Vector potential of current flowing to a point from all directions

  • #1
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Homework Statement:
This is from Griffith , Introduction to Electrodynamics 4th Edition question 10.7

A time-dependent point charge q(t) at the origin, ##\rho(r, t) = q(t)\delta^3(r)##, is fed by a current ##J(r, t) = −\frac{1}{4π} \frac{\dot q}{r^2}) \hat r##
(a) Check that charge is conserved, by confirming that the continuity equation is obeyed.
(b) Find the scalar and vector potentials in the Coulomb gauge. If you get stuck, try working on (c) first.
(c) Find the fields, and check that they satisfy all of Maxwell’s equations.
Relevant Equations:
$$A = \frac{\mu_0}{4\pi}\int \frac{ J}{|r-r'|}d^3r'$$
$$J = −\frac{1}{4π} \frac{\dot q}{r^2}) \hat r$$
I am having problem with part (b) finding the vector potential. More specifically when writing out the volume integral,
$$A = \frac{\mu_0}{4\pi r}\frac{dq}{dt}\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{?}\frac{1}{4\pi r'^2} r'^2sin\theta dr'd\theta d\phi$$
How do I integrate ##r'##?

The solution says because ##B## can only have ##r## component and only depend on ##r## and ##t## due to symmetry, and ##\nabla \cdot B = 0##, ##B = \nabla \times A = 0## also by using coulomb gauge we set ## \nabla\cdot A = 0##, ##A=0## or it could be any constant.
The argument makes sense to me, but I am not sure what is wrong with the integral, and how does it end up with 0.
 

Answers and Replies

  • #2
jasonRF
Science Advisor
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When you setup the integral,
$$
\mathbf{A }(\mathbf{r})= \frac{\mu_0}{4\pi}\int \frac{ \mathbf{ J}(\mathbf{r'})}{|\mathbf{r}-\mathbf{r'}|}d^3r'
$$
you left out the factor ##\frac{1}{|\mathbf{r}-\mathbf{r'}|}## in the integrand. The integral is messy.
 
Last edited:
  • #3
3
1
When you setup the integral,
$$
\mathbf{A }(\mathbf{r})= \frac{\mu_0}{4\pi}\int \frac{ \mathbf{ J}(\mathbf{r'})}{|\mathbf{r}-\mathbf{r'}|}d^3r'
$$
you left out the factor ##\frac{1}{|\mathbf{r}-\mathbf{r'}|}## in the integrand. The integral is messy.
Yes that's right. I'd fix it but I can't seem to edit it anymore. The correct integral should have an extra ##\frac{1}{r'}## factor in it, but I am still not sure what is the upper bound of integral so that it come out to be a constant.
 
  • #4
TSny
Homework Helper
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Homework Equations: $$A = \frac{\mu_0}{4\pi}\int \frac{ J}{|r-r'|}d^3r'$$
This equation is not valid in general when you have time-dependent fields and currents. In the Lorentz gauge (also called Lorenz gauge) there is a similar formula for A except the integrand is evaluated at "retarded times". But, you want A in the Coulomb gauge. In this gauge, it can be shown that $$\mathbf{A}(\mathbf r) = \frac{\mu_0}{4\pi}\int \frac{\mathbf{J_t}(\mathbf{r}', t_r)}{|\mathbf{r}-\mathbf{r'}|}d^3r'$$where again you use retarded times ##t_r##, but the integrand involves only the so-called "transverse component" ##\mathbf{J_t}## of ##\mathbf{J}##. If you want to read more about this, see section 5.2 here. For your problem ##\mathbf{J_t}=0##, which is consistent with ##\mathbf{A} = 0## in the Coulomb gauge for this particular problem.
 
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