# Coldness of Space

1. Nov 21, 2008

### Mentallic

From what I've been told, the temperature in space is approx. 3oK if not in direct contact with infrared rays from the sun/stars etc. This temperature is also decreasing as the universe expands. Space is also a vacuum, with very few particles per cubic unit.
My question is: If space is a vacuum and a point in space is completely blocked by objects so that no infrared radiation is in direct or indirect contact (reflections etc.) and there are no particles in the vicinity, why is it that the temperature is not absolute zero? Where does this tiny temperature come from? Also, if a vacuum were created here on Earth, would it also be the same temperature as in space?

2. Nov 21, 2008

### Jonathan Scott

Whatever is blocking off the sun/stars or surrounding the vacuum will be radiating some heat itself unless it too is near absolute zero. The effect may be reduced if its emissivity is designed to be much lower than that of a black body.

3. Nov 21, 2008

### Mentallic

Ahh so what I think you're implying is that all bodies of mass (even black holes?) radiate their own heat at the atomic level after being bombarded by heat from the other end of that body?
So the temperature in this point in space that I specified will actually be lower than 3oK which is the average of the universe?
What if, lets say we did attempt to bring that spherical body of mass that is isolating this point close to absolute zero also, by surrounding that mass with yet another large sphere. Would this mean that point in space will approach even closer to absolute zero, but never actually reach (apparently absolute zero can never be reached)?

4. Nov 21, 2008

### Jonathan Scott

Black holes have their own special rules relating to temperature and thermodynamics, so let's forget about them here.

Yes, all massive bodies radiate heat. The maximum rate is primarily determined (through Stefan's Law for "black bodies") by their temperature but the rate may be less than that depending on the emissivity of the surface.

I'm sorry, but I don't remember enough from my student days about how thermal equilibrium of radiation works to know whether passive shielding of a volume could actually make it colder than the shield in the long term.

If the inside surface of the shield had low emissivity, it would initially radiate less than a black body into the internal volume, but that energy would presumably last longer with reflection inside the shield before being reabsorbed, because lower emissivity means higher reflectivity.

I suspect that the long-term stable temperature inside a passive shield is likely to be not very different from the temperature of the shield itself. However, a highly reflective passive shield should prevent something already cool from warming up for a long time.

5. Nov 21, 2008

### D H

Staff Emeritus
This is somewhat of a misnomer. Space is not a pure vacuum. The space between stars and even galaxies has some stuff in it, called the interstellar and intergalactic medium. The interstellar and intergalactic medium intergalactic medium have all of the characteristics of an ionized gas, albeit a very, very tenuous one. Like any other gas, these media have a temperature, and this temperature can be extremely high, up to 108 Kelvins or more! So in one sense space is anything but "cold".

People say that space is "cold" because a macroscopic object such as a thermometer in deep space will never go into thermal equilibrium with the surrounding medium. Radiative heat transfer will complete dominate conductive heat transfer because the interstellar/intergalactic medium is so very tenuous.
It comes from the big bang, or more precisely, from the radiation released when the universe became transparent about 300,000 years after the big bang. This cosmic microwave background radiation pervades all of space. A macroscopic object well-shielded from or very far from any star will eventually come into thermal equilibrium with this background radiation: 2.725 Kelvin.

6. Nov 21, 2008

### Mentallic

And this microwave background radiation is everywhere in the universe? I don't see how this is possible. If the point where the big bang began released with it very high frequency radiation, wouldn't this radiation now be on the edges of the universe? Say for e.g. the point of the big bang is rather our sun. When both begin expanding (the sun explodes), all the radiation will move outwards and eventually our entire solar system will be starved of infrared, light, gamma rays etc. I highly doubt the point where the sun used to be, or even our Earth will still be bombarded by the sun's rays.

7. Nov 21, 2008

### D H

Staff Emeritus
That is correct.
You have a common misperception of the big bang: That it was an explosion in space. A much better way to look at the big bang is that it was an explosion of space.

Here are some introductory-level articles on the big bang and the cosmic microwave background radiation:
http://www.totse.com/en/technology/space_astronomy_nasa/300000.html [Broken]
http://www.astro.ubc.ca/people/scott/cmb_intro.html
http://nedwww.ipac.caltech.edu/level5/Guth/Guth_contents.html

Last edited by a moderator: May 3, 2017
8. Nov 21, 2008

### Integral

Staff Emeritus
You really cannot speak of the temperature of open space, you need something there to have a temperature. So lets assume a arbitrary object located in empty space. Initial temp = $T_0$ and emissivity 1.

Since the object is isolated in deep space the only path of energy exchange is radiation.

From Stephan Boltzmann law the Energy exchange between the surroundings and the body is :

$$E= \alpha (T_s^4 - T_0^4)$$

Where $T_s$ is the temperature of the suroundings. In deep space $T_s$ is 3.7K. The rate of heat loss will decrease as the temperature of the object decreases. So the equilbrium temperature approached over time will be 3.7K. That is when the energy exchanged is zero.

A simple fact is that there is no place in know space where you can see a uniform 3.7K in all directions. Any hot object, such as a star or planet, visible from the object will be exchanging energy at a rate determined by its temperature.

For a body in low earth orbit the side of the body facing the sun to be recieveing energy at one rate, the side faceing the earth will recieve energy at a different rate, while the side faceing deep space loses energy. This is one reason that virtually every satilte is rotating. The rotation ensures that the satilite maintains a uniform surface temperatue.

9. Nov 22, 2008

### D H

Staff Emeritus
There is something there, even in the space between galaxies. That something might be very tenuous, but it is not nothing at all. Astronomers regularly discuss the attributes of the intergalactic medium -- including its temperature. What you can't do is think that this interplanetary/interstellar/intergalactic medium temperature will have any bearing on the temperature of macroscopic objects imbedded in that medium.

10. Nov 22, 2008

### Vanir

Excuse me, just a tourist passing through.

Do you mean in the sense, of the duality of wavefunctions to be defined as a medium, in this case albeit quite tenuous? I've no idea for correct terminology...zero point energy field? Quantum field? Electromagnetic field? Gravitational field?

11. Nov 22, 2008

### D H

Staff Emeritus
No reason to get so exotic. The interplanetary/stellar/galactic medium is just plain old baryonic matter, mostly ionized hydrogen and helium. The intergalactic medium has a density of about one atom per cubic meter. This ionized gas has a random component to its kinetic energy, aka temperature.

12. Nov 22, 2008

### jambaugh

Actually you can. A given volume of space is a quantum system with excitations (particles/fields). You can directly talk about its entropy, energy and temperature.
If the volume is assumed to be truly empty you are asserting entropy=0 and energy =0 thus temperature = 0. (This is however impossible to actualize.)

13. Nov 22, 2008

### Integral

Staff Emeritus
DH has told us that the temperature is well defined because we may have 1 particle per cubic meter. Now you tell me that there is no such thing as empty space??? You guys need to get together and work this out. Seems that I could easily find a volume of deep space with NO particles. What is the temperature of that volume? O??? does it change in a step function when a stray particle drifts through? As I said you need something to have a temperature.

I think I will stick with my initial idea, slightly reworded to satisfy the nit pickers.

Temperature of deep space is a complicated issue and needs to be addressed carefully. You really need an "ensemble" of particles to define the temperature.

14. Nov 22, 2008

### marcus

Hello Integral, DH, Mentallic, and everybody.

In case it might help for me to butt in, simply to emphasize what DH already said, here goes:

DH is right. Deep intergalactic space is absolutely swarming with CMB photons. I forget how many per cubic meter but it is a huge number. That defines the temperature quite precisely. Mentallic please listen to what DH says. He is giving you the absolute straight dope.

An otherwise empty vacuum created on earth would be full of photons radiated off the walls of the box containing the vacuum. The temperature of a vacuum is the temperature of thermal radiation in the vacuum. It could be whatever, depending on the temp of the walls. If you use refrigeration to make the walls of the box 5 kelvin, then the temp inside will be 5 kelvin. If you make the walls 1 kelvin, the temp of the photons inside will be 1 kelvin.

It just happens that the temp of otherwise empty space (far enough out not to be affected by stars and stuff) is 2.728 kelvin. That's the temp of the swarm of CMB photons out there (which have a welldefined thermal energy bar-chart.)
So if you put a piece of metal out in deep space it will radiate off photons and absorb CMB photons until it settles into equilibrium 2.7 with the CMB photons and with the rest of space.

If you put a metal box around the piece of metal, to try to shield it from the CMB then the box itself will settle into equilibrium 2.7 and then it will be filled with its own 2.7 kelvin radiation. and then the original piece of metal will settle to the same 2.7 temp.

You can't shield against the 2.7 CMB temp unless you use some kind of refrigeration to cool the box.

Last edited: Nov 22, 2008
15. Nov 22, 2008

### marcus

Mentallic, here is something you might like to learn to do, since the

You can calculate for yourself how many CMB photons are in a cubic meter of the space out between the stars.

I just calculated it, using Google calculator, and it came to about 400 million photons per cubic meter.

Here is what I typed into the google window
0.24*(k*2.728 kelvin/(hbar*c))^3
then I pressed return and it said 411 million per cubic meter.

The main thing is if you want to know how many thermal photons per cubic meter are in any space, like your room, or an oven, or a skating rink---wherever, it's very simple.

You just take the temperature in kelvin, say T (in this case T was 2.728 kelvin), multiply it by the k constant (Google knows this and will do it if you say to) and divide that by hbar and c.
Then you cube whatever you got, and you are done! or nearly so.

All that's left is to multiply by a number that is close to 1/4 but which is actually more like 0.2436. If you forget to do that, it's OK because the answer will still be about right. (At least to get a rough order of magnitude estimate.)

So the main thing is simply take the temp, multiply it by the k constant, divide by Planck's hbar constant and by c, the speed of light.

Google knows all these universal constants like Boltzmann k, hbar, c. So it will take care of it.

So then you have kT/(hbar*c) and all you have to do is cube it.

Can you do this? Can you find the number of photons per cubic meter in your room?
Your room is probably around 293 kelvin. If you want, type this into Google and press return

(k*293 kelvin/(hbar*c))^3

this will say how many photons are with you right now, in the room (as a number per cubic meter)

and if you like, to make it more accurate divide by 4, or multiply by that auxilliary number I mentioned that is about 0.24 and doesnt affect the gross size very much.

The temperature in a space is revealed by the thermal light in that space no matter where it is. I'm telling you this because it's a very neat fact and today is Max Planck's birthday. Every day is.

Last edited: Nov 22, 2008
16. Nov 22, 2008

### Mentallic

Ahh no one has ever expressed it to me in such a way before. So concise, yet so understandable.

This same problem is what was bothering me when I first created this thread. But then Jonathan Scott said it is rather the radiation being emitted that heats up that point in space, not the particles passing through it.

But what if the universe were to be given enough time to expand enough such that the density of CMB photons per cubic metre is low enough so that there can be distinct points in space that are completely free of these photons. Will this assume 0 temperature?

Thanks for clarifying this dilemma for me

The photon density is about 1.2 million times more! At roughly 1.4 quadrillion / m3.
Just out of curiosity, I went scrounging and found the critical temperature required for hydrogen bombs to begin the fusion process is approx 40M K. This corresponds with 3.5x1030 photons per cubic metre? This sounds incredibly high, and makes me wonder if photons even have their own volume.

17. Nov 22, 2008

### marcus

Mentallic, I'm really glad to see that you took hold with that formula and calculated some interesting photon densities with various temperature. Then, looking back over my post I noticed I had left out a factor of 2.701 or approximately 3, that I should have divided by.
So I misled you, sorry. All our answers are too big by about a factor of 3. (But they still have the right order of magnitude so they give the right idea.)
I went back and corrected. the CMB is only 400 million photons per cubic meter.

18. Nov 22, 2008

### D H

Staff Emeritus
If the universe continues its expansion indefinitely (note well: the jury is still out on the ultimate fate of the universe), the universe will eventually cool toward absolute zero. Google "ultimate fate of the universe", "heat death", "big freeze".

It's rarely a good idea to talk about points. That is where demons such as singularities lie. What you should be talking about instead are volumes. Even now there are at any given instant of time volumes of space void of CMB photons. So that is not a particularly good way to look at it, either.

The concept of temperature starts to lose meaning when one looking at too small an ensemble of matter or too short a duration of time. When talking about temperature you need to look at a sufficiently large ensemble of matter and a sufficiently long duration of time.

As the universe ages and expands, the energy and flux density of the CMBR photons will decrease. Per some fixed time interval, a macroscopic object in deep space will absorb an ever decreasing number of photons of an ever decreasing frequency. The object in turn will emit an ever decreasing number of photons of an ever decreasing frequency as it cools toward absolute zero.

19. Nov 22, 2008

### Integral

Staff Emeritus
Thank you. I think that is what I have been trying to say.

20. Nov 22, 2008

### marcus

Mentallic,
(I concur with what DH and Integral just said* and have some side remarks)
Now that you know how many thermal photons there are in a cubic meter at some temperature, you might what to know the average energy of each photon.

If T is the temp (like T = 2.728 kelvin) then the average photon carries this energy:

2.701 kT which google probably likes to see written 2.701*k*2.728 kelvin

and if you want to know the wavelength of that average energy photon,
it is h*c/(2.701*k*2.728 kelvin)

that is h*c divided by the energy freight it carries. (the more energy, the shorter the wavelength).

So I'm inviting you to find, should you care to, the wavelength of an average-energy CMB photon
=========================

This is just for fun. You can also calculate the radiant energy in a cubic meter, by multiplying the number of photons by the average energy of each one. So you can find the radiant energy per cubic meter in your room, or in an oven etc. The energy density of the CMB, small as it is, is analogous to these more familiar thermal energy densities---just a lower temperature.

*including CMB photons along with other matter. I guess the point being that since you have 400 million CMB photons per cubic meter there is more of a statistical ensemble, to give a meaningful idea of temperature, than there is with other species of matter. Other species, like hydrogen atoms, can be rather sparse.

Last edited: Nov 22, 2008