Which form of Maxwell's equations is fundamental, in integral form or differential form?

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

Integral: simply for the reason that they apply to a larger range of use-cases: not just the cases where differential equations apply, but also the cases where there is no underlying continuum or differential structure at all (e.g. "quantized" or even geometries, like cell complexes). It could even be used on geometries that aren't even differential manifolds at all - i.e. that have no differential structure and admit no calculus.

You already see that the integral formulation is the more fundamental of the two by the fact that (1) even the differential equations are naturally written in the language of differential forms, which are kernels to integral operators (in fact, the exterior derivative operator $d(\_)$ is defined in terms of integral operators as a certain kind of "contracting to point" limit) and (2) historically, Maxwell tended to write the equations that way, as well, particularly before the treatise.

There is one major proviso: the constitutive laws break the mold. These are the equations which admit natural formulation in integral form: $$dA = F, \hspace 1em dF = 0, \hspace 1em dG = J, \hspace 1em dJ = 0,$$ with the differential forms defined in terms of the fields by: $$\begin{align} A &= \left(A_x dx + A_y dy + A_z dz\right) - φ dt, \\ F &= \left(B^x dy ∧ dz + B^y dz ∧ dx + B^z dx ∧ dy\right) + \left(E_x dx + E_y dy + E_z dz\right) ∧ dt, \\ G &= \left(𝔇^x dy ∧ dz + 𝔇^y dz ∧ dx + 𝔇^z dx ∧ dy\right) - \left(H_x dx + H_y dy + H_z dz\right) ∧ dt, \\ J &= ρ dx ∧ dy ∧ dz - \left(𝔍^x dy ∧ dz + 𝔍^y dz ∧ dx + 𝔍^z dx ∧ dy\right) ∧ dt. \end{align}$$

In contrast, this is the part that breaks the mold: $$ \left(𝔇^x, 𝔇^y, 𝔇^z\right) = ε_0 \left(E_x, E_y, E_z\right), \\ \left(B^x, B^y, B^z\right) = μ_0 \left(ℌ_x, ℌ_y, ℌ_z\right). $$ You never see integral forms for these relations, though it is true that you can express the relation between $F$ and $G$ in terms of differential forms as $G = \star{F/Z_0}$, using the Hodge duality operator $\star{(\_)}$, where $Z_0 = \sqrt{μ_0/ε_0}$. But it's not a natural operation, certainly not "background-free" - as are the other equations. It is dependent on the underlying metric.

Were it not for that, you could literally do away with the whole calculus and write everything algebraically in terms of global objects, like the integrals of differential forms. That gets in the way.

To truly get something that works with the integral form would require stepping back and abstracting away from the action principle, itself. The action is expressed in integral form: $$S = \int L, \hspace 1em L = 𝔏 d^4 x,$$ which involves a Lagrangian 4-form $L$ and its component $𝔏$, a Lagrangian density. The constitutive laws are actually specializations adapted to the Maxwell-Lorentz Lagrangian density $$𝔏 = \frac{ε_0 \left({E_x}^2 + {E_y}^2 + {E_z}^2\right)}{2} - \frac{{B^x}^2 + {B^y}^2 + {B^z}^2}{2 μ_0},$$ of the following generic constitutive laws: $$\begin{align} 𝔇^x &= \frac{∂𝔏}{∂E_x}, & 𝔇^y &= \frac{∂𝔏}{∂E_y}, & 𝔇^z &= \frac{∂𝔏}{∂E_z}, \\ ℌ_x &= -\frac{∂𝔏}{∂B^x}, & ℌ_y &= -\frac{∂𝔏}{∂B^y}, & ℌ_z &= -\frac{∂𝔏}{∂B^z}, \end{align}$$ with the relations extending to the source fields as well: $$ρ = -\frac{∂𝔏}{∂φ}, \hspace 1em 𝔍^x = \frac{∂𝔏}{∂A_x}, \hspace 1em 𝔍^y = \frac{∂𝔏}{∂A_y}, \hspace 1em 𝔍^z = \frac{∂𝔏}{∂A_z}, $$ or in terms of the respective differential forms by: $$δ𝔏 = (δA)∧J - (δF)∧G.$$

So, in fact, there actually is an integral form for the constitutive laws - and one that admits generalization to other use-cases, like the above-mentioned discrete or "quantized" geometries - and that is: $$S_Ω = \int_Ω L, \hspace 1em δ\left(S_Ω\right) = \int_Ω (δA)∧J - \int_Ω (δF)∧G,$$ where the domain of integration $Ω$ is made explicit.

This shows that the response field $G$ and source $J$ are not actually differential forms, but operators $$\frac{δS}{δA} = \hat{J}_Ω = \int_Ω \_∧J, \hspace 1em -\frac{δS}{δF} = \hat{G}_Ω = \int_Ω \_∧G,$$ and the relations $dG = J$ and $dJ = 0$ can also be written as: $$\begin{align} \hat{J}_Ω(a) &= \int_Ω\left(a∧J\right) \\ &= \int_Ω\left(a∧dG\right) \\ &= -\int_Ω\left(d(a∧G)\right) + \int_Ω\left(da∧G\right) \\ &= -\int_{∂Ω}\left(a∧G\right) + \int_Ω\left(df∧G\right) \\ &= -\hat{G}_{∂Ω}(a) + \hat{G}_Ω(da), \\ 0 &= \int_Ω\left(sdJ\right) \\ &= \int_Ω\left(d(sJ)\right) - \int_Ω\left(ds∧J\right) \\ &= \int_{∂Ω}\left(sJ\right) - \int_Ω\left(ds∧J\right) \\ &= \hat{J}_{∂Ω}(s) - \hat{J}_Ω(ds), \end{align}$$ where $∂Ω$ denotes the boundary of $Ω$, $a$ is an odd-degree differential form and $s$ is a scalar.

Although the equation $dG = J$ (with $dJ = 0$ being derived from it) is expressed as a differential equation - the Euler-Lagrange equation for the action $S$ - the actual underlying problem is not analytic at all, but potentially algebraic. It is an optimization problem; or more generally, a "stationary value" problem, if expressing the action principle as a stationary action principle. In a use-case where the underlying geometry isn't even a continuum, this would translate into an algebraic problem, not a calculus problem.