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Validating Third Law of Thermodynamics

  1. Aug 15, 2012 #1
    1) Does anyone have a general idea of how to experimentally validate or falsify the Third Law of Thermodynamics in its strong form?
    In another thread, I suggested a procedure for validating the third law of thermodynamics in its weak form. Basically, one determines the specific heat in the limit as the temperature goes to absolute zero. However, this could not validate the strong form of the third law.
    Just as a review, the Third Law of Thermodynamics in it weak form is:
    0 = lim[T→0] ∂S(T, ...)/∂T.
    Basically, this says that the specific heat of the system approaches 0 in the limit of the temperature approaching absolute zero.
    where T is absolute temperature of a system, and S is the entropy of the system.
    The Third Law of Thermodynamics in its strong form is:
    0= lim[T→0] S(T, ...)
    and,
    0 = lim[T→0] ∂S(T, ...)/∂T.
    The second equation is just the weak form of the third law. However, the first equation makes the third law even more specific.

    2) Has anyone done any experiments validating the Third Law of Thermodynamics in either its weak or strong form?
    -Perhaps someone has a citation and reference!
    3) Does the Third Law in either form put a limit on thermal conductivity?
     
    Last edited: Aug 15, 2012
  2. jcsd
  3. Aug 15, 2012 #2

    Simon Bridge

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    You mean that the entropy (of a perfect crystal) is zero at absolute zero?

    It is physically equivalent to the statement: "It is impossible by any procedure, no matter how idealized, to reduce the temperature of any system to zero temperature in a finite number of finite operations" ... so any experiment that achieves this would falsify the third law. A lit review for the coldest temperature reached should give you a lot of examples of attempts. People can be more subtle than that (see below)

    Also you can leverage the 2nd law in an experiment ... from the following lecture:
    http://www.scribd.com/doc/49208432/26/Various-statements-of-third-law
    ... looking for a better writeup.

    you'll find sections on experimental verification in textbooks, eg.
    http://www1.tecnun.es/biblioteca/2009/feb/libmtr1.pdf ch6.6
    ... google books version ... details the same experiment.
     
  4. Aug 15, 2012 #3
    No. I think that I am asking whether the entropy of an amorphous material can be zero at absolute zero temperature.
    I did some more reading since I posted the question. The strong form of the Third Law implies that the ground state of the system is degenerate. The strong form of the Third Law is broken if there is more than one state of minimum energy.
    For instance, suppose that there are two distinct crystalline forms that have identical energies. As a concrete example, consider hexagonal close packed crystal and a face center cubic close packed crystal. If the binding forces are isotropic, both types of crystal will have identical energies. According to the strong form, both crystals will have zero entropy at zero temperature.
    Now, make a mixture of these two distinct crystalline forms. Make a random mixture of these two forms of crystal. Now you have an amorphous material. This amorphous material will not violate the weak form of the Third Law of Thermodynamics.
    This amorphous material will violate the strong form of the Third Law of Thermodynamics. You can vary the relative proportions of the two types of crystal at absolute zero and get different levels of disorder. Therefore, you can get different values for entropy. However, the strong form says that there can be only one value of entropy at absolute zero temperature.
    Maybe there is an amorphous state that has less entropy than the equivalent crystal state! Maybe there is a liquid state that reaches absolute zero entropy at absolute zero temperature!
    Another way to say it is this. Is there any physical state other than a perfect crystal that has no more entropy than a perfect crystal?

    That is actually equivalent to the weak form of the Third Law of Thermodynamics. If the weak form of the Third Law is true, then the specific heat of a material approaches zero as the absolute temperature approaches zero. In that case, it is impossible from a positive temperature to reach absolute zero in a finite number of steps. So what you just said was the weak form of the Third Law.
    I said that I have not problem with the weak form of the Third Law. I am asking about the strong form of the Third Law.

    I did some additional reading. I now realize that I am asking about degeneracy of ground states. If there is no degeneracy in the ground state, the strong form is satisfied. If there is the ground state is degenerate, then the strong form is not satisfied. In either case, the weak form can be satisfied.
    1) Is there some test, preferably using calorimetry, which can test the degeneracy of a ground state?
    2) What are some of the important experimental consequences if the ground state is not degenerate?
     
  5. Aug 16, 2012 #4

    Simon Bridge

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    I don't think that is predicted by the third law. Or ...
    ... which would mean that the experimental evidence would show that the "strong form" as you appear to have defined it is false.

    You don't have to believe the actual experimental evidence or even have the same definitions as everyone else.

    Did you read the reference I gave you?

    eg. (2) the entropy tends to a constant as T tends to 0.
     
  6. Aug 16, 2012 #5
    However, the constant doesn’t have to be universal. The weak form of the Third Law implies that the system has to approach a ground state as the temperature goes to zero. However, if there is more than one ground state then the system can go into any one of them as the temperature goes to zero. The details of cooling will determine which ground state the system goes into.
    I did read the links that everyone gave me. None of the links addressed the possibility of degeneracy of the ground state. Some of the links that I posted made broad statements about the effect of degeneracy of a ground state. If you gave me a reference that wasn't a link, then I didn't read it. I don't buy textbooks unless I think they can explain some things better than the ones that I have. However, I did look through my library and find some textbooks that were somewhat helpful. I can't give a link, but I can give a reference.
    I found a textbook on thermodynamics which goes many of the details left out of other books. The book is,
    “Thermodynamics: Foundations and Applications” by Elias P. Gyftopoulos and Gian Paolo Berreta (Dover, 2005).
    I am reading it now. However, the book goes into detail concerning the possible consequences of a degenerate energy state. It may take me a little time to absorb it all. However, maybe I can explain what the problem is with degeneracy right now.
    On page 139 (section 9.9), it quotes the Third Law:
    “For each given set of values of the amounts of constituents and the parameters of a system, there exists one stable equilibrium state with zero temperature.”
    You are claiming that this is the only accepted version of the Third Law. I believe this definition is what other authors call the “weak form” of the third law. You are not considering the idea that there may be more than one ground energy state. I want to know what this multiplicity of ground states would do to thermodynamics.

    However, the authors have a later section on the third law (pages 201-203). This chapter (section 13.17) restates the second and third laws to be consistent with the possibility that there may be degeneracy in “the ground state” of the system.
    On page 201 (section 13.17), the Third Law is restated as follows:
    “To resolve this question without resorting to the formalism of quantum theory, we introduce the third law. It asserts that for each given set of values of the amounts of constituents and the parameters, the ground-energy stable equilibrium state has zero temperature (Section 9.9).”
    Consistent with these statements of the third law is the following. The ground-energy stable equilibrium state may vary with the constituents and parameters. By definition, each of these different ground-energy stable equilibrium states have the same energy. However, these different ground-energy stable equilibrium states may have different values for entropy.
    The book shows on page 203 a plot of energy versus entropy. The slope of this plot is the temperature. However, the hypothesis of this plot is that the ground state is degenerate. Therefore, there is a range of entropy between 0 and S_g>0 in which energy of the state doesn’t change. These are all ground states and they are all at zero absolute temperature!
    The discussion in this chapter is rather abstract. I am just wondering whether there is any experimental system with a degenerate ground state that has been investigated at colder temperatures. I am thinking that a ferromagnetic "perfect crystal" may be just such a system, or at least approximate it in some way.
    Or maybe not! Maybe there is no material with a degenerate ground state. That is what I am thinking. Maybe there is a rule that forbids ground states that are degenerate. Maybe as you get close to absolute zero, some type of symmetry breaking mechanism raises the energy of all states except the one and only ground state. I don’t know how one can prove such a statement.
    I think that is what the strong form of the Third Law implies. The Third Law in its strong form implies that there is no degeneracy in the ground state under any circumstances. It implies that there is one and only one state with a lowest energy, and that state is also the one with the lowest entropy at absolute zero. Then the entropy of a system at absolute zero in a crystal can be taken as a universal constant, which can be taken as zero.
    I just want to know what some of the experimental thermodynamics consequences would be if the ground energy state of a system is degenerate. I am wondering if anyone did any studies taking known degenerate states to very low absolute temperatures.
     
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