I've been trying to follow this post on deriving the Biot-Savart law from Maxwell's Equations but am getting stuck on this step:

$$-\frac{\mu_0}{4\pi}\iiint{\nabla\times\frac{J}{|r-r'|} d^3r'}.$$

Mainly the curl of current density part which would expand to:

$$\nabla\times\frac{J}{|r-r'|}=\frac{\nabla\times J}{|r-r'|}+\nabla\frac{1}{|r-r'|}\times J$$

The only way this would reduce to $\frac{|r-r'|\times J}{|r-r'|^3}$ would be if $\nabla\times J=0$.

So I tried showing that $\nabla\times J = 0$ in a steady-state situation, however, I'm getting stuck.

Assumption: Steady-state current which implies that for all of space and time $\frac{\partial B}{\partial t}=0$

We start with Ampere's law,

$$\nabla\times B=\mu_0J+\mu_0\epsilon_0\frac{\partial E}{\partial t}$$

Curl of both sides

$$-\nabla^2B=\mu_0\nabla\times J+\mu_0\epsilon_0\frac{\partial(\nabla\times E)}{\partial t}$$

And because $\nabla\times E = -\frac{\partial B}{\partial t}$

$$-\nabla^2B=\mu_0\nabla\times J-\mu_0\epsilon_0\frac{\partial^2 B}{\partial t^2}$$

$$\mu_0\epsilon_0\frac{\partial^2 B}{\partial t^2}=\mu_0\nabla\times J+\nabla^2B$$

However, this is where I'm getting stuck, there are two scenarios for $\partial^2 B / \partial t^2 = 0$: Either $\nabla^2 B$ is zero meaning that $\nabla\times J$ would also be zero, or that $\nabla^2 B$ is non-zero meaning that $\nabla\times J=-\nabla^2 B$.

So now I would need to show that $\nabla^2 B=0$ which I'm not sure is true (For example, wouldn't it be $\infty$ at the wire - assuming thin wire?)

So my question is: is $\nabla\times J=0$ in steady-state? If so, where am I going wrong in my proof?


1 Answer 1


This is a consequence of bad notation.

In writing the formula:

$$\nabla × \vec{A} = \vec{B}$$

You need to be very careful about which variables you are taking the curl of.

You should write:

$$\nabla_{\vec{r}} × \vec{A}(\vec{r}) = \vec{B}(\vec{r})$$

This is because we are taking the curl with respect to the variables $\vec{r} = x\hat i + y\hat j + z\hat k$

When we apply the $\nabla_{\vec{r}} ×$ operator on $\vec{J}$ we need to keep track of what $\vec{J}$ is a function of.

Your confusion lies in the fact that yes, in maxwells equations, $\vec{J}$ and $\vec{B}$ are both functions of $\vec{r}$.

However, when solving poissons equation, $\vec{J}(\vec{r})$ changes to $\vec{J}(\vec{r}')$


$\nabla_{\vec{r}} × \vec{J}(\vec{r}') = 0$

As $\vec{J}(\vec{r}')$ is independant on $\vec{r}$

So your conclusions you've drawn are presumably correct if $\nabla_{\vec{r}'} × \vec{J}(\vec{r}') = 0$, but that's not what we are saying when deriving biot savart.

  • $\begingroup$ Oh wow, I'm a newbie to this whole area of math and physics, and I keep making dumb mistakes like this, thanks so much for your help :) So curl of J is zero in the Biot-Savart law derivation because the curl with respect to r on a function of r' is zero. Math is hard ;-; $\endgroup$
    – nreh
    Commented Jun 4, 2022 at 12:47
  • $\begingroup$ Yup, which is why notation is very important! The curl is basically a fancy derivative, d/dx (x') = 0 as x' is a constant as far as x is concerned! $\endgroup$ Commented Jun 4, 2022 at 12:52
  • $\begingroup$ Hey I ran through the problem again and now I'm confused about something else, I used the rule for 'curl of a scalar field times a vector field' to expand $\nabla_r \times J(r')/|r'|$. However, because $J(r')$ is a constant as far r is concerned, it would become a scalar field times a constant vector. Wouldn't this make $\nabla_r \times J(r')/|r'|=0$? Which isn't done by the other derivations I've seen. $\endgroup$
    – nreh
    Commented Jun 4, 2022 at 14:40
  • $\begingroup$ I'm not sure I follow.you are forgetting the expression we are taking the curl of has 1/|r-r'| in the denominator, which IS a function of r, and not 1/|r'| $\endgroup$ Commented Jun 4, 2022 at 14:51
  • $\begingroup$ Sorry yeah you're right, I got the equation wrong. Thanks a lot for your help. $\endgroup$
    – nreh
    Commented Jun 4, 2022 at 15:01

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