In the 1860's Maxwell formulated what are now called Maxwell's equation, and he found that they lead to a remarkable conclusion: the existence of electromagnetic waves that propagate at a speed $c$, which turns out to be the speed of light, implying that light is an electromagnetic wave. Now the fact that Maxwell's equations predict speed of light is $c$ suggested to Maxwell and others that Maxwell's equations are not actually true in all frames of reference. Instead, they thought, Maxwell's equations only exactly true in one frame, the rest frame of the aether, and in all other frames they would have to be replaced by other equations, equations that were invariant under Galilean transformations in order to conform to the principle of relativity. These other equations implied that the speed of light in other frames was actually $c+v$ or $c-v$, where $v$ is the speed of the aether. But then the Michelson-Morley experiment, which was intended to find the speed $v$ of the aether, ended up showing that the speed of light was $c$ in all frames, apparently contradicting the principle of relativity. But Einstein showed that this doesn't contradict the principle of relativity at all, it's just that you need to rethink your notions of space and time.

But my question is, what are the equations that people thought were true in frames other than the aether frame? To put it another way, what are the equations you obtain if you apply a Galilean transformation to Maxwell's equations? (As opposed to a Lorentz transformation which leaves Maxwell's equations unchanged.)

I've actually seen the equations obtained before. They were formulated by some 19th century physicist, maybe Hertz or Heaviside, and they involve adding velocity-dependent terms to the Ampere-Maxwell law and Faraday's law. (Dependent on the velocity of aether, that is.) But I don't remember the details.


1 Answer 1


I'm no expert on the historical development of the subject, however I will offer a derivation.

Consider two frames of reference $S$ and $S'$, and suppose that $S'$ moves with speed $\textbf v$ with respect to $S$. Coordinates in $S$ and $S'$ are related by a Galileian transformation: $$\begin{cases} t' = t \\ \textbf x' = \textbf x-\textbf vt\end{cases}$$ To find how the fields transform, we note that a Lorentz transformation reduces to a Galileian transformation in the limit $c \to \infty$. In fact, under a Lorentz transformation the fields transform like: $$ \begin{cases} \textbf E' = \gamma (\textbf E + \textbf v \times \textbf B) - (\gamma-1) (\textbf E \cdot \hat{\textbf{v}}) \hat{\textbf{v}}\\ \textbf B' = \gamma \left(\textbf B - \frac{1}{c^2}\textbf v \times \textbf E \right) - (\gamma-1) (\textbf B \cdot \hat{\textbf{v}}) \hat{\textbf{v}}\\ \end{cases}$$ Taking the limit $c\to \infty$ so that $\gamma\to 1$, we obtain the Galileian transformations of the fields: $$ \begin{cases} \textbf E' = \textbf E + \textbf v \times \textbf B\\ \textbf B' = \textbf B\\ \end{cases}$$ We can then invert the transformation by sending $\textbf v \to -\textbf v$: $$ \begin{cases} \textbf E = \textbf E' - \textbf v \times \textbf B'\\ \textbf B = \textbf B'\\ \end{cases}$$ By the same reasoning, can obtain the Galileian transformation of the sources: $$ \begin{cases} \textbf J = \textbf J' + \rho' \textbf v\\ \rho = \rho'\\ \end{cases}$$ We know that the fields and sources satisfy Maxwell's equations in $S$: $$ \begin{cases} \nabla \cdot \textbf E = \rho/\epsilon_0\\ \nabla \cdot \textbf B = 0\\ \nabla \times \textbf E = -\frac{\partial \textbf B}{\partial t}\\ \nabla \times \textbf B = \mu_0 \left(\textbf J +\epsilon_0 \frac{\partial \textbf E}{\partial t} \right)\\ \end{cases}$$ Replacing the fields and sources in $S$ with those in $S'$ we obtain: $$ \begin{cases} \nabla \cdot \textbf (\textbf E' - \textbf v \times \textbf B') = \rho'/\epsilon_0\\ \nabla \cdot \textbf B' = 0\\ \nabla \times \textbf (\textbf E' - \textbf v \times \textbf B') = -\frac{\partial \textbf B'}{\partial t}\\ \nabla \times \textbf B' = \mu_0 \left(\textbf J' + \rho' \textbf v +\epsilon_0 \frac{\partial (\textbf E' - \textbf v \times \textbf B')}{\partial t} \right)\\ \end{cases}$$ As a last step, we need to replace derivatives in $S$ with derivatives in $S'$. We have: $$\begin{cases} \nabla = \nabla' \\ \frac{\partial }{\partial t} = \frac{\partial }{\partial t'} - \textbf v \cdot \nabla\end{cases}$$ Substituting and removing the primes and using vector calculus, we obtain: $$ \begin{cases} \nabla \cdot \textbf E + \textbf v \cdot (\nabla \times \textbf B) = \rho/\epsilon_0\\ \nabla \cdot \textbf B = 0\\ \nabla \times \textbf E = -\frac{\partial \textbf B}{\partial t}\\ \nabla \times \textbf B = \mu_0 \left(\textbf J + \rho \textbf v +\epsilon_0 \frac{\partial}{\partial t}( \textbf E - \textbf v \times \textbf B) - \epsilon_0 \textbf v \cdot \nabla (\textbf E - \textbf v \times \textbf B) \right)\\ \end{cases}$$

In a vacuum, we can take the curl of the fourth equation to obtain: $$c^2\nabla^2 \textbf B = \frac{\partial^2 \textbf B}{\partial t^2} + (\textbf v \cdot \nabla)^2 \textbf B - 2 \textbf v \cdot \nabla \left(\frac{\partial \textbf B}{\partial t}\right)$$ Substituting a wave solution of the form $\textbf B \sim \exp{i(\textbf k \cdot \textbf x -\omega t)}$ We obtain an equation for $\omega$, which we can solve to obtain: $$\omega = -\textbf v \cdot \textbf k \pm c |\textbf k|$$ Therefore the speed of propagation is the group velocity: $$\frac{\partial \omega}{\partial \textbf k} = -\textbf v \pm c \hat{\textbf{ k}}$$ which gives you the expected $c\pm v$ with an appropriate choice of $\textbf v$ and $\textbf k$.

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    $\begingroup$ wow, very impressive, +1. -NN $\endgroup$ Commented Sep 4, 2020 at 6:21
  • $\begingroup$ Let S(at rest) and S'(moving with velocity u) be the inertial reference frames. Let E and B be represent the electric and magnetic fields in the S frame and similarly, let E' and B' be the electric and magnetic fields in the S' frame. Therefore, the force acting on a charged particle with charge q in both frames should be the same. In the S frame, force is given by qE + q(v×B), where v is the velocity of charge in the S frame. Similarly, the force in the S' frame is qE'+q((v-u)×B'). Now, I can actually verify that E'=E+(v×B) and B'=B. $\endgroup$
    – Infinite
    Commented Feb 8 at 8:19
  • $\begingroup$ However, I can't confirm why B' and B should be the same in this way. So, can you give some insights on why the magnetic field is same in all inertial reference frames (without including relativistic effects)? $\endgroup$
    – Infinite
    Commented Feb 8 at 8:20

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