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Straight timelike geodesics of the flat model: constant coordinate velocity, four-velocity and the clock-reading identity

Example

Let (M,g,N) be the flat model of The flat model spacetime: the Minkowski metric on R^4 and its inertial chart with inertial chart x, and let c>0. Take the initial event p with x(p)=0 and the tangent vector

U0:=c v,v:=(23,13,0,0),η(v,v)=−1,v0>0,

so that gp(U0,U0)=−c2 and U0 is future timelike. Then:

  1. The maximal geodesic of the metric-selected connection with x(γ(0))=0 and velocity v at 0 is x(γ(s))=s v on R; it is a future timelike worldline, and τ(s)=s/c is its proper time up to an additive constant, with dγ/dτ=c v=U0 and g(dγ/dτ,dγ/dτ)=−c2.
  2. Its coordinate velocity in the parametrization t=x0/c is the constant w=(c/2,0,0) and ∣w∣=c/2<c.
  3. Between coordinate times t0<t1 the proper time elapsed on the worldline is Δτ=1−∣w∣2/c2 (t1−t0)=32(t1−t0), and this agrees with the direct parametrization t=v0τ.

Facts & Assumptions

Given: The flat model (M,g,N) with inertial chart x, the event p with x(p)=0, and the vector v=(2/3, 1/3, 0, 0).

[L1]

In the inertial chart, gab=ηab=diag⁡(−1,1,1,1) and gp(c ∂0,c ∂0)=−c2; timelike, null and causal classifications are those of η (The flat model spacetime: the Minkowski metric on R^4 and its inertial chart).

[L2]

A smooth curve γ:I→M is a geodesic of the metric-selected connection if and only if there are x0,v∈R4 with x(γ(s))=x0+sv; the speed g(γ′,γ′)=η(v,v) is constant; the maximal geodesic with x(γ(0))=x0 and velocity v at 0 has domain R; if η(v,v)<0 and v0>0 it is a future timelike worldline and τ(s)=−η(v,v)(s−s0)/c is proper time up to an additive constant, with dγ/dτ the physical four-velocity U and g(U,U)=−c2. Its coordinate velocity in the parametrization x(γ(t))=(ct,q(t)) is w=(c/v0)(v1,v2,v3) and ∣w∣<c (The flat model: vanishing connection and curvature, translation Killing fields and straight geodesics clauses (d)-(e), Timelike worldlines, proper time, length and four-velocity on a Lorentzian four-manifold).

[L3]

For a C1 future timelike worldline γ on an interval with nonempty interior with proper-time function τ, one has τ′(t)=1c−gγ(t)(γ′(t),γ′(t))>0, and τ is strictly increasing; reparametrizing by τ gives tangent U with g(U,U)=−c2 (Proper time is well defined and reparametrization invariant; the four-velocity is its proper-time tangent clauses 1 and 3, Intervals of R: the nine order-convex forms, nondegeneracy, and length).

[L4]

For v=(2/3,1/3,0,0): η(v,v)=−(4/3)+(1/3)=−1, v0=2/3>0, 1/v0=3/2, and 1−1/(v0)2=1−3/4=1/4; also c2η(v,v)=−c2, and 3/4=3/2 (Rational powers ar of a positive base).

Verification

technique · direct; evaluate the flat-model geodesic, velocity and proper-time statements of [L2] and [L3] on the explicit vector $v$, and compare the clock reading with the direct affine parametrization
1.1L1L2L4algebra

(The worldline.) By [L4], η(v,v)=−1 and v0>0, so U0=c v satisfies gp(U0,U0)=c2η(v,v)=−c2 by [L1] and is future timelike. By [L2] the maximal geodesic with x(γ(0))=0 and velocity v has x(γ(s))=s v on R and is a future timelike worldline; its speed is the constant η(v,v)=−1; and τ(s)=−η(v,v) (s−0)/c=s/c is proper time up to an additive constant, with dγ/dτ=c v=U0 and g(dγ/dτ,dγ/dτ)=c2η(v,v)=−c2.

1.2L2L4algebra

(Coordinate velocity.) In the parametrization x(γ(t))=(ct,q(t)) of [L2] clause (e), the coordinate velocity is w=(c/v0)(v1,v2,v3)=(3/2) c (1/3,0,0)=(c/2,0,0) using [L4]; hence ∣w∣=c/2<c and 1−∣w∣2/c2=1−1/4=3/4, so 1−∣w∣2/c2=3/2.

2.1step 1.2L1L3algebra

(Clock reading via the proper-time density.) Parametrize the worldline by t=x0/c; its tangent is γt′=(c,w) with components (c,c/2,0,0), so by [L1] g(γt′,γt′)=c2(−1+1/4)=−3c2/4 and [L3] gives τ′(t)=1c3c2/4=3/2>0. Integrating over [t0,t1] gives Δτ=(t1−t0)3/2=1−∣w∣2/c2 (t1−t0) by step 1.2.

3.1step 1.1step 1.2step 2.1L4algebra∎

(Cross-check with the affine parametrization.) On the parametrization γ(s)=sv one has x0(γ(s))=sv0, that is t=v0s/c, and step 1.1 identifies τ=s/c, so τ=t/v0=(3/2) t by [L4]; hence between t0 and t1 the elapsed proper time is (t1−t0)/v0=(t1−t0)3/2, in agreement with step 2.1. The identity 1/v0=1−∣w∣2/c2 is exactly η(v,v)=−1: 1−∣w∣2/c2=1−∣v1∣2/(v0)2=(1+∣v1∣2−∣v1∣2)/(v0)2=1/(v0)2.

Remarks

Depends on

Dependency tree · 2 levels

Sources