University Physics IV · Special Relativity I · 2.2

Electrodynamics Under a Galilean Boost

Every other equation in first-year physics changes frames without complaint. This one does not. Carry the substitution out honestly, keep the term everybody drops, and you can see exactly which piece of Maxwell breaks — and why the repair had to be more drastic than a better aether.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Newton's laws do not notice a Galilean boost: x′ = x − vt with t′ = t leaves the acceleration alone, so F = ma reads the same in every inertial frame. Maxwell's equations do not have that property, and the cleanest way to see it is to push the boost through the wave equation they imply. The chain rule sends ∂ₓ → ∂ₓ′ and ∂ₜ → ∂ₜ′ − v∂ₓ′, so ψₓₓ − ψₜₜ/c² = 0 becomes (1 − β²)ψₓₓ + (2v/c²)ψₓₜ − (1/c²)ψₜₜ = 0 in the new coordinates.

The damage is the middle term. It is first order in v, no wave equation has anything resembling it, and its characteristics are x′ = (c − v)t′ and x′ = −(c + v)t′ — light with a different speed each way. No further Galilean change of coordinates removes it, because a second boost w only replaces v by v + w, and the single choice that restores the symmetric form is the one that takes you back where you started.

So the equation is announcing that one frame is special, and c = 1/√(μ₀ε₀), assembled from two constants with no velocity anywhere in them, gives no hint which frame that is. Call it the aether's and you have a concrete, first-order prediction: a laboratory drifting at v measures c − v one way and c + v the other. If no such drift is found, the choice is between abandoning the aether frame and abandoning Galilean kinematics — and it took fifty years and three separate experiments, from Michelson–Morley in 1887 to Ives–Stilwell in 1938, to establish that the second was the one to give up.

Simple definition: Maxwell's equations are not Galilean-covariant: substituting x′ = x − vt, t′ = t into the electromagnetic wave equation adds a cross term the original does not contain, so the equation can hold in its standard form in at most one frame.
Example: For a laboratory drifting at v = 30 km s⁻¹ — Earth's orbital speed, β = 1.0 × 10⁻⁴ — the boosted equation predicts light at 2.9977 × 10⁸ m s⁻¹ one way and 2.9983 × 10⁸ m s⁻¹ the other: a one-way split of 60 km s⁻¹ that nobody has ever found.

Galilean boost, standard configurationx′ = x − vt · y′ = y · z′ = z · t′ = t

One parameter, and it declares time absolute by fiat. That declaration is what eventually has to go.

v constant, in m s⁻¹, along x; axes parallel and origins together at t = 0

The chain rule the boost forces∂/∂x = ∂/∂x′ · ∂/∂t = ∂/∂t′ − v ∂/∂x′

Everything follows from the lopsidedness: time derivatives pick up a space derivative, space derivatives do not.

Partial derivatives, each at fixed value of the other coordinate; v constant, so the operators commute

The wave equation after the boost(1 − β²) ψₓₓ + (2v/c²) ψₓₜ − (1/c²) ψₜₜ = 0

The cross term is first order in v and belongs to no wave equation. That single term is the whole failure.

β = v/c; derivatives now taken in the moving frame's x′ and t′; ψ in V m⁻¹ or T

Characteristic speeds in the boosted frameu² + 2vu − (c² − v²) = 0 → u = c − v or u = −(c + v)

A one-way anisotropy of 2v, first order in the drift — but only a genuinely one-way method can detect it.

u in m s⁻¹, measured in the moving laboratory; roots of the trial substitution ψ = f(x′ − ut′)

Maxwell's speed, built from the constants alonec = 1/√(μ₀ε₀) = 2.998 × 10⁸ m s⁻¹

The equations hand you a speed with nothing to measure it against, which is exactly what invites a medium.

μ₀ = 4π × 10⁻⁷ H m⁻¹, ε₀ = 8.854 × 10⁻¹² F m⁻¹; no velocity appears in either

Round-trip times, the two interferometer armst∥ = (2L/c)γ² · t⊥ = (2L/c)γ · Δt ≈ Lβ²/c

First order cancels there and back, so a drift search hunts β²: 0.37 fringe at L = 11.0 m and 589 nm.

L arm length in m; γ = (1 − β²)^(−½); the expansion needs β ≪ 1

Covariance is a property of the equation, not the physics

Covariance is a property of an equation, not of a phenomenon. An equation is covariant under a transformation when it has the same form after you change variables — same terms, same coefficients, primes everywhere. Newton passes easily. Under x′ = x − vt and t′ = t with v constant, one differentiation gives u′ = u − v and a second gives a′ = a; forces built from separations, F(x₁ − x₂), are untouched because the separation is untouched; so F = ma reads identically in both frames. That is why no mechanics experiment performed below decks can tell you the ship's speed, and it is the relativity principle exactly as Galileo stated it. Electromagnetism was expected to fall into line. The honest way to find out is not to argue about it but to carry the substitution through and look at what comes out the other side.

Push the boost through the wave equation

Take a single Cartesian component ψ of E or B in vacuum and let it depend on x and t alone, so that Maxwell gives ψₓₓ − (1/c²)ψₜₜ = 0. Change variables to x′ = x − vt, t′ = t. Because x′ depends on both x and t while t′ depends only on t, the chain rule is lopsided: ∂/∂x = ∂/∂x′, but ∂/∂t = (∂x′/∂t)∂/∂x′ + (∂t′/∂t)∂/∂t′ = ∂/∂t′ − v ∂/∂x′. Space derivatives pass through untouched; time derivatives pick up a space derivative. Square the second, v being constant so the operators commute: ∂²/∂t² = ∂²/∂t′² − 2v ∂²/∂x′∂t′ + v² ∂²/∂x′². Substitute and collect: (1 − v²/c²) ∂²ψ/∂x′² + (2v/c²) ∂²ψ/∂x′∂t′ − (1/c²) ∂²ψ/∂t′² = 0, which we abbreviate from here as (1 − β²)ψₓₓ + (2v/c²)ψₓₜ − (1/c²)ψₜₜ = 0 with all derivatives in the primed coordinates. Two things happened, and only one of them matters. The ψₓₓ coefficient shrank by β², a second-order change you might be tempted to ignore. And a cross term appeared, first order in v, of a kind no wave equation contains.

Read the damage off the characteristic speeds

Ask the new equation what speeds it supports. Try ψ = f(x′ − ut′): then ψₓₓ = f″, ψₓₜ = −uf″ and ψₜₜ = u²f″. Divide out f″, multiply through by c², and you are left with u² + 2vu − (c² − v²) = 0, whose roots are u = −v ± c. The forward disturbance runs at c − v and the backward one at c + v — precisely what a wind does to a sound wave. Can another change of coordinates repair it? Not a Galilean one. A second boost of velocity w simply replaces v by v + w, so the speeds become c − (v + w) and c + (v + w), and the only w making them equal in size is w = −v: undoing the first boost and returning to the frame you started in. The map that does leave ψₓₓ − ψₜₜ/c² = 0 alone is x′ = γ(x − vt) with t′ = γ(t − vx/c²), and it is not in the Galilean family at all. It is worth naming the culprit inside Maxwell: delete the displacement current and what remains has a consistent Galilean limit — but it also has no waves. The term that makes light exist is the term that breaks the symmetry.

Maxwell's c has nothing to be measured against

Maxwell's equations contain exactly one speed, and it is built out of two constants of empty space: c = 1/√(μ₀ε₀). With μ₀ = 4π × 10⁻⁷ H m⁻¹ and ε₀ = 8.854 × 10⁻¹² F m⁻¹ that is 2.998 × 10⁸ m s⁻¹, the number Fizeau and Foucault had already measured for light. Nothing in the expression mentions a source, a detector or an observer, so the equations never say what c is measured with respect to. Every other wave known in 1865 answered that question with a medium: sound travels at 343 m s⁻¹ through still air, and if the air streams past you at v then a pulse sent upwind closes at 343 − v while the downwind pulse makes 343 + v. Read Maxwell the same way and c is the speed in the rest frame of a luminiferous aether, which makes the boosted equation above the one a moving laboratory is obliged to use. The prediction is then perfectly concrete, and first order in the drift: c − v for a pulse sent along the laboratory's own motion, c + v for one sent back the other way.

Two classical escapes, and what each one gives up

There were two ways out, and they exclude each other. Aether models keep the wave equation and give up the relativity principle for light: one frame is preferred, the laboratory's velocity through it is a real quantity, and Earth's orbital motion alone guarantees at least v = 30 km s⁻¹, β ≈ 1.0 × 10⁻⁴. Emission or ballistic models keep the relativity principle and give up the fixed wave equation: light leaves at c relative to its source, the way a bullet does, so no frame is special and there is no drift to find. Each died in its own way. A fully dragged aether, carried along with the Earth, would explain a null drift result, but stellar aberration displaces a star's apparent position by up to 20.5 arcseconds — which requires the Earth to move through the aether, not with it — and Fizeau's flowing water gave a partial drag of 1 − 1/n², about 0.43 for water, not the total drag a carried aether needs. Emission theory failed on de Sitter's binary stars in 1913, and finally in 1964, when γ-rays from π⁰ mesons decaying at 0.99975c were timed at c to within one part in 10⁴.

What a single null result can and cannot exclude

Michelson and Morley compared two perpendicular round trips in 1887, and the round trip is the whole difficulty: t∥ = L/(c − v) + L/(c + v) = (2L/c)/(1 − β²), so the first-order terms cancel against each other and only β² survives. With folded arms of L = 11.0 m and β = 1.0 × 10⁻⁴, rotating the apparatus through 90° should shift the pattern by 2Lβ²/λ = 0.37 fringe at 589 nm; the observed bound was about 0.01 fringe, some forty times smaller. That is a null result, and it is not the end of the aether. It tests only the isotropy of the two-way speed, and Lorentz and FitzGerald met it head-on by supposing the parallel arm contracts by exactly γ⁻¹ — an ad hoc hypothesis that reproduces the null result while keeping the preferred frame. Closing the remaining gaps took two more experiments: Kennedy–Thorndike in 1932, with deliberately unequal arms, tests whether the two-way speed depends on the laboratory's velocity, and Ives–Stilwell in 1938, on a fast ion beam, tests time dilation directly. Even all three together leave the one-way speed fixed by a synchronisation convention rather than by measurement.

Change one variable at a time

Make the relationship visible.

Interactive model

Frame speed through the aether v/c0.30 c

Extra Galilean boost w/c0.00 c

Push v/c to 0.40 and the forward arrow slows to 0.60 c while the backward one runs at 1.40 c. Then hunt for an extra boost w/c that squares the cone up again: only w/c = −0.40 does it, and that setting is the aether frame itself.

FORWARD FRONT (1 − β)c0.70 c

BACKWARD FRONT (1 + β)c1.30 c

ONE-WAY SPLIT 2β0.60 c

ROUND-TRIP EXCESS β²/(1−β²)0.099

Live interpretationFORWARD FRONT (1 − β)c: 0.70 c. BACKWARD FRONT (1 + β)c: 1.30 c. ONE-WAY SPLIT 2β: 0.60 c. ROUND-TRIP EXCESS β²/(1−β²): 0.099

Catch the common trap

Explain before calculating.

Misconception check

The usual move is to look at (1 − β²)ψₓₓ + (2v/c²)ψₓₜ − (1/c²)ψₜₜ = 0, note that β ≈ 10⁻⁴ in any earthbound laboratory, and conclude that Maxwell is Galilean-covariant to any accuracy that matters. Two things are wrong with that. Covariance is not an approximation: the transformed equation either has the same form or it does not, and this one carries a term the original does not contain at all, so no member of the Galilean family restores it. And the offending term is the large one — the cross term is first order in β while the ψₓₓ correction is second — so the predicted anisotropy is 2β = 2.0 × 10⁻⁴, a 60 km s⁻¹ split, not 10⁻⁸ of c. The effect looked tiny in the laboratory only because every round-trip method cancels first order and can see nothing below β². Mistaking the apparatus's blind spot for smallness in the theory is the actual error.

A laboratory drifts at speed v through the aether, so the fields inside it obey (1 − β²)ψₓₓ + (2v/c²)ψₓₜ − (1/c²)ψₜₜ = 0, with the derivatives taken in the laboratory's own coordinates. Which statement about that equation is correct?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA plane wave in the aether frame is ψ = f(x − ct). Rewrite it in the coordinates of a laboratory moving at v = 30 km s⁻¹ along +x, using x′ = x − vt and t′ = t, and find the speed that laboratory measures for a pulse sent each way. Take c = 2.998 × 10⁸ m s⁻¹.

Invert the boost so the old coordinates can be eliminated: x = x′ + vt′ and t = t′.
Substitute into the phase: x − ct = (x′ + vt′) − ct′ = x′ − (c − v)t′. The waveform f is unchanged, but its argument now advances at c − v, so the laboratory measures the forward pulse at c − v.
Do the same for a backward wave g(x + ct): x + ct = x′ + vt′ + ct′ = x′ + (c + v)t′, a pulse running in the −x′ direction at speed c + v.
Numbers: c − v = 2.998 × 10⁸ − 3.0 × 10⁴ = 2.9977 × 10⁸ m s⁻¹, and c + v = 2.9983 × 10⁸ m s⁻¹.
The split between the two is 2v = 6.0 × 10⁴ m s⁻¹, a fraction 2β = 2.0 × 10⁻⁴ of c — first order in the drift, not second.

AnswerForward: c − v = 2.9977 × 10⁸ m s⁻¹. Backward: c + v = 2.9983 × 10⁸ m s⁻¹. The one-way split is 2v = 60 km s⁻¹, or 2.0 × 10⁻⁴ of c.

MediumA Michelson interferometer with folded arms of length L = 11.0 m sits in that same drifting laboratory (v = 30 km s⁻¹) and uses sodium light of wavelength 589 nm. Using the c ∓ v kinematics the boosted equation predicts, find the round-trip time difference between the arm along the drift and the arm across it, the optical path difference, and the fringe shift produced by rotating the whole apparatus through 90°.

β = v/c = 3.00 × 10⁴ / 2.998 × 10⁸ = 1.00 × 10⁻⁴, so β² = 1.00 × 10⁻⁸.
Along the drift the pulse goes out at c − v and returns at c + v: t∥ = L/(c − v) + L/(c + v) = 2Lc/(c² − v²) = (2L/c)/(1 − β²). Notice the first-order terms have already cancelled between the two legs.
Across the drift the light must be aimed upstream to reach the mirror, leaving a crossing speed √(c² − v²): t⊥ = 2L/√(c² − v²) = (2L/c)(1 − β²)^(−1/2).
Expand both to order β²: t∥ ≈ (2L/c)(1 + β²) and t⊥ ≈ (2L/c)(1 + β²/2), so Δt ≈ (2L/c)(β²/2) = Lβ²/c = 11.0 × 1.00 × 10⁻⁸ / 2.998 × 10⁸ = 3.67 × 10⁻¹⁶ s.
Optical path difference cΔt = Lβ² = 1.10 × 10⁻⁷ m = 110 nm. Rotating by 90° exchanges the arms, so the pattern moves by twice that: Δ = 2Lβ²/λ = 2.20 × 10⁻⁷ / 5.89 × 10⁻⁷ = 0.37 fringe.

AnswerΔt = 3.7 × 10⁻¹⁶ s, an optical path difference of 110 nm, and a shift of 0.37 fringe on rotation. Michelson and Morley bounded the shift at roughly 0.01 fringe — about forty times smaller than the drift model demands.

HardLorentz's repair. Keep x″ = x′, but let the moving laboratory's clocks read a position-dependent local time t″ = t′ + αx′. Find the α that removes the cross term from (1 − β²)ψₓₓ + (2v/c²)ψₓₜ − (1/c²)ψₜₜ = 0, state what the equation then becomes, and identify the rescaling that finishes the job.

The shear gives ∂/∂x′ = ∂/∂x″ + α ∂/∂t″ and ∂/∂t′ = ∂/∂t″, so ψₓ′ₓ′ = ψₓ″ₓ″ + 2αψₓ″ₜ″ + α²ψₜ″ₜ″, ψₓ′ₜ′ = ψₓ″ₜ″ + αψₜ″ₜ″, and ψₜ′ₜ′ = ψₜ″ₜ″.
Collect the ψₓ″ₜ″ coefficient and kill it: 2α(1 − β²) + 2v/c² = 0, so α = −(v/c²)/(1 − β²) = −γ²v/c². To first order in β this is exactly Lorentz's local time, t″ = t′ − vx′/c².
Collect the ψₜ″ₜ″ coefficient: (1 − β²)α² + (2v/c²)α − 1/c² = (v²/c⁴)/(1 − β²) − 2(v²/c⁴)/(1 − β²) − 1/c² = −(1/c²)[β²/(1 − β²) + 1] = −γ²/c².
So the equation is now γ⁻²ψₓ″ₓ″ − (γ²/c²)ψₜ″ₜ″ = 0, that is ψₓ″ₓ″ − (γ⁴/c²)ψₜ″ₜ″ = 0. The cross term has gone, but the wave now propagates at c/γ² rather than c.
Rescale the axes: put x_L = γx″ and t_L = t″/γ. Then ∂²ψ/∂x_L² − (1/c²)∂²ψ/∂t_L² = 0 exactly, and unwinding gives x_L = γ(x − vt) and t_L = γ(t − vx/c²) — the Lorentz boost, reached in three steps from the Galilean one.

Answerα = −γ²v/c², so t″ = γ²(t − vx/c²): the cross term goes, but the wave then runs at c/γ². Rescaling by x_L = γx″ and t_L = t″/γ completes the Lorentz boost. The patch costs absolute time — clocks must read differently at different x.