The Maxwell Displacement Current

Maxwell's Equations (ME) consist of two inhomogeneous partial differential equations and two homogeneous partial differential equations. At this point you should be familiar at least with the ``static'' versions of these equations by name and function:

$$\mbox{\boldmath$\nabla$}\cdot \mbox{\boldmath$D$} = \rho\ \ {\rm Gauss's\ Law\ for\ Electrostatics}$$ (8.1)

$$\mbox{\boldmath$\nabla$}\times \mbox{\boldmath$H$} = \mbox{\boldmath$J$}\ \ {\rm Ampere's\ Law}$$ (8.2)

$$\mbox{\boldmath$\nabla$}\cdot \mbox{\boldmath$B$} = 0\ \ {\rm Gauss's\ Law\ for\ Magnetostatics}$$ (8.3)

$$\mbox{\boldmath$\nabla$}\times \mbox{\boldmath$E$} + \frac{\partial \mbox{\boldmath$B$}}{\partial t} = 0\ \ {\rm Faraday's\ Law}$$ (8.4)

in SI units, where $\mbox{\boldmath$D$} = \epsilon\mbox{\boldmath$E$}$ and $\mbox{\boldmath$H$} = \mbox{\boldmath$B$}/\mu$.

The astute reader will immediately notice two things. One is that these equations are not all, strictly speaking, static - Faraday's law contains a time derivative, and Ampere's law involves moving charges in the form of a current. The second is that they are almost symmetric. There is a divergence equation and a curl equation for each kind of field. The inhomogenous equations (which are connected to sources in the form of electric charge) involve the electric displacement and magnetic field, where the homogeneous equations suggest that there is no magnetic charge and consequently no screening of the magnetic induction or electric field due to magnetic charge. One asymmetry is therefore the presence/existence of electric charge in contrast with the absence/nonexistence of magnetic charge.

The other asymmetry is that Faraday's law connects the curl of the $\mbox{\boldmath$E$}$ field to the time derivative of the $\mbox{\boldmath$B$}$ field, but its apparent partner, Ampere's Law, does not connect the curl of $\mbox{\boldmath$H$}$ to the time deriviative of $vD$ as one might expect from symmetry alone.

If one examines Ampere's law in its integral form, however:

$$\oint_C \mbox{\boldmath$B$}\cdot d\mbox{\boldmath$\ell$}= \... ...t_{S/C} \mbox{\boldmath$J$}\cdot \hat{\mbox{\boldmath$n$}}\ dA$$ (8.5)

one quickly concludes that the current through the open surface $S$ bounded by the closed curve $C$ is not invariant as one chooses different surfaces. Let us analyze this and deduce an invariant form for the current (density), two ways.

Figure 8.1: Current flowing through a closed curve $C$ bounded by two surfaces, $S_1$ and $S_2$.

Consider a closed curve $C$ that bounds two distinct open surfaces $S_1$ and $S_2$ that together form a closed surface $S = S_1 + S_2$. Now consider a current (density) ``through'' the curve $C$, moving from left to right. Suppose that some of this current accumulates inside the volume $V$ bounded by $S$. The law of charge conservation states that the flux of the current density out of the closed surface $S$ is equal to the rate that the total charge inside decreases. Expressed as an integral:

$$\oint_S \mbox{\boldmath$J$}\cdot\hat{\mbox{\boldmath$n$}}\ dA = -\frac{d }{d t}\int_{V/S}\rho\ dV$$ (8.6)

With this in mind, examine the figure above. If we rearrange the integrals on the left and right so that the normal $\hat{\mbox{\boldmath$n$}}_1$ points in to the volume (so we can compute the current through the surface $S_1$ moving from left to right) we can easily see that charge conservation tells us that the current in through $S_1$ minus the current out through $S_2$ must equal the rate at which the total charge inside this volume increases. If we express this as integrals:

$$\int_{S_1} \mbox{\boldmath$J$}\cdot\hat{\mbox{\boldmath$n$}}_1\ dA - \int_{S_2} \mbox{\boldmath$J$}\cdot\hat{\mbox{\boldmath$n$}}_2\ dA = \frac{d Q}{d t}$$

$$= \frac{d }{d t}\int_{V/S} \rho\ dV$$ (8.7)

In this expression and figure, note well that $\hat{\mbox{\boldmath$n$}}_1$ and $\hat{\mbox{\boldmath$n$}}_2$ point through the loop in the same sense (e.g. left to right) and note that the volume integral is over the volume $V$ bounded by the closed surface formed by $S_1$ and $S_2$ together.

Using Gauss's Law for the electric field, we can easily connect this volume integral of the charge to the flux of the electric field integrated over these two surfaces with outward directed normals:

$$\int_{V/S} \rho\ dV = \epsilon \oint_S \mbox{\boldmath$E$}\cdot\hat{\mbox{\boldmath$n$}}\ dA$$

$$= -\epsilon \int_{S_1} \mbox{\boldmath$E$}\cdot\hat{\mbox{\boldmath... ... dA + \epsilon \int_{S_2} \mbox{\boldmath$E$}\cdot\hat{\mbox{\boldmath$n$}}\ dA$$ (8.8)

Combining these two expressions, we get:

$$\int_{S_1} \mbox{\boldmath$J$}\cdot\hat{\mbox{\boldmath$n$}}_1\ dA - \int_{S_2} \mbox{\boldmath$J$}\cdot\hat{\mbox{\boldmath$n$}}_2\ dA = \hfill$$

$$\frac{d }{d t}\left\{-\epsilon\int_{S_1} \mbox{\boldmath$E$}\cdot... ...silon\int_{S_2} \mbox{\boldmath$E$}\cdot\hat{\mbox{\boldmath$n$}}_2\ dA\right\}$$ (8.9)

$$\int_{S_1} \mbox{\boldmath$J$}\cdot\hat{\mbox{\boldmath$n$}}_1\ dA + \frac{d }{d t}\epsilon \int_{S_1} \mbox{\boldmath$E$}\cdot\hat{\mbox{\boldmath$n$}}_1\ dA = \hfill$$

$$\int_{S_2} \mbox{\boldmath$J$}\cdot\hat{\mbox{\boldmath$n$}}_2\ d... ...d t}\int_{S_2} \epsilon \mbox{\boldmath$E$}\cdot\hat{\mbox{\boldmath$n$}}_2\ dA$$ (8.10)

$$\int_{S_1} \left\{\mbox{\boldmath$J$}+ \epsilon \frac{d \mb... ...oldmath$E$}}{d t}\right\}\cdot\hat{\mbox{\boldmath$n$}}_2\ dA$$ (8.11)

From this we see that the flux of the ``current density'' inside the brackets is invariant as we choose different surfaces bounded by the closed curve $C$.

In the original formulation of Ampere's Law we can clearly get a different answer on the right for the current ``through'' the closed curve depending on which surface we choose. This is clearly impossible. We therefore modify Ampere's Law to use the invariant current density:

$$\mbox{\boldmath$J$}_{\rm inv} = \mbox{\boldmath$J$}+ \epsilon \frac{d \mbox{\boldmath$E$}}{d t}$$ (8.12)

where the flux of the second term is called the Maxwell displacement current (MDC). Ampere's Law becomes:

$$\oint_C \mbox{\boldmath$B$}\cdot d\mbox{\boldmath$\ell$} = \mu \int_{S/C} \mbox{\boldmath$J$}_{\rm inv} \cdot \hat{\mbox{\boldmath$n$}}\ dA$$

$$= \mu \int_{S/C} \left\{\mbox{\boldmath$J$}+ \epsilon \frac{d \mbox{\boldmath$E$}}{d t}\right\}\cdot \hat{\mbox{\boldmath$n$}}\ dA$$ (8.13)

or

$$\oint_C \mbox{\boldmath$H$}\cdot d\mbox{\boldmath$\ell$}= \... ...boldmath$D$}}{d t}\right\}\cdot \hat{\mbox{\boldmath$n$}}\ dA$$ (8.14)

in terms of the magnetic field $\mbox{\boldmath$H$}$ and electric displacement $\mbox{\boldmath$D$}$. The origin of the term ``displacement current'' is obviously clear in this formulation.

Using vector calculus on our old form of Ampere's Law allows us to arrive at this same conclusion much more simply. If we take the divergence of Ampere's Law we get:

$$\mbox{\boldmath$\nabla$}\cdot (\mbox{\boldmath$\nabla$}\tim... ...$H$}) = 0 = \mbox{\boldmath$\nabla$}\cdot \mbox{\boldmath$J$}$$ (8.15)

If we apply the divergence theorem to the law of charge conservation expressed as a flux integral above, we get its differential form:

$$\mbox{\boldmath$\nabla$}\cdot \mbox{\boldmath$J$}- \frac{\partial \rho}{\partial t} = 0$$ (8.16)

and conclude that in general we can not conclude that the divergence of $\mbox{\boldmath$J$}$ vanishes in general as this expression requires, as there is no guarantee that $\frac{\partial \rho}{\partial t}$ vanishes everywhere in space. It only vanishes for ``steady state currents'' on a background of uniform charge density, justifying our calling this form of Ampere's law a magnetostatic version.

If we substitute in $\rho = \mbox{\boldmath$\nabla$}\cdot\mbox{\boldmath$D$}$ (Gauss's Law) for $\rho$, we can see that it is true that:

$$\mbox{\boldmath$\nabla$}\cdot (\mbox{\boldmath$\nabla$}\tim... ...J$} + \frac{\partial \mbox{\boldmath$D$}}{\partial t}\right\}$$ (8.17)

as an identity. A sufficient (but not necessary!) condition for this to be true is:

$$\mbox{\boldmath$\nabla$}\times \mbox{\boldmath$H$} = \mbox{\boldmath$J$} + \frac{\partial \mbox{\boldmath$D$}}{\partial t}$$ (8.18)

or

$$\mbox{\boldmath$\nabla$}\times \mbox{\boldmath$H$} - \frac{\partial \mbox{\boldmath$D$}}{\partial t} = \mbox{\boldmath$J$}.$$ (8.19)

This expression is identical to the magnetostatic form in the cases where $\mbox{\boldmath$D$}$ is constant in time but respects charge conservation when the associated (displacement) field is changing.

We can now write the complete set of Maxwell's equations, including the Maxwell displacement current discovered by requiring formal invariance of the current and using charge conservation to deduce its form. Keep the latter in mind; it should not be surprising to us later when the law of charge conservation pops out of Maxwell's equations when we investigate their formal properties we can see that we deliberately encoded it into Ampere's Law as the MDC.

Anyway, here they are. Learn them. They need to be second nature as we will spend a considerable amount of time using them repeatedly in many, many contexts as we investigate electromagnetic radiation.

$$\mbox{\boldmath$\nabla$}\cdot \mbox{\boldmath$D$} = \rho \quad{\rm (GLE)}$$ (8.20)

$$\mbox{\boldmath$\nabla$}\times \mbox{\boldmath$H$} - \frac{\partial \mbox{\boldmath$D$}}{\partial t} = \mbox{\boldmath$J$} \quad{\rm\ (AL)}$$ (8.21)

$$\mbox{\boldmath$\nabla$}\cdot \mbox{\boldmath$B$} = 0 \quad{\rm (GLM)}$$ (8.22)

$$\mbox{\boldmath$\nabla$}\times \mbox{\boldmath$E$} + \frac{\partial \mbox{\boldmath$B$}}{\partial t} = 0 \quad{\rm\ (FL)}$$ (8.23)

(where I introduce and obvious and permanent abbreviations for each equation by name as used throughout the rest of this text).

Aren't they pretty! The no-monopoles asymmetry is still present, but we now have two symmetric dynamic equations coupling the electric and magnetic fields and are ready to start studying electrodynamics instead of electrostatics.

Note well that the two inhomogeneous equations use the in-media forms of the electric and magnetic field. These forms are already coarse-grain averaged over the microscopic distribution of point charges that make up bulk matter. In a truly microscopic description, where we consider only bare charges wandering around in free space, we should use the free space versions:

$$\mbox{\boldmath$\nabla$}\cdot \mbox{\boldmath$E$} = \frac{1}{\epsilon_0} \rho$$ (8.24)

$$\mbox{\boldmath$\nabla$}\times \mbox{\boldmath$B$} - \mu_0\epsilon_0 \frac{\partial \mbox{\boldmath$E$}}{\partial t} = \mu_0 \mbox{\boldmath$J$}$$ (8.25)

$$\mbox{\boldmath$\nabla$}\cdot \mbox{\boldmath$B$} = 0$$ (8.26)

$$\mbox{\boldmath$\nabla$}\times \mbox{\boldmath$E$} + \frac{\partial \mbox{\boldmath$B$}}{\partial t} = 0$$ (8.27)

It is time to make these equations jump through some hoops.