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First question here. I'm really confused at the moment. An electron moves at constant velocity, no acceleration

Wikipedia says here Lorentz: $$\mathbf E=\frac{q}{4\pi\epsilon_0}\frac{1-v^2/c^2}{1-v^2\sin^\theta/c^2}\frac{\hat{\mathbf r}}{r^2},$$ which yields something like this:

enter image description here


Whereas here, Wikipedia says this and this, $$ \frac{E'_y}{E'_x} = \frac{E_y}{E_x\sqrt{1-v^2/c^2}} = \frac{y'}{x'}, $$ which yields something like this:

enter image description here

Which one is correct? If you could explain me exactly the reason why one of them is correct, I give you a big imaginary hug.

Last question: In none of those fields is there any radiated energy, since there is no acceleration, correct?

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    $\begingroup$ The first picture is wrong. Where did you get it? $\endgroup$ – Ján Lalinský Jan 12 '14 at 20:50
  • $\begingroup$ Just verified via Mathematica.(picload.org/image/lgicigd/field.gif) Yes, it may be a wrong denominator, but it looks the same, just mirrored. The more important thing are the formulas. ATM I'm reading in Jackson again... $\endgroup$ – simon.reiger Jan 12 '14 at 21:02
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    $\begingroup$ The field lines should be mirror symmetric with respect to plane of particle perpendicular to the velocity. The second picture looks fine. $\endgroup$ – Ján Lalinský Jan 13 '14 at 1:06
  • $\begingroup$ @JánLalinský, so this one phet.colorado.edu/sims/radiating-charge/… is incorrect? $\endgroup$ – user205695 Sep 13 '18 at 17:43
  • $\begingroup$ @Artur, unfortunately I am not able to view that page since it uses Flash. But based on the address, it seems the pictured field there is that of radiating charge, which is generally not as simple as the one here. For radiating charge the field lines are not straight. $\endgroup$ – Ján Lalinský Sep 13 '18 at 18:38
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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 be symmetric about $\theta=\pi/2$, the direction perpendicular to the velocity.

The second equation just says that the field lines are still radial for the moving charge, although they're no longer isotropic. Again echoing Jan Lalinsky, the second figure (another plot of the first equation) looks fine.

Finally, no acceleration does indeed mean no radiation.

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Liénard-Wiechert:

$$ E = \frac{q}{4\pi\epsilon_0} \frac{(1-\beta^2)}{(1-n\cdot\beta)^3}\frac{n-\beta}{|r-r_s|} $$

where $ n =\overline{r-r_s} $

$ \beta = v/c $

This is the first term of $E(r,t)$ from Wikipedia.

That thing in the denominator $ \frac{1}{1-n \centerdot\beta} $ is not symmetric in the direction of motion. the dot product is positive when n and beta are the same direction, and negative when they are opposite directions. We divide by a number that's less than one in one direction and bigger than one in the other. It will give a graph that looks a lot like the first picture in the original question (flipped around if the signs of the source and target charges are different) and not like the second picture at all.

This is a picture of the magnitude of the force at the red dot from particles moving to the right at $v = 0.5$.

magnitude of L-W force

This is a picture of the magnitude and direction of the force, $v = 0.5$.

magnitude and direction of L-W force

SO:

The Griffiths derivation and the Liénard-Wiechert derivations are different as follows:

$$ E = \frac{q}{4\pi\epsilon_0} \frac{(1-\beta^2)}{(1-n\cdot\beta)^3}\frac{n-\beta}{|r-r_s|} $$

$$ E = \frac{q}{4\pi\epsilon_0} \frac{(1-\beta^2)}{(1-\beta^2\sin^2\theta)^\frac{3}{2}}\frac{n-\beta}{|r-r_s|} $$

I eventually got a copy of Griffiths and found out what he's talking about. L-W uses retarded time. Griffiths does not. His distance $|r-r_s|$ is the distance between the charges in present time. His angle $\theta$ is the angle between the constant velocity and the location line in current time.

Basicly he's saying "This is what the force would be if it acted instantaneously between charges right now."

That works for constant velocity because the retarded-time position can be computed easily and directly from the current time position (and constant velocity), and vice versa.

I'm not clear what this result is good for, but this is what it's about.

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Both results seem correct to me. They are different because they represent fields measured in different frames of reference. Your first diagram represents the field due to a moving charge measured by a stationary observer (E). Your second diagram represents the field due to a moving charge measured by an observer moving with the charge (E').

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  • $\begingroup$ Zoé say the second diagram show the field lines of an observer moving with the charge. Since the motion of the observer/charge frame is uniform SR would require he should see the field lines just as if he and the charge were stationary? $\endgroup$ – Jens Mar 7 '16 at 23:14
  • $\begingroup$ I don't think this is right. The second diagram doesn't represents the field due to a moving charge measured by an observer moving with the charge. If that was the case, then the field lines would have been radial and isotropic as opposed to just radial. @Jens yes you're correct. Zoé goofed I believe. $\endgroup$ – thermomagnetic condensed boson Mar 7 '16 at 23:17
  • $\begingroup$ Ok, yes, my mistake sorry about that. $\endgroup$ – user110591 Mar 9 '16 at 14:15

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