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Entropy of 1d harmonic oscillator

  1. Sep 6, 2011 #1
    Hi. I want to write the entropy of a 1d harmonic oscillator as a function of energy, but for each energy there is only one possible configuration. So is the entropy zero? I mean, the energy is E=hw(n+1/2), so there is only one microstate for each energy.
  2. jcsd
  3. Sep 6, 2011 #2
    The entropy of a single harmonic oscillator?
    You may calculate (and define) the entropy of a system of harmonic oscillators. For one single oscillator does not make much sense. Unless you consider the internal states of the particles composing the oscillator. The entropy is defined for thermodynamic systems, right?

    For a system of oscillators, a given total energy may be obtained by various combinations of the individual energies.
  4. Sep 6, 2011 #3
    nasu: Well, you CAN define the partition function of a single harmonic oscillator and calculate the entropy this way. Then you get the entropy as a function of temperature.
  5. Sep 6, 2011 #4
    You need to evaluate the partition function:
    Z = \sum_{n = 0}^{\infty}{e^{-\frac{\hbar \, \omega}{k_{B} \, T} \left(n + \frac{1}{2}\right)}}

    For this, you will need the formula for a geometric series.

    Then, calculate the free energy:
    F = - k_{B} T \, \ln(Z)
    The entropy is given by:
    S = -\frac{\partial F}{\partial T}
  6. Sep 6, 2011 #5
    The energy is defined as:
    E = F + T \, S
    Eliminate the temperature from this and use it in the expression for [itex]S[/itex].
  7. Sep 6, 2011 #6
    Depends what you mean by "CAN". You may formally do it, but is this meaningful?
    How would you associate a (meaningful) temperature with a system composed from a SINGLE harmonic oscillator?
  8. Sep 6, 2011 #7
    Dickfore: I was just wondering, why can't I count the number of microstates and use this to calculate the entropy ?
  9. Sep 6, 2011 #8
    Because the number of microstates is equal to [itex]\Omega(\epsilon) = \int_{H(q, p) \le \epsilon}{\frac{dq \, dp}{2 \pi \hbar}}[/itex] in the semiclassical limit. If you do this integral, then the entropy is approximately:
    S = k_{B} \, \ln{\Omega}
    Last edited: Sep 6, 2011
  10. Sep 6, 2011 #9
    By making the harmonic oscillator in contact with a thermal reservoir at the given temperature.
  11. Sep 6, 2011 #10
    Are you talking about the actual temperature of the oscillator (spring, attached mass)? This will take into account the internal structure of the oscillator.

    I understand that the OP is about a system containing a single entity (one oscillator) with a specific energy state. What does it mean that will be in thermal equilibrium with the reservoir? What parameter of the single oscillator will decide the equilibrium state with the reservoir?
  12. Sep 6, 2011 #11
    A harmonic oscillator is an idealized system that has one degree of freedom. This degree of freedom couples to the heat bath. The internal degrees of freedom of the oscillator are 'frozen'. If you do the calculation, you will see that quantum effects are important in the regime where:
    k_{B} T \ll \hbar \omega
    For macroscopic objects, the r.h.s. is negligible, so we can make the assumption of high temperatures everywhere, even though the actual temperature might be quite low.

    The oscillator natural frequency [itex]\omega[/itex].
  13. Sep 6, 2011 #12

    Ken G

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    Gold Member

    Actually, I think the entropy is indeed zero, of a single harmonic oscillator that is excited to a known energy, but I don't think that is what is being asked, it's kind of pointless. If it is supposed to be in contact with a reservoir, then we can ask what is the entropy of the subsystem that is the harmonic oscillator, but that is a function of T not E. However, what is also a function of T is <E>, the expectation value of the energy of the oscillator. So one might ask, what is the entropy of the oscillator subsystem as a function of its expected energy, given that it is in contact with a reservoir. Perhaps that's equivalent to the answer Dickfore already gave.
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