General Relativity and Cosmology
415
allow for a unified approach. In a sense, the distinctions between the different
domains are becoming blurred and ultimately there may be only the core of
basic laws of physics in terms of which all observations can be explained.
11.7 EXAMPLES
In this section, some examples related to the ideas discussed earlier are given.
Example 1
Olbers argued that in an infinite, static universe, every part of the sky must have
a brightness comparable to that of the sun.
In an infinite, static universe, every part of the sky will be covered by a star.
Let the visible part of one such star subtend a solid angle ∆Ω at the earth. Now,
the intensity of radiation received at the earth, due to this star at distance d, is
proportional to the exposed are d
2
∆ω of the star, and decreases as 1/d
2
. Therefore
it is proportional to d
2
∆Ω (1/d
2
) ~ ∆Ω which is independent of the distance of
the star. Therefore, an equivalent area (d
2
sun
∆Ω) of the star might as well be at
a distance equal to that of the sun. If we make the reasonable assumption that
the stars in general have approximately the same inherent brightness as that of
the sun, we would then expect every part of the sky to have the brightness of the
sun, day or night. Any attempt at an explanation in terms of absorption does not
succeed since at equilibrium the absorbing material must emit as much energy
as it absorbs.
Example 2
The red shift of radiation in the presence of a gravitational field, to the leading
order in the field, can be shown to follow from energy conservation.
Consider an atom of mass m 2 which goes to a state of lower mass m 1 , with
the emission of a photon of frequency v 0 . Then energy conservation implies
m 2 c
2
= m 1 c
2
+ hv 0
(11.99)
If the atom is placed in a gravitational potential φ, its initial energy is
m 2 a
2
+ m 2 φ while the final energy is m 1 c
2
+ m 1 φ. Then the frequency v of the
photon, which comes out of the potential, is given by
hv = (m 2 c
2
+ m 2 φ) – (m 1 c
2
+ m 1 φ)
= (m 2 c
2
– m 1 c
2
) (1 + φ/c
2
)
(11.100)
Using Eq. (11.99)
v = v 0 (1 + φc
2
)
(11.101)
which agrees with the general expression in Eq. (11.39) to the leading order.
For emission from the sun, φ/c
2
≈ – 2 × 10
–6
.
415
allow for a unified approach. In a sense, the distinctions between the different
domains are becoming blurred and ultimately there may be only the core of
basic laws of physics in terms of which all observations can be explained.
11.7 EXAMPLES
In this section, some examples related to the ideas discussed earlier are given.
Example 1
Olbers argued that in an infinite, static universe, every part of the sky must have
a brightness comparable to that of the sun.
In an infinite, static universe, every part of the sky will be covered by a star.
Let the visible part of one such star subtend a solid angle ∆Ω at the earth. Now,
the intensity of radiation received at the earth, due to this star at distance d, is
proportional to the exposed are d
2
∆ω of the star, and decreases as 1/d
2
. Therefore
it is proportional to d
2
∆Ω (1/d
2
) ~ ∆Ω which is independent of the distance of
the star. Therefore, an equivalent area (d
2
sun
∆Ω) of the star might as well be at
a distance equal to that of the sun. If we make the reasonable assumption that
the stars in general have approximately the same inherent brightness as that of
the sun, we would then expect every part of the sky to have the brightness of the
sun, day or night. Any attempt at an explanation in terms of absorption does not
succeed since at equilibrium the absorbing material must emit as much energy
as it absorbs.
Example 2
The red shift of radiation in the presence of a gravitational field, to the leading
order in the field, can be shown to follow from energy conservation.
Consider an atom of mass m 2 which goes to a state of lower mass m 1 , with
the emission of a photon of frequency v 0 . Then energy conservation implies
m 2 c
2
= m 1 c
2
+ hv 0
(11.99)
If the atom is placed in a gravitational potential φ, its initial energy is
m 2 a
2
+ m 2 φ while the final energy is m 1 c
2
+ m 1 φ. Then the frequency v of the
photon, which comes out of the potential, is given by
hv = (m 2 c
2
+ m 2 φ) – (m 1 c
2
+ m 1 φ)
= (m 2 c
2
– m 1 c
2
) (1 + φ/c
2
)
(11.100)
Using Eq. (11.99)
v = v 0 (1 + φc
2
)
(11.101)
which agrees with the general expression in Eq. (11.39) to the leading order.
For emission from the sun, φ/c
2
≈ – 2 × 10
–6
.