Derivative of a potential function that depends on several position vectors

[andresordonez]

SOLVED

Homework Statement

Show that:
<br /> \sum_i \vec{\nabla_i}V \cdot \frac{\partial \vec{r_i}}{\partial q_j} = \frac{\partial V}{\partial q_j}<br />

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&lt;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:
<br /> \frac{\partial V}{\partial q_j}=\sum_i \frac{\partial V}{\partial \vec{r_i}} \frac{\partial \vec{r_i}}{\partial q_j}<br />
However, the partial derivative of a scalar function with respect to a vector is something I'm not used to. Thanks in advance

 

[Fredrik]

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]

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?

 

[andresordonez]

Thanks, I solved it.