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Griffiths 5.7

  • Thread starter ehrenfest
  • Start date
1,996
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1. Homework Statement
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. Homework Equations



3. The Attempt at a Solution
 
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Answers and Replies

511
1
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:
1,996
1
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.
I did all of that before I opened the thread. Can someone answer my questions?
 
511
1
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.
 
1,996
1
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.
Hmmm...that makes some sense. Can someone say more about that please?
 
clem
Science Advisor
1,308
15
r is an integration variable.
Nothing outside the integral can affect it.
Go back to your calculus book on differentiating an integral.
 
Ben Niehoff
Science Advisor
Gold Member
1,864
160
r is an integration variable.
Nothing outside the integral can affect it.
Go back to your calculus book on differentiating an integral.
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.
 
clem
Science Advisor
1,308
15
[tex]d\tau[/tex] in this notation stands for [tex]d^3 r[/tex].
 
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