Electric Vector Potential: Detect Non-Conservative Fields

Main Point

The electric vector potential offers a means of determining the non-conservative component of a mixed stationary or quasi-stationary electric field.

Background: magnetic vector potential

Many readers will be familiar with the static magnetic vector potential

$\bf A = (\mu/4\pi) \int \bf j~ dv/s$

with $\bf j$ = current (area) density within differential volume $dv$ and $s$ = distance from $dv$ to the point of observation. The magnetic field is then derivable from

$\bf B =  \nabla \times \bf A$.

Vector potentials for general vector fields

Finite vector potentials exist in any vector field $F$ for which $\nabla \times \bf F \neq 0$. If the field is wholly scalar, the vector potential is zero.

Helmholtz decomposition for the electric field

The Helmholtz decomposition theorem for any electric field states there are two kinds of E field:

• the irrotational (scalar) field with $\nabla \times \bf E = 0$, which we call $E_s$, and
• the non-conservative field with $\nabla \times \bf E \neq 0$, which we call $E_m$.

An electric field can be a mixture of both; we call that a mixed field:

$\bf E = \bf E_m + \bf E_s$.

The vector potential can serve to identify the $\bf E_m$ part of any static or quasi-static mixed E field at any observation point. Doing so ipso facto identifies both the irrotational and non-conservative components at that point.

In fact, a concise name for the electric vector potential is “$E_m$ detector.”

Generalized vector potential and gauge choice

The generalized vector potential for a stationary or quasi-stationary field $\bf F$ obeys

$\bf F = -\nabla U + \nabla \times \bf W$

where the scalar potential is $U$ and the vector potential is $\bf W$.

From vector calculus we also have

$\nabla^2 \bf W = -\nabla (\nabla \cdot \bf W)~ – \nabla \times (\nabla \times \bf W)$

For our vector potential $\bf W$ we are free to choose the gauge condition $\nabla \cdot \bf W = 0$ (by gauge invariance or direct verification). Then

$-\nabla^2 \bf W = \nabla \times \bf F$

which are three scalar equations. By analogy to the known expression for the scalar electric potential

$U = (1/4\pi)  \int \rho/\epsilon~ dv/s$

if we replace $\rho / \epsilon$ with $(\nabla \times \bf F)_z$ we obtain the general relation for a vector field $\bf F$ with vector potential $\bf W$:

$\bf W = (1/4\pi) \int (\nabla \times \bf F)~ dv/s.$

and therefore

$\bf F = \nabla \times \bf W – \nabla U$.

Electric vector potential for electromagnetic fields

For an electromagnetic E field we also know that $\nabla \times \bf E  = -\partial \bf B/\partial t$, so the electric vector potential $\bf W$ is

$$ \bf W = (1/4\pi) \int (-\partial \bf B/\partial t) ~dv/s $$

with

$\bf E_m = \nabla \times \bf W$  and  $\bf E_s = -\nabla U$.

Example: circular B field (infinite length)

As a simple example, apply this formula to a circular B field of infinite length ($-\infty < z < +\infty$) with cross-sectional area $a$ so flux $\phi = aB$. We seek the E field outside the B field. We already know a priori that in cylindrical coordinates

$\bf E = -(\partial \phi/\partial t)/2\pi~ r ~ \hat {\bf \theta}$

forming circles of radius r.

Using the electric vector potential,

$\bf W = (1/4\pi) \int (\nabla \times \bf E)~dv/s$

$= -(1/4\pi) \dot B\int dv/s.$

With $dv = a ~dz$, $\phi = aB$ and $s = (z^2 + r^2)^{1/2}$, we find $\bf W = W_z \hat {\bf k}$ since $\bf W$ is in the direction of $\bf B$. Thus

$W_z(r,d)  =~ – (1/4\pi) \dot \phi \int_{-d}^d ~ dz/(z^2 +r^2)^{1/2}$

Evaluating the integral gives

$W_z(r,d) = ~ – (1/4\pi) \dot \phi ~~ \ln~ \frac {(z^2 + r^2)^{1/2} + z} {(z^2 + r^2)^{1/2} – z}$ (evaluated at z = d and z = -d).

Now $\bf E = \nabla \times \bf W$, so

$E_{\theta} =  \lim_{d \to \infty} \partial {W_z(r,d)}/\partial r$

which yields

$=~ – (1/4\pi) \dot \phi   ~~ \lim_{d \to \infty} 2d/[r(r^2+ d^2)^{1/2}]$

$= ~- \dot \phi/2\pi~r$

i.e., the expected result.

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Resistive ring example and split fields

Now consider a resistive ring concentric with a time-varying B field of infinite extent at radius $a$. Assume first a uniform ring of resistance $r$. The formula gives the same result as without the ring:

$E_m = -\partial \phi/\partial t/2\pi a$ all around the ring.

Next, assume half the ring has resistance $r$ and the other half $2r$ (see Fig. 1):

Fig. 1 Non-uniform resistive ring surrounding time-varying B field

The electric vector potential formula still gives the same $E_m$, as it should. Yet applying Faraday’s law together with Ohm’s law shows the low-resistance ($r$) side has half the E field of the high-resistance ($2r$) side. How can this be?

The answer is that the fields are a mixture of scalar and non-conservative fields. The electric vector potential finds only the part of the E field ascribable directly to the time-varying B field — the $E_m$ part. The electrostatic part $E_s$ is of course $-\nabla U$ and accounts for the remaining field distribution (equal and opposite $E_s$ fields in this example).

See also:

https://www.physicsforums.com/threads/how-to-recognize-split-electric-fields-comments.984872/

Why split fields matter

You may ask why bother identifying split fields when measurable parameters (voltmeter readings, Joule heating, current, net E field, and satisfaction of Faraday’s law) can be obtained assuming a single kind of E field. My answer is twofold:

1. The E fields inside and outside the ring are strongly affected by the free charges that set up the $E_s$ fields. In Fig. 1, net positive free charge accumulates at A with an equal negative charge at B. With a uniform ring there are no such free charges, so $E_s = 0$ and the field is simply $E_m$.
2. Do we want only practical results, or do we want to understand what’s really happening? As with the ring laser gyro and the Sagnac effect, some phenomena admit simple pragmatic explanations but a fuller understanding requires distinguishing the underlying contributions.

Final note

The electric vector potential is more powerful than the Faraday rule

$\oint \bf E \cdot \bf {dl} = – \partial \phi/\partial t$

because Faraday’s integral law does not distinguish between electrostatic and emf-generated E fields, whereas the vector potential gives a point-by-point accurate reading for the emf ($E_m$) part of the field around any closed loop. The remaining part must be electrostatic, by Helmholtz decomposition.

Further reading:

• http://farside.ph.utexas.edu/teaching/em/lectures/node37.html
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