The problem now is: given a solenoidal (or vortical, rotational) field vs, how to find a potential field A such that ? 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).
Table of contents |
2 Measuring the Speed Differential of Parallel Velocities 3 More Evidence 4 Calculating the Vector Potential 5 Reference |
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:
Finally, the curl of the curl of A is
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 can be parallel to A, i.e. that the field A necessary to generate a solenoidal field v can be estimated from
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.
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.
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:
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 has 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 . In the +k direction, the strength of A is constant. Therefore
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:
(i = East, j = North, k = up towards the sky).More Evidence
Now we will examine more evidence, this time much more general, that 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.
Since
this means that
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.
Find the divergence of this:
If function Az(x,y) is concave downwards (as should be near Az 's maximum) then
so points in the -k direction ( is like the curvature of Az, pointing towards concavity, therefore points in the +k direction, same as A itself:
But (see Poisson's equation). If a potential A always points in the same direction, then its divergence is zero, so
We want to find an A field such that for a given solenoidal field vs. Then
Calculating the Vector Potential
Start out with an example from the electric field. A charge q generates a scalar potential:
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
where dτ is a differential volume element [Cheng, p. 95].
therefore, by analogy with the scalar case:
and this equation should be true up to a constant factor.
This equation should be true for any solenoidal velocity field vs.Reference
See also: solenoid, magnetic potential, retarded potential.