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 stan with (2-29),
based on the Doppler effect in the acceleraing frame. Consider two elevations,
the lower at \(r_{1}\) and the upper at \(r_{1}+d r\). Equation \((2-29)\) becomes
$$
\frac{f\left(r_{1}+d
r\right)}{f\left(r_{i}\right)}=\left(1-\frac{g\left(r_{1}\right) d
r}{c^{2}}\right)
$$
Similarly, if we consider elevations \(r_{1}+d r\) and \(r_{1}+\) \(2 d r\). we have
$$
\frac{f\left(r_{1}+2 d r\right)}{f\left(r_{1}+d
r\right)}=\left(1-\frac{g\left(r_{1}+d r\right) d r}{c^{2}}\right)
$$
We continue the process, incrementing \(r\) by \(d r\), until we reach \(r_{2}\)
$$
\frac{f\left(r_{2}\right)}{f\left(r_{2}-d
r\right)}=\left(1-\frac{g\left(r_{2}-d r\right) d r}{c^{2}}\right)
$$
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 \(g d r / c^{2}\) 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
$$
1-\frac{1}{c^{2}} \int g(r) d r
$$
(c) Deduce \(g(r)\) from Newton's universal law of gravitation, then argue that
equation \((2-31)\) follows from the result, just as \((2 \cdot 30)\) does from
\((2-29)\).
