Talk:Maxwell-Boltzmann distribution
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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.
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)
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
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:
- <math>
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] <math> (7)
Equation 4:
- <math>
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] <math> (4)
Equation 6:
- <math>
c = q (2 \pi mkT)^{3/2} <math> (6)
If I perform the act of substituting 6 into 4 and substituting p_i=mv_i I get:
- <math>
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] <math>
Please comment. --snoyes 22:17 Mar 9, 2003 (UTC)
You haven't texified (6) correctly. It's supposed to be
- <math>
c = q (2 \pi mkT)^{-3/2} <math>
not
- <math>
c = q (2 \pi mkT)^{3/2} <math>
-- 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(px2 + py2 + pz2)]
- 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[-(px2 + py2 + pz2)/(2mkT)]
and the others should changed analogously. -- Derek Ross
Excellent. However, there remain problems with (7); it would now have to be:
- <math>
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] <math> (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 ...
- <math>
( 2 \pi mkT) ^ {-3/2} = (1 / 2 \pi mkT) ^ {3/2} <math>
... 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 ex. It would be nice if you could change that -- Derek Ross
- Hmm, unfortunately it looks like this:
- <math>
e^{\left( \frac{-(p_x^2 + p_y^2 + p_z^2)}{2mkT} \right)} <math>--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:
- <math>
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] <math> (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
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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 <math>length^{-\frac{3}{2}}<math>. 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 <math>\int_x^{x+\Delta x} F_{M-B}(\xi)d\xi<math> evaluated at certain x is not the same as the M-B function <math>F_{M-B}(x)<math> 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:
- <math>
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)<math>
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:
<math>\lim_{ds \to 0}{\frac{P(s \to s+ds)}{ds}}<math>
where s is speed, ds is an infinitesimal speed interval, and <math>P((s \to s+ds)<math> 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 <math>f(v)<math> and has these units because <math>f(v)<math> has these units.
Notice that I changed the notation form <math>F(v)<math> to <math>f(v) dv<math>. this is because the cpd, pdf pair are defined as
<math>F(x) = \int_{-\infty}^{x}{f(\xi)d\xi}<math>
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 <math>dv<math> to <math>dE<math>. 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)
- <math>v_p = \sqrt { \frac{2kT}{m} }<math>
- <math>v_p = \sqrt { \frac{2 * 1.38E-23 J/K * 298 }{4 amu * 1.66 E-27 Kg/amu} }<math>
- <math>v_p = \sqrt { \frac{8.22E-21 m^2 s^{-2}}{6.64 E-27 } }<math>
- <math>v_p = 1113 m s^{-1}<math>
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)