Derivation of the formula of the vector potential

Question

I'm trying to derive the formula for vector potentials, $$\vec{A} = \frac{\mu_{0}}{4{\pi}} \int \frac {\vec{j} (\vec{r}')} {\lvert \vec{r} - \vec{r}' \rvert} \mathrm{d} \vec{r}' $$ from the Biot and Savart formula $$\vec{B} = \frac{\mu_{0}}{4{\pi}} \int \frac {\vec{j} (\vec{r}') \times ( \vec{r} - \vec{r}' ) } {{\lvert \vec{r} - \vec{r}' \rvert}^{3} } \mathrm{d} \vec{r}' .$$

I saw that the key of the derivation is to write $$ {\vec{\nabla}}_{\vec{r}} \times \frac {\vec{j} (\vec{r}')} {\lvert \vec{r} - \vec{r}' \rvert} = \frac { {\vec{\nabla}}_{\vec{r}} \times \vec{j} (\vec{r}') } { \lvert \vec{r} - \vec{r}' \rvert } - \vec{j} (\vec{r}') \times \left ( {\vec{\nabla}}_{\vec{r}} \times \frac {1} { \lvert \vec{r} - \vec{r}' \rvert } \right ) = \vec{j} (\vec{r}') \times \frac {\vec{r} - \vec {r}' } {{\lvert \vec{r} - \vec{r}' \rvert}^{3} } ,$$ but I don't understand why $$ \frac { {\vec{\nabla}}_{\vec{r}} \times \vec{j} (\vec{r}') } { \lvert \overrightarrow{r} - \vec{r}' \rvert } - \vec{j} (\vec{r}') \times \left ( {\vec{\nabla}}_{\vec{r}} \times \frac {1} { \lvert \vec{r} - \vec{r}' \rvert } \right ) = \vec{j} (\vec{r}') \times \frac {\vec{r} - \vec {r}' } {{\lvert \vec{r} - \vec{r}' \rvert}^{3} }. $$

Also, what is the difference between ${\vec{\nabla}}_{\vec{r}}$ and $ {\vec{\nabla}}_{\vec{r}'}$ ?

Answer 1

The difference between ${\vec{\nabla}}_{\vec{r}}$ and $ {\vec{\nabla}}_{\vec{r}'}$ is that the former represents derivatives with respect to $\vec r$, and the latter differentiates with respect to $\vec r'$.

The formula that puzzles you, $$ \frac { {\vec{\nabla}}_{\vec{r}} \times \vec{j} (\vec{r}') } { \lvert \overrightarrow{r} - \vec{r}' \rvert } - \vec{j} (\vec{r}') \times \left ( {\vec{\nabla}}_{\vec{r}} \color{\red}{\times} \frac {1} { \lvert \vec{r} - \vec{r}' \rvert } \right ) = \vec{j} (\vec{r}') \times \frac {\vec{r} - \vec {r}' } {{\lvert \vec{r} - \vec{r}' \rvert}^{3} }, $$ has two things going on.

• One is the term in ${\vec{\nabla}}_{\vec{r}} \times \vec{j} (\vec{r}')$, which vanishes because $\vec{j} (\vec{r}') $ does not depend on $\vec r$, and therefore under the action of $\vec\nabla_\vec{r}$ it gives zero.
• The other is the gradient (not the curl!) of the relative separation, $${\vec{\nabla}}_{\vec{r}} \frac {1} { \lvert \vec{r} - \vec{r}' \rvert } = -\frac {\vec{r} - \vec {r}' } {{\lvert \vec{r} - \vec{r}' \rvert}^{3} },$$ which reduces to a simple calculation. As an example, for the $x$ component of the gradient, and setting $\tilde r=\lvert \vec{r} - \vec{r}' \rvert$, we have $$ \frac{\partial}{\partial x}\frac{1}{\tilde r} = -\frac{1}{\tilde r^2} \frac{\partial \tilde r}{\partial x}, $$ where \begin{align} \frac{\partial \tilde r}{\partial x} & = \frac{\partial }{\partial x}\sqrt{(x-x')^2+(y-y')^2+(z-z')^2} \\& = \frac{2(x-x')}{2\sqrt{(x-x')^2+(y-y')^2+(z-z')^2}} \\& = \frac{(x-x')}{\tilde r}, \end{align} so $$ \frac{\partial}{\partial x}\frac{1}{\tilde r} = -\frac{x-x'}{\tilde r^3}, $$ as claimed.