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Worked example: invariance of elapsed proper time under a cubic reparametrization

Example

Worked computation. In an inertial chart with $η_{0} = diag (- 1, 1, 1, 1)$ let $d : = (5, 3, 0, 0)$ , a future timelike displacement with $- η (d, d) = 25 - 9 = 16$ and $\sqrt{- η (d, d)} = 4$ .

(a) The worldline $x (s) : = s d$ on the parameter interval $[1, 8]$ has $x^{'} (s) = d$ , integrand $\sqrt{- η (d, d)} = 4$ , and elapsed proper time $τ (8) - τ (1) = c^{- 1} \int_{1}^{8} 4 d s = 28 / c$ .

(b) The reparametrization $φ (u) : = u^{3}$ is a strictly increasing $C^{1}$ surjection of $[1, 2]$ onto $[1, 8]$ whose derivative $3 u^{2}$ is strictly positive throughout $[1, 2]$ ; the reparametrized curve $y (u) : = x (u^{3}) = u^{3} d$ has $y^{'} (u) = 3 u^{2} d$ , hence $\sqrt{- η (y^{'}, y^{'})} = 3 u^{2} \sqrt{- η (d, d)} = 12 u^{2}$ , and elapsed proper time $c^{- 1} \int_{1}^{2} 12 u^{2} d u = (4 / c) (8 - 1) = 28 / c$ . The two elapsed times agree, as Length-integral invariance under monotone C^1 maps and proper-time invariance under regular increasing maps requires, and the proper-time function of $x$ is $τ (s) = (4 / c) (s - s_{0})$ for the chosen base point $s_{0} \in [1, 8]$ , whose inverse is $ρ (τ) = s_{0} + c τ / 4$ on the corresponding proper-time interval.

(c) Boundary check: instead take the same straight curve $x (s) = s d$ on $[- 1, 1]$ and $φ (u) = u^{3}$ on $[- 1, 1]$ . This is a strictly increasing $C^{1}$ surjection, but $y^{'} (0) = 0$ , so $y$ is not a worldline of Curves, regular causal curves and future timelike worldlines in affine Minkowski space. The broad length-integral identity nevertheless holds: $\int_{- 1}^{1} 12 u^{2} d u = 8 = \int_{- 1}^{1} 4 d s$ . No proper-time function of $y$ is invoked outside its defining worldline domain.

Facts & Assumptions

Given: An inertial chart with $η_{0} = diag (- 1, 1, 1, 1)$ ; the displacement $d = (5, 3, 0, 0)$ ; the straight worldline $x (s) = s d$ on $[1, 8]$ and on $[- 1, 1]$ ; the map $φ (u) = u^{3}$ on $[1, 2]$ and on $[- 1, 1]$ ; and $c > 0$ .

[L1]

Inertial charts (Inertial frames and inertial charts of the affine Minkowski model): in an inertial chart the matrix of $η$ is $diag (- 1, 1, 1, 1)$ , so $η (v, v) = - (v^{0})^{2} + (v^{1})^{2} + (v^{2})^{2} + (v^{3})^{2}$ for $v = (v^{0}, v^{1}, v^{2}, v^{3})$ .

[L2]

Curves (Curves, regular causal curves and future timelike worldlines in affine Minkowski space): the curve $x (s) = s d$ has difference quotients $d$ identically, hence tangent $x^{'} (s) = d$ ; it is a $C^{1}$ future timelike worldline exactly when $d$ is future timelike; a curve with a zero tangent at some point is not a worldline, the zero vector having no causal character (Timelike, null, spacelike and causal vectors of a Lorentz vector space).

[L3]

Proper time (Proper time along a future timelike worldline): $τ (s) - τ (s_{0}) = c^{- 1} \int_{s_{0}}^{s} \sqrt{- η (x^{'} (u), x^{'} (u))} d u$ for a $C^{1}$ future timelike worldline; a constant integrand $A$ gives the increment $c^{- 1} A (s - s_{0})$ .

[L4]

Reparametrization (Length-integral invariance under monotone C^1 maps and proper-time invariance under regular increasing maps): for a strictly increasing $C^{1}$ surjection with positive derivative, the reparametrized curve is a worldline with the same elapsed proper time; the broad length-integral identity holds even where the derivative vanishes.

[L5]

Bilinearity (Lorentz forms of inertia (3,1,0) on a 4-dimensional real vector space): $η (λ v, λ w) = λ^{2} η (v, w)$ .

Verification

technique · direct

1.1givenL1L2

By [L1], $η (d, d) = - 25 + 9 = - 16$ , so $d$ is timelike by [L2]'s classification and $- η (d, d) = 16$ with $\sqrt{- η (d, d)} = 4$ .

1.2givenL2

The map $φ (u) = u^{3}$ is strictly increasing on $[1, 2]$ with derivative $3 u^{2} \geq 3 > 0$ there, and it maps $[1, 2]$ onto $[1, 8]$ ; the difference quotients of $y (u) = u^{3} d$ are $((u + h)^{3} - u^{3}) / h \cdot d = (3 u^{2} + 3 u h + h^{2}) d$ , which tend to $3 u^{2} d$ , so $y^{'} (u) = 3 u^{2} d$ .

2.1step 1.1L2L3

On $[1, 8]$ the difference quotients of $x$ are identically $d$ , so $x^{'} (s) = d$ and $x$ is a $C^{1}$ future timelike worldline by [L2]; by [L3] its elapsed proper time over $[1, 8]$ is $c^{- 1} \int_{1}^{8} 4 d s = 28 / c$ .

2.2step 1.2L2

On $[- 1, 1] ∖ {0}$ the difference quotients of $y (u) = u^{3} d$ are again $(3 u^{2} + 3 u h + h^{2}) d$ , with limit $3 u^{2} d$ , and at $u = 0$ the quotient is $h^{2} d \to 0$ ; hence $y^{'} (0) = 0$ and $y$ is not a worldline, since a worldline tangent must be future timelike and the zero vector has no causal character.

3.1step 1.1step 2.1step 1.2L4L5

By bilinearity [L5], $η (y^{'} (u), y^{'} (u)) = (3 u^{2})^{2} η (d, d)$ , so $\sqrt{- η (y^{'}, y^{'})} = 3 u^{2} \sqrt{- η (d, d)} = 12 u^{2}$ by step 1.1; hence the elapsed proper time of $y$ over $[1, 2]$ is $c^{- 1} \int_{1}^{2} 12 u^{2} d u = (4 / c) (2^{3} - 1^{3}) = 28 / c$ , which agrees with step 2.1 as [L4] requires.

3.2step 2.1L3

With base point $s_{0} \in [1, 8]$ the proper-time function of $x$ is $τ (s) = c^{- 1} \cdot 4 \cdot (s - s_{0}) = (4 / c) (s - s_{0})$ for $s \in [1, 8]$ , a strictly increasing affine function with $τ (s_{0}) = 0$ ; solving $τ = (4 / c) (s - s_{0})$ for $s$ gives its inverse $ρ (τ) = s_{0} + c τ / 4$ on the image interval.

4.1step 1.1step 3.1step 2.2L4∎

The broad length-integral identity still holds on $[- 1, 1]$ : $\sqrt{- η (y^{'}, y^{'})} = 3 u^{2} \sqrt{- η (d, d)} = 12 u^{2}$ for $u \neq 0$ and both sides vanish at $u = 0$ , so $\int_{- 1}^{1} \sqrt{- η (y^{'}, y^{'})} d u = \int_{- 1}^{1} 12 u^{2} d u = 4 - (- 4) = 8$ , while $\int_{- 1}^{1} \sqrt{- η (d, d)} d s = 4 \cdot 2 = 8$ ; equality holds without any worldline structure for $y$ .

Remarks

Two independent computations of the same number. The integrals $c^{- 1} \int_{1}^{8} 4 d s$ and $c^{- 1} \int_{1}^{2} 12 u^{2} d u$ are computed separately, and their agreement is the concrete instance of Length-integral invariance under monotone C^1 maps and proper-time invariance under regular increasing maps; the cubic's derivative, rather than an inverse function, produces the second computation.
The boundary case is deliberate. On $[- 1, 1]$ the derivative $3 u^{2}$ vanishes at $u = 0$ , so the reparametrized curve has a zero tangent there and is not a worldline; the broad integral identity survives, and no proper-time function of $y$ is claimed.

Worked example: invariance of elapsed proper time under a cubic reparametrization

Example

Facts & Assumptions

Verification

Remarks

Depends on

Dependency tree · 2 levels