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Finding the function from ds=[2/(1-r^2)]√ (dr^2+(rd∅)^2)

  1. Dec 10, 2013 #1
    in relativity theory, in a certain "rapid space" the distance between two neighboring points is given by-
    ds=[2/(1-r^2)]√ (dr^2+(rd∅)^2)

    (considering a 2D space )

    use the euler lagrange equation to show that the shortest distance from origin to any point is straight line.


    i don't know what's a rapid space, but i took the function in the euler lagrange equation as
    f= [2/(1-r^2)]√ (+(r∅')^2) [∅'=d∅/dr]

    then since ∂f/∂∅=0, so ∂f/∂∅'= const= r^2/ [(1-r^2)]√ (dr^2+(rd∅)^2)]

    but how does the integration lead to a straight line ? could anybody do that for me ?

    p.s: this isn't a homework !
  2. jcsd
  3. Dec 10, 2013 #2


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    rapidity space?
    http://abacus.bates.edu/~msemon/RhodesSemonFinal.pdf [Broken]
    Last edited by a moderator: May 6, 2017
  4. Dec 10, 2013 #3


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    I think all you need is to show that ##\ddot{\phi}=0##. With ##L=(1/2)(g_{rr}\dot{r}^2+g_{\phi\phi}\dot{\phi}^2)##. ##m## is set to 1 since it cancels out of the EOM.

    From ##\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\phi}}\right)=\frac{\partial L}{\partial \phi}## is evident that ##\ddot{\phi}=0##.
    Last edited: Dec 10, 2013
  5. Dec 11, 2013 #4

    i didn't understand. could you elaborate, please ?
  6. Dec 11, 2013 #5


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    Which piece did you not understand ?

    I have to admit that the extremization here is to find equations of motion and I don't know if that corresponds to what you are trying to do. Also, my assertion about the sufficiency of ##\ddot{\phi}=0## is probably not right.

    I need help too. Maybe someone else can put us right.

    Did you read the reference given by robphy ? The section on geodesics may be relevant.
  7. Dec 11, 2013 #6


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    See page 949 in Robphy's link. The derivation given is clear and simple.
  8. Dec 12, 2013 #7
    yes found it on rolphy. thank you everyone.
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