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A vector field is solenoidal iff its divergence is zero:
where v is the vector field.

This requires an explanation. A solenoid can be a helical coil of wire with a certain inductance. In the case of fluid dynamics, the movement of a fluid is solenoidal when the fluid moves in a circle, or any kind of closed loop. Fluid movement around a loop cannot be due to a scalar potential.

If a fluid moves due to a scalar potential φ, then it moves from higher potential to lower potential:

If a fluid moves in a circle, then some point of the circle must have the highest potential. Then the fluid moves around the circle (but in which direction?) and then reaches the point of lowest potential, except that this point is adjacent to the point of highest potential. There is a sharp discontinuity between the point of low potential and high potential, which is not characteristic of potential fields. If such low potential is so close to such high potential, then the fluid's movement would be short circuited; it would move directly to its adjacent low point, without bothering to move in a circle. Therefore cyclical movements of fluid cannot be described by means of a scalar potential.

One property of solenoidal fluid flow, though, is that it has a non-zero curl:

If it does have zero curl, then it is degenerate, not really solenoidal, or not solenoidal at all.

The curl operator can be used in two ways: (1) to measure the amount and direction of rotational (solenoidal) flow, or (2) to generate rotational flow. In the second case. In inequation (1), the curl is being used in the first way; to measure rotational flow. But solenoidal flow is also related to the curl in the second way, by means of a vector potential.

This means that for any fluid movement that is purely solenoidal (no lamellar movement), there is a vector potential A such that

This is not quite obvious though. It is not nearly as evident as inequation (1).

There is some evidence, though, that the curl of the curl of A might be somewhat similar to A. Imagine a fluid v moving in the x-y plane clockwise in a circle around the z-axis. Evidently the curl of v should be pointing downwards, in the negative z direction, near the origin. What about the curl of v just outside, at the edges of its circular movement? Curl does not just measure rotation: it can also measure differences in speed. Let traffic along a wide highway be thought of as a fluid, and let the velocities of cars be represented by a vector field. If the cars in the left lanes are moving faster than the cars in the right lanes, then the curl of the velocity field points downwards to the ground, even if the highway is straight and not turning. But if the situation were reversed, and the cars in the right lanes were to move faster than the cars on the left lanes, then the curl of their velocity field would point upwards towards the sky.

Back to the clockwise movement of v: suppose that at a farther distance from the origin, the velocity decays towards zero. Then the curls just outside of the solenoidal movement will point upwards in the +z direction even though inside the solenoidal movement, the curl points downwards in the -z direction. For example, in the positive x axis v points in the -y direction. But moving further up the x axis past the solenoidal movement, the magnitude of v weakens and no longer points so strongly in the -y direction. This implies a curl which points upwards.

It is as if the solenoidal movement were a gear, moving clockwise. Any other gear connected to this gear must move counterclockwise. As the solenoidal movement decays with distance, it is as if the central gear were connected to several smaller surrounding gears moving counterclockwise, each with a positive curl, but none of them with a curl as strong as the central curl inside the solenoidal movement.

Now imagine what kind of potential vector field A would be necessary, so that its curl would equal v. It would be shaped like a torus. At the outer edges of the torus, the potential points upwards (+z). At the top edge of the torus, the potential points inwards (towards the origin). At the inner edge of the torus, the potential points downwards. At the bottom edges of the torus, the potential points outwards. The potential field lines form circles, which are perpendicular to the solenoidal movement of v.

It is necessary to check to see that the curl of A yields v. Cut the torus in half by means of the x-z axis. Two circular potential field lines are now seen: one on the left side (-x) which is clockwise and one on the right side (+x) which is counterclockwise. The curl of the potential on the left side therefore points in the +y direction, and the curl of the potential on the right side points in the -y direction. If v moves clockwise in the x-y plane, then it should be expected to move in the -y direction on the right side and in the +y direction on the left side.

Back to the vector potential and the curl of v. The curl of v points downwards inside of the solenoidal movement, but so does the potential A in the inner side of the torus, which surrounds the solenoidal movement. The curl of v points upwards in the region where the solenoidal movement decays. But this is the same region as the outer side of the torus, where the potential A points upwards. Therefore it is possible to imagine that

therefore it is possible that
even as

Here more evidence in this for this argument: at the top of the torus the potential is moving inwards and then downwards. The potential field lines become denser, A is compressing, therefore divergence of A above the origin is negative. But close the the bottom of the torus the potential field lines start to diverge again: their divergence is positive below the origin. Then there is an identity in vector calculus, which states that

Consider the first term on the right side. Divergence of A is negative above the origin, and positive below the origin. So what is the gradient of the divergence of A at the origin? It must be pointing in the direction of increasing divergence, which is downwards. But this is the same direction as A and as the curl of v. The second term on the right side of the identity is too hard to examine here: the gradient of A is a tensor. (The divergence of this tensor turns it back into a vector, so the identity remains consistent.)

Anyway, the point is that not only does solenoidal (or vortical) velocity have non-zero curl, but it can itself be the curl of a potential (and the potential might not be too different from the curl of the velocity). So, let us say that all solenoidal movement can be generated from a vector potential: there exists a potential A such that


It is known identity that the divergence of a curl is always zero. Therefore if v is solenoidal, so that it is generated from a potential A, then

This is why solenoidal movements can simply be described as having zero divergence.