EM Four-Potential and INDUCED Electric Fields

[LarryS]

TL;DR Summary

Can EM Four-Potential Model INDUCED Electric Fields?

Say we have a long solenoid with a current that is fluctuating in time. Then the changing magnetic flux in the solenoid will induce an electric field around the outside of the solenoid (Faraday's Law). This induced electric field is not conservative and therefore cannot have a corresponding electric potential function, under any gauge. On the surface, this would seem to be a major shortcoming of the EM Four Potential. If not, why not?

As always, thanks in advance.

 

[Dale]

Summary: Can EM Four-Potential Model INDUCED Electric Fields?

This induced electric field is not conservative and therefore cannot have a corresponding electric potential function

It has a vector potential. The four potential is the scalar potential and the vector potential together.

I am not exactly sure why you think this is any kind of a shortcoming, let alone a major one.

 

[LarryS]

It has a vector potential. The four potential is the scalar potential and the vector potential together.

I am not exactly sure why you think this is any kind of a shortcoming, let alone a major one.

So, the induced electric field has a vector potential instead of an electric scalar potential. Interesting.

Is the vector potential for the magnetic field inside the solenoid distinguishable from the vector potential for the induced electric field?

 

[Dale]

It is the same vector potential that contributes to both. The curl is the magnetic field and the time derivative contributes to the electric field.

I must admit that your question confuses me. How are you even asking about the four potential if you don’t already know about the vector potential? What do you think the four potential is if you don’t know the vector potential already? Can you give some context because this order of discovery is very unusual and I don’t know what to assume is known in my answers here.

 

[vanhees71]

You can use the two homogeneous Maxwell equations (SI units),
$$\vec{\nabla} \times \vec{E} + \partial_t \vec{B}=0, \quad \vec{\nabla} \cdot \vec{B}=0$$
to introduce the electromagnetic potentials.

From Helmholtz's theorem we know that due to the 2nd equation ("no magnetic monopoles") that there is a vector potential for the magnetic field, i.e.,
$$\vec{B}=\vec{\nabla} \times \vec{A}.$$
It is defined up to an arbitrary gradient field, i.e.,
$$\vec{A}'=\vec{A}-\vec{\nabla} \chi \qquad (*)$$
describes the same physics as $\vec{A}$ for an arbitrary scalar field, $\chi$. That's called gauge invariance.

Now use this in the 1st equation ("Faraday's Law"):
$$\vec{\nabla} \times \vec{E} + \partial_t \vec{\nabla} \times \vec{A}=0$$
or
$$\vec{\nabla} \times (\vec{E}+\partial_t \vec{A})=0.$$
This implies that the field in the paranthesis can be written as the gradient of a scalar potential,
$$\vec{E} + \partial_t \vec{A} = -\vec{\nabla} \Phi$$
or
$$\vec{E}=-\vec{\nabla} \Phi - \partial_t \vec{A}.$$
Changing the gauge via (*) needs also a change for $\Phi$
$$\vec{E} = -\vec{\nabla} \Phi' - \partial_t \vec{A}'=-\vec{\nabla} \Phi' - \partial_t (\vec{A}-\vec{\nabla} \chi) = - \vec{\nabla}(\Phi'-\partial_t \chi) - \partial_t \vec{A},$$
i.e.,
$$\Phi=\Phi'-\partial_t \chi \; \Rightarrow \; \Phi'=\Phi + \partial_t \chi.$$

 

[LarryS]

It is the same vector potential that contributes to both. The curl is the magnetic field and the time derivative contributes to the electric field.

I must admit that your question confuses me. How are you even asking about the four potential if you don’t already know about the vector potential? What do you think the four potential is if you don’t know the vector potential already? Can you give some context because this order of discovery is very unusual and I don’t know what to assume is known in my answers here.

Hi Dale. Yes, I already knew that the EM four potential consists of the electric scalar potential in the first (time) position followed by the 3 components of the magnetic vector potential in the 3 spatial positions of the four-vector.

By "the same vector potential" you are referring to the magnetic vector potential not the entire EM four-vector, right?

 

[LarryS]

I had forgotten that the electric field consisted of more than just the gradient of the electric scalar potential. It also includes the partial time derivative of the magnetic vector potential.

 

[Dale]

By "the same vector potential" you are referring to the magnetic vector potential not the entire EM four-vector, right?

There is no magnetic vector potential. There is just the vector potential. (Three vector). The vector potential contributes to both the magnetic field and the electric field (three vectors).

I had forgotten that the electric field consisted of more than just the gradient of the electric scalar potential. It also includes the partial time derivative of the magnetic vector potential.

Yes.

 

[Vanadium 50]

There is no magnetic vector potential.

Well...you will find the electromagnetic vector potential called the "magnetic vector potential" in many places - e.g. https://en.wikipedia.org/wiki/Magnetic_vector_potential I agree it's not a very good name, but it is a "thing".

 

[Dale]

Well...you will find the electromagnetic vector potential called the "magnetic vector potential" in many places - e.g. https://en.wikipedia.org/wiki/Magnetic_vector_potential I agree it's not a very good name, but it is a "thing".

Oh, I didn’t know that. That makes the OP’s confusion more understandable. I just learned it as “the vector potential” with immediate connection to both E and B.

 

[vanhees71]

One should also avoid confusion by stressing that one should regard not $(\Phi,\vec{A})$ as describing the electromagnetic field but an entire equivalence class of such fields, with each member differing by a gauge transformation with an arbitrary scalar field, $\chi$.

In other words: the two potentials $(\Phi,\vec{A})$ and $(\Phi',\vec{A}')$ should be considered to be equivalent, if there's a scalar field, $\chi$, such that
$$\Phi'=\Phi+\partial_t \chi, \quad \vec{A}'=\vec{A}-\vec{\nabla} \chi.$$

Then it becomes clear that a complete solution of the Maxwell equations will determine the potentials only up to a gauge transformation, and you can significantly simplify this task by choosing some pretty arbitrary constraint to fix or at least partially fix the gauge.

Indeed the potentials define the electromagnetic field by
$$\vec{E}=-\partial_t \vec{A} - \vec{\nabla} \Phi, \quad \vec{B}=\vec{\nabla} \times \vec{A}.$$
Then the two homogeneous Maxwell equations are automatically fulfilled (see my previous posting #5).

From now on let's work in natural Heaviside-Lorentz units, i.e., setting $\mu_0=\epsilon_0=1$, which also implies that the speed of light $c=1$. In these very convenient units we get rid of the disturbing constants $\epsilon_0$ and $\mu_0$, which make the entire business much less beautiful:
$$\vec{\nabla} \times \vec{B} - \partial_t \vec{E} = \vec{j}, \quad \vec{\nabla} \cdot \vec{e}=\rho,$$
where $\rho$ is the electric-charge density and $\vec{j}$ is the electric-charge-current density. One should note that a necessary condition to make the inhomogeneous Maxwell equations solvable is the conservation of charge, i.e., the continuity equation,
$$\partial_t + \vec{\nabla} \cdot \vec{j}=0.$$
Now we plug in the fields in terms of the potentials:
$$\vec{\nabla} \times (\vec{\nabla} \times \vec{A}) - \partial_t (-\partial_t \vec{A} -\vec{\nabla} \Phi)=\vec{j}, \quad \vec{\nabla} \cdot (-\partial_t \vec{A} - \vec{\nabla} \Phi)=\rho.$$
These are pretty awful equations, mixing all components of the potential with each other. But we must not forget that we have pretty much freedom by exploiting the gauge invariance, i.e., we can constraint the potentials by some equations that make the equations simpler. Take the first equation, which we can transform a bit further,
$$\vec{\nabla} (\vec{\nabla} \cdot \vec{A}+\partial_t \Phi) - (\Delta-\partial_t^2) \vec{A} = \vec{j}.$$
From this we see that we can decouple the components completely if we use the constraint
$$\partial_t \Phi + \vec{\nabla} \cdot \vec{A}=0,$$
which is the socalled Lorenz-gauge condition. It only partially fixes the gauge, i.e., you can always find a field $\chi$ to gauge transform an arbitrary solution to one that fulfills the Lorenz-gauge condition, but this $\chi$ is only determined up to a solution of the homogeneous wave equation $(\partial_t^2-\Delta) \chi=\Box \chi=0$, but that doesn't matter too much (in classical electrodynamics), because we just need one solution for the potentials, i.e., the gauge freedom must not be completely fixed by the gauge constraint. In any case, if the Lorenz-gauge condition is fulfilled our equation for $\vec{A}$ simplifies to
$$\Box \vec{A}=\vec{j}.$$
Now the remaining equation,
$$-\partial_t \vec{\nabla} \cdot \vec{A}-\Delta \Phi=\rho,$$
can be simplified by using the Lorenz-gauge condition, which says that $\vec{\nabla} \cdot \vec{A}=-\partial_t \Phi$ and thus you get a wave equation for $\Phi$ too:
$$\Box \Phi=\rho.$$
You just need a Green's function of the d'Alembert operator, $\Box$, to find a solution. The one which is usually needed in classical field theory is the retarded propagator, which ensures that the field at time $t$ depends on its sources $\rho$ and $\vec{j}$ only at times $t'<t$. This Green's function is
$$D_{\text{ret}}(t,\vec{r})=\frac{\delta(t-r)}{4 \pi r}, \quad r=|\vec{r}|.$$
Then the solutions are
$$\vec{A}(t,\vec{x}) = \int_{\mathbb{R}} \mathrm{d} t' \int_{\mathbb{R}^3} \mathrm{d}^3 x' D(t-t',\vec{r}-\vec{r}') \vec{j}(t',\vec{r}') = \int_{\mathbb{R}^3} \mathrm{d}^3 x' \frac{\vec{j}(t-|\vec{x}-\vec{x}'|,\vec{x}')}{4 \pi |\vec{x}-\vec{x}'|}$$
and in the same way
$$\Phi(t,\vec{x})= \ldots = \int_{\mathbb{R}^3} \mathrm{d}^3 x' \frac{\rho(t-|\vec{x}-\vec{x}'|,\vec{x}')}{4 \pi |\vec{x} - \vec{x}'|}.$$
Which shows that the fields at time $t$ are given as the superposition of contributions from the sources at times earlier by the time light needs to run from the source point $\vec{x}'$ to the point $\vec{x}$. In Lorenz gauge thus the potentials are simply given by Huygens's principle, i.e., the potentials at a point $\vec{x}$ are given by the superposition of spherical waves spreading with the speed of light from each "source point" $\vec{x}'$.