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Average total energy of 3D harmonic oscillator in thermal equilibrium

  1. Oct 26, 2004 #1
    Hi,
    From knowing that the 3D harmonic oscillator has 3 degrees of freedom, how do you conclude that the average total energy of the oscillator has energy 3kT?
    Thanks,
    Ying
     
  2. jcsd
  3. Oct 26, 2004 #2

    Tide

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    The theorem of Equipartition of Energy says that the total energy of a system is shared equally by all the degrees of freedom. Each degree of freedom has (1/2)kT of energy per molecule.
     
  4. Oct 26, 2004 #3
    How many degrees of freedom does a 3D harmonic oscillator have? 6? so 6*1/2kT = 3kT?

    What exactly is degrees of freedom? How many dof does a 2D harmonic oscillator have?
     
  5. Oct 27, 2004 #4

    Tide

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    I am not sure what you mean by a "3D harmonic oscillator." If you mean a diatomic molecule that behaves like a harmonic oscillator then the number of degrees of freedom will be 6: there are 3 degrees of freedom associated with 3 dimensions of movement and 2 rotational degrees of freedom and one vibrational degree of freedom.

    Perhaps you could be more specific.
     
  6. Oct 27, 2004 #5
    Yes that is what I meant by 3D harmonic oscillator. How many degrees of freedom would a 2D harmonic oscillator have?
    Thanks!
     
  7. Oct 27, 2004 #6

    Tide

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    It would have 2 spatial degrees of freedom, 1 rotational and 1 vibrational.
     
  8. Dec 16, 2004 #7
    Sorry if this seems obvious, but what about a 1D "oscillator"? Is there such thing?
     
  9. Dec 16, 2004 #8

    Tide

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    That would be an oscillator that does not rotate.
     
  10. Dec 17, 2004 #9
    In this context, "degree of freedom" means a unique way for the system to increase its kinetic energy. The 3D harmonic oscillator has six degrees of freedom. The energy depends on the three components of position and of momentum. So it is 6 for the oscillator and 3 for a free particle.

    Hence the Dulong&Petit law for the specific heat of solids.
     
  11. Nov 3, 2008 #10
    There are 2 degrees of freedom associated with a one-dimensional harmonic oscilator - one for the potential energy (U=½kx^2) and one for the kinetic energy (½mv^2) (*) associated with the oscilation. Thus for a 3-dimensional oscilation we have 2x3=6 degrees of freedom!

    - Trolle

    (*): NOT the translational motion of the oscilator as a whole
     
  12. Sep 18, 2009 #11
    Trolle's explanation for the number of degrees of freedom of a 3D harmonic oscillator seems to be different from Tide's.

    Tide says that in 3D, you have 3 translational, 2 rotational, and 1 vibrational degree of freedom. 3+2+1=6

    Trolle says that in 3D, you have 1 potential degree and 1 kinetic degree for each dimension. 2x3=6

    These answers seem quite distinct from each other, so I ask: Is both their reasoning correct? And if so, could someone explain how both these explanations account for the same answer. For example, if they were both right, how would one account for the single vibrational degree of freedom mentioned by Tide, using Trolle's 3 pairs of degrees?

    Thanks!
     
  13. Sep 19, 2009 #12
    I believe we Tide and I are discussing two different topics.

    In general, a degree of freedom is an independent quantity, which must be specified to determine the total energy of a molecule.

    If the molecule is modelled as a particle, it is without any moment of inertia, and thus without any rotational energy. Furthermore, no vibration is expected (obviously) - we are thus only left with the kinetic energy term, which depends upon the three velocity components of the velocity vector, thus giving three degrees of freedom.

    If we model a diatomic molecule as two atoms connected by a spring-like potential we still have our three degrees of freedom as before, due to the translational motion of the center of mass of our molecule. In addition the moment of inertia is now different from zero along two rotational axes, giving two additional degrees of freedom (we need to know two components of angular velocity vector to determine the total energy of the molecule). In addition we have the spring-like potential, which gives two additional degrees of freedom (the distance between the two atoms is needed to specify the potential energy conserved in the spring, additionally the vibrational velocity is needed to specify the kinetic energy of the oscillator). This results in 3+2+2 = 7 degrees of freedom
    [Sears & Sallinger; Thermodynamics, Kinetic Theory, and Statistical Mechanics].

    Mathemathically the total energy [tex]H[/tex] is

    [tex] H = \frac{1}{2}m (v_x^2 + v_y^2 + v_z^2) + \frac{1}{2} I (\omega_x^2 + \omega_y^2) + \frac{1}{2}kd^2 + mv_{vib}^2, [/tex]
    the [tex]v[/tex]'s refering to velocities, the [tex]\omega[/tex]'s to angular velocity and [tex]d[/tex] to the distance betweem the atoms.
    The equation is clearly seen to contain 7 independent terms giving 7 degrees of freedom...

    - Trolle
     
  14. Sep 19, 2009 #13
    The 3 degrees of freedom contribute an energy of k T, not 1/2 kT, because for each degree of freedom, you have two quadratic terms in the Hamiltonian, a kinetic energy term and a potential energy term. It is wrong to say that there are 6 degrees of freedom.
     
  15. Sep 19, 2009 #14
    Trolle is not correct, just consider what happens to his argument if you let the strength of the harmonic potential tend to zero. Also, consider that in real gasses there is always an interaction potential between molecules. The concept of "degree of freedom" does not depend on such interactions.

    Indeed the notion of degrees of freedom is a useful tool precisely because you can consider a bound state of N particles to have 3 N independend degrees of freedom. So, the bound state has 3 center of mass degrees of freedom and 3 rotational degrees of freedom, so if there are X vibrational degrees of freedom, then:

    6 + X = 3 N --------->

    X = 3 N - 6

    And at room temperature (or significantly above the Debye temperature) these X degrees of freedom will contrubute k_b, not 1/2 k_b, to the heat capacity.
     
  16. Sep 28, 2009 #15
    sorry but I fail to understand: a molecule and a 3d harmonic oscillator are two different things: the classic 1dimensional harmonic oscillator has a potential equal to[tex] kx/2[/tex], for a 3d harmonic oscillator in cartesian coordinates the potential is a function of the distance from the origin so
    [tex]k(x^2+y^2+z^2)/2[/tex]
    this can be used to model many things, atoms in a lattice and so on, it has 3 degrees of freedom, but can store energy in both kinetic and potential energy, so it's mean energy is not [tex]3/2KT[/tex] but [tex]3KT[/tex], it does not have rotational or vibrational energy of course. polyatomic gases are a different story.
     
  17. Sep 29, 2009 #16
    I count seven degrees of freedom in this case: the vibrational energy has kinetic and potential.
     
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