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Unformatted text preview: Homework 3 William D. Linch III March 11, 2005 1.a The rate of change of a vector v a flowing along the integral curves of u a is given by v a = u b b v a = v b b u a = B a b v b = 1 3 v a + a b v b + a b v b for B a b := b u a where we have used the fact that the vector fields u and v commute. It follows from this that the length of v changes by d d | v | = 1 | v | ( 1 3 v 2 + ab v a v b ) = 1 3 | v | + ab v a v b / | v | . Now suppose that v a is one of a set of three spacial vector fields spanning the t = constant sections in a Robertson-Walker spacetime, i. e. the co-moving frame. By isotropy, the shear for the flow of this vector must vanish, for if it did not, a round sphere at t = t would evolve into a squashed sphere for t > t which is certainly not invariant under rotations about its center. Hence d d | v | = 1 3 | v | . Now the metric for a Robertson-Walker spacetime can be put in the form ds 2 = dt 2- a 2 ( t ) d 2 , (1) where d 2 is the volume element of a unit (pseudo)sphere or Euclidean 3-space. In other words, all lengths on are determined by a ( t ). In particular, the spacial sections are expanding uniformly in all directions at the rate a = 1 3 a (2) as we have just seen. 1 Consequently, a = 1 3 ( a + 1 3 2 a ). Then, by the shearless, twistless, Raychaudhuri equation, a/a =- 1 3 R ab u a u b . The trace-reverse of the standard Einstein equation, R ab- 1 2 g ab R = 8 T ab , is R ab = 8 ( T ab- 1 2 g ab T ). Futhermore, the stress-energy of a perfectly homogeneous and isotropic co-moving cosmological fluid is given in the coordinate system of (1) as ( T ) = diag( ,- p,- p,- p ), where is the energy density and p is the pressure. Plugging this in gives a a =- 8 3 - 1 2 ( - 3 p ) =- 4 3 ( + 3 p ) , (3) 1 It is worth emphasizing here that the expansion represented by is that of the co-moving frame or cosmological fluid (also known as the (in)famous aether)....
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