Clearly a necessary but not sufficient condition that a variation a of y(t) should yield a timelike curve from q top is that A should become positive. One has i 8s / /a 8\\ 82 / /_a_ 8\\__8_i /_a_ Dan

2eusdu^yBt'dt)} 8u28t[du^etjj du^ydn^ët dtjj 8s

8u28t and so 1 8sA

28u2 du1

This formula is very similar to that for the variation of the length of a timelike curve. It can be seen that the variation of A is zero for a variation vector proportional to the tangent vector 8/8t since 8/8t is null and R(8/8t, 8/8t) {8/8t) = 0 as the Riemann tensor is antisymmetric. Such a variation would be equivalent to simply repara-metrizing y(t). Thus if one wants a variation which will give a timelike curve one need consider only the projection of the variation vector into the space SQ at each point q of y(t). In other words, introducing a pseudo-orthonormal basis Ex, E2, E3, E4 along y(t) with E4 = 8/dt, the variation of A will depend only on the components Zm of the variation vector (m = 1,2).

If there is no point in [q,p\ conjugate to q along y(t) then d2A/dw2|u=0 will be negative for any variation a of y(t) whose variation vector d/8u\u=0 is orthogonal to the tangent vector 8/8t on y(t) andis not everywhere zero or proportional to d/dt. In other words, if there is no point in [q,p] conjugate to q then there is no small variation of y(t) which gives a timelike curve from q top.

The proof is similar to that for proposition 4.5.8, using instead the 2x2 matrix Amn of § 4.2. □

Proposition 4.5.12

If there is a point r in (q,p) conjugate to q along y(t) then there will be a variation of y(t) which will give a timelike curve from q top.

The proof is a bit finicky since one has to show that the tangent vector becomes timelike everywhere. Let Wm be the components in the space S (see §4.2) of the Jacobi field which vanishes at q and r. It obeys d2

where for convenience t has been taken to be an affine parameter. Since Wm will be at least C3 and since dWmldt is not zero at a and r, one can write Wm = fWm where Wm is a unit vector and/ and Ware C2. Then da a?f+hf=0>

where h=Wm^Wm + Rminl WmWn.

Let x e [r, p~] be such that Wm is not zero in [r, x]. Let hx be the minimum value of A in [r, x]. Let a > 0 be such that a2 + h1 > 0 and let b = {-/(eot-1)-1}!*. Then the field

Zm = {b{eat-i)+f}Wm will vanish at q and x and will satisfy

We shall choose a variation a(u, t) of y(t) from q to x such that the components in S of its variation vector 8/8u|u=0 equals Zm and such where tx is the value of t at x, and e > 0 but less than the least value of Zm RminiZn) in the range \tx < t < ftx. Then by (4.49)

(d2/8u2)g(8ldt, 8fit) will be negative everywhere in [g.a;] and so for sufficiently small u, a will give a timelike curve from q to x. If one joins this curve to the section of y from a; to p, one will obtain a non-spacelike curve from qtop which is not a null geodesic curve. Thus there will be a variation of this curve which gives a timelike curve from q top. □

By similar methods one can prove: Proposition 4.5.13

If y(t) is a null geodesic curve orthogonal to a spacelike two-surface«^ fromy top and if there is no point in [y,p] conjugate to along y, then no small variation of y can give a timelike curve from to p . □

Proposition 4.5.14

If there is a point in (SP,p) conjugate to ¿f along p, then there is a variation of y which gives a timelike curve from S? to p. □

These results on variations of timelike and non-spacelike curves will be used in chapter 8 to show the non-existence of longest geodesies.

that satisfies satisfies

-et for Outfits, e(t~itx) for e(tx-t) for | tx^t^tx,

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