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I'm working using the standard FRW metric,

$$ds^2=dt^2-a^2\left [\frac{dr^2}{1-kr^2}+r^2(d\theta^2+\sin^2\theta d\phi^2)\right ]$$

Using the definition of the Christoffel symbols,

$$\Gamma^c_{ab}=\frac{1}{2}g^{cd}(g_{ad,c}+g_{bd,a}-g_{ab,d})$$

I've found the non-zero Christoffel symbols for the FRW metric, using the notation $(t,r,\theta,\phi)=(0,1,2,3)$,

$$\Gamma^{0}_{11}=\frac{a\dot{a}}{1-kr^2} \ \ ,\ \ \Gamma^{0}_{22}=a\dot{a}r^2 \ \ ,\ \ \Gamma^{0}_{33}=a\dot{a}r^2 \sin^2\theta \ \ , \ \ \Gamma^{1}_{11}=\frac{kr}{(1-kr^2)}$$

$$\Gamma^{1}_{01}=\Gamma^{1}_{02}=\Gamma^{1}_{03}=\frac{\dot{a}}{a} \ \ ,\ \ \Gamma^{1}_{22}=-r(1-kr^2) \ \ ,\ \ \Gamma^{1}_{33}=-r(1-kr^2) \sin^2\theta$$

$$\Gamma^{2}_{12}=\Gamma^{3}_{13}=\frac{1}{r} \ \ ,\ \ \Gamma^{2}_{33}=-\sin\theta \cos\theta \ \ ,\ \ \Gamma^{3}_{23}=\frac{\cos\theta}{\sin\theta}$$

Now I'm trying to derive the geodesic equations for this metric, which are given as,

$$\frac{d^2x^\mu}{ds^2}+\Gamma^{\mu}_{\nu\lambda}\frac{dx^\nu}{ds}\frac{dx^\lambda}{ds}=0$$

For example, for $\mu=0$, I get that,

$$\frac{d^2t}{ds^2}+\frac{a\dot{a}}{1-kr^2}\left (\frac{dr}{ds}\right )^2+a\dot{a}r^2\left (\frac{d\theta}{ds}\right )^2+a\dot{a}r^2 \sin^2\theta\left (\frac{d\phi}{ds}\right )^2=0$$

However, when I checked with this document in the section of geodesics (http://popia.ft.uam.es/Cosmology/files/02FriedmannModels.pdf), they get,

$$\frac{|u|}{c}|\dot{u}|+\frac{\dot{a}}{a}|u|^2=0$$

where $\frac{du^0}{ds}=\frac{|u|}{c}|\dot{u}|$ and $u^\mu=\frac{dx^\mu}{dt}$.

I'm not sure what I'm doing wrong or if it's just a matter of convention. I also checked this question (Geodesics for FRW metric using variational principle) but the FRW metric is slightly different, so it didn't help.

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  • $\begingroup$ You missed a symbol: $\Gamma^1_{11}=\frac{kr}{1-kr^2}$. Also, aren't the symbols defined as $\Gamma^\sigma_{\mu\nu}=\frac{1}{2}g^{\sigma\rho}(\partial_\mu g_{\nu\rho}+\partial_\nu g_{\rho\mu}-\partial_\rho g_{\mu\nu})$? $\endgroup$ – Tesseract Nov 14 '19 at 21:15
  • $\begingroup$ Indeed, sorry I forgot to include that when typing. However, the definition of the symbols is equivalent when you contemplate their symmetries, as the one I put is the one used by my professor. In any case, the Christoffel symbols are correct as I checked with the following document (p. 22): icc.ub.edu/~liciaverde/Cosmology.pdf $\endgroup$ – Charlie Nov 14 '19 at 21:31
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Well, it seems the two are the same. First, I assume we are not working in natural units, because of your final equation. Thus, the metric changes to $$ds^2=(cdt)^2-a^2\left [\frac{dr^2}{1-kr^2}+r^2(d\theta^2+\sin^2\theta d\phi^2)\right ]$$ Thus, your equation changes to $$c\frac{d^2t}{ds^2}+\frac{a\dot{a}}{1-kr^2}\left(\frac{dr}{ds}\right)^2+a\dot{a}r^2\left(\frac{d\theta}{ds}\right)^2+a\dot{a}r^2\sin^2\theta\left(\frac{d\phi}{ds}\right)^2$$ Second, note that, according to the resource, the velocity in the term $\frac{\dot{a}}{a}|u|^2$ is the three velocity. So, I'm going to denote the four velocity with $u^\mu$ and the three velocity with $u^i$. So, the equation given becomes $$\frac{du^0}{ds}+\frac{\dot{a}}{a}u^iu_i=0$$ where I have taken the liberty of changing out the first term, as the derivation is given within said resource, and you have already noted it. We multiply by negative one, because the resource uses a different metric signature: $$-\frac{du^0}{ds}-\frac{\dot{a}}{a}u^iu_i=0$$ Your first term is equal to $$-\frac{du^0}{ds}=-\frac{c}{c}\frac{du^0}{ds}=c\frac{d}{ds}\left(-\frac{1}{c}\frac{dt}{d\tau}\right)=c\frac{d}{ds}\left(\frac{dt}{ds}\right)=c\frac{d^2t}{ds^2}$$ For the second, we have $$\begin{align}-\frac{\dot{a}}{a}(u^i u_i)&=\frac{\dot{a}}{a}\left(\frac{a^2}{1-kr^2}\left(\frac{dr}{ds}\right)^2+a^2r^2\left(\frac{d\theta}{ds}\right)^2+a^2r^2\sin^2\theta\left(\frac{d\phi}{ds}\right)^2\right) \\ &=\frac{a\dot{a}}{1-kr^2}\left(\frac{dr}{ds}\right)^2+a\dot{a}r^2\left(\frac{d\theta}{ds}\right)^2+a\dot{a}r^2\sin^2\theta\left(\frac{d\phi}{ds}\right)^2 \end{align}$$ Therefore, $$\frac{du^0}{ds}+\frac{\dot{a}}{a}u^iu_i=0=c\frac{d^2t}{ds^2}+\frac{a\dot{a}}{1-kr^2}\left(\frac{dr}{ds}\right)^2+a\dot{a}r^2\left(\frac{d\theta}{ds}\right)^2+a\dot{a}r^2\sin^2\theta\left(\frac{d\phi}{ds}\right)^2=0$$ and they are the same.

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