# Given *an* expression for the current and *the* expression for a density, can you determine the velocity field?

Suppose you a density field $$\rho(\vec r)$$ and, due to some equation of motion, you have derived that $$\partial_t \rho + \nabla \cdot \vec J = 0, \tag{1}$$ where $$\partial_t = \frac{\partial}{\partial t}$$, and you have found an explicit expression for $$\vec J$$ as a function of $$\rho$$ and $$\partial_t \rho$$. This continuity equation for $$\rho$$ shows that it is globally conserved. Comparing it to an equation of motion where $$\rho$$ is being advected by a velocity field $$\vec v$$ $$\partial_t \rho + \nabla \cdot (\rho \vec v) = 0, \tag{2}$$ it is tempting to identify $$\vec v = \frac{\vec J}{\rho}.$$ However, since Eq. (1) only determines $$\vec J$$ up to a divergence-free contribution $$\vec K$$, where $$\nabla \cdot \vec K=0$$, one can really only say that $$\vec v = \frac{\vec J + \vec K}{\rho}. \tag{3}$$ As I understand it, this stems from the ambiguity in a current as derived from the density field alone. E.g., a constant current of $$\vec K=(1,0,0)$$ would not be visible in the time evolution of $$\rho$$.

However, are there reasonable assumptions one can make in order to identify the velocity field as in Eq. (3)?

As context, what I am trying to understand is exemplified in this Physical Review Letters paper from 1997, where the author seems to do this identification between Eqs. (11) and (12).

You can go one step further: If you substract equation (1) from equation (2), you get $$\vec\nabla\cdot(\rho\vec v-\vec J)=\vec 0$$. Because of the Helmholtz decomposition (only if you consider the fields in $$\mathbb R^3$$ without a boundary, otherwise there will be a boundary term), there exists a vector field $$\vec L$$ with: $$\rho\vec v-\vec J =\vec\nabla\times\vec L \Leftrightarrow \vec v =\frac{\vec J+\vec\nabla\times\vec L}{\rho}.$$ Of course you could also directly apply the Helmholtz decomposition to $$\vec K$$, which you already mentioned is divergence-free, and get $$\vec K=\vec\nabla\times\vec L$$. Basically, the same is done in electrodynamics by writing the divergence-free magnetic field $$\vec B$$ as $$\vec B=\vec\nabla\times\vec A$$.

Using Stokes' theorem here won't help much as it just transfers a differential equation into an integral equation, but in fluid dynamics an incompressible fluid leads to the additional condition $$\vec\nabla\cdot\vec v=\vec 0$$, which you could also use here as the interpretation in Ginsburg-Landau theory is probably quite similar. If we put the upper equation into this condition and using a few identities of vector analysis, we get: \begin{align*} \vec\nabla\cdot\vec v =\vec\nabla\cdot\left(\frac{\vec J+\vec\nabla\times\vec L}{\rho}\right) =-\frac{\vec J+\vec\nabla\times\vec L}{\rho^2}\cdot\vec\nabla\rho +\frac{\overbrace{\vec\nabla\cdot\vec J}^{=-\partial_t\rho}+\overbrace{\vec\nabla\cdot(\vec\nabla\times\vec L)}^{=0}}{\rho} \stackrel ! =0 \\ \Rightarrow -\vec J\cdot\vec\nabla\rho +\vec\nabla\cdot(\vec L\times\vec\nabla\rho) -\rho\partial_t\rho=0 \\ \Rightarrow \vec\nabla\cdot(\vec L\times\vec\nabla\rho-\rho\vec J) -2\rho\partial_t\rho=0, \end{align*} and therefore another continuity equation: $$\partial_t(\rho^2) -\vec\nabla\cdot(\vec L\times\vec\nabla\rho-\rho\vec J)=0.$$

• You suggest adding the additional condition $\nabla \cdot \vec v=0$. You use this to derive another continuity equation, now with an unknown vector potential $\vec L$. But can you use the condition to $\nabla \cdot vec v=0$ to derive a unique expression for $\vec L$, and therefore $\vec v$? Sep 21, 2022 at 13:34
• I'm very sure that's not possible, as the vector space of (divergence-free) vector fields $\vec K=\vec\nabla\times\vec L$ has infinite dimension. Show this for example by finding a divergence-free vector field with compact support, then you can just shift it outside its compact support to get a new linear independent (divergence-free) vector field. You can add such vector fields (with scaled or shifted compact support) to a given vector field however you like without changing the divergence our any boundary term at all. Sep 21, 2022 at 13:46
• $\vec\nabla\cdot\vec v=0$ is only a single equation with not enough information to determine $\vec v$ (or $\vec L$) uniquely. There may be other conditions you can get from a suitable physical interpretation, but that may depend on the theory you consider. The velocity being divergence-free is the best I could come up with. Sep 21, 2022 at 13:48