madhadron

The material derivative

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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

density⋅∂tvelocity=−∇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,

density⋅∂tvelocity=−∇pressure\mbox{density} \cdot \partial_t \mbox{velocity} = -\nabla \mbox{pressure}
→\rightarrow { compose with a coordinate transform }
density∘trans⋅∂tvelocity∘trans=−∇pressure∘trans\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 }
density⋅∂tvelocity∘trans=−∇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 ℝn\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→ℳ\chi : X \rightarrow \mathcal{M} be coordinates fixed in matter, where ℳ⊂ℝ4\mathcal{M} \subset \mathbb{R}^4, and ϕ:X→ℱ\phi : X \rightarrow \mathcal{F} be coordinates fixed in space, where ℱ⊂ℝ4\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 ℳ\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. PℳP_\mathcal{M} for the physical quantity PP on the domain ℳ\mathcal{M} works nicely. In this notation,

ρX=ρℳ∘χ=ρℱ∘ϕ\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:X→ℝ3\mathcal{V} : X \rightarrow \mathbb{R}^3, u:ℳ→ℝ3u : \mathcal{M} \rightarrow \mathbb{R}^3 and v:ℱ→ℝ3v : \mathcal{F} \rightarrow \mathbb{R}^3, all related by

𝒱=u∘χ=v∘ϕ\mathcal{V} = u \circ \chi = v \circ \phi

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

ρℳ⋅∂3uk=−∂kPℳ\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 ℱ\mathcal{F} is χ∘ϕ−1\chi \circ \phi^{-1}, so we compose the differential equation with it.

ρℳ∘χ∘ϕ−1⋅(∂3uk)∘χ∘ϕ−1=−∂kPℳ∘χ∘ϕ−1\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 ρℳ∘χ∘ϕ−1=ρℱ\rho_\mathcal{M} \circ \chi \circ \phi^{-1} = \rho_\mathcal{F}, so

ρℱ⋅(∂3uk)∘χ∘ϕ−1=−∂kPℱ\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
= { $u \circ \chi = v \circ \phi }$_3 (v {-1})k$
= { chain rule }
∑j=03(∂jvk)∘ϕ∘χ−1⋅∂3(ϕ∘χ−1)j\sum_{j=0}^3 (\partial_j v^k) \circ \phi \circ \chi^{-1} \ \cdot \partial_3 (\phi \circ \chi^{-1})^j
= { ∂3(ϕ∘χ−1)k=(δk3−(1−δk3)uk)\partial_3 (\phi \circ \chi^{-1})^k = (\delta_{k3} - (1-\delta_{k3})u^k) }
∑j=03(δk3−(1−δk3)uk)⋅(∂jvk)∘(ϕ∘χ−1)\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)∘(ϕ∘χ−1)−∑j=02uk⋅(∂jvk)∘(ϕ∘χ−1)(\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 ϕ∘χ−1\phi \circ \chi^{-1} }
[∂0vk−(∑j=02uk⋅∂jvk)]∘ϕ∘χ−1\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(ϕ∘χ−1)k=(δk3−(1−δk3)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(ϕ∘χ−1)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?

ϕ∘χ−1\phi \circ \chi^{-1} and its derivatives are functions on ℳ\mathcal{M}, so we must use uu. The χ\chi coordinates move with the fluid, so χ∘ϕ−1\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(ϕ∘χ−1)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.

ρℱ⋅(∂3uk)∘χ∘ϕ−1=−∂kPℱ\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 }
ρℱ⋅[∂0vk−(∑j=02uk⋅∂jvk)]∘ϕ∘χ−1∘χ∘ϕ−1=−∂kPℱ\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 }
ρℱ⋅[∂0vk−(∑j=02uk⋅∂jvk)]=−∂kPℱ\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.