# Relativistic Coulomb Force

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

1. ### przyksquishyValued Senior Member

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3,171
No, the potential according to the first line would be "$\frac{q}{r} \delta(0)$", which is meaningless.

You're already making two mistakes here. First, the argument appearing in the Dirac delta in my post was $t' - t + \frac{1}{c} \lVert \boldsymbol{r} - \boldsymbol{v} t' \rVert$. Even if you set $t = r / c$ this generally doesn't simplify to $t' (1 - v/c)$.

Second, if $\delta(x) \approx \frac{1}{\varepsilon \sqrt{\pi}} e^{- x^{2} / \epsilon^{2}}$ then the corresponding approximation with $x = t' (1 - v/c)$ is just

$\displaystyle \delta \bigl( t' (1 - v/c) \bigr) \approx \frac{1}{\varepsilon \sqrt{\pi}} e^{- \bigl(t' (1 - v/c) \bigr)^{2} / \varepsilon^{2}} \,.$​

You seem to have invented an extra multiplication by $1 - v/c$ out of nowhere.

On this latter point you are, by the way, contradicting your own reference. You are effectively trying to claim that $\delta(\alpha x) = \delta(x)$, but the very same Wikipedia page that you linked to contradicts you:
https://en.wikipedia.org/wiki/Dirac_delta_function#Scaling_and_symmetry

Last edited: Jun 17, 2017

3. ### tsmidRegistered Senior Member

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368
No, the potential would incorporate a delta function at t=r/c at distance r if the charge density is a delta function t=0 at the origin (unless you propose instantaneous interaction)

If you write the equation as a scalar expression(i.e. one dimensional), you have

τ=t'*(1-v/c) -t+r/c

so

τ=t'*(1-v/c)

if t=r/c.

As I said, this function is not normalized (if you integrate over t' it does not give 1). You need the addititonal factor 1-v/c to normalize it.
Distribution functions in general must be properly normalized, otherwise you get wrong results.

This scaling factor is nothing else than the factor that comes in from the change of the integration variable in the derivation I gave in my previous post #140. But that cancels exactly with the normalization factor that should be there in the first place (but isn't in the derivation of the Lienard-Wiechert potential).

5. ### przyksquishyValued Senior Member

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You are not making any sense. You said the potential was:
If you substitute $t = r/c$ then you get $\frac{q}{r} \delta(0)$, which isn't a distribution at all.

So? The distribution $\delta(t' - t_{\mathrm{R}})$ with $t_{\mathrm{R}} = t - \frac{1}{c} \lVert \boldsymbol{r} - \boldsymbol{v}t' \rVert$ appears as a result of the derivation I did up to the point where it appears in post #137. You haven't found any error in that derivation and there is no reason it should be normalised any differently than it is. In fact it is precisely because it is not normalised that the resulting integration produces the correct potential for a moving charge.

The only issue with normalisation concerns total charge. The distribution $q \, \delta(\boldsymbol{r} - \boldsymbol{v}t)$ that I started with is normalised so that the total charge at any fixed time $t$ is $q$:

$\displaystyle \int \mathrm{d}^{3}r \, q \, \delta(\boldsymbol{r} - \boldsymbol{v}t) = q \,.$​

The rest of post #137 is just substituting this distribution into the equation for the potential and doing mathematics, which produces the correct result.

On this, you don't insert some "normalisation factor" that you invented into the middle of a derivation just because you wanted the result to be different.

Last edited: Jun 18, 2017

7. ### tsmidRegistered Senior Member

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368
That's not what I said. I said if

ρ(r,t) = q*δ(r)*δ(t)

then

Φ(r,t) = q/r *δ(t-r/c)

Are you questioning this?

A distribution that isn't normalized isn't a distribution, so writing

$\displaystyle \delta \bigl( t' (1 - v/c) \bigr) \approx \frac{1}{\varepsilon \sqrt{\pi}} e^{- \bigl(t' (1 - v/c) \bigr)^{2} / \varepsilon^{2}} \,.$

is mathematically wrong because by definition a δ-function must be normalized to 1 and this isn't normalized to 1 (as you can easily find by integrating it over t'.) You need a further factor 1-v/c on the right hand side to make it a distribution and justify calling it δ(t'*(1-v/c)) . And that factor will disappear again if you change the integration variable subsequently from t' to τ=t'*(1-v/c) ending up with

δ(τ) = 1/sqrt(π)/ε * exp(-τ^2/ε^2)

which is then correctly normalized, in contrast to when you leave out the factor 1-v/c and take δ(t'*(1-v/c)) as you did above.

The symbolic δ-function notation is clearly misleading here. You have to carry through the integration with the defining exponential function, and then it is clearly evident.

So quite evidently, the Lienard-Wiechert potential is just the result of incorrect mathematical operations here, which then of course resolves the issues with regard to the macroscopic consequences we have been discussing initially.

Last edited: Jun 18, 2017
8. ### przyksquishyValued Senior Member

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3,171
No, I am questioning what you wrote immediately after this:
Substituting $t = r/c$ into $\frac{q}{r} \, \delta(t - r/c)$ does not give $q / r$.

You can certainly decompose $\delta(\boldsymbol{r} - \boldsymbol{v} t)$ as an integral over different times, as in

$\displaystyle q \, \delta(\boldsymbol{r} - \boldsymbol{v} t) = q \, \int \mathrm{d}t' \, \delta(\boldsymbol{r} - \boldsymbol{v} t') \, \delta(t' - t) \,,$​

but if you try to calculate the potential based on this you're just going to end up doing exactly the derivation I already posted here.

No, you're making this up based on nothing (except what sounds like an etymological fallacy, taking the name "distribution" to mean something it doesn't). The Dirac distribution is by definition normalised so that

$\displaystyle \int \mathrm{d}t' \, \delta(t') = 1 \,.$​

But this says nothing about $\delta \bigl( t' (1 - v/c) \bigr)$. That is different and the integral of it with respect to $t'$ is not 1, as any good reference on Dirac deltas or on distribution theory in general will tell you. Like I already pointed out, your own Wikipedia reference directly contradicts you on this:
For $\alpha = 1 - v/c$ (and since it doesn't matter whether the integration variable is called $x$ or $t'$) you just get

$\displaystyle \int \mathrm{d}t' \, \delta \bigl( t' (1 - v/c) \bigr) = \frac{1}{1 - v/c} \,.$​

It is perfectly consistent with this that integrating the Gaussian approximation likewise gives

$\displaystyle \int \mathrm{d}t' \, \delta_{\varepsilon} \bigl( t' (1 - v/c) \bigr) = \int \mathrm{d}t' \, \frac{1}{\varepsilon \sqrt{\pi}} e^{- \bigl( t' (1 - v/c) \bigr)^{2} / \varepsilon^{2}} = \frac{1}{1 - v/c} \neq 1 \,.$​

Your extra multiplication by $1 - v/c$ is an invention that has no place in any of this.

Last edited: Jun 18, 2017
9. ### tsmidRegistered Senior Member

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368
But the expression for the potential

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

contains no time integral in the first place. The δ-function with regard to time is merely additionally put in in an ad-hoc manner in order to make the integration easier (as explained in detail for instance on the Wikipedia page https://en.wikipedia.org/wiki/Liénard–Wiechert_potential ). And in order for this operation to be valid we must obviously require that the added integral equates to 1. And for that you need to add a factor 1-v/c to δ(t'*(1-v/c)).

Last edited: Jun 20, 2017
10. ### przyksquishyValued Senior Member

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3,171
You seem to have misunderstood the derivation. The extra delta distribution is just introduced via

$\displaystyle \rho(\boldsymbol{r}, t) = \int \mathrm{d}t' \, \rho(\boldsymbol{r}, t') \, \delta(t' - t) \,.$​

There's nothing "ad hoc" about this. It's just applying the definition of the delta distribution. Substitute that into the equation for the potential and you get

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

Aside from an obvious typo (an $\boldsymbol{r}$ that should have been an $\boldsymbol{r}'$), this is exactly what I started with in post #137.

If you don't like the extra Dirac delta and integral over $t'$ then you can just as well calculate the potential without it. Of course, you just get the same result:

$\begin{eqnarray} \displaystyle \Phi(\boldsymbol{r}, t) &=& \frac{1}{4 \pi \epsilon_{0}} \int \mathrm{d}^{3}r' \, \frac{\delta \bigl( \boldsymbol{r}' - \boldsymbol{v} t + \boldsymbol{\beta} \lVert \boldsymbol{r} - \boldsymbol{r}' \rVert \bigr)}{\lVert \boldsymbol{r} - \boldsymbol{r}' \rVert} \\ &=& \frac{1}{4 \pi \epsilon_{0}} \, \frac{1}{R} \, \frac{1}{\lvert \det J \rvert} \end{eqnarray}$​

where $\boldsymbol{\beta} = \boldsymbol{v} / c$ and $\det J$ is the determinant of the Jacobian matrix

$\begin{eqnarray} J_{ij} &=& \frac{\partial}{\partial r'_{j}} \Bigl[ r'_{i} - v_{i} t + \beta_{i} \lVert \boldsymbol{r} - \boldsymbol{r}' \rVert \Bigr] \\ &=& \delta_{ij} - \frac{\beta_{i} (r_{j} - r'_{j})}{\lVert \boldsymbol{r} - \boldsymbol{r}' \rVert} \end{eqnarray}$​

evaluated at $\boldsymbol{r}' = \boldsymbol{r}'_{\mathrm{R}}$ such that $\boldsymbol{r}'_{\mathrm{R}} - \boldsymbol{v} t + \boldsymbol{\beta} \lVert \boldsymbol{r} - \boldsymbol{r}'_{\mathrm{R}} \rVert = 0$. It's not too difficult to show that

$\begin{eqnarray} \det J &=& 1 - \frac{\boldsymbol{\beta} \cdot (\boldsymbol{r} - \boldsymbol{r}'_{\mathrm{R}})}{\lVert \boldsymbol{r} - \boldsymbol{r}'_{\mathrm{R}} \rVert} \\ &=& \frac{R - \boldsymbol{\beta} \cdot \boldsymbol{R}}{R} \,. \end{eqnarray}$​

Last edited: Jun 20, 2017
11. ### tsmidRegistered Senior Member

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368
Since

$\displaystyle \rho(\boldsymbol{r}, t) = \int \mathrm{d}t' \, \rho(\boldsymbol{r}, t') \, \delta(t' - t) \,.$

is essentially an identity, it is valid if and only if

$1= \int \mathrm{d}t' \, \delta(t' - t) \,.$

for all possible values of t. If that normalization condition is not satisfied in the context you are using it, you are not entitled to insert it, as retrospectively your identity would become invalid.

Think about it for a moment what the potential equation

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

asks us to to do: it simply instructs us to sum up the contribution of all charges q_i according to q_i/R_i from those distances R_i where the light travel time equals R_i/c So there can not possibly anything else than q/R come out of it for a single charge (for which there is nothing else to sum up anyway, so the delta function approach is completely redundant if not bogus here). If you get q/R/(1-v/c) as a result it means that effectively you are calculating the potential according to the present location of the particle, which is clearly a contradiction in terms.

12. ### przyksquishyValued Senior Member

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The identity is valid for all $t$ at the point I introduced it. You're obviously grasping at straws here, making up some story about normalisation despite not being able to explain why anything appearing in either derivation I posted should be normalised any differently than it is.

The important thing to understand here is that $\int \mathrm{d}^{3}r' \, \rho(\boldsymbol{r}', t)$ 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 $\boldsymbol{r}'$. In particular, even if $\int \mathrm{d}^{3}r' \, \rho(\boldsymbol{r}', t) = q$ independently of time it does not follow that $\int \mathrm{d}^{3}r' \, \rho(\boldsymbol{r}', t - t_{\mathrm{R}}) = q$. They're generally different, whether $\rho$ is an ordinary function or a Dirac delta. In the particular derivation I posted, this difference gets transferred to the Dirac delta $\delta(t' - t_{\mathrm{R}})$ in the middle of the derivation, where it happens to be a little bit easier to deal with. But whichever way it is done, this different normalisation is consistently there in the integral somewhere just as a consequence of the charge density being evaluated at the retarded time, according to the equation for the potential. It is mathematically wrong to ignore this.

You're making up a story where there is no need for one. The integral above means to calculate exactly that integral as it is given following the same procedures that you would use to calculate any other integral. If you can't understand why the result is what it is then that points to a problem with your understanding of integrals.

Last edited: Jun 22, 2017
13. ### tsmidRegistered Senior Member

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368
The identity

$\displaystyle \rho(\boldsymbol{r}, t) = \int \mathrm{d}t' \, \rho(\boldsymbol{r}, t') \, \delta(t' - t) \,.$

is valid only as written, that is if t and t' are indeed independent variables. If you make t a function of t' (as you are doing), it is obviously not valid anymore. Take for simplicity the situation considered by me earlier with the moving charge passing through the origin (r=0) at t=0. The identity becomes in this case

$\displaystyle \rho(0, 0) = \int \mathrm{d}t' \, \rho(0, t') \, \delta(t'(1-v/c)) \, = \frac{\rho(0, 0)}{1-v/c}$

which is obviously a contradiction in terms unless v=0.

Last edited: Jun 23, 2017
14. ### przyksquishyValued Senior Member

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They are independent variables at the point I used the identity.

But I didn't make $t$ a function of $t'$. It's only applying the equation for the potential to the specific problem under consideration and then taking the integral over $\boldsymbol{r}'$ that results in a delta with an argument that is a nontrivial function of $t'$ appearing. It is not some random choice I made. If you actually read the derivation you would see this.

But I never used such an identity in the first place. You've just written something that has nothing to do with the derivation I posted.

15. ### przyksquishyValued Senior Member

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Like I said before, if you don't like the $\delta(t' - t)$ and extra integration over $t'$ then the calculation can just as well be done without it. You only need to know how to do coordinate substitution in an integral of more than one variable in that case.

A slightly more general problem than the moving point charge is to consider a charge density of the form

$\displaystyle \rho(\boldsymbol{r}, t) = \rho_{0}(\boldsymbol{r} - \boldsymbol{v} t) \,,$​

representing some distribution of charges moving together with velocity $\boldsymbol{v}$. In that case direct substitution into the equation for the potential gives

$\displaystyle \Phi(\boldsymbol{r}, t) = \frac{1}{4 \pi \epsilon_{0}} \int \mathrm{d}^{3}r' \, \frac{\rho_{0}(\boldsymbol{r}' - \boldsymbol{v} t + \boldsymbol{\beta} R)}{R}$​

(with $\boldsymbol{\beta} = \boldsymbol{v} / c$ and $\boldsymbol{R} = \boldsymbol{r} - \boldsymbol{r}'$). One way to approach evaluating this is to change the integration variable to $\boldsymbol{u} = \boldsymbol{r}' - \boldsymbol{v} t + \boldsymbol{\beta} R$:

$\displaystyle \Phi(\boldsymbol{r}, t) = \frac{1}{4 \pi \epsilon_{0}} \int \mathrm{d}^{3}u \, \Biggl\lvert \det \biggl(\frac{\partial u_{i}}{\partial r'_{j}} \biggr) \Biggr\rvert^{-1} \, \frac{\rho_{0}(\boldsymbol{u})}{R}$​

(with the $R$ appearing in the integral seen as a function of $\boldsymbol{u}$). Like I pointed out in an earlier post,

$\displaystyle \det \biggl(\frac{\partial u_{i}}{\partial r'_{j}}\biggr) = \frac{R - \boldsymbol{\beta} \cdot \boldsymbol{R}}{R} \,,$​

so the potential is equivalently given by

$\displaystyle \Phi(\boldsymbol{r}, t) = \frac{1}{4 \pi \epsilon_{0}} \int \mathrm{d}^{3}u \, \frac{\rho_{0}(\boldsymbol{u})}{R - \boldsymbol{\beta} \cdot \boldsymbol{R}} \,.$​

This should make it clear that your intuition about the integral, as just summing contributions proportional to $1/R$ for each bit of charge, is wrong.

Last edited: Jun 23, 2017
16. ### tsmidRegistered Senior Member

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368
You did you use it implicitly when you wrote

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

and insisted that the δ-function is not normalized here. As I pointed out above, this non-normalization would in turn imply

$\displaystyle \rho(0, 0) = \int \mathrm{d}t' \, \rho(0, t') \, \delta(t'(1-v/c)) \, = \frac{\rho(0, 0)}{1-v/c}$

17. ### tsmidRegistered Senior Member

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368

Yes, obviously you get the same result here because you now have r'*(1-v/c) in the integrand rather than t'*(1-v/c), which results in the same factor when you change integration variables.

The point is that it is not legitimate here to interpret r' as the path of the charged particle. When you do the integration, r' merely tags the locations of the individual charged particles, that is by integrating over r' you are summing up the contributions of different particles at the corresponding location. r' is not a variable that tracks an individual particle. If you do the latter you effectively count particles twice (with a certain weight). That's why you have to re-normalize the whole distribution.

You don't actually need the integral or δ-function to show that you can't assume r'=vt' here. Just take the one dimensional light equation

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

with (r',t') denoting the location and time of emission and (r,t) the location and time of detection. Assuming again t=r/c, you get r'=ct', so a particle would have to travel with the speed of light if you wanted to interpret this as a particle path. The only possible interpretation of r' is that it is the location of a different particle which is detected at the same time at location r as e.g. a particle at the origin at t'=0. This can not be the same particle. Each particle can only contribute from one specific location and time (r',t') in order to be observed at (r,t).

18. ### przyksquishyValued Senior Member

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No. The Dirac delta in what you quoted above is normalised with respect to $t'$:

$\displaystyle \int \mathrm{d}t' \, \delta \bigl( t' - t + \tfrac{1}{c} \lVert \boldsymbol{r} - \boldsymbol{r}' \rVert \bigr) = 1 \,.$​

It is only later in the derivation that you get a non-normalized Dirac delta. Ultimately the cause of this is that $\int \mathrm{d}^{3}r' \, \rho(\boldsymbol{r}', t) \neq \int \mathrm{d}^{3}r' \, \rho(\boldsymbol{r}, t_{\mathrm{R}})$, as I told you before.

There is certainly nothing in the derivation I wrote that would let you deduce $\rho(0, 0) = \frac{\rho(0, 0)}{1 - v/c}$.

19. ### przyksquishyValued Senior Member

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I didn't interpret anything at all. I simply substituted the charge density $\rho(\boldsymbol{r}, t) = \rho_{0}(\boldsymbol{r} - \boldsymbol{v}t)$ into the general equation for the scalar potential in the Lorentz gauge. The point was to show that the potential equation gives the Liénard–Wiechert potential in the case of a moving charge, just like everyone says it does.

You're proposing to introduce an extra normalisation factor that simply isn't there in the potential equation. This is just wishing the potential equation were different than what it is. Wishing is not going to change the equation.

There isn't any variable $t'$ or any interpretation about "particle paths" in the derivation of post #152.

20. ### tsmidRegistered Senior Member

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368
In post #143 above you claimed the opposite

So? The distribution $\delta(t' - t_{\mathrm{R}})$ with $t_{\mathrm{R}} = t - \frac{1}{c} \lVert \boldsymbol{r} - \boldsymbol{v}t' \rVert$ appears as a result of the derivation I did up to the point where it appears in post #137. You haven't found any error in that derivation and there is no reason it should be normalised any differently than it is. In fact it is precisely because it is not normalised that the resulting integration produces the correct potential for a moving charge.

At the point where you substitute r'=vt' (which is pretty much the implied next step here) the delta function will become non-normalized.

Then explain how you get from this

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

to this

$\displaystyle \Phi(\boldsymbol{r}, t) = \frac{1}{4 \pi \epsilon_{0}} \int \mathrm{d}^{3}r' \, \frac{\rho_{0}(\boldsymbol{r}' - \boldsymbol{v} t + \boldsymbol{\beta} R)}{R}$

21. ### przyksquishyValued Senior Member

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$\delta \bigl( t' - t + \frac{1}{c} \lVert \boldsymbol{r} - \boldsymbol{v} t' \rVert \bigr)$ is not the same thing as $\delta \bigl( t' - t + \frac{1}{c} \lVert \boldsymbol{r} - \boldsymbol{r}' \rVert \bigr)$. There's no contradiction here.

No, the next step is to take the integral over $\boldsymbol{r}'$ which results in $\boldsymbol{r}'$ being replaced with $\boldsymbol{v}t'$ and the elimination of one of the Dirac deltas. What matters for the derivation is that

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

This does not assume, require, or imply that $\delta \bigl( t' - t + \frac{1}{c} \lVert \boldsymbol{r} - \boldsymbol{r}' \rVert \bigr) = \delta \bigl( t' - t + \frac{1}{c} \lVert \boldsymbol{r} - \boldsymbol{v} t' \rVert \bigr)$ or that they are normalised the same way.

Direct substitution of a function definition into an equation. I don't see what you find difficult about this. If

$\rho(\boldsymbol{r}, t) = \rho_{0}(\boldsymbol{r} - \boldsymbol{v}t)$​

then

$\begin{eqnarray} \rho \bigl( \boldsymbol{r}',\, t - \tfrac{1}{c} \lVert \boldsymbol{r} - \boldsymbol{r}' \rVert \bigr) &=& \rho_{0} \Bigl( \boldsymbol{r}' - \boldsymbol{v} \bigl( t - \tfrac{1}{c} \lVert \boldsymbol{r} - \boldsymbol{r}' \rVert \bigl) \Bigr) \\ &=& \rho_{0} \bigl( \boldsymbol{r}' - \boldsymbol{v} t + \tfrac{\boldsymbol{v}}{c} \lVert \boldsymbol{r} - \boldsymbol{r}' \rVert \bigr) \\ &=& \rho_{0}(\boldsymbol{r}' - \boldsymbol{v} t + \boldsymbol{\beta} R) \end{eqnarray}$​

with $\boldsymbol{\beta} = \boldsymbol{v} / c$, $R = \lVert \boldsymbol{R} \rVert$, and $\boldsymbol{R} = \boldsymbol{r} - \boldsymbol{r}'$.

Last edited: Jun 26, 2017
22. ### Neddy BateValued Senior Member

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So, just out of curiosity... We have only two charged particles, the +1 positive one stationary at the origin, and the -1 negative one moving relative to the origin, just grazing past the origin at one instant of time. I think we ascertained that these charges cancel each other and produce a zero charge at that moment, in that location, even though one particle moves relative to the other?

If so, would that zero charge then sort of spherically 'propagate outward' at speed c, like an expanding spherical wavefront of light, so that at the appropriate retarded time there would be a zero charge further out from the origin, even though the charges would not really be co-located anymore by that retarded time? Or is the shape not a sphere, but somehow dependent on the direction of motion of the moving charge?

Or is that line of thinking just the same kind of conceptual mistake that tsmid made?

Last edited: Jun 28, 2017
23. ### Neddy BateValued Senior Member

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Hmm, I think what I just said (above) is certainly wrong for locations where the moving charge becomes closer after passing the origin.