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Griffiths 5.7

  1. Mar 19, 2008 #1
    1. The problem statement, all variables and given/known data
    This question refers to Griffiths E and M book.

    To finish the problem, I just need to show that

    [tex]\frac{d}{dt}\int_V \vec{r} \rho d\tau = \int_V \vec{r} \frac{\partial \rho}{\partial t} d\tau [/tex]

    When you apply the product rule to the integral, why do you get

    [tex]\frac{\partial \vec{r}}{\partial t} = 0 [/tex]


    Actually I am really not really sure what that derivative even means here. My instinct would say you just get [itex]\dot{\vec{r}} [/itex], but what does that mean??? I am so confused!! So, r is a position vector and position vectors can be functions of time and Griffiths says there is current in the region... But what is it the position vector of?!?! Ahhh, this is insane--can someone just explain the whole situation regarding that integral and the time derivative of it?

    2. Relevant equations

    3. The attempt at a solution
    Last edited: Mar 19, 2008
  2. jcsd
  3. Mar 19, 2008 #2
    To do this problem,simply evaluate [tex]\int\ div( x \vec J ) dV[/tex] using the product rule.Also this integral is also equal to a closed surface integral using divergence theorem. Using the volume as all space,you get the boundary at infinity where this surface integral vanishes.Rearranging the other side of the integral, you will get the result for x direction.Compute for all x,y, and z directions.This gives you the result.
    Last edited: Mar 19, 2008
  4. Mar 19, 2008 #3
    I did all of that before I opened the thread. Can someone answer my questions?
  5. Mar 19, 2008 #4
    You are confusing between the trajectory concept and co-ordinate concept.In this problem r is a function of x,y,z and not t.You are not worried about the trajectory of the charge distribution.
  6. Mar 19, 2008 #5
    Hmmm...that makes some sense. Can someone say more about that please?
  7. Mar 19, 2008 #6


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    r is an integration variable.
    Nothing outside the integral can affect it.
    Go back to your calculus book on differentiating an integral.
  8. Mar 19, 2008 #7

    Ben Niehoff

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    No, it's not. The integration variable is tau. It looks similar to r in this font.

    To the OP: [itex]\vec r[/itex] is just a static position; the location at which you are measuring the field. It doesn't change with time.
  9. Mar 19, 2008 #8


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    [tex]d\tau[/tex] in this notation stands for [tex]d^3 r[/tex].
    Last edited: Mar 19, 2008
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