I've probably made mathematical mincemeat out of this but,
suppose I have a product of scalar and vector functions, such as the momentum $\mathbf{p} = m \mathbf{v}$. To keep it reasonably simple but sufficiently general, let each term depend on time $t$ and the 2D space $(x,y)$ so
$$\mathbf{p}(x,y,t) = m(x,y,t) \mathbf{v}(x,y,t)$$
I am trying to write down the 'total differential'. To begin, omitting the function arguments for notational simplicity, applying product rule gives
$$d(m \mathbf{v}) = \mathbf{v}d(m) + m d(\mathbf{v}) $$
the total differential of the scalar function is clearly
$$d(m) = \frac{\partial m}{\partial t}dt + \frac{\partial m}{\partial x}dx
+ \frac{\partial m}{\partial y}dy $$
Now for the vector term… I believe we can treat each scalar component $(v_x(x,y,t), v_y(x,y,t))$ individually as above for the scalar, is that right? then,
$$d(\mathbf{v}) = d(v_x) \mathbf{\hat x} + d(v_y) \mathbf{\hat y}$$
where each of the total differential (scalar) terms $d(v_x)$ and $d(v_y)$ would expand just as for $d(m)$ above.
Is this making any sense??
Is there a simpler/cleaner notation for this, such as tensor index notation??
(or perhaps using vector calculus notation? I am familiar with the vector-calculus divergence identity for the product of a scalar & vector ($\nabla \cdot (m \mathbf{v}) $), but I think what I'm attempting here is something different, but perhaps related although I can't figure out how the 'del' ('nabla' $\nabla$) operator notation can be used?
Thanks.
Best Answer
The total derivative of a vector quantity is itself not a vector quantity. So you have to make a choice as to what you mean by total derivative. Here's one way. Instead of thinking of $\mathbf v$ as the vector $\mathbf v=v_x\mathbf {\hat x}+v_y\mathbf {\hat y}$, you can think of it as the $1$-form $\mathbf v= v_xdx + v_ydy$.
Then the "total differential" is just the exterior derivative. To calculate the exterior derivative just use these rules: If $\alpha$ is a $k$-form, $\beta$ is a $j$-form, $du$ and $dv$ are $1$-forms, $f=f(x,y,z)$ is a $0$-form (a function) , and $\wedge$ is the exterior product, then
Then we can calculate $d(m\mathbf v)$ as:
$$\begin{align}d(m\mathbf v)&= d(mv_xdx + mv_ydy) \\ &= d(mv_x)\wedge dx + d(mv_y)\wedge dy \\ &= \left[\frac{\partial (mv_x)}{\partial x}dx + \frac{\partial (mv_x)}{\partial y}dy\right]\wedge dx + \left[\frac{\partial (mv_y)}{\partial x}dx + \frac{\partial (mv_y)}{\partial y}dy\right]\wedge dy \\ &= \left[\frac{\partial(mv_y)}{\partial x} - \frac{\partial(mv_x)}{\partial y}\right]dx\wedge dy \\ &= \left[m\left(\frac{\partial v_y}{\partial x}-\frac{\partial v_x}{\partial y}\right) +v_y\frac{\partial m}{\partial x}- v_x\frac{\partial m}{\partial y}\right]dx\wedge dy \\ &= md\mathbf v + dm\wedge \mathbf v\end{align}$$
Or if $m$ and $\mathbf v$ are explicitly functions of $t$, then you would take the exterior derivative of $m\mathbf v = m(v_xdx + v_ydy+v_tdt)$.