. The formula is invariant with respect to translating both $\mathbf{r}$ and $P$ by the same displacement, but not with respect to boosting them by the ...

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The idea is that it takes time for a signal to travel from a source to where it is being observed--so the field here and now doesn't depend on the charge distribution now, it depends on the value that the charge distribution had $t - \frac{\ell}{c}$ ago, since information cannot travel instantaneously.

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This is the usual argument for explaining retarded time - Consider a charge moving with a constant velocity along a straight line. If the charge suddenly comes to a halt, there will be a change in the electric field due to the acceleration. But this change in the electric field isn't communicated instantaneously through the whole universe, that's ...

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Actually since charge is at rest $u_{\nu}r^{\nu} = u_0 r^0 = ct'$ where $t'$ is retarded time, $t'=r/c$, where $r$ is the (constant) distance to the charge.

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I won't try to defend Feynman's derivation, which seems strangely non-relativistic. (A similar argument is used by Schwartz in his "Principles of Electro-Dynamics".) However, I will defend the result (the Lienard-Wiechert potentials), and specifically claim that they are not in conflict with your discrete charge example, at least for the case of uniform ...

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Both equations (for the instantaneous field of a charge moving with constant velocity $v$) are correct. (Well, maybe the primes should be swapped in the second equation, so that the unprimed frame is that in which the charge is moving.) The first figure is not an accurate representation of the first equation: as Jan Lalinsky stated, the field lines should ...

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There are many ways to do approximations. If you want to make mathematically reliable approximation instead of an "easy" one, I would use Taylor expansion in $\mathbf v/c$ around zero. Then you have to decide on the order of accuracy you want to use - if you want to have velocity dependence, you need to preserve constant and first order terms in $\mathbf v ... 1 To say that we assume that the mass$M$of the top particle is so large that its acceleration due to force$F$is negligible. means to take the limit$M \to \infty$(since$a = \frac{F}{M}$shall produce$a = 0$), since only infinitely heavy things do not accelerate when a force is applied. Under this assumption,$$\lim_{M\to\infty} E_\text{top} = ... 1 Your thought experiment stumbles upon an important idea in electrodynamics which is quite counter-intuitive.The EM field produced as radiation due to the charge in fact produces a reaction force on the charge itself. This is known as the Abraham–Lorentz force which is proportional to rate of change of acceleration of the charge. In SI units it is given by, ... 1 An Electric Field is only conservative if it is static. The propagation of E with a L-W field contradicts this, so it is not conservative. 1 Firstly, I think you've confused several points in your analysis of the situation. Working with the first picture, by standard electrostatic analysis, we see that there must be a standard Coulomb force between the plates for all times$t\ge t_2$, not just at$t=t_2$. This point is not too important, but I thought I'd mention it. Continuing to work with the ... 1 There is no obvious inconsistency, whether we use retarded, advanced, or any other field. If we use only retarded fields, things go as follows. At the time$t=0$, we begin to exert force$\mathbf{F}$on the charge$q$. It will move with acceleration$\mathbf{F}/m$for the time interval$R/c$, where$R$is the radius of the sphere. At the time$t = R/c\$, ...

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