I Conservation of Mass for Compressible Flow

[vanhees71]

Let $\omega$ be a k-form on m-dimensional manifold $M,\quad m\ge k.$ Let $v$ be a vector field on $M$ and $g^t$ be its flow.

Let $S\subset M$ be a k-dimensional compact submanifold. Then
$$\frac{d}{dt}\Big|_{t=0}\int_{g^t(S)}\omega=\int_SL_v\omega.$$ Here $L_v$ is a Lie derivative and the formula follows directly from the change of variables.

Particularly if k=m then $\omega=\rho(x)dx^1\wedge\ldots\wedge dx^m$ and
$$L_v\omega=\frac{\partial (\rho v^i)}{\partial x^i} dx^1\wedge\ldots\wedge dx^m.$$

From the group property it is not hard to show that the following conditions are equivalent

1) $\frac{d}{dt}\Big|_{t=0}\int_{g^t(S)}\omega=0$ for any k-dimensional compact submanifold S;
2) $\frac{d}{dt}\int_{g^t(S)}\omega=0,\quad \forall t$ and for any k-dimensional compact submanifold S;
3) $L_v\omega=0$

This method is covering a lot of different circulations theorems in electrodynamics and hydro dynamics as well

In the fluid-mechanics literature those are known also as "Reynolds transport theorems". This is the modern Cartan-calculus formulation of them.

 

[wrobel]

Yes! Elie Cartan's integral invariants.

Df. A form $\omega$ is called relative integral invariant iff $L_v\omega=d\Omega\quad (\exists \Omega)$.
A form $\alpha$ is called absolute integral invariant iff $L_v\alpha=0$.

If $\omega$ is a relative integral invariant then $d\omega$ is an absolute integral invariant: $L_vd\omega=dL_v\omega=dd\Omega$
and
$$\frac{d}{dt}\Big|_{t=0}\int_{g^t(S)}\omega=\int_SL_v\omega=\int_{\partial S}\Omega.$$

For further play with these formulas use
$$L_v\omega=i_vd\omega+di_v\omega;$$

De Rham's theorem is also near here
etc.

For example in the extended phase space of a Hamiltonian system one has: $dp_i\wedge dx^i-dH\wedge dt$ is an absolute integral invariant; $p_idx^i-Hdt$ is a relative integral invariant.

 

[erobz]

In the fluid-mechanics literature those are known also as "Reynolds transport theorems". This is the modern Cartan-calculus formulation of them.

The funny thing is, I know about Reynolds Transport Theorems, but I never would have recognized that(and frankly still can’t)! @wrobel said “k form on m dimensional manifold” and my head exploded.

 

[vanhees71]

The advantage of the modern Cartan calculus is that it becomes in a way very natural, and it's very general too, because it works not only in (pseudo)-Riemannian manifolds but in general differentiable manifolds, i.e., the Lie derivative and Stokes's theorem for alternating (differential) forms are independent of any additional structures like connections and/or (pseudo-)metrics.