University Physics IV · Special Relativity II · 3.6
Relativistic Momentum & Dynamics
Newton's momentum is not so much wrong as not portable: a collision that balances in one frame stops balancing in another. This lesson builds the repair, follows it into a force law that splits in two, and watches a constant push chase c forever without arriving.
Build the model
Connect the measurement to the mechanism.
Momentum earns its place by being conserved, and a conservation law is only worth having if every inertial observer agrees on it. Newton's p = mv fails that test the moment velocities compose relativistically: the boost sends u to (u − v)/(1 − uv/c²), a nonlinear map, so a sum Σmu that balances before and after a collision in one frame has no reason to balance in another. The repair is small and forced.
Keep the mass and the displacement; replace the observer's clock with the particle's own. Differentiating by proper time τ, which every frame agrees on, gives p = m dx/dτ = γmv — the spatial part of a four-vector that boosts linearly, so a balance struck in one frame is automatically struck in all of them. The cost is paid in dynamics.
Because γ now depends on speed, F = dp/dt no longer factors into a mass times an acceleration: push along the motion and F = γ³ma, push across it and F = γma, the same force producing accelerations that differ by γ². No single speed-dependent mass serves both, which is why relativistic mass is abandoned and m means the one invariant number a particle carries into every frame. What is left is an asymptote: under a constant force the momentum grows without bound and linearly in t, while the speed climbs towards c and never arrives.
- Simple definition
- Relativistic momentum is mass times the rate at which position changes per unit of the particle's own proper time, p = m dx/dτ = γmv — the only definition of momentum whose conservation survives a change of inertial frame.
- Example
- A proton at 0.600c has γ = 1.250, so its momentum is 3.76 × 10⁻¹⁹ kg m s⁻¹ — 25% above the Newtonian mv of 3.01 × 10⁻¹⁹ kg m s⁻¹, or 704 MeV/c in accelerator units.
The γ is not an extra force-like factor bolted on: it is exactly dt/dτ, the conversion between the observer's clock and the particle's own.
m invariant mass in kg, v the frame's velocity in m s⁻¹, τ the particle's own proper time in s
A nonlinear map applied term by term, so Σmu conserved in S says nothing at all about Σmu′ in S′.
uᵢ the velocities in S and v the boost speed, both m s⁻¹; the denominator differs particle by particle
At 0.900c the same force is γ² = 5.26 times more effective sideways than forwards.
F in N and a in m s⁻²; ∥ and ⊥ are taken relative to v at that instant, not to fixed axes
Shows where the extra powers come from: the force must change γ as well as v.
Holds for any angle between F and v, and then F and a are not parallel
Momentum climbs linearly and for ever; the speed it corresponds to saturates at c.
F constant in N, t the lab time in s, starting from v = 0 at t = 0; β = v/c
A constant lab force is a constant proper acceleration — the γ³ is a clock-and-ruler effect.
α the acceleration read in the instantaneous rest frame, m s⁻²; at 1 g, c²/α = 0.968 light-year
Newton's momentum does not survive a boost
Take a frame S in which two identical particles of mass m approach along x at +0.600c and −0.600c, collide and stick, leaving the composite at rest. The Newtonian ledger reads m(+0.600c) + m(−0.600c) = 0 before and 0 after: conserved. Now watch the same event from S′, the frame in which the left-moving particle is initially at rest, which moves at −0.600c relative to S. Velocities compose as u′ = (u + 0.600c)/(1 + 0.600u/c), so the incoming particle is seen at 1.200c/1.360 = 0.882c, its target at rest, and the composite at 0.600c. The Newtonian ledger now reads 0.882mc before and 2m × 0.600c = 1.200mc after — a 36% mismatch produced by nothing but a change of observer. The fault is structural rather than numerical: u′ is a nonlinear function of u, so a linear sum that balances in one frame has no reason to balance in another. Either momentum conservation is not a law, or p = mv is not momentum.
Differentiate by proper time, not by the frame's clock
p = m dx/dt has three ingredients and only one of them is fatally frame-dependent. The mass is a property of the particle; the displacement is what the motion actually does; the dt is a reading on the observer's clock. Replace that clock with the particle's own, whose ticks every frame agrees on, and you get p = m dx/dτ. Since dt/dτ = γ, this is p = γmv. The payoff is not that a γ has appeared but that this object is the spatial part of the four-vector (E/c, p), which boosts by the same matrix as (ct, x). A four-vector transforms linearly, so if the total four-momentum of a system is unchanged by an interaction in one frame, the boosted totals are equal too — automatically, in every frame. Run the boosted collision again with γmv: 2.125 × 0.882mc + 0 = 1.875mc before, and 1.250 × 2.500m × 0.600c = 1.875mc after. It balances, provided the composite is assigned the invariant mass 2.500m that the energy ledger gives it, not the 2m Newton would have written.
One force, two acceleration laws
Keep F = dp/dt as the definition of force and differentiate p = γmv. The mass is constant; γ is not. From γ = (1 − v²/c²)(−1/2), dγ/dt = γ³(v⋅a)/c². For a push along the motion, v⋅a = va, so dp/dt = m(γa + vγ³va/c²) = mγa(1 + γ²β²) = γ³ma, using γ²β² = γ² − 1. For a push across the motion the speed is unchanged to first order, γ is momentarily constant, and dp/dt = γma. In three dimensions the two cases are the single relation F = γma + γ³m(v⋅a)v/c². Numbers make the split concrete: an electron at 0.900c has γ = 2.294, so a 1.00 × 10⁻¹⁵ N force delivers 9.09 × 10¹³ m s⁻² forwards and 4.79 × 10¹⁴ m s⁻² sideways. Same particle, same speed, same force, and the response differs by γ² = 5.26 — because only the forward push has to change γ.
Why relativistic mass has no job left
Historically the γ³ and the γ were swallowed into direction-dependent 'longitudinal' and 'transverse' masses, γ³m and γm, so that F = ma could be kept. It is a bookkeeping trick that fails as soon as the force sits at a general angle to the velocity, where F and a are not even parallel and no scalar can relate them. Modern practice takes the opposite route: hold m invariant, defined by E² − (pc)² = (mc²)², and let γ live where it belongs, in p = γmv and E = γmc². The reward is that m becomes a label on the particle — 938.272 MeV/c² for a proton sitting in a bottle and for one circulating at 7 TeV — while E and p carry every trace of the speed. It also removes a standing source of nonsense: saying that mass increases with speed invites the conclusion that a fast particle's gravity grows or that it might collapse, when in its own rest frame nothing whatever has changed.
A constant force never reaches c
Newton's answer to a constant force is v = (F/m)t, unbounded. The relativistic account starts identically, because F = dp/dt integrates to p = Ft from rest whatever the speed — that part is exact and has no γ in it. But p = γmv, so γβ = Ft/mc, and solving for the speed gives v = c(Ft/mc)/√(1 + (Ft/mc)²), which rises towards c and stops. Momentum is what grows without limit; speed is what saturates. Set F/m = g and the natural scale is c/g = 3.06 × 10⁷ s, just under a year. After one year of lab time the momentum has reached 1.03mc, so Newton would claim v = 1.03c; the true speed is 0.718c with γ = 1.437, and the same force is now buying only 1/γ³ = 34% of the acceleration it bought at the start. The worldline is a hyperbola, (x + c²/α)² − (ct)² = (c²/α)², whose scale c²/g = 0.968 light-year sets how sharply it bends.
The low-speed limit, and two pieces of fine print
Expanding the factor gives p = γmv = mv(1 + ½β² + ⅜β⁴ + …), so the relativistic correction to momentum is fractionally ½β² at leading order. It reaches 1% at β ≈ 0.14, about 4.2 × 10⁴ km s⁻¹, and 10% at β = 0.417 — which is why ballistics, chemistry and every laboratory before the electron never noticed. Two pieces of fine print travel with the result. First, the F here is the three-force dp/dt measured in one inertial frame; unlike the four-force it does not transform simply between frames, so a force law quoted in one frame must be re-derived rather than relabelled in another. Second, everything above is for a single particle of fixed invariant mass. Once a system can turn kinetic energy into mass, as the sticking collision did, the total momentum is still γMV — but M is the system's invariant mass and is not the sum of its parts.
Change one variable at a time
Make the relationship visible.
Hold F/m at 1.00 g: by 1.0 yr the momentum has reached 1.03mc, so Newton would claim v = 1.03c, but β is only 0.718 and the same force now buys 34% of its original acceleration. Push F/m to 1.50 g and drag t to 2.0 yr, and that share collapses to 3%.
SPEED β = v/c0.718
MOMENTUM p/mc = γβ1.033
LORENTZ FACTOR γ1.437
ACCELERATION a/(F/m)0.337
Live interpretationSPEED β = v/c: 0.718. MOMENTUM p/mc = γβ: 1.033. LORENTZ FACTOR γ: 1.437. ACCELERATION a/(F/m): 0.337
Catch the common trap
Explain before calculating.
A particle of mass m is moving at 0.800c. Two forces of equal magnitude F are applied at that instant — one along its velocity, one perpendicular to it. What is the ratio a∥/a⊥ of the accelerations they produce?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA proton (m = 1.673 × 10⁻²⁷ kg) travels at 0.600c. Find its relativistic momentum in SI units and in MeV/c, and compare it with the Newtonian mv.
- β = 0.600, so γ = 1/√(1 − 0.600²) = 1/√0.640 = 1/0.800 = 1.250.
- The speed itself is v = 0.600 × 2.998 × 10⁸ = 1.799 × 10⁸ m s⁻¹, so the Newtonian product is mv = 1.673 × 10⁻²⁷ × 1.799 × 10⁸ = 3.009 × 10⁻¹⁹ kg m s⁻¹.
- Relativistic momentum: p = γmv = 1.250 × 3.009 × 10⁻¹⁹ = 3.762 × 10⁻¹⁹ kg m s⁻¹, a 25% increase — the whole excess is the factor γ, since m and v are unchanged.
- In accelerator units use pc = βγ mc² with mc² = 938.3 MeV: pc = 0.600 × 1.250 × 938.3 = 703.7 MeV, so p = 704 MeV/c. Check in SI: 3.762 × 10⁻¹⁹ × 2.998 × 10⁸ = 1.128 × 10⁻¹⁰ J = 704 MeV.
Answerp = 3.76 × 10⁻¹⁹ kg m s⁻¹, equivalently 704 MeV/c — exactly γ = 1.25 times the Newtonian 3.01 × 10⁻¹⁹ kg m s⁻¹.
MediumAn electron (m = 9.109 × 10⁻³¹ kg) is moving at 0.900c. A force of magnitude 1.00 × 10⁻¹⁵ N is applied, first along its velocity and then perpendicular to it. Find both accelerations and compare them with the Newtonian F/m.
- γ = 1/√(1 − 0.900²) = 1/√0.190 = 2.294, so γ² = 5.263 and γ³ = 12.07.
- Newtonian benchmark: F/m = 1.00 × 10⁻¹⁵ ÷ 9.109 × 10⁻³¹ = 1.098 × 10¹⁵ m s⁻².
- Along the motion the force must change γ as well as v, so F = γ³ma and a∥ = 1.098 × 10¹⁵ ÷ 12.07 = 9.09 × 10¹³ m s⁻².
- Across the motion the speed — and therefore γ — is momentarily unchanged, so F = γma and a⊥ = 1.098 × 10¹⁵ ÷ 2.294 = 4.79 × 10¹⁴ m s⁻².
- The same push is γ² = 5.26 times more effective sideways than forwards, while the Newtonian estimate overshoots the forward answer by γ³ = 12.1. No single 'relativistic mass' can reproduce both numbers at once.
Answera∥ = 9.09 × 10¹³ m s⁻² and a⊥ = 4.79 × 10¹⁴ m s⁻², against a Newtonian F/m = 1.10 × 10¹⁵ m s⁻² for both.
HardA proton (m = 1.673 × 10⁻²⁷ kg) moves at 0.950c when a force of magnitude 1.00 × 10⁻¹² N is applied at 45.0° to its velocity. Find the two components of the acceleration, the magnitude of a, and the angle a makes with v.
- γ = 1/√(1 − 0.950²) = 1/√0.0975 = 3.203, so γ² = 10.26 and γ³ = 32.85.
- Resolve the force onto the velocity direction: F∥ = F⊥ = 1.00 × 10⁻¹² × cos 45.0° = 7.071 × 10⁻¹³ N, and either component divided by the mass gives 7.071 × 10⁻¹³ ÷ 1.673 × 10⁻²⁷ = 4.227 × 10¹⁴ m s⁻².
- The two components answer to different laws. Along the motion F∥ = γ³ma∥, so a∥ = 4.227 × 10¹⁴ ÷ 32.85 = 1.287 × 10¹³ m s⁻²; across it F⊥ = γma⊥, so a⊥ = 4.227 × 10¹⁴ ÷ 3.203 = 1.320 × 10¹⁴ m s⁻².
- Magnitude: |a| = √((1.287 × 10¹³)² + (1.320 × 10¹⁴)²) = 1.326 × 10¹⁴ m s⁻², where the Newtonian estimate F/m = 5.98 × 10¹⁴ m s⁻² is 4.51 times too large.
- Direction: tan θₐ = a⊥/a∥ = γ² tan 45.0° = 10.26, so θₐ = 84.4° from the velocity while the force sits at 45.0°. The general rule is tan θₐ = γ² tan θF, so only a purely parallel or purely perpendicular push leaves a lined up with F — which is exactly why no scalar mass can be written into F = ma.
Answera∥ = 1.29 × 10¹³ m s⁻² and a⊥ = 1.32 × 10¹⁴ m s⁻², giving |a| = 1.33 × 10¹⁴ m s⁻² at 84.4° to the velocity: the acceleration is dragged 39.4° away from the 45.0° force.