pmb_phy said:
You can find a derivation I did on this web page. Its pretty much a standard derivation.
http://www.geocities.com/physics_world/gr/grav_red_shift.htm
There is, actually, a simpler derivation, which doesn't involve general relativity at all.
Fundamentally, any clock is a non-stationary physical system that
exhibits a periodic process with a stable frequency. In quantum
mechanics, a state of any non-stationary system is represented by a
superposition of two or more stationary states. For simplicity, we
will assume that these states have distinct discrete energy levels
E_i and E_f with masses m_i = E_i/c^2 and m_f = E_f/c^2,
respectively. Then the rate of the corresponding "clock" is
proportional to the energy difference E_i - E_f. For example, the
frequency of the electromagnetic radiation emitted in an
atomic transition is
<br />
\nu(0) = \frac{2 \pi}{\hbar} (E_i - E_f) = \frac{2 \pi<br />
c^2}{\hbar} (m_i - m_f)<br />
Total energies of the system "stationary clock + Earth" in the
ground and excited states are
<br />
\mathcal{E}_f \approx Mc^2 + m_fc^2 + \phi m_f <br />
<br />
\mathcal{E}_i \approx Mc^2 + m_ic^2 + \phi m_i
where M is the Earth's mass, \phi = - GMR^{-1} is the
value of the gravitational potential on the Earth's surface,
and R is the Earth's radius. Then the frequency of the clock on
Earth can be found as
<br />
\nu(\phi) = \frac{2 \pi}{\hbar} (\mathcal{E}_i - \mathcal{E}_f)<br />
\approx \nu(0)(1 + \frac{\phi}{c^2})
This means that a clock on the Earth surface is running
slower than an identical clock at a higher elevation (higher
gravitational potential \phi).
Eugene.