I'm working through example 3.2 of Zangwill's Modern Electrodynamics and have come across a change in integration variables that I just can't seem to get. The example has two different change distributions related by $ \rho_b(\lambda \mathbf{r}) = \rho_a(\mathbf{r})$ and asks for the relationship between the electric potentials. According to the book, $$ \int d^3 r'= \frac{1}{\lambda^2} \int d^3 (\lambda r'). $$ When I try to work through the Jacobian, I get $$ d^3 r' = \begin{vmatrix} \frac{1}{\lambda} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{vmatrix} d^3(\lambda r') = \frac{1}{\lambda} d^3(\lambda r'),$$ where I'm assuming changing from $ r'$ to $ \lambda r'$ is like going from $\vec{x}$ to $\vec{y}$ where $$ y_1 = \lambda x_1, \hskip{1cm} y_2 = x_2\hskip{0.5cm} \text{and} \hskip{0.5cm} y_3 = x_3 . $$ What am I missing here?
-
$\begingroup$ The change $\mathbf{r}\rightarrow\lambda\mathbf{r}$ means that you should make the change $x\rightarrow\lambda x$, $y\rightarrow\lambda y$ and $z\rightarrow\lambda z$. $\endgroup$– TheAverageHijanoCommented Feb 15, 2019 at 8:33
2 Answers
If $$ \boldsymbol r = \begin{vmatrix} x \\ y \\ z \\ \end{vmatrix} = \begin{vmatrix} r \sin \theta \cos \phi \\ r \sin \theta \sin \phi \\ r \cos \theta \\ \end{vmatrix} $$ the associated volume element is $$ d^3 \boldsymbol r = \begin{vmatrix} r \sin \theta \cos \phi & r \cos \theta \cos \phi & - r \sin \theta \sin \phi \\ r \sin \theta \sin \phi & r \cos \theta \sin \phi & r \sin \theta \cos \phi \\ r \cos \theta & - r \sin \theta & 0 \\ \end{vmatrix} dr d \theta d \phi = r^2 \sin \theta dr d \theta d \phi $$ Similarly, if $$ \lambda \boldsymbol r = \begin{vmatrix} \lambda x \\ \lambda y \\ \lambda z \\ \end{vmatrix} = \begin{vmatrix} \lambda r \sin \theta \cos \phi \\ \lambda r \sin \theta \sin \phi \\ \lambda r \cos \theta \\ \end{vmatrix} $$ the associated volume element is $$ d^3 ( \lambda \boldsymbol r ) = \begin{vmatrix} \lambda r \sin \theta \cos \phi & \lambda r \cos \theta \cos \phi & - \lambda r \sin \theta \sin \phi \\ \lambda r \sin \theta \sin \phi & \lambda r \cos \theta \sin \phi & \lambda r \sin \theta \cos \phi \\ \lambda r \cos \theta & - \lambda r \sin \theta & 0 \\ \end{vmatrix} dr d \theta d \phi = ( \lambda r)^2 \sin \theta dr d \theta d \phi = \lambda^2 ( r^2 \sin \theta dr d \theta d \phi ) $$ So, comparing the two expressions above for $d^3 \boldsymbol r$ and $d^3 ( \lambda \boldsymbol r )$, you get $$ d^3 ( \lambda \boldsymbol r ) = \lambda^2 d^3 \boldsymbol r $$ or $$ d^3 r = \frac{1}{\lambda^2} d^3 ( \lambda \boldsymbol r ) $$
I think you forget pieces of the integral that gives the potential :
${{d}^{3}}(\lambda \overrightarrow{r'})={{\lambda }^{3}}{{d}^{3}}(\overrightarrow{r'})$ as indicated in the comment.
But : $\int{{{d}^{3}}(\overrightarrow{r'})\frac{\rho (\lambda \overrightarrow{r'})}{\left| \overrightarrow{r}-\overrightarrow{r'} \right|}}=\frac{1}{{{\lambda }^{2}}}\int{{{d}^{3}}(\lambda \overrightarrow{r'})\frac{\rho (\lambda \overrightarrow{r'})}{\left| \lambda \overrightarrow{r}-\lambda \overrightarrow{r'} \right|}}$
