Retarded potential of a moving point charge

• I
• Kashmir
In summary, the potential of a moving point charge is given by the equation ##V(\mathbf{r},t) = \frac{1}{4\pi\epsilon_0}\int \frac{\rho(\mathbf{r'},t_r)}{|\mathbf{r}-\mathbf{r'}|} \mathrm d^3\mathbf{r'}##, where ##\rho(\mathbf{r'},t_r)## is the charge density of the point charge located at ##\mathbf{r'}## and moving with a trajectory given by ##\mathbf{w}(t)##. The denominator ##|\mathbf{r}-\mathbf{r'}|## comes outside
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?

Kashmir
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.

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!

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?

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).

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.

1. What is the retarded potential of a moving point charge?

The retarded potential of a moving point charge is a concept in electromagnetism that describes the potential energy of a point charge as it moves through space. It takes into account the time delay of electromagnetic interactions between the charge and its surroundings.

2. How is the retarded potential calculated?

The retarded potential is calculated using the Lienard-Wiechert formula, which takes into account the charge's velocity, acceleration, and position relative to an observer. It also considers the speed of light and the distance between the charge and the observer.

3. What is the significance of the retarded potential?

The retarded potential is significant because it helps explain the behavior of electromagnetic waves and their interactions with moving charges. It is also used in the study of radiation and the propagation of electromagnetic fields.

4. How does the retarded potential differ from the advanced potential?

The retarded potential and the advanced potential are two solutions to the same equation, but they differ in their time dependence. The retarded potential takes into account the past positions and velocities of the charge, while the advanced potential takes into account the future positions and velocities.

5. What are some real-world applications of the retarded potential?

The retarded potential has numerous applications in various fields, including telecommunications, radar technology, and particle accelerators. It is also used in the study of astrophysics and the behavior of celestial bodies. Additionally, it has practical applications in the design and optimization of electronic circuits and devices.

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