91影视

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

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

and the entropy of the photons is:

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

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

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

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

$V{T}^3\left(u\left(T\right)+f\left(T\right)+\frac{2{蟺}^4}{45}\right)k=cons\tan t鈥�鈥�\left(1\right)$

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

$U_v鈭�V{T}_v^4$

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

$S_v鈭�V{T}_v^3$

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):

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

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

$u\left(T\right)=\frac{7{蟺}^4}{120}f\left(T\right)=\frac{7{蟺}^4}{360}鈮�1.894$

Then,

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

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

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

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:

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

And from part (b):

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

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

The graph is: