Chapter 2, Problem 98E

Equation (2-30) is an approximation correct only if the gravitational time-dilation effect is small. In this exercise, it is also assumed to be small but we still allow for a nonuniform gravitational field. We start with (2-29), based on the Doppler effect in the accelerating frame. Consider two elevations, the lower at $r_1$ and the upper at $r_1+dr$ . Equation (2-29) becomes
$\frac{f\left(r_1+dr\right)}{f\left(r_1\right)}=\left(1-\frac{g\left(r_1\right)dr}{c^2}\right)$

Similarly, if we consider elevations $r_1+dr$ and $r_1+2dr$ , we have
$\frac{f\left(r_1+2dr\right)}{f\left(r_1+dr\right)}=\left(1-\frac{g\left(r_1+dr\right)dr}{c^2}\right)$

We continue the process, incrementing r by dr, until we reach $r_2$ .

$\frac{f\left(r_2\right)}{f\left(r_2-dr\right)}=\left(1-\frac{g\left(r_2+dr\right)dr}{c^2}\right)$

Now imagine multiplying the left sides of all the equations and setting the product equal to the product of all the right sides. (a) Argue that the left side of the product is simply $f\left(r_2\right)/f\left(r_1\right)$ . (b) Assuming that the term $gdr/c^2$ in each individual equation is very small, so that products of such terms can be ignored, argue that the right side of the product is $1-\frac{1}{c^2}{\displaystyle \int g\left(r\right)dr}$ (c) Deduce g(r) from Newton’s universal law of gravitation, then argue that equation (2-31) follows from the result, just as (2-30) does from (2-29).