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.

  • $\begingroup$ This seems strange - isn't $t$ proper time for comoving observers? $\endgroup$ – Javier Apr 21 '19 at 12:29
  • $\begingroup$ Yes I believe so. $\endgroup$ – kevint Apr 21 '19 at 16:12
  • $\begingroup$ The exercise is wrong, it's mentioned in the errata: preposterousuniverse.com/spacetimeandgeometry $\endgroup$ – Javier Apr 21 '19 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

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

leading to the Euler-Lagrange equation

$$\ddot{t} = - \frac{H}{2} e^{Ht} v^2$$

(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 $t = \lambda + \text{const}$.

These coordinates do in fact only cover a portion of de Sitter space, but you can't get that from comoving geodesics.


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.