# Conditions on expressing magnetic field in terms of curl of current density

Given a current density distribution $\mathbf J(\mathbf x)$ inside a closed bounded region $\Omega$, the magnetic field at any point $\mathbf y$ outside of $\Omega$ can be expressed as \begin{aligned}\mathbf B(\mathbf y)&=\frac{\mu_0}{4\pi}\int_\Omega\mathbf J(\mathbf x)\times\nabla_{\mathbf x}\frac{1}{|\mathbf x-\mathbf y|}d^3\mathbf x\\ &=\frac{\mu_0}{4\pi}\int_\Omega\left[\frac{1}{|\mathbf x-\mathbf y|}\nabla_{\mathbf x}\times\mathbf J(\mathbf x)-\nabla_{\mathbf x}\times\left(\frac{\mathbf J(\mathbf x)}{|\mathbf x-\mathbf y|}\right)\right]d^3\mathbf x\\ &=\frac{\mu_0}{4\pi}\int_\Omega\frac{1}{|\mathbf x-\mathbf y|}\nabla_{\mathbf x}\times\mathbf J(\mathbf x)d^3\mathbf x-\frac{\mu_0}{4\pi}\int_{\partial\Omega}\mathbf n(\mathbf x)\times\left(\frac{\mathbf J(\mathbf x)}{|\mathbf x-\mathbf y|}\right)d^2 S(\mathbf x) \end{aligned} where $\partial\Omega$ is the boundary of $\Omega$, $n(\mathbf x)$ is the unit normal of $\partial \Omega$ and $S(\mathbf x)$ is the area of the surface element. Now, if the current density $\mathbf J(\mathbf x)$ is zero at the boundary $\partial\Omega$ (this can be achieved by slightly enlarging $\Omega$ if $\mathbf J(\mathbf x)$ is not zero at $\partial\Omega$) we can then drop the second term on the last line. Now we simply have \begin{aligned}\mathbf B(\mathbf y)&=\frac{\mu_0}{4\pi}\int_\Omega\frac{1}{|\mathbf x-\mathbf y|}\nabla_{\mathbf x}\times\mathbf J(\mathbf x)d^3\mathbf x \end{aligned}.

If the current density $\mathbf J(\mathbf x)$ is continuous and differentiable, the above conclusion should be correct. However, $\mathbf J(\mathbf x)$ might not be continuous in $\Omega$, e.g., infinite thin coils inside $\Omega$ carrying electrical current. Is the above derivation correct for $\mathbf J(\mathbf x)$ containing delta functions? What kind of singularities in $\mathbf J(\mathbf x)$ is permitted?

• I think that the above is always true, simply because the the definition of the derivative of a distribution (such as a delta-function or a step function, which is how we describe the current configurations you're talking about) is done via a similar formula. The Mathworld article on distributions (aka "generalized functions") might be worth a look on your part. – Michael Seifert Oct 6 '15 at 23:47
• Thanks for bringing the reference. As you said, it is correct even if $\mathbf J$ contains delta functions, since it can be verified that $\int_\Omega f(\mathbf x)\nabla\delta(\mathbf x-\mathbf x_0)d^3\mathbf x=-\nabla f(\mathbf x)|_{\mathbf x=\mathbf x_0}$ for $\mathbf x_0$ in the interior of $\Omega$ for any differentiable function $f(\mathbf x)$. – Jasper Oct 7 '15 at 18:32

• Can you give an example of singularity of current? The current is considered as a result of charge movement. As long as you can still talking about the continuous movements of charges, $\mathbf{J}$ may always be treated as continuous. Unless you are thinking of quantum processes when sudden jumps happen at the microscopic level and quantum electrodynamics should serve your purpose, otherwise I think you are safe. – Xiaodong Qi Oct 6 '15 at 21:44
• Sure. Let's assume $\mathbf J(\mathbf x)$ is a segment of line current, e.g., $\mathbf J(\mathbf x)=I_c\int_{\tau_1}^{\tau_2}\delta(\mathbf x-\mathbf x^\prime(\tau))\frac{d\mathbf x^\prime(\tau)}{d\tau}d\tau$, where $\delta(\mathbf x-\mathbf x^\prime)$ is the Dirac delta function, $I_c$ is the amplitude of the current, $\mathbf x^\prime(\tau)\in c\in \mathbb R^3$ is the parametric representation of the line segment $c$ with parameter $\tau\in[\tau_1,\tau_2]$. – Jasper Oct 6 '15 at 21:50
• In your case, the integral $\int_{t_1}^{t_2}\delta(x-x'(\tau))\frac{dx'}{d\tau}d\tau=\int_{x'(t_1)}^{x'(t_2)}\delta(x-x'(\tau))dx'=\mathrm{constant}$. Is that correct? – Xiaodong Qi Oct 6 '15 at 22:02