What is the Geodesic Equation for FRW Metric's Time Component?

[unscientific]

Taken from Hobson's book:

Metric is given by
ds^2 = c^2 dt^2 - R^2(t) \left[ d\chi^2 + S^2(\chi) (d\theta^2 + sin^2\theta d\phi^2) \right]

Thus, $g_{00} = c^2, g_{11} = -R^2(t), g_{22} = -R^2(t) S^2(\chi), g_{33} = -R^2(t) S^2(\chi) sin^2 \theta$.

Geodesic equation is given by:
\dot u_\mu = \frac{1}{2} \left( \partial_\mu g_{v\sigma} \right) u^v u^\sigma

The coordinates are given by $u^0 = \dot t, u^1 = \dot \chi, u^2 = \dot \theta, u^3 = \dot \phi$.

For the temporal component,
\dot u_0 = \frac{1}{2} (\partial_0 g_{v\sigma})u^v u^\sigma

Photons

u^0u_0 = 0
u^0 g_{00} g^0 = 0
g_{00}\dot t^2 = 0
\dot t = 0

This doesn't make any sense. For massive particles, $\dot t = 1$.

 

[Orodruin]

Why do you think $u_0 u^0 = 0$? It is not true. Remember the Einstein summation convention.

 

[unscientific]

Why do you think $u_0 u^0 = 0$? It is not true. Remember the Einstein summation convention.

For a photon, it is 0, as shown in the text. (null vector)

 

[Orodruin]

For a photon, it is 0, as shown in the text. (null vector)

No, it is not $u^\mu u_\mu = 0$ does not imply $u^0 u_0 = 0$.

 

[unscientific]

No, it is not $u^\mu u_\mu = 0$ does not imply $u^0 u_0 = 0$.

So, $u^0u_0 + u^1u_1 + u^2u_2 + u^3u_3 = 0$?

 

[Orodruin]

So, $u^0u_0 + u^1u_1 + u^2u_2 + u^3u_3 = 0$?

Yes.