- #1
ehrenfest
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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?
Homework Equations
The Attempt at a Solution
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