Is the LHC colder than space?

By Phil Plait | July 23, 2008 2:00 pm

I’ve received the odd email and comment or two about a news headline claiming that some parts of the Large Hadron Collider are now "colder than space". Surely, these emails ask, space has no temperature, so the headline must be wrong.

Well, it’s complicated.

It depends on how you define temperature. It’s really just an average measure of how much energy matter has. All the molecules in an object are wiggling and jiggling, and the amount of that motion is what we call temperature. The more energy they have, the more they move, and the higher the temperature.

So, if empty space has no matter in it, it’s a vacuum, and therefore has no temperature. Right?

Well, it’s complicated.

Sure, if there’s no there there, then there is no temperature. Of course, space isn’t really empty, it’s only mostly empty. Near the Earth, space actually has lots of subatomic particles per cubic centimeter. Even between galaxies, there are one or two particles per cc. But still, space is so empty that these hardly count when talking temperature. Right?

Well, it’s complicated.

When the Universe formed, it was hot and dense. As it expanded, it cooled off. At first, your average photon — a particle of light — had lots of energy. But fighting the expansion of space itself takes its toll, and the photons lost energy. Flash forward 13.7 billion years, and today we see these photons have lost a lot of energy. What used to be a raging fireball from the moment the Universe formed is now a chilled brew, with the photons in the microwave part of the spectrum; very low energy indeed. You can convert that energy into a kind of temperature, and the number you get is about 2.7 Kelvin (-270 Celsius, or about -450 F).

Marshmallow of Science, sacrificing allThat means that even if space is totally empty of matter, there are still photons at that energy flowing through it. If you took a mini marshmallow (and why not) and stuck it in deepest space, it would drop in temperature until it got to 2.7 Kelvin. It wouldn’t get any colder than that, because those photons would warm it up to 2.7 K.

So in a sense space does have a temperature. It may not satisfy purists, but realistically it’s not too bad to think of it this way.

So when parts of the LHC are cooled below 2.7 K — and you can do this locally by pumping away any errant heat, but it takes a lot of energy to do — it is in fact colder than space.

So long story short, I don’t have any real issues with the headline. And hey, it gave me a chance to explain some things, and to use the Mini Marshmallow of Science. So I’m glad it came up.

CATEGORIZED UNDER: Astronomy, Science

Comments (62)

  1. Tim G

    Would that marshmallow resemble a “blackbody”? If it does not absorb most of the photons at around one millimeter in wavelength, wouldn’t it get colder than 2.7 Kelvins?

  2. Quatguy

    Phil, what is the scientific definition of the highly technical terms “wiggling and jiggling”?

    Here is another technical question….wouldn’t a mini marshmallow explode if it were suddenly placed in the “vacuum” of space?

  3. Thomas Siefert

    That begs another question, how big would the marshmallow become before it froze?

  4. Greg in Austin

    If you throw in a piece of Hershey bar and two little graham crackers, you could have a miniature S’more. And then, it would be just like some of the camping trips I’ve been on. ;)

    8)

  5. How big would the marshmallow get? It would get this big. And it wouldn’t be happy about it.

  6. And hey, it gave me a chance to explain some things, and to use the Mini Marshmallow of Science.

    And you can do that without paying royalties to Bill Nye? Now that’s power…

    Actually, I think I saw a YouTube video of marshmallows in a vacuum: they shrank! Presumably, the gasses diffuse out of the marshmallow due to the decreased outer air pressure and its inherent stretchiness causes it to collapse on itself.

  7. I don’t think purists have any right to be upset. Out in space you are surrounded by isotropic 2.7 K blackbody radiation — that’s about as close to thermal equilibrium as you will find in a natural setting. Much closer than, say, your local airport where they take temperature readings. Space is not a vacuum, it’s just a good approximation.

    And (to Tim G) the nice thing about thermal equilibrium is that, given time to equilibrate, *everything* is a blackbody. If you give the marshmallow enough time, it will be giving off exactly the same spectrum as the microwave background.

  8. Thomas Siefert

    ~ahem~ let me rephrase: That begs another question, how small would the marshmallow become before it froze? :-)

  9. So, how close do you have to move the Mini Marshmallow of Science to a start like the Sun before it turns a nice golden brown, so we can make the Mini S’more of Science?

  10. BMcP

    Nice to see the Mini Marshmallow of Science is getting some play time and not completely eclipsed by it’s more famous co-host, the Squishy Brain of Science :)

  11. [Homer] MMMmmmMarshmallow of Science MMMmmmmaahaghshghahhh…[/Homer]

  12. But thermodynamicists have every reason to be upset. What’s this nonsense about average energy of matter?! Thermodynamics needs no such direct physical meaning for defining temperature! Temperature is the property that two things share when they’re in thermal equilibrium, pure and simple. I’ll have none of your statistical mechanics malarkey!

    Ho hum. ;-)

  13. *Curmudgeonly*

    Ah, I suppose I’ll call myself a purist and agree to disagree.

    Interesting question just came to my mind, will this background radiation ever diminish or disappear?

  14. Sully

    Who cares whether the LHC is colder than space, the real question is whether either one is colder than a brass bra on a witch’s. . .

  15. TMB

    Chemist – if it said “colder than a vacuum”, then you’d have a point. But it doesn’t, it says “colder than space”, and space is definitely not a vacuum.

    Because of the expansion of the universe, both the energy of the background photons and their space density goes down with time, resulting in a lower temperature, but it will only really “disappear” once cosmic acceleration pulls the average distance between CMB photons out beyond the horizon scale. (hmmm… interesting question how long that would take… might have to work that out later).

  16. Thermodynamics is such a mess that “purists” wouldn’t be satisfied by any of it. What you’ve described is as good a definition as anything can be for “being 2.7 K”: thermal equilibrium.

  17. Hugo

    Just a clarification, the photons at ~2.7K are what’s usually referred to as the cosmic background radiation, correct? Or am I missing something obvious?

  18. Robert

    Good explanation! Thanks for that, very understandable. I’ve always wondered about that. It does make sense, because if there was no “temperature” in space, astronauts – and spacecraft – wouldn’t need heat, correct?

    Robert

  19. Try spending 9 years between stars. All alone… You don’t know the meaning of the word “cold” till you’ve done that.

    Not that I’m bitter.

  20. Torbjörn Larsson, OM

    what is the scientific definition of the highly technical terms “wiggling and jiggling”?

    Well, it’s complicated.

    “the molecules in an object are wiggling and jiggling” ⇒ “wiggling and jiggling” := the amount of freedom of a molecule (velocity, rotation, vibration if a crystal, et cetera, all separable into the three dimensions of space – i.e. three components of velocity et cetera.)

    Now, I get that molecules can wiggle, I’ve seen some such action late at night on big birthday parties, but I dunno if they jigg much.

    will this background radiation ever diminish or disappear?

    Good question, but BA’s post actually answered that already – it is equi-distributed in all of space so it can’t disappear “by leaking outside” [into what, he asked with a gravelly Major Marshmallow Man Voice of Science Fiction Horror Movies], but it diminishes as the universe expands.

  21. Torbjörn Larsson, OM

    but it will only really “disappear” once cosmic acceleration pulls the average distance between CMB photons out beyond the horizon scale.

    Oops, there’s that of course.

  22. SLC

    How about the “temperature” of dark matter in the galaxies and dark energy?

  23. But thermodynamicists have every reason to be upset.

    Very true. I would be upset if I had to do thermodynamics.

  24. Ken B: “So, how close do you have to move the Mini Marshmallow of Science to a start like the Sun before it turns a nice golden brown, so we can make the Mini S’more of Science?”

    The surface of the Moon, 1 AU from the Sun, gets as hot as 120 C.

    Sugar starts to caramelize at 150 C or so. So it would be closer than 1 AU, I’d guess.

  25. oldamateurastronomer

    “what is the scientific definition of the highly technical terms “wiggling and jiggling”?”

    Jello, of course or at least until it assumed the temperature of space and then it would be a jello-pop.

  26. I remember reading somewhere (it might have been Arthur C. Clarke) that if you found yourself floating in space without a spacesuit, the first thing to kill you would not be asphyxiation or freezing, but sunburn (assuming you were not in the shadow of the Earth or Moon).

    Raw unfiltered direct sunlight at 1 A.U. distance can kill you in less than 30 seconds according to A.C.C. Can anyone verify this?

  27. madge

    And you claim to do all your own stunts on this blog! The Mini Marshmallow of Science does all the dangerous stuff. Maybe he should have his own show :)

  28. I dunno Elwood, if you found yourself floating in space without a spacesuit I think the least of your problems would be sunburn.

    I know it is winter down here in Oz but yesterday I went out in the sun for a few minutes and I didn’t even feel warm and we’re about 1 AU from the sun.
    ;-)

  29. Mentioned this in the other thread yesterday but I thought it was interesting enough to repeat but the coldest naturally occurring place in the Universe we’ve found so far is the Boomerang Nebula at about 1 Kelvin.

  30. Jend

    This is unrelated, but I’m usually notified by email when this blog updates, and I haven’t recieved any of those emails lately. I didn’t see them in my spam folder, so I’m wondering what’s going on with that.

  31. TMB

    SLC – dark matter in galaxies and galaxy clusters has a fairly well-defined temperature in the sense of having random velocities that correspond to some “typical” kinetic energy… however, without knowing the mass of the dark matter particle, we can’t calculate it. On the other hand, the gas that comes to hydrostatic equilibrium within those dark matter halos (at least in galaxy clusters) do reach a well defined temperature (for a typical galaxy cluster, that might be 1 million Kelvin), and so one can think of that as the characteristic temperature of the dark matter.

    Dark energy, on the other hand, does not have a well-defined temperature (at least, not for a cosmological constant… maybe someone else can comment on other dark energy models?), since there’s no motion associated with it.

  32. Ryan

    Hey Phil, what’s your take on Apollo 14 astronaut Dr. Edgar Mitchell’s recent “expose” of NASA’s work with aliens? Granted, it did appear in the Daily Mail, which was talking about something he said on a late night radio show, but still, it’s interesting to wonder why he would make those comments.

    http://www.dailymail.co.uk/sciencetech/article-1037471/Apollo-14-astronaut-claims-aliens-HAVE-contact–covered-60-years.html

  33. mmmmm marshmallows.

  34. “Raw unfiltered direct sunlight at 1 A.U. distance can kill you in less than 30 seconds according to A.C.C. Can anyone verify this?”

    Waitaminit – be right back.

    30 seconds later….

    I’m back. Got a good tan. +250F in the sun, -250F in the shade. Funny, put your hand in front of your face and your hand burns while your face freezes. You pass out right away because there’s no oxygen…and you can’t hold your breath in a vacuum. But you may be able to survive the temperature and lack of oxygen for 30 seconds. You could even survive the fizzyness of your blood at vacuum for that long (remember, you will already have passed out). So, it’s either the radiative flux from the sun destroying your cells directly or the -250F freezing your blood, leading to massive cell death. And I’m leaving out numerous other possibilities that I’m sure I have yet to considered.

    Just be sure to use good suntan lotion with a high SPF, like SPF Googol.

  35. HvP

    The Alpha Centaurian said: “Try spending 9 years between stars. All alone… You don’t know the meaning of the word “cold” till you’ve done that.
    Not that I’m bitter.”

    Your name wouldn’t happen to be Marvin would it? You know, ‘brain the size of a planet’ and all that.

  36. Supernova

    @Andrew:

    “Thermodynamics is a funny subject. The first time you go through it, you don’t understand it at all. The second time you go through it, you think you understand it, except for one or two small points. The third time you go through it, you know you don’t understand it, but by that time you are so used to it, it doesn’t bother you any more.”

    — Arnold Sommerfeld

  37. @HvP:
    I won’t deny the “brain the size of a planet” part ;) But no, a depressive android I am not. It’s just a long ass trip to get here!
    And as for my name– well, humans don’t have enough tongues to properly pronounce it. Nor are your ears sensitive to the 30k+ Hz vocalizations required.

  38. So Alpha Centaurian you’re a member of the dog family (30k+ Hz hearing )? ;-)

  39. wadef

    I’m curious, what would a space feel like that had no temperature? Would it feel like 0 degrees Fahrenheit?

  40. Ronn! Blankenship

    # The Alpha Centaurian Says:
    July 23rd, 2008 at 4:43 pm

    Try spending 9 years between stars. All alone… You don’t know the meaning of the word “cold” till you’ve done that.

    Not that I’m bitter.

    – – –

    Good. We’d hate to learn that you were bitterly cold . . .

    *rimshot*

  41. Ronn! Blankenship

    # Sully Says:
    July 23rd, 2008 at 3:33 pm

    Who cares whether the LHC is colder than space, the real question is whether either one is colder than a brass bra on a witch’s. . .

    . . . clothes line?

  42. I wonder how the woo believers will spin this?

    Can you be a moon hoaxer and a believer in little green men? If so they can’t give Dr Mitchell any credibility as an astronaut because to them he is no more than an actor.

  43. Tim G

    Sean Carroll,

    I stand by my original statement. If that marshmallow has a high albedo, say 0.9, then its radiating equilibrium temperature would be 1.5 Kelvins–NOT 2.7 Kelvins . Far more energy would be reflected than radiated.

  44. Tim G, you are simply wrong. That more energy will be reflected is irrelevant; what matters is the amount of energy absorbed compared to the amount radiated. A high albedo object absorbs less, but it also radiates less. The equilibrium temperature is the same.

    Were this not so, you could use the difference in temperature between a low albido “source” and a high albido “sink” to drive a heat engine, deriving work from the thermal background, in contravention of the second law of thermodynamics.

  45. Charles

    Phil, I just love it when you talk about science.

  46. Mike

    Phil, I’ve got a question about the term “subatomic particles” after your second “Well, it’s complicated.” I’m sure there are actually subatomic particles (let’s limit them to electrons, neutrons, protons, and pieces of nuclei–I have no clue how to determine the temperature of a quark or a neutrino) in near-Earth space, but shouldn’t there also be “atomic particles”, i.e., atoms, and maybe even some molecules out there too at measurable density? High vacuum may sound cool, but there’s still plenty of stuff there (relatively speaking).

  47. Ginger Yellow

    “Granted, it did appear in the Daily Mail, which was talking about something he said on a late night radio show, but still, it’s interesting to wonder why he would make those comments.”

    Well, given that it’s the Mail, you might want to check that he actually did.

  48. Tim G

    Daran,

    How does that contradict the second law? You can derive work from a solar cell. The sun’s energy supply is large but not inexhaustible. The power won’t last forever. Wouldn’t the same be true for the background radiation? I don’t see how you cannot derive work from it.

  49. How does that contradict the second law? You can derive work from a solar cell. The sun’s energy supply is large but not inexhaustible. The power won’t last forever. Wouldn’t the same be true for the background radiation? I don’t see how you cannot derive work from it.

    The background radiation will last forever, becoming ever colder as the universe expands. Eventually I suppose it will reach the point where the photon density is less than one photon within the radius of the theoretically visible universe.

    But none of that is relevant. Uniform black body radiation is by definition of maximal entropy. You can only derive work from a temperature difference.

    And that is how solar cells work. Cells need to be cooler than the radiation they harness, which means they have to be able to lose heat to their environment, initially their physical environment – the earth or a spaceship – which ultimately radiates it away toward the colder regions of space. (Technically it is radiated in all directions, but there is only a net loss toward the colder regions of space.)

  50. Or look at this this way, one formulation of the second law is the statement “heat shall not of its own accord move from a colder to a hotter body”. Now suppose you placed your two objects of differing albido, initially at the same positive absolute temperature, inside an completely evacuated, perfectly reflective container of negligible volume at zero K, such that no part of the system is in conductive contact with another part. The two bodies will radiate and absorb, eventually coming into thermal equilibrium, at different temperatures according to you. Heat cannot have come from, or have been lost to the container. Moreover the energy due the photon density within will be negligible because the volume is. The only significant heat transfers are between the bodies, from the colder to the hotter according to you, in violation of the second law.

  51. John Phillips, FCD

    Tim G: but while a high albedo means it is reflecting the majority of the energy striking it it is also absorbing some of that energy, i.e. a net gain. Thus over time it will reach an equilibrium temperature the same as its surroundings. All a high albedo effects is the length of time it will take to absorb enough energy to match the surrounding temperature.

  52. Phil, I’ve got a question about the term “subatomic particles” after your second “Well, it’s complicated.” I’m sure there are actually subatomic particles (let’s limit them to electrons, neutrons, protons, and pieces of nuclei–I have no clue how to determine the temperature of a quark or a neutrino) in near-Earth space, but shouldn’t there also be “atomic particles”, i.e., atoms, and maybe even some molecules out there too at measurable density? High vacuum may sound cool, but there’s still plenty of stuff there (relatively speaking).

    Well, it’s complicated.

    Firstly a subatomic particle cannot have a temperature. Temperature is a statistical property possessed by collections of particles.

    Perhaps an analogy will help to explain how even a pure vacuum (defined as containing no matter) can have a temperature:

    When I was a kid, I didn’t didn’t understand the idea of numbers less than zero. My idea of a number was that it was a measure of how much of something you had, and the least amount of something you could have was to have none of it.

    Then later, I started to think of numbers differently. Instead of thinking of them as how much of something I had, I started to think of them as shifts along a line. Shifts along a line aren’t the same thing as how much of something you could have, but they work the same way, i.e., they obey the same mathematical laws, and of course, you can have a shift along a line in either direction, i.e., both positive and negative numbers.

    The idea of temperature as a measure of how much the particles in a physical body are jiggling about is like the kid’s view of numbers. A more adult (I don’t mean that to be insulting) way to think of it is as a statistical property of collections of particles, any particles, not just the particles which make up a physical body.

    By convention the “temperature of space” refers to the temperature of the photons in a region of space, which may differ from the temperature of the matter particles in the same region. Since matter can interact with photons, the second law guarantees that, unless some process drives them in another direction, the matter particles will eventually come into thermal equilibrium with the photons. Thus if you place a thermometer into space, it will eventually achieve an equilibrium reading – the temperature of space.

  53. Horse

    Of course, space isn’t really empty, it’s only mostly empty.

    Inconceivable!

  54. Tim G

    Thanks Daran and John Phillips.

    I am now convinced that you are correct.

    I think my problem was that I took phenomenon such as reflection-absorbtion and radiation as stand-alone independent factors.

    I will still have to mull this over when I’m refreshed until I understand all issues on a fundamental level.

  55. andyo

    Can anyone explain something for me?

    I thought all particles were the same and didn’t change without converting into another type of particle (or splitting, or merging, whatevah). How is it that photons can lose energy and remain photons?

  56. David Ratnasabapathy

    In Afterglow of Creation Marcus Chown mentions that the Cosmic Background Radiation warms gas clouds in deep space. Years before the CBR was discovered, astronomers measured the clouds’ temperature and wondered why they weren’t stone cold.

  57. Buzz Parsec

    andyo – A photon’s energy is directly proportional to its frequency (or equivalently, inversely proportional to its wavelength.) If it changes its frequency it will change its energy. But here’s the kicker, the *observed* frequency of a photon depends on the velocity of the observer with respect to the source (a blue shift or increased frequency if the observer is approaching the source and a red shift or decreased frequency if the observer is moving away from the source. Since the microwave background is receding from us a high velocity due to the expansion of the universe, the photons, originally representative of a very high temperature (ultraviolet) now appear as those of an extremely cold black body (microwaves.) If you want to know more about this, look up black body radiation, Planck’s constant, the Doppler effect and the good Dr. Einstein. :-)

  58. andyo

    Thanks Buzz, I was about to ask the same question in the more recent post about colors. Didn’t realize you answered here. I have read about those subjects a bit (just layman stuff), but I get it.

    Thanks again.

  59. Anne

    Er, Sean Carrol, I have to argue with you on one point: not everything at thermal equilibrium puts out a blackbody spectrum. It puts out the same amount of power as it absorbs, but the spectrum can look rather different – that’s why you get emission lines. If you have enough gas – if it becomes “optically thick” – it will start to have a blackbody spectrum, but that’s a somewhat different issue.

    In response to the original article, I’d like to point out that space has at least two “temperatures”: there is the temperature of the photons in some fixed volume, which is normally pretty much 2.7 K, but you also have the temperature of the particles, which is often *not* 2.7 K. Considering, say, a cubic AU of “space” as a whole, it doesn’t have a well-defined temperature because it’s not in thermodynamic equilibrium: the photons are not in equilibrium with the particles. If you put it in a perfectly insulating and reflecting container for long enough, the photons and the particles would randomly exchange energy and it would settle down to equilibrium; its temperature would probably be not too far from 2.7 K, since there’s a lot of energy in the CMB.

    In fact, this issue of thermodynamic equilibrium is a tricky one: under most terrestrial conditions systems come to thermodynamic equilibrium very rapidly; it takes moderately heroic measures to move a system out of thermodynamic equilibrium (this is why it’s fairly difficult to build lasers). In “space”, though, it’s very common for systems to be out of thermodynamic equilibrium; in fact we see naturally-occurring masers and lasers from all sorts of strange places. And a system that’s not in thermodynamic equilibrium doesn’t have a well-defined temperature.

    Practically speaking, saying that the temperature of space is 2.7 K may be accurate but is misleading for most purposes. It doesn’t carry away heat like a solid, liquid, or gas (of reasonable density) at 2.7 K does, so you don’t feel cold in the same way. You can still radiate heat, and for those purposes it behaves like a system at 2.7 K – except that there’s the Sun, which produces another radiation field which is much hotter. And if you put an object in space, it will be heated – a lot – by sunlight on one side. So thinking of space as “cold” is a bit misleading. (You can’t really think of it as “hot” either, since the object will lose a lot of heat from the non-illuminated side.)

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