Talk:Boltzmann constant

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 Definition The physical constant relating energy at the particle level with temperature observed at the bulk level. [d] [e]
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There's nothing wrong with liking physics! --Robert W King 09:03, 31 December 2007 (CST)

And what's wrong with a little self deprecation then? :-)
Seriously tho, Paul Wormer taught me a couple things this morning when he fixed some formatting issues and identified the Equipartition Theorem equation I had used as an example of the Boltzmann constant. So here, just by trying to share something I thought I already understood, I learned something new...and that _is_ a good thing.
Here's the problem, in investigating the Equipartition Theorem in more depth, I raised more questions for myself. Having done so, and since this seems to be the place that those who feel liking Physics is a good thing tend to frequent, hopefully someone can now help me understand the thing I just here goes.

The formula that is listed in the article and is apparently uncontested is:

Where KEavg is the average kinetic energy of the particle, k is the Boltzmann Constant, and T is the temperature in kelvin.

The thing is when I investigated the Equipartition Theorem in more detail this doesn't seem to add's why:

According to this article

"A result from classical statistical mechanics is the equipartition theorem: When a substance is in equilibrium, there is an average energy of kT/2 per molecule or RT/2 per mole associated with each degree of freedom."

Now, with three degrees of freedom (like in 3-d space for instance) everything seems to add up...kT/2 * 3 is exactly what we have. However as I look closer, we can have potential energy in rotation and apparently perhaps in oscillation as well. (My book does include the possibility of rotational kinetic energy in it's accounting of degrees of freedom but neglects the oscillatory but, whatever)

The bottom line here is that here if we want to call this the "average kinetic energy of the particle in a general sense, (i.e. if this is really to be the Equipartition Theorem) it seems as tho we need to determine the degrees of freedom the particle actually has and modify the formula accordingly. Perhaps I picked too complex an example simply to illustrate the Boltzmann constant.

--David Yamakuchi 19:29, 31 December 2007 (CST)


Usually I don't like to advertise my own work, but here it goes: in Journal Chemical Physics vol. 122 , p. 184301, (2005) I wrote about the stat. mech. of rigid rotors. Table I in this paper gives deviations from equipartition for rotation. If the values in this table would satisfy F_x=F_y=F_z=0 (which they don't) then equipartition would hold. Conclusion: for rotation (three degrees of freedom) equipartition holds approximately, but not exactly. Happy New Year! --Paul Wormer 02:49, 1 January 2008 (CST)