Meaning of potentials in Helmholtz decomposition

Question

I am trying to understand the meaning of the potentials $U$ and $A$ arising in the Helmholtz decomposition of a vector field $\vec{V}$:

$\vec{V} = \nabla U + \nabla \times A$

Let's focus on the curl-free potential $U$ and please excuse me for being very informal (I do not have a strong maths background):

$\vec{V} = \nabla U$

What does the potential tell me about the vector field? I know extrema are related to sink and source behavior in the vector field. Are there any circumstances in which the magnitude, i.e. the specific value of $U$, says something about the vector field? For example, are there any conditions under which the potentials tell me "how much" the vector field behaves like a source or sink (vortex) at a given location?

Answer 1

First, the absolute value of $U$ does not tell you anything in general, since $U$ and $U+c$ are equivalent potentials, i.e. they lead to exactly the same vector field $V$, for any constant $c$. This is what one may call a gauge transformation; you may fix the constant by requiring e.g. $U=0$ at some fixed position.

Note that this all holds for a 3D vector field depending on 3D spatial coordinates. If you are considering some aditional variable, such as time, the constant $c$ can be any function of time, $c(t)$, and while you can, e.g., impose $U(\vec r_0,t)=0$ for some fixed $\vec r_0$ and all $t$, this does not in general provide any obvious intuition (AFAICT).

Next, the source-or-sink character of $\vec V$ is captured by the divergence $\vec\nabla \cdot \vec V$, which you can calculate in terms of the potentials as $\vec\nabla \cdot \vec V=\Delta U$ (note that this holds even for nonzero curl, since the divergence of a curl vanishes). Clearly, this is invariant under gauge transformations.

Thus, to evaluate the "source strength", you would have to evaluate $\Delta U$. On the other hand, you do not need to lok at the potential; you can just compute $\phi:=\vec\nabla \cdot \vec V$ (which is a scalar function) and work with that.

Incidentally, the "vorticity" of $\vec V$ is given by the curl, i.e. $\vec\nabla\times\vec V=\vec\nabla\times\vec\nabla\times A=\vec\nabla \left(\vec\nabla\cdot \vec A\right)-\Delta \vec A$.

(In your linked related question, you mention a discrete setting. I know nothing about that.)

Answer 2

The specific value of U says nothing unless you define it such that it does. A handy thing about physics is that you can set the 0 of your potential energy to whatever is convenient. For instance, in a spring we tend to set it to 0 at the point of equilibrium, meaning the potential is non-negative everywhere. However, with gravity we tend to set it at 0 for a point at infinity, meaning the potential is negative everywhere. All that actually matters is that the sum of kinetic and potential energy of a system remain constant, and the gradient of the potential tells you about how potential energy is traded with kinetic energy and vice versa.