I'm working on a problem from Carroll's Spacetime and Geometry. Supposedly I should be able to use the geodesic equatio

$$\frac{d^2x^\mu}{d\lambda^2}+\Gamma^\mu_{\rho\sigma}\frac{dx^\sigma}{d\lambda}\frac{dx^\rho}{d\lambda}=0$$ to show that the flat de Sitter metric

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

fails to cover the entire manifold known as 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.

  • 2
    $\begingroup$ 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. $\endgroup$ – Michael Apr 23 '13 at 3:19
  • $\begingroup$ Which exercise? $\endgroup$ – Qmechanic Sep 10 '15 at 21:10

We consider the metric

$$\mathrm{d}s^2=-\mathrm{d}t^2+a^2(t)\mathrm{d}\vec x^2 $$

where $a(t):= a_0e^{Ht}$. To show that these coordinates do not cover the entire spacetime manifold, we consider the trajectory of a freely falling observer, which of course extremizes the proper time $$\tau=\int\mathrm{d}t\sqrt{1-a^2\dot{\vec x^2}} $$ Performing the variation is quite straightforward. After applying the chain rule a few times, we obtain: $$\frac{\delta \tau[\vec x(t)]}{\delta \vec x(t)}=0\implies -\frac{\mathrm{d}}{\mathrm{d}t}\frac{a^2\dot{\vec x}}{\sqrt{1-a^2\dot{\vec x^2}}}=:-\frac{\mathrm{d}}{\mathrm{d}t}\vec p=0$$ where we have now introduced the momentum (per unit mass) $\vec p$. Note that this makes sense, because a freely falling observer experiences no forces, hence the momentum should be constant. From our definition of the momentum we derive the useful identity $$\frac{a^4\dot{\vec x^2}}{1-a^2\dot{\vec x^2}}=\vec p^2\implies \vec p^2+a^2=\frac{a^2}{1-a^2\dot{\vec x^2}} \implies \frac{\vec p^2+a^2}{\vec p^2}=\frac{1}{a^2\dot{\vec x^2}} $$ Now, we consider with a nonzero velocity at $t=0$, i.e. $\dot{\vec x}\neq 0$ so that $|p|\neq 0$, and we explicitly evaluate the proper time elapsed between $t=-\infty$ and $t=0$. It yields $$\tau_0=\int_{-\infty}^0\int\mathrm{d}t\sqrt{1-a^2\dot{\vec x^2}}=\int_{-\infty}^0\mathrm{d}t\frac{a(t)}{\sqrt{\vec p^2+a^2(t)}} $$ Using---once again---the chain rule, we can change to an integral over $a(t)$, which is doable with "the usual" type of change of variables involving inverse trigonometric functions. This is left as an exercise (partially to prevent this answer from being abused by lazy students), and we just quote the result: $$\tau_0=H^{-1}\sinh^{-1}\frac{1}{|\vec p| }$$ which is obviously finite for nonzero momenta (the zero-momentum case yielding infinity is also physically very reasonable: If you're standing still you'll never reach the edge!). Thus, an observer comes "from infinity" in finite proper time. This cannot be true, unless the coordinates do not cover the entire manifold, as we were supposed to show.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.