Talk:Boltzmann distribution

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Contents 1 Cut from Maxwell-Boltzmann distribution 2 Re exponential decay.. 3 Difficult Subject 4 Illustration 5 Notation

[edit] Cut from Maxwell-Boltzmann distribution

Also it may be expressed as a discrete distribution over a number of discrete energy levels, or as a continuous distribution over a continuum of energy levels.

The Maxwell-Boltzmann distribution can be derived using statistical mechanics (see the derivation of the partition function). As an energy distribution, it corresponds to the most probable energy distribution, in a collisionally-dominated system consisting of a large number of non-interacting particles in which quantum effects are negligible.

[edit] Re exponential decay..

(relevant to user jheald) I've read your comments and added some more of my own (see below) You mention that the equation I supplied below might/should become like the Boltzmann distribution when there is infinite energy - One way I can get it to resemble the Boltzmann distribution is to take the probability of any given particle having energy (n) OR MORE this is the same as the probability that any given particle has the n th energy level filled with an energy quanta. The equation for this would be: probality that n th energy level is occupied =

    sum from x=n to x=q (q is the maximum energy) of p(x)

There fore using this (the above equation) does give a drop off in probability as n increases but the shape of the curve is not the same as inverse exponential (at least when q the total energy is finite)HappyVR 19:40, 12 February 2006 (UTC)

[edit] Difficult Subject

Difficult Subject but...

A correct equation is 'hard' to derive but making an assumption of quantised energy levels equally spaced it is easily shown that for total energy quanta q distributed randomly over N particles/atoms the probability that any given particle/atom has n energy quanta is:

   ( (1/N)^n x ((N-1)/N))^(q-n) x q! )/(n! x (q-n)!) (taking 0! to equal 1 instead of
                                                      using the longer form (x+1)!/(x+1)  )

This gives the correct distribution - which shows a maxima in the energy distribution ( a hump ) as opposed to the exponential decay given by the incorrectly derived partition function. HappyVR 23:12, 10 February 2006 (UTC)

That's all very interesting. But the Boltzmann distribution is the probability distribution for the system at a fixed, sharply defined temperature; for example, the distribution occurring if the system could exchange energy with an infinite heat reservoir; i.e. a canonical distribution.

In your example there is no infinite heat reservoir, the temperature is not sharply defined, it's not a canonical distribution, so ... surprise, surprise, it's not the Boltzmann distribution.

I'm somewhat at a loss as to what it is you think your example is supposed to show that's relevant to this article. -- Jheald 22:07, 11 February 2006 (UTC).

Thanks for your interest. However the above equation also only applies at a fixed temperature i.e. at a fixed total energy (to quote 'a closed thermally isolated system')- so I will ignore your second point. In the above equation q gives the total energy.

—No; this is exactly the point. A system with fixed total energy (ie in a microcanonical situation) does not have a single fixed well-defined temperature. Statistically it has a spread-out / fuzzy / not well-defined temperature. It is only in the limit of such a system having an infinite energy content that each sub-system can be considered to be in equilibrium with an infinite heat bath, and the system and all its parts can be considered to have a single well-defined temperature.

If the energy content is fixed, but not (reasonably near to being) infinite, then you can't consider the system to have a well-defined single temperature, so you can't expect a Boltzmann distribution. -- Jheald 03:17, 12 February 2006 (UTC).

Thanks again. I agree that as a statistical treament it may seem that the situation is 'fuzzy' or 'spread out' - but if you consider all the possibilities of this sytem (ie N particles q energy) i.e. each possible distribution of energy - and sum them giving an average distribution of energy - this gives a well defined partition function - as a above - I honestly don't fully understand you point about infinite energy - using a v.simple analogy - a cup of tea - at a given moment the tea has a given temperature - now the tea could theoretically be totally isolated from its surroundings ie very good insulation - therefore not in contact with an infintite heat resevoir at all - and not having infinite energy in its own system either - but still having a well defined temperature (measured). Is their a fundamental difference between measured temp and the theorectical temp?HappyVR 18:24, 12 February 2006 (UTC)

The point I was making was that for a random distribution of energy amongst a number of particles-

The distribution of - the number of particles with a given energy (eg molecules with total energy n) vs the energy (n in the above example)

is different from

The distribution of - the relative occupancy of an energy level of enegy Ei (in the text of the article) vs the energy (Ei in the text of the article)

Therefore I would change the first line of this article from

In physics, the Boltzmann distribution predicts the distribution function for the fractional number of particles Ni / N occupying a state i with energy Ei:

to

In physics, the Boltzmann distribution predicts the distribution function for the fractional number of particles Ni / N with state i of energy Ei occupied to be:

(followed by rest of text)

(however I am not sure if the following equations in the text absolutely match this discription.)

As I see it the current text describes a distribution similar to the equation I gave above. Perhaps the text should/could be clearer?

—Fair point. I've made some changes to some of the first two paragraphs, and I hope this is now clearer. -- Jheald 03:43, 12 February 2006 (UTC).

Also further on in the article it states:

Alternatively, for a single system at a well-defined temperature, it gives the probability that the system is in the specified state.

Should this not read (for example):

Alternatively, for a single system (of particles) at a well-defined temperature, it gives the probability that one particle of the system is in the specified state.

—No. The word 'alternatively' is supposed to quietly put the reader on alert, that we're moving away from Boltzmann's picture of the frequencies of energies of single particles, to Gibbs's picture of probabilities for the energy of an entire thermodynamic system (in energy equilibrium with its surroundings). -- Jheald 03:43, 12 February 2006 (UTC).

I feel that in its current state the text does not adequately describe the situation descibed in the given equations. That is the text and equations are at odds. The equation I supplied was one matching the situation given in the first line of the text. Albeit a simplified expression - equally spaced energy levels are clearly assumed.HappyVR 22:53, 11 February 2006 (UTC)

—Yes. I hope between us the text is now clearer, and it's clearer that your equation does not reflect the underlying physical set-up required for a Boltzmann distribution.

It's only in the limit that the total energy in your set-up tends to infinity that one would expect to start finding Boltzmann-like distributions. And I think this is what happens - your 'hump' energy tends to zero; and the fall-off of probability above the hump becomes closer and closer to exponential. -- Jheald 03:43, 12 February 2006 (UTC).

Unfornately my distribution does not appear to do this - the hump tends to move up in energy as total energy increases - however I will check more carefully... However this page is about the Boltzmann distribution.

However if is fair to say that beyond the hump (the peak probability) the fall off is similar to exponential decay.

I have put a more important point below:

(Note that in my supplied equation - the energy is assumed to be discretely quantised - if the size of the quanta decreases - giving more energy quanta the shape of the distribution stays the same but becomes smoother. At infintitely small quanta size there are infinite quanta but not of course infinite energy - this info probably not relevant?)

Thanks for your changes - I'm still reading the text very slowly and carefully but I assume it is clearer now.Thanks again. However your comment above about the system needing to have infinite energy has intrigued me - is it that you are saying that the boltzmann distribution only applies in this case if so I must look at other pages - surely this feature should be stated VERY CLEARLY in say - pages relating to derivation or use of the Boltzmann distribution - in my experience this necessesity is not stated at all...

I assumed that the boltzmann distr. would apply to isolated systems of finite energy.HappyVR 18:24, 12 February 2006 (UTC)

[edit] Illustration

If a picture paints a thousand words... Could someone find a graph to illustrate the text. It makes it much easier to visualise. Thanks.--King Hildebrand 19:06, 5 August 2006 (UTC)

[edit] Notation

What does E' mean in the probability distribution near the bottom? Is it the derivative of E? With respect to what? Is it the total energy? --aciel 22:52, 2 April 2007 (UTC)

I think it's just a dummy energy variable to integrate over. Dicklyon 23:53, 2 April 2007 (UTC)