How do you calculate the cool-down rate of the Universe?

[bbbl67]

TL;DR Summary

Trying to figure out, on my own, how to calculate the universe's rate of cooling. Used the Ideal Gas Law, but I'm not sure if that's the right way.

Okay, just for kicks, I was wondering how do you calculate the the rate of cooling of the universe? All I could think of is using the Ideal Gas Law from Thermodynamics:

Ideal Gas Law:

P V = n R T |
V | volume
P | pressure
n | amount
T | temperature
R | molar gas constant (≈ 8.314 J/(mol K))

I assumed P & n stay pretty constant, then temperature (T) varies directly with volume (V). So the change in volume is equal to the change in temperature.

ΔV = R ΔT

So assuming that the Hubble Parameter (I know this number is controversial, but I just picked one of them) is:

H_0 ≈ 68 km/s/Mpc (kilometers per second per megaparsec)
≈ 2.2×10^-18 per second

So the volumetric version of the Hubble Parameter would just be the cube of H_0:

H_0^3 ≈ (2.2×10^-18)^3 per second
≈ 1.065×10^-53 per second

So the universe's volume would grow by a factor of 1.065×10^-53 per second.

ΔT = ΔV/R
= 1.065×10^-53 / 8.314
= 1.281x10^-54 K per sec

So is my reasoning okay here?

 

[PeterDonis]

I was wondering how do you calculate the the rate of cooling of the universe?

There isn't just one rate. There are (at least) three, one for each of the three distinct kinds of "stuff" that the universe is composed of: radiation, matter, and dark energy. Each of these three kinds of stuff has a different density, temperature, and pressure, and is affected differently by expansion.

All I could think of is using the Ideal Gas Law from Thermodynamics

Unfortunately, this won't work, because none of the three different kinds of stuff behaves like an ideal gas:

Radiation is inherently relativistic and has an equation of state that does not involve temperature at all (and therefore can't be an ideal gas because the ideal gas equation of state does involve temperature).

Matter has zero pressure so the ideal gas equation is useless for describing it.

Dark energy has a constant density and pressure, and to the extent "temperature" even makes sense as a concept for it, its temperature is therefore constant as well.

 

[bbbl67]

So how do you calculate any of this?

 

[Orodruin]

So how do you calculate any of this?

You have to compute the new distribution function. You can do this for a non-interacting gas by computing the redshift of the momentum. Only for the relativistic gas is the new distribution going to be a Maxwell Boltzmann distribution if you start from one.

 

[kimbyd]

Summary:: Trying to figure out, on my own, how to calculate the universe's rate of cooling. Used the Ideal Gas Law, but I'm not sure if that's the right way.

Okay, just for kicks, I was wondering how do you calculate the the rate of cooling of the universe? All I could think of is using the Ideal Gas Law from Thermodynamics:

Ideal Gas Law:

P V = n R T |
V | volume
P | pressure
n | amount
T | temperature
R | molar gas constant (≈ 8.314 J/(mol K))

I assumed P & n stay pretty constant, then temperature (T) varies directly with volume (V). So the change in volume is equal to the change in temperature.

ΔV = R ΔT

So assuming that the Hubble Parameter (I know this number is controversial, but I just picked one of them) is:

H_0 ≈ 68 km/s/Mpc (kilometers per second per megaparsec)
≈ 2.2×10^-18 per second

So the volumetric version of the Hubble Parameter would just be the cube of H_0:

H_0^3 ≈ (2.2×10^-18)^3 per second
≈ 1.065×10^-53 per second

So the universe's volume would grow by a factor of 1.065×10^-53 per second.

ΔT = ΔV/R
= 1.065×10^-53 / 8.314
= 1.281x10^-54 K per sec

So is my reasoning okay here?

This doesn't work because the overall temperature of the universe is a gas of photons, not an ideal gas.

But the good news is it's much simpler. The temperature of the universe, currently 2.725K, is proportional to one over the scale factor. That is:

$T = {T_0 \over a}$

Here $T_0$ is the temperature when $a=1$. If we define $a=1$ to be now, then $T_0 = 2.725$K. The rate of change in the temperature can then be determined using a derivative:

$${dT \over dt} = T_0 {d \over dt} {1 \over a} = -{T_0 \over a^2} {da \over dt}$$

Now we can use the definition of the Hubble parameter to put this into terms that relate to observations:

$$H(t) = {1 \over a}{da \over dt}$$

So:

$${dT \over dt} = -{T_0 \over a} H(t)$$

At the current time, $H(t) = H_0$, and $a = 1$, so the present rate of change in temperature is $-T_0 H_0$.

In units of $1/s$, $H_0$ is approximately equal to $2 \times 10^{-18}$, so the rate of change in temperature is roughly $-6 \times 10^{-18}$K/s.

 

[PeterDonis]

the overall temperature of the universe is a gas of photons

The temperature of the CMB is that of a gas of photons. But it's not necessarily the case that this is "the overall temperature of the universe". It depends on how you define the latter term.

 

[kimbyd]

The temperature of the CMB is that of a gas of photons. But it's not necessarily the case that this is "the overall temperature of the universe". It depends on how you define the latter term.

If you're far away from any local structures, it's the temperature you'd equilibrate to.

I don't think there's another definition that would be reasonably useful.

 

[vanhees71]

The Friedmann equations from a matter-point of view are just ideal-hydro equations, which follows already from the symmetry of the FLRW solution alone. From them you can derive explicitly the equation
$$\mathrm{d}_t (\epsilon a^3) + P \mathrm{d}_t a^3,$$
where $\epsilon$ is the internal energy density and $P$ the pressure of the "substrate". To close the equation you need an equation of state. Usually one considers "dust", i.e., $P=0$ or ultrarelativistic matter (and of course electromagnetic radiation) with $\epsilon=3P$. Leading to $\epsilon a^3=\text{const}$ for dust and $\epsilon a^4=\text{const}$ for ultrarelativistic matter.