# [Physics] Derivation of Ampere’s Law from Biot-Savart

electric-fieldselectromagnetismmagnetic fieldsmagnetostaticsmaxwell-equations

Our aim is to derive $$\nabla\times \mathbf B=\mu_0(\mathbf J+\epsilon_0\frac{\partial E}{\partial t})$$.

To begin with,
let $$\mathbf A=\frac{\mu_0}{4\pi}\int_{\mathbb R^3}\frac{\mathbf J(\mathbf r')}{|\mathbf r-\mathbf r'|}dV'$$. Then
$$\nabla\times \mathbf A=\mathbf B(\mathbf r)= \int_{\mathbb R^3}\frac{\mathbf J(\mathbf r')}{|\mathbf r-\mathbf r'|^3}\times (\mathbf r-\mathbf r')dV',\\ \nabla^2 \mathbf A(\mathbf r)=-\mu_0 \mathbf J.$$
Therefore, using the identity, $$\nabla\times(\nabla\times \mathbf A)=\nabla(\nabla\cdot \mathbf A)-\nabla^2 \mathbf A$$,
$$\nabla\times \mathbf B=\frac{\mu_0}{4\pi}\nabla \left[\int \mathbf J(\mathbf r')\cdot \nabla\left(\frac{1}{|\mathbf{r-r'}|}\right)dV' \right]+\mu_0\mathbf J(\mathbf r).$$
Therefore, we have to prove that
$$\nabla \left[\int \mathbf J(\mathbf r')\cdot \nabla\left(\frac{1}{|\mathbf{r-r'}|}\right)dV' \right]=4\pi\epsilon_0 \frac{\partial \mathbf E}{\partial t}.$$
If we let $$\mathbf E=-\nabla \phi$$, then we only need to prove that
$$\int \mathbf J(\mathbf r')\cdot \nabla\left(\frac{1}{|\mathbf{r-r'}|}\right)dV'=-4\pi \epsilon_0\frac{\partial \phi}{\partial t}.$$

But I don't know how to handle this at all – I have not experienced $$\frac{\partial \phi}{\partial t}$$ before.

How can I continue?

EDIT: I can vaguely feel what's going on:
$$\frac{1}{4\pi \epsilon_0}\int \mathbf J(\mathbf r')\cdot \nabla\left(\frac{1}{|\mathbf{r-r'}|}\right)dV'\\ =\frac{1}{4\pi \epsilon_0} \int \mathbf \rho \mathbf v \cdot \left(\frac{\mathbf{r-r'}}{|\mathbf{r-r'}|^3}\right)dV'\\=\int \mathbf{F(\mathbf r',\mathbf r)\cdot v(\mathbf r')} dV',$$
where $$\mathbf{F}(\mathbf r',\mathbf r)$$ is the electric force $$r$$ exerts on $$r'$$. So the integral represents the rate of work done on other part of the electric field by $$r$$, hence it equals to the rate of decrease on potential at $$r$$.

But how can I make this more clear and formal?

The Biot-Savart law says that under magnetostatic conditions ($$\frac{\partial}{\partial t}\rightarrow 0$$),

$$\mathbf B(\mathbf r) = \frac{\mu_0}{4\pi}\int \frac{\mathbf J(\mathbf r') \times (\mathbf r - \mathbf r')}{|\mathbf r - \mathbf r'|^3} dV'$$

Noting that $$\frac{\mathbf r - \mathbf r'}{|\mathbf r - \mathbf r'|^3} = \nabla \left(\frac{1}{|\mathbf r - \mathbf r'|}\right)$$ where $$\nabla$$ refers to differentiation by the unprimed coordinates, this can be written

$$\mathbf B(\mathbf r) = \nabla \times\frac{\mu_0}{4\pi} \int \frac{\mathbf J(\mathbf r')}{|\mathbf r - \mathbf r'|}dV'$$

Taking the curl of this and using the fact that $$\nabla \times (\nabla \times \mathbf F) = \nabla(\nabla \cdot \mathbf F) - \nabla^2 \mathbf F$$,

$$\nabla \times \mathbf B(\mathbf r) = \nabla\left(\int J(\mathbf r')\cdot \nabla\left[\frac{1}{|\mathbf r - \mathbf r'|}\right]dV'\right) - \nabla^2 \frac{\mu_0}{4\pi} \int \frac{\mathbf J(\mathbf r')}{|\mathbf r - \mathbf r'|}dV'$$

Noting that $$\nabla\left[\frac{1}{|\mathbf r - \mathbf r'|}\right] = -\nabla'\left[\frac{1}{|\mathbf r - \mathbf r'|}\right]$$

we can integrate the first term by parts to obtain

$$\nabla\left(\int \nabla' \cdot \left[\frac{\mathbf J(\mathbf r')}{|\mathbf r - \mathbf r'|}\right]dV' - \int\frac{\nabla' \cdot \mathbf J(\mathbf r')}{|\mathbf r - \mathbf r'|} dV'\right)$$

The first term is a surface term, and vanishes if we assume that $$\mathbf J(\mathbf r') \rightarrow 0$$ as $$|\mathbf r'| \rightarrow \infty$$. The second term vanishes because according to the continuity equation, $$\nabla\cdot\mathbf J = -\frac{\partial \rho}{\partial t} = 0$$ in magnetostatics. This leaves us with

$$\nabla \times \mathbf B(\mathbf r) = \nabla^2 \frac{\mu_0}{4\pi} \int \frac{\mathbf J(\mathbf r')}{|\mathbf r - \mathbf r'|}dV'$$

and since

$$\nabla^2 \frac{1}{|\mathbf r - \mathbf r'|} = 4\pi \delta^{(3)}(\mathbf r - \mathbf r')$$

we have

$$\nabla \times \mathbf B = \mu_0 \mathbf J$$.

Again, Biot-Savart is valid only under magnetostatic conditions, and therefore so is this version of Ampere's law. It would be nice to relax these conditions and re-do this derivation more generally, but we don't yet know what to replace Biot-Savart with.

Instead, let's see how this version of Ampere's law fails when we go to general electrodynamics. Clearly since $$\mathbf J \propto \nabla \times \mathbf B$$, we have that $$\nabla \cdot \mathbf J = 0$$. However, according to the general continuity equation, $$\nabla \cdot \mathbf J = -\frac{\partial \rho}{\partial t}$$.

To fix this, let's assume that we need a new term, so

$$\nabla \times \mathbf B = \mu_0 \mathbf J + \mathbf G$$

for some vector field $$\mathbf G$$. Taking the divergence of both sides yields

$$0 = - \mu_0 \frac{\partial \rho}{\partial t} + \nabla \cdot \mathbf G$$

$$\implies \nabla \cdot \mathbf G = \mu_0 \frac{\partial \rho}{\partial t}$$

From Gauss' law for electric fields, we know that $$\rho = \epsilon_0 \nabla \cdot \mathbf E$$, and so

$$\nabla \cdot \mathbf G = \epsilon_0 \mu_0 \frac{\partial}{\partial t} \nabla \cdot \mathbf E = \nabla \cdot \left(\epsilon_0 \mu_0 \frac{\partial}{\partial t}\mathbf E\right)$$

and so we can simply postulate that

$$\mathbf G = \epsilon_0\mu_0 \frac{\partial}{\partial t} \mathbf E$$

so

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

This was Maxwell's correction to Ampere's law, and it has been validated over and over by experiment.

In summary, magnetostatics + Biot-Savart gives us $$\nabla \times \mathbf B = \mu_0 \mathbf J$$. Predictably, this fails when we leave the domain of magnetostatics, and in particular is inconsistent with the continuity equation. We don't know how to generalize Biot-Savart, but patching up the problem with the continuity equation in the simplest possible way yields the correct Ampere's law, $$\nabla \times \mathbf B = \mu_0 \mathbf J + \epsilon_0 \mu_0 \frac{\partial}{\partial t}\mathbf E$$.

From this, we can work backward to find the correct generalization of Biot-Savart; this is one of Jefimenko's equations.

EDIT:

Returning to the original derivation after eliminating the surface term (but before sending $$\nabla'\cdot \mathbf J(\mathbf r')\rightarrow 0$$), we have

$$\nabla \times \mathbf B(\mathbf r) = \mu_0 \mathbf J(\mathbf r) - \frac{\mu_0}{4\pi}\int \frac{\nabla '\cdot \mathbf J(\mathbf r')}{|\mathbf r - \mathbf r'|}$$

Under the conditions for which Biot-Savart holds, the latter term is equal to zero; however, we can be bold and throw those restrictions aside just to see what happens. Under general conditions, $$\nabla \cdot \mathbf J = -\frac{\partial \rho}{\partial t}$$, so that term becomes $$\frac{\mu_0}{4\pi} \nabla \int \frac{\partial \rho}{\partial t} \frac{1}{|\mathbf r - \mathbf r'|} dV'= \frac{\partial}{\partial t} \frac{\mu_0}{4\pi} \nabla \int\frac{\rho(\mathbf r')}{|\mathbf r - \mathbf r'|}$$

Defining $$\phi(\mathbf r) = \int \frac{\rho(\mathbf r')}{4\pi \epsilon_0 |\mathbf r-\mathbf r'|}$$ and letting $$\mathbf E = -\nabla\phi$$, this becomes

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

What we did here - simply disregarding the conditions under which Biot-Savart is applicable and plugging in the more general continuity equation - is morally the same as Maxwell's addition of the extra term to compensate for the nonzero divergence of $$\mathbf J$$.

Note also that we have glossed over how to go from magnetostatics to electrodynamics $$\big(\mathbf J(\mathbf r) \rightarrow \mathbf J(\mathbf r,t), \rho(\mathbf r)\rightarrow \mathbf \rho(\mathbf r,t)\big)$$. Simply plugging in a $$t$$ to Biot-Savart and letting it "go along for the ride" is insufficient; working backwards from the full Maxwell's equations demonstrates the need to introduce the retarded time $$t_r = t - \frac{|\mathbf r - \mathbf r'|}{c}$$, indicating that Biot-Savart is genuinely wrong for electrodynamics.