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Derivative of a potential function that depends on several position vectors

  1. Mar 29, 2010 #1
    SOLVED
    1. The problem statement, all variables and given/known data
    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]

    2. Relevant 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]

    3. 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: Mar 30, 2010
  2. jcsd
  3. Mar 29, 2010 #2

    Fredrik

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    Staff Emeritus
    Science Advisor
    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

    [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]
     
  4. Mar 29, 2010 #3

    gabbagabbahey

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    Homework Helper
    Gold Member

    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?
     
  5. Mar 30, 2010 #4
    Thanks, I solved it.
     
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