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Negative temperature and the amazing power of SM

  1. May 1, 2010 #1
    Imagine an ideal two-state paramagnet. I mean a solid consisting of miniscule magnetic dipoles which can only align upward or downward (I have no idea why this latter restriction turns out to be physically feasible -- I am aware the spin of an electron is either up or down, but I was under the impression that the atom bearing the electron can rotate whichever way, but I haven't had any quantum mechanics yet, so this will not be my question). Say the material consists of N dipoles (and I think there is the unmentioned assumption that other intermolecular interactions are negligble). According to the Statistical Mechanical definiton of Entropy:

    [tex]S = k ln\left( \stackrel{N}{N_{\uparrow}} \right)[/tex] where [tex]N\uparrow[/tex] is the total number of dipoles pointing up.

    From electromagnetism, we can say [tex]E = - B\mu (N\uparrow)[/tex] if we say B is pointing up.

    Now using [tex]1/T = \frac{\partial S}{\partial E} [/tex] and a Sterling approximation (or more rigidly using a Boltzmann distribution) we get:

    [tex]\frac{N\uparrow}{N} = \frac{1}{2} \left( 1 + tanh \left( \frac{\mu B}{kT} \right) \right)[/tex] (the details aren't important for this discussion, the main fact is that this formula follows from the stated assumptions/details of the system according to SM)

    Does this mean that we can know the fraction of the dipoles pointing upward simply by measuring the temperature? That seems quite amazing!

    So say we surround this ideal paramagnet with an ideal gas at 300K. Does it matter that in essence the temperature of the resevoir and the system are of a different species? I mean, the gas-temperature is a measure for the average kinetic energy, while that of the paramagnet is a measure of (positive, B-wise) magnetization. But SM-wise, they are defined by the same principle, namely [tex]\frac{\partial S}{\partial E}[/tex] so the gas and the paramagnet reach equilibrium and you know right away the fraction of dipoles pointing up? (taking the ideal gas to be a reservoir, thus having constant T)

    Now a very interesting thing
    , is that the temperature of the magnet is NEGATIVE for the macrostates where more than half of the dipoles are pointing downward/against B. Is this simply a sort of (logical) statement that we can never (no matter how hot our reservoir is) make more than half of the dipoles point against the upward magnetic field B by heating the magnet (because something at a negative temperature, interpreting it as the derivate mentioned earlier, will always give off energy to anything at a positive temperature, i.e. the reservoir)?

    Now another very interesting thing, is that this derivation indeed turns out to be physically feasible, as it is a sort of "ideal proof" for Curie's law (or a prediction thereof, anyway) which states that the magnetization of a paramagnetic solid (at high T, low 1/T) is proportional to 1/T. But it seems odd: in a physical paramagnet you can imagine the gas molecules (from the reservoir) bumping into the dipoles, making them vibrate more so they give less of a contribution to the total upward magnetization. But in this ideal model, the dipoles are immobile, they can only switch up or down, yet somehow bumping gas molecules deliver work on them. Does this simply mean my intuitive idea of a real paramagnet is wrong and that Curie's law is much more quantummechanical than I think?

    Open to all reactions,
    mr. vodka
    Last edited: May 2, 2010
  2. jcsd
  3. May 2, 2010 #2
    Yes, you cannot get the negative temperatures with a "conventional" heat reservoir. You need laser pumping.
    Btw, which way does heat flow between negative and positive temperatures?
    What has to be true for the density of states, so that the total internal energy is proportional to temperature?

    The fundamental reason is that there is energy transfer between vibration energy and spin energy. The act of vibration isn't strictly necessary. In the ideal model too, we assume that there is some way that energy is transfered between vibration and spin energy. Basically whenever you have unconstrained energy transfer, temperatures equilibrate.
    Some quantum mechanics comes in when you consider discrete spin states instead of continuous directions. But that's a different issue.
  4. May 2, 2010 #3

    Andy Resnick

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    I think many of us have had this same reaction when seeing a SM description of a two-level system for the first time (for me it was in the context of lasers).

    Unfortunately, rather than recognize that the assumptions of SM don't hold in this case (that is, a two-state *isolated* system with an inverted population is not an equilibrium state), the student is presented with this "amazing" idea that seems to lie outside of thermodynamics.

    The truth is, the (absolute) temperature of a system T cannot be negative. If anyone is convinced otherwise, then please explain why heat flow does not occur- or explain why Fourier's law of heat conduction should not apply, either.

  5. May 2, 2010 #4
    It is well an equilibrium. If you place a small spin system in contact with a large system at negative temperature, too, then you will find they stay in equilibrium.

    The reservoir is completely fine being at negativ temperature. Just think there are N spins, but so much energy that most spins are bound to be in the highest energy states.
    Take an isolated system, 10 spins, 20 energy and two states with energy 1 and energy 2. Is this reservoir going to be at negativ temperature and static over time, i.e. equilibrium? Right?

    A similar spin system in contact with it will also be at negativ temperature, right?

    Oh, it does occur. Place an inverted spin population in contact with a common reservoir and you will find the the spins go back to a positive temperature. They lose energy and this is exactly heat! I suspect that you are not aware that spins also have "heat" in the general sense.
    There is a reason that you find articles about negativ temperature everywhere. It only doesn't exist to those people who think temperature is restricted to molecule motion.
  6. May 2, 2010 #5

    Andy Resnick

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    Really? Can you show this, because it doesn't make sense to me. Especially the part about having a thermal reservoir at negative absolute temperature- I have never heard of such a thing existing.
  7. May 2, 2010 #6


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    As long as you can set up a system such that the spin degrees of freedom interact with one another on a timescale faster than they interact with the other degrees of freedom in the system then the subsystem of spins is "isolated" on such timescales and can be shown experimentally to have a negative temperature as predicted by statistical mechanics. You can call this an effective temperature if you want, but the predictions of statistical mechanics within this subsystem are realized. See the following references:

    N.F. Ramsey, "Thermodynamics and statistical mechanics at negative absolute temperature," Phys. Rev. 103, 20 (1956) Describes the theory being spins systems in which negative temperatures may be realized.

    A. S. Oja and O. V. Lounasmaa, "Nuclear magnetic ordering in simple metals at positive and negative nanokelvin temperatures", Rev. Mod. Phys. 69, 1 - 136 (1997) A lengthy review article which discusses negative temperatures in it, including experimental realizations

    E. M. Purcell and R. V. Pound, "A Nuclear Spin System at Negative Temperature", Phys. Rev. 81, 279 - 280 (1951) The first, according to the above review article, realization of negative temperature in a nuclear spin system.
  8. May 2, 2010 #7
    That's why I gave an example which you have skipped. Any finite system with enough energy has a negative temperature. An inverted population is usually a negative temperature. So if I pump into a system of 10 spins 20 energy units, where the spin system has level energy 1 and energy 2, then you will necessarily have a negative temperature, when this system is kept as an isolated reservoir. All spins will be in the higher energy state.
  9. May 2, 2010 #8

    Andy Resnick

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    Is that true? I think, if you take the limit of T as E -> +inf, starting from an initially positive T_0, T tends to +inf.

    Is this incorrect?

    If I am correct, then T must pass from +inf to -inf without passing back though 0. Is that physical? Does that agree with our experience?

    Now *that's* an interesting thing to say- if you prepare a 2-state system with an inverted population, then isolate it from the environment, does it remain in that state? I think it does, unfortunately (by definition -the system is isolated), we can't measure it to see.
  10. May 2, 2010 #9
    It's probably hard to isolate a simple finite state system from the usual vibrations. That's why it's a challenge to create an negative temperature system, I guess.
    But of course temperature applies to any abstract system and if you manage to take a real finite system, then you can have negative temperatures.

    I'm not sure how to analyze your statements yet, but I wish to point out one thing. The parameter to consider for continuous transforms is [tex]\beta=1/T[/tex]. In fact the second law can be written as [tex]\operatorname{sgn}Q_{1\to 2}=\operatorname{sgn}(\beta_2-\beta_1)[/tex].
    Only for positive temperatures this reduces to the statement that heat flow from higher temperature to lower. That explains why negative temperatures are hotter than any positive temperature.
    I believe considering [itex]\beta[/itex] resolves the confusion?!
  11. May 2, 2010 #10

    Andy Resnick

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    That's not what I am objecting to. Statistical mechanics is a perfectly fine theory.

    But 'temperature' (and 'heat') are concepts that lie outside of mechanics. Those (and other, related ones) cannot be satisfactorily explained using a mechanical theory.

    In this view, 'negative temperatures' are a non-physical result of extrapolating a theory beyond it's particular domain of applicability.

    But, thanks for the references. Ramsey's paper is quite good.
  12. May 2, 2010 #11
    I just meant that T=+0 to T=+inf to T=-inf to T=-0 corresponds naturally to 1/T=+inf to 1/T=+0 to 1/T=-0 to T=-inf which is nice.
  13. May 2, 2010 #12

    Andy Resnick

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    That's not true at all- you probably have several devices nearly that display this behavior. Every laser (especially semiconductor lasers) operates on the principle of population inversion.

    When I turn on one of my lasers, I see no behavior consistent with part of the device possessing an infinite temperature (or energy, for that matter).
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