Change of Variables in Double Volume Integral

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 2K views
pherytic
Messages
7
Reaction score
0
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?
 
on Phys.org
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].
 
Reply
  • Like
Likes   Reactions: pherytic
Thank you, that really helped. If the two original variables are independent wrt each other, that has to still be true after substitution.