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!

  • 1
    $\begingroup$ Possible duplicate: link. $\endgroup$
    – secavara
    Commented May 15, 2021 at 13:38
  • $\begingroup$ Instead of (\dfrac12) =$(\dfrac12)$ use \left(\dfrac12\right)=$\left(\dfrac12\right)$ to adjust to the height of the content. $\endgroup$
    – Frobenius
    Commented May 15, 2021 at 19:19

1 Answer 1


Figured it out.

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.