Proper Time and the Four-Velocity

Derivation

Proper Time and the Four-Velocity

D-097 Home PU-105 Threads symmetry · matter Depends on Invariance of the Spacetime Interval, Time Dilation

Statement

Along any timelike worldline the invariant quantity c dτ defined by c² dτ² = ds² measures elapsed proper time — the time read by a clock carried on that worldline. Differentiating the coordinates with respect to this invariant scalar yields the four-velocity U^μ = dx^μ/dτ = γ(c, v), a timelike four-vector whose Minkowski norm is fixed for every observer at U^μU_μ = c², independent of the state of motion.

Why it matters

Coordinate time t is frame-dependent and therefore useless as a parameter for building objects that must transform cleanly between inertial frames. Proper time is the unique parameter that every observer agrees on, so quantities differentiated with respect to it inherit definite tensor character. The four-velocity is the seed of all of relativistic kinematics: four-momentum is p^μ = mU^μ, four-acceleration is A^μ = dU^μ/dτ, and the constant-norm constraint U·U = c² propagates into the mass-shell condition p·p = m²c².

Physically, the constancy of the norm encodes a deep fact: nothing "moves faster or slower through spacetime" — every timelike object advances through spacetime at the fixed rate c, merely re-apportioning that motion between the time direction and the space directions as its speed changes.

Assumptions

Flat Minkowski spacetime with metric signature (+,−,−,−).In curved spacetime the flat metric η_μν is replaced by g_μν(x); proper time and the four-velocity survive locally but the norm becomes g_μνU^μU^ν = c² and partial derivatives must become covariant. The worldline is timelike, so ds² > 0 everywhere on it.If any segment is null or spacelike then dτ is zero or imaginary, no comoving clock exists, and U^μ = dx^μ/dτ is undefined. The worldline is smooth (C¹): the tangent dx^μ/dτ exists and is continuous.At an idealised instantaneous "kink" (an infinite-acceleration event, as in the naive twin turnaround) the four-velocity is discontinuous and must be handled as a limit of smooth boosts. Invariance of the interval ds² and the time-dilation relation dτ = dt/γ are already established (prior results).Without interval-invariance there is no observer-independent scalar to differentiate against, and the four-velocity loses its tensor character.

Derivation

ds² = c²dt² − dx² − dy² − dz²

The invariant interval in Minkowski coordinates; invariance is a prior result. A

c² dτ² ≡ ds² (defined only where ds² > 0)

Definition of proper time as interval arc length divided by c; legal because ds² is a scalar, so τ is the same number in every frame. B

c² dτ² = c²dt² (1 − (dx²+dy²+dz²)/(c²dt²))

Algebraic factoring of c²dt² out of the spatial terms — pure rearrangement, no physics added. A

dτ² = dt² (1 − v²/c²), v² ≡ (dx/dt)² + (dy/dt)² + (dz/dt)²

Identify the coordinate three-speed v = |dx/dt|; cancel c². A

dτ = dt √(1 − v²/c²) = dt/γ, γ ≡ 1/√(1 − v²/c²)

Take the positive root (proper time increases with coordinate time); this reproduces the prior time-dilation result and fixes dt/dτ = γ. B

U^μ ≡ dx^μ/dτ

Definition of the four-velocity. Because τ is an invariant scalar and dx^μ is a four-vector displacement, the ratio transforms as a contravariant four-vector — this is the whole point of using τ rather than t. C

U^μ = (dx^μ/dt)(dt/dτ) = γ dx^μ/dt

Chain rule with the invariant parameter, substituting dt/dτ = γ from Step 5. B

U⁰ = γ d(ct)/dt = γc, Uⁱ = γ dxⁱ/dt = γvⁱ

Evaluate components using x⁰ = ct and xⁱ = (x,y,z). A

U^μ = γ(c, v_x, v_y, v_z) = γ(c, v)

Collect the components into the four-velocity. A

U^μU_μ = η_μνU^μU^ν = (U⁰)² − |U|² = γ²c² − γ²v²

Contract with the Minkowski metric η_μν = diag(+,−,−,−); the spatial part carries the minus sign. B

U^μU_μ = γ²c²(1 − v²/c²) = γ²c²·γ⁻² = c²

Substitute 1 − v²/c² = γ⁻² from the definition of γ; the speed dependence cancels exactly. C

Result

U^μ = γ(c, v), U^μU_μ = c² (constant, timelike)

Reading. Proper time is the invariant arc length dτ = ds/c along a timelike worldline — the reading of a clock carried by the object. The four-velocity is the unit tangent to that worldline (up to the factor c): its time component γc measures how fast coordinate time flows relative to the clock, and its spatial component γv is the momentum-per-unit-mass direction. The norm is fixed at c² for every observer and every speed — accelerating changes the direction of U^μ in spacetime but never its length.

Units check. τ has units of s (from ds/c, m divided by m/s). U^μ = dx^μ/dτ has units m/s; γ is dimensionless and both c and v are m/s, so γ(c,v) is consistent. The norm U·U has units (m/s)² = m²/s², matching c². ✓

Limiting cases

Non-relativistic (v ≪ c): γ → 1, so U^μ → (c, v) — the spatial part reduces to the ordinary Newtonian velocity and dτ → dt (clocks agree).
Rest frame: v = 0 ⇒ U^μ = (c, 0, 0, 0) — a body at rest still moves through the time axis at rate c. This is the frame in which τ = t by definition.
Ultra-relativistic (v → c): γ → ∞, both U⁰ and |U| diverge while their difference of squares stays exactly c²; the worldline approaches a null line and dτ → 0.
Photon limit: for a genuinely null worldline dτ = 0, the construction fails, and one parametrises with an affine parameter λ instead of τ.

Breaks when

Null (lightlike) worldlines. Massless particles have ds² = 0 so dτ = 0 everywhere: no comoving clock exists and dx^μ/dτ is a 0/0 form. Four-velocity is undefined; use an affine parameter and a null tangent k^μ with k·k = 0.
Spacelike separations. If a putative worldline is spacelike (ds² < 0, i.e. v > c), then c²dτ² < 0 and τ is imaginary — such a curve cannot be a physical trajectory, and the timelike-norm result does not apply.
Non-smooth worldlines. At an idealised instantaneous velocity reversal the tangent is discontinuous; U^μ jumps and A^μ = dU^μ/dτ is a delta-function — the differential construction must be replaced by a matching of one-sided limits.
Strong gravity / curved spacetime. The flat-metric contraction η_μνU^μU^ν must be replaced by g_μνU^μU^ν = c², and dU^μ/dτ by the covariant derivative; the naive component formula γ(c,v) holds only in a local inertial frame.

Failure modes

Dropping the γ in the time component. Writing U⁰ = c instead of γc — forgetting that x⁰ = ct is also differentiated with respect to τ, not t. This breaks the norm.
Confusing the spatial four-velocity with the three-velocity. Believing |U| = γv ≤ c. In fact γv is unbounded and exceeds c for v > c/√2; only the coordinate speed v is capped at c.
Sign errors from the signature. Using η = diag(−,+,+,+) but keeping the (+,−,−,−) sign in the contraction, giving U·U = −c² with the wrong sign attached — always state the convention first.
Parametrising with coordinate time. Defining "four-velocity" as dx^μ/dt = (c,v). This object is not a four-vector (t is frame-dependent) and its norm c² − v² is not invariant.
Applying proper time to a photon. Trying to compute "the proper time experienced by light"; dτ = 0 makes the question ill-posed.

Discussion

The single most important structural fact is the constraint U·U = c². Because a four-vector in four dimensions has four components but this scalar equation removes one degree of freedom, the four-velocity really carries only three independent numbers — exactly the three of the ordinary velocity v. Nothing has been lost or gained in information; the relativistic object simply repackages the three-velocity into a manifestly covariant form whose length is a fixed invariant.

Geometrically, U^μ/c is the unit tangent to the worldline in the Minkowski metric. Proper time is arc length measured with that metric, so the four-velocity is "unit-speed" parametrisation of the curve, precisely analogous to parametrising a curve in Euclidean space by its arc length so that the tangent has unit length. The peculiarly relativistic twist is that "unit length" here means c and the metric is indefinite, which is why the spatial part γv can grow without bound while the invariant length stays fixed.

Differentiating the constraint gives an immediate and powerful corollary. Since U·U = c² is constant along the worldline, d/dτ(U^μU_μ) = 2U_μ(dU^μ/dτ) = 2U_μA^μ = 0. The four-acceleration is therefore always Minkowski-orthogonal to the four-velocity, U·A = 0. In the instantaneous rest frame U^μ = (c,0,0,0), so orthogonality forces A⁰ = 0 there: proper acceleration is purely spatial, and its magnitude √(−A·A) is the invariant felt by an accelerometer. This is the structural reason a constant-norm four-velocity can nonetheless accelerate indefinitely — acceleration rotates U^μ in spacetime rather than lengthening it, exactly as a force perpendicular to a velocity changes direction but not speed.

Common misconceptions. "Moving fast slows your motion through time" is loosely right but the sharper statement is that everything moves through spacetime at the invariant rate c; increasing spatial speed borrows from the time component (U⁰ = γc grows too — it is the proper-time rate of coordinate time, not a slowing). Likewise, the four-velocity of a photon does not "have magnitude c" — it does not exist; only a null tangent with zero norm does.

Worked examples

Example 1 — Four-velocity of a fast muon. A muon moves along the x-axis at v = 0.990c. Find γ, the four-velocity components, and verify the norm. Take c = 2.998 × 10⁸ m/s.

γ = 1/√(1 − v²/c²) = 1/√(1 − 0.990²) = 1/√0.0199 = 7.089

Evaluate the Lorentz factor at v/c = 0.990. A

U⁰ = γc = 7.089 × (2.998×10⁸) = 2.125 × 10⁹ m/s

Time component from U⁰ = γc. A

U¹ = γv = 7.089 × 0.990 × (2.998×10⁸) = 2.104 × 10⁹ m/s

Spatial component from U¹ = γv; note U¹ ≫ c. A

U·U = (U⁰)² − (U¹)² = (4.516 − 4.427)×10¹⁸ = 8.99 × 10¹⁶ m²/s²

Contract with η_μν; compare with c² = 8.988×10¹⁶ m²/s². B

U^μ ≈ (2.125, 2.104, 0, 0) × 10⁹ m/s, U·U ≈ c² ✓

Reading. The spatial four-velocity 2.10×10⁹ m/s is seven times c, yet the invariant norm reduces to c² to within rounding — the hallmark of a timelike unit tangent.

Units check. All components in m/s; norm in m²/s² matching c². ✓

Example 2 — Proper time of an interstellar cruise. A ship travels at constant v = 0.600c for a coordinate (Earth-frame) time Δt = 10.0 yr. Find the elapsed proper time and the four-velocity.

γ = 1/√(1 − 0.600²) = 1/√0.640 = 1/0.800 = 1.250

Lorentz factor at v/c = 0.600. A

Δτ = Δt/γ = 10.0 yr / 1.250 = 8.00 yr

Integrate dτ = dt/γ over constant v. A

U^μ = γ(c, v, 0, 0) = 1.250(c, 0.600c, 0, 0) = (1.250c, 0.750c, 0, 0)

Components from U^μ = γ(c,v). A

U·U = (1.250c)² − (0.750c)² = (1.5625 − 0.5625)c² = c²

Verify the invariant norm exactly. B

Δτ = 8.00 yr, U^μ = (1.250c, 0.750c, 0, 0), U·U = c²

Reading. The traveller ages 8 years while Earth clocks advance 10 — the missing 2 years is exactly the deficit encoded in dτ = dt/γ, and the four-velocity's fixed norm confirms the kinematics are self-consistent.

Units check. Δτ in years (same as Δt, since γ is dimensionless); U^μ in multiples of c (m/s). ✓

Problems

A particle moves at v = 0.800c along x. Compute γ and all four components of U^μ in units of c, and verify U·U = c².
Solution
γ = 1/√(1−0.64) = 1/√0.36 = 1/0.600 = 1.667. Then U⁰ = γc = 1.667c, U¹ = γv = 1.667×0.800c = 1.333c, U² = U³ = 0. Norm: (1.667c)² − (1.333c)² = (2.778 − 1.778)c² = 1.000c² = c² ✓.
In the lab a particle exists at v = 0.995c for a lab time Δt = 1.00 μs. Find its proper lifetime.
Solution
γ = 1/√(1 − 0.995²) = 1/√(1 − 0.990025) = 1/√0.009975 = 1/0.09987 = 10.01. Proper time Δτ = Δt/γ = 1.00 μs / 10.01 = 0.0999 μs ≈ 99.9 ns. The comoving clock records about 100 ns while the lab clock records 1000 ns.
Show that the magnitude of the spatial four-velocity, |U| = γv, first exceeds c at a definite speed, and find it.
Solution
Require γv = c, i.e. v/√(1−v²/c²) = c. Square: v² = c²(1 − v²/c²) = c² − v², so 2v² = c² and v = c/√2 ≈ 0.707c. For v > c/√2 the spatial four-velocity exceeds c, even though the coordinate speed v never can — the two are different objects.
A four-velocity is measured to be U^μ = (5c/3, 4c/3, 0, 0). Verify it is a valid (timelike, correctly normalised) four-velocity and extract the ordinary speed v.
Solution
Norm: (5c/3)² − (4c/3)² = (25 − 16)c²/9 = 9c²/9 = c² ✓, so it is timelike and correctly normalised. Since U⁰ = γc = 5c/3, we get γ = 5/3. From U¹ = γv = 4c/3, v = (4c/3)/(5/3) = 4c/5 = 0.800c. (Check: γ = 1/√(1−0.64) = 5/3 ✓.)
Using only the constancy of the norm, prove that the four-acceleration A^μ = dU^μ/dτ is always Minkowski-orthogonal to U^μ.
Solution
The norm is constant along the worldline: U^μU_μ = c². Differentiate with respect to proper time: d/dτ(U^μU_μ) = (dU^μ/dτ)U_μ + U^μ(dU_μ/dτ) = 2U_μA^μ (the two terms are equal because the metric is symmetric and constant). Since the left side is d(c²)/dτ = 0, we get U_μA^μ = U·A = 0. Hence four-acceleration is always orthogonal to four-velocity — acceleration turns U^μ in spacetime without changing its length.