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Definitions Emitter v.s. Observer

  1. Mar 9, 2013 #1
    Let's say we are working with the Schwarzschild metric and we have an emitter of light falling into a Schwarzschild black hole.

    Suppose we define the quantity [itex]u=t- v[/itex] where [itex]dv/dr= 1/(1-r_{s}/r)[/itex] where [itex]r_s[/itex] is the Schwarzschild radius. What is the [itex]u[/itex] as observed by the emitter? I just need a *definition of [itex]u_e[/itex]*. I have problems identifying the quantities as measured by an observer at large [itex]r[/itex] and that of the emitter. Would I be right at least to say that [itex]t_{e}=\tau[/itex] the proper time? Many thanks.

    _____

    In fact, I've been told that

    $$du_o/d\tau=du_e/d\tau$$

    Why is it?
     
    Last edited: Mar 9, 2013
  2. jcsd
  3. Mar 9, 2013 #2

    Dale

    Staff: Mentor

    I don't understand. You already gave the definition of u as u=t-v.
     
  4. Mar 9, 2013 #3
    @DaleSpam: You are right. I have missed out the subscript [itex]e[/itex]. I meant [itex]u[/itex] as measured by the emitter.
     
    Last edited: Mar 9, 2013
  5. Mar 9, 2013 #4

    Dale

    Staff: Mentor

    Well, v is just some function of r and so u is just some function of t and r. I don't see how it is something measured by anyone.

    Please fix your LaTeX.
     
  6. Mar 9, 2013 #5
    @DaleSpam: LaTex fixed. I suppose [itex]u_e[/itex] would be in terms of [itex]t_e,r_e[/itex] measured by the emitter, no? But then what is its explicit form?
     
  7. Mar 9, 2013 #6

    Dale

    Staff: Mentor

    Well, what coordinate system do you want to use for the emitter? Then you would just have to substitute in the transforms for the emitter's coordinates.
     
  8. Mar 9, 2013 #7
    @DaleSpam: the proper time is that of the emitter. I am still not sure how to do the substitution though. (Sorry about my daftness. :( )
     
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