What does a Galilean transformation of Maxwell's equations look like?

Question

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.

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 $\textbf k$.

Answer 2

The equations $$ ∇· = σ,\quad ∇× + \frac{∂}{∂t} = -,\\ ∇· = ρ,\quad ∇× - \frac{∂}{∂t} = , $$ where $$∇ = \left(\frac{∂}{∂x},\frac{∂}{∂y},\frac{∂}{∂z}\right) $$ are invariant under the 4-D diffeomorphism group, which contains the Galilei group as a subgroup ... as well as the Lorentz group, 4D Euclidean group.

They can be written as $$dF = J,\quad dG = -K,$$ where $$ F = ·d + ·d∧dt,\quad J = ρ dV - ·d∧dt,\\ G = ·d - ·d∧dt,\quad K = σ dV - ·d∧dt,\quad $$ where $$d = (dx, dy, dz),\quad d = (dy∧dz, dz∧dx, dx∧dy),\quad dV = dx∧dy∧dz,$$ which shows off the diffeomorphism invariance explicitly.

By "invariance" I literally mean invariant. The coordinates $(x,y,z,t)$ can be anything, not just Cartesian + time ... as long as they have an invertible transform from the space-time coordinates ... and with a suitable transform on the fields and their components, the equations will be exactly the same.

Apart from $σ ≠ 0$ and $ ≠ $ (which Heaviside used, though he stated they were fictitious), they are essentially what Maxwell had, as well as Hertz, Lorentz, etc.

To find the transforms under a Galilei boost, $$ → - t,\quad t → t,$$ take the differential $$d → d - dt,\quad dt → dt,$$ and, from this, infer the following transforms $$ d → d + ×d∧dt,\quad d∧dt → d∧dt,\\ d∧dt → d∧dt,\quad dV → dV - ·d∧dt. $$

Finally, requiring that $F$, $J$, $G$, $K$ all remain invariant, one then infers $$ → ,\quad → + ×,\quad ρ → ρ,\quad → - ρ,\\ → ,\quad → - ×,\quad σ → σ,\quad → - σ. $$ One of these lines corresponds to the "electric limit" and the other to the "magnetic limit" of Lévi-Leblond. They're misnamed: they're just the transform behaviors of $(,,ρ,)$ and $(,,σ,)$, not any kind of limit, per se.

The "velocity vector" you refer to was what distinguished the "stationary" versus "moving" versions of Maxwell's theory. That distinction still exists even when passing to the relativistic form, but more on that later. Lorentz used $-$, Heaviside used $$, Maxwell used $$ or $$ variously, but mostly $$ ... which confused the issue, because he also used $$ generically to denote relative velocities. To keep with Maxwell's "alphabet soup" lettering, use $$.

The diffeomorphism invariance is broken by the set of relations that link $(,)$ to $(,)$ - the constitutive relations. In fact, if you write them in generalized form like this: $$ + α× = ε( + β×),\quad - α× = μ( - β×).$$ These are the relations suitable for a constitutive law that is isotropic in at least one frame of reference. The frame of isotropy occurs where $ = $. The equations in this frame are the "stationary" form.

The symmetry group selected out by this form of the constitutive law is that which has the following as its invariants: $$β|d|^2 - α {dt}^2,\quad β\left(\frac{∂}{∂t}\right)^2 - α∇^2,\quad d·∇ + dt \frac{∂}{∂t}.$$ This includes the following:

• $αβ > 0$: the Poincaré group, with $c = \sqrt{β/α}$,
• $αβ < 0$: the 4-D Euclidean group,
• $α = 0, β ≠ 0$: the Galilean group, for $c → ∞$,
• $α ≠ 0, β = 0$: the Carroll group, for $c → 0$,
• $α = 0, β = 0$: the Static group, in which case space and time each become absolute and the three invariants split into six invariants $$|d|^2,\quad ∇^2,\quad d·∇,\quad {dt}^2,\quad \left(\frac{∂}{∂t}\right)^2,\quad dt \frac{∂}{∂t}.$$

For the Galilei group, setting $β = 1$, the constitutive laws reduce to the form $$ = ε( + ×),\quad = μ( - ×).$$ Maxwell did not include the $-×$ term, which was a mistake ... and one he should have caught, because he actually did carry out the analysis of the transform behavior of the fields under Galilei in his treatise! It was later corrected by Thomson. Heaviside is probably to be credited with the correction, since he introduced it anew, in his version of Maxwell's theory. Lorentz had the correction, as well.

These equations are covariant with respect to the Galilei transforms, provided that one adopts the obvious transform for the velocity vector: $$ → - .$$

Both Maxwell and Heaviside used the "covariant" forms of $$ and $$, which I'll label $\bar{}$ and $\bar{}$, respectively: $$\bar{} = + ×,\quad \bar{} = - ×,$$ except that in Maxwell's case, owing to his omission of $-×$, his $\bar{}$ was just his $$ ... which (therefore) was also a mistake. Under Galilei, these are invariant $$\bar{} → \bar{},\quad \bar{} → \bar{}.$$ In Maxwell's case, the $(,)$ fields were expressed in terms of the potentials $(,φ)$, via the relations $$ = ∇×,\quad = -∇φ - \frac{∂}{∂t},$$ which can also be written in diffeomorphism-invariant form as $$F = dA,\quad A = ·d - φ dt.$$ This precludes any non-zero $K$, so no such terms appear in Maxwell's formulation of electromagnetism. Therefore, since Maxwell used the Galilean-invariant form $\bar{}$ of $$, he actually wrote: $$\bar{} = × - ∇φ - \frac{∂}{∂t}.$$

For Special Relativity, $αβ > 0$, with $c = \sqrt{β/α}$. Then the generic form of the constitutive laws become equivalent to those expressed by Minkowski in 1908 and Einstein / Laub in 1907-1908, which are the "Maxwell-Minkowski" equations. There continues to be a distinction between the "stationary" version $ = $ and "moving" version $ ≠ $, in general. However, in isotropic media where $βεμ = α$ (i.e. where $εμ = 1/c^2$), when $|| < c$ (i.e. the vacuum), the equations are independent of $$ and the stationary and moving versions of the Maxwell-Minkowski form of the constitutive laws are equivalent. In the vacuum, there is no longer any dependence on $$ ... unless $|| ≥ c$ which is ironic, if you recall that Einstein's foray into Special Relativity began with the question "what would it be like to move alongside a light wave?".

It's awkward to state the transform laws in finite form, except for the Galilei, Static and Carroll cases, but they can be stated in infinitesimal form in such a way that covers all the cases together. For an infinitesimal boost $$, one has $$δ = -βt,\quad δt = -α·.$$ Taking differentials, one infers $$δ(d) = -βdt,\quad δ(dt) = -α·d.$$ From this, in turn, one infers $$ δ(d) = β×d∧dt,\quad δ(d∧dt) = -α×d,\\ δ(dV) = -β·d∧dt,\quad δ(d∧dt) = -αdV. $$

The requirement that $F$, $J$, $G$, $K$ and $A$ remain invariant then leads to the following transforms: $$ δ = -α×,\quad δ = β×,\quad δρ = -α·,\quad δ = -βρ,\\ δ = α×,\quad δ = -β×,\quad δσ = -α·,\quad δ = -βσ,\\ δ = -αφ,\quad δφ = -α·. $$ For the Maxwell-Minkowski relations to be covariant, this requires $$δ = - + αβ·.$$ For Special Relativity, $β$ transforms as a velocity under an infinitesimal boost given by an infinitesimal velocity $β$: $$δ(β) = -(β) + \frac{α}{β} (β)·(β)(β) = -(β) + \frac1{c^2} (β)·(β)(β).$$ So, for Special Relativity, as well as the Euclidean 4D and Galilei cases, without any real loss of generality, since you could absorb $β$ into velocities, you could just take $β = 1$ and $α = 1/c^2$. The $β$ factor is really only there to handle the $c → 0$ Carroll case and the case of the Static group.