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What is a good way to think of the displacement current? Maxwell imagined it as being movements in the aether, small changed of electric field producing magnetic field. I don't even understand that definition-assuming there is aether. (On the topic of which, has aether actually been disproved? I read that even with the Michelson-Morley experiment the aether wasn't disproved.)

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Re the final, parenthesized part of the question, there is no way to answer this. It depends on what you mean by "aether" and what you mean by "disproved." If you have some written source that says the existence of the aether wasn't disproved, tell us what you source is -- and, preferably, ask it in a separate question. –  Ben Crowell May 12 '13 at 22:32

4 Answers 4

Maxwell's equations in a vacuum have induction terms. (1) There is a term saying that a time-varying magnetic field produces an electric field. (2) There is a term saying that a time-varying electric field produces a magnetic field.

Among people who insist on giving hard-to-remember names to all the terms in Maxwell's equations, #2 is called the displacement current. The name is a bad one, because it's not a current, i.e., it has nothing to do with the motion of charged, material particles. The only reason it has the misleading name is that it adds to the current term, and Maxwell, who made up the name, wasn't sure what its ultimate origin was.

The importance of term #2 is mainly that it allows the existence of electromagnetic waves. In an electromagnetic wave, the changing E field induces the B field, and the changing B field induces the E field.

There are elementary reasons that term #2 has to exist. For example, suppose you have a circular, flat Amperian surface $S_1$ and you shoot a single charged particle perpendicularly through its center. In this situation, Maxwell's equations without term #2 predict that the magnetic field at the edge of the surface will be zero, then infinite for an instant, and then zero again after that. But if we construct a similar Amperian surface $S_2$ with the same boundary but an interior surface that is bowed out rather than flat, we get a prediction that the infinite field occurs at a different time. This proves that we can't get away with leaving Maxwell's equations in a form with all the terms except term #2.

The deeper reason for term #2 is that it's required by relativity. Only with term #2 do Maxwell's equations have a form that is the same in all frames of reference.

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Contra Ben Crowell's answer, I'll argue that "displacement current" is a fine name, because Maxwell's equations treat the rate of change of electric flux density (the "displacement current") exactly the same as a charge current. When you hear the term, you know exactly what's being discussed, and even how it fits into the equations.

Without the displacement current, Maxwell's equations are inconsistent. The standard example involves a circuit with current flowing through a wire to charge a capacitor. (See the Wikipedia article for displacement current for a figure of this geometry.)

The task is to calculate the line integral of magnetic field around a loop surrounding the wire. By Stokes' theorem, one converts this line integral to a surface integral of $\boldsymbol{\nabla \times B}$ over a surface whose boundary is the loop. By Maxwell, this quantity is:

$$\boldsymbol{\nabla \times B = J + J_d}$$

where $\boldsymbol{J}$ is the usual charge current density and $\boldsymbol{J_d} = \partial \boldsymbol{D} / \partial t$ is the displacement current density.

Of course, many different surfaces have the loop as a boundary:

  • If the particular surface chosen is intersected by the circuit's wire, its integrated current density (the wire current) gives a non-zero (and correct) answer for the surface (and therefore the line) integral.
  • But what if the surface passes through the interior of the capacitor, where there is no charge current density ($\boldsymbol{J}=0$)? The value of the line integral should be the same, no matter what surface is chosen to span it and apply Stokes' theorem. Displacement current to the rescue... When the rate of change of the capacitor electric field is determined and its surface integral calculated, the displacement current turns out to exactly equal the wire current, giving the same result for the $\boldsymbol{B}$ line integral.
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I'm having trouble interpreting your descriptin of the capacitor circuit with the test loop. Can you point us to a diagram or something? –  Ben Crowell May 13 '13 at 2:50
    
@BenCrowell: Yes, it's an awkward description. See the 2nd figure in the Wikipedia article for displacement current: en.wikipedia.org/wiki/Displacement_current , where S1 and S2 are the 2 surfaces. –  Art Brown May 13 '13 at 2:53
    
Electrodynamics is not limited by the Maxwell's equations only. There are other laws and equations too, for example, the continuity equation. And here the "displacement current" becomes a confusing name. –  firtree May 13 '13 at 11:32
    
@firtree: The continuity equation can be proved from Maxwell's equations. It's not logically independent. Maxwell's equations are a complete description of classical electrodynamics, excluding only statements like Ohm's law that are really descriptions of the behavior of materials. –  Ben Crowell May 13 '13 at 15:48
    
@BenCrowell The continuity equation can be proved from Maxwell's equations. - yes, and the name 'displacement current' is not used in that proof. Only formulas. Maxwell's equations are a complete description of classical electrodynamics - not complete. You would also need the Lorentz force law at least. It comes from the same interaction Langrangian as the right part of the Maxwell's equations. And yes, material equations could be needed as well. By the way, they are also needed to define and measure fields, as a matter of fact. –  firtree May 13 '13 at 16:04

The displacement current is the 'phantom' current that passes through a capacitor in a circuit, since no real current runs between two plates of a capacitor. This is given by finding the rate of change of the electric flux with respect to time, and multiplied by epsilon nought. A great video on this can be found here: http://www.learner.org/resources/series42.html

Go to 'Maxwell's Equations' and watch the video about 20 minutes in.

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Saying "The displacement current is..." and then giving a statement about capacitors is like saying that a car is a device that exists because it has a rear-view mirror, and the rear-view mirror gives us something to hang a pine-scented air freshener from. –  Ben Crowell May 12 '13 at 23:00

Let's look at Ampere's law, $\nabla\times\vec{H} = \vec{J}_f + \partial\vec{D}/\partial t$, where $\vec{J}_f$ is the free current density. The partial derivative of electric displacement vector $\vec{D}$ is the displacement current density. It arises because of a non-steady, that is time-dependent, free charge density $\rho_f$.

One example of a non-steady $\rho_f$ is a capacitor getting charged (or discharged).

To see how $\partial\vec{D}/\partial t$ term is related to time variation of $\rho_f$, apply the divergence operator to Ampere's law to get $0 = \nabla\cdot\vec{J}_f + \partial\rho_f/\partial t$, where we used Gauss law $\nabla\cdot\vec{D} = \rho_f$.

A more direct way of seeing the role of $\rho_f$ is by using the integral form of Ampere's law. \begin{equation} \oint \vec{H}\cdot\hat{n}dA = I_f + \frac{\partial}{\partial t}\oint\vec{D}\cdot\hat{n}dA, \end{equation} $I_f$ being the free current.

The second term on the right hand side can be immediately written in terms of $\rho_f$ using Gauss law to get \begin{equation} \oint \vec{H}\cdot\hat{n}dA = I_f + \int_V \frac{\partial\rho_f}{\partial t}dV \end{equation} The second term on the right hand side is the displacement current. It vanishes when we have a steady current.

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