Differential forms and partial derivatives

Question

If I have a vector valued function $f(\vec R)$ and I imagine that $R$ is also a function of time, so that $R(t)$ Then I'm struggling to prove that:

$$ df \left({\partial R \over \partial t}\right) = {df \over dt}$$

Intuitively, it makes some sense, that if df is a differential form, and a covector; and $\partial R \over \partial t$ is the velocity vector, then a covector and vector product will give a scalar. Since $df \over dt $ is also a scalar output function differentiated in terms of a scalar, it too would be a scalar.

What I'm struggling to see is why this should be true.

Answer 1

The function $f$ takes in a point $(x,y,z)$ in space and outputs a real number $f(x,y,z)$; provided that it is regular enough, you can write by the total differential theorem that $$df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy + \frac{\partial f}{\partial z} dz \tag1$$

On the other hand, $\mathbf R$ is a function that takes in a real number (the time $t$) and outputs a point $\mathbf R(t)$ in space; if it is regular enough, its derivative is well-defined and we may say that the vector $\mathbf R$ defines a trajectory in space. The derivative of $\mathbf R$ (total or partial is the same) is the vector $$\frac{d\mathbf R}{dt} = \left(\frac{dR_x}{dt}, \frac{dR_y}{dt}, \frac{dR_z}{dt} \right)\tag2$$ If you imagine formally dividing $(1)$ by $dt$, you get $$\frac{df}{dt} = \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y} \frac{dy}{dt} + \frac{\partial f}{\partial z} \frac{dz}{dt}, $$ and you can think of “evaluating” this expression on the trajectory of $\mathbf R$ by plugging in the components of $(2)$ into the total derivatives of $x$,$y$, and $z$. I think that is what is meant by the LHS of your formula.

Answer 2

Given the functions $f:\Bbb R^n \to \Bbb R \ $, and $\vec R: \Bbb R \to \Bbb R^n$ also:

$$df(\vec v) = \nabla f \cdot \vec v \ \ \text{ and}\ \ {\partial \vec R\over \partial t} = \sum_{i \in n}{\partial \vec R\over \partial x_i}{\partial x_i\over \partial t}$$

So:

$$ df\left({\partial \vec R\over \partial t}\right ) = df\left(\sum_{i \in n}{\partial \vec R\over \partial x_i}{\partial x_i\over \partial t}\right ) = \sum_{i\in n}{\partial x_i\over \partial t} df\left({\partial \vec R\over \partial x_i}\right )$$

Given that

$$df\left({\partial \vec R\over \partial x_i}\right ) = df(e_i) = {\partial f\over \partial x_i}$$

So:

$$ \sum_{i\in n}{\partial x_i\over \partial t} df\left({\partial \vec R\over \partial x_i}\right ) = \sum_{i\in n}{\partial x_i\over \partial t}{\partial f\over \partial x_i} = {\partial f\over \partial t}$$

Now, because $f$ maps from $\Bbb R^n$ to $\Bbb R$, and $\vec R$ from $\Bbb R$ to $\Bbb R^n$, then $f(\vec R): \Bbb R \to \Bbb R^n \to \Bbb R = \Bbb R \to \Bbb R$, hence:

$${\partial f \over \partial t} = {df\over dt} $$

because $f$ is a function of $t$.