Let's consider a water flow in two dimensional space, assume the velocity field is $\vec{v}(t,x,y)$, we can define: $$\rho=\hat{z}\cdot(\partial_x\vec{v}\times \partial_y\vec{v})$$ $$j^x=\hat{z}\cdot(\partial_y\vec{v}\times \partial_t\vec{v})$$ $$j^y=\hat{z}\cdot(\partial_t\vec{v}\times \partial_x\vec{v})$$ we can prove: $$\partial_t \rho+\partial_xj^x+\partial_yj^y=0$$ as long as the velocity field is smooth. My question is:

1) What's the physics corresponding to this conservation law?

2) Is there any important consequence due to this conservation law?

  • $\begingroup$ This just encapsulates the conservation of mass (or it's 2D analogue). en.wikipedia.org/wiki/Continuity_equation $\endgroup$ – Sayan Mandal Jun 14 '18 at 18:19
  • $\begingroup$ @SayanMandal The notation here is somewhat misleading - as far as I can tell, those quantities have nothing to do with mass density or current density. As a simple example, if the flow is uniform, then all of those quantities are equal to zero. $\endgroup$ – J. Murray Jun 14 '18 at 18:28
  • $\begingroup$ @SayanMandal Please don't post answers as comments -- you're correct of course, but it should be made into a full answer (with a little more detail probably). $\endgroup$ – tpg2114 Jun 14 '18 at 18:28
  • 1
    $\begingroup$ @SayanMandal How did you reach that conclusion? OP clearly defines $\rho,j^x,$ and $j^y$, and they are not what those symbols usually mean. $\endgroup$ – J. Murray Jun 14 '18 at 18:36
  • 1
    $\begingroup$ @OP, can you provide some references? $\endgroup$ – Sayan Mandal Jun 14 '18 at 18:38

Note that

$$\partial_t \rho = \hat z \cdot (\partial_t\partial_x \vec v \times \partial_y \vec v + \partial_x \vec v \times \partial_t\partial_y \vec v)$$ $$\partial_x j^x = \hat z \cdot (\partial_x \partial_y \vec v \times \partial _t \vec v + \partial_y \vec v \times \partial_x\partial_t \vec v)$$ $$\partial_y j^y = \hat z\cdot (\partial_y\partial_t \vec v \times \partial _x \vec v + \partial_t \vec v + \partial_y\partial_x \vec v)$$

Because you can switch the order of partial derivatives and because the cross product is antisymmetric, the first term on line 1 cancels the second term on line 2; the second term on line 1 cancels with the first term on line 3; and the first term on line 2 cancels with the second term on line 3.

This would be true for any smooth function $\vec v$ (whether it is vector-valued or not) so I cannot see how it would encode any meaningful physics.

As an aside, using the symbols $\rho$ and $j$ for this is probably a bad idea. My initial thought was that you were trying to cast this into the form of a conservation law, but this is misleading. Conservation laws have physical content - they are either taken to be true by assumption, or they arise from more fundamental considerations (e.g. declaring mass conservation to be a necessary component of fluid mechanics vs. deriving the continuity equation as the zeroth moment of the Boltzmann transport equation).

In this case, however, your equation is actually an identity. The fact that it is true does not rely on the Navier-Stokes equations or on any other physics, in the same way as the vector identity $\vec v \times \vec v=0$.

| cite | improve this answer | |
  • $\begingroup$ I totally agree with what you said. The original idea is: if we regard uniform flow as a vacuum and smooth fluctuation as excitation, how can we describe the transport of this fluctuations. In fact we came up with the question above in other context instead of fluid dynamics. Do you know is there any relevant discussion in fluid dynamics? $\endgroup$ – Ji Zou Jun 14 '18 at 20:18
  • $\begingroup$ Thanks for the answer. Things now make much more sense. $\endgroup$ – Sayan Mandal Jun 14 '18 at 21:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.