Relativistic momentum and dynamics

Momentum is worth having only because its total is conserved, and a law of physics must hold for every inertial observer. Test Newton's mv on a collision built to be symmetric. Two identical balls, A and B, each of mass m. In frame S, A moves straight up the y-axis at speed u, glances off B and comes straight back down at u. B's own frame S′ moves along x at speed v relative to S, and in S′ the story is the mirror image: B moves straight down at u and rebounds straight up. Seen from S, B's velocity follows from the velocity transformation of the previous topics: uₓ = v and u_y = −u/γ, with γ = 1/√(1 − v²/c²). Take v = 0.800c, so γ = 1.667 and B's transverse speed in S is only 0.600u. Newton's y-momentum in S before the collision is mu − 0.600μ = +0.400mu; afterwards both y-velocities have reversed, so it is −0.400mu. The total has changed by 0.800mu with no external force anywhere. Nothing about the collision is exotic; the mismatch comes entirely from how velocities transform. Either momentum conservation is not a law, or mv is not momentum.

p = m dx/dt has three ingredients. The mass belongs to the particle. The displacement dx is what the motion does in the frame doing the measuring. The dt is the one dubious ingredient: a reading on the observer's clock, which another observer disputes. Replace it with the particle's proper time τ, which every frame agrees on, and p = m dx/dτ. Since dt/dτ = γ evaluated at the particle's own speed, this is p = γmv. Return to the collision. A's factor is γ_A = 1/√(1 − u²/c²). B's speed squared in S is v² + u²/γ² = v² + u²(1 − v²/c²), so 1 − v_B²/c² = 1 − v²/c² − u²/c² + u²v²/c⁴ = (1 − v²/c²)(1 − u²/c²), which gives γ_B = γγ_A exactly. B's y-momentum is therefore γ_B m(u/γ) = γ_A μ, A's own value, up against down: the total is zero before and zero after, in S, in S′ and in every other inertial frame. The γ is not a factor bolted on to make the sums close. It is the conversion between the two clocks, and no other timekeeper would have transformed the same way. Nothing here uses energy; the next topic shows that γmc² does for energy what γmv has just done for momentum.

Keep F = dp/dt as the definition of force and differentiate p = γmv with m constant: F = m(γ dv/dt + v dγ/dt). The second term is the new physics. From γ = (1 − v²/c²)^(−1/2), the chain rule gives dγ/dt = γ³(v/c²)(dv/dt), so for a push along the motion F = mγa(1 + γ²v²/c²) = γ³ma, using γ²β² = γ² − 1. For a push at right angles to the velocity the speed is unchanged to first order, dγ/dt = 0, and F = γma. The same force is γ² times more effective sideways than forwards. At 0.600c, γ = 1.250: a forward push produces a = F/(1.953m), a sideways one F/(1.250m). At 0.950c the two are F/(32.8m) and F/(3.20m). No single speed-dependent mass can be written into F = ma to reproduce both, which is why this course keeps m invariant and lets γ live inside p. The sideways law is the one that runs bending magnets: with F = qvB always perpendicular to v, qvB = γmv²/r gives r = γmv/(qB) = p/(qB), and the orbital frequency qB/(γm) falls with speed. A fixed-frequency cyclotron loses step with its protons once γ has grown by a few per cent.

Integrate F = dp/dt for a constant force on a particle starting from rest: p = Ft, exact at every speed, with no γ anywhere in it. Relativity enters when you ask what speed that momentum represents. From γmv = Ft, γβ = Ft/(mc), and solving for β gives β = (Ft/mc)/√(1 + (Ft/mc)²). The natural time is t₁ = mc/F, when Newton would announce v = c. An electron in a uniform field of 1.00 MV m⁻¹ feels F = eE = 1.602 × 10⁻¹³ N, and its mc is 2.731 × 10⁻²² kg m s⁻¹, so t₁ = 1.70 ns. At t₁ the real speed is β = 1/√2 = 0.707; at 2t₁ it is 2/√5 = 0.894; at 10t₁, 17.0 ns into the flight, it is 10/√101 = 0.995. The momentum keeps to its straight line, 10mc after 17 ns and 100mc after 170 ns, while the speed flattens against c. The forward acceleration at 10t₁ is F/(γ³m) with γ = √101 = 10.05, so γ³ = 1015 and the electron is gaining speed at a thousandth of its starting rate. c is not a wall the electron hits. It is the asymptote of a curve whose slope the γ³ keeps stealing, and the momentum, which the force actually delivers, never notices.

Expand for small speeds: p = γmv = mv(1 + ½β² + ⅜β⁴ + …). The fractional correction is ½β² at leading order, 0.1% at β = 0.045, 1% at β = 0.14 and 10% at β = 0.42, so mv is recovered exactly where it was ever tested. Every result above must pass the same check: γ³ma → ma as β → 0, and p = Ft gives v = Ft/m for t ≪ mc/F. Dimensions pass too, because γ is a pure number and γmv, γ³ma carry the units of mv and ma. Now the boundaries. The mass m is invariant: a proton at 0.950c is the same 1.673 × 10⁻²⁷ kg it was at rest, and a mass that grows with speed is simply the γ of p filed under the wrong symbol. The three-force F = dp/dt is measured in one frame and, unlike position and momentum, does not transform by a simple rule, so a force law must be applied in the frame where it is stated. And everything here is for a single particle whose mass does not change. A collision that absorbs kinetic energy, or a decay, changes the rest masses of the bodies taking part, and momentum can only be balanced there once energy has been brought into the ledger, which is the business of the next topic.