total differential of product of scalar & vector functions

Question

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.

Answer 1

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

1. $du \wedge dv = -dv \wedge du$
2. $\alpha \wedge \alpha = 0$
3. $d(d\alpha) = 0$
4. $df = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy + \frac{\partial f}{\partial z}dz$
5. $f(\alpha \wedge \beta) = (f\alpha)\wedge \beta = \alpha \wedge (f\beta)$
6. $d(fdu) = df\wedge du$
7. $d(\alpha + \beta) = d\alpha + d\beta$

Then we can calculate $d(m\mathbf v)$ as:

$$\begin{align}d(m\mathbf v)&= d(mv_xdx + mv_ydy) \\ &= d(mv_x)\wedge dx + d(mv_y)\wedge dy \\ &= \left[\frac{\partial (mv_x)}{\partial x}dx + \frac{\partial (mv_x)}{\partial y}dy\right]\wedge dx + \left[\frac{\partial (mv_y)}{\partial x}dx + \frac{\partial (mv_y)}{\partial y}dy\right]\wedge dy \\ &= \left[\frac{\partial(mv_y)}{\partial x} - \frac{\partial(mv_x)}{\partial y}\right]dx\wedge dy \\ &= \left[m\left(\frac{\partial v_y}{\partial x}-\frac{\partial v_x}{\partial y}\right) +v_y\frac{\partial m}{\partial x}- v_x\frac{\partial m}{\partial y}\right]dx\wedge dy \\ &= md\mathbf v + dm\wedge \mathbf v\end{align}$$

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Or if $m$ and $\mathbf v$ are explicitly functions of $t$, then you would take the exterior derivative of $m\mathbf v = m(v_xdx + v_ydy+v_tdt)$.

Answer 2

Since you have a coordinate system here, we can identify the differential with the Jacobian matrix of partial derivatives. For the example in your question, this is a $2\times3$ matrix. Writing $\mathbf p=(p_x,p_y)^T=m\mathbf v=(mv_x,mv_y)^T$, we have $$\mathrm d\mathbf p=\begin{bmatrix}{\partial p_x\over\partial x}&{\partial p_x\over\partial y}&{\partial p_x\over\partial t}\\{\partial p_y\over\partial x}&{\partial p_y\over\partial y}&{\partial p_y\over\partial t}\end{bmatrix}.$$ Each element of this matrix can be expanded via the product rule. For example, the upper-left entry is $${\partial p_x\over\partial x}=m{\partial v_x\over\partial x}+{\partial m\over\partial x}v_x.$$ Separating these terms into individual matrices, $$\begin{align} \mathrm d\mathbf p &= \begin{bmatrix}m{\partial v_x\over\partial x}&m{\partial v_x\over\partial y}&m{\partial v_x\over\partial t}\\m{\partial v_y\over\partial x}&m{\partial v_y\over\partial y}&m{\partial v_y\over\partial t}\end{bmatrix} + \begin{bmatrix}{\partial m\over\partial x}v_x&{\partial m\over\partial y}v_x&{\partial m\over\partial t}v_x\\{\partial m\over\partial x}v_y&{\partial m\over\partial y}v_y&{\partial m\over\partial t}v_y\end{bmatrix} \\ &= m\begin{bmatrix}{\partial v_x\over\partial x}&{\partial v_x\over\partial y}&{\partial v_x\over\partial t}\\{\partial v_y\over\partial x}&{\partial v_y\over\partial y}&{\partial v_y\over\partial t}\end{bmatrix} + \begin{bmatrix}v_x\\v_y\end{bmatrix}\begin{bmatrix}{\partial m\over\partial x}&{\partial m\over\partial y}&{\partial m\over\partial t}\end{bmatrix} \\ &=m\,\mathrm d\mathbf v+\mathbf v\,\mathrm dm. \end{align}$$ So, you were on the right track by starting with the product rule. Being a special case of the chain rule, it’s widely applicable. Where you appear to have run into trouble, though, is that the differential is a linear map that, for a vector-valued function, outputs a vector, which means that it can’t be identified with a vector as can the differential of a scalar function.