What is the fundamental explanation for electromagnetic induction from the relativistic point of view?

Question

Background

Faraday's law of induction gives a quantitive formula for the induced emf in a loop with changing magnetic influx through it:

$$\epsilon =-\frac{d\Phi_B}{dt}$$

Although this law is entirely correct, from modern perspective there are two different phenomena whose quantity is captured by this formula: a loop which changes its area $A(t)$ in a constant magnetic field $\vec{B}$, and a loop of constant area in a changing magentic field $\vec{B}(t)$. The explanation in each case is different - in the first case the origin of the emf is that due to the changing area of the loop, an asymmetric Lorenz force (force by the external magnetic field on the (bound or free) charges in the loop material) is generated in the loop and this in turn induces non-zero emf.

A fundamental explanation of the second case is, however, much less trivial, and involves some difficulty. In an electromagnetism course I studied recently, the lecturer gave a relativistic argument for the correctness of the rule $ \vec{\nabla} \times \vec{E} = - \frac{ \partial\vec{B}}{\partial t}$, which is based of the fundamental observation that this equation is a result of the requirement that Gauss's law of magnetism (which states that $\nabla \cdot \vec{B} =0$) should be invariant under change of frame of reference.

That is, this observation uses the relativistic transformation formulas for the electromagnetic field (both the electric $\vec{E}$ and the magnetic $\vec{B}$) and the Lorenz transformations of time and space coordinates to derive Faraday's law in the case of changing magnetic field. I will describe a simplified version of this derivation, in which the Lorenz factor is $\gamma \approx 1$.

Derivation of Faraday's law

Suppose we have one frame of reference $S$ in which there is a static (not changing in time) magnetic field, and that the magnetic field satisfies Gauss's law for magnetism; that is: $\frac{\partial B_x}{\partial x}+\frac{\partial B_y}{\partial y}+\frac{\partial B_z}{\partial z}=0$. Now suppose we have a second moving frame of reference $S'$ which moves along the x axis at speed $v$ relative to $S$. Since the Lorenz factor is approximately $1$, the spatial and time differences $dx,dy,dz,dt$ remain unaltered when switching between the frames of referneces; that is: $dx'=dx,dy'=dy,dz' = dz,dt'=dt$. However, simultanueity is broken, so if the values $B_x(x),B_x(x+\Delta x)$ were measured simultaneuosly in $S$, than they will be measured in time lag $\Delta t = \frac{v}{c^2}\Delta x$ in $S'$. Since the magnetic field in $S'$ is changing in time (due the motion of this frame of reference), an observer in $S'$ will have to take into account both the time lag $\Delta t$ and the measured rate of change of the magnetic field $\frac{\partial B'_x}{\partial t'}$ in his calculation of $\nabla \cdot \vec{B'} = \frac{\partial B'_x}{\partial x'}+\frac{\partial B'_y}{\partial y'}+\frac{\partial B'_z}{\partial z'}$. More specifically:

$$\frac{\partial B'_x}{\partial x'} = \frac{\partial B'_x}{\partial x} = -\frac{v}{c^2}\frac{\partial B'_x}{\partial t'} + \frac{\partial B_x}{\partial x}$$

Now take the formulas of transformation of electric and magnetic fields:

$$B_y = B'_y + \frac{v}{c^2}E'_z \implies \frac{\partial B'_y}{\partial y'} = \frac{\partial B_y}{\partial y}-\frac{v}{c^2}\frac{\partial E'_z}{\partial y'}$$ $$B_z = B'_z - \frac{v}{c^2}E'_y \implies \frac{\partial B'_z}{\partial z'} = \frac{\partial B_z}{\partial z}+\frac{v}{c^2}\frac{\partial E'_y}{\partial z'}$$

Now the requirement that Gauss's law will be valid in $S'$ demands that:

$$0=(-\frac{v}{c^2}\frac{\partial B'_x}{\partial t'} + \frac{\partial B_x}{\partial x})+(\frac{\partial B_y}{\partial y}-\frac{v}{c^2}\frac{\partial E'_z}{\partial y'})+(\frac{\partial B_z}{\partial z}+\frac{v}{c^2}\frac{\partial E'_y}{\partial z'})\implies \frac{\partial B'_x}{\partial t'}= \frac{\partial E'_y}{\partial z'}-\frac{\partial E'_z}{\partial y'}$$

Similar procedure enables to write the full three dimensional Maxwell's third equation (Faraday's law):

$$ \vec{\nabla} \times \vec{E} = - \frac{ \partial\vec{B}}{\partial t}$$

Questions

The derivation presented here is very technical and I find it not very transparent. Despite being able to follow the developement, I still don't get the basic idea - what is the underlying principle of this derivation? can someone present a more illuminating and intuitively clear explanation of this derivation?

And is it justified to say that special relativity together with Gauss's law of magnetism are more fundamental than Faraday's law (since the letter is derived from them)?

Answer 1

(What is considered "fundamental" depends on what starting points you begin with. From the spacetime viewpoint below, Faraday's Law and Gauss-for-B are on an equal footing.)

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From a spacetime formulation of electromagnetism, the Maxwell Equations are encoded in two equations involving the electromagnetic field tensor $F_{ab}=\nabla_a A_b-\nabla_b A_a$:

(following Wald, General Relativity, p. 70, using the (-,+,+,+) signature convention, where the above expression for $F_{ab}$ is based on (4.2.28) ) $$\nabla^a F_{ab}=-4\pi j_b \qquad(4.3.12)$$ $$\nabla_{[a}F_{bc]}=0.\qquad(4.3.13),$$ where for an observer moving with 4-velocity $v^a$, we define $E_a=F_{ab}v^b$ (4.2.21) as the electric field measured by that observer and $B_a=-\frac{1}{2}\epsilon_{ab}{}^{cd}F_{cd}v^b$ (4.2.22) as the magnetic field. $\epsilon_{abcd}$ is the totally antisymmetric Levi-Civita tensor.

Ampere's Law is the vector-component of (4.3.12) and Gauss-for-E is the scalar component of (4.3.12).

Faraday's Law is the vector-component of (4.3.13) and Gauss-for-B is the scalar component of (4.3.13). Equation (4.3.13) is sometimes called the [first] Bianchi identity.

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For details, refer to https://en.wikipedia.org/wiki/Electromagnetic_tensor#Significance

Calculations (using possibly different conventions) are given in Equivalent form of Bianchi identity in electromagnetism , Electromagnetism Bianchi Identity , Expressing Maxwell's equations in tensor form using Electromagnetic field strength tensor

(When I find a good resource for giving intuition (via differential forms), I'll update this answer.)