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Homework Help: Entropy physics homework

  1. Mar 12, 2006 #1
    Ok i thought i had this under control, but now i come to think about it, i get more confused. I know d sigma/d U for any system must be positive because the temperature must be positive (sigma is the entropy and U is the energy). But what about this? Consider the model of the paramagnetic system where theres a constant N particles in a constant volume. The energy is proportional to how many spin down - spin up particles there is. So lets say for system 1, N= 30. 29 are spin down, 1 is spin up (Lets just say energy is 28). Lets say this system comes in thermal contact with another system (system 2). (the systems are allowed to exchange energy). So initially the entropy of the 1st system is 30 choose 1 = 30. Now lets say system 1 exchanges its spin up for spin down (Change in energy is +2). But now entropy is 30 choose 0 = 1. So change is entropy is -29. so change is entropy wrt to change in energy is negative? And temperature which is 1 over that is negative?
    Last edited: Mar 12, 2006
  2. jcsd
  3. Mar 12, 2006 #2

    Physics Monkey

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    Don't worry! As it turns out, negative temperature is a perfectly reasonable thing that has been realized in the laboratory in nuclear spin systems. Negative temperature is possible provided three conditions are met. First, the system must have bounded energy. Second, the system must be able to reach internal equilibrium quickly. Third, the system must be isolated (i.e. much longer relaxation time) from systems not obeying the first two conditions. Basically, systems with negative temperature are "hotter than infinity" since they will give up energy to any system with positive temperature. The standard example consists of a spin system where more than half the spins point anti-parallel to the applied field: as you increase the energy of the system, fewer and fewer microstates are available. Try googling for an article by Ramsey from the 50's where he talks about thermodynamics with negative temperature.

    Hope this helps.
    Last edited: Mar 12, 2006
  4. Mar 12, 2006 #3
    wow didnt know. Im gonna have to read up about that, but since i have a test tomorrow, im just trying to understand the more basic cases. Like for the example I gave, the multiplicity function is N choose how many up spins there is, and each up spins means negative energy. So doesnt this concept of negative temperature apply to anything that satisfies those properties (i.e, Multiplicity function g(N,k) N!/(k!(N-k!)) where k is proportional to -U? Or is the case i presented unrealistic because a system with so many down spins will never accept to switch for more down spins, it will instead switch a down spin to gain an upspin so that it can increase the joint multiplicity of the joint system(1+2)?

    Edit: actually even if it switches its down spin to gain an upspin, Change in U will be negative and change in entropy will be positive, so its seems like its a general property of any system will the given type of multiplicity function regardless of how it switches its spins. Am i right?
    Last edited: Mar 12, 2006
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