Deriving Ampere's law from Biot-Savart equation [closed]

Question

As an exercise, I've been trying to derive Ampere's law from the Biot-Savart equation (in the static case). So basically I'm trying to prove:

\begin{equation} \nabla \times \vec{B}(\vec{r}) = \mu_0\vec{J}(\vec{r}) \end{equation}

starting from:

\begin{equation} \vec{B}(\vec{r}) = \frac{\mu_0}{4 \pi} \int \frac{\vec{J}(\vec{r}')\times(\vec{r} - \vec{r}')}{\left| \vec{r} - \vec{r}' \right|^3} d^3r' \end{equation}

For the moment, here's what I've been able to do. First, I've expressed the magnetic field as the curl of a vector potentiel:

\begin{equation} \vec{B}(\vec{r}) = \nabla \times \underbrace{\left[ \frac{\mu_0}{4 \pi} \int \frac{\vec{J}(\vec{r}')}{\left| \vec{r} - \vec{r}' \right|} d^3r' \right]}_{=\vec{A}(\vec{r})} \end{equation}

Then I have applied a curl on both side of the equation:

\begin{equation} \nabla \times \vec{B}(\vec{r}) = \nabla \times \nabla \times \underbrace{\left[ \frac{\mu_0}{4 \pi} \int \frac{\vec{J}(\vec{r}')}{\left| \vec{r} - \vec{r}' \right|} d^3r' \right]}_{=\vec{A}(\vec{r})} \end{equation}

From there I used the vectorial identity:

\begin{equation} \nabla \times \nabla \times \vec{A}(\vec{r}) = \nabla(\nabla \cdot \vec{A}) - \nabla^2\vec{A} \end{equation}

So my equation becomes:

\begin{align} \nabla \times \vec{B}(\vec{r}) &= \nabla \left[ \nabla \cdot \left( \frac{\mu_0}{4 \pi} \int \frac{\vec{J}(\vec{r}')}{\left| \vec{r} - \vec{r}' \right|} d^3r' \right) \right] - \nabla^2\left( \frac{\mu_0}{4 \pi} \int \frac{\vec{J}(\vec{r}')}{\left| \vec{r} - \vec{r}' \right|} d^3r' \right) \\ &= \nabla \left[ \nabla \cdot \left( \frac{\mu_0}{4 \pi} \int \frac{\vec{J}(\vec{r}')}{\left| \vec{r} - \vec{r}' \right|} d^3r' \right) \right] - \left( \frac{\mu_0}{4 \pi} \int \nabla^2\frac{\vec{J}(\vec{r}')}{\left| \vec{r} - \vec{r}' \right|} d^3r' \right) \\ &= \nabla \left[ \nabla \cdot \left( \frac{\mu_0}{4 \pi} \int \frac{\vec{J}(\vec{r}')}{\left| \vec{r} - \vec{r}' \right|} d^3r' \right) \right] + \left( \frac{\mu_0}{4 \pi} \int \vec{J}(\vec{r}') 4 \pi \delta(\vec{r} - \vec{r}') d^3r' \right) \\ &= \nabla \left[ \nabla \cdot \left( \frac{\mu_0}{4 \pi} \int \frac{\vec{J}(\vec{r}')}{\left| \vec{r} - \vec{r}' \right|} d^3r' \right) \right] + \mu_0 \vec{J}(\vec{r}) \end{align}

,where I used the identity:

\begin{equation} - \nabla^2 \left(\frac{1}{\left|\vec{r} - \vec{r}' \right|} \right) = 4\pi\delta(\vec{r} - \vec{r}') \end{equation}

So the second integral gives me exactly the wanted expression. The problem then only consist of proving that the first integral is equal to zero:

\begin{equation} \nabla \left[ \nabla \cdot \left( \frac{\mu_0}{4 \pi} \int \frac{\vec{J}(\vec{r}')}{\left| \vec{r} - \vec{r}' \right|} d^3r' \right) \right] = 0 \end{equation}

I have already tried a couple things but I am unable to make this integral vanish. I'm clearly missing an obvious mistake, but I haven't been able to locate it. This is similar to other questions that have been asked before, but I have a specific question about a step in the derivation which is not answered elsewhere.

Answer 1

Hint: The integral itself does not vanish, but its divergence does. Bringing the divergence inside the integral, we have $$ \nabla_r \cdot \left( \frac{\mu_0}{4 \pi} \int \frac{\vec{J}(\vec{r}')}{\left| \vec{r} - \vec{r}' \right|} d^3r' \right) = \frac{\mu_0}{4 \pi} \int \vec{J}(\vec{r}') \cdot \left[ - \frac{\vec{r} - \vec{r}'}{\left| \vec{r} - \vec{r}' \right|^3} \right] d^3r', $$ where $\nabla_r$ corresponds to the gradient with respect to $\vec{r}$. But we also know that $$ \nabla_{r'} \left( \frac{1}{\left| \vec{r} - \vec{r}' \right|} \right) = \frac{\vec{r} - \vec{r}'}{\left| \vec{r} - \vec{r}' \right|^3}, $$ where $\nabla_{r'}$ is the gradient with respect to $\vec{r}'$.

Proceed from here. You will need to make an additional assumption concerning $\vec{J}(\vec{r})$ to obtain Ampere's Law.