Coldness of Space: Exploring 3oK Temperature & Vacuum Effects

[Mentallic]

From what I've been told, the temperature in space is approx. 3^oK 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?

 

[Jonathan Scott]

From what I've been told, the temperature in space is approx. 3^oK 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?

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.

 

[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 3^oK which is the average of the universe?
What if, let's 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)?

 

[Jonathan Scott]

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 3^oK which is the average of the universe?
What if, let's 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)?

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.

 

[D H]

From what I've been told, the temperature in space is approx. 3^oK if not in direct contact with infrared rays from the sun/stars etc.

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.

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?

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.

 

[Mentallic]

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.

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.

 

[D H]

And this microwave background radiation is everywhere in the universe?

That is correct.

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.

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
http://www.astro.ubc.ca/people/scott/cmb_intro.html
http://nedwww.ipac.caltech.edu/level5/Guth/Guth_contents.html

 

[Integral]

You really cannot speak of the temperature of open space, you need something there to have a temperature. So let's assume a arbitrary object located in empty space. Initial temp = [itex] T_0 [/itex] 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 :

[tex] E= \alpha (T_s^4 - T_0^4) [/tex]

Where [itex] T_s [/itex] is the temperature of the suroundings. In deep space [itex] T_s [/itex] 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 receive 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.

 

[D H]

You really cannot speak of the temperature of open space, you need something there to have a temperature.

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.

 

[Vanir]

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.

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?

 

[D H]

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?

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.

 

[jambaugh]

You really cannot speak of the temperature of open space...

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.)

 

[Integral]

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.)

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.

 

[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:

...Where does this tiny temperature come from? ...

This cosmic microwave background radiation pervades all of space. A macroscopic object ...[far enough from other sources]... will eventually come into thermal equilibrium with this background radiation: 2.725 Kelvin.

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.

... Also, if a vacuum were created here on Earth, would it also be the same temperature as in space?

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.

 

[marcus]

Mentallic, here is something you might like to learn to do, since the
CMB radiation interests you.

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 doesn't 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.

 

[Mentallic]

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.

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

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.

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.

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.

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?

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

Thanks for clarifying this dilemma for me

Mentallic, here is something you might like to learn to do, since the
CMB radiation interests you.

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 1.11 billion photons per cubic meter.
...
Can you find the number of photons per cubic meter in your room?...

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.

 

[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.

 

[D H]

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?

Short answer:
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".

Long answer:
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.

 

[Integral]

Short answer:
...
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.
...

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

 

[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.

 

[Mentallic]

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.

While I calculated 3x more photons than there should be, since I cannot even begin to fathom this magnitude in numbers, 1/3 of that won't be change my perspective on the topic These magnitudes are enough to tell me "yes there are lots and lots of photons everywhere".

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.

Thank you D H. This has really opened my eyes in understanding just a little bit more about the universe and the way it works.

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.

The fun here begins to lose all meaning when I have no idea how these answers are being derived. Even then, I probably wouldn't have the mathematical/physics knowledge required to follow such derivations.
While I enjoy dealing with magnitudes of a high degree of largeness/smallness (since it is quite interesting), unless I can comprehend the meaning behind the number, I don't see any point just yet to delve too deeply into the topic.

Don't get me wrong, I love this sort of stuff! But just a few days ago, before creating this thread, I always thought of light as a "sea" of massless radiation that will never have 0 intensity no matter how distant the light travels. Now that I hear of photons, it seems possible that for a human with an infinitely strong telescope, that is very, very far from a star might get a blurry(?) image on the telescope. As a photon bombards the human's eye every so often, for that instant, light from the sun is seen again.

Marcus, just if you like to know, my class has been learning about gravity in physics and the effects large masses have on the surrounding space. Apparently the mass of an object depends on the speed a satellite must have in order to orbit at a certain distance from the mass. We were able to apply this idea and its formulas to find the mass of Jupiter by analysing the effect it has on its moons (actually we just took the moon information off the net) and for every moon the results are amazingly all the same. We calculated Jupiter's mass to be 1.9E27 kg. Now that is interesting!

 

[marcus]

... We calculated Jupiter's mass to be 1.9E27 kg. Now that is interesting!

That is interesting for sure! Compliments to your teacher. I like hearing about calculator physics exercises that you can use the Google calculator for. This looks like a good one.

 

[jambaugh]

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.

I don't see any inconsistency! ?

Seems that I could easily find a volume of deep space with NO particles.

If by particles you're including photons then you'd find it impossible to have exactly 0 particles. At best you might get the expected particle number down below 1, i.e. a quantum superposition of 0 and 1 (2, 3, ...).

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.

Given the assumption of true vacuum you've specified the system maximally and so the entropy is exactly 0. The temperature then too would be 0. Said another way the only way to find such a piece of space is to physically cool it down to absolute 0... i.e. suck up all the stray photons with a zero degree black body. Not quite physically possible however it makes a good limit point for a series of possible but increasingly harder situations.

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.

A theoretical definition is not that complicated at all. You begin with the entropy defined for the density operator of the system. It needn't be an ensemble. You can talk about the energy, entropy and thus temperature of a single electron.

BTW, the issues are not nit picking trivialities. "Empty space" is by no means empty even if you extract all the electrons, protons, and neutrons from it. The Bose gas of photons is there, exerts pressure and affects the curvature of space-time via GR. It must be accounted for when considering, for example, the mean free paths of very high energy cosmic rays through "empty" space.

 

[questions4all]

Its simple, All things vibrate and thus generates some type of heat. Even in voids of space there are virtual particles and they vibrate.

 

[Sir_Arthur]

Isn't it true that temperature is a property of particles (be they the interstellar medium or the radiation traveling through space) in space? Isn't talking about the temperature "of space" inherently meaningless because even if you could create a void with no particles (even virtual ones) it would simply not have a temperature because there would be nothing there to measure...?

 

[marcus]

Isn't it true that temperature is a property of particles (be they the interstellar medium or the radiation traveling through space) in space? Isn't talking about the temperature "of space" inherently meaningless because even if you could create a void with no particles (even virtual ones) it would simply not have a temperature because there would be nothing there to measure...?

If you want to be picky then I shouldn't ever refer to the temperature of the room I'm sitting in. I shoud always say the temperature of the air in the room

But in fact we do talk about the temperature of our room, and we do talk about the temperature of space.

Admittedly it's imprecise, and something of a shorthand expression. But it's not completely meaningless. In astro context it usually (or at least often) means the temperature in space out away from other sources, interstellar, intergalactic. And in that case it certainly would be more correct to say "the temperature of the CMB" because that is what's usually intended.

To recap what was said earlier in the thread, space has about 400 million CMB photons per cubic meter. And those photons have a near-perfect thermal distribution, with a temperature of 2.728 kelvin.

 

[jambaugh]

Isn't it true that temperature is a property of particles (be they the interstellar medium or the radiation traveling through space) in space? Isn't talking about the temperature "of space" inherently meaningless because even if you could create a void with no particles (even virtual ones) it would simply not have a temperature because there would be nothing there to measure...?

You are correct in the context of classical particle mechanics. But once you consider say the electromagnetic field in a region of space then (without invoking a medium per se) you have mass-energy (even if its zero) in that region and entropy and a meaningful definition of temperature. To say the e-m field is zero (or specified exactly whatever the field) in a region is equivalent to saying its temperature is zero.

Operationally to effect a desired e-m field exactly you must cool the region of space down, absorbing all the random thermal fluctuations.

And this of course carries over into quantum mechanics and QFT where you then may describe the e-m field as a distribution of photons (some virtual & some physical).

Now conceptually you can think of this as the temperature of the "luminiferous ether" permeating space. Nothing terribly wrong with that but this ether is not a physically observable substance and in an operational context we should excise that which we cannot observe empirically. But in so doing you move a way from thinking of temperature as a property of an object and into thinking of temperature as a property of the modes of interaction with/between physical systems.

 

[flatmaster]

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.

Q=mCpT

 

[taitae25]

I didn't read all the post, but it must also be understood that the "temperature" that we measure in the room is simply the "average" temperature of the air molecules. In reality, the temperature (or energy of the air molecules) is a distributed spectrum :) 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).

Now your original questions of:

"Where does this tiny temperature come from?"

Sorry I'm not knowledgeful enough to answer that question.

 

[flatmaster]

I now remember from a chemistry paper I once heard that there are some subtlties to temperature when you get to near vaccumes. We understand that temperature is the average kinetic energy of the moles of a gass. However, kinetic energy is not only translation (moving through space), but also rotational and vibrational.

In the lab, you can create a gass that has one translational temperature and another vibrantional temperature. For example, you can shoot benzene out of a cold molecular beam with a carrier gass molecule size much smaller than the beem. The larger benzene molecules move more slowly than the carrier gass, so they orient themselves to allow the quickly moving gas to pass. The benzene is "rotationally cold". Interestingly, the researchers called this the "frisbee method"

 

[marcus]

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.

 

[D H]

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.

 

[MuggsMcGinnis]

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?

 

[marcus]

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".