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Angle of deviation general relativity, trigo

  1. May 5, 2013 #1
    1. The problem statement, all variables and given/known data
    Hello, I have a little problem with this figure, this is not an assignment or something else I just don't understand how we can find that d∅=gocos3(α)/c² *dz



    2. Relevant equations
    go=G*M/R² with R the sun's radius

    3. The attempt at a solution
    I wrote all what I can deduce
    ge=gr*cos(α)
    r=R/cos(α)
    dt=dz/c
    tg(∅)≈∅= R/dz
    tg(α)=z/R

    Please I really need to understand this :frown:
     

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    Last edited: May 5, 2013
  2. jcsd
  3. May 5, 2013 #2
    What is this all about, exactly?
     
  4. May 5, 2013 #3
    In the course the teacher advised us to do it and just gave us the expression, so I'm trying to, my exam is Tuesday and I want to understand how we find dphi because it's from this equation, by integration that we find that phi=2*G*M/(c²*R)
    Thanks
     
  5. May 5, 2013 #4
    To do "it" - what is "it"? Describe what the drawing represents and what you are supposed to find.
     
  6. May 5, 2013 #5
    "It" is to prove that d∅=go*cos3(α)/c² *dz
    It's the angle of deviation of a light beam in the gravitational field of the sun.
    What I'm trying to find is this expression d∅=go*cos3(α)/c² *dz from the drawing
    Thanks
     
  7. May 5, 2013 #6
    Are you going to feed us one spoon at a time? Unless you explain what all those symbols mean - except ##R## and ##g_0## which you did explain - you won't get much help here.
     
  8. May 5, 2013 #7
    dz is the width of the reference frame
    r is the distance between the reference frame and the center of sun
    gr gravitational interaction
    ge is the effective gravitational acceleration
     
    Last edited: May 5, 2013
  9. May 5, 2013 #8
    None of this is making sense. There cannot be a distance to a "reference frame" because a reference frame fills up the entire space, and there is no width of it, for the same reason.

    In the context of curving a light beam by the Sun, I would expect a distance source (star) and a receiver (observer) on the Earth. I cannot see any of that on your diagram.
     
  10. May 5, 2013 #9
    In the example in the class the teacher explained that the little squares were local inertial reference frame (as elevators) and as the photon of a star were crossing them they fall in the direction of the sun.
    I'm sorry for the lack of informations, I'm trying to remember because my notes are not complete..
     
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