Possible error in Griffiths, Intro. to Electrodynamics, Problem 7.60

Question

Probelm 7.60 of Griffiths' Introduction to Electrodynamics, 4th ed, says:

Suppose $\mathbf J(\mathbf r)$ is constant in time but $\rho(\mathbf r, t)$ is not—conditions that might prevail, for instance, during the charging of a capacitor.

(a) Show that the charge density at any particular point is a linear function of time: $$ \rho(\mathbf r, t) = \rho(\mathbf r, 0) + \dot \rho(\mathbf r, t)t, $$ where $\dot \rho(\mathbf r, 0)$ is the time derivative of ρ at t = 0.

(b) Show that $$ \mathbf B(\mathbf r) = \frac{\mu_0}{4\pi} \int \frac{\mathbf J(\mathbf r')\times \hat{\boldsymbol{\ell}}}{\ell^2} d \tau' $$ obeys Ampère’s law with Maxwell’s displacement current term.

Where $ \ell = |\boldsymbol{\ell}|, \quad \boldsymbol{\ell} = \mathbf {r - r'}, \quad \boldsymbol{\hat \ell} = \boldsymbol{\ell}/\ell, \quad $ (since I don't know how to type his cursive $r$ in LateX.)

So the (a) is solved by invoking (as hinted) the continuity equation. But for (b) I have a different answer, namely $$ \mathbf B(\mathbf r) = \frac{\mu_0}{4\pi} \int \frac{(\mathbf J(\mathbf r') + \varepsilon_0 \dot{\mathbf E}(\mathbf r',0) )\times \hat{\boldsymbol{\ell}}}{\ell^2} d \tau' $$ and this is my justification:

Since the charge density is as above and $\mathbf E(\mathbf r,t)$ is given (as in the manual) by $$ \mathbf E(\mathbf r,t) = \frac{1}{4\pi\varepsilon_0} \int \frac{\rho(\mathbf r',t) \hat{\boldsymbol{\ell}}}{\ell^2} d\tau' = \mathbf E(\mathbf r,0) + \dot{\mathbf E}(\mathbf r,0)t $$ then the two Maxwell's equations \begin{align*} \nabla \cdot \mathbf B &= 0 & \nabla \times \mathbf B = \mu_0 \mathbf J + \mu_0\varepsilon_0 \partial_t \mathbf E, \end{align*} with $\mathbf J$ and $\partial_t \mathbf E$ functions only of position (but not of time) and with $\nabla\times \mathbf E = 0$ (by direct computation), are decoupled from the other two and we have the solution for $\mathbf B$ as the Bio-Savart Law with $\mathbf J(\mathbf r')$ replaced with $\mathbf J(\mathbf r') + \varepsilon_0 \dot{\mathbf E}(\mathbf r')$.

Otherwise where is my mistake ?

Answer 1

Griffiths wants you to realize that that integral of $\partial_t \mathbf E$, apparently contributing to magnetic field $\mathbf B$, is zero.

Any field $\mathbf B$ whose divergence vanishes everywhere can be expressed as

$$ \mathbf B = \nabla \times \mathbf A $$ where one possible vector potential $\mathbf A$ is

$$ \mathbf A(\mathbf x) = \frac{1}{4\pi}\int \frac{\nabla_{\mathbf x'} \times \mathbf B(\mathbf x')}{|\mathbf x - \mathbf x'|}d^3 \mathbf x'. $$

This follows from Helmholtz's theorem. We assume the integral exists.

Using Maxwell's equations, we can expresss the vector potential in this way:

$$ \mathbf A = \frac{1}{4\pi}\int \frac{\mu_0\mathbf j(\mathbf x')+ \mu_0\epsilon_0 \partial_t \mathbf E(\mathbf x')}{|\mathbf x - \mathbf x'|}d^3 \mathbf x'. $$

There are two terms in the integrand. If the latter term contributes zero to magnetic field $\mathbf B$, it can be dropped and we arrive at the conclusion that magnetic field, obeying Maxwell's equations with displacement current, is given by the Biot-Savart formula.

We have to assume (in addition to what is stated as assumptions in the assignment) that magnetic field is constant in time, so electric field is a conservative field, so it can be expressed as

$$ \mathbf E = - \nabla \varphi $$

for some function $\varphi$. We will show this means that contribution of $\partial_t \mathbf E$ to the vector potential is a conservative field, hence a gradient of something, hence has zero curl, hence does not contribute to magnetic field.

We start with the integral

$$ I = \int \frac{\partial_t \mathbf E(\mathbf x')}{|\mathbf x - \mathbf x'|}~d^3\mathbf x' $$ and express it using electric potential: $$ I = -\int \frac{\partial_t \nabla_{\mathbf x'} \varphi(\mathbf x')}{|\mathbf x - \mathbf x'|}~d^3\mathbf x' $$

Using substitution of variables in the integral and switching the order of differentiation and integration, we can express this quantity also in this way: $$ I = -\nabla_{\mathbf x} \int \frac{\partial_t \varphi(\mathbf x')}{|\mathbf x - \mathbf x'|}~d^3\mathbf x' $$ so this is a conservative field with zero curl. Hence it does not contribute to magnetic field.

Hence under the assumption (magnetic field is constant in time), the Biot-Savart formula gives correctly magnetic field obeying the Maxwell equations with the displacement current term.

Answer 2

Your $\dot {\bf E}$ term corresponds to a gradient added to the vector potential, and when integrated is zero. Basically, your calculation is equivalent to calculating in Coulomb gauge where ${\bf \nabla} \cdot {\bf A} =0$ while Griffiths calculation is equivalent to calculating the vector potential in Lorenz Gauge. That is, if you had used the retarded potential for ${\bf A}$, since ${\bf J}$ doesn't change with time, the Lorenz gauge retarded potential would be Griffiths result.

Griffiths is asking you to show that his result is correct.