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I am not sure which form of Maxwell's equations is fundamental, integral form or differential form. Imagine an ideal infinitely long solenoid. When a current is changing in time, can we detect classical effects outside a solenoid, for example generating a circular current around solenoid by Faraday's law. If the differential form is fundamental, we won't get any current, but the integral form is fundamental we will get a current.

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In QED all I've seen is the differential form. I would say they are more fundamental. I've never really thought about if the integral form has any meaning in QED. –  Love Learning Mar 17 at 8:06

3 Answers 3

If the differential form is fundamental, we won't get any current, but the integral form is fundamental we will get a current.

I'm not sure how you came to that conclusion, but it's not true. Both the differential and integral forms of Maxwell's equations are saying exactly the same thing. Either can be derived from the other, and both of them predict the exact same physical consequences in any situation.

Most physicists would say the differential form is more fundamental, but that's just an artifact of how we think about modern physics, in terms of fields which interact at specific points. It's really a philosophical issue, not a physical one, because it doesn't matter for the purpose of doing calculations which form you consider to be more fundamental.

In the specific situation you're asking about, with the solenoid, you will get a current in the loop around the solenoid. It may be easier to see that by using the integral form of Faraday's law, but the differential form makes the exact same prediction.

Let me demonstrate this explicitly. Suppose you have an ideal solenoid of radius $r_0$, with $n$ turns per unit length, oriented along the $z$ axis. Its magnetic field is given by

$$\vec{B} = \begin{cases}\mu_0 n I\hat{z} & r < r_0 \\ 0 & r > r_0\end{cases}$$

As you've noticed, this implies that $\nabla\times\vec{E} = 0$ outside the solenoid. Now, you might think that implies the integral $\oint\vec{E}\cdot\mathrm{d}\ell$ around a loop outside the solenoid, which gives the EMF, must be zero. But that's not actually the case. The relationship between $\nabla\times\vec{E}$ and $\oint\vec{E}\cdot\mathrm{d}\ell$ comes from Stokes' theorem, and it says

$$\oint_{\mathcal{C}}\vec{E}\cdot\mathrm{d}\ell = \iint_{\mathcal{S}}(\nabla\times\vec{E})\cdot\mathrm{d}^2\vec{A}$$

So the line integral around the loop is determined by the curl of $\vec{E}$ everywhere inside the loop, including inside the solenoid where

$$\nabla\times\vec{E} = -\mu_0 n \frac{\partial I}{\partial t}\hat{z}\quad(r < r_0)$$

Performing the integral gives you

$$\mathcal{E} = \oint_{\mathcal{C}}\vec{E}\cdot\mathrm{d}\ell = \iint_{\mathcal{S}}(\nabla\times\vec{E})\cdot\mathrm{d}^2\vec{A} = -\int_0^{2\pi}\int_{0}^{r_0}\mu_0 n \frac{\partial I}{\partial t}\hat{z}\cdot r\mathrm{d}r\,\mathrm{d}\theta\,\hat{z} = -\mu_0 \pi r_0^2 n\frac{\partial I}{\partial t}$$

so you can see that any time-varying current in the solenoid will create an EMF and induce a current.

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Agreed. The locality of the differential form makes it nicer and more in line with the current points of emphasis in modern physics, but other than that it's the same thing –  Danu Mar 17 at 7:24
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On the other hand, the integral form handles things like point charges and line currents somewhat better, without requiring as much language of distributions as the differential form. –  Emilio Pisanty Mar 17 at 7:36
    
Thank you for your explanations. I am sorry not for clarifying my question well. I might be confused. For a specific situation, an infinitely long solenoid, $\nabla\times E$ will be zero outside a solenoid and hence there will be no electric field outside even if there is a changing current since the differential form consider only local behavior, while the integral form will give a voltage around a solenoid. Does it make a sense? –  Traveler Mar 17 at 11:33
    
@Traveler if $E = 0$ everywhere outside the solenoid for a nontrivially changing current, you can't satisfy the differential form at every point on a circle around the solenoid. I'll try to update my answer with a derivation when I have time. –  David Z Mar 17 at 16:19
    
Yes, I think I was wrong. $\nabla\times E=0$ leads to $E=\nabla\chi$ generally. I would like to see your explanation. Thank you so much. –  Traveler Mar 18 at 17:18

Neither the integral or differential representation are more fundamental; one can arrive at either via vector calculus theorems. The most elegant formulation of Maxwell's equations employs differential forms (in the differential geometry sense). With the potential 1-form $A$, we may construct a field strength tensor $F = \mathrm{d} A = \partial_\mu A_\nu - \partial_\nu A_\mu$ and Maxwell's equations become:

$$\mathrm{d} F =0, \, \, \, \, \delta F =0$$

where $\delta = \star \mathrm{d} \star$, normally called the codifferential and the operator $\star$ is the Hodge dual.

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One can work out which is fundemental or derived, according to whether they constitute a definition of a quantity.

$\nabla\cdot D = \rho$ is fundemental, since this is the modern way of defining a flux-like field. One adds to this, the equations $F=EQ$ and $D=\epsilon E$, which although are not part of the classical four equations, are variously added by way of aside.

$\nabla\cdot B=0$ is a statement that there are no magnetic charges, was derived by first assuming magnetic charge, and then proving it does not exist.

$\nabla\times E = -\tau B$ is a derived relation, since nothing is defined in this relation. This is faraday's induction law.

$\nabla\times H = \tau D + J$ is likewise derived, since all of $H$, $D$, and $J$ are derived elsewhere. Hint, equations with additions are usually a derived equality.

Leo Young (System of Units in Electricity and Magnetism), tells us that one needs eight equations to make Maxwell's equations work as a base for electromagnetism. Six have been shown above. One needs also $B = \mu H$ and $F = I \times B $, in order to derive electromagnetism.

Since Leo Young was addressing a theory which is coherent to both C.G.S. Gaussian and SI, he makes use of additional constants S and U, which are set to unity in SI, but take the values of $S=4\pi$ and $U=1/c$ in C.G.S. One simply adds into an SI equation these numbers, such that when the substitution is made, c.g.s. formulae arise.

Oliver Heaviside, who first gave an account of EM theory starting off with Maxwell's equations, did not automatically suppose $\nabla\cdot B=0$, but $\nabla\cdot B = m$, where $m$ is the point-density of magnetic charge. This affects $\nabla\times E$ as well.

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