# Relativistic Coulomb Force

Discussion in 'Physics & Math' started by tsmid, May 16, 2017.

1. ### tsmidRegistered Senior Member

Messages:
368
You thought wrongly then. You integrated from -L/2/(1-v/c) to +L/2/(1-v/c) which is the width of the retarded distribution. The unretarded distribution extends only from -L/2 to +L/2. Of course, your integration limits are correct (provided the propagation speed for the static electric field has indeed a finite value c), but you failed to take into account the correspondingly reduced charge density over this interval.

Anyway, I believe I have heard enough now in this matter, and I think I'll be able to progress from here and develop this further on my own. I thank you for your contributions which have helped to clarify the issue for me.

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3. ### przyksquishyValued Senior Member

Messages:
3,171
It is also the range of coordinates $x'$ for which $\rho(x', t_{\mathrm{R}})$ is nonzero, where $\rho(x, t)$ is the unretarded charge density function commonly used in electromagnetism that you are supposed to use in the integral, as you would know if you were familiar with how the integral is derived from Maxwell's equations.

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5. ### tsmidRegistered Senior Member

Messages:
368
No, it isn't. The function ρ is zero for x'<-L/2 and x'>L/2 , so it does not give the extent of the retarded distribution. The latter is not only wider in this case but also has a smaller density over its whole range, as shown below (and as you agreed just a few posts ago). As I indicated earlier, you are confusing yourself here by using the same variable ρ for the unretarded and retarded distributions.

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7. ### przyksquishyValued Senior Member

Messages:
3,171
I have no idea where you're getting this from, particularly since I explained in quite some detail how $\rho$ is defined in post #171 (for the moving line) and elsewhere (for how $\rho$ is defined in general).

For the moving line $\rho(x, t)$ is zero for $x < -L/2 + vt$ and $x > L/2 + vt$. Note that these limits depend on $t$.

$\rho(x',\, t_{\mathrm{R}})$ is zero for $x' < -L/2 + v \, t_{\mathrm{R}}$ and $x' > L/2 + v \, t_{\mathrm{R}}$. Under the conditions described in post #171 these correspond to $x' < -\frac{L/2}{1 - v/c}$ and $x' > \frac{L/2}{1 - v/c}$, taking into account that $t_{\mathrm{R}}$ itself depends on $x'$.

The charge density depends on what the density is defined relative to. You seem to have been assuming that $\rho$ must mean the charge density with respect to the integration variable $\boldsymbol{r}'$ in the integral $\int \mathrm{d}^{3}r' \, \frac{\rho(\boldsymbol{r}',\, t_{\mathrm{R}})}{R}$. But that would be the retarded charge density, which I brought up specifically to say is not what is used in the integral. I don't know where you got the idea it is. The integral expresses the potential in terms of the (unretarded) charge density function $\rho(\boldsymbol{r}, t)$ normally used in electromagnetism, though in the integral it is evaluated at the retarded time $t_{\mathrm{R}}$.

I did agree with you that the retarded charge density is different from the unretarded charge density, but I have no idea why you think this is noteworthy since I told you this and related points multiple times before. It was me, not you, that first referred to a "retarded charge density" in this thread (in post #179, where I even denoted it $\tilde{\rho}$ instead of $\rho$) and even your figure from post #195 is only illustrating information I'd already stated or derived in posts #171, #190, and #192.

Last edited: Aug 20, 2017