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Derivation of ideal gas law by Hamiltonian mechanics

  1. Jun 6, 2013 #1

    I am trying to understand the statistical mechanics derivation of the ideal gas law shown at: http://en.wikipedia.org/wiki/Ideal_gas_law inder "Derivations".

    First of all, the statement "Then the time average momentum of the particle is:
    [itex] \langle \mathbf{q} \cdot \mathbf{F} \rangle= ... =-3k_BT.[/itex]". Isn't this wrong? For sure, the dimension of [itex] \langle \mathbf{q} \cdot \mathbf{F} [/itex] is energy, as is the dimension of [itex]-3k_BT[/itex] and not momentum.

    Secondly, I do not understand the equation
    [itex] -\langle \sum_{i=1}^{N}\mathbf{q}_k \cdot \mathbf{F}_k \rangle = P \oint_{\mathrm{surface}} \mathbf{q} \cdot d\mathbf{S} [/itex].
    I dot not understand what [itex]\mathbf{q} [/itex] is, since all position vectors have been subsricpted. Nor do I understand the physical interpretation of [itex] \mathbf{q} \cdot d\mathbf{S}[/itex]: the scalar product of [itex] \mathbf{q} [/itex] (a position vector?) and the vector area element.

    If anybody could shed some light on this, I would be very grateful.
  2. jcsd
  3. Jun 7, 2013 #2

    Andrew Mason

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    I agree with you there. It should be "time average kinetic energy of the particle". You should post a talk comment in the Wikipedia article.

    Pressure is the average force/unit area of the gas. So the time average of the force created by a single particle on the container is F = PΔS where ΔS is the area of the container wall that is interacting with the particle (ie. in a collision between the particle and the wall). So [itex]\oint PdS[/itex] is the total force on all the particles of the gas.

    Where it gets a little unclear is setting the total energy to P∫qds. This does not explain the physics. A better explanation can be found on the Wikipedia page on Kinetic theory.

  4. Jun 8, 2013 #3
    Thank you for your reply!

    I have been discussing this on another forum as well. There it was explained to me that q and F arise from approximation of the discrete model with a continuous one, so that the derivation with additional steps, would look something like
    [itex] -\langle \sum_{k=1}^{N} \mathbf{q}_k \cdot \mathbf{F}_k \rangle \approx \int_{\mathrm{volume}}\mathbf{q}\cdot \mathbf{F}dV=
    \oint_{\mathrm{surface}}\mathbf{q}\cdot (Pd\mathbf{S})=P\oint_{\mathrm{surface}}\mathbf{q} \cdot d\mathbf{S}... [/itex],
    where the second is valid to due F vanishing everywhere except for at the surface.

    I have looked at the more intuitive derivation on the wiki page on kinetic theory. However, I wanted to understand the more rigorous statistical mechanics proof as well.
  5. Jun 9, 2013 #4

    Andrew Mason

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    I don't follow this. It does not appear to be dimensionally correct. This has dimensions of Force x distance x volume.

  6. Jun 9, 2013 #5


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    This is somewhat misleading since the equipartition theorem holds only true for quadratic forms of the canonical variables, i.e., for the harmonic oscillator. The classical grand-canonical partition function is given by
    [tex]Z=\sum_{N=0}^{\infty} \exp(\alpha N) \frac{z^N}{N!} = \exp(z \exp \alpha),[/tex]
    with the single-particle partion function
    [tex]z=\int \mathrm{d}^3 \vec{x} \mathrm{d}^3 \vec{p} \exp \left [-\beta \left (\frac{\vec{p}^2}{2m}+\frac{\omega^2}{2} \vec{x}^2 \right ) \right ]=\frac{8 \pi m^{3/2}}{\beta^3 \omega^3}.[/tex]
    Then you get
    [tex]\langle N \rangle=\frac{\partial \ln Z}{\partial \alpha}=\frac{8 \pi^3 m^{3/2} \exp \alpha}{\beta^3 \omega^3}[/tex]
    [tex]\langle H \rangle = -\frac{\partial \ln Z}{\partial \beta} = 3 \frac{\langle N \rangle}{\beta}.[/tex]
    As you see only in a harmonic oscillator potential you get the mean energy in the form given in Wikipedia.
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