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I Concept q -- affine parameter and coordinate time geodesic

  1. Jan 30, 2017 #1
    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. jcsd
  3. Feb 6, 2017 #2
    I think you can't use proper time as the affine parameter for null geodesics.
     
  4. Feb 6, 2017 #3

    PeterDonis

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    You can't, but I don't see the OP doing that.
     
  5. Feb 6, 2017 #4

    PeterDonis

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    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.
     
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