Applying Reynold's Theorem to Infinitesimal Element: Fluid Dynamics

[nonequilibrium]

So Reynold's transport theorem states that \frac{\mathrm d}{\mathrm d t} \int_{V(t)} f \; \mathrm d V = \int_{V(t)} \partial_t f \; \mathrm d V + \int_{V(t)} \nabla \cdot \left( f \mathbf v \right) \; \mathrm d V.

Now I would expect (on basis of conceptual reasoning) that if I were to apply this to an infinitesimal element around \mathbf r(t), I should get the well-known (the so-called convective derivative) result \frac{\mathrm d}{\mathrm d t} f(\mathbf r(t)) = \partial_t f + \left( \mathbf v \cdot \nabla \right) f

However, it's straight-forward to see that one gets \frac{\mathrm d}{\mathrm d t} f(\mathbf r(t)) = \partial_t f + \nabla \cdot \left( f \mathbf v \right)

I get a term f \; \nabla \cdot \mathbf v too much. What gives?

 

[AlephZero]

I think your first equation is wrong. Compare Eq 8 in http://www.eos.ubc.ca/~ehearn/EOSC_453/Reynolds_Transport.pdf

 

[nonequilibrium]

You're correct, thank you. To keep it clean I fixed my mistake in the OP. However, it seems the problem remains...

 

[nonequilibrium]

On further thought, both results might be correct and they might indicate a fundamental difference between a "point" and an "infinitesimal volume": the extra term indicates the volume change of the infinitesimal volume, which apparently is not negligible.

 

[AlephZero]

The transport equation you wrote applies to any finite volume which moves with the fluid (and changes size and shape as it moves, in general).

If you want to work with a fixed volume in space, you get a different form of the transport equation. The difference between the two (i.e. the Lagrangian and Euleran formulations) is fundamental. Of course you can transform one equation into the other. In one form you the boundary moves. In the other, there is fluid flow through the fixed boundary.

But you can't chop a finite sized moving volume into a sum of "infinitesimal" parts that are fixed in space, and just hope things will work out OK!

 

[nonequilibrium]

Oh but I wasn't keeping it fixed in space, I had simply not entertained the thought that an infinitesimal volume would also change in size, which it does, by a factor of \nabla \cdot \mathbf v. Keeping this in mind, for the infinitesimal volume we get \frac{\mathrm d}{\mathrm d t}\left( f \mathrm d V \right) = \left( \frac{\mathrm d}{\mathrm d t}f \right) \mathrm d V + f \; \nabla \cdot \mathbf v \; \mathrm d V where the second derivative w.r.t. to time is the convective derivative.

Alternatively, this leads to a different way to prove Reynold's transport theorem, i.e. one can show on grounds of physical reasoning that an infinitesimal volume changes in size with factor \nabla \cdot \mathbf v. Reynold's transport theorem follows from this.

 

[AlephZero]

I had simply not entertained the thought that an infinitesimal volume would also change in size

That's one reason why I try to stick with finite volumes, whenever possible

One way to dig yourself out of this sort of hole is to back off and think about an analogous 1-D problem. You can convince yourself of the difference between say $$\frac{d}{dt}\int_a^b f(x,t)\,dx\quad\mathrm{and}\quad\frac{d}{dt}\int_{a(t)}^{b(t)} f(x,t)\,dx$$ by drawing pictures of the area under the curve at times $t$ and $t + \delta t$. But drawing pictures of 3-D vector fields in arbrtrary regions is hard!