# Geodesic equation & Four-velocity

I've been studying Kolb & Turner's "The Early Universe", and came across an equation that somehow I can't understand. Given the equation:

$$\frac{1}{u_0} \frac{d |\vec u|}{ds} + \frac{\dot a(t)}{a(t)}|\vec u| = 0$$

They write:

$$\frac{|\dot{ \vec{u}}|}{|\vec u|} = -\frac{\dot a(t)}{a(t)}$$

Stating that

$$u^0 \equiv \frac{dt}{ds}$$

This last expression is quite simple, but I can't derive it... Can someone help me?

Just plug $u_0= \frac{d t}{d s}$ into the first equation $$\frac{1}{u_0} \frac{d |\vec u|}{ds} + \frac{\dot a(t)}{a(t)}|\vec u| = \frac{d s}{d t} \frac{d |\vec u|}{ds} + \frac{\dot a(t)}{a(t)}|\vec u| = 0 0$$

Identify $\frac{d s}{d t} \frac{d |\vec u|}{ds}= \frac{d |\vec u|}{dt}=\dot u(t)$ through chain rule and you are ready

• No, what I don't understand is the last expression, $$u^0 = \frac{dt}{ds}$$ Not the replacement. But thanks! Commented Sep 2, 2016 at 16:47
• I dont have the book but I would guess it is just the definiton of the zeroth four-velocity component Commented Sep 2, 2016 at 16:49
• And in FLRW metric $u_0=u^0$ Commented Sep 2, 2016 at 16:49
• $u^0\equiv dt/ds$ is the definiton of the time component of the four velocity, that is why there is an "equivalent" sign not an "equal". It is the same as in special relativity.
– N0va
Commented Sep 2, 2016 at 18:45

Well simply put, a metric can be written as $$ds^2 = g_{\mu\nu}dx^{\mu}dx^{\nu}$$

and if you rewrite it a little, it gives $$1 = g_{\mu\nu}u^{\mu}u^{\nu} = u_{\nu}u^{\nu}$$ with Einstein summation implied.

Here we have defined $u^{\mu}$ and $u_{\mu}$ such that it equals $\frac{dx^{\mu}}{ds}$ and $\frac{dx_{\mu}}{ds}$ respectively where ${\mu}$ spans {0,1,2,3} in most cases. Denoting the zeroth index in {0,1,2,3} for time, while other three usually reserved exclusively for the spatial indexes (some authors identify the time index with the number 4 as in {1,2,3,4} rather than 0)

Thus we get the relation

$$u^0 = \frac{dt}{ds}$$