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0 K and S_max

  1. Jul 13, 2004 #1

    turin

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    Is there a theoretical maximum entorpy? I just read on Wikipedia that there are certain conditions in which the absolute temperature can be negative, based on the definition: T-1 = dS/dE. If the entropy has nowhere to go but down (from Smax), can temperature still be defined?
     
  2. jcsd
  3. Jul 14, 2004 #2
    Temperature has not a top limit... so entropy has not a top limit as well ...

    Nernst's enunciate of the 3rd principle says that the limit as T goes to 0 of entropy is 0...

    Lets take the first principle: [tex]dE = TdS + y_{i} dx_{i}[/tex]

    T is defined as:

    [tex] T = \frac{ \partial E}{\partial S} [/tex] and the variation of energy with entropy is usually positive.

    If "E" is the ocupation of energy levels, and "S" is disorder...

    [tex] \frac{1}{T} = \frac{ \partial S}{\partial E} > 0 [/tex]

    If we have a system with a finite number of energy levels, if E increases the order increases as well and entropy decreases, so that partial is < 0 and T < 0 !

    That can occur for example in a population inversion:

    I can draw a little scheme:

    E2:
    E1: oooooo

    T = 0, S = min

    E2: o
    E1: ooooo

    T > 0, S > 0

    E2: ooo
    E1: ooo

    T = ± infinity
    S = max

    E2: ooooo
    E1: o

    E increases
    S decreases
    so T < 0

    Negative temperatures are over infinite temperature.

    Bye

    -

    I forgot to say that this can only occur in unstable systems, and a macroscopic object can not be at a negative temperature, only can occur in a few particles...
     
    Last edited: Jul 14, 2004
  4. Jul 14, 2004 #3

    Clausius2

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    This puzzled me much time ago. I have read a lot of books talking about what migui has explained about inverting population.

    My question is: Are negative temperatures "hotter" than positives ones?. I mean, does it have smaller levels of thermal energy?. I was wondering if it would be employable in classical engines or something like that. All of you know what happens if we substitute for negative temperatures in the Carnot efficiency or in Kelvin relation. Engines would generate work without any energy added, and freezers would produce mechanical work. The classical thermodynamics fall down automatically with negative temperatures. So what happens with this contradiction?.
     
  5. Jul 14, 2004 #4

    turin

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    How can T = (+/-)infinity at that midpoint in in the 6 particle example? Shouldn't T = (+/-)infinity correspond to a change in entropy requiring an infinite change in energy (which it doesn't seem to in the 6 particle example)? Do I have to extend your example to an infinite number of particles (or a continuum) so that the inflection point has perfectly zero slope?

    I find that Smax = S3,3 = kB ln(6!/(3!)2) = 3.00 kB

    but

    S2,4 = S4,2 = kB ln(6!/2!4!) = 2.71 kB

    (It's been a while, so please correct me if I'm disrespecting the entropy calculation)

    So, the way I'm seeing it (in this discrete example, at least), the temperature is not infinite, it is just not defined, because it should equal a positive and negative finite number according to the definition (depending on the direction of ΔE). In that sense, I would expect the limit, rather than approaching some infinite value, to simply be undefined (even mathematically).

    I guess I am wondering if this is really the correct (sufficient) way to define temperature:

    T = (∂S/∂E)-1

    since it doesn't seem to be definable in this manner in all cases. Are there other parts of the definition besides this mathematical statement that I have left out?
     
    Last edited: Jul 14, 2004
  6. Jul 15, 2004 #5
    "Are negative temperatures "hotter" than positives ones?".

    Yes
     
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