How to Derive the Shapiro Time Delay?

[stephen cripps]

Homework Statement

The step I am trying to follow is detailed here where I am trying to get from equation 6.26:
t=\int_{r_1}^{r}(1+\frac{2M}{r}+\frac{b^2V(r)}{2}+\frac{Mb^2V(r)}{r})dr
to equation 6.30
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}

Homework Equations

V(r)=r^{-2}(1-\frac{2M}{r})
b=r_z(1-\frac{2M}{r_1})^{-1/2}\approx r_1(1+M/r_1)

The Attempt at a Solution

I tried simplifying the equation by subbing in V, however my integral:
t=\int_{r_1}^{r}(1+\frac{2M}{r}+\frac{b^2}{2})(\frac{1}{r^2}-\frac{2M}{r^4})dr
seems to get nowhere near the required answer when integrated.

 

[TSny]

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 $M$.

 

[stephen cripps]

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 t=\int_{r_1}^r(1+\frac{2M}{r}+\frac{b^2}{2}[\frac{1}{r^2}-\frac{4M^2}{r^4}])dr

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?

 

[TSny]

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?

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 $M$.

 

[stephen cripps]

So starting from 6.24, by bringing the b inside the square root, I can replace (1-b^2V(r))^{1/2} With (1-\frac{r_1^2}{r^2}(1-\frac{2M}{r}+\frac{2M}{r_1}))^{1/2}. Ignoring higher orders of M. Expanding this again however doesn't seem to get me near the right answer.

 

[TSny]

So starting from 6.24, by bringing the b inside the square root, I can replace (1-b^2V(r))^{1/2} With (1-\frac{r_1^2}{r^2}(1-\frac{2M}{r}+\frac{2M}{r_1}))^{1/2}. Ignoring higher orders of M. Expanding this again however doesn't seem to get me near the right answer.

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?