# Deriving the Biot-Savart Law from the Helmholtz theorem?

The Helmholtz theorem tells us that any vector function $$\vec{F(\vec{r})}$$ that goes to zero sufficiently fast can be expressed as $$\vec{F(\vec{r})}=\nabla(\frac{-1}{4\pi}\int\frac{\nabla'\cdot\vec{F(\vec{r}')}}{R}dV')+\nabla\times(\frac{1}{4\pi}\int\frac{\nabla'\times\vec{F(\vec{r}')}}{R}dV')$$

where $$R=|\vec{r}-\vec{r'}|$$ is the magnitude of the separation vector. Now from Maxwell's equations (assuming magnetostatics) we have that $$\nabla \cdot \vec{B}=0$$ and $$\nabla \times \vec{B}=\mu_0\vec{J}$$. Combining these facts with the Helmholtz theorem above, we get that the magnetic field (within the domain of magnetostatics) should take the form $$\vec{B(\vec{r})}=\nabla\times\frac{\mu_o}{4\pi}\int\frac{\vec{J(\vec{r}')}}{R}dV' \tag{1}$$ Now the above should be equivalent to the Biot-Savart law for volume currents which is $$\vec{B(\vec{r})}=\frac{\mu_0}{4\pi}\int\frac{\vec{J(\vec{r}')\times\hat{R}}}{R^2}dV'\tag{2}$$ But these two equations are noticeably different (at least in their present form). Is there any way to manipulate equation (1) such that we end up with equation (2) ?

Any help would be most appreciated!

• Possible duplicate: link. Commented May 15, 2021 at 13:38
• Instead of (\dfrac12) =$(\dfrac12)$ use \left(\dfrac12\right)=$\left(\dfrac12\right)$ to adjust to the height of the content. Commented May 15, 2021 at 19:19

Using the product rule $$\nabla \times(f\vec{J})=f(\nabla \times \vec{J})-\vec{J}\times (\nabla f)$$ with $$f=1/R$$ we get $$\vec{B(\vec{r})}=\nabla\times\frac{\mu_o}{4\pi}\int\frac{\vec{J(\vec{r}')}}{R}dV'$$ $$\vec{B(\vec{r})}=\frac{\mu_o}{4\pi}\int\nabla\times(\frac{\vec{J(\vec{r}')}}{R})dV'$$ $$\vec{B(\vec{r})}=\frac{\mu_o}{4\pi}\int \frac{1}{R}(\nabla \times \vec{J(r')})-\vec{J(r')}\times (\nabla \frac{1}{R})dV'$$ But now the curl operator is with respect to the un-primed $$\vec{r}$$ coordinates, not the primed $$\vec{r'}$$ coordinates. So the first term in the above is zero. So then we have $$\vec{B(\vec{r})}=\frac{-\mu_o}{4\pi}\int \vec{J(r')}\times (\nabla \frac{1}{R})dV'$$ Now using the identity $$\nabla \frac{1}{R}=\nabla \frac{1}{|\vec{r}-\vec{r'}|}=-\frac{\hat{R}}{R^2}$$ , we get that $$\vec{B(\vec{r})}=\frac{\mu_o}{4\pi}\int\frac{\vec{J(\vec{r}')\times \hat{R}}}{R^2}dV'$$ which is precisely the Biot-Savart Law as required