# FRW Metric: time component

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1. Apr 7, 2015

### 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, $\dot t = 1$.

2. Apr 7, 2015

### Orodruin

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

3. Apr 7, 2015

### unscientific

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

4. Apr 7, 2015

### Orodruin

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

5. Apr 7, 2015

### unscientific

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

6. Apr 7, 2015

### Orodruin

Staff Emeritus
Yes.

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