bei48482_FM

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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