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Derivative of a potential function that depends on several position vectors

  • #1
SOLVED

Homework Statement


Show that:
[tex]
\sum_i \vec{\nabla_i}V \cdot \frac{\partial \vec{r_i}}{\partial q_j} = \frac{\partial V}{\partial q_j}
[/tex]

Homework Equations


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

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:
[tex]
\frac{\partial V}{\partial q_j}=\sum_i \frac{\partial V}{\partial \vec{r_i}} \frac{\partial \vec{r_i}}{\partial q_j}
[/tex]
However, the partial derivative of a scalar function with respect to a vector is something I'm not used to. Thanks in advance
 
Last edited:

Answers and Replies

  • #2
Fredrik
Staff Emeritus
Science Advisor
Gold Member
10,851
407
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

[tex]\frac{\partial V}{\partial (r_i)_k}\frac{\partial (r_i)_k}{\partial q_j}[/tex]

where [itex](r_i)_k[/itex] is component k of vector i.

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

[tex](f\circ g)_{,i}(x)=\sum_j f_{,j}(g(x))g_{j,i}(x)[/tex]

where ",i" denotes partial differentiation with respect to the ith variable, and [itex]g_j[/itex] 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

[tex](f\circ g)_{,i}(x)=f_{,j}(g(x))g^j_{,i}(x)[/tex]

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.

[tex](f\circ g)^k_{,i}(x)=(f^k\circ g)_{,i}(x)=f^k_{,j}(g(x))g^j_{,i}(x)[/tex]
 
  • #3
gabbagabbahey
Homework Helper
Gold Member
5,002
6
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.:wink:

Let's look at a simpler example. Let's say that [itex]V=V(\textbf{r}_1)[/itex] only and that [itex]\textbf{r}_1=\textbf{r}_1(q_1,q_2,q_3)[/itex]. You then have [itex]V=V(\textbf{r}_1)=V(\textbf{r}_1(q_1,q_,q_3))[/itex] which is the same thing as saying [itex]V=V(q_1,q_,q_3)[/itex]...what do you get when you apply the chain rule to that?
 
  • #4
Thanks, I solved it.
 

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