# Relativistic Coulomb Force

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

1. ### tsmidRegistered Senior Member

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368
It is clearly a contradiction if you require that the general distribution

$\delta \bigl( t' - t + \frac{1}{c} \lVert \boldsymbol{r} - \boldsymbol{r}' \rVert \bigr)$

is normalized, but not the specific one obtained by inserting r'=vt'

$\delta \bigl( t' - t + \frac{1}{c} \lVert \boldsymbol{r} - \boldsymbol{v} t' \rVert \bigr)$.

The same holds regarding what you said in an earlier post

$\int \mathrm{d}^{3}r' \, \rho(\boldsymbol{r}', t)$ is generally different from $\int \mathrm{d}^{3}r' \, \rho(\boldsymbol{r}', t - t_{\mathrm{R}})$ with $t_{\mathrm{R}}$ itself being a nontrivial function of $\boldsymbol{r}'$.

$\int \mathrm{d}^{3}r' \, \rho(\boldsymbol{r}', t)$ is simply a general requirement whatever the specific form of the arguments is. If that specific form does not result in a normalized distribution, then the general requirement is violated, so a corresponding re-normalization has to be performed.

Having said this, for the present purpose, you can in fact eliminate r' from the problem altogether by assuming r'<<r. So from

$\begin{eqnarray} \Phi (\boldsymbol{r}, t) &=& \frac{1}{4 \pi \epsilon_{0}} \int \mathrm{d}^{3}r' \, \frac{\rho \Bigl(\boldsymbol{r}',\, t - \frac{\lVert \boldsymbol{r} - \boldsymbol{r}' \rVert}{c} \Bigr)}{\lVert \boldsymbol{r} - \boldsymbol{r}' \rVert} \,, \end{eqnarray}$

you simply get then

$\begin{eqnarray} \Phi (\boldsymbol{r}, t) &=& \frac{1}{4 \pi \epsilon_{0}} \int \mathrm{d}^{3}r' \, \frac{\rho \Bigl(\boldsymbol{r}',\, t - \frac{\lVert \boldsymbol{r}\rVert}{c} \Bigr)}{\lVert \boldsymbol{r} \rVert} \,, \end{eqnarray}$

i.e.

$\begin{eqnarray} \Phi (\boldsymbol{r}, t) &=& \frac{1}{4 \pi \epsilon_{0}} *q/r \end{eqnarray}$

Alternatively, assume we are just considering the potential due to a sub-volume through which a larger cloud of charges is moving such that the density is constant in this sub-volume for all relevant times t-(r-r')/c. Again, the integration is trivial as you can even take ρ out of the integral altogether, being constant.

Last edited: Jun 28, 2017

3. ### przyksquishyValued Senior Member

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This is disingenuous. I just reminded you in the very post you are quoting from that the step in the derivation in question does not consist of just substituting $\boldsymbol{v} t'$ in place of $\boldsymbol{r}'$.

Maybe you have a different idea of how a derivation works than everyone else does. Normally the requirement is that the expression at each step in a derivation is equivalent to the previous step. So if you want to claim that there is something wrong with the derivation I posted then you should be able to explain why

$\displaystyle \frac{q}{4 \pi \epsilon_{0}} \int \mathrm{d}t' \, \frac{1}{\lVert \boldsymbol{r} - \boldsymbol{v} t' \rVert} \, \delta \bigl( t' - t + \tfrac{1}{c} \lVert \boldsymbol{r} - \boldsymbol{v} t' \rVert \bigr)$​

is not equal to

$\displaystyle \frac{q}{4 \pi \epsilon_{0}} \int \mathrm{d}^{3}r' \int \mathrm{d}t' \, \frac{\delta(\boldsymbol{r}' - \boldsymbol{v} t')}{\lVert \boldsymbol{r} - \boldsymbol{r}' \rVert} \, \delta \bigl( t' - t + \tfrac{1}{c} \lVert \boldsymbol{r} - \boldsymbol{r}' \rVert \bigr) \,.$​

It is not enough that $\delta \bigl( t' - t + \frac{1}{c} \lVert \boldsymbol{r} - \boldsymbol{v} t' \rVert \bigr)$ is different from $\delta \bigl( t' - t + \frac{1}{c} \lVert \boldsymbol{r} - \boldsymbol{r}' \rVert \bigr)$ on its own since these are only parts of the expressions they appear in alongside other terms and operations. That would be like saying $\alpha \times 2 \times 3 = \alpha \times 6$ is invalid because 6 is different from 3.

$\int \mathrm{d}^{3}r' \, \rho(\boldsymbol{r}', t)$ is a number. It is not a requirement on anything.

This is disingenuous. You're just repeating your belief that retardation can be neglected while the derivations I've already posted, which are exact, prove that it cannot.

If you do an exact derivation for a moving point charge you get a scalar potential proportional to $(R - \boldsymbol{\beta} \cdot \boldsymbol{R})^{-1}$. If you ignore $\boldsymbol{r}'$ in some places in the equation as an "approximation" then you only get $R^{-1}$. The only thing this proves is that your approximation is a bad one.

Last edited: Jun 29, 2017

5. ### tsmidRegistered Senior Member

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368
What would be the conceptual mistake with this? If you assume the speed of light does not depend on the speed of the emitter, then the field of the two co-located charges should always propagate together, that is it should always be zero at any distance for two equal but opposite charges.

7. ### tsmidRegistered Senior Member

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368
But you can make the approximation arbitrarily accurate by making r'/r small enough. The point is that your solution still differs by the same factor independently of r'/r. So something must be wrong here.

What is wrong with your derivation is that you fail to take into account the length contraction of the volume in the direction of motion caused by the retardation: if you have a constant density ρ>0 extending from r'=-a to r'=+a (and ρ=0 otherwise) in its rest frame, then in the moving frame this volume will only extend from r'=-a*(1-v/c) to r'=+a*(1-v/c) in the direction of motion. This exactly cancels the increase of ρ that you have in your integrand. Otherwise, the total charge would not be invariant anymore. This is exactly why the distribution function must be normalized (and this is in fact what you get if you do the derivation correctly)

Anyway, you can actually have it much easier if you consider the whole situation in the rest frame of the charges: assume ρ(r) to be independent of time, then at point r you have a potential Φ(r) = q/r independent of time throughout space, whether you have a test charge at rest or moving at that point. After all, there is no explicit velocity dependence for the electrostatic potential.

8. ### przyksquishyValued Senior Member

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That doesn't justify ignoring it. $\boldsymbol{r}'$ is the integration variable of the problem. You can't just do whatever you like with it.

So why don't you look carefully at the two exact derivations I've already posted here? Particularly the second one, since the first version with two Dirac deltas appearing just seems to be confusing you more. Comparing with them, or at least making an honest attempt at doing an exact derivation yourself, would tell you exactly why your approximation is wrong.

Retardation does not "cause" length contraction. You are not making any sense here at all.

The integral equation I started with relates the scalar potential in a given reference frame to the charge density function in the same reference frame, and this is exactly the way I applied it. I am not making any comparisons between different reference frames here. They are simply not pertinent to the potential equation I am applying or generally to this thread for that matter.

(An aside: the Lorentz length contraction factor is $\sqrt{1 - v^{2}/c^{2}}$ and not $1 - v/c$. I have no idea where you got that from.)

There is no "increase of $\rho$" in the integrand. Just the mathematical fact that $\int \mathrm{d}^{3}r' \, \rho(\boldsymbol{r}', t)$ taken with constant $t$ is generally different from $\int \mathrm{d}^{3}r' \, \rho \bigl( \boldsymbol{r}',\, t - \tfrac{1}{c} \lVert \boldsymbol{r} - \boldsymbol{r}' \rVert \bigr)$. The former gives the total charge. The latter is just the integral of a term that happens to appear in an equation for the scalar potential (a different problem from calculating the total charge) and has no direct physical significance of its own.

What does this have to do with the Liénard-Wiechert potential applying in reference frames other than the source charge's rest frame?

Last edited: Jul 2, 2017
9. ### Neddy BateValued Senior Member

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I think you mean that it should be zero at any distance 'at the appropriate retarded time'? We have to account for c being finite, and if that information propagates outward at the speed of c, then it would be at a specific (retarded) time that the zero field would arrive at any given distance.

If that is what you meant, then what I was asking was whether or not that is a conceptual mistake. I don't understand this well enough to know. In my next post, I thought I found a flaw in the concept, but now I am not so sure it is a flaw. I would like to hear some non-mathematical explanation as to what is really going on in this case.

10. ### tsmidRegistered Senior Member

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368
Well, at times when there is no signal, the field is zero anyway.

11. ### tsmidRegistered Senior Member

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368

I wasn't referring to the Lorentz contraction but the apparent contraction by the time-layering effect caused by the retardation here:

Take again the retardation equation connecting the emission event (r',t') and the detection event (r,t)

t-t' = (r-r')/c

and assume for simplicity r=ct (this is merely a convention adjusting t' such that t'=0 for r'=0). Then you have the condition

t'=r'/c

So if you have a stationary rectangular charge density extending from r'=-a to r'=a

t'= ±a/c

is the time offset required for signals from the front or back of the charged volume to reach the detector at the same time t as the signal emitted from the origin at t'=0.

However, if the volume is moving, this does not hold anymore as during the t' the volume has moved a distance vt'=v/c*a closer to the detector, so the signal from the front edge will arrive at r earlier than t. You have to move the whole configuration back by an amount Δr'=-v/c*a in order for the signal from to arrive at time t. So compared to the stationary case, the front of the cloud is actually seen at an earlier instant at location a*(1-v/c), whereas the back is seen at a later instant at location -a*(1-v/c). Correspondingly for all other values of r'. This means the whole cloud appears compressed in the direction of motion due to this time-layering effect (if the speed approaches c, the charged cloud would apear to be compressed to a point in the direction of motion; you could compare this then probably to what happens when a jet breaks the sound barrier (after all, this is here a completely classical problem as well)).
The point is that in your derivation you took the increase of the charge density due to this compression into account (through the factor 1/(1-v/c) that came in through the variable change), but not the simultaneous reduction of the size of the cloud by the factor 1-v/c (which effectively changes the lower and upper bounds of the integral and thus the result of the integration).

Let me do the derivation correctly in this sense. For simplicity, I keep the problem one-dimensional and assume, as above, x=ct. Also I set the constant in front of the integral=1 (i.e. cgs units)

The equation for the potential is then

$\begin{eqnarray} \Phi (x, x/c) &=& \int_{-a}^{a} dx' \frac{\rho(x'*(1-v/c))}{|x-x'|} \,,\end{eqnarray}$

Assuming a constant linear density ρ between x'=-a and x'=a and ρ=0 otherwise (i.e. ρ*2a = q) , and changing the integration variable to u'= x'*(1-v/c)

we can write

$\begin{eqnarray} \Phi (x, \frac{x}{c}) &=& \frac{\rho}{1-\frac{v}{c}} \int_{-a*(1-\frac{v}{c})}^{a*(1-\frac{v}{c})} du' \frac{\rho(u')}{|x-\frac{u'}{1-\frac{v}{c}}|} \end{eqnarray}$

We take out the constant factor x from the denominator in the integral and do a Taylor expansion assuming u'/(1-v/c) << x (after all, we want the potential of a point charge)

$\begin{eqnarray} \Phi (x, \frac{x}{c}) \approx \frac{\rho}{x*(1-\frac{v}{c})} \int_{-a*(1-\frac{v}{c})}^{a*(1-\frac{v}{c})} du' ( 1+ \frac{u'}{x*(1-\frac{v}{c}) } ) \end{eqnarray}$

The linear term with u' obviously integrates do zero with the integration symmetrical sbout the origin, leaving us with

$\begin{eqnarray} \Phi(x,\frac{x}{c}) &=& \frac{\rho}{x*(1-\frac{v}{c})} *2a*(1-\frac{v}{c}) = \frac{\rho *2a}{x} = \frac{q}{x} \end{eqnarray}$

Last edited: Jul 4, 2017
12. ### przyksquishyValued Senior Member

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Then you're describing an error I never made. Your line of charge of length $2 a$ is something you brought up yourself just a few posts ago. None of the derivations I posted involve integration bounds so there wasn't any place I could have made any error with them in the first place.

For a moving line of charge (and considering the problem in just 1D), you're correct to say that the range of retarded coordinate $x'$ over which the charge density is nonzero is different from the length of the line due to the retardation. But your discussion of this and how to account for it is wrong and you make exactly the kind of mistake you have just accused me of making.

Your derivation and resulting multiplicative factor are wrong. Considering just the $x$ axis, the limits to use for the retarded coordinate $x'$ for a moving line of charge are just where the line $x' = c t'$ intersects with the two edges $x' = v t' - a$ and $x' = v t' + a$. This happens for

$\displaystyle x' = - \frac{a}{1 - v/c} \text{ and } x' = + \frac{a}{1 - v/c} \,,$​

and not $x' = \pm a (1 - v/c)$.

No. Like I pointed out, you got the boundaries wrong, and the effect this has on the integral is exactly the same factor $\frac{R}{R - \boldsymbol{\beta} \cdot \boldsymbol{R}}$ that appears in the derivations I posted, just done in a different way.

This has nothing to do with charge conservation or "increasing charge density". Like I have previously pointed out, $\int \mathrm{d}^{3}r' \, \rho(\boldsymbol{r}',\, t_{\mathrm{R}})$ does not equal the total charge.

No, you're making exactly the kind of error you just accused me of making. In particular, you are wrong to take the integral from $-a$ to $+a$. You're only justified to restrict the integral to values of $x'$ for which the integrand is nonzero. This is the same as values of $x'$ for which what you've written as $\rho \bigl( x' (1 - v/c) \bigr)$ is nonzero. For a moving line of length $2 a$ the boundaries are (up to a common offset) $x' (1 - v/c) = \pm a$, or $x' = - \frac{a}{1 - v/c}$ and $x' = \frac{a}{1 - v/c}$, as I already point out above.

Your derivation is wrong and missing the factor $1 / (1 - v/c)$ as a result.

Last edited: Jul 5, 2017
13. ### tsmidRegistered Senior Member

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368
No, you didn't give any integration bounds at all. That's why you actually haven't given a valid derivation. Obviously, the integration bounds (or the non-zero range of the function you are integrating) is vital for the value of the integral. You haven't addressed this at all.

No, you are having it the wrong way around: the condition is that the location of the leading edge x' coincides with location 'a' at time t_a=a/c (the time it takes for the signal from the origin to arrive at 'a'). Only this way will both signals (from the center and edge of the charged line) be able to eventually travel together and be detected at the same time t at x. This requires obviously x'+v*t_a=a , from which you get the retarded location x'=a*(1-v/c) and (from x'=ct') the retarded time t'=t_a*(1-v/c).

So the retardation condition is given by eliminating the detection time (t_a above) not the emission time t' as you did.

P.S.: I am taking a break now for a couple of weeks, so you have ample time to think this all through again.

Last edited: Jul 5, 2017
14. ### przyksquishyValued Senior Member

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There aren't any integration bounds in the equation for the potential to begin with (the integral is from $-\infty$ to $+\infty$ in all the variables integrated over). Introducing integration bounds is not logically required and if it is done it is only justified where it doesn't change the value of the integral (if a function is zero in some region it makes no difference if you integrate over that region or not). This is a non-issue for the derivations I posted and you did it incorrectly in the derivation you posted.

You haven't pointed out any error in either derivation I posted. Concerning these:

1) I gave two versions of the derivation where the charge density was represented by a Dirac distribution. In both cases I evaluated the integral using standard and routine methods for evaluating integrals with Dirac delta distributions in them. This does not involve determining integration bounds.

2) I gave a more general derivation (up to a point) for a charge density $\rho(\boldsymbol{r}, t) = \rho_{0}(\boldsymbol{r} - \boldsymbol{v}t)$. There aren't integration bounds in that case for the simple reason that I left the function $\rho_{0}$ unspecified. However if you introduce a function that is zero outside a small region then the variable change makes it easy to determine what range the integral can be restricted to: it is simply the range of values $\boldsymbol{u}$ for which $\rho_{0}(\boldsymbol{u})$ is nonzero.

This is just the result I posted but with different definitions. If you make $\pm a$ the limits on the retarded coordinate $x'$ by definition then you determine that the ends of the line are located at $x = \pm a (1 - v/c)$ at time $t = 0$. In other words, this would mean your line is of length $2 a (1 - v/c)$. Which is confusing, since in post #168 you wrote $\rho \times 2 a = q$, suggesting you think your line is of length $2 a$ (or you defined $q$ to be something different than the total charge). Basically, you don't seem to be clear on what your own variables $a$ and $q$ are supposed to represent.

You would avoid this kind of confusion if you actually posed the problem properly from the beginning, starting by writing explicitly what the charge density function is instead of just writing a random integral. Considering just the $x$ axis, a line of charge of length $L$ moving with velocity $v$ could be represented by a charge density function of the form

$\displaystyle \rho(x, t) = \begin{cases} \frac{q}{L} & \text{ if } vt - \frac{L}{2} \leq x \leq vt + \frac{L}{2} \\ 0 & \text{ otherwise} \end{cases} \,.$​

Obviously, this can equivalently be written $\rho(x, t) = \rho_{0}(x - vt)$ with

$\displaystyle \rho_{0}(x) = \begin{cases} \frac{q}{L} & \text{ if } -\frac{L}{2} \leq x \leq \frac{L}{2} \\ 0 & \text{ otherwise} \end{cases} \,.$​

The important thing here is that this is normalised so that the total charge is $q$, independently of $t$:

$\displaystyle \int \mathrm{d}x \, \rho(x, t) = \int_{vt - \frac{L}{2}}^{vt + \frac{L}{2}} \mathrm{d}x \, \frac{q}{L} = \frac{q}{L} \biggl[ \biggl( vt + \frac{L}{2} \biggr) - \biggl( vt - \frac{L}{2} \biggr) \biggr] = q \,.$​

Now we insert this charge density into the integral equation for the potential for $t = x / c$ (and taking $4 \pi \varepsilon_{0} = 1$ like you did):

$\begin{eqnarray} \Phi(x, x/c) &=& \int_{-\infty}^{+\infty} \mathrm{d}x' \, \frac{\rho \bigl( x',\, \tfrac{1}{c} (x - \lvert x - x' \rvert ) \bigr)}{\lvert x - x' \rvert} \\ &=& \int_{-\infty}^{+\infty} \mathrm{d}x' \, \frac{\rho_{0} \bigl( x' - \tfrac{v}{c} (x - \lvert x - x' \rvert ) \bigr)}{\lvert x - x' \rvert} \,. \end{eqnarray}$​

As written the integral is taken from $x' = - \infty$ to $x' = + \infty$. This can be restricted, without changing the result, to just coordinates $x'$ where the integrand is nonzero. This is the range of values $x'$ such that

$\displaystyle \rho_{0} \bigl( x' - \tfrac{v}{c} (x - \lvert x - x' \rvert ) \bigr) \neq 0 \,,$​

or $x'$ satisfying

$\displaystyle - \frac{L}{2} \leq x' - \frac{v}{c} \Bigl( x - \lvert x - x' \rvert \Bigr) \leq \frac{L}{2} \,.$​

If $x$ is taken to satisfy $x > v t + L/2 = \tfrac{v}{c} x + L/2$, (i.e., $x$ is outside and in front of the moving line of charge) then the above inequalities can only be satisfied with $x' < x$, in which case we can replace $\lvert x - x' \rvert = x - x'$. The constraint on $x'$ in this case simplifies to

$\displaystyle -\frac{L}{2} \leq \Bigl( 1 - \frac{v}{c} \Bigr) x' \leq \frac{L}{2} \,,$​

or equivalently

$\displaystyle -\frac{L / 2}{1 - v / c} \leq x' \leq \frac{L / 2}{1 - v / c} \,.$​

Since the charge density is given by the constant $\rho_{0} \bigl( (1 - v/c) x' \bigr) = q / L$ in this range and zero outside of it, the potential is given by (using again that $\lvert x - x' \rvert = x - x'$ in this range)

$\begin{eqnarray} \Phi(x, x/c) &=& \int_{- \frac{L/2}{1 - v/c}}^{\frac{L/2}{1 - v/c}} \mathrm{d}x' \, \frac{q}{L} \, \frac{1}{x - x'} \\ &=& \frac{q}{L} \Biggl[ - \log \biggl( x - \frac{L/2}{1 - v/c} \biggr) + \log \biggl( x + \frac{L/2}{1 - v/c} \biggr) \Biggr] \\ &=& \frac{q}{L} \Biggl[ - \log \biggl(1 - \frac{L/2}{(1 - v/c) x} \biggr) + \log \biggl( 1 + \frac{L/2}{(1 - v/c) x} \biggr) \Biggr] \end{eqnarray}$​

where $\log$ is the natural logarithm function. Finally, using the approximation $\log(1 + x) \approx x$, we get

$\displaystyle \Phi(x, x/c) \approx \frac{q}{L} \, \frac{L}{(1 - v/c)} = \frac{q}{(1 - v/c) x}$​

if $L$ is small relative to $(1 - v/c) x$.

This is the correct result in terms of the total charge $q$ if you write and solve the problem properly.

I already understood all of this weeks ago.

Last edited: Jul 6, 2017
15. ### przyksquishyValued Senior Member

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I made up an illustration for the situation where a moving line of negative charge passes a stationary line of positive charge. Apart from the addition of the stationary line of positive charge this is the same kind of situation I dealt with in my previous post. If you have two point charges this can be understood as the limit where the lengths are taken to zero.

This shows a point (P) and regions where charges are located in time and space:
• The (stationary) region where the positive charge is located over time is shown in cyan.
• The (moving) region where the negative charge is located over time is shown in red.
• Where these regions intersect the net charge density is zero. This is shown in white.
• The dotted line labelled $t = 0$ shows that the net charge density is zero everywhere at this time.
The electromagnetic field at the point marked $P$ depends on the charge at earlier times along the dashed line shown on the figure. (Some aspects of the electromagnetic field also depend explicitly on the current, which isn't shown.) As you can see in the figure, the dashed line crosses a significant white area where the net charge is zero. This contributes nothing to the scalar potential. But it also crosses a bit of the red area, where the negative charge isn't cancelled out by any positive charge, so you do get some contribution from negative charges to the electromagnetic field at the point $P$.

The amount of this contribution stays roughly constant as you make the lengths of the lines shorter (keeping the total negative and positive charge constant), so the contributions from the positive and negative charge don't cancel out even in the limit where the lines become point charges.

Neddy Bate likes this.
16. ### Neddy BateValued Senior Member

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Wow, thank you for that diagram, przyk. I think I understand it pretty well, but I still have a small question about it.

Let me add a different point, Q, located at the intersection of the horizontal t=0 line and the right-hand endpoints of the two charge lines. Would the dotted line passing through point Q still be on a 45 degree angle, or would it then be horizontal like the t=0 line? I assume it would have to remain on a 45 degree angle, because I think that represents the speed of information being c. If so, that would mean that, at time t=0, the electromagnetic field at point Q would have some contributions from the negative charge (red) which are not cancelled out by any positive charge, correct?

The reason I ask is because the bullet which reads, "The dotted line labelled t=0 shows that the net charge density is zero everywhere at this time," made me think for a moment that my point Q should experience the net charge density being zero. So am I right to assume what I wrote above? Thanks again.

17. ### przyksquishyValued Senior Member

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Like this:

?

(The electromagnetic field also depends on points along the right dotted line. I drew both this time, even though there's nothing interesting to the right of $Q$. The full region represented by the dotted lines in 3 dimensions is a cone -- the past light cone associated with $Q$.)

The electromagnetic field at any point $Q$ depends directly on the charges and currents on the past light cone of $Q$, i.e., the positions and times $(x', t')$ in the past such that information travelling in a straight line at the speed of light from $(x', t')$ could reach $Q$. This is represented by the 45 degree dotted lines on the figures I've posted.

Yes -- you can see this on the figure above.

Yes. This was the point: the net charge density is zero everywhere at a particular time $t = 0$, but you get a nonzero electromagnetic field because the electromagnetic field does not depend on the charge density at any one constant time.

18. ### Neddy BateValued Senior Member

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Yes.

I'm glad you drew both dotted lines. As I was thinking about what would have happened if I had chosen my t=0 point to be located on the left endpoint of the two charge lines, (rather than the right endpoint), I realized that the 45 degree dotted line in that case would have to be mirrored in the other direction. Then I thought about what would have happened if I had chosen my t=0 point to be located at the midpoint of the two charge lines, and I realized that there would have to be both types of 45 degree dotted lines in that case. So now it makes even more sense to see that there could be two such 45 degree dotted lines in any case.

This all makes perfect sense now, thanks.

19. ### tsmidRegistered Senior Member

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368
Just another quick one before my break:
If you distribute the same amount of charges over a larger region then the density ρ has to decrease correspondingly. Otherwise the total charge would not be q anymore. This then cancels out the factor that you get due the larger integral bounds.

the only effect the retardation in general has is that you see each charge at a slightly different time instant. If the whole volume is stationary, this wouldn't even produce any observable consequence compared to an instantaneous interaction, as it is irrelevant for the potential at what time instant we see the charge; the only relevant factors are the distance and the magnitude of the latter (as clearly stated by the definition for the potential), and if those are invariant, the time-layering caused by the retardation is not observable. Now the only thing that happens when the charges are not stationary is that in addition to seeing the charge at different times we see them also in slightly different places compared to instantaneous interaction, but we still see the same charges. If there are N charges, then you will still see the potential of N charges as in the stationary case. Of course, this apparent change of the spatial configuration will in general cause slight corrections to the potential, but these can be made arbitrarily small by letting the size of the volume go to zero (which is the case we actually were discussing initially).

Last edited: Jul 7, 2017
20. ### przyksquishyValued Senior Member

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I defined the charge density function in such a way that the charges are distributed, with charge density $q / L$, over a distance $L$ at any given time. So the total charge is $q$.

It is the formula for the scalar potential that requires us to integrate not with $\rho_{0}(x')$ but instead $\rho_{0} \bigl( (1 - v/c) x' \bigr)$, which is why the integration limits on $x'$ are $\pm \frac{L/2}{1 - v/c}$ instead of $\pm L/2$. There is nothing wrong with this, because:
1. The charge density, $q / L$, is by definition the amount of charge per unit $x$ at some fixed time and not per unit of the retarded coordinate $x'$ at the retarded time. Thus there is no reason to expect that $\int \mathrm{d}x' \, \rho(x', t_{\mathrm{R}})$ should equal the total charge and, except in certain special cases, generally it does not.
2. The purpose of the equation for the scalar potential is to give the scalar potential and not to count the total charge.

The effect of the retardation here is that the formula for the scalar potential requires us to integrate a term involving the charge density, which is $q / L$ in this case, over a coordinate range of $L / (1 - v/c)$ instead of the length $L$ that the charges are spread out over at any fixed time.

There is nothing subtle or difficult in what I am doing. It is simply a question of writing what the charge density function is and then substituting it into the equation for the potential and calculating the resulting integral. I've described 2 or 3 different ways of doing this now for a point charge.

The results I've already posted correctly reflect this. If $\boldsymbol{v} = 0$ then $q / (R - \boldsymbol{\beta} \cdot \boldsymbol{R}) = q / R = q / r$ (assuming the charge is located at the coordinate origin).

You are interpreting the integral in a way that contradicts the result you get when you actually calculate the integral. That was the point of my properly writing out the integral and evaluating it in detail in the first place: actually doing the calculation shows that the integral does not count charges in the way you have been assuming it does.

Last edited: Jul 8, 2017
21. ### tsmidRegistered Senior Member

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368
I am back from my break and ready to resume the discussion.

Only that you are not integrating over $\rho_{0}(x')$ but over $\rho(x', t_{\mathrm{R}})$. And the latter charge density must be different as the charges are distributed over a different length. Whether this changed length exists at a fixed time or as a result of the retardation doesn't matter. You would see exactly the same potential if you actually redistribute the charges over the same length and then assume instantaneous interaction. For the observed potential at a given moment, it is irrelevant whether the apparent location of the charges is the result of retardation or not.
Of course, as I mentioned before, any spatial redistribution of charges will in general lead to a different potential (even though total charge is conserved), but this difference can be made arbitrarily small here:

assume just two particles, each of charge q/2, separated by a fixed distance L i.e. at locations x'=-L/2 and x'=+L/2 . For an observer moving with speed v on the x-axis, these locations would change to x'=-L/2/(1-v/c ) and x'=+L/2/(1-v/c) as per your previous argument, so the potential at a location x would be

Φ(x,L) = q/2/(x-L/2/(1-v/c)) + q/2/(x+L/2/(1-v/c))

and it is immediately obvious that if you let L→0 , you obtain the usual potential of a point charge

Φ(x,L=0)=q/x

If you somehow obtain Φ(x,L=0)=q/x/(1-v/c) instead, you have violated charge conservation in the process.

Last edited: Aug 3, 2017
22. ### przyksquishyValued Senior Member

Messages:
3,171
No, I integrated $\rho(x',\, t_{\mathrm{R}})$, which is the same as $\rho_{0}(x' - v t_{\mathrm{R}})$ for the definition of $\rho_{0}$ I used, which is the same as $\rho_{0} \bigl( (1 - v/c) x' \bigr)$ for $t = x/c$ for suitable values of $x$, as I explained in post #171. Did you even read the derivation in that post? It's basically the same thing you tried to do in post #168 except, unlike you, I actually formulated the problem properly, starting by specifying what the charge density function was.

You're just doing what you've already done before. You claim or imply I made some sort of error, but you make no reference to the derivation I posted and don't say at which step this error supposedly occurs or how it would change the result. You were doing the same last month. You claimed I'd made some error with integration limits, but you were unable to say at what step integration limits should be introduced in any derivation I'd actually posted, what those integration limits should be, why they should be introduced, or how they would changed the result.

Note that at this point I've posted 2 or 3 exact derivations (depending on how you want to count them) showing that the potential around a moving point charge is $\propto q / (R - \boldsymbol{\beta} \cdot \boldsymbol{R})$ according to the integral equation for the potential I started with, and you can't actually point to any error in any of them.

This is nonsense. The act of calculating an integral over the retarded time, as the integral equation for the potential says to do, does not physically change the charge distribution. It does, however, change the result of the calculation, which is why the retarded time is there in the formula and why I've been telling you it is important to take it into account since nearly the beginning of this thread.

There is no "changed length". The length of a line of charge is simply the distance between its endpoints at a given time $t$ and not different retarded times $t_{\mathrm{R}}$. I defined this to be $L$ in post #171. The charge density $\rho$ is also by definition the amount of charge per unit distance (generally, volume) at a given time $t$ and not for the varying retarded time.

You could in principle define a "retarded charge density" $\tilde{\rho}$ as the amount of charge per unit $x'$ at the retarded time (so that integrating $\int \mathrm{d}x' \, \tilde{\rho}(x', t_{\mathrm{R}})$ would give the total charge). But this is not what is used in the integral formula for the potential. That formula uses the same function $\rho$ that appears elsewhere in electrodynamics, including in Maxwell's equations. (In fact, the integral equation for the potential is derived as a formal solution to Maxwell's equation for the scalar potential in the Lorentz gauge, for the same function $\rho$.) If you change the definition of $\rho$ then you are not applying the formula correctly.

You're doing at least two things wrong here.

First, your approach is circular. In order to calculate the potential for two point charges you first need to know the potential for one point charge, which is the very thing we are arguing here.

Second, you are attempting to derive the potential for moving charges essentially by changing the reference frame. This is possible but the way you're doing it is completely wrong. To begin with, $\Phi$ is generally not invariant under Lorentz transformations.

On this, I don't know why you say "as per your previous argument" since I never changed the reference frame in any of my posts. In post #171 for instance I substitute some function or variable definitions for convenience but there is never any change of reference frame. I start with the line of charge already moving with velocity $v$ and end up with the potential in that same reference frame in which it is moving with velocity $v$.

You're also changing the subject, since you earlier specifically claimed that $\int \mathrm{d}^{3}r' \, \frac{\rho(\boldsymbol{r}',\, t_{\mathrm{R}})}{\lVert \boldsymbol{r} - \boldsymbol{r}' \rVert}$ would give $q / R$ for a moving point charge. That is mathematically wrong, independently of the context of electromagnetism.

How so? $\Phi$ is the scalar potential, not the total charge. You understand that these are different quantities, right?

Last edited: Aug 4, 2017
23. ### tsmidRegistered Senior Member

Messages:
368
First of all, you should not be calling the expression for the potential an 'integral equation' (I noticed this already a couple of times before), as people might get wrong ideas here. In an integral equation the unknown variable appears under the integral (see https://en.wikipedia.org/wiki/Integral_equation ), which is not the case here (the unknown variable here is the potential Φ, which does not appear under the integral).

Secondly, you seem to be missing the whole point here, which is the fact that the positional change due to the retardation is the only issue here. And this is only an implicit velocity dependence but not an explicit one. The contribution of point x' to the potential at point x depends only on the charge density ρ(x') and the distance |x-x'|. This is what the potential equation states, nothing more and nothing less. So a charge q at point x' at time t will result in the same potential Φ(x,t) in the case of instantaneous interaction as a charge q at the same point at time t-|x-x'|/c/ in the case of a propagation speed c. It is completely irrelevant for the test charge whether the received signal is transmitted instantaneously from point x' or retarded as long as it is detected at time t. If you have worked out the retarded times and positions of all particles relating to (x,t), you would obtain the same potential with all those positions applied at time t and assuming instantaneous interaction. Anything else you interpret into this is simply the result of a circular conclusion based on the incorrect formulae you seem to take for granted.

I could say more regarding the other points in your post, but I think we should clarify the basic concepts first before going further.