Talk:Boltzmann constant/Archive 1

k is not one of the Planck units - it is one of the physical constants on which the Planck units are based. At least, that's what our current article says - I'm not arguing the physics of this situation. To maintain the information content, I'll put in links from energy, temperature, etc. to Planck units. -- Tim

Sorry. I wanted to put a link in to Planck units; I was hoping someone would corrent my phrasing rather than just remove ... :) -- Tarquin

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I have to say, CYD, I disagree with just about every change you made to my text in your latest edit. Apparently k is now "a fundmental quantity which relates temperature to energy". What were the other quantities you were thinking of? And is k really a quantity? What is it a quantity of? It seems to me that it's a proportionality constant -- you can't produce 3.6k of something, but you can use it to say 1K is equivalent (in some sense) to 1.38×10^-23J.

Why did you change the HTML specification of the value to <math>? The only visible difference is that it's not wikified anymore.

In my section on characteristic energy, I was attempting to hint at nomenclature used in solid state physics. I deliberately left out a specific example of what the characteristic energy is, because it shows up in both statistical mechanics and solid state/quantum. I wanted to give a sense that k is used in almost any field of physics involving temperature (which it is).

What's wrong with meV? 25.9 meV is so much nicer than 0.00259 eV.

Entropy is important, and I appreciate your new section on it.

"Historically, the units of energy and temperature were defined before it was discovered that they were related." True. "As a result, the Boltzmann constant is expressed using the unit of temperature, and not the other way around." You've lost me. Which other way around? "Conceptually, however, the Boltzmann constant is the more fundamental quantity." Nope, still lost. More than what?

I think I can see what you're trying to get at with those last few paragraphs. My interpretation of the physical situation is that the relationship between energy and temperature is similar to the relationship between distance and time. In both cases, there are two sets of units originally defined in terms of physical quantities (e.g. orange-red radiation and triple points). Theoretically though, it would be much more sensible to define c=1, or c=α (the fine structure constant), in just the same way as it would be more sensible to set k = 1 and use Planck temperature. We can't necessarily argue that physically, k should be 1, but it should definitely be dimensionless.

-- Tim Starling 05:34 May 3, 2003 (UTC)

I'm not particularly concerned about most of the changes, so I have tweaked them to suit your objections. --CYD

Thanks. -- Tim

As for the relationship between energy and temperature, the key idea is this: energy is a microscopic (mechanical) quantity, and temperature is a macroscopic (thermodynamic) one. The main reason Boltzmann's definition of the entropy is so important is that it shows us how this macroscopic property of a system emerges from the microscopic details of its constituents. If you know enough about the mechanical properties of atoms, Boltzmann's equation shows you how to define a quantity, the "temperature", of large assemblies of atoms. In this sense, Boltzmann's constant defines the temperature, and not the other way around. -- CYD

The distinction between Boltzmann's constant defining temperature and temperature defining Boltzmann's constant seems rather arbitrary to me. Temperature is defined either macroscopically, by the properties of gases, or microscopically, with 1/(dS/dE) and Boltzmann's formalism. But in either case, k is merely a proportionality constant, and tells you about the units used to keep the books, not the physics involved. By itself, it tells you very little about the concept of temperature. It is Boltzmann's formalism (or the properties of gases), not k itself, that defines temperature.

But I'm getting distracted by a discussion of physics, and forgetting our main goal. Would it be fair to say that if your final paragraph confused me, it's going to confuse the general public? If you can think of a more eloquent way to put your ideas, in such a way that seems intuitive and correct to me, I'll be happy. I know my current final paragraph says something completely different to what you were getting at, but I think some reference to Planck units is warranted. Tarquin tried to include something about them a while back, but I cut it out. -- Tim Starling 00:15 May 5, 2003 (UTC)

Relation between charge and thermal energy

k is a relation between temperature and energy for a single atom (or single mole). There are a lot of equations in electrochemistry where you want to relate temperature to energy for a unit of charge (because in electrics we are more interested in units of charge than in number of atoms). So I was wondering if there is any defined symbol for the quantity e / k or k / e (equivalently F / R or R / F)? --Chinasaur 01:51, 14 Sep 2004 (UTC)

About this part of the article

quote With 20:20 hindsight however, it is perhaps a pity that Boltzmann did not choose to introduce a rescaled entropy such that

${S^{\,'}=\ln \Omega }\;;\;\;\;\Delta S^{\,'}=\int {\frac {dQ}{kT}}$

These are rather more natural forms; and this (dimensionless) rescaled entropy exactly corresponds to Shannon's subsequent information entropy, and could thereby have avoided much unnecessary subsequent confusion between the two. quote end

I know soem physics researchers have jumped the gap regarding this, and as a natural consequence, they will express temperature in the same unit as energy. I'll try to find a reference for that.

I am of the opinion that k is arbitrary, precisely that cutting kT in k and T is arbitrary, so that T is just as arbitrary. Because temperature is an energy scale. Think of it this way: I don't understand why temperature is given in here as a base unit of SI system. Because, if you get rid of charge, or mass, then you can't have the same physics anymore, you just lose the corresponding parts. But, if you count temperature as energy, and remove all occurences of Boltzman's constant in expressions where you meet it, then this doesn't change anything to the objective meaning of your equations. You restore exactly the same physics.ThorinMuglindir 01:28, 5 November 2005 (UTC)