# Geodesic quation coordinate time

1. Nov 11, 2014

### Svendsen

Hi guys

So im having trouble reparameterizing the geodesic equation in terms of coordinate time.
Normally you have:

$$\frac{d^2 x^{\alpha}}{d \tau^2} + \Gamma_{nm}^{\alpha} \frac{d x^{n}}{d \tau}\frac{d x^{m}}{d \tau} = 0$$

Where $\tau$ is the proper time. I class we were told to express the above in terms of coordinate time and so i reasoned that one would use the chain rule:

$$\frac{d }{d \tau} = \frac{d t}{d \tau} \frac{d }{d t}$$

When i do so i get the following:

$$\frac{d^2 x^{\alpha}}{d t^2} + \Gamma_{nm}^{\alpha} \frac{d x^{n}}{d t}\frac{d x^{m}}{d t} = - \frac{d^2t/d \tau^2}{dt/d\tau} \frac{d x^{\alpha}}{dt}$$

Which - i guess - has the form that one would expect because t is non-affine.

However i dont know how to proceed from here. I´ve tried to use $d\tau ^2 = g_{nm}dx^ndx^m$ to find $dt/d \tau$, but i can´t seem to get anything meaningful.

2. Nov 12, 2014

### pervect

Staff Emeritus
I did something related to this once upon a time, but it's been a while. Are you looking to find the form of the geodesic equations given that you have $x^1$, $x^2$, $x^3$ as functions of t? If so I might try to reconstruct what I did if you haven't already figured it out by the time I get to it., As I recall it was a matter of using the chain rule, plus some algebra involving the geodesic equation for $x^0$ (which is another name for "t") in it's original form as a function of $\tau$, i.e. the standard geodesic equation for $x^0(\tau)$ or $t(\tau)$.

3. Nov 14, 2014

### WannabeNewton

You cannot simplify it any further but you can express it in terms of meaningful quantities: $\frac{dt}{d\tau} = u^0 = \gamma$ where $\gamma$ is the "time dilation factor" of the particle in the coordinate system that it is moving through. In stationary space-times, that is those with a time-like Killing field, you can also express $\frac{dt}{d\tau}$ in terms of the conserved energy per unit mass $e$. For example in Schwarzschild space-time we have $e = (1 - \frac{2M}{r})\frac{dt}{d\tau}$. Furthermore you can use the geodesic equation for $\alpha = t$ to get rid of $\frac{d^2 t}{d\tau^2}$.

Last edited: Nov 14, 2014
4. Nov 14, 2014

### ZealScience

What did you mean by meaningful? This is the expression for geodesic parametrized by non-affine parameter and any actual physical manifestation might require information on the physical system. If you mean to eliminate the τ derivatives; then I think it is possible to set α to 0th component.

In case you have never done so, physical interpretations usually rely on the Newtonian weak field limit, where the τ derivatives are eliminated by the original geodesic equation under that limit (again by setting α to 0 and making suitable assumptions about the Christoffel connection coefficients).