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I have the following relation for a function $\phi_A(\textbf{r})$ and it is known that $\rho_B(\lambda \textbf{r}) = \rho_A(\textbf{r})$

$\phi_A(\textbf{r}) = \int{d^3r' \frac{\rho_A(\textbf{r}')} {|\textbf{r - r}'|}} = \int{d^3r' \frac{\rho_B(\lambda \textbf{r}')} {|\textbf{r - r}'|}}$.

Now if I change the integration variable to $\lambda\textbf{r}'$, then

$\phi_A(\textbf{r}) = \int{d^3(\lambda r') \frac{\rho_B(\lambda^2 \textbf{r}')} {|\textbf{r} - \lambda \textbf{r}'|}}$.

My book says that

$\phi_A(\textbf{r}) = \frac{1}{\lambda^2}\int{d^3(\lambda r') \frac{\rho_B(\lambda \textbf{r}')} {|\lambda\textbf{r} - \lambda \textbf{r}'|}}$.

I am guessing that this relation is trivial, but I am unable to get the third relation from the second one.

Paddy
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1 Answers1

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I'm assuming $\bf{r}$ and $\bf{r}'$ are vectors in $\mathbb{R}^{3}$, so: $$|{\bf{r}}-{\bf{r'}}| = \sqrt{(x-x')^{2}+(y-y')^{2}+(z-z')^{2}} =\sqrt{\frac{1}{\lambda^{2}}[(\lambda x-\lambda x')^{2}+(\lambda y-\lambda y')^{2}+(\lambda z-\lambda z')^{2}]} = \frac{1}{\lambda}|\lambda{\bf{r}}-\lambda{\bf{r}'}|$$ Here, I assumed $0< \lambda$, ${\bf{r}}=(x,y,z)$ and ${\bf{r}}'=(x',y',z')$. Thus, we have: $$\int d^{3}{\bf{r}'}\frac{\rho_{B}(\lambda {\bf{r}}')}{|{\bf{r}}-{\bf{r}}'|} =\lambda\int d^{3}{\bf{r}'}\frac{\rho_{B}(\lambda {\bf{r}}')}{|\lambda{\bf{r}}-\lambda{\bf{r}}'|} $$ Finally, $d^{3}{\bf{r}}' = \frac{1}{\lambda^{3}}d^{3}(\lambda{\bf{r}}')$, from where it follows that: $$\phi_{A}({\bf{r}}) = \frac{1}{\lambda^{2}}\int d^{3}(\lambda{\bf{r}'})\frac{\rho_{B}(\lambda {\bf{r}}')}{|\lambda{\bf{r}}-\lambda{\bf{r}}'|}$$

IamWill
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