Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 310 / 3 / c

Table of contents
- Solution c

Solution

At

T_{ν} / a ≪ m_{ν}

, the Taylor expansion of the relativistic energy is

p^{2} + m_{ν}^{2} = m_{ν} + p^{2} / (2 m_{ν}) + O (p^{4} / m_{ν}^{3})

. The number density is

n_{ν} = \frac{1}{π ^{2}} (\frac{T _{ν}}{a})^{3} \int_{0}^{\infty} \frac{x ^{2}}{e ^{x} + 1} d x = \frac{3 ζ ( 3 )}{2 π ^{2}} (\frac{T _{ν}}{a})^{3} .

The ratio of the next momentum moment to this one is

\frac{\int _{0}^{\infty} x ^{4} / ( e ^{x} + 1 ) d x}{\int _{0}^{\infty} x ^{2} / ( e ^{x} + 1 ) d x} = \frac{15 ζ ( 5 )}{ζ ( 3 )} .

Therefore

ρ_{ν} ≃ n_{ν} m_{ν} [1 + \frac{15 ζ ( 5 )}{2 ζ ( 3 )} (\frac{T _{ν}}{a m _{ν}})^{2}],

so

F = 15 ζ (5) / (2 ζ (3)) ≃ 6.47

.

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