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Negative Kelvin temperature? (Recent Science paper)

  1. Jan 3, 2013 #1
    Chemistry student saying hi to everyone

    Just read http://www.nature.com/news/quantum-gas-goes-below-absolute-zero-1.12146 about negative temperature. I knew I wouldn't understand a single sentence of the actual Science paper, but I took a look at the Science perspective (Lincoln D. Carr) but I didn't understand it either.

    So I understand nothing about physics. But as far as I know, K=0 is defined as when all energy levels (vibrational movements etc) are in their respective ground state? So as little movements as possible. Then I don't really understand how you can get negative temperature.

    If I understand the Perspective (http://www.sciencemag.org/content/339/6115/42.summary if you have access) correctly, at temperatures above 0, particles have a distribution where only a few are in a high-energy state. I get that.
    But what's crazy is that to achieve negative temperatures, they mess with the particles so that MOST are in a high-energy state.
    But then I really don't follow what happens next. K=0 is already as little movement as possible (as there is no excited state), so how could you get below that?

    How would you explain this phenomena to someone with no physic knowledge?

    Thanks!!!
     
  2. jcsd
  3. Jan 3, 2013 #2

    Bill_K

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    Well if you have a collection of atoms or molecules which each have two energy levels Eexc and Egnd, where Eexc > Egnd, and the number of atoms in these states are Nexc and Ngnd respectively, then assuming everything is in thermal equilibrium, the Maxwell distribution says the ratio Nexc/Ngnd = exp-(Eexc - Egnd)/kT.

    Normally Nexc < Ngnd and of course T is positive. But what if the excited state is overpopulated? That is, Nexc > Ngnd. Then for the same formula to work, and give a ratio greater than one, T must be negative.
     
  4. Jan 4, 2013 #3

    mfb

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    In thermodynamics, temperature is defined via the relation between entropy and energy:
    $$\frac{1}{T}=\frac{dS}{dE}$$

    Usually, entropy increases with energy, and temperature is positive. With population inversion (see Bill_K), this can change, and an increased energy corresponds to a lower entropy - which gives a negative temperature.

    A negative temperature is not cold - it is hotter than everything with positive temperature.
     
  5. Jan 4, 2013 #4
    Thanks mfb - I read about this in New Scientist and was confused, but now I see
     
  6. Jan 4, 2013 #5

    f95toli

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    Note that this is not entirely a new concept since population inversion is such a common phenomenon.
    E.g. the noise temperature of a maser based amplifier can -at least in theory- be negative for exactly same reason.
     
  7. Jan 4, 2013 #6

    D H

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    This is getting an undue amount of hype, partly thanks to misleading headlines such as "Atoms Reach Record Temperature, Colder than Absolute Zero." This is just wrong. Negative temperatures are hotter than hot rather than colder than cold.

    The concept of negative temperatures arises from the thermodynamic definition of temperature, [itex]\frac 1 T = \frac{\partial S}{\partial E}[/itex] (see previous posts). It is possible to construct systems where adding energy decreases entropy, making 1/T (and hence T) negative.

    This newest result has received a huge amount of hype. It's not new. It's older than me.

    E. M. Purcell and R. V. Pound, A Nuclear Spin System at Negative Temperature, Phys. Rev. 81, 279 - 280 (1951)
     
  8. Jan 5, 2013 #7
    Baez is always a good route to check up on things to get a better idea of what is going on, I've found: http://math.ucr.edu/home/baez/physics/ParticleAndNuclear/neg_temperature.html

    Now, besides the hype, I don't recall hearing of a large ensemble of atoms being placed in a negative temperature state before, it seems like it IS interesting... but not "WORLD SHATTERING" interesting... more like "so did ya hear about..." type stuff.
     
  9. Jan 5, 2013 #8

    Hurkyl

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    The usual extended number line looks like this:

    [tex](-\infty) --- (-1) --- (0) --- (1) --- (+\infty)[/tex]

    (where I've rescaled the line so that I can draw it in finite space)

    However, for the temperature extended number line, it's organized like this instead:

    [tex](0^+) --- (1) --- (\infty) --- (-1) --- (0^-)[/tex]

    where I've affixed a decoration to indicate the two endpoints are different.
     
  10. Jan 5, 2013 #9
    Great reply! Most people think temperature as movement, but they just exploited a thermodynamic definition :uhh:

    Could you please clarify the lower number line? I have problems understanding it :frown: Are you talking about how hot it is in the temperature number line?
     
  11. Jan 5, 2013 #10

    Hurkyl

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    Both number lines are ordered from smallest to largest. So 1 is a colder temperature than -1.

    *: In the extended sense. I don't think
     
  12. Jan 6, 2013 #11
    It's easier to understand if one looks at the more fundamental beta. Heat flows from lower to higher beta and that handles negative temperatures fine.
     
  13. Jan 7, 2013 #12
    So T=-1 is hotter than T=infinity????
     
  14. Jan 7, 2013 #13

    D H

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    Yes. That's essentially what I said in post #6 that "negative temperatures are hotter than hot."

    Heat flows from an object with a finite negative temperature to an object with a finite positive temperature.
     
    Last edited: Jan 7, 2013
  15. Jan 7, 2013 #14

    mfb

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    @D H:
    I think you mixed positive and negative, otherwise I don't understand your post.

    ##\beta=\frac{1}{T}##-scale:
    (+∞) ... (1) .... (0) ... (-1) ... (-∞)
    Absolute zero . . . . . . . . . . hottest possible object
     
  16. Jan 7, 2013 #15

    D H

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    You're right. I don't know what made me type that exactly backwards, but I obviously did just that. Thanks. I fixed my previous post.
     
  17. Jan 8, 2013 #16
    what was the bit about defying gravity and an analouge for the cosmological constant?

    just more hype or soemthing future researchers in other fields might begin probing?
     
  18. Jan 8, 2013 #17
    For people not versed in statistical mechanics, the statistical mechanics definition of temperature may seem kind of weird. But really, all it is saying is that temperature is basically defined by which way energy (heat) will spontaneously flow if you put two items in contact. It doesn't say anything about the total energy in the items.

    For positive temperatures, energy (heat) will spontaneously flow from a higher temperature to a lower temperature. This is also true for negative temperatures. But if you put a negative temperature item in contact with a positive temperature item, then heat will flow from the former to the latter. Ergo, all negative temperatures are hotter than all positive temperatures.

    Absolute zero is still the coldest of temperatures. With negative temperatures, there is also an Absolute hot, which is the limit of temperature 0 approaching from the negative side. But since it is just a mathematical limit, it isn't something that can be achieved, just like absolute zero.
     
  19. Jan 8, 2013 #18

    I like Serena

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    Okay, so I get that the scale runs a bit weird due to the choice of definition for thermodynamic temperature.

    But I'm still missing something here.
    We already know that physics starts behaving weird when temperature and entropy both approach zero.
    So what happens if they approach zero "from the other side".
    Are there any weird or otherwise unexplained phenomena there?
    Should we expect any?
     
  20. Jan 11, 2013 #19
    Entropy is positive, so it can't approach zero from the other side. Entropy tends to zero as you approach zero temperature from either side. If you plot entropy versus energy for a "finite system" (that is, a system with a maximum energy), the graph looks qualitatively like a parabola or a sine half period. See top plot of attachment. The maximum energy that the system can hold is 1.

    At low energy, the entropy is near 0, because with little energy, there are few ways to arrange the energy. (Think of it this way. You have n energy quanta and m bins to put them into. You have fewer options if n is small.) But at high energy, entropy is also near 0, because the system is filling up with energy, and there are fewer ways to arrange the holes. (Since this is a finite system, you can only stick at most, say, 2 quanta of energy into each bin. Then, when they are almost all filled up, you have few options of ways to arrange the vacancies.) The vacancies or holes behave more or less analogously to energy quanta.

    If you plot temperature versus energy, it looks like bottom plot of attachment. For this example, temperature doesn't ever reach 0. But for a larger system, temperature can get very close to 0, and entropy will get relatively close to 0.

    So, in a system where there is a maximum energy capacity, a low energy (positive temperature) and high energy system (negative temperature) can behave very similarly. But the positive temperature system deals with quanta of energy and the negative temperature system deals with quanta of holes.
     

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  21. Jan 15, 2013 #20
    I can't understand the following from Baez:

    " We define the temperature, T, by 1/T = dS/dE, so that the equilibrium condition becomes the very simple T1 = T2.

    This statistical mechanical definition of temperature does in fact correspond to your intuitive notion of temperature for most systems. So long as dS/dE is always positive, T is always positive. For common situations, like a collection of free particles, or particles in a harmonic oscillator potential, adding energy always increases the number of available microstates, increasingly faster with increasing total energy. So temperature increases with increasing energy, from zero, asymptotically approaching positive infinity as the energy increases. "

    Isnt S(E) a concave function in "common" situations?
    Why does it say "faster with increasing total energy" ? If S increases faster and faster with increasing E, then the derivative dS/dE is larger and larger, then 1/T is larger and larger, than T is smaller and smaller . What am I missing?

    thankx

    Wentu
     
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