About the uniqueness of the displacement current

Question

In the Maxwell-Ampère equation, i.e.: \begin{equation} \nabla\times\vec{B} = \mu_0 \vec{J} + \mu_0\epsilon_0 \frac{\partial \vec{E}}{\partial t} \end{equation} the $\vec{J}_d$ term: $$ \vec{J}_d := \epsilon_0 \frac{\partial \vec{E}}{\partial t} $$ was derived by taking the divergence of the left hand side of the equation. Explicitly, before Maxwell's addition of the $\vec{J}_d$ term Ampère's law was: $\nabla\times\vec{B} = \mu_0 \vec{J}$, but when acting with $\nabla \cdot$ we had: $$ 0 \equiv \nabla \cdot \left(\nabla\times\vec{B}\right) = \mu_0 \nabla\cdot\vec{J} = -\mu_0 \frac{\partial \rho}{\partial t} $$ from the $\text{div}(\text{curl}\ \bullet)$ identity and the continuity equation. But $\frac{\partial \rho}{\partial t} $ is not necessarily zero so we need to add a new term let's call it $\vec{J}_d$. And now comes my question. We need $\vec{J}_d : \nabla\cdot\left( \vec{J} + \vec{J}_d \right) = 0 \Rightarrow \nabla \cdot \vec{J}_d = \frac{\partial \rho}{\partial t} $. And indeed $\vec{J}_d = \epsilon_0 \frac{\partial \vec{E}}{\partial t}$ is a solution, but for this "test of the divergence" $$\vec{J'}_d = \epsilon_0 \frac{\partial \vec{E}}{\partial t} + \vec{k}$$ where $\vec{k}$ is a constant vector, or even $$\vec{J''}_d = \epsilon_0 \frac{\partial \vec{E}}{\partial t} + \nabla \times \vec{T}$$ where $\vec{T}$ is any vector, satisfy $\nabla\cdot\left(\nabla\times\vec{J}_d\right) = 0$. Why, then does $\vec{J}_d$ has the form it has and not any of the other possible solutions presented above?

Thanks in advance.

Answer 1

But surely that is not the only constraint.

If $$\vec{J'}_d = \epsilon_0 \frac{\partial \vec{E}}{\partial t} + \vec{k}$$ then $$ \nabla\times\vec{B} = \mu_0 \vec{J} + \mu_0\epsilon_0 \frac{\partial \vec{E}}{\partial t} + \mu_0 \vec{k} $$

This implies that even without any current or time-dependent electric field there is a non-conservative magnetic field. But without currents or time-dependent electric fields we know that the B-field is curl-free.

Answer 2

First off, you can't define $$\vec{J'}_d = \epsilon_0 \frac{\partial \vec{E}}{\partial t} + k$$ because you'd be adding a vector to a scalar, so we're left with your second idea:

$$\vec{J}_d = \epsilon_0 \frac{\partial \vec{E}}{\partial t} + \nabla \times \vec{T}$$

If we take this to be the value of $\vec{J}_d$, then Ampere's law becomes

$$\nabla\times\vec{B} = \mu_0 \vec{J} + \mu_0\epsilon_0 \frac{\partial \vec{E}}{\partial t} + \nabla \times \vec{T}$$

Consider a circumstance where the electric field is constant ($\frac{\partial \vec{E}}{\partial t} = 0$) and there are no currents ($\vec{J} = 0$). Your modified Ampere's law predicts that

$$\nabla\times\vec{B} = \nabla \times \vec{T}$$

But experimentally, we find that in such circumstances, $\nabla \times \vec{B} = 0$, so we conclude that $\nabla \times \vec{T} = 0$

Answer 3

The other answers say some interesting things about the consequences of this term, but the biggest reason it's not there is that all experiments have confirmed Maxwell's equations as we know them, and there is no evidence that a modification would be necessary.

After all, the equations are not unique: for example, they could easily be modified to include magnetic monopoles and they would be made more symmetrical. But nobody has ever seen a magnetic monopole, and so they don't appear in Maxwell's equations.

Answer 4

Explicitly, before Maxwell's addition of the $\vec{J}_d$ term Ampere's law was: $\nabla\times\vec{B} = \mu_0 \vec{J}$

The important thing to understand with all this is that Ampere's law is about conduction current, and conduction current is not the only current. Have a look at Taming Light at the Nanoscale:

"Look around, and you will probably see numerous electronic and optical gadgets, such as mobile phones, personal digital assistants, laptops, TVs and digital cameras. These may all do different things but they have one thing in common: in the electronic circuits that drive these devices, charged particles flow through components and impart power via what is known as the conduction current. But is the motion of charged particles the only current we have available?"

The answer is no, because we also have displacement current. It's "a time-varying electric field", and that's exactly what we see when an electromagnetic wave passes us by. There is no charged particle present, but the displacement current is present, and it's alternating: the field-variation increases to a maximum, then decreases back to zero. Note that we could put this wave through pair production, so we can convert displacement current into charged particles. Then when we move them, we call the phenomenon conduction current. Also note that because of this, displacement current is more fundamental than conduction current. And that knowing all this makes it clear that the original version of Ampere's law doesn't go far enough:

"Ampère's law determines the magnetic field associated with a given current, or the current associated with a given magnetic field, provided that the electric field does not change over time".

Why, then does $\vec{J}_d$ has the form it has and not any of the other possible solutions presented above?

Because of what displacement current is. Maxwell effectively worked backwards from Ampere and conduction current, and ended up saying "light consists of transverse undulations in the same medium that is the cause of electric and magnetic phenomena". When people read this they tend to think of sinusoidal E and B waves:

_{Image courtesy of mathematica}

However that doesn't go far enough either. See the Wikipedia electromagnetic radiation article:

"Also, E and B far-fields in free space, which as wave solutions depend primarily on these two Maxwell equations, are in-phase with each other. This is guaranteed since the generic wave solution is first order in both space and time, and the curl operator on one side of these equations results in first-order spatial derivatives of the wave solution, while the time-derivative on the other side of the equations, which gives the other field, is first order in time, resulting in the same phase shift for both fields in each mathematical operation."

E is the spatial derivative of the wave, and B is the time derivative. So the real wave is the integral of the E and B sine waves. I'll illustrate this with a canoe analogy: imagine you're in a canoe as a ten-metre ocean wave* approaches. Your canoe starts to slope upwards, slowly at first, then faster, then the slope starts to flatten out, and your canoe is momentarily horizontal on the top of the wave. At this point displacement current is at a maximum, at the mid-point of the E and B sine waves. Then the process is reversed, something like this:

The slope of your canoe denotes E, and the rate of change of slope denotes B. One is the spatial derivative, the other is the time derivative. The displacement current is represented by the current of water that lifted you bodily up ↑ by ten metres, then let you down ↓ again. It has a vector nature, and the |direction| is the polarization direction. We write it as:

$$\vec{J}_d = \epsilon_0 \frac{\partial \vec{E}}{\partial t}$$

Adding a constant k would be like repeating the canoe scenario in deeper water. The wave is 10m high, so the deeper water won't alter the tilt of your canoe or how high you go. You can't add something to one side of the expression because each side is telling you about one aspect of the wave, and you can't change one side without changing the other.

^{* without a trough}