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Deriving the Shapiro time delay

  1. Dec 24, 2015 #1
    1. The problem statement, all variables and given/known data
    The step I am trying to follow is detailed here where I am trying to get from eqation 6.26:
    [tex]t=\int_{r_1}^{r}(1+\frac{2M}{r}+\frac{b^2V(r)}{2}+\frac{Mb^2V(r)}{r})dr[/tex]
    to equation 6.30
    [tex]t=\sqrt{r^2-r_1^2}+2Mln(\frac{r+\sqrt{r^2-r_1^2}}{r_1})+M(\frac{r-r_1}{r+r_1})^{1/2}[/tex]
    2. Relevant equations
    [tex]V(r)=r^{-2}(1-\frac{2M}{r})[/tex]
    [tex]b=r_z(1-\frac{2M}{r_1})^{-1/2}\approx r_1(1+M/r_1)[/tex]


    3. The attempt at a solution
    I tried simplifying the equation by subbing in V, however my integral:
    [tex]t=\int_{r_1}^{r}(1+\frac{2M}{r}+\frac{b^2}{2})(\frac{1}{r^2}-\frac{2M}{r^4})dr[/tex]
    seems to get nowhere near the required answer when integrated.
     
  2. jcsd
  3. Dec 25, 2015 #2

    TSny

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    Equation 6.26 in the link is misleading. It seems to imply that all of the terms that are linear in ##M## have been written out explicitly. But if you look at 6.25 there are terms of the form ##b^4V_{eff}^2##, ##b^6V_{eff}^3##, and so on that each contribute terms linear in ##M##. So, 6.25 generates an infinite number of terms that are linear in ##M##. Not a good way to go!

    Try going back to 6.24 and don't expand the square root factor until you first simplify the argument of the square root to first order in ##M##.
     
  4. Dec 27, 2015 #3
    Hi, thanks for you reply
    Sorry I made a mistake in my first statement, where I put my integral in section 3 tit should show [tex]t=\int_{r_1}^r(1+\frac{2M}{r}+\frac{b^2}{2}[\frac{1}{r^2}-\frac{4M^2}{r^4}])dr[/tex]

    Referring back to your solution though, is the argument not already in first order of M before it is expanded, just in a different form?
     
  5. Dec 27, 2015 #4

    TSny

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    Consider the term ##\frac{3}{8}b^4V_{eff}^2## that occurs in 6.25 but is not written out in 6.26. If you use the expressions for ##b## and ##V_{eff}## in terms of ##M## and expand ##\frac{3}{8}b^4V_{eff}^2##, you will see that you get a term independent of ##M##, a term linear in ##M##, as well as terms of higher order in ##M##.
     
  6. Dec 27, 2015 #5
    So starting from 6.24, by bringing the b inside the square root, I can replace [tex](1-b^2V(r))^{1/2}[/tex] With [tex](1-\frac{r_1^2}{r^2}(1-\frac{2M}{r}+\frac{2M}{r_1}))^{1/2}[/tex]. Ignoring higher orders of M. Expanding this again however doesn't seem to get me near the right answer.
     
  7. Dec 27, 2015 #6

    TSny

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    Your expression looks correct. It works out for me. Could you show the next few steps of how you expanded the square root to first order in M?
     
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