Geodesic quation coordinate time

[Svendsen]

Hi guys

So I am 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 don't 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.

Thanks for your time!

 

[pervect]

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 $t(\tau)$.

 

[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 $\frac{d^2 t}{d\tau^2}$.

 

[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).

 

[sardar]

Hi there,

It seems like you are on the right track with reparameterizing the geodesic equation in terms of coordinate time. One thing to keep in mind is that the proper time, \tau, is related to the coordinate time, t, by the equation:

d\tau = \sqrt{g_{nm}dx^ndx^m} \, dt

So in order to find dt/d\tau, you can simply take the inverse of this equation:

\frac{dt}{d\tau} = \frac{1}{\sqrt{g_{nm}dx^ndx^m}}

Once you have this, you can substitute it into your original equation and simplify to get the geodesic equation in terms of coordinate time. I hope this helps! Let me know if you have any further questions.