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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?

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    $\begingroup$ Biot-Savart holds only for magnetostatic conditions, does it not? You may want to look at Jefimenko's equations instead. $\endgroup$ – J. Murray Aug 7 at 3:01
  • $\begingroup$ @J.Murray But how to derive Jefimenko? $\endgroup$ – Ma Joad Aug 7 at 3:30
  • $\begingroup$ In any derivation, you need to have a starting point. Historically, Maxwell understood that the magnetostatic Ampere's law ($\nabla \times \mathbf B = \mu_0 \mathbf J$) was insufficient, and added a correction term which has been endlessly validated by experiment since then. Are you asking about how he did this? $\endgroup$ – J. Murray Aug 7 at 3:50
  • $\begingroup$ @J.Murray Yes, exactly. I am just trying to make sense of the term $\partial E/\partial t$. $\endgroup$ – Ma Joad Aug 7 at 3:58
  • $\begingroup$ You might want to take a look at this Wikipedia article for an example of why Ampere's law without the displacement current term doesn't work. $\endgroup$ – Puk Aug 7 at 4:33
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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.

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  • $\begingroup$ So, is my formula for $\dot \phi$ above correct? $\endgroup$ – Ma Joad Aug 7 at 6:01
  • $\begingroup$ @MaJoad No. Note that for magnetostatic conditions, the right hand side is zero while the left is generically not. $\endgroup$ – J. Murray Aug 7 at 14:18
  • $\begingroup$ Thank you very much! $\endgroup$ – Ma Joad Aug 7 at 14:20
  • $\begingroup$ @MaJoad Please see my latest edit. $\endgroup$ – J. Murray Aug 7 at 14:54
  • $\begingroup$ Oh, nice. Thank you. $\endgroup$ – Ma Joad Aug 7 at 15:34
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Our aim is to derive $\nabla\times \mathbf B=\mu_0\left(\mathbf J+\epsilon_0\frac{\partial \mathbf E}{\partial t}\right)$.

You cannot derive the complete Ampere's law above (including $\epsilon_0\frac{\partial \mathbf{E}}{\partial t}$) from the Biot-Savart law $$ \mathbf A(\mathbf r)=\frac{\mu_0}{4\pi}\int\frac{\mathbf J(\mathbf r')}{|\mathbf r-\mathbf r'|}dV'.$$

You can only derive the incomplete Ampere's law (without $\epsilon_0 \frac{\partial \mathbf{E}}{\partial t}$) $$ \nabla\times \mathbf B=\mu_0\mathbf J $$ from it.


I am just trying to make sense of the term $\partial \mathbf{E}/\partial t$

The term $\partial \mathbf{E}/\partial t$ can best be motivated by following Maxwell's road.

Before Maxwell the conservation of charge was already a well-established experimental fact. Written as differential equation this is $$\frac{\partial\rho}{\partial t}+\mathbf{\nabla}\mathbf{J}=0. \tag{1}$$

Maxwell also knew these laws: $$\epsilon_0\mathbf{\nabla}\mathbf{E}=\rho \tag{M1}$$ $$\mathbf{\nabla}\mathbf{B}=0 \tag{M2}$$ $$\mathbf{\nabla}\times\mathbf{E}=-\frac{\partial}{\partial t}\mathbf{B} \tag{M3}$$ $$\frac{1}{\mu_0}\mathbf{\nabla}\times\mathbf{B}=\mathbf{J} \tag{M4}$$ All of these apparently were well-established experimental facts, too.

Now from (M1) Maxwell could derive $$ \frac{\partial\rho}{\partial t}=\epsilon_0\frac{\partial}{\partial t}\mathbf{\nabla}\mathbf{E}$$ and from (M4) $$\nabla{\mathbf{J}}=0.$$ Hence $$\frac{\partial\rho}{\partial t}+\mathbf{\nabla}\mathbf{J}\neq 0$$ which is obviously in contradiction to (1).

So one of the laws (1), (M1) and (M4) must be wrong. Nevertheless, all of them must be at least very good approximations, because all experiments done until his time didn't find any deviations.

Maxwell's solution was to modify equation (M4): $$\frac{1}{\mu_0}\mathbf{\nabla}\times\mathbf{B}=\mathbf{J}+\epsilon_0\frac{\partial}{\partial t}\mathbf{E} \tag{M4'}$$

This has two consequences:

  1. The additional term $\epsilon_0\frac{\partial}{\partial t}\mathbf{E}$ is extremely small because of the small value $\epsilon_0$. And therefore the unmodified equation (M4) is still a very good approximation, and the change is unmeasurable small in all experiments done so far with currents and magnetic fields.
  2. The additional term salvages the conservation of charge. Now from (M1) we again get $$ \frac{\partial\rho}{\partial t}=\epsilon_0\frac{\partial}{\partial t}\mathbf{\nabla}\mathbf{E}$$ and from (M4') now we get $$\nabla{\mathbf{J}}=-\epsilon_0\mathbf{\nabla}\frac{\partial}{\partial t}\mathbf{E}.$$ Hence $$\frac{\partial\rho}{\partial t}+\mathbf{\nabla}\mathbf{J}=0$$ which is now in agreement with (1).

The correctness of Maxwell's modification was experimentally confirmed soon afterwards by the discovery of high-frequency electromagnetic waves. Here the term $\epsilon_0\frac{\partial\mathbf{E}}{\partial t}$ is essential.


Maxwell's equations (M1), (M2), (M3) and (M4') can be solved and result in the retarded potentials (with $c:=1/\sqrt{\epsilon_0\mu_0}$): $$\begin{align} \mathbf A(\mathbf r,t)&=\frac{\mu_0}{4\pi} \int\frac{\mathbf J\left(\mathbf r',t-\frac{|\mathbf r-\mathbf r'|}{c}\right)}{|\mathbf r-\mathbf r'|}dV' \\ \Phi(\mathbf r,t)&=\frac{1}{4\pi\epsilon_0} \int\frac{\rho\left(\mathbf r',t-\frac{|\mathbf r-\mathbf r'|}{c}\right)}{|\mathbf r-\mathbf r'|}dV'. \end{align}$$

Obviously these potentials are not identical to the Biot-Savart and Coulomb potentials, because of the time-retarding term $-\frac{|\mathbf r-\mathbf r'|}{c}$.

But - because of the very large value of $c$ (the speed of light) - it is often legitimate to neglect this time-retarding term, especially when the densities $\mathbf{J}$ and $\rho$ vary only slowly with time $t$. This way we recover the Biot-Savart and Coulomb potentials as an approximation of the exact retarded potentials. $$\begin{align} \mathbf A(\mathbf r,t)&\approx\frac{\mu_0}{4\pi} \int\frac{\mathbf J(\mathbf r',t)}{|\mathbf r-\mathbf r'|}dV' \\ \Phi(\mathbf r,t)&\approx\frac{1}{4\pi\epsilon_0} \int\frac{\rho(\mathbf r',t)}{|\mathbf r-\mathbf r'|}dV'. \end{align}$$

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