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Derivation

Relativistic Force and Four-Acceleration

D-103 Home PU-105 Threads force · matter · symmetry Depends on The Energy-Momentum Four-Vector, Proper Time and the Four-Velocity
Statement

We define the four-force as the proper-time derivative of four-momentum, Fμ = dPμ/dτ, show that for a particle of constant rest mass it equals mass times four-acceleration, Fμ = mAμ, and reduce it to the three-force law f = mγa + mγ3(u·a)u/c2, whose parallel and perpendicular projections give f∥ = γ3ma∥ and f⊥ = γma⊥.

Why it matters

In Newtonian mechanics force and acceleration are strictly parallel: f = ma. Relativity breaks this. The same force produces different accelerations depending on whether it acts along or across the velocity, and in general a is not even parallel to f. Recovering the correct relation is essential for accelerator design, the dynamics of charged particles in fields, and any dynamical problem at speeds approaching c.

Casting force as a four-vector also makes the dynamics manifestly Lorentz-covariant: Fμ = mAμ is the relativistic replacement for Newton's second law, and it transforms cleanly between frames while the three-force law does not.

Assumptions
Constant rest mass.If m varies (radiation reaction, an object shedding or absorbing mass), the product rule adds a dm/dτ term and Fμ ≠ mAμ; the four-force acquires a component along Uμ.
Flat spacetime, inertial frame.In curved spacetime or a non-inertial frame the ordinary derivative d/dτ must be promoted to a covariant derivative and Christoffel terms appear, changing the four-acceleration.
The four-force is a "pure" mechanical force (orthogonal to four-velocity).If the interaction changes rest energy, the timelike projection UμFμ no longer vanishes; the power theorem dE/dt = f·u then fails and the neat parallel/perpendicular split must be re-derived.
Proper time is well defined along the worldline.For a massless particle dτ = 0 and the parametrisation by τ collapses; the whole construction must be replaced by an affine parameter.
Derivation
1
Pμ = mUμ = (E/c, p),   E = γmc2,  p = γmu
Four-momentum from the prior result, with Uμ = γ(c, u) the four-velocity. A
2
Fμ ≡ dPμ/dτ
Definition of the four-force (Minkowski force): differentiate the four-momentum with respect to proper time, the one invariant parameter along the worldline. A
3
Fμ = m dUμ/dτ = mAμ
Product rule on Pμ = mUμ with dm/dτ = 0 (constant rest mass); Aμ ≡ dUμ/dτ is the four-acceleration. B
4
dτ = dt/γ  ⟹  Fμ = γ dPμ/dt
Time dilation relates proper time to coordinate time; the chain rule converts the τ-derivative into a t-derivative, exposing the ordinary (lab-frame) rates. B
5
Fμ = γ ( (1/c) dE/dt,  dp/dt ) = γ ( (f·u)/c,  f )
Write components of dPμ/dt. The three-force is f ≡ dp/dt, and the power theorem dE/dt = f·u supplies the time component. B
6
UμUμ = c2 = const  ⟹  UμAμ = 0  ⟹  UμFμ = 0
Differentiating the normalisation of the four-velocity along the worldline forces four-acceleration (hence four-force) to be Minkowski-orthogonal to Uμ; component-wise this is exactly the power theorem γ2(f·u − dE/dt) = 0. C
7
f = dp/dt = m d(γu)/dt = m ( γ̇ u + γa )
Insert p = γmu and apply the product rule, with a ≡ du/dt the ordinary three-acceleration and γ̇ ≡ dγ/dt. A
8
γ̇ = d/dt (1 − u2/c2)−1/2 = γ3 (u·a)/c2
Chain rule on γ(u) using d(u2)/dt = 2u·a. This factor of γ3 is the whole source of the anisotropy. C
9
f = mγa + mγ3 (u·a)u/c2
Substitute γ̇ from step 8 into step 7. The force splits into a term along a and a correction along u; the two coincide only when a ∥ u or u·a = 0. B
10
a ∥ u:  f∥ = ma∥ (γ + γ3u2/c2) = γ3ma∥
With a along u, u·a = ua∥ and both terms point along u. The identity γ2u2/c2 = γ2 − 1 collapses the bracket to γ3. C
11
a ⊥ u:  u·a = 0  ⟹  f⊥ = γma⊥
The correction term vanishes, leaving only the mγa piece. This defines the transverse response; contrast with the γ3 longitudinal response of step 10. B
Result
Fμ = dPμ/dτ = mAμ = γ ( (f·u)/c,  f )  ;   f = mγa + mγ3(u·a)u/c2  ;   f∥ = γ3ma∥,  f⊥ = γma⊥

Reading. The four-force is what the four-momentum feels per unit of the particle's own clock; for constant rest mass it is exactly mass times four-acceleration. Projected into the lab, the spatial part is γ times the ordinary force f = dp/dt, and the time part is γ times the power delivered, divided by c. Because momentum carries the factor γ(u), a force along the motion must fight the γ3 stiffness of longitudinal inertia, while a force across the motion sees only γ. A general force is therefore not parallel to the acceleration it produces.

Units check. Pμ is kg·m·s−1 and τ is s, so Fμ = dPμ/dτ is kg·m·s−2 = N in every component. In the time slot, (f·u)/c is (N·m·s−1)/(m·s−1) = N, matching. In f∥ = γ3ma∥, γ is dimensionless so the right side is kg·m·s−2 = N.

Limiting cases
  • Newtonian limit u ≪ c: γ → 1, the γ3(u·a)u/c2 term → 0, and f → ma with f∥ = f⊥ = ma.
  • Pure longitudinal push: a ∥ u gives f = γ3ma — the "longitudinal mass" γ3m diverges as u → c, so a fixed force yields ever-smaller acceleration.
  • Pure transverse push: a ⊥ u (e.g. magnetic bending) gives f = γma — the "transverse mass" γm, softer by a factor γ2 than the longitudinal case.
  • Zero power: when f·u = 0 (force always perpendicular to velocity), the time component of Fμ vanishes and speed — hence γ and E — is constant; only direction changes.
Breaks when
  • Variable rest mass. If the particle radiates, fragments, or absorbs energy into internal degrees of freedom, dm/dτ ≠ 0 and Fμ = mAμ fails; a term (dm/dτ)Uμ parallel to the four-velocity reappears and the force is no longer orthogonal to Uμ.
  • Massless particles. With m = 0 the proper time dτ = 0 and both the four-velocity and this parametrisation are undefined; dynamics must use an affine parameter and the three-force decomposition is meaningless (no rest frame exists).
  • Non-inertial frames / gravity. In an accelerating frame or curved spacetime the plain derivative dUμ/dτ is not a tensor; it must become the covariant derivative DUμ/dτ = dUμ/dτ + ΓμαβUαUβ, and free fall (geodesic motion) has zero four-force despite non-zero coordinate acceleration.
  • Radiation reaction / self-force. For an accelerating charge the Abraham–Lorentz–Dirac self-force adds a term in the jerk (dAμ/dτ); the simple f–a relations above are only the leading, non-radiative part.
Failure modes
  • Using a single "relativistic mass" in f = mrela. There is no one factor: it is γ3m along u and γm across it. Treating mrel = γm as universal gives the wrong longitudinal acceleration by γ2.
  • Assuming a ∥ f always. For oblique force the (u·a)u term tilts the acceleration away from the force; students who set a = f/(something) misdirect it.
  • Differentiating with respect to t when the definition needs τ. Forgetting the extra γ from dτ = dt/γ gives a three-vector, not the four-force, and loses covariance.
  • Dropping the time component of Fμ. Ignoring (f·u)/c hides the power theorem and the orthogonality UμFμ = 0.
  • Using γ̇ = γa/c2 (missing the γ3 and the dot product). The correct chain rule gives γ̇ = γ3(u·a)/c2; the error propagates into both projections.
Discussion

The deep content is a single geometric fact: the four-velocity has fixed Minkowski length, UμUμ = c2. Differentiating that constraint forces the four-acceleration to be orthogonal to the four-velocity, so a "pure" force can never change the length of Uμ — only rotate it in spacetime. That rotation, projected into the lab, is what we perceive as the anisotropic response: the timelike projection of Fμ is the rate of energy gain, and the spacelike projection is the rate of momentum change, but momentum already carries the velocity-dependent factor γ(u), so its rate of change is not simply ma.

The old language of "longitudinal mass" γ3m and "transverse mass" γm is historically important (it emerged from early electron-deflection experiments) but is best read as bookkeeping for the two projections, not as evidence that mass is a direction-dependent quantity. The modern view keeps a single invariant rest mass m and locates all velocity dependence in the kinematic factors γ and γ3. This is why accelerator physicists speak of momentum and energy, not "relativistic mass": p = γmu and E = γmc2 are unambiguous, whereas "mass" is not.

The split has direct engineering consequences. A linear accelerator applies a longitudinal force and pays the γ3 penalty: near light speed almost all the added energy goes into momentum along the beam with negligible speed gain. A magnetic bending magnet applies a transverse force and pays only γ; this is why circular machines steer ultra-relativistic beams with fields that would barely deflect a slow particle, and why the cyclotron frequency drops as ω = qB/(γm).

At the covariant level, Fμ = mAμ is the template into which specific interactions are inserted. For the Lorentz force the four-force is Fμ = qFμνUν, and the antisymmetry of the field tensor Fμν = −Fνμ guarantees UμFμ = qFμνUμUν = 0 automatically — the orthogonality we derived from mass conservation is here enforced by the symmetry of electromagnetism. This is a first glimpse of how the structure of forces is dictated by symmetry: the field must be antisymmetric precisely so that it preserves rest mass. The four-acceleration invariant AμAμ = −α2 defines the proper acceleration α felt in the instantaneous rest frame, with α = γ3a∥ for one-dimensional motion — the quantity a comoving accelerometer actually reads.

Common misconceptions. (i) "Nothing can be pushed past c because it gets heavier." Rest mass is invariant; what grows without bound is the momentum, and the longitudinal inertia γ3m, so a finite force yields vanishing speed gain — the speed limit is kinematic, not a mass increase. (ii) "Force and acceleration are always parallel." Only when the force is purely along or purely across the velocity; obliquely they diverge. (iii) "Four-acceleration is zero for uniform speed." Only if velocity is also constant in direction; circular motion at fixed speed still has non-zero Aμ.

Worked examples
1
Longitudinal drive of a proton. A force acts along the velocity of a proton (m = 1.673×10−27 kg) moving at u = 0.900c. Find the acceleration produced by f∥ = 1.00×10−12 N.
Longitudinal case: a∥ = f∥/(γ3m). Set up symbolically first. A
2
γ = (1 − 0.9002)−1/2 = (0.190)−1/2 = 2.294,   γ3 = 12.07
Evaluate the Lorentz factor and its cube at u = 0.9c. A
3
a∥ = 1.00×10−12 / (12.07 × 1.673×10−27) = 1.00×10−12 / 2.019×10−26
Insert numbers into a∥ = f∥/(γ3m). A
a∥ ≈ 4.95×1013 m·s−2

Reading. The same force applied to a proton at rest would give a = f/m = 5.98×1014 m·s−2, a factor γ3 = 12.07 larger — the longitudinal stiffness at 0.9c.

Units check. N/(kg) = kg·m·s−2/kg = m·s−2. ✓

1
Transverse bending of an electron. An electron (m = 9.109×10−31 kg, q = 1.602×10−19 C) moves at u = 0.990c perpendicular to a magnetic field B = 1.00 T. Find its transverse acceleration.
Magnetic force is perpendicular to u, so u·a = 0 and a⊥ = f⊥/(γm) with f⊥ = quB. B
2
f⊥ = quB = 1.602×10−19 × (0.990 × 3.00×108) × 1.00 = 4.758×10−11 N
Compute the Lorentz force magnitude with u = 2.97×108 m·s−1. A
3
γ = (1 − 0.9902)−1/2 = (0.0199)−1/2 = 7.089,   γm = 6.457×10−30 kg
Lorentz factor at 0.99c and the transverse inertia γm. A
4
a⊥ = 4.758×10−11 / 6.457×10−30
Insert into a⊥ = f⊥/(γm). A
a⊥ ≈ 7.37×1018 m·s−2

Reading. Only one factor of γ softens the response here, versus γ3 longitudinally. The corresponding orbit radius is r = γmu/(qB) = u2/a⊥ ≈ 1.20×10−2 m, and no energy is gained because f·u = 0.

Units check. N/kg = m·s−2. ✓ For f⊥ = quB: C·(m·s−1)·T = A·s·m·s−1·kg·A−1·s−2 = kg·m·s−2 = N. ✓

Problems
  1. Show explicitly that γ2u2/c2 = γ2 − 1, and use it to collapse f∥ = ma∥(γ + γ3u2/c2) to γ3ma∥.
    Solution From γ = (1−u2/c2)−1/2, square to get γ2 = 1/(1−u2/c2), so γ2(1−u2/c2) = 1, i.e. γ2 − γ2u2/c2 = 1, giving γ2u2/c2 = γ2 − 1. Then γ + γ3u2/c2 = γ(1 + γ2u2/c2) = γ(1 + γ2 − 1) = γ·γ2 = γ3. Hence f∥ = γ3ma∥.
  2. A proton is accelerated from rest along a line by a constant force f = 3.20×10−13 N. What is its instantaneous acceleration when γ = 5.00? (Proton m = 1.673×10−27 kg.)
    Solution Motion is one-dimensional, force parallel to velocity, so a = f/(γ3m). Here γ3 = 125. Denominator = 125 × 1.673×10−27 = 2.091×10−25 kg. So a = 3.20×10−13 / 2.091×10−25 = 1.53×1012 m·s−2. (At rest the same force would give 1.91×1014 m·s−2, a factor 125 = γ3 greater.)
  3. A particle moves at u = 0.600c along x̂. A force f = (fx, fy, 0) with fx = fy = F0 acts. Find the ratio ay/ax of the resulting acceleration components, and comment on whether a is parallel to f.
    Solution Along the velocity (x) the response is fx = γ3max; across it (y) the response is fy = γmay. So ax = fx/(γ3m) and ay = fy/(γm). Ratio ay/ax = (fy/fx)·γ2 = 1×γ2. With γ = (1−0.36)−1/2 = 1.25, γ2 = 1.5625. So ay/ax = 1.56 even though fy/fx = 1. The acceleration is tilted toward the transverse direction — not parallel to f — because longitudinal inertia (γ3m) exceeds transverse inertia (γm).
  4. Starting from UμUμ = c2, prove UμFμ = 0 for constant rest mass, and translate the result into the lab-frame statement dE/dt = f·u.
    Solution Differentiate the invariant with respect to proper time: d/dτ(UμUμ) = 2Uμ(dUμ/dτ) = 2UμAμ = 0, since c2 is constant. Multiply by m: UμFμ = 0. Now write components with signature (+,−,−,−): Uμ = γ(c, u) and Fμ = γ((f·u)/c, f). Then UμFμ = γ2[c·(f·u)/c − u·f] = γ2[(f·u) − (f·u)] = 0 — an identity once we use dE/dt = f·u. Equivalently, the time component of Fμ = γdPμ/dt is γ(dE/dt)/c; equating it to γ(f·u)/c gives the power theorem dE/dt = f·u.
  5. An electron (m = 9.109×10−31 kg) circulates at constant speed u = 0.950c in a uniform magnetic field, radius r = 2.00 cm. Find (a) γ, (b) the transverse acceleration a⊥ = u2/r, (c) the three-force magnitude, and (d) the required field B.
    Solution (a) γ = (1−0.9502)−1/2 = (1−0.9025)−1/2 = (0.0975)−1/2 = 3.203. (b) u = 0.950×3.00×108 = 2.85×108 m·s−1; a⊥ = u2/r = (2.85×108)2/0.0200 = 8.12×1016/0.0200 = 4.06×1018 m·s−2. (c) Perpendicular force: f = γma⊥ = 3.203 × 9.109×10−31 × 4.06×1018 = 1.18×10−11 N. (d) From f = quB: B = f/(qu) = 1.18×10−11/(1.602×10−19 × 2.85×108) = 1.18×10−11/(4.566×10−11) = 0.259 T. Equivalently B = γmu/(qr), the relativistic gyroradius relation.