# De Sitter spacetime affine parameter

I am reading Chapter 8 in Carroll's "Spacetime and geometry " textbook and I was lead to exercise 8.2, given as:

Consider de Sitter space in coordinates where the metric takes the form $$ds^{2} = -dt^{2} + e^{Ht}[dx^{2} + dy^{2} + dz^{2}].$$ Solve the geodesic equation for co-moving observers ($$x^{i} = constant$$) to find the affine parameter as a function of $$t$$. Show that the geodesics reach $$t = -\infty$$ in a finite affine parameter, demonstrating that these coordinates fail to cover the entire manifold.

By solving the geodesic equation I got $$\lambda = \frac{1}{H}\exp^{Ht}$$. Though I am not sure whether its correct as I am having trouble solving the last part.

• This seems strange - isn't $t$ proper time for comoving observers? – Javier Apr 21 at 12:29
• Yes I believe so. – kevint Apr 21 at 16:12
• The exercise is wrong, it's mentioned in the errata: preposterousuniverse.com/spacetimeandgeometry – Javier Apr 21 at 23:52

The exercise is wrong (the errata says so), but it's easy to see why. The geodesic equations can be derived from the Lagrangian

$$L = \frac12 g_{\mu\nu}(x) \dot{x}^\mu \dot{x}^\nu,$$

where a dot is a derivative with respect to the affine parameter. For our metric (which should say $$e^{2Ht}$$ instead of $$e^{Ht}$$ but let's leave it like it is), the Lagrangian is

$$L = -\frac12 \dot{t}^2 + \frac12 e^{Ht} (\underbrace{\dot{x}^2 + \dot{y}^2 + \dot{z}^2}_{\equiv v^2})$$

and we have

$$\frac{\partial L}{\partial t} = \frac{H}{2} e^{Ht} v^2, \qquad \frac{\partial L}{\partial \dot{t}} = -\dot{t},$$

$$\ddot{t} = - \frac{H}{2} e^{Ht} v^2$$
(and more for the spatial coordinates). Now, comoving observers are precisely those that don't move with respect to the comoving coordinates, so $$v^2 = 0$$ and $$t$$ is just a linear function of the affine parameter. To determine the constant of proportionality, we require that $$g_{\mu\nu} \dot{x}^\mu \dot{x}^\nu = -1$$, which is the condition that our wordline be parametrized by proper time. For our simple case with $$\dot{x}^i = 0$$, this gives that $$\dot{t}^2 = 1$$, so $$t = \lambda + \text{const}$$.