6
$\begingroup$

Feynman's proof utilizes a geometrical and fundamental integration argument. I like it, except this bit:

FLP

What makes me unconfortable somehow is that in (c) we are counting in some of the charge we counted at (b). It seems to me that it is this extra counting which makes the potential to be larger than expected, and I am uncomfortable with it. To see why, consider the following situation with discrete charges:

enter image description here

Here, the yellow line represents the light cone (the observer being, of course at its apex), and the blue dots, the places where the observer "sees" each of the constituent charges. However, it is clear that if the charge cloud was small enough, or if we were far enough, the potential would be just the potential for a point charge of charge equal to the total charge of the cloud, as no charge is "overcounted" (something which is also due to the cloud's speed being less than c).

This is in disagreement with what Feynman derives and with the Liénard-Wiechert's potential of a moving charge. So why is my argument wrong? Is the continuity of the cloud, somehow crucial for the proof? If so why?

$\endgroup$
  • 2
    $\begingroup$ You're not the only one who's noticed this double counting: arxiv.org/abs/0704.1574. What matters is if it gives a correct solution to Maxwell's equations and Feynman's derivation does. $\endgroup$ – Larry Harson Dec 10 '14 at 0:13
2
+50
$\begingroup$

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 Liénard-Wiechert potentials), and specifically claim that they are not in conflict with your discrete charge example, at least for the case of uniform motion.

The argument proceeds in two steps: 1) calculation of the potentials via relativity, 2) comparison with the Liénard-Wiechert potentials.

1) For a discrete charge in uniform motion, the key is that $(\phi, \mathbf {A})$ form a relativistic 4-vector potential, which transforms like any other 4-vector under a Lorentz transformation. In particular, $\phi$ transforms like the time $t$.

Consider, in the "primed" coordinates, a stationary discrete charge at the origin. Its potential $\phi'$ is simply:

$$ \phi'(x',y',z',t') = \frac{q}{4 \pi \epsilon_0 r'} = \frac{q}{4 \pi \epsilon_0} \frac{1}{\sqrt{x'^2 + y'^2 + z'^2}} \, , \,\, \mathbf {A'}=0 $$

Now define the "unprimed" coordinates as boosting the charge into motion with speed $v$ and velocity $\mathbf{v}$ parallel to the $x$-axis, so that the coordinate Lorentz transformation reads:

$$ t'= \gamma \left( t - \frac{v}{c^2}x \right) \, , \, x' = \gamma (x - vt ) \, , \, y' = y \, , \, z' = z \, , \, \gamma=\frac{1}{\sqrt{1-(v/c)^2}}$$

Then, applying this same Lorentz transformation to the 4-potential, we get:

$$ \phi(x,y,z,t) = \gamma \phi' (x', y', z', t') = \gamma \frac{q}{4 \pi \epsilon_0} \frac{1}{\sqrt{\gamma^2(x-vt)^2 + y^2 + z^2}} \, , \, \mathbf{A} = \gamma \frac{\mathbf{v}}{c^2} \phi'= \frac{\mathbf{v}}{c^2} \phi$$

as the 4-potential of a charge with constant velocity $v$ in the $x$-direction. Actually, Feynman performs this same calculation in Section 25-5.

2) The kicker is that this expression is precisely what one gets from the Liénard-Wiechert potentials, as Feynman demonstrates in Section 21-6.

As I said, I'm not going to try to defend Feynman's derivation. I suspect a relativistic alternative could be developed that gives the same, correct, Liénard-Wiechert potentials as a result (but I don't have one in hand).


Update: Ah, Jackson (sections 12.11 and 14.1) develops a general expression for the potentials in the Lorenz gauge that is directly applicable to your point charge example. In Feynman's notation:

$$A^\mu(1,t_1) = \frac{1}{4 \pi \epsilon_0 c^2} \int {\mathrm dt_2~\mathrm dV_2 \frac{\delta(t-t_2-r_{12}/c)}{r_{12}} J^\mu(2,t_2)} $$

where the four-vectors $A^\mu = (\phi/c, \mathbf{A})$ and $J^\mu=(\rho c, \mathbf{J})$. The $\delta$-function picks out exactly the retarded points on the past light cone of the observation point. [It will also turn out to introduce the non-obvious "stretching factor" that arises in Feynman's development.]

For a point charge, $ \rho(\mathbf{x},t)=q \, \delta[\mathbf{x}-\mathbf{r}(t)] $, the four-current is: $$ J^\mu(x) = qc \int { \mathrm d \tau \,U^\alpha (\tau) \delta^{(4)}[x-r(\tau)] } $$

where $ \tau $ is the particle's proper time, $x^\mu=(ct,\mathbf{x})$, $r^\mu(\tau) = (c\tau, \mathbf{r}(\tau))$, the four-velocity $U^\mu = ( \gamma c , \gamma \mathbf{v} )$, and $ \gamma = 1/\sqrt{1-(v/c)^2} $. (The "extra" $c$ at the front of $J^\mu$ is needed to balance the $ct$ within the $\delta$-function; these two $c$'s cancel when the integration over $t$ is performed.)

Specializing to the potential $\phi$ and integrating over the 4-volume, we get: $$ \phi(1,t) = \frac{q}{4 \pi \epsilon_0 c} \int { \mathrm d \tau \, U^0(\tau) \frac{\delta(t - t'(\tau) - r_{12}(\tau)/c)}{r_{12}(\tau)} } $$ where $t'$ is the retarded time.

Now to evaluate that delta function, we use the rule: $$\delta(f(\tau)) = \sum_i \frac{\delta(\tau-\tau_i)} {\left| \left( \frac{\mathrm df}{\mathrm d\tau} \right)_{\tau=\tau_i} \right|} $$ where $\tau_i$ are the zeros of $f(\tau)$. Here there is only one zero (that is, one contributing point from the past light cone), and $$ \frac{df}{d\tau} = \frac{\mathrm d}{\mathrm d\tau}{\left( t - t'(\tau) - r_{12}(\tau)/c) \right)} = - \gamma (\tau) + \gamma (\tau) v_r (\tau) /c = - \gamma (\tau) ( 1 - v_r (\tau) /c ) $$ where $v_r$ is the component of the charge's velocity towards the observer (same definition as Feynman's). Note the appearance of the "stretching" or "over-counting" factor.

Well, the $\gamma$ in $U^0$ in the numerator cancels the $\gamma$ arising in the denominator from the $\delta$-function, and we are left with $$ \phi(1,t) = \frac{1}{4 \pi \epsilon_0} \frac{q}{r_{12}'} \frac{1}{1 - v_r'/c }$$ where the prime denotes evaluation at the retarded time.

So, there's no contradiction: a correct relativistic treatment of a point charge gives the same result as Feynman's analysis. The "over-counting" that concerned you in Feynman's development is just an approximation to the exact behavior of the light-cone delta-function, reducing to it in the limit. (I think it's akin to calculating a function's derivative by changing the argument by a small $\delta$, calculating the rise/run ratio, and then letting $\delta$ go to 0.)

$\endgroup$
  • $\begingroup$ Thanks, but that doesn't really explain why Feynman's approach gives the right result while the method seems wrong to me. $\endgroup$ – guillefix Oct 15 '14 at 18:48
  • $\begingroup$ Btw, a minor thing, but I think that your 4-current should have an extra factor of c (Jackson agrees with me). I think it then goes when you integrate the temporal part of the 4-delta function over dt_2 as the delta function has a factor of c inside. $\endgroup$ – guillefix Nov 6 '14 at 17:57
  • $\begingroup$ @guillefix, thanks! I see now I got sloppy with $c$ in a couple places. (I understand better now why people set $c=1$!) I will patch it today or tomorrow. (I think I need more time than I've got right now, to avoid making another goof.) $\endgroup$ – Art Brown Nov 6 '14 at 18:28
  • $\begingroup$ @guillefix, think I got it. Thanks again for the catch. $\endgroup$ – Art Brown Nov 7 '14 at 1:24
  • 1
    $\begingroup$ @MAFIA36790, thank you. I have corrected the section reference. $\endgroup$ – Art Brown Jun 15 '16 at 18:11
0
$\begingroup$

I don't think the increase in potential due to the moving charge leading to an "overcounting" IS in disagreement with Feynman's result. In fact, the "overcounting" is what leads to the 1/(1-v/c) enhancement factor in the potential equation precisely accounts for the fact that the slow moving charge has some of it's charge contributions "overcounted" because the count is over the space-time distribution of the charge, not just it's spatial distribution.

When you say "it is clear that if the charge cloud was small enough, or if we were far enough, the potential would be just the potential for a point charge of charge equal to the total charge of the cloud" you've also implicitly made the assumption that the charge is moving slowly enough that it's distribution may be integrated over at a single time co-ordinate. When the charge moves relativistically, we gain the 1/(1-v/c) factor which you can consider leading to an effective charge q' = q * 1/(1-v/c) which DOES increase so that the "point like charge" equation form is retained, but that ignored the relativistic nature of the charge. Feynman highlights this when he says the equation preceding 21.28 is wrong.

So think the way to think of it, is that the terms in the denominator for the potential act as an "enhancement factor" to the charge because of the integration over it's history where the same spatial element of the charge may contribute over a period of time to the potential felt at any given moment to the observer.

At least, that's how it seems to me...

$\endgroup$
  • $\begingroup$ I said " It seems to me that it is this extra counting which makes the potential to be larger than expected", so I agree with what you say in the first paragraph. The rest, you seem to just be repeating the results, but not really address my argument. $\endgroup$ – guillefix Oct 4 '14 at 10:26
  • $\begingroup$ Hmmm, I think I may have misinterpreted your question somewhat. As to why this we may apply this reasoning to the case of discrete point charges, Feynman provides: "Also, the analysis we have made goes exactly the same way for a charge distribution of any shape—it doesn't have to be a cube. Finally, since the “size” of the charge q doesn't enter into the final result, the same result holds when we let the charge shrink to any size—even to a point." So the discrete charge case is a limit as "a" (the length of the square) goes to 0, which appears nowhere in the expression for the potential. $\endgroup$ – user3821372 Oct 5 '14 at 10:55
0
$\begingroup$

According to CK Whitney in multiple papers starting in 1987, the Lienard-Wiechert potential of electrodynamics does not exhibit conservation of electric charge, similar to what the author of this question points out. None of her papers can be found on the internet. She points out that the Lienard-Wiechert potentials are a solution to Maxwell's equations, but do not satisfy the appropriate boundary conditions for highly-relativistic sources. According to her, this is due to a Jacobian factor missed, as many physicists do not apply a rigorous treatment of distribution theory and work with the Dirac delta function. JH Field, Chubykalo et al, and others have found the same and/or similar corrections as Whitney, both suggesting the necessity of action-at-a-distance or instantaneous phenomena.

Jackson refutes Chubykalo's argument by claiming that Lienard-Wiechert potentials are indeed a solution of Maxwell's equations, but Chubykalo did not state the issue as precisely as Whitney, which is related to boundary conditions rather than solutions to the differential equations. Before Field or Chubykalo, Harold Aspden seemed to suggest that instantaneous fields were only needed if the internal structure of hadrons was different than leptons, which I think might be true in Einstein-Cartan theory. Aspden has worked on various aspects of aether theories, and Whitney is also against notions of Einsteinian relativity, which led to their work being largely disregarded by the physics community at large.

I find it interesting how M-theory is stating that everything can be described by noncommutative D0-branes, which to my limited knowledge seems to exhibit an SO(9) Gallilean symmetry and/or Little group in the transverse space, suggesting the possibility of instantaneous phenomena for strongly-interacting systems that may cause ultra-relativistic processes. Jackson also points out that other gauges in classical electrodynamics lead to instantaneous dynamics, but is not needed in the Lorenz gauge. Whitney's solution is geometrically simpler than Lienard and Wiechert, and resolves the issue pointed out by the author. My thesis briefly discusses aspects of Whitney's argument and cites many relevant references for further study.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.