I have a follow-up question from my answer to a previous question.

The electric field of a point charge moving at a constant velocity, as derived from the Liénard-Wiechert equation, reads $$ E = \frac{q}{4\pi\epsilon_0} \frac{(1-\beta^2)}{(1-n\cdot\beta)^3}\frac{n-\beta}{|r-r_s|^2},$$ where $ n =\overline{r-r_s}$ and $ \beta = v/c $.

There is nothing controversial about this, it is repeated in various textbooks and the derivation gets taught. For example, these notes or these ones.

In contrast, here is the electric field for the same point charge moving at constant velocity derived from the relativistic field tensors, $$ E = -\frac{e}{4\pi\epsilon_0} \frac{\gamma}{(1+u_r^2\gamma^2/c^2)^{3/2}}\mathbf r,$$ from this link.

Here is the E field for constant velocity derived by Griffiths, $$ E = \frac{q}{4\pi\epsilon_0} \frac{(1-\beta^2)}{(1-\beta^2\sin^2\theta)^\frac{3}{2}}\frac{n-\beta}{|r-r_s|},$$ as reviewed in Wikipedia.

Griffith's version is symmetric about $y$, Liénard-Wiechert is not.

If $n \cdot \beta = |\beta| \sin\theta$, then

  • $ \sqrt{1-\beta^2\sin^2\theta} = \sqrt{(1-|\beta|\sin\theta)(1+|\beta|\sin\theta)} $ in one case, and

  • $ (1-n \cdot \beta) = (1-|\beta|\sin\theta) = \sqrt{(1-|\beta|\sin\theta)^2} $ in the other case.

Presumably someone has made a sign error. One or the other of these formulas is wrong.

Have I misunderstood this? Are they both right?

  • 1
    $\begingroup$ Please use LaTeX formatting for formulas instead of images. $\endgroup$ Oct 31 '17 at 10:50
  • $\begingroup$ No worries. Just take note of the changes and follow suit on your future posts ;-). $\endgroup$ Oct 31 '17 at 11:10
  • $\begingroup$ Your expression for $n\cdot\beta$, if that's just a dot product it should have (the even in $\theta$) $\cos\theta$, not $\sin$. $\endgroup$
    – CDCM
    Oct 31 '17 at 11:31
  • $\begingroup$ Yes, agreed! I haven't actually seen Griffiths's derivation. If he chooses to call the velocity direction I call 0 instead pi/2, then it works out. Otherwise I get something like this: $$\sqrt{1 -\beta^2\sin^2\theta^2}=\sqrt{(1 -\beta^2)+\beta^2 -\beta^2\sin^2\theta^2} = \sqrt{(1 -\beta^2)+\beta^2 (1-\sin^2\theta^2)} = \sqrt{(1 -\beta^2)+\beta^2 cos^2\theta^2} $$ And it comes out somewhat more different. $\endgroup$
    – J Thomas
    Oct 31 '17 at 23:17

I bought Griffifths's book and looked at his derivation.

I thought he had something wrong because I didn't read it carefully enough. His angle theta is not the angle I thought it was. We have the direction from the source at retarded time to the target, $n$. And we have the velocity vector, $v$. I thought theta was the angle between $n$ and $v$.

But instead his theta is the angle between $n-v$ and $v$.

Also, he does not consider the distance between the source at retarded time and the target, $R$. He instead considers the distance between the source at present time and the target at present time, $R'$.

So when he accounts for distance, he uses not $\frac{n-v}{R^2}$ but instead $\frac{\overline{n-v}}{R'^2}$

When I make those changes, the result is exactly the same as L-W.

It's possible to make a physical interpretation for both graphs. In the first case, we have a bunch of source charges which are equidistant from a target at retarded time, and they all travel at velocity V. Since they are equidistant, their force arrives at the target at the same time, now, while the charges themselves each move. The direction each force pushes is tilted from its direction at retarded time, and that direction happens to coincide with the charge's location at present time. The magnitude of the force is also changed.

forces from circle of charges at retarded time

With constant velocity, we can instead consider the following physical interpretation: Consider a bunch of source charges which are equidistant from a target at present time, and they all travel at velocity V. Imagine that the speed of light is infinite. Then the forces will all arrive at the same time, now. None of the forces are tilted relative to their sources, but their magnitudes are reduced inline with velocity, and increased normal to it.

This fanciful interpretation works only for constant velocity. If the sources do not reach their proper place at present time, the illusion is destroyed because in fact the forces depend on the location and velocity at retarded time.

forces from circle of charges at present time


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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