Concept q -- affine parameter and coordinate time geodesic

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  • #1
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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
 

Answers and Replies

  • #2
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let sss be the affine paramater I parameterise the geodesics with
I think you can't use proper time as the affine parameter for null geodesics.
 
  • #3
PeterDonis
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I think you can't use proper time as the affine parameter for null geodesics.
You can't, but I don't see the OP doing that.
 
  • #4
PeterDonis
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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}##.
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##.

And not ##\frac{dy}{dt}=\frac{dz}{dt}=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}##.

is one always free to choose initially that ##s=t=0##.
Yes.

if the answer to question 1 is yes, doesn't this imply the answer to question 2 is 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.
 

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