Talk:Boltzmann constant

Order of magnitude

The statement:-

Given a thermodynamic system at an absolute temperature T, the thermal energy carried by each microscopic "degree of freedom" in the system is on the order of magnitude of kT/2 (i. e., about 2.07×10−21 J, or 0.013 eV, at room temperature).

is unsupported, neither is it self evident. The part about the order of magnitude is far to generalised, and probably incorrect, to be allowed to remain as it is. --Damorbel (talk) 17:00, 9 December 2012 (UTC)

Well, here is the calculation for room temperature at about 25°C:

${\frac {1}{2}}kT\approx {\frac {1}{2}}\times (1.38\times 10^{-23})\times (273+25)=2.0355\times 10^{-21}$

What's your point about it being far too "generalized"? Maschen (talk) 14:07, 10 December 2012 (UTC)

Maschen, if your calculation of the energy in a particle (1/2kT - above ) gave a result within an order of magnitude it would read like this:-

${\frac {1}{2}}kT\approx {\frac {1}{2}}\times (1.38\times 10^{-23})\times (273+25)=2.0355\times 10$ ^-21(±1)

i.e. there would be a factor of uncertainty of 10 in the value of 1/2kT, and yet you quote it to four decimal places. --Damorbel (talk) 11:44, 11 December 2012 (UTC)

What I did was approximate the answer using the numbers shown - the answer to 4 dp is the exact answer using those numbers, but still an approximation, not far off the quote you give, which does not say "within an order of magnitude", it says "on (of?) the order of magnitude", hence your "±1" is irrelevant. Maschen (talk) 13:54, 11 December 2012 (UTC)

Maschen, Is this what you mean by order of magnitude, where it has:-

"We say two numbers have the same order of magnitude of a number if the big one divided by the little one is less than 10. For example, 23 and 82 have the same order of magnitude, but 23 and 820 do not."? --Damorbel (talk) 15:58, 11 December 2012 (UTC)

Yes, although it's simpler to see if two numbers are of the same magnitude if written in scientific notation, and comparing the powers of 10 - the power of 10 is the order of magnitude.

Back to the main topic of this section, there doesn't seem to be anything to fix (WP:CALC). Maschen (talk) 17:21, 11 December 2012 (UTC)

Reorganize article?

This article places too much emphasis on gases. It would be better to start with a definition of temperature from statistical mechanics (as in Statistical Physics by Landau and Lifshitz):

$1/T=dS/dE,$

point out that T has dimensions of energy, and identify k as the conversion factor between temperature and other energy units. Much of the material in the section "Role in the statistical definition of entropy" could go here. The rest of the article could be organized on these lines:

• Model systems that are used to measure k: ideal gases, black-body radiation and spin paramagnetism (Appendix B of Kittel's Thermal Physics)
• Equipartition of energy
• Role in other systems
• History

Comments? RockMagnetist (talk) 18:42, 14 December 2012 (UTC)

Sounds good, but shouldn't that be a partial derivative? Ref:F. Mandl (2008). Statistical physics (2nd ed.). Wiley. p. 88. ISBN 9-780-471-915331. (... also by considering dE = TdS − PdV). M∧Ŝc²ħεИ_τlk 19:00, 14 December 2012 (UTC)

I just transcribed an equation in L&L. Yes, in general it should be a partial derivative. RockMagnetist (talk) 19:08, 14 December 2012 (UTC)

Since the Boltzmann constant is defined as the energy of a particle per kelvin or perhaps energy per degree of freedom per kelvin, I don't see the relevance to multiparticle energy partitions as in equipartition? Surely this is the domaine of the Equipartition theorem? --Damorbel (talk) 19:21, 14 December 2012 (UTC)

Except "multiparticle energy partitions" are already the case by taking averages over the number of particles to get average energy (which equates to multiples of kT per degree of freedom, which is equipartition) etc... The point RockMagnetist is trying to say is to state the connection between energy and entropy first (since energy and entropy are fundamental), then discuss equipartition and so on... M∧Ŝc²ħεИ_τlk 19:38, 14 December 2012 (UTC)

I don't know who defines the Boltzmann constant as energy of a particle per Kelvin. Landau and Lifshitz don't - in the edition I have, they define k as I describe above on page 35 and don't get to equipartition of energy until page 131. RockMagnetist (talk) 19:49, 14 December 2012 (UTC)

Similarly Kittel defines k in chapter 2 and discusses equipartition in chapter 3. Both books discuss the Boltzmann constant when they first define the temperature. RockMagnetist (talk) 19:53, 14 December 2012 (UTC)

maschen you write:-

The point RockMagnetist is trying to say....

I'd rather hear it from RockMagnetist, if you don't mind.

Oh, and while you are at it, would you mind putting:-

Except "multiparticle energy partitions" are already the case by taking averages over the number of particles to get average energy (which equates to multiples of kT per degree of freedom, which is equipartition) etc..

in better language?

I mean ..."which equates to multiples of kT per degree of freedom, which is equipartition"... is meaningless as it stands. --Damorbel (talk) 20:00, 14 December 2012 (UTC)

"Except "multiparticle energy partitions" are already the case..." As in, we're not just talking about one particle, we take averages over many particles...

"Multiples of kT for each degree of freedom" meaning the energy of each degree of freedom is proportional to kT.

...F. Mandl discusses the ideal gas law in the first introduction, giving the definition k_B = R/N_A (at the very beginning of this article - which is presumably what Damorbel was referring to) and pV = nRT, then mentions equipartition (but explains this in detail later)... So Damorbel is correct in terms of units (if referring to k_B = R/N_A).

We seem to be conflicted with sources, Mandl's approach is more or less in the style of the current article...M∧Ŝc²ħεИ_τlk 20:11, 14 December 2012 (UTC)

There is some pedagogical value in discussing the ideal gas law first. I'd be fine with using it as an introductory example and then moving on to a more general definition. RockMagnetist (talk) 20:18, 14 December 2012 (UTC)

Re: I don't know who defines the Boltzmann constant as energy of a particle per Kelvin.

The Boltzmann constant will (probably) replace the kelvin as the fundamental standard of energy density because it can be determined more accurately (than the kelvin) by measuring Johnson noise, see here section 2.3.5 3rd para. where it has:-

Thus we have the exact relation k = 1.380 x 10⁻²³ J/K. The effect of this definition is that the kelvin is equal to the change of thermodynamic temperature that results in a change of energy per degree of freedom kT by 1.380 x 10⁻²³ J.

The quotation is from a document of the International Bureau of Weights and Measures where international agreement is sought for physical standards. --Damorbel (talk) 20:24, 14 December 2012 (UTC)

Ok... So does this mean reorganizing reduces to rewriting the section Role in the statistical definition of entropy? Then some possible shuffling of sections? M∧Ŝc²ħεИ_τlk 20:28, 14 December 2012 (UTC)

The Boltzmann constant pops up all over the place where particle energy is involved, what is needed is to show how it is unique and that it has many applications; but in no way is it defined by multi particle averages!

PS The reason Johnson noise is used to define the Boltzmann constant is that electrons are all identical but the freezing point of water is dependent on the isotopic purity of the water, something that is exceptionally difficult to establish.--Damorbel (talk) 20:49, 14 December 2012 (UTC)

<<Personal attack by Damorbel removed>> M∧Ŝc²ħεИ_τlk 21:05, 14 December 2012 (UTC)