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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).

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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: $$\partial_t(\rho^2) -\vec\nabla\cdot(\vec L\times\vec\nabla\rho-\rho\vec J)=0.$$

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  • $\begingroup$ 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 $\vec v$? $\endgroup$ Sep 21, 2022 at 13:34
  • $\begingroup$ 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. $\endgroup$ Sep 21, 2022 at 13:46
  • $\begingroup$ $\vec\nabla\cdot\vec v=0$ 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. $\endgroup$ Sep 21, 2022 at 13:48

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