# I Concept q -- affine parameter and coordinate time geodesic

1. Jan 30, 2017

### binbagsss

I am asked a question about how far a light ray travels, the question is to be solved by solving for the null goedesic.

I am given the initial data: the light ray is fired in the $x$ direction at $t=0$.

The relvant coordinates in the question are $t,x,y,z$, let $s$ be the affine paramater I parameterise the geodesics with.

MY QUESTION

Q1)Concerning the intial data, does this mean that at $t=0$, $\dot{y}=\dot{z}=0$ and $\dot(z)\neq 0$? where $\dot{x}$ denotes $\frac{dx}{ds}$ .

Q2)And not $\frac{dy}{dt}=\frac{dz}{dt}=0$? can we only make conclusions on the change with respect $s$, or is it with respect to the coordinate time $t$ too?

Q3)Also, is one always free to choose initially that $s=t=0$. If so, then if the answer to question 1 is yes, doesn't this imply the answer to question 2 is yes?

Many thanks

2. Feb 6, 2017

### davidge

I think you can't use proper time as the affine parameter for null geodesics.

3. Feb 6, 2017

### Staff: Mentor

You can't, but I don't see the OP doing that.

4. Feb 6, 2017

### Staff: Mentor

It means $\dot{y} = \dot{z} = 0$, since that's what "in the $x$ direction" is supposed to mean. I'm not sure what the third thing is that you are asking if it's not equal to zero; but at $t = 0$, you will have $\dot{t} \neq 0$ and $\dot{s} \neq 0$.

Those will be true too. The proof is easy: you have $\frac{dy}{dt} = \frac{dy}{ds} \frac{ds}{dt}$ by the chain rule, so the fact that $\frac{dy}{ds} = 0$ implies that $\frac{dy}{dt} = 0$. Similarly for $\frac{dz}{dt}$.

Yes.

No, because choosing $s = t = 0$ at the initial event doesn't tell you anything by itself about their derivatives, or derivatives with respect to them. But as above, you can answer question 2 anyway.