Special relativity · Electrodynamics §8

A packet of light does not transform like a rigid material body.

How do the energy and volume of a bounded light complex transform between frames?

The finite light complex

Static worked example

CurrentThese numbers match the current settings.

Model note

Primary outputs frameSpeed, propagationAngleStationary, propagationAngleMoving, lightComplexEnergyStationary, lightComplexEnergyMoving, lightComplexVolumeStationary, lightComplexVolumeMoving, lightAmplitudeStationary, lightAmplitudeMoving, dopplerFactor, lorentzFactor, energyDensityFactor, volumeFactor, materialVolumeFactor: Host calculation (waves). Owner waves.
Accepted input revision 1.
Snapshot version 1.
Not modeled: media and dispersion; quantum photon structure; finite pulse dispersion in dielectric; gravitational redshift; boundary diffraction at packet edges.

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A packet of light seen from a moving frame changes both its energy and the volume it fills, and its energy changes exactly as its frequency does. Neither change is the shortening of a moving rod.

§8 follows a bounded packet of plane light waves, a light complex. In K, a sphere whose surface moves with the light at speed V along the wave normal lets no energy through, so it always encloses the same light. Seen from k, that sphere is an ellipsoid, and §8 computes its volume as S′/S = √(1 − (v/V)²)/(1 − (v/V) cos φ), where φ is the angle between the wave normal and the motion. Since A²/8π is the energy per unit volume and the amplitude transforms as in §7, the enclosed energy is E′/E = (1 − (v/V) cos φ)/√(1 − (v/V)²), which for φ = 0 becomes √((1 − v/V)/(1 + v/V)). §8 remarks that the energy and the frequency of a light complex change with the observer's motion by the same law. At the lab's default, 0.6c with the light moving along the motion, the energy factor is 0.5: 1 J becomes 0.5 J and 1 m³ becomes 2 m³, and the energy per unit volume falls to a quarter. At φ = 180° the energy doubles and the volume halves. The lab also runs a wrong model, labelled as one and kept out of the numbers above: it keeps light's energy per unit volume but gives the packet a rigid body's volume, 1/γ = 0.8, so its energy scales by q²/γ. Along the motion it predicts 0.2 J against the packet's 0.5 J, and against the motion 3.2 J against 2 J. The decisive ray is one at right angles to the motion in k, cos φ = 0.6 in K: the wrong model gives 0.512 J and 0.8 m³, the packet 0.8 J and 1.25 m³. At φ = 90° in K the two agree on both counts, 1.25 J and 0.8 m³, so that ray cannot tell them apart.

Take γ first: at 0.6c, γ = 1/√(1 − 0.36) = 1.25. Write q = γ(1 − β cos φ), with β = 0.6; §8's two factors are then E′/E = q and S′/S = 1/q. For light moving along the direction of motion, φ = 0 and q = 1.25 × (1 − 0.6) = 0.5, so the energy halves, 1 J to 0.5 J, and the volume doubles, 1 m³ to 2 m³. The energy per unit volume is energy divided by volume, 0.5/2 = 0.25 of what it was. That is q², the square of the amplitude factor, as it must be, since energy per unit volume goes as the amplitude squared. For light moving against the motion, φ = 180° and q = 1.25 × 1.6 = 2, so the energy doubles to 2 J and the volume halves to 0.5 m³. Why does the volume change at all, and why not by the rod's factor? A rod's ends are at rest in K, and measuring it in k at one instant shortens it by 1/γ. The surface of the light complex is not at rest in K; it moves at the speed of light along the wave normal. The instant τ = 0 in k corresponds to different times in K at different places, and in those intervals the surface moves, so its shape in k is an ellipsoid whose volume depends on the direction of the light, not on γ alone. Now the wrong model. It keeps the energy per unit volume, which does change by q², but takes the volume from a rigid body, 1/γ = 0.8, so its energy changes by q² × 0.8 = q²/γ. At φ = 0 that is 0.25 × 0.8 = 0.2 J against the packet's 0.5 J, and its volume is 0.8 m³ against 2 m³. At φ = 180°, 4 × 0.8 = 3.2 J against 2 J. For a ray at right angles to the motion in k, cos φ = β = 0.6 in K, so φ = 53.13° and q = 1.25 × (1 − 0.36) = 0.8: the wrong model gives 0.64 × 0.8 = 0.512 J and 0.8 m³, the packet 0.8 J and 1/0.8 = 1.25 m³, so both numbers expose it. At φ = 90°, cos φ = 0 and q = γ = 1.25, so q²/γ = 1.5625 × 0.8 = 1.25 J, exactly the packet's energy, and the packet's volume 1/q = 0.8 m³ is exactly the rigid body's. There the wrong model and the right one agree on everything the lab measures, which is why a test of this idea must use a ray that is not transverse in K.

Predict before the numbers

When a spherical light complex travels in the same direction as an observer moving at 0.6c (receding along x), what happens to its volume in the moving frame?

Three relations the model could have Expands to 2.0 times original volume (1/q = 2.0) Remains unchanged at 1.0 Contracts to 0.8 times volume like a rigid rod (1/gamma = 0.8)

The result appears when you choose, say you have one in mind, or skip.

Set observer speed and packet parameters

A bounded pulse of light does not transform like a solid rod. Its energy follows q = γ(1 − β cos φ) and its volume 1/q. The tempting mistake keeps light’s energy density, q², but gives the packet a rod’s volume, 1/γ, so its energy becomes q²/γ. Along the axis that is 0.2 against 0.5 at 0.6c, and for a ray transverse in the moving frame, cos φ = β, it is 0.512 against 0.8. At φ = 90° in K, q = γ and the two agree at 1.25, so that ray cannot tell them apart.

Observer speed β = v/c

Propagation angle φ in K (degrees)

Experiment settings initial energy and volume, the material rod countermodel

Initial energy in K (Joules)

Initial volume in K (m³)

Show the rigid-rod countermodel (volume 1/γ)

Changes here apply with Apply settings.

Worked example: seen from the frame moving at 0.6c, a light complex of 1 J filling 1 m³ in the stationary frame carries 0.5 J in 2 m³, its ray still at 0°.

Energy ratio E′/E (q)

0.5

γ(1 − β cos φ) = ν′/ν

Volume ratio V′/V (1/q)

2

1 / [γ(1 − β cos φ)]

Energy density ratio u′/u

0.25

q² = (A′/A)²

Lorentz factor γ

1.25

1 / √(1 − β²)

Countermodel Comparison: “Treat the packet like a rigid rod”

If the packet kept light’s energy density but had a rigid rod’s volume, its volume would scale by 1/γ = 0.8 and its energy by q²/γ = 0.2 (E′_wrong = 0.2 J). The true energy factor is q = 0.5 and the true volume factor is 1/q = 2. At cos φ = β they differ; at φ = 90° in K they agree, because there q = γ.

Values at these settings

Relativistic light complex transformation factors compared to the naive material contraction countermodel.
Energy in K (E)	1 J
Physical energy in k (E′)	0.5 J
Volume in K (V)	1 m³
Physical volume in k (V′)	2 m³
Energy ratio E′/E (q)	0.5
Volume ratio V′/V (1/q)	2
Material volume factor 1/γ	0.8
Lorentz factor γ	1.25
Transformed angle in k (φ′)	0 °
Countermodel E′ (q²/γ)	0.2 J (wrong model)
Countermodel V′ (1/γ)	0.8 m³ (wrong model)

Remarkably, the energy factor E′/E equals the Doppler frequency ratio ν′/ν = q across all angles and speeds. This exact proportionality between light energy and wave frequency holds invariantly for any bounded light packet under Lorentz transformations.

Not modeled: media and dispersion; quantum photon structure; finite pulse dispersion in dielectric; gravitational redshift; boundary diffraction at packet edges.

Worked case (readable without JavaScript)

Consider a spherical light complex of initial volume V = 1.0 m³ and total energy E = 1.0 J propagating along the x-axis (φ = 0°) in the stationary system K. An observer moves along the x-axis at speed v = 0.6c (β = 0.6, γ = 1.25).

The Doppler factor is q = γ(1 − β cos φ) = 1.25(1 − 0.6) = 0.5. Because the moving observer's simultaneous spatial plane cuts across a moving wave front, the volume of the complex in k transforms as:

\begin{gathered}\frac{V'}{V} = \frac{1}{q} = \frac{\sqrt{1 - \beta^2}}{1 - \beta\cos\varphi} = \frac{1}{0.5} = 2.0 \\ \implies V' = 2.0\text{ m}^3\end{gathered}

Meanwhile, the energy density transforms with the square of the amplitude ratio, $u'/u = q^2 = 0.25$ . The total energy in the moving frame is:

\begin{aligned}E' &= u'V' \\ &= (u q^2)(V / q) \\ &= u V q \\ &= E q \\ &= 0.5\text{ J}\end{aligned}

Notice the contrast with a rigid material body: a solid rod of volume V would undergo ordinary Lorentz contraction to V′_rod = V/γ = 0.8 m³. Treating the light complex like that rod would give E′_wrong = E/γ = 0.80 J, not 0.50 J. The case that isolates this mistake is a ray transverse in K (φ = 90°): q = γ = 1.25 while 1/γ = 0.80, so they differ by γ². A ray transverse in k (cos φ = β) gives q = 1/γ and cannot catch the mistake.

Einstein observed: “It is remarkable that the energy and the frequency of a light complex vary with the state of motion of the observer in accordance with the same law.”

\frac{E'}{E} = \frac{\nu'}{\nu} = \gamma(1 - \beta\cos\varphi)

Return to paper section 8

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