Question

The results of the previous problem can be used to explain why the current temperature of the cosmic neutrino background (Problem 7.48) is 1.95 K rather than 2.73 K. Originally the temperatures of the photons and the neutrinos would have been equal, but as the universe expanded and cooled, the interactions of neutrinos with other particles soon became negligibly weak. Shortly thereafter, the temperature dropped to the point where kT/c² was no longer much greater than the electron mass. As the electrons and positrons disappeared during the next few minutes, they "heated" the photon radiation but not the neutrino radiation.

(a) Imagine that the universe has some finite total volume V, but that V is increasing with time. Write down a formula for the total entropy of the electrons, positrons, and photons as a function of V and T, using the auxiliary functions u(T) and f(T) introduced in the previous problem. Argue that this total entropy would have ben conserved in the early universe, assuming that no other species of particles interacted with these.

(b) The entropy of the neutrino radiation would have been separately conserved during this time period, because the neutrinos were unable to interact with anything. Use this fact to show that the neutrino temperature Tv and the photon temperature T are related by

${\left(\frac{T}{T_{\nu }}\right)}^3\left[\frac{2{\pi }^4}{45}+u\left(T\right)+f\left(T\right)\right]=\text{constant}$

as the universe expands and cools. Evaluate the constant by assuming that T=T_v when the temperatures are very high.

(c) Calculate the ratio T/T_v, in the limit of low temperature, to confirm that the present neutrino temperature should be 1.95 K.

(d) Use a computer to plot the ratio T/T_v, as a function of T, for kT/mc²ranging from 0 to 3.*

Short answer

Therefore,

The formula for total entropy of electrons is $S_{\text{tot.}}=\frac{16\pi V(kT{)}^3}{(hc{)}^3}\left(u\left(T\right)+f\left(T\right)+\frac{2{\pi }^4}{45}\right)k$

The neutrino temperature and photon temperature is relates by:

${\left(\frac{T}{T_v}\right)}^3\left(u\left(T\right)+f\left(T\right)+\frac{2{\pi }^4}{45}\right)=\frac{11{\pi }^4}{90}$

And$T_v=1.948\mathrm{K}$

Step-by-step solution

Given information

The results of the previous problem can be used to explain why the current temperature of the cosmic neutrino background (Problem 7.48) is 1.95 K rather than 2.73 K. Originally the temperatures of the photons and the neutrinos would have been equal, but as the universe expanded and cooled, the interactions of neutrinos with other particles soon became negligibly weak. Shortly thereafter, the temperature dropped to the point where kT/c² was no longer much greater than the electron mass. As the electrons and positrons disappeared during the next few minutes, they "heated" the photon radiation but not the neutrino radiation.

Explanation

(a) From problem 749, the entropy of the positron-electron pairs is:

$S_e=\frac{16\pi V(kT{)}^3}{(hc{)}^3}\left(u\right(T)+f(T\left)\right)k$

and the entropy of the photons is:

$S_{\gamma }=\frac{32{\pi }^{5}V(kT{)}^{3}k}{45(hc{)}^3}$

the total entropy of the photons, electrons, positrons is therefore:

$$\begin{gathered} S_{tot.}=S_e+S_{\gamma } \\ {S}_{\text{tot.}}=\frac{16\pi V(kT{)}^3}{(hc{)}^3}\left(u\right(T)+f(T\left)\right)k+\frac{32{\pi }^{5}V(kT{)}^{3}k}{45(hc{)}^3} \\ {S}_{\text{tot.}}=\frac{16\pi V(kT{)}^3}{(hc{)}^3}\left(u\left(T\right)+f\left(T\right)+\frac{2{\pi }^4}{45}\right)k \end{gathered}$$

Because the radiation is in internal equilibrium during the expansion, it happens adiabatically, which means the entropy is conserved, therefore we may write:

$V{T}^3\left(u\left(T\right)+f\left(T\right)+\frac{2{\pi }^4}{45}\right)k=cons\tan t\hspace{0.17em}\hspace{0.17em}\left(1\right)$

(b) From problem 7.48, the total energy of the neutrino-antineutrino background radiation is given by:

$U_v\propto V{T}_v^4$

but the entropy is the partial derivative of the total energy at constant volume, so:

$S_v\propto V{T}_v^3$

and since the universe expands adiabatically, then this entropy is constant:

$V{T}_v^3=cons\tan t\left(2\right)$

Divide the equation (2) by (1):

${\left(\frac{T}{T_v}\right)}^3\left(u\left(T\right)+f\left(T\right)+\frac{2{\pi }^4}{45}\right)=\text{constant}$

At high temperature limit T=T_v, and from the previouS problem at high temperature limit we have:

$u\left(T\right)=\frac{7{\pi }^4}{120}f\left(T\right)=\frac{7{\pi }^4}{360}\approx 1.894$

Then,

$$\begin{gathered} (1{)}^3\left(\frac{7{\pi }^4}{120}+\frac{7{\pi }^4}{360}+\frac{2{\pi }^4}{45}\right)=\text{constant} \\ \text{constant}=\frac{11{\pi }^4}{90} \end{gathered}$$

Explanation

(c) At low temperature limit $T\to 0\text{both}f\left(T\right)\text{and}u\left(T\right)$go to zero, so equation (3) will become:

$$\begin{gathered} {\left(\frac{T}{T_v}\right)}^3\left(0+0+\frac{2{\pi }^4}{45}\right)=\frac{11{\pi }^4}{90} \\ {\left(\frac{T}{T_v}\right)}^3=\frac{11}{4} \\ \frac{T}{T_v}={\left(\frac{11}{4}\right)}^{1/3} \end{gathered}$$

given that the present temperature of the photos T'= 2.73 K, then the temperature of the neutrinos is:

$$\begin{gathered} T_v={\left(\frac{4}{11}\right)}^{1/3} \\ {T}_v=1.948\mathrm{K} \end{gathered}$$

(d)From the previous problem f(T) and u(T), are given by:

$$\begin{gathered} u\left(T\right)={\int }_0^{\infty }\frac{x^2\sqrt{x^2+(1/t{)}^2}}{e^{\sqrt{x^2+(1/t{)}^2}}+1}dx \\ f\left(T\right)={\int }_0^{\infty }{x}^2\ln \left(1+e^{-\sqrt{x^2+(1/t{)}^2}}\right)dx \end{gathered}$$

And from part (b):

$$\begin{gathered} {\left(\frac{T}{T_v}\right)}^3\left(u\left(T\right)+f\left(T\right)+\frac{2{\pi }^4}{45}\right)=\frac{11{\pi }^4}{90} \\ \frac{T}{T_v}={\left(\frac{11{\pi }^4}{90}\right)}^{1/3}{\left(u\left(T\right)+f\left(T\right)+\frac{2{\pi }^4}{45}\right)}^{-1/3} \end{gathered}$$

To plot the function, integrate it with respect to x and matlab was used to solve it, the code is:

The graph is: