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This represents a confused attempt to work through a problem in Carroll's Spacetime and Geometry. Supposedly I should be able to use the geodesic equation,

$$\frac{d^2x^\mu}{d\lambda^2}+\Gamma^\mu_{\rho\sigma}\frac{dx^\sigma}{d\lambda}\frac{dx^\rho}{d\lambda}=0$$ to show why this metric

$$ ds^2=-dt^2+e^{2Ht}[dx^2 + dy^2 + dz^2] $$

fails to cover the entire manifold of de Sitter space. I am urged to combine these two equations to solve for the affine parameter $\lambda$ as a function of $t$, and then show that the geodesics of the space reach $t=-\infty$ in a finite value of the affine parameter, demonstrating what I sought to prove. I begin by parametrizing the metric in terms of the proper time, $\tau$ $(\lambda = \tau)$, to yield:

$$-1 =-\frac{dt^2}{d\lambda^2}+e^{2Ht}[\frac{dx^2}{d\lambda^2} + \frac{dy^2}{d\lambda^2} + \frac{dz^2}{d\lambda^2}] $$

And from the geodesic equation, I derive:

$$\frac{d^2 x^i}{d\lambda^2} = -2H\left(\frac{dt}{d\lambda}\right)\left(\frac{dx^i}{d\lambda}\right) $$

To solve any one of these three equations, say, for example, the first one, I make the subsitution:

$$\nu = \frac{dx}{d\lambda}, t'=\frac{dt}{d\lambda} $$


$$\nu'=-2H\nu t'\\\nu'/t'=\frac{d\nu}{d\lambda}\frac{d\lambda}{dt}=-2H\nu=\frac{d\nu}{dt}\\\frac{d\nu}{dt}=-2H\nu\Rightarrow\nu = C_1e^{-2Ht}$$

Substituting back into the metric, I'm left with:

$$1 = \left(\frac{dt}{d\lambda}\right)^2 - e^{2Ht}\left(\sum_i C_i^2 e^{-4Ht}\right)$$

Define $\sum_i C_i^2 = \alpha$, yielding: $$ \frac{dt}{d\lambda} = \sqrt{1 + \alpha e^{-2Ht}} $$

I can't seem to find a nice analytical solution for $\lambda(t)$ that demonstrates that $t \rightarrow -\infty$ in a finite value of the affine parameter without relying on a computer, and this doesn't leave me with any nice physical (or mathematical) intuition for what's going on in the problem. Have I made an error somewhere, or is this, as Walter Cronkite (and my GR professor) would say, "the way it is"? Can anyone point me in the right direction, or show me how to proceed? Thanks in advance.

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Just looking at your last equation (haven't check through the derivation): in the $t\to -\infty$ limit the exponential dominates and you can simplify the square root then separate variables. –  Michael Brown Apr 23 '13 at 3:19

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