(Revised 5 Aug 2021. See my previous post for notes and comments on Callen's Section 3.6.)
3.6-1 The universe is considered by cosmologists to be an
expanding electromagnetic cavity containing radiation that now is at a
temperature of 2.7 K. What will be the temperature of the radiation when the
volume of the universe is twice its current value? Assume the expansion
to be isentropic (this being a nonobvious prediction of cosmological model
calculations).
So the entropy will be the same when the volume V1 becomes V2 and when T1 becomes T2 :
S(U, V) = constant = (4/3) (bV2)1/4 (U2 )3/4 = (4/3) (bV1 )1/4 (U1 )3/4 ,
where U2 =
bV2 (T2 ) 4 and U1 = b V1 (T1 ) 4 .
Canceling the 4/3 and b factors on each side of the above entropy equation and
putting in temperature T in place of the energy U gives
(exponent on V's becomes 1/4 + 3/4 = 1, and exponent
of T's becomes 4 x 3/4 = 3)
V2 (T2 )3 = V1 (T1 )3
(T2 /T1 ) 3 = V1/V2
=
(T2 /T1 ) 3
= 1/2
T2 =
(1/2)1/3 T1 =
0.79 (2.7 K) = 2.1
K .
Question for further research: in how many years will this doubling in size occur? The accepted answer to this question is not the same now as it was in 1985 when Callen’s book was published, due to the discovery in 1998 of the acceleration of the expansion (dark energy).
3.6-2 Assuming
the electromagnetic radiation filling the universe to be in equilibrium
at T = 2.7 K, what is the pressure associated with this
radiation? Express the answer in both pascals and atmospheres.
From Stefan-Boltz law, we have U/V = bT4 , and the pressure equation for isotropic radiation is PR = U/3V. So PR = (b/3)T4
= (7.56x10-16 Joule · meter-3 · Kelvin-4))(2.7 K)4/3
= 134 x 10-16 Newton/meter2
PR = 1.34 x 10-14 Pa
PR = 1.32 x 10-19 atm.
3.6-3 The
density of matter (primarily hydrogen atoms) in intergalactic space is such
that its contribution to the pressure is of the order of 10-23 Pa.
(a) What is the
approximate density of matter (in atoms/meter3) in intergalactic
space?
(b) What is the ratio of
the kinetic energy of matter to the energy of radiation in intergalactic space?
(Recall Problems 3.6-1 and 3.6-2.)
(c) What is the
ratio of the total matter energy (i.e., the sum of the kinetic energy and the
relativistic energy mc2) to the energy of radiation in
intergalactic space?
(a) We want to calculate N/V from PM, and, assuming (maybe incorrectly) the hydrogen gas in intergalactic space is in equilibrium with the CMB radiation, we use the ideal gas law, PMV = NkT, where T = 2.7 K. Rearranging gives
N/V = PM/kT = 10-23 Pa/[(1.38 x 10-23 J/K) · 2.7K]
=
(1/3.7) atoms/meter3
= 0.27 atoms/meter3
The answer is actually about one per cubic centimeter, or one million per cubic meter, alas. That means the temp of the very diffuse hydrogen gas is about 4 million degrees Kelvin. Here's the first paragraph from a good discussion of the subject:
There are some missing details that are often skipped over in popular science articles/documentaries, which can make this confusing. First: the temperature of space, 2.7 K or -270 C, refers to the remaining radiation from the Big Bang, not to the temperature of any matter. If you were in space and somehow prevented any matter from touching you, you would still absorb microwave radiation equivalent to being surrounded by matter with a temperature of 2.7 K (since all matter with temperatures above 0K emits radiation). Of course, with all the stars and such, you also receive a lot of other radiation, so you wouldn't necessarily cool down to 2.7 K. That temperature refers specifically, and only, to the background microwave radiation left over from the early history of the universe.
--------
(b) For our intergalactic atomic hydrogen gas, the only energy is kinetic energy (well, ignoring the spin flipping the electron can undergo relative to the spin of the proton, resulting in the famous 21 cm spectral line). (Also ignoring, until part (c), the relativistic mass-energy.) So we have (KE)H = (3/2) NkT = (3/2)PMV for the kinetic energy of matter, which we want to divide by the Urad = bVT4 = 3PRV of radiation:
(KE)H
/ Urad = (3/2) PMV
/ 3PRV
= PM / 2PR
= 10-23 Pa
/ 2(1.34 x 10-14 Pa)
= 3.7 x 10-10
(c) The relativistic energy is Urel
= Mc2, where M is the mass of all the hydrogen atoms in
intergalactic space. We could do this calculation by estimating the size of the
universe which we’d multiply by the density of hydrogen atoms. But volume
cancels out in a ratio of energy density calculations, and energy density is the determining factor for the total energy of radiation, of
hydrogen atoms’ KE, and of relativistic energy. And we've calculated number density of H atoms, so we should use it. We divide by total volume, V, of
the universe to get density:
Urel/V = (Mc2 /V) = m(N/V)c2
= (1.67 x 10-27 kg for one H atom) (0.27 H atoms per meter3)(3 x 108
m/s)2
= 9 x 1016 · 0.45 x 10-27 joule/meter3
= 4 x 10-11 joule/meter3
Now I need to divide this by Urad/V
for radiation, and add the result to the answer to (b). I didn’t do the Urad/V
calculation separately for radiation energy density in part (b)—but we know
U/3V = P for the radiation pressure, so we take 3P as our radiation energy
density
Urad/V = 3P = 3 x 1.34 x 10-14 Pa = 4 x 10-14 joule/meter3
so
Urel/Urad = 1000.
The KE/Urad ratio of 3.7 x 10-10 is
negligible in comparison with the factor-of-1000 ratio of
relativistic-to-radiation energy. No point in adding them! But we’d expect the relativistic energy to be much greater
than the total KE of atoms in a gas at temperature 2.7 K.
Questions for further research:
At what temperature does the KE of
the H atoms equal their mass-energy?
At what temp does the radiation
energy equal the H atoms mass-energy?
Does the mass-energy of all the
loose H atoms in the universe have any significance beyond the fact that
it has to be included when we account for all the energy released (created) in
the Big Bang?
