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Given a DeSitter-space metric from the line element:

$$ ds^2=\left(1-\frac{r^2}{R^2}\right)dt^2-\left(1-\frac{r^2}{R^2}\right)^{-1}dr^2-r^2d\Omega^2 $$

Where $R=\sqrt{\frac{3}{\Lambda}}$, and $\Lambda$ is a positive cosmological constant, I am trying to derive the equations for radial null geodesics. I derived the geodesic equations from the definition and Christoffel symbols, but I'm a little suspicious of my solution to these equations. So, the differential equations I derived are ($\lambda$ is an "affine" parameter):

$$ \frac{d^2r}{d\lambda^2}-\frac{r}{R^2}\left(1-\frac{r^2}{R^2}\right)\left(\frac{dt}{d\lambda}\right)^2+\frac{r}{R^2-r^2}\left(\frac{dr}{d\lambda}\right)^2=0 $$

$$ \frac{d^2t}{d\lambda^2}+\frac{2r}{r^2-R^2}\frac{dt}{d\lambda}\frac{dr}{d\lambda}=0 $$

Now, I tried to use the identity $\textbf{u}\cdot\textbf{u}=0$ where $\bf u$ is the four velocity (or I guess a vector tangent to the light ray's world line if I'm thinking about this correctly). Thus, because $u^2=u^3=0$:

$$g_{\mu\nu}u^{\mu}u^{\nu}=\left(1-\frac{r^2}{R^2}\right)(u^0)^2-\left(1-\frac{r^2}{R^2}\right)^{-1}(u^1)^2=0 $$

$$\implies -\frac{r}{R^2}\left(1-\frac{r^2}{R^2}\right)(u^0)^2+\frac{r}{R^2-r^2}(u^1)^2=0$$

Then, substituting this into the first (radial) geodesic equation, I get:

$$\frac{d^2r}{d\lambda^2}=0$$

This is what I am suspicious of. It seems too easy and simple. Do you think this is correct? If I carry on anyways and integrate to find $u^0(r)$ then substitute $r(\lambda)$ I get the following for $t(\lambda)$:

$$ r(\lambda)=A\lambda+B $$

$$ t(\lambda)=\frac{D}{AR}\tanh^{-1}\left[\frac{A\lambda+B}{R} \right]+E $$

Where, $A,B,D,E$ are constants of integration. To sum up, is this a viable way to solve these equations or am I missing something? If this is correct, how does one define initial conditions for these solutions. Specifically, the derivative conditions $t'(0)$ and $r'(0)$.

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    $\begingroup$ I didn't check your work, but there is a trick for easily finding null geodesics in 1+1 dimensional problems (which this is because of symmetry): just set $ds=0$ in the metric. This gives you $(dr/dt)^2 = (1-r^2/R^2)^2$ or $t = A \pm R \tanh^{-1} (r/R)$, which is at least close to what you got. (Note that you can substitute $r(\lambda)$ into your second equation and eliminate $\lambda$.) $\endgroup$ – benrg Aug 9 '14 at 7:47
  • $\begingroup$ Hey thanks for the reply, sorry I've been away from a computer for a little. This makes it so much easier! I find it almost strange how we can solve without ever having to touch the geodesic equation, but I guess it is tied up in there with the line element. It even looks like my answer is extremely close like you said. Just a $1/R$ instead of a $R$. Maybe those constant could take care of it though. @benrg $\endgroup$ – ClassicStyle Aug 10 '14 at 15:26
  • $\begingroup$ Also, something weird might be going on...if $r>R$ then we get imaginary time values. @benrg $\endgroup$ – ClassicStyle Aug 10 '14 at 16:31
  • $\begingroup$ There's a coordinate singularity at $r=R$, and beyond that the $r$ coordinate is timelike and the $t$ coordinate is spacelike. It's similar to the $r=2m$ singularity in the Schwarzschild metric. I don't know whether $r>R$ has a physical interpretation (it does in the Schwarzschild case) but normally you would use different coordinates if you wanted to cover the rest of the de Sitter space, so don't worry about strange behavior when $r\ge R$. $\endgroup$ – benrg Aug 10 '14 at 18:32
  • $\begingroup$ Setting $ds=0$ works because the light cone is discrete—the light can only go "left" or "right" and switching directions would make the geodesic nondifferentiable, so ds=0 plus differentiability is enough. There's a way to get non-null geodesics in n+1 dimensions from the metric with a variational principle, but I don't know how to get null geodesics in n+1 dimensions without calculating the Christoffel symbols. $\endgroup$ – benrg Aug 10 '14 at 18:36

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