A differential geometry question from continuum mechanics

[wrobel]

Consider the flow of a continuous medium with a smooth velocity field $\boldsymbol v(x)$ in $\mathbb{R}^3\ni x$. The density of the medium is $\rho(x)$. Here $\mathbb{R}^3$ is equipped with the standard inner product $\delta_{ij}$.

By definition, the momentum of a material volume $D$ (where $D$ is a bounded domain in $\mathbb{R}^3$) is given by the formula:
$$\boldsymbol P=\int_D\boldsymbol v\mu, \qquad(1)$$
where $\mu=\rho\sqrt g dx^1\wedge dx^2\wedge dx^3$ is a differential form (the infinitesimal mass).

And what about formula (1) if we replace $\mathbb{R}^3$ with some other Riemannian manifold with non-zero curvature?
In this case, formula (1) becomes senseless, and I have no idea how to fix it—or if it is even possible.
Any opinions on this?

 

[QuarkyMeson]

On a curved manifold, velocities at different points belong to different tangent spaces, so they cannot be added directly. You need a chosen way to transport vectors or a symmetry of the manifold to define a total momentum.

So I don't think it's generally possible, but you might be able to rescue (1) in some cases.

Side note, even when a manifold has no symmetries and momentum is lost, ideal flows still have conserved quantities of a different kind, they're just now topological ones. This would be bleed into some of the braiding stuff I've been reading, but it's very tangential to this.

 

[Matterwave]

And what about formula (1) if we replace $\mathbb{R}^3$ with some other Riemannian manifold with non-zero curvature?
In this case, formula (1) becomes senseless, and I have no idea how to fix it—or if it is even possible.
Any opinions on this?

I am not aware of a way to in general fix this integral in the sense that there is a unique 1:1 correspondence between this integral and a "fixed version" that works in general Riemannian manifolds. And so, I might not have too many intelligent things to say about this. So take the following with a huge grain of salt.

One thing I might try, could be to turn the vector into a 1-form using the metric $g(v,\,\,)$ and then contracting with some Killing Field $\xi$ to perhaps obtain some conserved aggregate quantity. You'd get an integral like:

$$ I(\xi) = \int_D g(v, \xi)\mu$$

Your form of $\mu$ seems to me to work on a general manifold. This integral that I constructed is going to come out with just number though and not a vector.

Trying this integral from a transport-related view seems gnarly to me since the transport would be path dependent.

 

[wrobel]

I feel that if we want to generalize the equations of a continuous medium's motion to an arbitrary Riemannian manifold, then the only way is to start with the Lagrange–d'Alembert principle or with some other variational-like argument. The naive approach that I tried above and that is employed in most textbooks for $\mathbb{R}^3$ is, in general, hopeless.

 

[martinbn]

Why is 1 meaningless in the Riemannian case? Can you not take $\mu$ to be $\rho$ times a chosen volume form?

 

[wrobel]

Why is 1 meaningless in the Riemannian case?

On a curved manifold, velocities at different points belong to different tangent spaces, so they cannot be added directly.

 

[martinbn]

Oh, I see, it is the integral of the velocity.

 

[jbergman]

Well since you already added the restriction of a bounded domain $D$, then I don't see the restriction of adding the restriction of requiring the neighborhood to be parellelizable as problematic. Then, yes, you can define one using the vector fields basis.

It probably doesn't make sense anyways beyond that, physically.