How is partial time derivative $\frac{\partial}{\partial t}$ defined for vector flows?

Question

This question emerged when I was thinking about Liouville's theorem of classical mechanics. As far as I understand, the change of any function along the integral curves of Hamiltonian vector field is

$$\frac{d \rho}{d t} = \{\rho, H \} = \mathcal L_X \rho $$

where $\{\,,\}$ is Poisson bracket, $X$ is a vector field associated with $dH$ and $\mathcal L$ is Lie derivative. I have the following understanding of $\frac{d f}{d t}$ --- as we move on the manifold $M$ with a speed $X$, the landscape ($f$ visioned as mountains on $M$) changes. The mountains are still, it us who is moving.

On the other hand, in certain formulations of Liouville's theorem $\frac{\partial}{\partial t}$ is used:

$$\frac{\partial \rho}{\partial t} = - \{\rho, H \} $$

(The question is about the definition of $\frac{\partial}{\partial t}$, not about the contents of Liouville's theorem). I am familiar with the notion of material derivative and partial time derivative from usual formulations of fluid dynamics, but I am not sure of their counterparts in differential geometry.

What is partial time derivative $\frac{\partial}{\partial t}$ and how is it defined invariantly for functions, vector fields and differential forms? How is it related to $X$ and $\mathcal L_X$?

Answer 1

The thing I have missed is that phase space distribution is by definition time-dependent function. Indeed, it is defined as a propagation of the initial distribution with the Hamiltonian flow:

$$\rho_t = \varphi^*_{-t} \, \rho_0$$

thus, partial time derivative makes sense at any point regardless of any flows:

$$\frac{\partial \rho_t}{\partial t} = \lim_{s \to 0} \frac{\rho_{t+s} - \rho_t}{s}$$

On the other hand, the change of a time-dependent function along some flow, that is $\frac{d}{dt}$, is given by:

$$\frac{d}{dt} \left(\varphi^*_t f_t \right) = \varphi^*_t \left[\frac{\partial f_t}{\partial t} + \mathcal L_X \, f_t \right]$$

see, for example, Online Supplement for Jeffery Lee's "Manifolds and Differential Geometry". This expression, by the way, is valid for tensors of all ranks.

Returning to the Liouville's theorem, we get:

$$0 = \frac{d \rho_0}{d t} = \frac{d}{dt} \left(\varphi^*_t \rho_t \right) = \varphi^*_t \left[\frac{\partial \rho_t}{\partial t} + \mathcal L_X \, \rho_t \right],$$

or, in concise notation

$$\frac{d \rho_t}{d t} = \frac{\partial \rho_t}{\partial t} + \{ \rho_t, H \} = 0.$$

Answer 2

I think that this is what is meant:

We are working in a Hamiltonian system $(M,\omega, H)$, where $M$ is a $2n$ dimensional symplectic manifold and $H\in C^\infty(M)$. Equip $M$ with local coordinates $x^1,\dots,x^n,y^1,\dots,y^n$. We have the Hamiltonian vector field $X_H\in\Gamma(TM)$. Also, this phase distribution function $\rho$ is in in $C^\infty(M)$.

By definition $\{\rho, H\}\in C^\infty (M)$. Now $\rho$ alone is a function on $M$, but we only care about what it's doing to solutions of Hamilton's equations, i.e. maps $\gamma:\mathbb{R}\to M$ which are integral curves of $X_H$. So in this sense $\rho\circ\gamma:\mathbb{R}\to \mathbb{R}$. Thus on solutions to Hamilton's equations, $\rho$ is a function of $x^i,v^i$ and $t$. Taking the partial derivative with respect to $t$ and evaluating at $\gamma(t)$, which is $\frac{\partial\rho}{\partial t}(\gamma(t))$, gives you a real number, which Liouville says is $-\{\rho,H\}(\gamma(t))$

As for your second question, I am not sure if this is what you are looking for but one way in which $\frac{\partial}{\partial t}$ is related to $\mathcal L$ is through the fact that if $\mathcal L_{X_H}f=0$, for $f\in C^\infty (M)$, then $f$ is constant on the integral curves of $X_H$. That is, $\frac{\partial}{\partial t}(f\circ\gamma)=0$, or equivalently, $f\circ\gamma=f$. In particular, it is always the case that $\mathcal L_{X_H}H=0$ so that if $\theta$ is the flow of $X_H$ then $H$ is constant on the flow of $X_H$ (i.e. $H\circ\theta_t=H)$.

Also it's related to forms since we also have that $\mathcal L_{X_H}\omega=0$ so that $\theta_t^\ast\omega=\omega$, meaning the flow of $X_H$ preserves $\omega$. I mean a $\frac{\partial}{\partial t}$ appears since $\left.\frac{\partial}{\partial t}\right|_{t=0}\theta^{(p)}(t)=(X_H)_p$?