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Explicit static clock-rate ratios of the Schwarzschild exterior at r = 2s and r = 4s

Example

Let (M,g,N) be the Schwarzschild exterior datum with Schwarzschild radius s>0, f(r)=1−s/r (The Schwarzschild exterior model datum: domain, metric, mass parameter and time orientation) and static observer congruence n=f−1/2∂0 (The static observer congruence of the Schwarzschild exterior model: unit field, integral curves, acceleration and clock-rate ratios clause (a), Observer congruences, their observer curves and adapted charts). By The static observer congruence of the Schwarzschild exterior model: unit field, integral curves, acceleration and clock-rate ratios clause (c), a static observer worldline at radius r0∈(s,∞) has proper-time increment

Δτ(r0)=f(r0) Δt=1−sr0  Δt

over a coordinate-time increment Δt=Δx0/c>0. Then:

  1. Exact values at the two radii. For r1=2s and r2=4s one has f(r1)=12, f(r2)=34, so Δτ(r1)=Δt2,Δτ(r2)=32 Δt,Δτ(r1)Δτ(r2)=23=63<1.
  2. Monotonicity in the radius. The function r↦f(r)=1−s/r is smooth and strictly increasing on (s,∞) with derivative ddrf(r)=s2r2f(r)>0; consequently a static observer at the larger radius accumulates strictly more proper time over the same coordinate interval: Δτ(r2)>Δτ(r1) for s<r1<r2, and the ratio of clause 1 is the exact instance at (r1,r2)=(2s,4s).
  3. Large-radius limit. f(r)→1 as r→+∞; the ratios Δτ(r1)/Δτ(r2) are invariant under the common scaling of Δt and tend to 1 as min⁡(r1,r2)→+∞ with r2>r1.
  4. The static observers are accelerated. At r1=2s the acceleration of the congruence is a=s2r2∂r (The static observer congruence of the Schwarzschild exterior model: unit field, integral curves, acceleration and clock-rate ratios clause (e)), whose coefficient at r1=2s is s2(2s)2=18s>0: the observer held at r1=2s has nonzero acceleration pointing in the direction of increasing r, so it is not freely falling.

Facts & Assumptions

Given: The exterior datum with s>0 and f=1−s/r, a fixed Δt>0, the radii r1=2s, r2=4s, and the static observer congruence n=f−1/2∂0.

[L1]

The static congruence n=f−1/2∂0 is a smooth future-directed unit timelike field, its integral curves are the worldlines at constant r, and along each such worldline Δτ=f(r0) Δt with Δt=Δx0/c; for two radii over the same nonzero coordinate interval, Δτ1/Δτ2=f(r1)/f(r2) (The static observer congruence of the Schwarzschild exterior model: unit field, integral curves, acceleration and clock-rate ratios clauses (a)-(d), Observer congruences, their observer curves and adapted charts, The Schwarzschild exterior model datum: domain, metric, mass parameter and time orientation).

[L2]

The acceleration of the congruence is a=∇nn=s2r2∂r (The static observer congruence of the Schwarzschild exterior model: unit field, integral curves, acceleration and clock-rate ratios clause (e)).

[L5]

A function continuous on [a,b] and differentiable on (a,b), for a<b, satisfies h(b)−h(a)=h′(ξ)(b−a) for some ξ∈(a,b) (The mean value theorem, as the case g(x)=x of Cauchy's: for f continuous on [a,b] with a<b and differentiable on (a,b) there is c∈(a,b) with f(b)−f(a)=f′(c)(b−a)).

Proof

technique · direct; evaluate the clock identity at the two explicit radii, differentiate the square root to get monotonicity, and substitute $r_1=2s$ in the acceleration coefficient
1.1L1L3algebra

(Values at 2s and 4s.) By [L3], f(2s)=1−s2s=12 and f(4s)=1−s4s=34, both positive, so [L1] gives Δτ(r1)=1/2 Δt=Δt2 and Δτ(r2)=3/4 Δt=32Δt. The ratio is 1/23/4=23=63, and 6/3<1 because 6<9. This is clause 1.

2.1L1L3L4L5step 1.1algebra

(Monotonicity.) On (s,∞) the chain rule gives ddrf(r)=f′(r)2f(r)=s/r22f(r)=s2r2f(r), using f′(r)=s/r2 and f>0 by [L3]. The numerator and denominator are positive, so the derivative is positive. On every [a,b]⊂(s,∞) with a<b, f is continuous and differentiable by [L3], and [L5] gives f(b)−f(a)=(b−a)s/(2ξ2f(ξ))>0 for some ξ∈(a,b). Thus f, hence r↦Δτ(r) at fixed Δt>0, is strictly increasing; with f(r1)<f(r2) by [L4], r1<r2 gives Δτ(r1)<Δτ(r2) and the ratio of clause 1 is the instance r1=2s, r2=4s. This is clause 2 and the inequality Δτ(r2)>Δτ(r1) of clause 1.

2.2L1L3step 1.1algebra

(Large-radius limit.) As r→+∞, s/r→0, so f(r)→1 and f(r)→1 by [L3]; hence Δτ(r)=f(r) Δt→Δt and the ratio Δτ(r1)/Δτ(r2)=f(r1)/f(r2)→1 whenever min⁡(r1,r2)→+∞ with the radii positive. The ratio does not involve Δt, since the common factor cancels. This is clause 3.

3.1L1L2L3algebra∎

(Acceleration at 2s.) By [L2] the acceleration field is a=s2r2∂r, whose coefficient function is smooth and positive on (s,∞); at r1=2s it is s2(2s)2=18s>0. Since the coefficient is nonzero, the worldlines of the congruence do not satisfy the geodesic equation, so the observer held at radius 2s is accelerated with the stated outward-coefficient 1/(8s) in the coordinate basis (∂0,∂r,∂ϑ,∂φ). This is clause 4.

Remarks

  • What the example computes. All four clauses are exact evaluations of the clock identity and the acceleration formula of the cited lemma at explicit radii; the coordinate-time increment Δt is an arbitrary positive coordinate difference, not a measurement, and no value of c, GN, m∗ or s in any unit system is asserted. In particular clause 1 does not report a clock comparison or a redshift observation; it is the conditional identity of the model evaluated at two radii.

Depends on

Dependency tree · 2 levels

Sources