Expansion of the universe

Classical Mechanics Level 3

The cosmic microwave background of the universe can treated as photon gas with a current temperature $T = 2.725\,\text{K}$ . Due to the expansion of the universe the background temperature will decrease steadily. What will be the temperature $T$ of the background radiation in 5 billion years in the future? Round the result to three decimal places.

Details and assumptions :

The current age of the universe is $t_0 = 13.8 \cdot 10^9 \,\text{a}$ . Assume, that the length of the universe $l = \alpha t$ is proportional to the time $t$ .
The thermodynamic equations for a photon gas read $\begin{aligned} U &= \beta V T^4 \\ p &= \frac{1}{3} \beta T^4 \end{aligned}$ with the internal energy $U$ , the volume $V$ , the pressure $p$ and a constant $\beta = \frac{\pi^2 k^4}{15 c^3 \hbar^3}$
Assume an adiabatic expansion (constant entropy), so that no heat is exchanged with the environment. (Because there is no enviroment outside our universe, right?)

The answer is 2.000.

1 solution

Markus Michelmann
Oct 22, 2017

The first law of thermodynamics reads $\begin{aligned} & & dU &= \underbrace{T dS}_{=0} - p dV \\ \Rightarrow & & \beta T^4 dV + 4 \beta V T^3 dT &= - \frac{1}{3} \beta T^4 dV \\ \Rightarrow & & \frac{dT}{T} &= - \frac{1}{3} \frac{dV}{V} \\ \Rightarrow & & \int_{T_0}^T \frac{dT}{T} &= - \frac{1}{3} \int_{V_0}^V \frac{dV}{V} \\ \Rightarrow & & \ln \frac{T}{T_0} &= - \frac{1}{3} \ln \frac{V}{V_0} \\ \Rightarrow & & T &= T_0 \left( \frac{V_0}{V} \right)^{1/3} = T_0 \frac{l_0}{l} = T_0 \frac{t_0}{t} \end{aligned}$ Therefore, $T = 2.725 \,\text{K} \frac{13.8 \cdot 10^9 \,\text{a}}{18.8 \cdot 10^9 \,\text{a}} \approx 2\,\text{K}$

Expansion of the universe

The answer is 2.000.

1 solution

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