To see what our equations look like physically in the post-big-bang region Vv, we must rewrite things in terms of the 'reverse-hatted' quantities, with the metric gab = to2gab, with Q = to-1. As mentioned earlier, I shall write the total post-big-bang energy tensor as Uab, in order to avoid confusion with the conformally rescaled energy tensor of the (massless) matter entering Vv from VA:

Since Tab is traceless and divergence-free, this must hold for fab also (the scalings being in accordance with A8):

We shall find that the full post-big-bang energy tensor must involve two additional divergence-free components, so that

Here, Vab refers to a massless field, which is to be the phantom field Q, having now become an actual self-coupled conformally invariant field in the g-metric, since w = Q now satisfies the w-equation in the g-metric

(□+1)w=§A w3, which it must do, since the w-equation is conformally invariant and is satisfied by w = -1 in the g-metric, this becoming w = - to-1 = Q in the g-metric. This is reading things the opposite way around from what we did for VA, where the 'phantom field' Q was taken to be a solution of the w-equation in the g-metric, and interpreted merely as the scale-factor that gets us back to the physical Einstein g-metric. In that metric the phantom field is simply '1', and so it has no independent physical content. Now, we are looking at Q as an actual physical field in the Einstein physical metric gab and its interpretation as a conformal factor is the opposite, since it tells us how to get back to the g-metric, where in that metric the field would be '1'. For this interpretation, it is indeed essential that the conformal factors w and H are reciprocals of one another—although we need also to incorporate the minus sign, so that it is really - H that provides the scaling from gab back to gab. This reverse interpretation is consistent with the equations because it is H, not w, that has to satisfy the ra-equation in the appropriate metric.

Accordingly, the tensor Vab is the energy tensor of this field H in the g-metric:

We find

Note that the trace-free and divergence properties hold:

It is important to notice that the equation satisfied by w in the g-metric is not the ra-equation, for we have seen that it is H, namely ( -1 times) the inverse of w, that satisfies this equation, whence

Accordingly, the scalar curvature of the g-metric is not constrained to be equal to 4A. Instead (see B2, P&R 6.8.25, A4), we have

R=4A + 8TCGM, with w2R - R = 6w-1^w, whence w2(4A+8tcGm) - 4A=6 w-1{2 w-1(VawVaw - §A)+§Aw}, from which we deduce (see B6)

M=2^w-4 (1 -w2)2 (3nana- A) = {3VaHVaH-A(H2-1)2} =(H2-1)2(3nana - A).

The full energy tensor Uab is to satisfy Einstein's equations, so we have, in addition to R=4A+8tcGm,

Since neither fab nor Vab has a trace, it falls to Ma to pick it up:

and assuming the expressions for Ua, fab , and Vab given above, we can calculate Wa from

to obtain the following expression for 4nGWab

!(3nana+A)(H2-1)2,gab+(2H2+1)HVA(A-VB-)BH -2(3H2+1)Va(a'HVb-)bH - H4Oab which is in need of further interpretation.

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