# Vector potential

In vector calculus, a vector potential is a vector field which generates a solenoidal vector field. Any solenoidal velocity field vs should have a potential A such that

[itex] \mathbf{v_s} = \nabla \times \mathbf{A}. [itex]

Note that a solenoidal field cannot be described as having a scalar potential. This is explained in the solenoidal article.

The problem now is: given a solenoidal (or vortical, rotational) field vs, how to find a potential field A such that [itex] \nabla \times \mathbf{A} = \mathbf{v_s} [itex] ? The answer given here to this question will emphasize two points:

(1) (a) It is evident that: a curl can be used to measure a solenoidal field (the curl of the field is proportional to the angular velocity).

(b) It is not so evident that: the curl of a field can also generate a solenoidal field. This generative capacity of the curl tends to be the reverse of its measuring capacity (if [itex] \nabla \times \mathbf{A} = \mathbf{v_s} [itex], then [itex] \nabla \times \mathbf{v_s} \approx \mathbf{A} [itex]).

(2) (a) it is evident that: the curl measures rotational motion in a circle or a closed loop,

(b) it is not so evident that: the curl also measures the speed differential between parallel and adjacent velocities.
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## First the sentence, then the evidence

Let potential A be defined by

This vector potential always points downwards (-z direction) but its magnitude is a radial Gaussian function, centered at the z axis, with inflection points around in a circle of radius 1. A points downwards most strongly at the z-axis, and moving away from the z-axis, the magnitude of A quickly drops asymptotically to zero.

Now find the curl of A:

[itex] \nabla \times \mathbf{A} = 2 e^{-r^2} (y \mathbf{i} - x \mathbf{j}) = \mathbf{v_s}. [itex]

This is a clockwise movement around the z-axis, and the magnitude of this movement is a Gaussian function which depends on the distance away from the z-axis. v represents a solenoidal movement around the z-axis.

Finally, the curl of the curl of A is

[itex] \nabla \times \mathbf{v_s} = \nabla \times \nabla \times \mathbf{A} = 4 e^{-r^2} \mathbf{k} (-1 + r^2). [itex]

The curl of the velocity points downwards for r<1 and upwards when r>1 (r is the radius in cylindrical coordinates: [itex] r^2 = x^2 + y^2 [itex]). The curl of the velocity points most strongly downwards at the z-axis (r=0), with a magnitude of 4. It points most strongly upwards at [itex] r = \sqrt{2} [itex] with magnitude [itex] 4 e^{-2} = 0.541 [itex].

This shows that the curl of v points downwards 7.4 times more strongly at the central axis of the vortex than upwards at the periphery of the vortex. This is strong evidence that [itex] \nabla \times \nabla \times \mathbf{A} [itex] can be parallel to A, i.e. that the field A necessary to generate a solenoidal field v can be estimated from [itex] \nabla \times \mathbf{v}. [itex]

It also shows that A need not rotate in a circular motion (have field lines forming closed loops) around v in order to generate v: all that is needed is a succession of concentric cylindrical layers: with A pointing parallel to the layers, but with varying magnitude, so that A is stronger in the central layer and gets weaker in outward layers.

## Measuring the speed differential of parallel velocities

A similar phenomenon is the differential between the speed of the road and the speed of a car. Let us say the car drives towards the East. The Earth rotates towards the East, the road moves along with the Earth, and the car moves towards the East just a tiny bit faster than the road (proportionally) but this speed differential of parallel and adjacent velocities is enough to generate a curl:

[itex] {\partial v_x \over \partial z} \ne 0; \qquad \nabla \times \mathbf{v} = {\partial v_x \over \partial z} \mathbf{j}. [itex]

(i = East, j = North, k = up towards the sky).

As the car moves to the East faster than the road, the speed differential is proportional to the curl of the velocity, which points towards the North. But the speed difference is also proportional to the angular speed of the car's wheels, and the axis of rotation of these wheels points towards the North as well. This illustrates the point: about what kind of potential field may be used to generate a given vortical velocity.

## More evidence

Now we will examine more evidence, this time much more general, that [itex] \nabla \times \nabla \times \mathbf{A} [itex] can be parallel to A: let the potential field A be parallel to itself throughout an entire region, or through all space. Then A points always in the same direction. Let this direction be the +z direction. Let the magnitude of A change: the density of the field lines of A changes, but the field lines are straight parallel lines.

What is the divergence of A? Let there be a cylinder inside the A field, such that the axis of the cylinder is parallel to A. The curved wall of the cylinder is parallel to A, so the flux of A through the cylinder's curved wall is zero. Then, any field line of A going into the cylinder through one circular base must come out the other: the fluxes of A through the pair of circular bases of the cylinder cancel each other out. Let the cylinder become smaller so that its volume approaches zero as a limit. Then its balance of flux is still zero, so the divergence of the A field is zero:

[itex] \nabla \cdot \mathbf{A} = 0 \qquad \qquad \mbox{when} \ \mathbf{A} \ \mbox{points always in the same direction}. [itex]

Since

[itex] \operatorname{curl} \ \operatorname{curl}\ \mathbf{A} = \operatorname{grad} \ \operatorname{div} \ \mathbf{A} - \operatorname{div} \ \operatorname{grad} \ \mathbf{A} [itex]

or

[itex] \nabla \times \nabla \times \mathbf{A} = \nabla ( \nabla \cdot \mathbf{A} ) - \nabla \cdot ( \nabla \mathbf{A} ) [itex]

this means that

[itex] \operatorname{curl}\ \operatorname{curl}\ \mathbf{A} = - \operatorname{div} \ \operatorname{grad} \ \mathbf{A} = - \nabla \cdot \nabla \mathbf{A} [itex]

or

[itex] \nabla \times \nabla \times \mathbf{A} = - \nabla \cdot \nabla \mathbf{A} [itex]

We will estimate the gradient of A (a tensor). Imagine the A field on a plane x-y, which is perpendicular to A. The x-y plane is the ground. Let the magnitude of A be strongest at the z-axis and weaker outwards. Imagine the A field on the x-y plane as a group of vertical pillars rising from the ground: their height is the magnitude of A. These pillars form something similar to a mountain: e.g. it could be a paraboloid with its concavity facing downwards.

Start out by standing on the topmost pillar; walk down slightly in the +i direction. Then the A pillars decrease in height (along k) so the dyadic tensor [itex] \nabla \mathbf{A} [itex] has [itex] - \mathbf{i} \mathbf{k} \epsilon_1 [itex] as one of its dyadic terms. Now walk from that point towards the +j direction; if the potential is not saddle-shaped (assume it is not, because the starting point was the maximum) then the A pillars decrease in height further contributing a dyadic term [itex] - \mathbf{j} \mathbf{k} \epsilon_2 [itex]. In the +k direction, the strength of A is constant. Therefore

[itex] \nabla \mathbf{A} = - \mathbf{i} \mathbf{k} \epsilon_1 - \mathbf{j} \mathbf{k} \epsilon_2. [itex]

Find the divergence of this:

[itex] \nabla \cdot \nabla \mathbf{A} = \left( \mathbf{i} {\partial \over \partial x} + \mathbf{j} {\partial \over \partial y} + \mathbf{k} {\partial \over \partial z} \right) \cdot \left( - \mathbf{i} \mathbf{k} \epsilon_1 - \mathbf{j} \mathbf{k} \epsilon_2 \right) [itex]
[itex] = - \mathbf{k} {\partial \epsilon_1 \over \partial x} - \mathbf{k} {\partial \epsilon_2 \over \partial y}. [itex]

If function Az(x,y) is concave downwards (as should be near Az 's maximum) then

[itex] {\partial \epsilon_1 \over \partial x} > 0, \qquad {\partial \epsilon_2 \over \partial y} > 0, [itex]

so [itex] \nabla^2 \mathbf{A} [itex] points in the -k direction. [itex] \nabla^2 \mathbf{A} [itex] is like the curvature of Az, pointing towards concavity, therefore [itex] \nabla \times \nabla \times \mathbf{A} = - \nabla^2 \mathbf{A} [itex] points in the +k direction, same as A itself:

[itex] \nabla \times \nabla \times \mathbf{A} \| \mathbf{A}. [itex]

Note that [itex] \nabla^2 \mathbf{A} = 0 [itex] determines the locus of inflection points of the surface Az(x,y). Inside this locus and surrounding the maximum, Az is concave downwards. Outside of this locus Az is concave upwards.

## Calculating the vector potential

Start out with an example from the electric field. A charge q generates a scalar potential:

[itex] \phi(r) = {q \over 4 \pi \epsilon_0 r}. [itex]

where ε0 is permittivity, and r is distance away from the point charge. The total potential at a point is a sum of potential contributions of all charges. Let ρ be charge density, then

[itex] \phi (\mathbf{R'}) = {1 \over 4 \pi \epsilon_0} \int_{\tau} {\rho (\tau) \over \| \mathbf{R(\tau)} - \mathbf{R'} \| } \, d \tau [itex]

where is a differential volume element [Cheng, p. 95].

But [itex] \rho = - \nabla^2 \phi [itex] (see Poisson's equation). If a potential A always points in the same direction, then its divergence is zero, so

[itex] \nabla \times \nabla \times \mathbf{A} = - \nabla^2 \mathbf{A}, [itex]

therefore, by analogy with the scalar case:

[itex] \mathbf{A} (\mathbf{R}') = {1 \over 4 \pi} \int_{\tau} { \nabla \times \nabla \times \mathbf{A} (\tau) \over \| \mathbf{R} (\tau) - \mathbf{R'} \| } \, d\tau, [itex]

and this equation should be true up to a constant factor.

We want to find an A field such that [itex] \nabla \times \mathbf{A} = \mathbf{v_s} [itex] for a given solenoidal field vs. Then

[itex] \mathbf{A} (\mathbf{R}') = {1 \over 4 \pi} \int_{\tau} { \nabla \times \mathbf{v_s} (\tau) \over \| \mathbf{R} (\tau) - \mathbf{R'} \| } \, d\tau. [itex]

To generalize this equation to any solenoidal velocity field vs, move the curl out of the integral:

[itex] \mathbf{A} (\mathbf{R}') = {1 \over 4 \pi} \nabla \times \int_{\tau} { \mathbf{v_s} (\tau) \over \| \mathbf{R} (\tau) - \mathbf{R'} \| } \, d\tau. [itex]

This last equation is valid for any solenoidal field as long as it tends to vanish asymptotically to zero towards infinity (more specifically: if A decays faster than 1/r and if [itex] \nabla \times \mathbf{A} [itex] decays faster than 1/r2).

## Reference

• Fundamentals of Engineering Electromagnetics by David K. Cheng, Addison-Wesley, 1993.

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