Why Isn't Space at Absolute Zero Despite Its Vacuum and Low Temperature?

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Space is not at absolute zero due to the presence of cosmic microwave background radiation, which maintains a temperature of approximately 2.725 Kelvin. Even in a vacuum, objects radiate heat, and the temperature of space is influenced by the interstellar and intergalactic medium, which contains low-density particles that also contribute to thermal energy. Attempts to create a vacuum on Earth would not replicate the same temperature as space because the conditions and surrounding influences differ significantly. Additionally, while shielding an object can reduce heat absorption, it cannot lower the temperature below the background radiation level. Ultimately, the temperature of space is a complex interplay of radiation, particle density, and cosmic history.
  • #31
flatmaster said:
There was a discrepency before as how a near vacuum can have a temperature near T=0K, but also have no significant cooling effect on a macroscpoic body placed within the vaccum. I thought of it as if the vacuum has a tiny heat capacity. A tiny change in the amount of the thermal energy ov the near vacuum will result in a huge change in it's temperature.

Flatmaster, it sounds like you want to be able to calculate the heat capacity of a cubic meter of perfect vacuum at temperature T.
The heat capacity will depend on T, of course. Think of of a box with volume one meter containing nothing but EM radiation which is in equilbrium with the walls of the box.

The energy in the box will be E(T) = (pi^2/15)k^4 T^4/(hbar*c)^3
Let's see how much thermal radiation energy is in a cubic meter at 10 kelvin. I think all we need to do is paste this into google:
(pi^2/15)k^4 (10 kelvin)^4/(hbar*c)^3*1 meter^3
It gives the answer 7.6 picojoules.

To find the picojoules per degree heat capacity you simply need to take derivative in T.

dE/dT = ((4*pi^2)/15)(k^4)*(T^3)/(hbar*c)^3

So let's calculate this for T = 10 kelvin. The above is on a per volume basis, so multiply it by one cubic meter to get the heat capacity of the space in the box. I think we just need to paste this into google:
(4*pi^2/15)(k^4)*(10 kelvin)^3/(hbar*c)^3*1 meter^3 in picojoule per kelvin

It gives the answer 3.03 picojoule/kelvin

That would be the heat capacity of a cubic meter of vacuum at a temperature of 10 kelvin.
As you can see, to increase the temperature by one kelvin only involves putting in a tiny (3 picojoule) amount of energy.
This is the heat capacity of thermal radiation itself. If you don't like the walls of the box (with their own heat capacity) being there then think of a very large volume of empty space so that the walls can be neglected.
 
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  • #32
taitae25 said:
So my understanding of the average temperature of space is the average temperature, or energy of particles in a given space ? so obviously there will be particles in universe that has temperatures (kinetic energy) a lot smaller than 3K and a lot higher than 3K (as D H said as well).
Temperature is characteristic of a largish collection of objects. Asking "what is the temperature of a particle" is a bit of a nonsense question.

Temperature in space is also a bit of a slippery concept. The place to start is with the zeroth law of thermodynamics: Two things have the same temperature if the are in thermal equilibrium with one another. The problem is that a macroscopic object will never come into thermal equilibrium with the extremely tenuous plasma that occupies the space around the object. Radiative heat transfer is essentially the only heat transfer process that occurs for a macroscopic object in space. The few collisions with the ions that comprise the space environment do very, very little to change the object's temperature. Macroscopic objects come into thermal equilibrium with the local radiative environment rather than with the local physical medium.
Now your original questions of:

"Where does this tiny temperature come from?"
Short answer: From the big bang. Long answer: Read the thread.
 
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  • #33
Unruh Radiation makes empty space look like it has a non-zero temperature. Does it work the same way for a non-accelerating observer viewing inflating spacetime? The spacetime inflation pulling the universe apart should give the vacuum a non-zero temperature, right? This temperature would increase with distance, presuming that spacetime inflation is uniform throughout the universe.

Is there any evidence for this?
 
  • #34
MuggsMcGinnis said:
Unruh Radiation makes empty space look like it has a non-zero temperature. Does it work the same way for a non-accelerating observer viewing inflating spacetime?

There is something analogous to the Unruh temperature in a universe with positive cosmological constant (as ours is thought to be.)
According to the standard picture, there is a cosmological event horizon at a distance L of about 15 or 16 billion lightyears.

Everything that is now outside that horizon is analogous to stuff that has fallen into a BH. As of today it can't send signals to us.

This distance L determines a temperature T such that the associated energy kT is given by
kT = hbar*c/(2 pi L)
This is roughly analogous to an Unruh or BH temperature

Because the distance L is so huge, the temperature is very small and it would be insignificant even beside the already very low CMB temp of 2.728 kelvin.

But when people imagine the universe extremely far into the future, according to the usual LCDM picture, then expansion will have cooled the CMB way down (and stars will have burnt out etc). At that time, the model says, the distance L will still be approximately the same, and so the temperature determined by the horizon will begin to dominate. Even though a very very small temp, it won't have much competition. So it's part of the late universe picture.

Here's what google calculator tells me:
(hbar * c) / (2 * pi * k * (15 billion light years)) = 2.56818621 × 10^-30 kelvin

You've asked an interesting question. A keyword for finding out more is "de Sitter space".
 
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