H
KE = mgH
KE = 0 E = hv
E = hv +hv
c^2
gH = hv′
Figure 2.30A photon that falls in a gravitational field gains energy, just as a stone does. This gain in
energy is manifested as an increase in frequency from to .
Hence,
final photon energy initial photon energy increase in energy
hhmgH
and so
hhgH
hh 1 (2.28)
Example 2.8
The increase in energy of a fallen photon was first observed in 1960 by Pound and Rebka at
Harvard. In their work Hwas 22.5 m. Find the change in frequency of a photon of red light
whose original frequency is 7.3 1014 Hz when it falls through 22.5 m.
Solution
From Eq. (2.28) the change in frequency is
1.8 Hz
Pound and Rebka actually used gamma rays of much higher frequency, as described in Exercise 53.
(9.8 m/s^2 )(22.5 m)(7.3 1014 Hz)
(3.0 108 m/s)^2
gH
c^2
gH
c^2
h
c^2
86 Chapter Two
Photon energy after
falling through height H
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