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Does the 2nd law of thermodynamics need to be postulated?

  1. Dec 4, 2008 #1
    Taking [tex] S=kln\Omega [/tex] to define entropy, it follows immediately that [tex] \frac{dS}{dt} \geq 0 [/tex] holds as a statisical law so long as we assume equal a priori probabilities for each microstate. It seems that so long as we use the Bayesian interpretation of probability, the second law does not need to be postulated but rather can be proved.

    Also, would it be possible to prove Carnot's theorem in the following way: The area enclosed in a P,V graph for a material undergoing a cyclic process between heat reservoirs of given temperatures tells us the work done in the cycle, from which we can calculate the efficiency. Couldn't we set up a functional which tells us the efficiency of a given path in the P,V diagram, and maximise it using functional differentiation to tell us the most efficient path? And then from this couldn't the second law be deduced, since carnot's theorem is a statement of the second law?
  2. jcsd
  3. Dec 5, 2008 #2
    Im fairly ignorant about thermodynamics. (Something about it isn't very attractive...)

    So I chose a variable s=-t.

    [tex] \frac{dS}{ds} \geq 0 [/tex]


    [tex] \frac{dS}{dt} \leq 0 [/tex]

    Did I make any blunders?
  4. Dec 5, 2008 #3
    phrak, as Feynmann notes out, it is the entropy that makes time to move forward :)
    Last edited: Dec 6, 2008
  5. Dec 6, 2008 #4
    No nothing is wrong with that phrak, unless youre trying to claim that your reasoning invalidates the second law. You just did a change of variables. If s is now your time variable then sure entropy goes in the other direction, but then again so does everything else. You're just choosing the opposite direction for the arrow of time.
  6. Dec 6, 2008 #5
    I don't understand your reasoning then. Is omega dependent upon t?
    Last edited: Dec 6, 2008
  7. Dec 6, 2008 #6


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    Why does entropy increase in your interpretation? Shouldn't it reach its maximum value the moment you assume equal a priori probabilities?
  8. Dec 7, 2008 #7
    Omega is dependent on t. It is the number of microstates that correspond to a macrostate with given macroscopic variables. Each microstate has the same probability, but there is a small range of macrostates which have a vastly greater number of microstates associated with them in any other. Hence by statistics alone, the system will tend to evolve to and stay in such a macrostate. Once is it there entropy has reached a maximum, hence the greater than OR equal to sign. But the system can revert to a macrostate of lesser weight, thus entropy would decrease. This is just overwhelmingly unlikely to happen - the second law is a statistical law.
  9. Dec 8, 2008 #8
    I'm sorry, that's much to confusing to follow. Microstates have microstates?
  10. Dec 8, 2008 #9
    A microstate is a complete specification of the system on a microscopic level ie the position and velocity of all the particles in the system. A macrostate is a partial desription of the system using bulk variables like temperature, pressure, etc. There are in general many different possible microscopic arrangements that lead to the same macrostate. For example if my system consisted of two particles each of which could exist at one of two energy level, E=0,A then there would be two microstates corresponding to total energy=A; one where the first particles has energy A and the other has 0, and one where first particle has 0 and the other particle has A. As the number of particles and the energy of the system goes up, the number of microstates corresponding to a macrostate becomes vast.
  11. Dec 18, 2008 #10
    I guess the question becomes, why does a system move from a less probable state to a more probable one, instead of just staying where it is? The concept of probability does not include the passage of time; so maybe that's why the 2nd Law of Thermodynamics is a necessary postulate. Probability is purely mathematical, but time is something physical.
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