Wednesday 30 December 2020

Exercise 8.1 N+n+1-dimensional spacetime

My first exercise in almost five months!
Questions
Consider an ##\left(N+n+1\right)##-dimensional spacetime with coordinates ##\left\{t,x^I,y^i\right\}## where ##I## goes from 1 to ##N## and ##i## goes from 1 to ##n##. Let the metric be $$
{ds}^2=-{dt}^2+a^2\left(t\right)\delta_{IJ}{dx}^I{dx}^J+b^2\left(t\right)\gamma_{ij}\left(y\right){dy}^i{dy}^j
$$where ##\delta_{IJ}## is the usual Kronecker delta and ##\gamma_{ij}\left(y\right)## is the metric on an ##n##-dimensional maximally symmetric spatial manifold. Imagine we normalise the metric ##\gamma## such that the curvature parameter $$
k=\frac{R\left(\gamma\right)}{n\left(n-1\right)}
$$is either +1,0 or -1, where ##R\left(\gamma\right)## is the Ricci scalar for the metric ##\gamma_{ij}##.
(a) Calculate the Ricci tensor for this metric
(b) Define an energy-momentum tensor in terms of an energy density ##\rho## and pressure in the ##x^I## and ##y^i## directions, ##p^{\left(N\right)}## and ##p^{\left(n\right)}:##$$
T_{00}=\rho
$$$$
T_{IJ}=a^2p^{\left(N\right)}\delta_{IJ}
$$$$
T_{ij}=b^2p^{\left(n\right)}\gamma_{ij}
$$Plug the metric and ##T_{\mu\nu}## into Einstein's equation and to derive Friedmann like equations for ##a## and ##b## (three independent equations in all).
(c) Derive equations for the energy density and the two pressures at a static solution, where ##\dot{a}=\dot{b}=\ddot{a}=\ddot{b}=0##, in terms of ##k,n## and ##N##. Use these to derive expressions for the equation of state parameters ##w^{\left(N\right)}=\frac{p^{\left(N\right)}}{\rho}## and ##w^{\left(n\right)}=\frac{p^{\left(n\right)}}{\rho}##, valid at the static solution.
Answers
I got some answers and wonder if this is anywhere near string theory. The Ricci tensor was an exercise in tensor splitting and looked quite nice, the Friedmann like equations were less pretty and there was something very wrong with ##w^{\left(n\right)}## in part (c). 

My efforts are here. Seasons greetings: Ex 8.1 N+n+1 dimensions.pdf (7 pages) 

2 comments:

  1. Your "handwaving" derivation to get (10) is in fact not handwaving at all. It is quite easy to prove in linear algebra that the inverse of a block-diagonal matrix diag(A, B, C) is simply diag(A^{-1}, B^{-1}, C^{-1}).

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  2. There are quite a few errors in the subsequent calculation. For example, all (a-dot/a)^2 terms should cancel out in (27), and the sign of ρ is wrong in (36). A quick sanity check after each calculation is to compare your result with the calculations leading to the standard Friedmann equation. When you remove all those terms containing b and set N=3, you should get the same results (other that minor differences caused by the difference between cartesian x,y,z and polar ρ,θ,φ).

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