# Derivative of a potential function that depends on several position vectors

SOLVED

## Homework Statement

Show that:
$$\sum_i \vec{\nabla_i}V \cdot \frac{\partial \vec{r_i}}{\partial q_j} = \frac{\partial V}{\partial q_j}$$

## Homework Equations

$$V=V(\vec{r_1},\vec{r_2},\vec{r_3},...\vec{r_N},t)$$
$$\vec{r_i}=\vec{r_i}(q_1,q_2,q_3,...,q_n); i=1,2,3,...,N; n<N$$

## The Attempt at a Solution

I'm not sure how to apply the chain rule when the function depends on vectors. I guess is something like this:
$$\frac{\partial V}{\partial q_j}=\sum_i \frac{\partial V}{\partial \vec{r_i}} \frac{\partial \vec{r_i}}{\partial q_j}$$
However, the partial derivative of a scalar function with respect to a vector is something I'm not used to. Thanks in advance

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Fredrik
Staff Emeritus
Gold Member
You don't take the partial derivatives with respect to a vector, you do it with respect to the the vector's components (because they are all independent variables), so the result will contain terms like

$$\frac{\partial V}{\partial (r_i)_k}\frac{\partial (r_i)_k}{\partial q_j}$$

where $(r_i)_k$ is component k of vector i.

By the way, I like to write the chain rule as

$$(f\circ g)_{,i}(x)=\sum_j f_{,j}(g(x))g_{j,i}(x)$$

where ",i" denotes partial differentiation with respect to the ith variable, and $g_j$ is the jth component of g. This is both easy to remember and easy to use.

Normally I would use Einstein's summation convention as well, and write component indices upstairs, so I'd write the chain rule as

$$(f\circ g)_{,i}(x)=f_{,j}(g(x))g^j_{,i}(x)$$

and if f is vector-valued, the same formula still holds for the components of f, so we only have to put an index on f as well.

$$(f\circ g)^k_{,i}(x)=(f^k\circ g)_{,i}(x)=f^k_{,j}(g(x))g^j_{,i}(x)$$

gabbagabbahey
Homework Helper
Gold Member
However, the partial derivative of a scalar function with respect to a vector is something I'm not used to.
That's because its nonsense.

Let's look at a simpler example. Let's say that $V=V(\textbf{r}_1)$ only and that $\textbf{r}_1=\textbf{r}_1(q_1,q_2,q_3)$. You then have $V=V(\textbf{r}_1)=V(\textbf{r}_1(q_1,q_,q_3))$ which is the same thing as saying $V=V(q_1,q_,q_3)$...what do you get when you apply the chain rule to that?

Thanks, I solved it.