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Is it possible to define some average temperature of the universe? If yes, what fixes this temperature and how t estimate today's temperature? Is it different from the temperature of the black-body spectrum of the CMB today i.e $2.7$ K?

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  • $\begingroup$ Averaging temperature is free of sense, if the bodys "averaged" do not have the same specific heat. $\endgroup$
    – Georg
    Jan 23, 2015 at 15:57

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Temperature is a property of a system in equilibrium, when there is no net energy flow between the various degrees of freedom in the system.

The universe has a number of systems hat are roughly in equilibrium individually (and so have well-defined temperatures) but are not in equilibrium with one another. The CMB is one such system,1 and indeed its temperature is $2.7\ \mathrm{K}$.

On the other hand, most of the (baryonic, i.e. normal) matter in the universe is in the form of ionized atoms outside galaxies, concentrated wherever galaxies are clustered. This intracluster medium has temperatures in the $10^7\mathrm{-}10^8\ \mathrm{K}$ range.

If you wanted to average over the universe, you could try assigning weights to the different components, but it's not entirely clear what those should be. The fact is the universe is not in perfect equilibrium, and the only way to define a temperature is to place a thermometer (either real or hypothetical) in equilibrium with the thing you want a temperature for. You cannot place your thermometer in equilibrium with all parts of the universe without simultaneously equilibrating those parts (the "zeroth law of thermodynamics").

The best you can do is define a thermal coupling to all the different components, and ask what temperature your thermometer reaches in steady state. For example, a typical macroscopic, mercury-filled tube placed in intergalactic space would couple relatively very well to radiation (it would absorb CMB photons, and it would emit blackbody radiation according to the Stefan-Boltzmann law) and not so well to the translational kinetic energy of intergalactic hydrogen (because there would be so few collisions per unit time). As a result, this particular thermometer would reach steady state very close to $2.7\ \mathrm{K}$, during which it would be receiving a small trickle of heat from chance collisions with very fast hydrogen atoms, releasing this heat as radiation.


1 Interestingly the CMB doesn't have any good means of equilibrating, as CMB photons are no longer being scattered or absorbed and re-emitted. But the distribution of energies is nonetheless frozen in as that of blackbody radiation, so the temperature is well defined.

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  • $\begingroup$ I would probably measure by the dominating contribution to the mass-energy density? Meaning that we would need to know the properties of Dark Matter really well (not counting DE here, it is just too weird). $\endgroup$
    – Thriveth
    Jan 24, 2015 at 12:34
  • $\begingroup$ In fact, I guess the temperature of DM could be constrained in terms of the virial theorem. We know the Large Scale Structure reasonably well, I suppose an average energy could be found and from there some reasonable definition of temperature could be applied...? $\endgroup$
    – Thriveth
    Jan 25, 2015 at 19:30
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Using the word "Average" points to an exact number which means that you will have to use exact values, so ...

AverageTemp = Sum(TempOfAllPointsInUniverse) / VolumeOfUniverse

  • If the universe is expanding and the amount of heat is constant then there's no 'average' since the universe is constantly getting 'colder'.
  • If the universe is expanding and there is a constant average value then that means that the total heat is constantly increasing, which (to my knowledge) science thinks is absurd.
  • If there was a constant 'average' and the total heat is also constant then the universe is not expanding and ... there goes the big bang!
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    $\begingroup$ That is certainly not how an average is defined. At the very least, you would want to average over all points in space, but as pointed out by Floris, this requires some notion of equilibrium. $\endgroup$ Jan 23, 2015 at 21:14
  • $\begingroup$ Sorry about that. I guess it was the "programmer" side of my brain that was doing the thinking. Did fix it though. :-) $\endgroup$
    – micmanos
    Jan 23, 2015 at 21:27
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The temperature of a body is surprisingly tricky to define. Usually we say "if A is in thermal equilibrium with B, they have the same temperature". But if you are measuring the entire universe that doesn't work.

The more immediate problem is that the universe is not even in equilibrium with itself - far from it. As such there is no meaning to "average temperature" today. As the entropy of the universe increases, so will the degree of equilibrium. Right now, there is no sensible answer to your question. And by the time the question can be answered, worrying about the temperature of the universe will no longer be on our list.

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  • $\begingroup$ I think this answer is partly misleading. The second half is fine - there is no well defined temperature in a system which is not in equilibrium between energy states, as is better explained in @Chris Whites answer. But the idea that we would need a second Universe to define the temperature of ours is just plain wrong. $\endgroup$
    – Thriveth
    Jan 24, 2015 at 12:32
  • $\begingroup$ @Thriveth I have edited the answer - is it better like this? $\endgroup$
    – Floris
    Jan 25, 2015 at 12:41

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