Talk:Maxwell–Boltzmann distribution

Equation 7 (Energy distibution)

Can it be made clearer that $f_{E}dE=f_{p}{\frac {dp}{dE}}dE$ is actually $f_{E}dE=f_{p}{\frac {dp_{x}}{dE}}{\frac {dp_{y}}{dE}}{\frac {dp_{z}}{dE}}d^{3}E$?

While on the subject, is the numerical factor correct? I keep getting 4 (from $d^{3}E=4\pi dE$ instead of 2. Warrickball 11:14, 11 May 2007 (UTC)

Song etc..

I wrote a song about the Maxwell-Boltzmann distribution... and it goes a little like this:

Oh, baby, baby, baby...

We've had some hard times.

You and me.

We've had our differences.

Don't you see?

And maybe... it's just not meant to be. (not meant to be)

Where are we going? And how fast?

How long can this love last?

Our half-life is approaching rapidly... (ra-apidly)

There's only one solution,

the Maxwell-Boltzmann Distribution.

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The graph there needs some color coding!

Good point. I'll try to add it Pdbailey 15:22, 24 Oct 2004 (UTC)

How is that? Pdbailey 17:53, 24 Oct 2004 (UTC)

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The way it is presented at first it seems to refer to the general Boltzmann distribution? - then it can be used for any, arbitrarily strongly interacting system, as long as the subsystem considered is large enough. user:FlorianMarquardt

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Equation (7) in the article looks positively wrong. The author states Substituting Equation 6 into Equation 4 and using p_i=mv_i for each component of momentum gives:

$f_{p}(p_{x},p_{y},p_{x})=\left({\frac {\pi mkT}{2}}\right)^{3/2}\exp \left[{\frac {-m}{2kT(v_{x}^{2}+v_{y}^{2}+v_{z}^{2})}}\right]$ (7)

Equation 4:

$f_{p}(p_{x},p_{y},p_{x})=cq^{-1}\exp \left[{\frac {-1}{2mkT(p_{x}^{2}+p_{y}^{2}+p_{z}^{2})}}\right]$ (4)

Equation 6:

$c=q(2\pi mkT)^{3/2}$ (6)

If I perform the act of substituting 6 into 4 and substituting p_i=mv_i I get:

$f_{p}(p_{x},p_{y},p_{x})=(2\pi mkT)^{3/2}\exp \left[{\frac {-1}{2kT(mv_{x}^{2}+mv_{y}^{2}+mv_{z}^{2})}}\right]$

Please comment. --snoyes 22:17 Mar 9, 2003 (UTC)

You haven't texified (6) correctly. It's supposed to be

$c=q(2\pi mkT)^{-3/2}$

not

$c=q(2\pi mkT)^{3/2}$

-- Derek Ross 22:56 Mar 9, 2003 (UTC)

Thanks a lot for pointing that out ! I really must be more careful. But what about the second part of (7), where one substitutes p_i = mv_i ? --snoyes 23:03 Mar 9, 2003 (UTC)

I'm still lookin at it but I think the (4) is wrongly texified too. It makes much more sense if the sum of p's is on top of the fraction. But I'm just doing a little research to ensure that that's the right thing to do. -- Derek Ross

Which would mean that (3) is also wrongly texified. The problem is that:

exp[-1/2mkT(p_x² + p_y² + p_z²)]

is just so damn interpretable. That is partly the reason I'm going to all the trouble of texifying all these articles; To disambiguate them. --snoyes 23:21 Mar 9, 2003 (UTC)

Yep, (3) should be

exp[-(p_x² + p_y² + p_z²)/(2mkT)]

and the others should changed analogously. -- Derek Ross

Excellent. However, there remain problems with (7); it would now have to be:

$f_{p}(p_{x},p_{y},p_{x})=(2\pi mkT)^{-3/2}\exp \left[{\frac {-(v_{x}^{2}+v_{y}^{2}+v_{z}^{2})}{2kT}}\right]$ (7)

(ie. the m's cancel and the stuff in the first bracket is all to the power ^(-3/2)) ? --snoyes 23:42 Mar 9, 2003 (UTC)

And why is cq^(-1) not written as (c/q) in (4) ? (style?)--snoyes 23:44 Mar 9, 2003 (UTC)

Yep, style. They both mean the same thing. -- Derek Ross

Substituting p² with p=mv gives m²v². You can then pull all the m²'s out with the distributive law and cancel the m in the denominator leaving an m in the numerator which is what you want.

As for the other point ...

$(2\pi mkT)^{-3/2}=(1/2\pi mkT)^{3/2}$

... so there's no problem there, just a matter of personal taste about how you want to write the formula. -- Derek Ross 23:52 Mar 9, 2003 (UTC)

I stand corrected again, thank you Derek. As for the personal taste, I don't care which one - do you have a personal preference? I shall use that. --snoyes 23:57 Mar 9, 2003 (UTC)

Some people get confused by the q^-n notation. I think that it's often a better idea to change it to 1/(qⁿ) instead. Also I would separate out the relatively constant parts to make something like... well if I knew Tex I would do it myself. Sadly I don't. Guess I'll have to learn ! -- Derek Ross

I learnt it in a couple of hours solely for changeing all the stuff on wikipedia ;-) --snoyes 00:08 Mar 10, 2003 (UTC)

One thing I would like. exp[x] is actually supposed to be e^x. It would be nice if you could change that -- Derek Ross

Hmm, unfortunately it looks like this:

$e^{\left({\frac {-(p_{x}^{2}+p_{y}^{2}+p_{z}^{2})}{2mkT}}\right)}$--snoyes 00:27 Mar 10, 2003 (UTC)

Too bad, exp it will have to be then. -- Derek Ross

I see someone has learnt some tex. ;-) Couple of small things with (8). Corrected it is:

$f_{v}(v_{x},v_{y},v_{z})=\left({\frac {m}{2\pi kT}}\right)^{3/2}\exp \left[{\frac {-m(v_{x}^{2}+v_{y}^{2}+v_{z}^{2})}{2kT}}\right]$ (8) --snoyes 05:34 Mar 10, 2003 (UTC)

Not at all, just monkey see, monkey do, plus good ole cut'n'paste -- hence the mistakes. One other change which I think needs making is that the integral signs should actually be partial integral signs and likewise the differentiation operators should be partial differentiation operators but as I said, if only I knew Tex... -- Derek Ross

Distribution plot

I have to be pedantic regarding the newly added plot: There can not be any units at the axis for "Probability"!! A probability is always without units. This must be fixed. Awolf002 15:07, 26 Aug 2004 (UTC)

Well, I guess that axis should read "Probability density." The probability is obviously unitless, but since you integrate over a certain range on that plot to get the probability, it has to have the inverse units of the x-axis so the integral comes out unitless. If that sounds weird and you don't believe me, do a quick unit analysis on any of the Maxwell-Boltzmann equations on the article page. Ed Sanville 03:31, 27 Aug 2004 (UTC)

I was anticipating that answer. You are correct if you are "binning" the x-axis. Then the y-axis needs to have the inverse units. However, I always thought the M-B distribution function has as value the probability, not its density. If I am correct then it has no units, and that's what should be drawn without binning in order to not confuse people. Awolf002 14:03, 27 Aug 2004 (UTC)

This statement seems confused: "I always thought the M-B distribution function has as value the probability, not its density". Obviously the cumulative probability distribution function has probabilities as its values, and obviously the probability density function does not. Since these graphs are of density functions, their values are not probabilities. But I need to ask you: the probability of what event? If you're talking about the cumulative probability distribution function, the event would be that of being less than or equal to the argument to the function. Is that what you had in mind? Michael Hardy 20:36, 28 Aug 2004 (UTC)

Well, I don't see how you can graph a continuous probability distribution without giving the y axis inverse units to the x axis. Think about it, what's the probability that the particle will have EXACTLY 345.218417823... m/s as its speed. Zero, of course. It's kind of like the little-known fact that the quantum mechanical wavefunction is NOT unitless because in 3d space, it has to have units of $length^{-{\frac {3}{2}}}$. Ed Sanville 17:47, 27 Aug 2004 (UTC)

Okay, let me see if I remember this. The M-B distribution is *in fact* a single probability at one *point*. In real life, you will never ask what the probability of one value might be. Instead you "bin" your x-axis and calculate the integral of the M-B distribution between the boundaries, which is what you did to create the plot in the article. My point is that $\int _{x}^{x+\Delta x}F_{M-B}(\xi )d\xi$ evaluated at certain x is not the same as the M-B function $F_{M-B}(x)$ itself. Right? Also note, that the result of this integration indeed has a unit. Awolf002 18:44, 27 Aug 2004 (UTC)

Since I wanted to create a graph which would display the probability densities of the speeds for those noble gases, I actually graphed the following function which is right above the actual graph in the article:

$F(v)=4\pi v^{2}\left({\frac {m}{2\pi kT}}\right)^{3/2}\exp \left({\frac {-mv^{2}}{2kT}}\right)\qquad \qquad (11)$

So, as you can see, what I graphed is the probability density (units s/m), which must have units inversely proportional to the x axis (in this case speed). The chart is similar to charts found in many physics and physical chemistry books, for instance see Physical Chemistry by Laidler and Meiser. It's somewhere in the first few chapters. Basically the exact meaning of what I graphed is:

$\lim _{ds\to 0}{\frac {P(s\to s+ds)}{ds}}$

where s is speed, ds is an infinitesimal speed interval, and $P((s\to s+ds)$ represents the probability that a particle will have a speed between s and s+ds. So, the graph I made shows probabilities only as the areas under slices of that curve, (probability density). The only thing wrong with the graph, I think, is that it might be better to change the y-axis label to "Probability Density (s/m)" instead of just "Probability (s/m)". I think I'll do that now... Ed Sanville 20:21, 27 Aug 2004 (UTC)

Okay, the more I think about this, it seems to be the most reasonable thing to do. Thanks for going through this with me! Awolf002 23:19, 27 Aug 2004 (UTC)

speeds density function

just edited the section on speeds to show that the function is actually the pdf, not a "distribution" which is close but not quite right. the graph below should be changed to show that the y-axis is labeled probability density with units (s/m). It should be a graph of $f(v)$ and has these units because $f(v)$ has these units.

Notice that I changed the notation form $F(v)$ to $f(v)dv$. this is because the cpd, pdf pair are defined as

$F(x)=\int _{-\infty }^{x}{f(\xi )d\xi }$

This means that different differentials can have different meanings and different cpds associated with them. Case in point, when you convert from the speed distribution to the energy distribution, you will be looking at a new cpd, pdf pair and will have to convert $dv$ to $dE$. The (often minor) additional clarity does not distract but can prove a very useful reminder. (unsigned comment by User:Pdbailey)

quantum mechanics

Why the start with qm when Boltzmann derived this thing way before qm came to pass? One of the interesting things about the Maxwell-Boltzmann distribution is that it doesn't use qm and can be used to find avagadros number. Also it may be worth pointing out that Maxwell did the first derivation (before we had partials!) with a little hand waving and Boltzmann did the cleanup proof. I don't have the original papers so I can't say much more than that right now. Maybe I'll look into that. (unsigned comment by User:Pdbailey)

• Agreed. It looks like the Maxwell-Boltzmann distribution (MBD) derives from QM, while, quite the opposite, the first (?) complete demonstration of Planck's law of black body radiation - by Einstein and (?) - uses the MBD as a known fact. Planck himself didn't used it in his proof, but he did derived it again from scratch.-Nabla 13:42, 2004 Aug 30 (UTC)
• Okay, the problem is that I don't really want to have to do all of Maxwell's derivation here and this is much shorter. Perhaps we could just say that this newfangled derivation is quick but is not the one used by Maxwell or Boltzmann who derived this before qm. Unless someone wants to do a non-copyrighted derivation of this using clasical physics. --Pdbailey 22:24, 30 Aug 2004 (UTC)

Chart verified to be wrong

I looked at the chart and picked off the maximum of He-4 as about 1500 m/s but i get a different number (with minor rounding)

$v_{p}={\sqrt {\frac {2kT}{m}}}$

$v_{p}={\sqrt {\frac {2*1.38E-23J/K*298}{4amu*1.66E-27Kg/amu}}}$

$v_{p}={\sqrt {\frac {8.22E-21m^{2}s^{-2}}{6.64E-27}}}$

$v_{p}=1113ms^{-1}$

I'll upload a new one. Anyone care to check that I got this right?

Spelling in Picture

Gasses should be Gases!

Thanks for the catch. I tried to fix it, I hope I'm just seeing the cache now... Pdbailey 23:45, 18 Nov 2004 (UTC)

Adding energy distribution

I wanted to add the energy distribution because of some things I'm doing on statistics. The main revision besides adding the distribution function for the magnitude of momentum, and the energy, is to take out the quantum "particle in a box" references. We don't need it to go from equation 1 to equation 3, we just need p^2=2mE and theres some complications with a particle in a box, because energy eigenfunctions do not coexist with momentum eigenfunctions, so you can't talk about the momentum of a particle in a box while talking about its energy at the same time. (You can, however, talk about the absolute value of the momentum along any axis.).Paul Reiser 21:55, 22 Dec 2004 (UTC)

Boltzman distribution

Should Boltzmann distribution really redirect here? My memory says that that term describes the simple exponential distribution of energy for each degree of freedom; the Maxwell-Boltzmann distribution consists of three such degrees of freedom. Shimmin 13:28, Feb 17, 2005 (UTC)

Are you thinking of Boltzmann's equation for Entropy? I'm quite sure the Maxwell-Boltzmann distribution has two names, one less flattering to Maxwell. --Pdbailey 00:26, 14 October 2005 (UTC)

Blackbody radiation?

Does this distribution relate to blackbody radiation? That is, is there any relation between the distribution of mollecular energies and the emission spectrum? —BenFrantzDale 17:13, 13 October 2005 (UTC)

Just as the Maxwell-Boltzmann distribution describes the energy distribution for a bunch of massive particles at a particular temperature, the Planck's law of black body radiation specifies the energy distribution for a bunch of photons at a particular temperature. So in that sense, there is a conceptual relationship. However, the two distributions are different because the particles have different statistics. The two are physically related only through the temperature. If you bring a bunch of photons into thermal equilibrium with any other body that is maintained at some temperature T, they will take on the Planck distribution for that temperature and it doesn't matter what the distribution of energies are in the other body. So in that sense there is not a relationship. PAR 18:53, 13 October 2005 (UTC)

PDF Box

Someone should add a Probability distribution template and put the PDF at the beginning of the article. -anon

I actually don't like this because I think (and maybe I'm just wrong here) of the distribution of a mixture of a velocity component distribution, a speed distribution, and an energy distribution. Each of these is a distribution. Granted the last two are related to each other by a change of variables, but they have different summary stastics. --Pdbailey 03:02, 3 March 2006 (UTC)

Order of sections?

I'm not an expert on how technical articles about formulas work, but I think it's preferable to have some of the less-technical explanation earlier in the article. For example, move: "The Maxwell-Boltzmann distribution forms the basis of the kinetic theory of gases, which explains many fundamental gas properties, including pressure and diffusion. The Maxwell-Boltzmann distribution is usually thought of as the distribution of molecular speeds in a gas, but it can also refer to the distribution of velocities, momenta, and magnitude of the momenta of the molecules, each of which will have a different probability distribution function, all of which are related." up before the formula itself, so that non-specialists can get a sense of what the distribution is in a larger context. I'll defer to the people who know the topic. -- Epimetreus 12:43, 16 March 2006 (UTC)

Hi,

There is a problem with a conclusion at the end of the article. The variance of the normal distribution of speed in one direction isn't correct I think. What it is writed is the inverse of the variance. The correct expression is $kT/m$

If you think in units, it is correct, because the variance must be a the square of a speed.

Jonathan

Graph etc of molecular speeds

The graph at the end of the article for molecular speeds of noble gases seems wrong to me based on a maxwell-boltzmann distribution - why is it peeked - using the maxwell boltzmann distribution of energies and v=sqrt(2e/m) surely the graphs should be similar in shape to an exponential decay - like the energy distribution - is there an explanation for this?HappyVR 11:42, 15 April 2006 (UTC)

Yes, the explanation has to do with the number of states, (quantum mechanically), or the volume in phase space, (classically), as the velocity increases. Think of it this way, assuming that the velocity vector of a random particle is random, with the energy weighted as an exponential decay curve, that doesn't mean that the velocity will also follow an exponential decay curve. The reason is that there are far more vectors corresponding to high velocities than there are vectors corresponding to low velocities. In fact the "number," (or volume of space in "velocity space" if you will), increases with the square of the velocity. You can think of this as being the area of the sphere centered at the origin, with a radius equal to the particle's speed. So, when this quadratic increase in the number of states is multiplied by the Gaussian-shaped curve e^(-kv^2), (remember, energy is proportional to velocity squared), you end up with a function which has a non-zero peak velocity. Good eye, though... Ed Sanville 18:19, 15 April 2006 (UTC)

Thanks - so it could be described as the probability of finding a particle in a given volume with a specific velocity (time averaged) - that wasn't totally clear to me from the 'probability density (s/m)' y axis label - still not sure what m stands for but doesn't really matter.
I would have thought (simple model) that the equation would be of the type v x e^(-k x v x v) - also gives a 'bump' or peak - due to a volume proportional to molecular radius squared times velocity ie a cylinder along the direction of travel (very similar to what you said). It's more clear to me what probability density means compared to just probaility.Thanks.HappyVR 19:22, 15 April 2006 (UTC)

I think you have misunderstood my explanation... I was talking about "volume" in the space of velocity vectors, not real space. So the radius and other size characteristics of the moecule don't come into it. Let me try to give a more concrete example... If the particle has a velocity of between 1 and 2 m/s, the velocity vector can be anywhere in the spherical "shell" between radius 1 and 2 m/s in "velocity space." This shell has a volume of ${\frac {4}{3}}\pi (2^{3}-1^{3})=7{\frac {4}{3}}\pi$ in velocity space. However, if it has a velocity between 2 and 3 m/s, the volume of space it can be in is ${\frac {4}{3}}\pi (3^{3}-2^{3})=19{\frac {4}{3}}\pi$. So, my point is that even if all of the energies were equally probable, there would still be a higher chance that the particle has a velocity betwee 2 and 3, than there is that it is between 1 and 2! (This concept is closely related to entropy in fact, but that's another story...) Anyway, the energy of any point in velocity space is proportional to the sum of the squares of the three components $v_{x},v_{y},v_{z}$. According to Boltzmann, the probability that a particle has a velocity occuring in any given volume of velocity space is then proportional to $e^{\frac {-m(v_{x}^{2}+v_{y}^{2}+v_{z}^{2})}{2kT}}$. But, you have to integrate this probability over the velocity space, so you end up with a speed probability distribution like $\int e^{\frac {-mv^{2}}{2kT}}4\pi v^{2}\,dv$ (where v is the magnitude of the velocity, ${\sqrt {v_{x}^{2}+v_{y}^{2}+v_{z}^{2}}}$). If you solve this integral, you will get the cumulative probability with respect to speed, (magnitude of the velocity vector). The graph is actually of $e^{\frac {-mv^{2}}{2kT}}4\pi v^{2}$, the probability density function. Ed Sanville 22:30, 20 April 2006 (UTC) Sorry about all the mistakes before edit... I was tired.

The graph is of a pdf of molecular velocity for particles in a volume with a given energy and given mass. The moledular velocity distribution function is f(v)dv, it has units seconds/meter (s/m) as I've plotted it. This has come up several times, so I'll add a note on the image's page. Pdbailey 02:28, 18 April 2006 (UTC)

Distribution of momentum and velocity vectors are not (yet) M-B distributions

Distribution of momentum and velocity vectors are not (yet) M-B distributions. So formulas (6) and (8) are not M-B distributions. I think this should be more clearly stated, and then the work should be finished in order to actually give the formulas for the M-B distributions of p and v. Otherwise this can get quite confusing.

So I'd happy to see an equation f(p)dp, which would also ease the step between equations 7 and 8, and another equation f(v)dv illustrating the distribution of speeds such as the following: $f_{\mathbf {v} }(v)=N{\sqrt {{\frac {2}{\pi }}\left({\frac {m}{kT}}\right)^{3}}}v^{2}\exp \left[{\frac {-mv^{2}}{2kT}}\right],$ with N, the number of particles you consider in your box (I guess this factor can just be removed to have probabilities instead of number of molecules).

The $v^{2}$ that appears in that formula makes the difference and gives its nature to the MB formula.

B. Roy Frieden's anonymous POV-pushing edits

B. Roy Frieden claims to have developed a "universal method" in physics, based upon Fisher information. He has written a book about this. Unfortunately, while Frieden's ideas initially appear interesting, his claimed method is controversial:

• Binder, Philippe M. (2000). "Physics from Fisher Information: A Unification (a review)". American Journal of Physics. 68: 1064–1065.
• Kibble, T. W. B. (1999). "Physics from Fisher Information: A Unification (a review)". Contemporary Physics. 40: 1999. (the reviewer has some positive comments but concludes that Frieden's work is "misguided")
• Case, James (2000). "An Unexpected Union---Physics and Fisher Information". SIAM News. July 17. eprint (highly favorable)
• Matthews, Robert (1999). "Physics and Fisher Information (a review)". New Scientist. January. unauthorized electronic reprint
• Physics from Fisher Information: A Unification (a review) from Cosma Shalizi (Computer Science, University of Michigan) (highly critical)
• Physics from Fisher Information (a review) from R. F. Streater (Mathematics, Kings College, London) (highly critical)
• Physics from Fisher Information thread from sci.physics.research, May 1999 (mostly critical)
• Fisher Information - Frieden unification Of Physics thread from sci.physics.research, October 1999 (mostly critical)

Note that Frieden is Prof. Em. of Optical Sciences at the University of Arizona. The data.optics.arizona.edu anon has used the following IPs to make a number of questionable edits:

1. 150.135.248.180 (talk · contribs)
   1. 20 May 2005 confesses to being Roy Frieden in real life
   2. 6 June 2006: adds cites of his papers to Extreme physical information
   3. 23 May 2006 adds uncritical description of his own work in Lagrangian and uncritically cites his own controversial book
   4. 22 October 2004 attributes the uncertainty principle to the Cramer-Rao inequality, which is potentially misleading
   5. 21 October 2004 adds uncritical mention of his controversial claim that the Maxwell-Boltzmann distribution can be obtained via his "method"
   6. 21 October 2004 adds uncritical mention of his controversial claim that the Klein-Gordon equation can be "derived" via his "method"
2. 150.135.248.126 (talk · contribs)
   1. 9 September 2004 adds uncritical description of his work to Fisher information
   2. 8 September 2004 adds uncritical description of his highly dubious claim that EPI is a general approach to physics to Physical information
   3. 16 August 2004 confesses IRL identity
   4. 13 August 2004 creates uncritical account of his work in new article, Extreme physical information
   5. 11 August 2004 creates his own wikibiostub, B Roy Frieden

These POV-pushing edits should be modified to more accurately describe the status of Frieden's work.---CH 21:54, 16 June 2006 (UTC)

Too Complex

Hi this page is great but it is very complex - is it possible to give an overview on how this distribution works in simple terms and perhaps with an everyday life example??

boltzmann distribution is not energy distribution???

can anyone explain how one can go from the boltzmann distribution directly to the energy distribution? it is very confusing that the are not identical. —Preceding unsigned comment added by 84.191.237.160 (talk) 00:54, 17 December 2007 (UTC)

I have a big problem with equation (3)

Equation (3) has subscript i on the left handside but no i on the other side. How would you fix it? should it be $p_{x,i}^{2}+p_{y,i}^{2}+p_{z,i}^{2}$ instead of $p_{x}^{2}+p_{y}^{2}+p_{z}^{2}$ ???

Besides, what happened to $g_{i}$ in front of (3). $g_{i}$ should have been inherited from equation (1)?