91影视

In addition to the cosmic background radiation of photons, the universe is thought to be permeated with a background radiation of neutrinos (v) and antineutrinos ($\stackrel{-}{v}$), currently at an effective temperature of 1.95 K. There are three species of neutrinos, each of which has an antiparticle, with only one allowed polarisation state for each particle or antiparticle. For parts (a) through (c) below, assume that all three species are exactly massless.

(a) In two pairs of photons, neutrinos and antineutrinos can be annihilated as follows:

$v+\stackrel{炉}{v}鈫�2纬$

In terms of chemical potentials, the equilibrium condition for this reaction is:

${渭}_v+{渭}_{\stackrel{炉}{v}}=2{渭}_{纬}$

Assume that the number of neutrinos in each species equals the number of antineutrinos, and that their chemical potentials are equal ${渭}_v={渭}_{\stackrel{炉}{v}}$and that the chemical potential of the photon is zero, as explained on page 290.

${渭}_v={渭}_{\stackrel{炉}{v}}=0$

(b)The Fermi Dirac distribution, but with zero chemical potential, gives the likelihood of any single state being occupied by a neutrino:

${\stackrel{炉}{n}}_{\mathrm{FD}}=\frac{1}{e^{系/kT}+1}$

The total energy equals the sum over the energies multiplied by their probabilities, that is:

$U=3路2\underset{n_x}{鈭�}\underset{n_y}{鈭�}\underset{n_z}{鈭�}系{\stackrel{炉}{n}}_{\mathrm{FD}}$

Now, factor 2 comes from the fact that anti matter (neutrinos and antineutrinos) exists, and factor 3 comes from the fact that we have three spices. Consider a cubic box with a volume of V and a side width of L. The permitted energy for neutrinos is:

$系=\frac{hcn}{2L}$

Hence,

$$\begin{gathered} U=6\underset{n_x,n_y,n_z}{鈭�}\frac{系}{e^{系/kT}+1} \\ U=6\underset{n_x,n_y,n_z}{鈭�}\frac{hcn/2L}{e^{hcn/2LkT}+1} \end{gathered}$$

We must now convert the total to an integral in spherical coordinates, which we may do by multiplying it by the spherical integration factor $n^2\sin \left(胃\right)$so:

$U=6{鈭�}_0^{蟺/2}d桅{鈭�}_0^{蟺/2}\sin \left(胃\right)d胃{鈭�}_0^{鈭�}\frac{(hc/2L)n^3}{e^{hcn/2LkT}+1}dn$

The first two integrals are easy to evaluate, and they give a factor of $\mathrm{蟺}/2$, so:

$U=3蟺{鈭�}_0^{鈭�}\frac{(hc/2L)n^3}{e^{hcn/2LkT}+1}dn$

Let,

$x=\frac{hcn}{2LkT}鈫�dx=\frac{hc}{2LkT}dn$

Hence,

$$\begin{gathered} U=3蟺{\left(\frac{2LkT}{hc}\right)}^4\left(\frac{hc}{2L}\right){鈭�}_0^{鈭�}\frac{x^3}{e^x+1}dx \\ U=3蟺{\left(\frac{2L}{hc}\right)}^3(kT{)}^4{鈭�}_0^{鈭�}\frac{x^3}{e^x+1}dx \end{gathered}$$

The integral is given as:

${鈭�}_0^{鈭�}\frac{x^3}{e^x+1}dx=\frac{7{蟺}^4}{120}$

Thus,

$U=3蟺{\left(\frac{2L}{hc}\right)}^3(kT{)}^4\frac{7{蟺}^4}{120}$

Volume of box is V=L³

$$\begin{gathered} U=24蟺V\frac{(kT{)}^4}{(hc{)}^3}\frac{7{蟺}^4}{120} \\ \frac{U}{V}=\frac{7{蟺}^5(kT{)}^4}{5(hc{)}^3} \end{gathered}$$

(c)The number of neutrinos can be computed in the same way as the number of protons, but without the factor $蔚$, yielding:

$N=6{鈭�}_0^{蟺/2}d桅{鈭�}_0^{蟺/2}\sin \left(胃\right)d胃{鈭�}_0^{鈭�}\frac{(hc/2L)n^3}{e^{hcn/2LkT}+1}dn$

The first two integrals are easy to evaluate, and they give a factor of $\mathrm{蟺}/2$, so:

$N=3蟺{鈭�}_0^{鈭�}\frac{n^2}{e^{hcn/2LkT}+1}dn$

Let,

$x=\frac{hcn}{2LkT}鈫�dx=\frac{hc}{2LkT}dn$

Thus,

$N=3蟺{\left(\frac{2LkT}{hc}\right)}^3{鈭�}_0^{鈭�}\frac{x^2}{e^x+1}dx$

The value of the integral is 1.803, hence

$N=3(1.803)蟺{\left(\frac{2LkT}{hc}\right)}^3$

Volume of the box is V=L³

$$\begin{gathered} \frac{N}{V}=24(1.803)蟺{\left(\frac{kT}{hc}\right)}^3 \\ \frac{N}{V}=(135.94){\left(\frac{kT}{hc}\right)}^3 \end{gathered}$$

At temperature of 1 = 1.94 K, the number of neutrinos per unit volume is:

$$\begin{gathered} \frac{N}{V}=(135.94){\left(\frac{\left(1.38脳{10}^{-23}\mathrm{J}/\mathrm{K}\right)(1.94\mathrm{K})}{\left(6.626脳{10}^{-34}\mathrm{J}路\mathrm{s}\right)\left(3.0脳{10}^8\mathrm{m}/\mathrm{s}\right)}\right)}^3 \\ \frac{N}{V}=3.321脳{10}^8{\mathrm{m}}^{-3} \end{gathered}$$

(d)Assuming that there is only one type of neutrino, the number of neutrinos per unit volume is one-third of what we computed in the preceding section, i.e.

$\frac{N}{V}=1.1脳{10}^8{\mathrm{m}}^{-3}$

The average density of the universe is one proton per cubic metre, and the energy of this proton in eV is $m_p{c}^2=1.0脳{10}^9$eV per cubic metre, hence the neutrino's energy should be:

$$\begin{gathered} m_n{c}^2=\frac{1.0脳{10}^9{\mathrm{m}}^{-3}}{1.1脳{10}^8{\mathrm{m}}^{-3}} \\ {m}_n{c}^2=11\mathrm{eV} \end{gathered}$$