Photon energy, momentum and the massless limit

Relativistic momentum is usually met as p = γmv and energy as E = γmc². Put m = 0 and v = c into either and you get 0 × ∞: an indeterminate form, which is a refusal to answer, not an answer of zero. The relation that survives is the one with no mass buried in a denominator, E² = (pc)² + (mc²)². Read it as what it is — the statement that the four-vector (E/c, p⃗) has invariant length mc, a number every inertial observer agrees on. Nothing in relativity forbids that length from being zero. Take the m = 0 branch and you are left with E = pc: a particle whose energy and momentum are rigidly proportional, with mass appearing nowhere. Its speed follows from β = pc/E, which is exactly 1 — and 1 in every frame, because a boost multiplies E and pc by the same factor. A massless particle is not a massive particle with the mass switched off; it is a different solution of the same invariant.

Planck and Einstein supply the quantum condition E = hf; relativity supplies E = pc. Put them together with fλ = c and the whole photon kinematics is one line: p = E/c = hf/c = h/λ. Carry hc = 1239.84 eV nm and the arithmetic stops needing a calculator — a 500 nm photon has E = 1239.84/500 = 2.48 eV and p = h/λ = 6.63 × 10⁻³⁴ / 5.00 × 10⁻⁷ = 1.33 × 10⁻²⁷ kg m s⁻¹. Quote that momentum as 2.48 eV/c and the conversion disappears altogether. It is worth seeing how strange p = h/λ becomes once you leave photons. An electron with the same 500 nm wavelength carries the same momentum, 1.33 × 10⁻²⁷ kg m s⁻¹, but its energy is p²/2m = 6.0 μeV — four hundred thousand times smaller than the photon's 2.48 eV. Same de Broglie relation, same momentum, wildly different energy, because the two obey different dispersion relations: E = pc against E = p²/2m.

Because (E/c, p⃗) transforms exactly like (ct, x⃗), the shift needs no wave picture at all. For a boost of speed v along x, E′ = γ(E − v pₓ) and p′ₓ = γ(pₓ − vE/c²). Let the photon travel at angle θ to the boost axis, so pₓ = (E/c)cos θ, and the first equation collapses to E′ = γE(1 − β cos θ) — the relativistic Doppler formula, obtained as a change in one component of a four-vector. Longitudinally, with θ = 0, it factorises to E′ = E√((1 − β)/(1 + β)): at β = 0.80, γ = 5/3 and γ(1 − β) = 1/3 exactly, so a 500 nm line arrives at 1500 nm carrying a third of its energy. The second equation then gives p′ₓ = E′/c, so the boosted photon is still null — no frame can hand it a mass. The transverse case rewards care: θ = 90° in the source's frame gives E′ = γE, a blueshift, while θ′ = 90° in the receiver's frame gives E′ = E/γ, the familiar transverse redshift. Aberration is why both are true, and naming the frame the angle belongs to is the whole of the bookkeeping.

An atom of mass M at rest emits a photon; momentum conservation hands the atom −p and the photon +p, so the photon cannot carry the full rest-energy difference E₀. Square the four-momentum ledger rather than expanding square roots: with the photon moved to the other side, (Mc² − E_γ)² − E_γ² = (M_g c²)², giving E_γ = E₀[1 − E₀/(2Mc²)]. The missing piece, T_R = E_γ²/(2Mc²), is the recoil kinetic energy, and the emitted line sits that far to the red of line centre. Whether it matters is entirely a question of scale. Sodium's D line at 2.105 eV from a ²³Na atom (Mc² = 2.14 × 10¹⁰ eV) is displaced by 1.03 × 10⁻¹⁰ eV, a fractional shift of 4.9 × 10⁻¹¹ — five orders of magnitude below the 3.3 × 10⁻⁶ thermal Doppler width of a 500 K vapour. The 14.4 keV gamma of ⁵⁷Fe is displaced by 1.95 × 10⁻³ eV, about 4 × 10⁵ natural linewidths, so a free nucleus cannot absorb the photon its twin emitted. Mössbauer's answer was to deny the nucleus a free recoil: bound in a lattice, a definite fraction of nuclei emit without exciting a single phonon, and the momentum is taken up by the crystal as a whole, whose macroscopic M drives T_R below the linewidth.

A null four-vector stays null under every Lorentz transformation, and E′ = γE(1 − β cos θ) sends E′ → 0 only as β → 1, which no frame reaches. So there is no frame in which a photon is at rest, and the phrase "from the photon's point of view" names nothing — the boost that would take you there is singular, not merely extreme. The interval along a light path is zero, so a photon has no proper time to run and no clock to set against yours. One further absence is easy to miss: a massless helicity-1 particle admits no self-adjoint position operator that transforms as a position should, so there is no ψ(x) whose |ψ|² is a probability density for finding the photon at a point. What a detector registers is an absorption event somewhere in a finite volume, and what the field supplies is an energy density quadratic in the field amplitudes — not a probability of finding anything at a point. Photon number, energy and momentum are perfectly good observables; "where the photon is between emission and absorption" is not one.

That m_γ = 0 is a physical claim, and it is falsifiable. If the photon had mass the vacuum would be dispersive — β = √(1 − (m_γc²/E)²) would depend on frequency, so a sharp astrophysical pulse would arrive colour by colour, the reddest last. Coulomb's law would acquire a Yukawa factor e^(−r/λ_C) with λ_C = ħ/(m_γc), failing beyond that range, and the field would carry three polarisation states rather than two, the extra one longitudinal. Every one of these has been looked for. Solar-system tests of the interplanetary magnetic field bound m_γ below about 10⁻¹⁸ eV/c², some 5 × 10²³ times under the electron's 511 keV, which corresponds to a range ħ/(m_γc) of roughly 2 × 10¹¹ m — longer than the Earth–Sun distance of 1.5 × 10¹¹ m. So E = pc for light is not an idealisation to worry about at any energy in this course; knowing what would break if it failed is what makes it physics rather than notation.