WrQ Wr

Therefore at q, (8/8t)y will be timelike. This shows that the curve y(t) will enter the region Cq. But expff (Cq) is the region of on which a is negative and in which by the previous lemma the surfaces of constant <7 are spacelike. Thus cr must monotonically decrease along y(t) since (8[8t)y being timelike can never be tangent to the surfaces of constant cr and since at any non-differentiable point of y(t) the two tangent vectors point into the same half of the null cone. Therefore p e exp5(C5) which completes the proof for timelike curves. To prove that a non-spacelike curve y(t) remains in expff (Cq), one performs a small variation of y(t) which makes it into a timelike curve. Let Y be a vector field on Tg such that in the induced vector field expff*(Y) is everywhere timelike and such that g(Y, (8j8t)y\Q) < 0. For each e ^ 0 let f}(r, e) be the curve Tq starting at the origin such that the tangent vector (8l8r)fi equals (8/8t)y|t=,r + eY|/^r>c). Then /?(r, e) depends differ-entiably on r and e. For each e > 0, exp9 (/?(r, e)) is a timelike curve in W and so is contained in expff (Cq). Thus the non-spacelike curve expff (yQ(r, 0)) = y(r) is contained in expff (Cq) = exp9 (Cq). □

Corollary

IfpeW can be reached from q by a non-spacelike curve but not by a timelike curve, then p lies on a null geodesic from q. □

The length of a non-spacelike curve y(t) from q to p is a™,*) -£[-*(!. !)]**•

where the integral is taken over the differentiable sections of the curve.

In a positive definite metric one may seek the shortest curve between two points but in a Lorentz metric there will not be any shortest curve as any curve can be deformed into a null curve which has zero length. However, in certain cases there will be a longest non-spacelike curve between two points or between a point and a spacelike three-surface. We deal first with the situation when the two points are close together. We shall then derive necessary conditions in the general case when the two points are not close. The sufficient condition in this case will be dealt with in §6.7.

Proposition 4.5.3

Let q and p lie in a convex normal neighbourhood Then, if q and p can be joined by a non-spacelike curve in the longest such curve is the unique non-spacelike geodesic curve in from q top. Moreover, defining p(q,p) as the length of this curve if it exists, and as zero otherwise, p(q,p) is a continuous function on

By the definition of convex normal neighbourhoods (§2.5), there is a unique geodesic y(i) in with y(0) = y(l) = P- Since this geodesic depends differentiably on its endpoints, the function will be differentiable on x eSi. (This function cr is the same as that in lemma 4.5.2.) Thus p(q,p) will be continuous on since it equals [ — cr(q,p]$ if cr < 0 and is zero otherwise. It now remains to show that if q and p can be joined by a timelike curve in then the timelike geodesic y between them is the longest such curve. Let a(s, t) be exp9(sX(i)) as before where g(X(t),X(t)) = — 1. If A(i) is a timelike curve in from q to p, it can be represented as A(i) = a(f{t),t).

Since the two vectors on the right are mutually orthogonal by lemma 4.5.2. and since g((8f8s)a, (8l8s)a) = — 1, this gives the equality holding if and only if (8/8t)a = 0 and hence if and only if A is a geodesic curve. Thus the equality holding if and only if A is the unique geodesic curve in W

We shall now consider the case where q and p are not necessarily contained in a convex normal neighbourhood W. By considering small variations we shall derive necessary conditions for a timelike curve y(i ) from q to p to be the longest such curve from q to p. A variation a of y(t) is a Cmap a: ( - e, e) x [0, tp]J( such that

(2) there is a subdivision 0 = ij < i2... < tn = tp of [0, ip] such that a is C3 on each ( — e, e) x ii+1];

(4) for each constant u, a(u, t) is a timelike curve.

Was this article helpful?

0 0

Post a comment