<p><strong>Cosmic expansion wouldn't cool your can</strong>, at least directly. (Of course, the can would eventually equilibrate with the cosmic microwave background and thereby cool due to cosmic expansion indirectly.)</p>
<p>Indeed, in the case of a mirrored box that contains photons, <em>those photons won't cool either.</em></p>
<p>Why not?</p>
<p>The idea that photons &quot;stretch due to the expansion of space&quot; is kind of misleading. The fact that distant objects are receding from us does not change local particle dynamics. &quot;Expanding space&quot; is itself <a href="https://astronomy.stackexchange.com/questions/51774/how-do-we-know-were-not-getting-bigger/51787#51787">just a coordinate choice</a>.</p>
<p>Instead, the cosmological redshift is fundamentally just a Doppler shift due to the change in the reference frames of the comoving observers as a photon travels. Comoving observers represent reference frames that follow the Hubble flow of cosmic expansion with no further motion. Since comoving observers are receding from each other, a photon moving from one comoving observer to another will be redshifted in the latter's frame. See <a href="https://arxiv.org/abs/0808.1081" rel="nofollow noreferrer">Bunn &amp; Hogg (2009)</a> for a much more extended discussion of this point.</p>
<p>Another perspective is that a radiation fluid has a large positive pressure. Imagine containing a bunch of photons inside a mirrored box that is expanding with the Hubble flow. As the box expands, the radiation pressure causes it to do work on the walls, thus reducing its energy. This behavior is exactly an <a href="https://en.wikipedia.org/wiki/Adiabatic_process#Adiabatic_heating_and_cooling" rel="nofollow noreferrer">adiabatic cooling</a> process, which conserves <span class="math-container">$TV^{\gamma-1}$</span>, where <span class="math-container">$T$</span> is the temperature of the fluid, <span class="math-container">$V$</span> is the volume, and <span class="math-container">$\gamma=4/3$</span> is the <a href="https://physics.stackexchange.com/questions/405124/why-is-the-relativistic-adiabatic-index-4-3">adiabatic index of a relativistic gas</a>. Thus, <span class="math-container">$T\propto V^{1-\gamma}= V^{-1/3}$</span>, which is the same <span class="math-container">$T\propto a^{-1}$</span> associated with cosmic expansion, where <span class="math-container">$a$</span> is the <a href="https://en.wikipedia.org/wiki/Scale_factor_(cosmology)" rel="nofollow noreferrer">cosmic expansion factor</a>.</p>
<p>Your box full of photons isn't expanding, though. Its volume <span class="math-container">$V$</span> is fixed, and hence the temperature <span class="math-container">$T$</span> of the radiation is also fixed. (More generally, you could make an expanding box, but there's no reason to make it expand at the same rate as the universe. If it expands at a different rate, then the photons will cool at a different rate.)</p>
<p>Beer is more complicated, since it's a liquid. Still, it should be clear that if the can isn't expanding, there's no reason to expect the contents to cool.</p>

CC BY-SA 4.0
2023-03-07T17:53:05.957Z
2023-03-07T18:47:20.590Z
2023-03-07T18:47:20.590Z
Sten
<p>So the expansion of the universe stretches the light traveling through the void, as demonstrated by the cosmic microwave background radiation.  These photons are lower energy than when they are emitted.</p>
<p>Suppose I have a (very) well insulated container with a can of beer inside at slightly above room temperature (300 K) that is floating in the void, would the expansion of the universe cause the beer to cool off like it does cosmic microwave background photons?</p>
<p>If the beer starts to freeze at 270 K (10% cooler), how long would it take to start to freeze if it started floating in the void today?  10 billion years ago?  In 100 billion years?</p>
<p>Would water act differently?  (ie, does the heat capacity of the substance matter)  How about a box with perfect mirrors and photons in it?</p>
<p>And if it does start to freeze (despite being well insulated; an isolated system), what happens with conservation of energy?  It appears to be a closed system.</p>
<p>And if it doesn't start to freeze, what makes my free-floating beer different than free-floating photons?</p>

CC BY-SA 4.0
2023-03-07T16:33:32.937Z
Before expansion had any affect, the can would radiate energy in the infrared to lose heat.  After that, it would cool with the rest of the universe as it expands.  The difference from a photon is that the photons are the energy leaving the can, the expanding universe just lets them spread out more.
Greg Miller
<expansion><redshift><dark-energy>
Can the expansion of the universe cool my beer?
Yakk - Adam Nevraumont
# Can the expansion of the universe cool my beer?

(Asked by: username_0 on 2023-03-07)

So the expansion of the universe stretches the light traveling through the void, as demonstrated by the cosmic microwave background radiation. These photons are lower energy than when they are emitted.

Suppose I have a (very) well insulated container with a can of beer inside at slightly above room temperature (300 K) that is floating in the void, would the expansion of the universe cause the beer to cool off like it does cosmic microwave background photons?

If the beer starts to freeze at 270 K (10% cooler), how long would it take to start to freeze if it started floating in the void today? 10 billion years ago? In 100 billion years?

Would water act differently? (ie, does the heat capacity of the substance matter) How about a box with perfect mirrors and photons in it?

And if it does start to freeze (despite being well insulated; an isolated system), what happens with conservation of energy? It appears to be a closed system.

And if it doesn't start to freeze, what makes my free-floating beer different than free-floating photons?

**Tags:** expansion, redshift, dark-energy

> **Comments:**
> **username_1**: Before expansion had any affect, the can would radiate energy in the infrared to lose heat.  After that, it would cool with the rest of the universe as it expands.  The difference from a photon is that the photons are the energy leaving the can, the expanding universe just lets them spread out more.  

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## 1 answers

### Answer by username_2 on 2023-03-07. Score: 4

**Cosmic expansion wouldn't cool your can**, at least directly. (Of course, the can would eventually equilibrate with the cosmic microwave background and thereby cool due to cosmic expansion indirectly.)

Indeed, in the case of a mirrored box that contains photons, *those photons won't cool either.*

Why not?

The idea that photons "stretch due to the expansion of space" is kind of misleading. The fact that distant objects are receding from us does not change local particle dynamics. "Expanding space" is itself [just a coordinate choice](https://astronomy.stackexchange.com/questions/51774/how-do-we-know-were-not-getting-bigger/51787#51787).

Instead, the cosmological redshift is fundamentally just a Doppler shift due to the change in the reference frames of the comoving observers as a photon travels. Comoving observers represent reference frames that follow the Hubble flow of cosmic expansion with no further motion. Since comoving observers are receding from each other, a photon moving from one comoving observer to another will be redshifted in the latter's frame. See [Bunn & Hogg (2009)](https://arxiv.org/abs/0808.1081) for a much more extended discussion of this point.

Another perspective is that a radiation fluid has a large positive pressure. Imagine containing a bunch of photons inside a mirrored box that is expanding with the Hubble flow. As the box expands, the radiation pressure causes it to do work on the walls, thus reducing its energy. This behavior is exactly an [adiabatic cooling](https://en.wikipedia.org/wiki/Adiabatic_process#Adiabatic_heating_and_cooling) process, which conserves $TV^{\gamma-1}$, where $T$ is the temperature of the fluid, $V$ is the volume, and $\gamma=4/3$ is the [adiabatic index of a relativistic gas](https://physics.stackexchange.com/questions/405124/why-is-the-relativistic-adiabatic-index-4-3). Thus, $T\propto V^{1-\gamma}= V^{-1/3}$, which is the same $T\propto a^{-1}$ associated with cosmic expansion, where $a$ is the [cosmic expansion factor](https://en.wikipedia.org/wiki/Scale_factor_(cosmology)).

Your box full of photons isn't expanding, though. Its volume $V$ is fixed, and hence the temperature $T$ of the radiation is also fixed. (More generally, you could make an expanding box, but there's no reason to make it expand at the same rate as the universe. If it expands at a different rate, then the photons will cool at a different rate.)

Beer is more complicated, since it's a liquid. Still, it should be clear that if the can isn't expanding, there's no reason to expect the contents to cool.


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