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Schutz page 294 (Deflection of light)

  1. Jul 6, 2015 #1
    Hello Everyone,

    I am working through Schutz's A first Course in General relativity. On page 294 he defines the equations (11.52) to simplify equation (11.49) and calculated the deflection of light around the sun. I know that he wants to simplify it and also to preserve the effect of the mass M. I am just not sure how and why he defined (11.52) in this particular way.

    ( 11.52): y=u(1-Mu) and u=y(1+My)+O(M2u2)

    (11.49): ##\frac{d\phi}{du}={(b^{-2}-u^{2}+2Mu^{3})}^{-\frac{1}{2}}##

    Thanks.
     
  2. jcsd
  3. Jul 8, 2015 #2
    Anyone? :oops:
     
  4. Jul 10, 2015 #3

    pervect

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    I don't have Schutz, so it's hard to comment. I've seen other texts in which u=1/r, and b is given a name the "impact parameter". I can't confirm that's what the symbols mean in Schutz's text at this point though.
     
  5. Jul 10, 2015 #4

    pervect

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    I don't have Schutz, so it's hard to comment. I've seen other texts in which u=1/r, and b is given a name the "impact parameter". I can't confirm that's what the symbols mean in Schutz's text at this point though. The impact parameter was closely related to the distance of closest approach, IIRC - there was additional explanation of it's significance which I have since forgotten.
     
  6. Jul 10, 2015 #5
    Thanks for the reply pervect. Yes you are correct about the definitions. But I actually know what they mean and can do the calculation just fine. I am just not sure why he defined (11.52) in that specific way to simplify (11.49).
     
  7. Jul 12, 2015 #6

    vanhees71

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    That's just a substitution to get the approximation of the elliptic integral (11.49). Of course, you can just evaluate this integral to get the deflection of a light ray (not a photon, as unfortunately written all the time by Schutz; a photon has no trajectory, because you cannot even define a position operator in flat spacetime; I guess QED in the Schwartzschild metric is pretty complicated and fortunately not necessary for the question under investigation; see also the thread on "null geodesics").
     
  8. Jul 12, 2015 #7
    Thanks vanhees71. I have read your post in the thread you have mentioned. tbh I have always thought of a photon in GR text books as the classical light rays.
    I wish if Schutz stated why he uses some of those substitutions to avoid confusion. What is that second order term doing in (11.52)? Also how are the two equations IN (11.52) related ? (y and u)
     
  9. Jul 12, 2015 #8

    DrGreg

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    Are you familiar with Big O notation?

    If y is defined to be [itex]y = u(1-Mu)[/itex], then [itex]My =Mu + O(M^2u^2) \text{ as } Mu \rightarrow 0[/itex], i.e. [itex]Mu = My + O(M^2u^2)[/itex] and so[tex]
    \begin{align*}
    u &= \frac{y}{1-Mu} \\
    &= y(1 + Mu + O(M^2u^2)) \\
    &= y(1 + My) + O(M^2u^2) \, .
    \end{align*}
    [/tex]
     
  10. Jul 12, 2015 #9

    vanhees71

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    The 2nd Eq. in (11.52) is an approximate inverse of the first. You get the approximation as follows. You have
    $$y=u(1-Mu)$$
    or
    $$u=\frac{1}{2M} (1-\sqrt{1-4My})=y \left [1+M y+\mathcal{O}(M^2 y^2) \right].$$
    Since thus ##u \simeq y## up to linear order in ##My## you can as well write
    $$u=y[1+My+\mathcal{O}(M^2 u^2)].$$
     
  11. Jul 12, 2015 #10
    Thank you both DrGreg and vanhees71. Very easy to follow.
    I guess I will read up about elliptic integrals to get a better feel of the integral.

    Just one thing, I think that there are easier ways to calculate the deflection. Why did Schutz choose this way to do it? any conceptual/mathematical reason or listen behind it?
     
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