# Retarded potential of a moving point charge

• I
Kashmir
Potential of a moving point charge is given as
##V (\mathbf r,t)= \frac{1}{4\pi\epsilon_0}\int \frac{\rho (\mathbf r',t_r) }{|\mathbf{ (r-r')}|}d\tau'##

Griffiths says:
" It is true that for a point source the denominator ## |\mathbf{(r-r')}|## comes outside the integral..."

Why does it come outside the integral?

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• Kashmir
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In my opinion it doesn't just come outside the integral, but it is what we get after evaluating the integral. This is because the charge density for a point source located at ##\vec{r_0}##is given by $$\rho(\vec{r'},t_r)=\delta (\vec{r'}-\vec{r_0})q(t_r)$$ that is it involves a delta dirac function so after we do the integration with respect to ##\vec{r'}## what we get is $$V(\vec{r},t)=\frac{1}{4\pi\epsilon_0}\frac{1}{|\vec{r}-\vec{r_0}|}q(t-\frac{|\vec{r}-\vec{r_0}|}{c})$$

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That doesn't make sense to me. What do you mean by ##q(t)##? The four-current (in (1+3)-notation and using natural units, ##c=1##) is
$$j^{\mu}(t,\vec{x})=q \frac{\mathrm{d}}{\mathrm{d} t}{y}^{\mu}(t) \delta^{(3)}[\vec{x}-\vec{y}(t)]=q \delta^{(3)}[\vec{x}-\vec{y}(t)] \begin{pmatrix}1 \\ \mathrm{d} \vec{y}/\mathrm{d} t \end{pmatrix}.$$
In manifestly covariant and parametrization-invariant form it reads
$$j^{\mu}(t,\vec{x})=\int_{\mathbb{R}} \mathrm{d} \lambda q \frac{\mathrm{d}}{\mathrm{d} \lambda}{y}^{\mu}(\lambda) \delta^{(4)}[x-y(\lambda)].$$

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Potential of a moving point charge is given as
##V (\mathbf r,t)= \frac{1}{4\pi\epsilon_0}\int \frac{\rho (\mathbf r',t_r) }{|\mathbf{ (r-r')}|}d\tau'##

Griffiths says:
" It is true that for a point source the denominator ## |\mathbf{(r-r')}|## comes outside the integral..."
Why does it come outside the integral?
If you are working from the 4th edition and this is page 451, equation 10.40, then that statement comes with a footnote that explains why.

The following paragraph also explains why what @Delta2 did is not valid.

• Kashmir
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The following paragraph also explains why what @Delta2 did is not valid.
What I did was for a stationary point source with time varying charge ##q(t)##.

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In my opinion it doesn't just come outside the integral, but it is what we get after evaluating the integral. This is because the charge density for a point source located at ##\vec{r_0}##is given by $$\rho(\vec{r'},t_r)=\delta (\vec{r'}-\vec{r_0})q(t_r)$$ that is it involves a delta dirac function so after we do the integration with respect to ##\vec{r'}## what we get is $$V(\vec{r},t)=\frac{1}{4\pi\epsilon_0}\frac{1}{|\vec{r}-\vec{r_0}|}q(t-\frac{|\vec{r}-\vec{r_0}|}{c})$$
That's a good try, but Griffiths explains why this is wrong, for a very subtle reason: that a point charge must be considered as the limit of an extended charge, which introduces an additional factor into the equation.

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What I did was for a stationary point source with time varying charge ##q(t)##.
That's not what we have here, not to mention the violation of conservation of charge!

• vanhees71
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If you are working from the 4th edition and this is page 451, equation 10.40, then that statement comes with a footnote that explains why.
The scalar potential is:

##V = \frac{q}{4\pi\epsilon_0}\int \frac{\delta\left(\mathbf{r'} - \mathbf{w}\left(t - \frac{|\mathbf{r} - \mathbf{r'}|}{c}\right)\right)}{|\mathbf{r'}-\mathbf{r}|} \mathrm d^3\mathbf{r'}##

It can be shown that at most only ONE event on the trajectory of the charge produces the potential at the observation event. This is the event ##(\mathbf{w}(t_r),t_r)##, where ##t_r## is such that ##|\mathbf{r}-\mathbf{w}(t_r)| = c(t-t_r)##

Because the delta function is 0 apart from at one point, it seems to make sense that ##\mathbf{w}(t_r)## must be the point that it picks out. So the denominator comes out as:

##V = \frac{q}{4\pi\epsilon_0|\mathbf{r} - \mathbf{w}(t_r)|}\int \delta\left(\mathbf{r'} - \mathbf{w}\left(t - \frac{|\mathbf{r} - \mathbf{r'}|}{c}\right)\right) \mathrm d^3\mathbf{r'}\;?##

Is that correct?

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The way I know it regarding integrals of dirac delta function, is that it vanishes when it picks a point, that is $$\int\delta (x-a)f(x)dx=f(a)$$. The way you do it is some in between procedure, where you keep the delta function in the integral but give the value f(a) to the function f. Maybe it is correct , I am not sure because here the "a" (w in your post) , depends on x(r' in your post).