Change of Variables in Double Volume Integral

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  • #1
pherytic
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In Greiner's Classical Electromagnetism book (page 126) he has a derivation equivalent to the following.
$$\int_V d^3r^{'} \nabla \int_V d^3r^{''}\frac {f(\bf r^{''})}{|\bf r + \bf r^{'}- \bf r^{''}|}$$

$$ \bf z = \bf r^{''} - \bf r^{'} $$

$$\int_V d^3r^{'} \nabla \int_V d^3z \frac {f(\bf z + \bf r^{'})}{|\bf r - \bf z|}$$


$$\nabla\int_V d^3z \frac{1}{|\bf r - \bf z|} \int_V d^3r^{'} {f(\bf z + \bf r^{'})}$$

The book says that after the change of variables to z, it is okay to just move the 1/|r-z| term out of the r' integral. But this doesn't make sense to me, given z is still a function of r'. Am I missing something or is this mathematically incorrect?
 

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  • #2
pasmith
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The integration with respect to [itex]\mathbf{r}'[/itex] is with [itex]\mathbf{z}[/itex] held constant. You have replaced the independent variables [itex]\mathbf{r}'[/itex] and [itex]\mathbf{r}''[/itex] with the independent variables [itex]\mathbf{r}'[/itex] and [itex]\mathbf{z}[/itex] by replacing [itex]\mathbf{r}''[/itex] with [itex]\mathbf{r}' + \mathbf{z}[/itex]. It then makes sense to do the integral over [itex]\mathbf{r}'[/itex] first, precisely because you can regard [itex]\|\mathbf{r} - \mathbf{z}\|[/itex] as constant while doing so. The domain of integration will have changed because we require both [itex]\mathbf{r}' \in V[/itex] and [itex]\mathbf{r}' + \mathbf{z} \in V[/itex].
 
  • #3
pherytic
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Thank you, that really helped. If the two original variables are independent wrt each other, that has to still be true after substitution.
 

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