As an exercise, I've been trying to derive the Biot-Savart law from the second set of Maxwell's equations for steady-state current


I've been able to do this using the fact that an incompressible field has a vector potential $\mathbf{A}$, allowing me to rewrite the second equation as


which can be solved by components using the Green's function for the Laplacian, yielding


and since $\nabla\times\left(\psi\mathbf{J}\right)=\psi\nabla\times\mathbf{J}+\nabla\psi\times\mathbf{J}$, $$\nabla\times\mathbf{A}=\mathbf{B}(\mathbf{x})=\frac{\mu_0}{4\pi}\int\frac{\mathbf{J}\times(\mathbf{x}-\mathbf{x}')}{|\mathbf{x}-\mathbf{x}'|^3}\,d^{3}\mathbf{x}'$$

as desired. However, if instead I take the curl of both sides of Ampere's Law, and use the identity $\nabla \times \left( \nabla \times \mathbf{B} \right) = \nabla(\nabla \cdot \mathbf{B}) - \nabla^{2}\mathbf{B}$ initially, I find that

$$\nabla(\nabla \cdot \mathbf{B}) - \nabla^{2}\mathbf{B}=0-\nabla^2\mathbf{B}=\mu_0\nabla\times\mathbf{J}$$

which I can again solve like Poisson's equation, yielding


which can be simplified using the identity $\psi(\nabla\times\mathbf{J})=-\nabla\psi\times\mathbf{J}+\nabla\times\left(\psi\mathbf{J}\right)$, giving


The first integral is precisely the Biot Savart law, but I have no idea how to make the second integral vanish. I've exhausted any obvious vector calculus identities, and Stokes theorem doesn't help much. 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.

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    $\begingroup$ In your 2nd equation a minus sign is missing $$ \nabla^2\mathbf{A}=\boldsymbol{-}\mu_0\mathbf{J} \tag{01} $$ since $$ \boldsymbol{\nabla}\boldsymbol{\times}\mathbf{B}= \boldsymbol{\nabla}\boldsymbol{\times}(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf{A})=\boldsymbol{\nabla} \overbrace{(\nabla\boldsymbol{\cdot}\mathbf{A})}^{=0}-\nabla^{2}\mathbf{A}=-\nabla^{2}\mathbf{A} \tag{02} $$ $\endgroup$
    – Frobenius
    Jan 3, 2017 at 8:49
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    $\begingroup$ @AHB your remark is wrong. Maxwell's equation are the foundation of electrodynamics. They are fundamental, in the sense that all of electrodynamics can be derived from them (together with Lorentz force law). There is nothing "more fundamental" than them $\endgroup$
    – tomph
    Jan 3, 2017 at 14:24
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    $\begingroup$ @AHB: Neither law is derived, or, if you prefer, they both can be derived from each other. But you can't start a textbook deriving Biot-Savart's law; it's an experimental result. $\endgroup$
    – Javier
    Jan 3, 2017 at 14:35
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    $\begingroup$ @tomph, thanks for the clarification. I'm not a native speaker, I misunderstood your post. $\endgroup$ Jan 3, 2017 at 23:41
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    $\begingroup$ Hey @tomph! You know what I mean. Lorentz force law leads to Biot Savarat law. Then we come to Maxwell equations. It's the matter of sequence. One doesn't start form Maxwell equations to find the more basic laws. Ah. $\endgroup$
    – AHB
    Jan 4, 2017 at 6:11

2 Answers 2


As far as I can remember, the formula you obtain is right. You can make this "problematic" integral disappear by using the following identity, that we will call "curl theorem" :

$$\int\vec{\nabla}\times\vec{w}dV = -\int\vec{w}\times d\vec{S}$$

To show this is true, we are going to use the divergence or Green-Ostrogradski theorem, namely

$$\int\vec{\nabla}\cdot \vec{v}dV = \int \vec{v}\cdot d\vec{S}$$

Since the divergence theorem is a scalar identity while the curl theorem is a vector identity, we are going to need three distinct vector fields that we are going to denote $\vec{v}_i$. Now, we would want $\vec{\nabla}\cdot\vec{v}_i = (\vec{\nabla}\times\vec{w})_i$ to deduce an identity on the curl. Writing that in tensor notation :

$$\partial^k(v_i)_k=\epsilon_{ikl}\partial^k w^l$$

As we can see, it is sufficient to take $(\vec{v}_i)_k = \epsilon_{ikl}w^l$ and the relation will be satisfied. So, for such a vector field we have $\vec{\nabla}\cdot\vec{v}_i = (\vec{\nabla}\times\vec{w})_i$.

Applying the divergence theorem to $\vec{v}_i$ : $$\int(\vec{\nabla}\times\vec{w})_idV = \int\vec{\nabla}\cdot\vec{v}_idV = \int\vec{v_i}\cdot d\vec{S} = \int (v_i)_k(d\vec{S})^k = \int\epsilon_{ikl}w^l(d\vec{S})^k = -\int(\vec{w}\times d\vec{S})_i$$

Thus giving a proof of the "curl theorem". Using it on your problematic integral : $$-\frac{\mu_0}{4\pi}\int\nabla'\times\left(\frac{\mathbf{J}(\mathbf{x}')}{|\mathbf{x}-\mathbf{x}'|}\right)\,d^3\mathbf{x}' = -\frac{\mu_0}{4\pi}\int\left(\frac{\mathbf{J}(\mathbf{x}')}{|\mathbf{x}-\mathbf{x}'|}\right)\,\times d\vec{S}'$$

Now, the volume integral is done on all of space, and provided you suppose that $\lim_{x'\rightarrow\infty}\frac{\vec{J}(x')}{|x-x'|} = 0$, it gives a 0 contribution. Why does this not add any crazy assumptions ?

For this limit to be non-zero, we must necessarily have that $|J(x)|$ tend to infinity. Indeed, suppose $J(x)$ is finite. Then, there is a constant $C$ such that $|J(x)|<C$. Then, $lim_{x'\rightarrow\infty}\frac{|J(x')|}{|x-x'|}<\lim_{x'\rightarrow\infty}\frac{C}{|x-x'|} = 0$. Thus, if we were to have this "extra" integral not vanish, we would be required to have an infinite current density at infinity, which seems to be not so physical.

Of course, all my derivation where done in the context of well-behaved functions. It won't work say for an infinitely small wire, as the current density becomes a distribution (using the dirac delta $\delta(x)$). I am not qualified enough to tackle this case rigorously, but I hope the explanation above gives an idea to why setting this integral to 0 is sensible.

  • $\begingroup$ Thanks for the detailed explanation. I tried something like this earlier, but found the requirement that $\mathbf{J}(\mathbf{x}')\rightarrow 0$ at infinity a bit suspect. An infinite wire or solenoid can obviously produce a physically meaningful magnetic field. Of course, unless we care about the field at infinity, $1/|\mathbf{x}-\mathbf{x}'|$ will go to zero anyway, so this isn't a concern. Still, this is a kind of complication I wouldn't have expected. $\endgroup$
    – JAustin
    Jan 3, 2017 at 16:05
  • $\begingroup$ Yes, that is what I meant by "looser conditions". It seems to me also a quite constraining condition, but maybe (I haven't looked into it, to be honest), if you require that your magnetic field be finite (which is, I think, a sound assumption) then this will impose some conditions of the form of $J(x)$. Given this conditions, maybe we do have that $\frac{\mathbf{J}(\mathbf{x}')}{|\mathbf{x}-\mathbf{x}'|}\rightarrow 0$ necessarily. Indeed, for this term not to vanish, $J(x)$ cannot be finite at $|x|\rightarrow \infty$. This implies an infinite current, which seems to be unphysical. I'll edit. $\endgroup$
    – Frotaur
    Jan 3, 2017 at 16:11
  • $\begingroup$ Your edit makes sense. I'm still a little concerned about the boundary behavior, since the simultaneous limit as both $\mathbf{x}$ and $\mathbf{x}'$ go to infinity is potentially indeterminate. $\endgroup$
    – JAustin
    Jan 3, 2017 at 22:04
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    $\begingroup$ This should not be a problem, consider this : the limits are not taken "simultaneously". Indeed, you first choose a point of space $x$ in which you want to evaluate the magnetic field. In this calculation, you have to take $x'$ going to $\infty$ in the boundary integral, while $x$ remains finite. So there should be no problem. Now, if you'd want to determine $B(x)$ for $|x|$ going to infinity, you should compute B as a function of a FINITE x, and THEN take $x\rightarrow\infty$. So, if you will, the $x'$ "limit" is always taken for a finite $x$, even for arbitrarily big $x$. $\endgroup$
    – Frotaur
    Jan 3, 2017 at 22:09

A first observation is that this is not particular to magnetism. The exact same thing happens if you try to find Coulomb's law for the electric field; you get a term like

$$\int \nabla' \frac{\rho(\mathbf{x}')}{|\mathbf{x}-\mathbf{x}'|}\ d^3 \mathbf{x}'$$

which should be zero. Well, there are no fancy vector calculus identities involved, just plain old fundamental theorem of calculus. To see this, let's look at your version. The integral is a vector, and each component has two terms because of the curl. Let's concentrate on the first term of the first component:

$$\int \partial'_2 \left( \frac{J_3(\mathbf{x}')}{|\mathbf{x}-\mathbf{x}'|} \right)\ d^3\mathbf{x}'$$

By Fubini's theorem (assuming sufficiently well behaved functions), we can integrate the three variables in any order. The $x_2'$ integration is trivial because the integrand is a total derivative, so the result is just the thing inside the parenthesis evaluated at $x_2' = \pm \infty$, which we typically assume to be zero. Therefore this term vanishes, and so do all the others because they are essentially the same.


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