The material derivative

Math in this page not rendering? See the fix

Section 2 of Landau and Lifshitz’s Fluid Mechanics starts by writing down equations of motion for a fluid particle in coordinates attached to that particle, then transforms to coordinates fixed in space. The argument is physically clear, but mathematically opaque. This was bothering me, so I cleaned it up. Introducing a primal, underlying space, without coordinates simplifies everything nicely (of course, this is just introducing a proper manifold structure, but it’s a pretty example of its use).

In coordinates fixed in matter, the velocity of the fluid is governed by

densitytvelocity=pressure\mbox{density} \cdot \partial_t \mbox{velocity} = -\nabla \mbox{pressure}

When we compose each function with a coordinate transformation, density and pressure change in an elementary manner. The only work is in the "tvelocity\partial_t \mbox{velocity}" term. Essentially what we do is,

densitytvelocity=pressure\mbox{density} \cdot \partial_t \mbox{velocity} = -\nabla \mbox{pressure}
\rightarrow { compose with a coordinate transform }
densitytranstvelocitytrans=pressuretrans\mbox{density} \circ \mbox{trans} \cdot \ \partial_t \mbox{velocity} \circ \mbox{trans}\ =\ -\nabla \mbox{pressure} \circ \mbox{trans}
\rightarrow { density and pressure change coordinates simply }
densitytvelocitytrans=pressure\mbox{density} \cdot \partial_t \mbox{velocity} \circ \mbox{trans}\ =\ -\nabla \mbox{pressure}
\rightarrow { expand tvelocity\partial_t \mbox{velocity} in terms of the velocity on the new coordinates }
density[expansion of velocity]=pressure\mbox{density} \cdot \left[\mbox{expansion of velocity}\right] = -\nabla \mbox{pressure}

As always, good notation is priceless. First, coordinate systems. We’ll go the manifold route, and immediately invoke the metatheorem that "if I prove pp for arbitrary coordinates, and my space is a manifold, I’ve proved pp on the manifold."

The very clever might set up both coordinate systems and the transform between them directly. I’m too out of shape, so I’ll start with a primal space (we’ll call it XX) representing the physical space-time occupied by the fluid, with no coordinates or other structure. Then I construct coordinates as invertible functions from XX to open subsets of n\mathbb{R}^n. The metatheorem lets me brush all the worries about not covering the whole space under the rug. n=4n = 4 – three space dimensions (labelled 0, 1, 2) and one time dimension (labelled 3). It’s simpler to keep time with the other coordinates.

Let χ:X\chi : X \rightarrow \mathcal{M} be coordinates fixed in matter, where 4\mathcal{M} \subset \mathbb{R}^4, and ϕ:X\phi : X \rightarrow \mathcal{F} be coordinates fixed in space, where 4\mathcal{F} \subset \mathbb{R}^4. Naming the coordinate’s ranges distinctly often makes the question "what function goes here?" trivial: "it must have domain \mathcal{M}, and I have only one function like that which makes sense here."

For physical quantities that transform simply (density ρ\rho and pressure PP), I want to capture what physical quantity I am looking at, and the domain that this particular version of that quantity is defined on in my notation. PP_\mathcal{M} for the physical quantity PP on the domain \mathcal{M} works nicely. In this notation,

ρX=ρχ=ρϕ\rho_X = \rho_\mathcal{M} \circ \chi = \rho_{\mathcal{F}} \circ \phi

That’s fine for quantities that carry through without major changes, but velocity transforms in a more complicated manner, and we will be using its different forms heavily. Instead of inevitably miscopying a subscript, I assign the versions on different coordinate systems different letters: 𝒱:X3\mathcal{V} : X \rightarrow \mathbb{R}^3, u:3u : \mathcal{M} \rightarrow \mathbb{R}^3 and v:3v : \mathcal{F} \rightarrow \mathbb{R}^3, all related by

𝒱=uχ=vϕ\mathcal{V} = u \circ \chi = v \circ \phi

In this notation, the differential equation in coordinates fixed in matter is

ρ3uk=kP\rho_\mathcal{M} \cdot \partial_3 u^k = -\partial_k P_\mathcal{M}

The only function we have defined that will make this an equation on \mathcal{F} is χϕ1\chi \circ \phi^{-1}, so we compose the differential equation with it.

ρχϕ1(3uk)χϕ1=kPχϕ1\rho_\mathcal{M} \circ \chi \circ \phi^{-1} \cdot (\partial_3 u^k)\circ \chi \circ \phi^{-1} = -\partial_k P_\mathcal{M} \circ \chi \circ \phi^{-1}

Pressure and density transform like ρχϕ1=ρ\rho_\mathcal{M} \circ \chi \circ \phi^{-1} = \rho_\mathcal{F}, so

ρ(3uk)χϕ1=kP\rho_\mathcal{F} \cdot (\partial_3 u^k)\circ \chi \circ \phi^{-1} = -\partial_k P_\mathcal{F}

Now expand the kkth component of velocity fixed in matter 3uk\partial_3 u^k in terms of vv.

3uk\partial_3 u^k
= { $u = v }
3(vϕχ1)k\partial_3 (v \circ \phi \circ \chi^{-1})^k
= { chain rule }
j=03(jvk)ϕχ13(ϕχ1)j\sum_{j=0}^3 (\partial_j v^k) \circ \phi \circ \chi^{-1} \ \cdot \partial_3 (\phi \circ \chi^{-1})^j
= { 3(ϕχ1)k=(δk3(1δk3)uk)\partial_3 (\phi \circ \chi^{-1})^k = (\delta_{k3} - (1-\delta_{k3})u^k) }
j=03(δk3(1δk3)uk)(jvk)(ϕχ1)\sum_{j=0}^3 (\delta_{k3} - (1-\delta_{k3})u^k)\ \cdot\ (\partial_j v^k)\circ (\phi \circ \chi^{-1})
= { split sum over range (j=0 to 2, and j = 3) }
(0vk)(ϕχ1)j=02uk(jvk)(ϕχ1)(\partial_0 v^k) \circ (\phi \circ \chi^{-1})\ -\ \sum_{j=0}^2 u^k \cdot (\partial_j v^k) \circ (\phi \circ \chi^{-1} )
= { factor out ϕχ1\phi \circ \chi^{-1} }
[0vk(j=02ukjvk)]ϕχ1\left[ \partial_0 v^k - \left( \sum_{j=0}^2 u^k \cdot \partial_j v^k\right) \right] \circ \phi \circ \chi^{-1}

Where did the identity 3(ϕχ1)k=(δk3(1δk3)uk)\partial_3 (\phi \circ \chi^{-1})^k = (\delta_{k3} - (1-\delta_{k3})u^k) come from? 3\partial_3 is a time derivative. How do the the coordinates fixed in matter change over time with respect to coordinates fixed in space? Coordinate 3 (time) is the same in both, so 3(ϕχ1)3=1\partial_3 (\phi \circ \chi^{-1})^3 = 1. The change in coordinates 0, 1, 2 is how far the bit of matter we are looking at moves in an infinitesimal time, that is, the velocity. But is it vv or uu, and what is its sign?

ϕχ1\phi \circ \chi^{-1} and its derivatives are functions on \mathcal{M}, so we must use uu. The χ\chi coordinates move with the fluid, so χϕ1\chi \circ \phi^{-1} would be positive velocity. We are using the inverse, so it is negative velocity. So for k=0, 1, 2, we have 3(ϕχ1)k=uk\partial_3 (\phi \circ \chi^{-1})^k = -u^k. Combining these with Kronecker δ\delta’s yields the identity.

Finally, put the expansion of 3uk\partial_3 u^k into the equation of motion.

ρ(3uk)χϕ1=kP\rho_\mathcal{F} \cdot (\partial_3 u^k)\circ \chi \circ \phi^{-1} = -\partial_k P_\mathcal{F}
= { substitute for 3uk\partial_3 u^k }
ρ[0vk(j=02ukjvk)]ϕχ1χϕ1=kP\rho_\mathcal{F} \cdot\left[\partial_0 v^k-\left(\sum_{j=0}^2 u^k \cdot\partial_j v^k \right)\right]\circ \phi \circ \chi^{-1}\circ\chi\circ\phi^{-1}=-\partial_k P_\mathcal{F}
= { coordinates and inverses cancel }
ρ[0vk(j=02ukjvk)]=kP\rho_\mathcal{F} \cdot \left[ \partial_0 v^k - \left( \sum_{j=0}^2 u^k \cdot \partial_j v^k\right) \right] = -\partial_k P_\mathcal{F}

which we all know and love as the simplest equation of motion in fluid mechanics.