I Gauss' law seems to imply instantaneous electric field

[Bob44]

Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire.

We wish to find an expression for the horizontal component of the electric field at a distance $\mathbf{r}$ from the sphere as it discharges.

By using the Lorenz gauge condition:
$$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$
we find the following retarded solutions to the Maxwell equations

If we assume that $\rho(\mathbf{r},t)=q(t)\delta^3(\mathbf{r})$ and $\partial_t\,\mathbf{j}(\mathbf{r},t)$, $\partial_t\mathbf{A}(\mathbf{r},t)$ are purely vertical then the horizontal component of the electric field is given by the retarded expression
$$\mathbf{E}(\mathbf{r},t)=-\nabla\phi=\frac{q(t-r/c)}{4\pi\epsilon_0}\frac{\hat{\mathbf{r}}}{r^2}.\tag{2}$$
Now let us compare eqn.$(2)$ with the result that we get from Gauss' law
$$\nabla \cdot \mathbf{E} = \frac{\rho}{\epsilon_0}.\tag{3}$$
We integrate both sides of eqn.$(3)$ and use the divergence theorem over a sphere centered at the origin to obtain
$$\mathbf{E}(\mathbf{r},t)=\frac{q(t)}{4\pi\epsilon_0}\frac{\hat{\mathbf{r}}}{r^2}.\tag{4}$$
Eqn.$(2)$ is a causal retarded expression whereas eqn.$(4)$ is instantaneous.

Why do we have this discrepancy?

It seems that we have artificially obtained a causal solution $(2)$ by choosing the Lorenz gauge condition. The non-local expression $(4)$ obtained from Gauss' law did not assume any gauge condition.

 

[PeroK]

Have you assumed a uniform field across the surface of the sphere?

 

[Bob44]

I’ve just used a spherical Gaussian surface with the same center as the charged sphere but with a larger radius so that it encloses the charged sphere. By Gauss’ law the total electric flux through the Gaussian surface is equal to the total charge inside divided by $\epsilon_0$.

By symmetry I assume that the electric field is uniform over the charged sphere and normal to both its surface and the enclosing Gaussian spherical surface.

 

[PeroK]

I’ve just used a spherical Gaussian surface with the same center as the charged sphere but with a larger radius so that it encloses the charged sphere. By Gauss’ law the total electric flux through the Gaussian surface is equal to the total charge inside divided by $\epsilon_0$.

By symmetry I assume that the electric field is uniform over the charged sphere and normal to both its surface and the enclosing Gaussian spherical surface.

The discharge breaks the symmetry.

 

[TSny]

The potential at the field point $p$ at time $t$ is given by the retarded-time expression $\phi(r, t) = \large \frac{q(t-r/c)}{4\pi\epsilon_0 r}$.

When evaluating $\nabla \phi$, we need to take into account that the numerator and denominator of $\large \frac{q(t-r/c)}{4\pi\epsilon_0 r}$ depend on $r$.
$$\nabla \phi = \nabla \left[\frac{q(t-r/c)}{4\pi\epsilon_0 r}\right] = \frac{\nabla q(t-r/c)}{4\pi\epsilon_0 r} + \frac{ q(t-r/c)}{4\pi\epsilon_0 } \nabla \frac 1 r$$ Using the chain rule, you can show that $$\nabla q(t-r/c) = \dot q(t-r/c) \nabla(t-r/c) = -\frac 1 c \dot q(t-r/c) \hat{\mathbf{r}}$$ Here, $\dot q(t-r/c)$ is the rate at which $q$ is changing at the retarded time; that is, $\dot q(t-r/c) = -I(t-r/c)$.

So, we find, using $\nabla \large \frac 1 r = -\frac{\hat{\mathbf{r}}}{r^2}$, $$\nabla \phi =\frac{1}{4\pi\epsilon_0}\left[ \frac{I(t-r/c)}{rc} - \frac{q(t-r/c)}{r^2}\right] \hat{\mathbf{r}}$$

As a simple example, suppose the current is a constant, $I_0$, for all time. Then $q(t) = Q_0 - I_0t$ for all time t, where $Q_0$ is the charge at $t = 0$. Since the current is constant, the vector potential $\mathbf{A}$ is time independent. Thus $\partial_t \mathbf{A}$ does not contribute to the electric field. The charge at the retarded time can be written as $q(t-r/c) = Q_0 - I_0\cdot(t-r/c)$.

So, for this example, the electric field at point $p$ is $$\mathbf{E}_p = -\nabla \phi = -\frac{1}{4\pi\epsilon_0}\left[ \frac{I_0}{rc} - \frac{Q_0-I_0\cdot(t-r/c)}{r^2}\right] \hat{\mathbf{r}}$$ This simplifies to $$\mathbf{E}_p = \frac{Q_0-I_0t}{4\pi\epsilon_0r^2}\hat{\mathbf{r}} = \frac{q(t)}{4\pi\epsilon_0r^2}\hat{\mathbf{r}}$$ Thus, for this simple example, the field at $p$ at time t is the field of a point charge with $q$ equal to the charge at the present time $t$, not the retarded time $t -r/c$.

 

[tech99]

So far as I can see, and I may be incorrect here, when the switch is closed there is oscillation, which is gradually damped out by resistive and radiation losses. The energy we extract by the oscillation was contained within a radius of half a wavelength of the oscillation.
If you discharge a sphere through a resistor, then it takes for ever to discharge. This is because you are now obtaining the energy returning from an infinite radius.
When you charge a sphere, the field will expand outwards for the whole time it is connected. When you discharge it, the field collapses for the time it is connected, returning the portion of the energy contained within a radius time x c. In some cases the time is fixed by oscillation in the wire and the time for a half cycle of oscillation.

 

[JimWhoKnew]

Thus, for this simple example, the field at $p$ at time t is the field of a point charge with $q$ equal to the charge at the present time $t$, not the retarded time $t -r/c$.

A very nice demonstration.

In addition, and also following @PeroK's remark in #4, we can see that by considering a large "virtual" sphere of radius $R$ around the origin,$$\int\mathbf{E}\cdot d\mathbf{A}$$ on it will remain constant at the time between the the closing of the switch (assumed to be located near the charge) and the arrival of this information to the point where the wire exits the sphere (no charge goes in or out of the sphere during this interval). Therefore Equation (4) in OP is evidently incorrect during this interval. Since$$I(t-R/c)=0$$during this interval, equation (2) of OP is valid and $~\mathbf{E}(t,R)~$ is the same as before, as expected by causality.

 

[Bob44]

Thus, for this simple example, the field at $p$ at time t is the field of a point charge with $q$ equal to the charge at the present time $t$, not the retarded time $t -r/c$.

Interesting.

I guess as $q(t)$ is not arbitrary your example does not break causality as we cannot use it to signal faster than light.

 

[TSny]

Interesting.

I guess as $q(t)$ is not arbitrary your example does not break causality as we cannot use it to signal faster than light.

Yes. That's right. In my example, the electric and magnetic fields at any point are time independent. So, no signal propagation can occur.

 

[Demystifier]

If we assume that $\rho(\mathbf{r},t)=q(t)\delta^3(\mathbf{r})$

That's an illegitimate assumption unless $q(t)$ is $t$-independent. If $q(t)$ depends on $t$ then the charge is not conserved, which is not compatible with Maxwell equations which imply charge conservation.

 

[Bob44]

This argument is another version of my previous post.

Imagine that we have two long vertical wires connected either side of a charged sphere.

We connect the two wires to the charged sphere simultaneously so that it is discharged by equal and opposite currents.

Using the Lorenz gauge $\nabla\cdot\mathbf{A}+(1/c^2)\partial \phi/\partial t=0$, Maxwell's equations have the following retarded wave solutions in the scalar and vector potentials.

Starting from Gauss's law

$$\nabla \cdot \mathbf{E}=\frac{\rho}{\epsilon_0}.\tag{1}$$

We substitute in the potentials to give

$$\nabla \cdot (-\nabla \phi - \frac{\partial \mathbf{A}}{\partial t})=\frac{\rho}{\epsilon_0}.\tag{2}$$

Now for every contribution to $\frac{\partial \mathbf{A}}{\partial t}$ from an element of current density $\mathbf{j}$ at position $(3)$ on the upper wire there is an equal and opposite contribution from an element of current density $\mathbf{j}$ at position $(4)$ on the lower wire.

Therefore the scalar potential $\phi$ at point $(1)$ is simply due to the charge density $\rho$ on the sphere at position $(2)$:
$$\nabla^2 \phi=-\frac{\rho}{\epsilon_0}.\tag{3}$$

Poisson's equation $(3)$ has an instantaneous general solution given by:

$$\phi(1,t)=\int \frac{\rho(2,t) dV_2}{4\pi\epsilon_0 r_{12}}.\tag{4}$$

This is different from the expected retarded solution for the scalar potential wave equation:

$$\phi(1,t)=\int \frac{\rho(2,t-r_{12}/c) dV_2}{4\pi\epsilon_0 r_{12}}.\tag{5}$$

What is the reason for this discrepancy?

 

[JimWhoKnew]

Equation (4) is correct in Coulomb gauge, where equation (3) holds everywhere, not only on the equatorial plane.

 

[JimWhoKnew]

In my example, the electric and magnetic fields at any point are time independent.

Your last equation in post #5 seems to contradict this statement. What am I missing?

 

[TSny]

Your last equation in post #5 seems to contradict this statement. What am I missing?

You're not missing anything. When I posted #9, I was apparently missing something - my brain

The magnetic field in my example is static, but the electric field varies with time as given in post #5.

Of course, there is no instantaneous signaling going on between the charge and the field at a distant point.

Thanks for catching this.

 

[PeterDonis]

Moderator's note: Posts from the second thread the OP started on the same topic have been moved to this thread. @Bob44 there is no need to start a second thread on the same topic, just keep posting in this one.

 

[PeterDonis]

Thread closed for moderation.

 

[PeterDonis]

A post with an unacceptable reference has been deleted, and the thread has been reopened. @Bob44 please confine discussion to what standard electrodynamics says. Your question is perfectly answerable within that framework.

 

[PeterDonis]

Gauss' law

Is not applicable the way you are using it in scenarios where there is time variation in the fields. If you dig into the details of how solutions of Maxwell's Equations are obtained from initial data, you will find that Gauss's Law is a constraint on the initial data, not an evolution equation. So you can't expect Gauss's Law to give you correct answers by itself for scenarios where time evolution is significant. It only gives you correct answers by itself in static scenarios.

In your particular case, the key error is here:

We integrate both sides of eqn.$(3)$ and use the divergence theorem over a sphere centered at the origin to obtain
$$\mathbf{E}(\mathbf{r},t)=\frac{q(t)}{4\pi\epsilon_0}\frac{\hat{\mathbf{r}}}{r^2}.\tag{4}$$

No, that's not what you get when you do the integral you describe. You are integrating over a spacelike surface of constant time. That can't give you an equation that has any function of time in it. All it can do is tell you the relationship between $\mathbf{E}$ and $q$ on the particular spacelike surface of constant time that you integrated over, i.e., at one particular value of $t$, not as functions of $t$.