# Deriving the Shapiro time delay

1. Dec 24, 2015

### stephen cripps

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:
$$t=\int_{r_1}^{r}(1+\frac{2M}{r}+\frac{b^2V(r)}{2}+\frac{Mb^2V(r)}{r})dr$$
to equation 6.30
$$t=\sqrt{r^2-r_1^2}+2Mln(\frac{r+\sqrt{r^2-r_1^2}}{r_1})+M(\frac{r-r_1}{r+r_1})^{1/2}$$
2. Relevant equations
$$V(r)=r^{-2}(1-\frac{2M}{r})$$
$$b=r_z(1-\frac{2M}{r_1})^{-1/2}\approx r_1(1+M/r_1)$$

3. 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.

2. Dec 25, 2015

### 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 $M$.

3. Dec 27, 2015

### 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?

4. Dec 27, 2015

### TSny

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 $M$.

5. Dec 27, 2015

### 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.

6. Dec 27, 2015

### TSny

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?