I Retarded potential of a moving point charge

AI Thread Summary
The discussion centers on the potential of a moving point charge, specifically how the denominator in the integral for potential can be simplified. Griffiths notes that for a point source, the denominator can be factored out of the integral due to the nature of the delta function involved in the charge density. The conversation also touches on the implications of using a time-varying charge and the importance of considering the point charge as the limit of an extended charge, which affects the calculations. Participants debate the validity of various approaches to deriving the potential, emphasizing the need for a correct understanding of the delta function's role in the integration process. Overall, the dialogue highlights the complexities of electrodynamics and the subtleties in deriving potentials for moving charges.
Kashmir
Messages
466
Reaction score
74
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?
 
Physics news on Phys.org
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 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
$$j^{\mu}(t,\vec{x})=\int_{\mathbb{R}} \mathrm{d} \lambda q \frac{\mathrm{d}}{\mathrm{d} \lambda}{y}^{\mu}(\lambda) \delta^{(4)}[x-y(\lambda)].$$
 
Kashmir said:
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.
 
  • Like
Likes Kashmir
PeroK said:
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)##.
 
Delta2 said:
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.
 
Delta2 said:
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!
 
  • Like
Likes vanhees71
PeroK said:
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?
 
  • #10
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).
 
  • #11
Once more: Here is a very simple derivation of the Lienard-Wiechert potential and the corresponding electromagnetic field of an arbitrarily moving (massive) point charge, using relativistically covariant electrodynamics (a tautology by the way :-)), Sect. 4.5:

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf

The example of a uniformly moving charge is in Sect. 4.6. The latter case is of course more simply derived using the Lorentz transformation of the four-potential starting from the usual Coulomb potential in the rest frame, ##A^0=q/(4 \pi r)##, ##\vec{A}=0## and then performing the boost.
 
Back
Top