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Radial Null Geodesics in Static Maximally Symmetric DeSitter Space

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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 $r'(0)$.

This post imported from StackExchange Physics at 2014-08-09 08:48 (UCT), posted by SE-user TylerHG
asked Aug 9, 2014 in Theoretical Physics by TylerHG (10 points) [ no revision ]
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 $\lambda$.)

This post imported from StackExchange Physics at 2014-08-09 08:48 (UCT), posted by SE-user benrg

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