# Biot-Savart law from Ampère's (with multivariate calculus)

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1. Jan 24, 2016

### DavideGenoa

Let us assume the validity of Ampère's circuital law$$\oint_{\gamma}\mathbf{B}\cdot d\mathbf{x}=\mu_0 I_{\text{linked}}$$where $\mathbf{B}$ is the magnetic field, $\gamma$ a closed path linking the current of intensity $I_{\text{linked}}$.

All the derivations of the Biot-Savart law for a tridimensional distribution of current $\mathbf{B}=\frac{\mu_0}{4\pi}\int_V \frac{\mathbf{J}\times\hat{\mathbf{r}}}{r^2}dv$ that I have been able to find on line and on cartaceous resources use Dirac's $\delta$.

Can the Biot-Savart law, at least in the case of a linear distribution of current $\mathbf{B}=\frac{\mu_0}{4\pi}\oint\frac{Id\boldsymbol{\ell}\times\hat{\mathbf{r}}}{r^2}$ or, explicitly, $$\mathbf{B}(\mathbf{x})=\frac{\mu_0}{4\pi}\int_a^b I\boldsymbol{\ell}'(t)\times\frac{\mathbf{x}-\boldsymbol{\ell}(t)}{\|\mathbf{x}-\boldsymbol{\ell}(t)\|^3}dt$$ where $\boldsymbol{\ell}:[a,b]\to\mathbb{R}^3$ is a parametrisation of a closed (or infinite) wire carrying the current $I$, be inferred without using Dirac's $\delta$, only by using the tools of multivariate calculus and elementary differential geometry, if we assume the validity of Ampère law (at least assuming the validity of the Gauss law for magnetism or other of the Maxwell equations)?

2. Jan 25, 2016

### vanhees71

Of course, all derivations are equivalent to using the Dirac-$\delta$ distribution in the definition of the Green's function of the Laplace operator.

The fundamental equations for magnetostatics are the two Maxwell equations (in Heaviside-Lorentz units)
$\vec{\nabla} \times \vec{B}=0, \quad \vec{\nabla} \times \vec{B}=\frac{1}{c} \vec{j}.$
Now there are two ways to proceed

(a) Use the first equation to introduce the vector potential (introducing gauge dependence) via
$$\vec{B}=\vec{\nabla} \times \vec{A}.$$
Then the 2nd equation tells you (in Cartesian coordinates)
$$\vec{\nabla} \times (\vec{\nabla} \times \vec{A})=\vec{\nabla} (\vec{\nabla} \cdot \vec{A})-\Delta \vec{A}=\frac{1}{c} \vec{j}.$$
Now the gauge freedom allows for one subsidiary condition. In magnetostatics the Coulomb gauge is most convenient, i.e.,
$$\vec{\nabla} \cdot \vec{A}=0,$$
and then the Cartesian components decouple and are simply given by three independent Poisson equations,
$$-\Delta \vec{A}=\frac{1}{c} \vec{j}.$$
Now you can argue in a very simple way, because that's nothing as electrostatics for each of the components, for which you know the solution. It's just adding up Coulomb fields of point charges $\mathrm{d} q=\mathrm{d}^3 \vec{x} \rho$. Translated to the three components of the current, this leads immideately to
$$\vec{A}(\vec{x})=\frac{1}{4 \pi c} \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x}' \frac{\vec{j}(\vec{x}')}{|\vec{x}-\vec{x}'|}.$$
It's very important to note that the Coulomb-gauge condition is fulfilled, because of the continuity equation, which boils down in the static case to
$$\vec{\nabla} \cdot \vec{j}=0.$$
The Biot-Savart Law then follows from taking the curl of this solution.

(b) You don't introduce the vector potential, but take the curl of Ampere's Law and use Gauss's Law for the magnetic field:
$$\vec{\nabla} \times (\vec{\nabla} \times \vec{B})=\vec{\nabla} (\vec{\nabla} \cdot \vec{B})-\Delta \vec{B}=-\Delta \vec{B}=\frac{1}{c} \vec{\nabla} \times \vec{j}.$$
Now you get by the same argument as before directly
$$\vec{B}(\vec{x})=\frac{1}{4 \pi c} \int_{\mathbb{R}} \mathrm{d}^3 \vec{x}' \frac{\vec{\nabla}' \times \vec{j}(\vec{x}')}{|\vec{x}-\vec{x}'|},$$
and you find the usual Biot-Savart formula using integration by parts.

Of course you can use the analogous argument also for surface and line currents.

Mathematically the solution of the Poisson equation, of course involves the Green's function of the Laplace operator,
$$\Delta G(\vec{x},\vec{x}')=-\delta(\vec{x}-\vec{x}'),$$
which is quite easily solved by
$$G(\vec{x},\vec{x}')=\frac{1}{4 \pi |\vec{x}-\vec{x}'|},$$
using spherical coordinates around $\vec{x}'$ and radial symmetry of the problem.

3. Jan 26, 2016

### DavideGenoa

Thank you so much vanhees71! So, it seems that it is not possible to prove the Biot-Savart law without using functional analysis and the $\delta$ "function"...

4. Jan 27, 2016

Hi DavideGenoa,
Biot-Savart's law is often taught separate from Ampere's law, and Ampere's law in differential form is often converted to an integral form with B by using Stoke's theorem so that curl B=uo*J (M.K.S. units) when integrated over an area becomes (with Stokes theorem) integral B*dl (around a closed loop)=uo*I where "I" is the current through the loop. There is an alternative way to proceed when working with Ampere's/Maxwell's curl equation: It often isn't presented in the textbooks in such a fashion but I recently (about two years ago) made the observation that the integral solution of Ampere's curl equation is in fact Biot-Savart's law. i.e. the solution of the curl equation is in general found by putting it into an integral of the Biot-Savart form. (A similar type of integral solution exists with the divergence operator: div E=rho(x)/eo (M.K.S. units) ==>> E(x)=(1/(4*pi*eo))*integral((rho(x')(x-x')/|(x-x')|^3)d^3 x' (basically the inverse square law=Coulomb's law).) Quite useful solutions that seem to often be overlooked in the curriculum.

5. Feb 29, 2016

### DavideGenoa

I hope I have been able to reach a proof, with the essential help of Hawkeye18, whom I heartily thank again.

Let $\mathbf{J}:\mathbb{R}^3\to \mathbb{R}^3$ be of class $C^2 (\mathbb{R}^3)$ and let its support be contained in the bounded compact subset $V\subset\mathbb{R}^3$, $\mu$-measurable according to the usual Lebesgue measure $\mu$ defined in $\mathbb{R}^3$.
Let us define $$\mathbf{A}(\mathbf{x}):=\frac{\mu_0}{4\pi}\int_{V}\frac{\mathbf{J}(\mathbf{l})}{\|\mathbf{x}-\mathbf{l}\|}d\mu_{\mathbf{l}}=\frac{\mu_0}{4\pi}\int_{\mathbb{R}^3}\frac{\mathbf{J}(\mathbf{l})}{\|\mathbf{x}-\mathbf{l}\|}d\mu_{\mathbf{l}}$$where we notice that the point $\mathbf{l}=\mathbf{x}$ does not prevent the summability of $\mathbf{l}\mapsto \|\mathbf{x}-\mathbf{l}\|^{-1}$ on any bounded measurable subset of $\mathbb{R}^3$, as we can verify by using spherical coordinates around $\mathbf{x}$ and by taking the properties of the Lebesgue integral and its relationship with the Riemann integral into account.
I would definitely say that $\frac{\partial\mathbf{J}(\mathbf{y}_0+\mathbf{x}_0)}{\partial x_k}=\frac{\partial\mathbf{J}(\mathbf{y}_0+\mathbf{x}_0)}{\partial y_k}$and analogously with partial derivatives of $\mathbf{J}$ in place of $\mathbf{J}$. By using this result, I would say that, for any $m_1,m_2, m_3\in\mathbb{N}\cup\{0\}$ such that $m_1+m_2+m_3=2$, $$\frac{\mu_0}{4\pi}\frac{\partial^{m_1+m_2+m_3}}{\partial x_1^{m_1}\partial x_2^{m_2}\partial x_3^{m_3}}\int_{\mathbb{R}^3}\frac{\mathbf{J}(\mathbf{l})}{\|\mathbf{x}-\mathbf{l}\|}\,d\mu_{\mathbf{l}}=\int_{\mathbb{R}^3}\frac{\mu_0}{4\pi\|\mathbf{y}\|}\frac{\partial^{m_1+m_2+m_3}\mathbf{J}(\mathbf{y}+\mathbf{x})}{\partial y_1^{m_1}\partial y_2^{m_2}\partial y_3^{m_3}}\,d\mu_{\mathbf{y}}.$$

Therefore, if we define $\mathbf{B}(\mathbf{x}):=\frac{\mu_0}{4\pi}\int_V \mathbf{J}(\mathbf{l})\times\frac{\mathbf{x}-\mathbf{l}}{\|\mathbf{x}-\mathbf{l}\|^3}d\mu_{\mathbf{l}}=\frac{\mu_0}{4\pi}\int_{\mathbb{R}^3}\nabla_x\times\left[\frac{\mathbf{J}(\mathbf{l})}{\|\mathbf{x}-\mathbf{l}\|}\right]d\mu_{\mathbf{l}}$, I would say that (also under less strict assumptions on $\mathbf{J}$, as Hawkeye18 very brilliantly explains here) $\nabla\times\mathbf{A}=\mathbf{B}$ and

$$\nabla_x\times\mathbf{B}(\mathbf{x})=\frac{\mu_0}{4\pi}\nabla_x\times\left[ \nabla_x\times\left[ \int_{\mathbb{R}^3}\frac{\mathbf{J}(\mathbf{l})}{\|\mathbf{x}-\mathbf{l}\|}d\mu_{\mathbf{l}}\right]\right]=\frac{\mu_0}{4\pi}\nabla_x\times\left[ \nabla_x\times\left[ \int_{\mathbb{R}^3}\frac{\mathbf{J}(\mathbf{y}+\mathbf{x})}{\|\mathbf{y}\|}d\mu_{\mathbf{y}}\right]\right]$$ $$=\frac{\mu_0}{4\pi} \int_{\mathbb{R}^3}\nabla_x\times\left[ \nabla_x\times\left[\frac{\mathbf{J}(\mathbf{y}+\mathbf{x})}{\|\mathbf{y}\|}\right]\right]d\mu_{\mathbf{y}}=\frac{\mu_0}{4\pi} \int_{\mathbb{R}^3}\frac{\nabla_y\times[ \nabla_y\times[\mathbf{J}(\mathbf{y}+\mathbf{x})]]} {\|\mathbf{y}\|}d\mu_{\mathbf{y}}$$so, by using a known identity for the curl of the curl,$$\nabla_x\times\mathbf{B}(\mathbf{x})=\frac{\mu_0}{4\pi} \int_{\mathbb{R}^3}\frac{\nabla_y[\nabla_y\cdot\mathbf{J}(\mathbf{y}+\mathbf{x})]} {\|\mathbf{y}\|}d\mu_{\mathbf{y}}-\frac{\mu_0}{4\pi} \int_{\mathbb{R}^3}\frac{\nabla_y^2\mathbf{J}(\mathbf{y}+\mathbf{x})} {\|\mathbf{y}\|}d\mu_{\mathbf{y}}$$$$=\frac{\mu_0}{4\pi} \int_{\mathbb{R}^3}\frac{\nabla_x[\nabla_y\cdot\mathbf{J}(\mathbf{y}+\mathbf{x})]} {\|\mathbf{y}\|}d\mu_{\mathbf{y}}-\frac{\mu_0}{4\pi} \int_{\mathbb{R}^3}\frac{\nabla_y^2\mathbf{J}(\mathbf{y}+\mathbf{x})} {\|\mathbf{y}\|}d\mu_{\mathbf{y}}$$$$=\frac{\mu_0}{4\pi} \nabla_x\int_{\mathbb{R}^3}\frac{\nabla_y\cdot\mathbf{J}(\mathbf{y}+\mathbf{x})} {\|\mathbf{y}\|}d\mu_{\mathbf{y}}-\frac{\mu_0}{4\pi} \int_{\mathbb{R}^3}\frac{\nabla_y^2\mathbf{J}(\mathbf{y}+\mathbf{x})} {\|\mathbf{y}\|}d\mu_{\mathbf{y}}$$$$=\frac{\mu_0}{4\pi} \nabla_x\int_{\mathbb{R}^3}\frac{\nabla_y\cdot\mathbf{J}(\mathbf{y})} {\|\mathbf{y}-\mathbf{x}\|}d\mu_{\mathbf{y}}-\frac{\mu_0}{4\pi} \int_{\mathbb{R}^3}\frac{\nabla_y^2\mathbf{J}(\mathbf{y})} {\|\mathbf{y}-\mathbf{x}\|}d\mu_{\mathbf{y}}$$where the first addendum is null if $\forall \mathbf{x}\in\mathbb{R}^3\quad\nabla_x\cdot\mathbf{J}(\mathbf{x})=0$, which holds for a stationary current, and where the second addendum really is $$-\frac{\mu_0}{4\pi} \int_{\mathbb{R}^3}\frac{\nabla_y^2\mathbf{J}(\mathbf{y})} {\|\mathbf{x}-\mathbf{y}\|}d\mu_{\mathbf{y}}=\mu_0\mathbf{J}(\mathbf{x})=:\mu_0\int\delta(\mathbf{y}-\mathbf{x})\mathbf{J}(\mathbf{y})d^3y$$this time.