10
$\begingroup$

I am little confused, so please correct me if I am wrong.

Temperature of Universe is calculated using cosmic microwave background and it is around 2.7 K.

The cosmic microwave background is the relic radiation of recombination epoch of the Universe around 380,000 years after Big Bang. The temperature was then around 3000K.

But after that the space itself stretched and this caused the highly energetic photons to have their wavelengths stretched. Therefore we see the cosmic microwave background radiation and hence the temperature 2.7 K.

But I have the following question based on the above:

Why do we say that the temperature of the Universe is 2.7 K when actually it was 3000 K?

I would like to illustrate my problem using Sun in a thought experiment. Suppose there is an Alien civilisation living 13 billion light years away and they are measuring radiation coming from the Sun. The radiation leaving Sun now will reach the alien civilisation after 13 billion years. But the radiation wavelength will get stretched as the space itself is expanding and Alien civilisation will measure the wavelength far more larger than what was transmitted by the Sun now. They will deduce that the Sun has a temperature of say 0.4-0.5 K. But that is not the case. Sun’s temperature is 5772 K. Same applies to the Universe during the recombination epoch. The radiation which was left in Universe had the temperature of 3000 K and it is just that the emitted photon’s wavelength got stretched due to expansion of space and not due to the decrease in the temperature of the source of the radiation.

$\endgroup$
7
  • $\begingroup$ Careful writers also state the redshift, z, which is ~1090 for the CMBR. physics.stackexchange.com/q/548087/123208 $\endgroup$
    – PM 2Ring
    Dec 14, 2023 at 10:24
  • 12
    $\begingroup$ I think it is the difference between "is" and "was" in your statement: "is 2.7 K when actually it was 3000 K?" we live in the "is", and assume that the small portion of the universe we measure is indicative of "is". The microwave radiation is not coming from a star $\endgroup$
    – anna v
    Dec 14, 2023 at 11:58
  • 1
    $\begingroup$ @annav I know that the radiation is not coming from a star. I took an example of star only for illustration purpose of my logic. If the space time was filled with radiation then also the logic applies. $\endgroup$ Dec 14, 2023 at 13:13
  • 4
    $\begingroup$ Temperature of the universe is a questionable word group, because the whole universe is obviously not at single temperature now; already on Earth, temperature varies by many orders of magnitude. Temperature of the cosmic microwave background radiation is just that, temperature of that radiation. It's not temperature of the universe. $\endgroup$ Dec 14, 2023 at 20:12
  • 7
    $\begingroup$ Park your star ship a few hundred thousand light years away from any galaxy, and then wait. You will start to feel cold after your fuel is used up and your ship radiates its residual heat away. You will feel some warmth when you absorb cosmic microwaves and the light of distant galaxies, but it won't be much. Eventually, you will be chilled to the temperature at which the heat you radiate will be equal to the heat you absorb. That temperature is approximately 2.7K. $\endgroup$ Dec 15, 2023 at 4:33

6 Answers 6

30
$\begingroup$

If you plant yourself in intergalactic space, you will equilibrate at a temperature of 2.7 kelvin. That is the temperature of the radiation today. (It's not really the temperature of the universe -- just the temperature of the component of it that interacts most strongly with ordinary stuff.)

Your analogy about inferring the temperature of the Sun is more like inferring the temperature of the surface of last scattering of the cosmic microwave background. The last scattering surface indeed had a temperature of about 3000 kelvin.

$\endgroup$
10
$\begingroup$

If we are going to assign a temperature to the Universe as a whole then we also need to define an epoch (a time after the Big Bang) to which that temperature applies.

The temperature spectrum of the radiation in the CMBR when that radiation was emitted was centred at about $3000$ K (and was remarkably isotropic). So we can (loosely) say that at an epoch of $380,000$ years after the Big Bang the temperature of the universe was $3000$ K.

In earlier epochs, closer to the Big Bang, the temperature of the universe was higher. In later epochs the temperature of the universe becomes lower as the universe expands.

The temperature spectrum of the radiation in the CMBR as we observe it now is centred at about $2.7K$ (and is still remarkably isotropic once you subtract the effects of the local motions of the Earth and the Sun). So we can (loosely) say that the temperature of the universe in the current epoch (about $13.8$ billion years after the Big Bang) is about $2.7$ K.

$\endgroup$
8
$\begingroup$

A big difference between the CMB "temperature of the universe" and your example of looking at a distant star, is that all observers throughout the universe at the same epoch agree that the CMB has the same temperature. It therefore makes complete sense for all observers to assign and talk about a single, present-day temperature to the universal CMB, which decreases as the universe gets older and was much hotter in the past.

In your example, observers at different distances from the Sun would measure a different (redshifted) temperature for the Sun and so the only temperature they might agree on is the temperature of the radiation when and where it was emitted.

$\endgroup$
5
$\begingroup$

When we talk about the "temperature" of something there are a couple things we could mean. The normal definition is based on the Boltzmann (really Bose here) distribution with occupation numbers proportional to $e^{-E/T}$. This is equivalent to a black body spectrum. In this sense, the CMB temperature of $2.7K$ is correct.

However, that does not mean that the CMB is in thermal equilibrium for the purposes of cosmology. That would require quasi-static equilibrium; that is, the interaction strength is sufficient that kinetic equilibrium is reached quickly relative to the timescales under consideration (typically a Hubble time). That, of course, is not happening; indeed the whole point of decoupling is that the photon distribution is frozen in after recombination, and any perturbations are also frozen in and do not equilibrate. If they did, the CMB would be somewhat less useful for us. Depending on your definition, some would say that the CMB does not have a temperature because it is not in thermal equilibrium, but many would call this pedantic.

In that sense it is actually something of a "coincidence" of geometry and statistics that the CMB today has the distribution of a thermal distribution with $T = 2.7K$. This would fail if the expansion of space were not homogeneous and isotropic. Of course it isn't actually a coincidence. It's just the ideal gas law. We have $pV = NT$, and $N$ is constant. For photons $p = \rho/3 \propto a^{-4}$, and $V \propto a^3$, so we have $T \propto a^{-1}$. The CMB was in thermal equilibrium prior to decoupling, with $T = 3000K$ and $a^{-1} = (1+z) = 1100$, which leads to $T=2.7K$ for $a=1$ today. Indeed the ideal gas law is actually a free gas, and therefore not a gas which is in thermal equilibrium because a free gas has no interactions and does not reach equilibrium (though it can still have a temperature in the sense of the Boltzmann distribution).

As long as you're fine with saying that an ideal gas can have a temperature, then it's also fine to say that the CMB's temperature today is $2.7K$. But it is also true that the mechanism at play here is not thermal equilibrium today but merely dimensional analysis. Likewise if we measure the temperature of the photon gas emitted by some high redshift star, or if some far future civilization measures the temperature of the photon gas emitted from the Sun, the result will be lower than the temperature of the surface of the star.

$\endgroup$
3
$\begingroup$

Technically, $2.7K~$ CMB temperature identifies a hypothetical black body, which would radiate NOW at peak wavelengths of $1~mm$. But CMB afterglow is a past thing,- there is no real source which would radiate actively such EM waves over all universe. These waves are just the remains of some universe epoch when universe was opaque to photons.

Aliens, if any, would make similar conclusion about Sun temperature. Like, if sun would radiate NOW such and such EM waves with peak at $x~cm$, then it's temperature would be $y~K$. But similarly, they would recall that this radiation of Sun may be from epoch 13 billion years ago and so they will trace back Sun real temperature at the time Sun emitted original waves.

Hope that helps.

$\endgroup$
3
$\begingroup$

I'll give it to you simply, as follows.

The further back in time you go, the hotter the universe was. When the universe released the CMB, its temperature was about 3000K. That thermal radiation then got stretched out as the universe expanded and got cooler and is now 2.7K.

What does that 2.7K mean? just this: all hot objects emit a spectrum of electromagnetic radiation. If you are sitting in a room at 70 F (which corresponds to about 300K), your surroundings are beaming you continuously with the spectrum corresponding to 70 F (and you feel warm as a result). At the same time, you are radiating your own 70 F spectrum towards your surroundings, and you would get cooler except that the incoming radiation is the same as the outgoing so your temperature stays constant.

Now we place you into the airless vacuum of deep space. Your 70 F spectrum radiates away and you get colder because there's only the 2.7K spectrum coming to you from the surrounding space. you just keep on getting colder and colder, you freeze solid, and you stop getting colder when your temperature matches that of outer space.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.