University Physics IV · Photons and Matter Waves · 4.1
Photon Energy, Momentum & the Massless Limit
Every photon relation in this unit comes out of one move: set m = 0 in E² = (pc)² + (mc²)². This lesson makes that move carefully, converts it into p = h/λ, and then spends it — on Doppler shifts read as energy boosts, on the recoil that reddens an emitted line, and on saying plainly what a massless particle cannot have.
Build the model
Connect the measurement to the mechanism.
Ask Newtonian mechanics what momentum a massless particle carries and the answer is forced: p = mv with m = 0 leaves nothing, so light has to be exempted from mechanics — which is not an answer, only a note that the framework has run out. Relativity dissolves the problem by promoting a different relation to fundamental. Mass stops being a factor you multiply a velocity by and becomes a label attached to the four-momentum: the Minkowski length of (E/c, p⃗), fixed for a given particle whoever measures it and whatever it is doing.
Zero is a permitted value of that label, and the particle wearing it obeys E = pc — one equation with no m in it, which is also why light travels at c for everybody instead of at some speed a frame could argue about. Everything after that is bookkeeping. Planck's condition E = hf turns the fixed proportion into p = h/λ, the relation Compton and de Broglie will both run on.
Because four-momentum transforms exactly as (ct, x⃗) does, the Doppler shift stops being a wave argument and becomes one component of a boost, E′ = γE(1 − β cos θ), where the only thing to keep straight is which frame owns the angle θ. Turn the same ledger on an atom shedding a photon and the atom is obliged to keep part of the energy as recoil, so the line emerges red of the transition by about E₀²/(2Mc²): unmeasurable in a sodium lamp, and hundreds of thousands of linewidths wide of resonance for a 14.4 keV nuclear gamma. Set against all that, the null length costs the photon three things a massive particle keeps — a frame in which it sits still, a clock that ticks along its path, and any operator that would say where it is.
- Simple definition
- A photon is the massless quantum of the electromagnetic field: because m = 0, its energy and momentum are locked together as E = pc, so fixing E = hf fixes p = h/λ, and neither can change without the other.
- Example
- A 500 nm photon carries E = 1239.84 eV nm ÷ 500 nm = 2.48 eV and p = E/c = 1.33 × 10⁻²⁷ kg m s⁻¹; a 1.0 mW beam of them delivers momentum at P/c = 3.3 × 10⁻¹² N.
The invariant is a length, not a formula for p. Zero length is allowed — and it is what a photon has.
E and pc carry the same unit, J or eV; m is the invariant mass, identical in every frame
λ = 500 nm gives 2.48 eV and 1.33 × 10⁻²⁷ kg m s⁻¹, or simply 2.48 eV/c if momentum is kept in eV/c.
hc = 1239.84 eV nm = 1.9864 × 10⁻²⁵ J m; h = 6.626 × 10⁻³⁴ J s
Conserve all four components through a collision, then square the total to read off the invariant mass.
Components boost exactly like (ct, x⃗); the value 0 is the same number in every frame
θ = 0 gives E′ = E√((1 − β)/(1 + β)); θ = 90° in the source's frame gives γE, a blueshift.
θ is the photon's direction in the unprimed frame; the primed frame moves at +βc along x
⁵⁷Fe at 14.4 keV recoils by 1.95 meV — 4 × 10⁵ natural linewidths, so free nuclei miss resonance.
E₀ is the rest-energy gap between the two states; Mc² = 931.494 MeV × A for a nucleus of mass number A
The bound sits some 5 × 10²³ below the electron's 511 keV, so E = pc is safe at any energy you will meet.
A nonzero mγ makes the vacuum dispersive and gives Coulomb a finite range ħ/(mγc) ≈ 2 × 10¹¹ m
Set m = 0 in the invariant, not in p = γmv
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.
From E = pc to p = h/λ, with the numbers attached
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.
The Doppler shift is a Lorentz boost of the energy
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 emitting atom recoils, so the line moves
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 − Eγ)² − Eγ² = (Mg c²)², giving Eγ = E₀[1 − E₀/(2Mc²)]. The missing piece, TR = Eγ²/(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 TR below the linewidth.
No rest frame, no proper time, no position operator
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.
The massless limit is a measurement, not a definition
That mγ = 0 is a physical claim, and it is falsifiable. If the photon had mass the vacuum would be dispersive — β = √(1 − (mγc²/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 λC = ħ/(mγc), 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 mγ below about 10⁻¹⁸ eV/c², some 5 × 10²³ times under the electron's 511 keV, which corresponds to a range ħ/(mγc) 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.
Change one variable at a time
Make the relationship visible.
Set the rest energy to zero: the hyperbola collapses onto the dashed 45° line and β locks at 1. Now drag the boost slider — E and pc both change and the dot slides along the null line toward the origin, but never arrives, and the invariant readout never moves off zero.
LAB ENERGY E1.34 MeV
β = pc/E0.894
BOOSTED ENERGY E′1.03 MeV
√(E′² − (p′c)²)0.60 MeV
Live interpretationLAB ENERGY E: 1.34 MeV. β = pc/E: 0.894. BOOSTED ENERGY E′: 1.03 MeV. √(E′² − (p′c)²): 0.60 MeV
Catch the common trap
Explain before calculating.
A 500 nm photon (2.48 eV) travels in the +x direction. An observer moves in the +x direction at β = 0.60, so she is running away from the source. Using the four-momentum boost E′ = γ(E − v pₓ), what energy does she measure?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA green laser pointer emits 5.0 mW at 532 nm. Find the energy and momentum of one photon, the number of photons leaving the aperture each second, and the momentum the beam delivers per second.
- Energy from E = hc/λ with hc = 1239.84 eV nm: E = 1239.84/532 = 2.3305 eV, which is 2.3305 × 1.602177 × 10⁻¹⁹ = 3.734 × 10⁻¹⁹ J.
- Momentum comes from the massless branch, not from any mass: p = E/c = 3.734 × 10⁻¹⁹ ÷ 2.998 × 10⁸ = 1.25 × 10⁻²⁷ kg m s⁻¹. Cross-check with p = h/λ = 6.626 × 10⁻³⁴ ÷ 5.32 × 10⁻⁷ = 1.25 × 10⁻²⁷ kg m s⁻¹ — the same number, as E = pc guarantees.
- Photon rate: N = P/E = 5.0 × 10⁻³ ÷ 3.734 × 10⁻¹⁹ = 1.34 × 10¹⁶ photons per second.
- Momentum delivered per second: N p = 1.34 × 10¹⁶ × 1.25 × 10⁻²⁷ = 1.7 × 10⁻¹¹ N. That equals P/c = 5.0 × 10⁻³ ÷ 2.998 × 10⁸ = 1.7 × 10⁻¹¹ N exactly, which it must when every photon obeys E = pc.
AnswerE = 2.33 eV = 3.734 × 10⁻¹⁹ J; p = 1.25 × 10⁻²⁷ kg m s⁻¹; 1.34 × 10¹⁶ photons per second, carrying 1.7 × 10⁻¹¹ N.
MediumA hydrogen cloud emits the Lyman-α line, rest wavelength 121.57 nm, and recedes from Earth at β = 0.20. Using the photon four-momentum rather than any wave argument, find the energy, wavelength and momentum received on Earth, and the redshift z.
- In the cloud's frame the photon has E = hc/λ = 1239.84/121.57 = 10.199 eV, and since m = 0 its momentum is pc = 10.199 eV, i.e. p = 10.199 eV/c.
- Earth moves in the same direction as the photon (that is what receding means here), so take θ = 0 in E′ = γE(1 − β cos θ), giving E′ = γE(1 − β).
- γ = 1/√(1 − 0.20²) = 1/√0.96 = 1.0206, so γ(1 − β) = 1.0206 × 0.80 = 0.81650 — which is √((1 − β)/(1 + β)) = √(0.8/1.2), as it must be.
- E′ = 10.199 × 0.81650 = 8.327 eV, and λ′ = 1239.84/8.327 = 148.89 nm.
- Momentum: p′ = E′/c = 8.327 eV/c, so E′² − (p′c)² = 0 still. A boost stretches a photon's four-momentum but cannot give it a mass.
- Redshift: z = λ′/λ − 1 = 148.89/121.57 − 1 = 0.225, matching √((1 + β)/(1 − β)) − 1 = 1.2247 − 1.
AnswerE′ = 8.33 eV, λ′ = 148.9 nm, p′ = 8.33 eV/c, and z = 0.225.
HardA free ⁵⁷Fe nucleus at rest emits its 14.4 keV gamma ray. The excited state has natural linewidth Γ = 4.7 × 10⁻⁹ eV, and 1 u c² = 931.494 MeV. Find the recoil energy, the separation of the emission and absorption lines in linewidths, and the source speed that would close that gap.
- Rest energy: Mc² ≈ 57 × 931.494 MeV = 5.310 × 10⁴ MeV = 5.310 × 10¹⁰ eV. (The true mass 56.935 u changes TR by 0.1%, invisible here.)
- Squaring the four-momentum ledger gives Eγ = E₀[1 − E₀/(2Mc²)], so the photon falls short of the transition energy by TR = Eγ²/(2Mc²) = (1.44 × 10⁴)² ÷ (2 × 5.310 × 10¹⁰) = 1.95 × 10⁻³ eV. Treating the recoil non-relativistically is safe: TR/Mc² ≈ 4 × 10⁻¹⁴.
- Absorption is the mirror image — exciting an identical nucleus at rest costs E₀ + TR — so the emission and absorption lines sit 2TR = 3.9 × 10⁻³ eV apart.
- In linewidths: 3.9 × 10⁻³ ÷ 4.7 × 10⁻⁹ = 8.3 × 10⁵. There is no overlap whatever, which is why free nuclei will not resonantly scatter their own gamma ray.
- Closing the gap by moving the source is the same energy boost: β = ΔE/Eγ = 3.9 × 10⁻³ ÷ 1.44 × 10⁴ = 2.7 × 10⁻⁷, so v = 81 m s⁻¹. One linewidth alone needs only 9.8 × 10⁻⁵ m s⁻¹ = 0.098 mm s⁻¹.
- This is why the Mössbauer effect is worth having: a lattice-bound nucleus can emit recoil-free, the 2TR gap collapses below Γ, and a drive of a few mm s⁻¹ is then enough to sweep across the whole line.
AnswerTR = 1.95 meV ≈ 4.2 × 10⁵ Γ; the emission and absorption lines lie 3.9 meV = 8.3 × 10⁵ linewidths apart; 81 m s⁻¹ closes the gap, against 0.098 mm s⁻¹ per linewidth.