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Length monotonicity of the squared displacement: once a future causal curve in a normal neighborhood has a future timelike position vector, the displacement radius grows at least as fast as the accumulated Lorentzian length

Statement

Assume Countable Choice and Dependent Choice (The Axiom of Countable Choice (ACω), The axiom of dependent choice: a relation in which every element is related to something admits an N-indexed chain) for the absolutely continuous causal-curve calculus and the supplied causal results used below.

Let (M,g,N) be a time-oriented smooth Lorentzian four-manifold with metric-selected connection ∇ (Smooth time-oriented Lorentzian four-manifolds: the geometric objects of the model), let p∈M and let Up=exp⁡p(Wp) be a normal neighborhood of p as in Normal neighborhoods, convex normal sets, the position vector and the squared-displacement function clause (a). Let σ:[a,b]→Up be a future causal curve in the class of Causal curves, chronological and causal reachability, and inextendibility (locally absolutely continuous, with future causal nonzero velocity almost everywhere), and write v(t):=exp⁡p−1(σ(t)),P(t):=d(exp⁡p)v(t)(v(t)),Q(t):=gp(v(t),v(t)) for t∈[a,b]. Suppose there is t∗∈[a,b] such that v(t∗) is future timelike. For any locally absolutely continuous future causal curve η:[u,v]→M, define its Lorentzian length by Lg(η):=∫uv−gη(s)(η′(s),η′(s)) ds, where the derivative is taken almost everywhere and the integral is the Lebesgue integral. This integral is finite on a compact parameter interval: the curve image is compact, a finite chart subdivision has bounded metric coefficients on each piece, and each coordinate derivative is integrable by local absolute continuity. The radicand is nonnegative almost everywhere by causality. On piecewise-C1 future timelike curves this agrees with the length in Timelike worldlines, proper time, length and four-velocity on a Lorentzian four-manifold. Then:

(a) Propagation. For every t∈[t∗,b] with σ([t∗,t])⊆Up, the vector v(t) is future timelike; consequently Q(t)<0, and R(t):=−Q(t)>0.

(b) Length inequality. For all t∗≤t1<t2≤b with σ([t1,t2])⊆Up, R(t2)−R(t1) ≥ Lg(σ∣[t1,t2]). In particular R is nondecreasing on [t∗,b].

(c) Derivative formula. The map t↦Q(t) is absolutely continuous and ddtQ(t)=2 gσ(t)(P(t),σ′(t)) at almost every t∈[a,b].

Facts & Assumptions

Given: A time-oriented smooth Lorentzian four-manifold (M,g,N), a point p∈M, a normal neighborhood Up=exp⁡p(Wp), a future causal curve σ:[a,b]→Up and t∗ with v(t∗) future timelike.

[L1]

The length density −g(σ′,σ′) is measurable, nonnegative and integrable on each compact parameter interval. Indeed, subdivide into finitely many chart pieces with compact closure; bounded smooth metric coefficients bound the density by a constant times the sum of absolute values of the integrable coordinate derivatives. Its integral defines Lg and its indefinite integral is absolutely continuous and nondecreasing (Chart independence, concatenation, reparametrization and transitivity for causal curves clause (a), Absolutely continuous functions have integrable derivatives, Fundamental theorem of calculus for absolutely continuous functions, Timelike worldlines, proper time, length and four-velocity on a Lorentzian four-manifold).

[L2]

Reverse Cauchy-Schwarz, derived locally. Let T be future timelike, put e=T/−g(T,T), and decompose the nonzero future causal z as z=ae+u, with u⊥e. The observer projection and positivity give a=−g(e,z)>0 and 0≤g(u,u)≤a2. Therefore −g(T,z)=−g(T,T)a≥−g(T,T)a2−g(u,u)=−g(T,T)−g(z,z). Equality holds precisely when u=0, equivalently z is a positive multiple of T (The observer rest-space decomposition and the spatial projection, Pairing of an observer with future causal vectors).

[L3]

Q is absolutely continuous and ddtQ(t)=2gσ(t)(P(t),σ′(t)) at almost every t: indeed Dp2 is the smooth function of The differential of the squared-displacement function and the position vector of the reversed geodesic clause (a), σ is locally absolutely continuous, and the scalar smooth composition is absolutely continuous by Chart independence, concatenation, reparametrization and transitivity for causal curves clause (b); the chain rule gives the stated derivative at every differentiability point of σ (The chain rule for total derivatives: D(g∘f)(a)=Dg(f(a))∘Df(a), The differential of the squared-displacement function and the position vector of the reversed geodesic clause (a)).

[L4]

R is absolutely continuous on every subinterval where Q<0, with R′=−12RQ′=−1Rgσ(P,σ′) almost everywhere (Fundamental theorem of calculus for absolutely continuous functions).

[L5]

If v(t) is future timelike then P(t)≠0 is future timelike and gσ(t)(P(t),P(t))=Q(t) (Covariant differentiation along curves: metric pairing, parallel transport and geodesic speed, The Lorentzian Gauss lemma and the radial maximum of proper time clause (a)).

Proof

technique · direct
1.1L2L4L5givenalgebra

(Propagation.) Let T0:=sup⁡{t∈[t∗,b]: for all s∈[t∗,t] one has σ([t∗,s])⊆Up and v(s) future timelike}. The set is nonempty and contains a relative right neighborhood of t∗ in [t∗,b]: v(t∗) is future timelike, future-timelikeness is open in TpM while v is continuous, and σ([t∗,t])⊆Up for t near t∗. On (t∗,T0) define R=−Q>0. By [L5] and [L2] applied with T:=P(t) and z:=σ′(t) at every differentiability point with Q(t)<0, gσ(t)(P(t),σ′(t))≤−R(t)−g(σ′(t),σ′(t)); hence by [L4] R′(t)≥−g(σ′(t),σ′(t))≥0 a.e., so R is nondecreasing on [t∗,T0) (Fundamental theorem of calculus for absolutely continuous functions), and Q=−R2≤Q(t∗)<0 there. If T0<b and σ([t∗,T0])⊆Up, then Q(T0)=lim⁡t↑T0Q(t)≤Q(t∗)<0 by continuity, so v(T0) is future timelike and there is ε>0 with σ([t∗,T0+ε])⊆Up and v future timelike on [t∗,T0+ε], contradicting maximality of T0. Hence T0=b′ where b′=sup⁡{t∈[t∗,b]:σ([t∗,t])⊆Up}, proving (a).

1.2L3

(Derivative formula.) The map t↦Q(t) is the composition of the locally absolutely continuous curve σ with the smooth function Dp2, hence is absolutely continuous, and the chain rule at every differentiability point of σ gives ddtQ(t)=d(Dp2)σ(t)(σ′(t))=2gσ(t)(P(t),σ′(t)) by [L3]. This proves (c).

2.1step 1.1L1L4algebra

(Length inequality.) Let t∗≤t1<t2≤b with σ([t1,t2])⊆Up. By step 1.1, Q<0 and R>0 on [t1,t2], and, by the computation of step 1.1, R′≥−g(σ′,σ′) a.e. on [t1,t2]. Integrating with [L1] gives R(t2)−R(t1)=∫t1t2R′≥∫t1t2−g(σ′,σ′) dt=Lg(σ∣[t1,t2]). This proves (b).

3.1step 1.1step 2.1step 1.2∎

Claims (a), (b), (c) are steps 1.1, 2.1 and 1.2.

Remarks

Depends on

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Sources