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I Symmetry on the partition function

  1. Jun 4, 2017 #1
    We have a partition function

    ## \displaystyle Z=\frac{1}{N! \: h^{f}} \int dq\: dp \:e^{-\beta H(q,p)} ##

    And we obtain, for example, the pressure by ##\displaystyle p = \frac{1}{\beta} \frac{\partial\: \ln Z}{\partial V}##. So if we do the transformation ##Z \rightarrow a Z## where ##a >0## we obtain the same pressure. This is not a symmetry? It should exist a conserved quantity?
     
  2. jcsd
  3. Jun 6, 2017 #2

    Twigg

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    The transformation ##Z \rightarrow aZ## simply takes every Boltzmann factor and multiplies it by ##a##, which is equivalent to adding a constant amount of energy ##\frac{1}{\beta}ln(a)## to the energy of each state. Shifting every energy up or down consistently won't change any observables. There's no reason why that would result in any conservation laws. If you're referring to Noether's theorem, the Hamiltonian of the system isn't invariant and the partition function is a dynamical variable. Hope that helps!
     
  4. Jun 6, 2017 #3
    Thank you!
     
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