Talk:Maxwell–Boltzmann distribution

message: 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)
replies: text: 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)
replies: text: 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)

text: 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)

text: 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)

text: 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)
thread_title:  Distribution plot 

message: Why the start with Quantum Mechanic when Boltzmann derived this thing way before Q.M. came to pass? One of the interesting things about the Maxwell-Boltzmann distribution is that it doesn't use Q.M. 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 clean-up 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 Q.M., 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)
thread_title:  Quantum Mechanics 

message: 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-23 J/K * 298 }{4 amu * 1.66 E-27 Kg/amu} }
:v_p = \sqrt { \frac{8.22E-21 m^2 s^{-2}}{6.64 E-27 } }
:v_p = 1113 m s^{-1}
I'll upload a new one. Anyone care to check that I got this right?
Why u no doctor?  — Preceding unsigned comment added by 93.218.73.181 (talk) 15:12, 20 January 2012 (UTC)
replies: text: I agree with your numbers (although you mean kg not Kg). You can Google this to get the same answer: sqrt(((2*(Boltzmann constant)* 298 kelvin)/(4*1.66e-27 * kg))). But the graph looks right to me. —Ben FrantzDale (talk) 18:01, 20 January 2012 (UTC)
thread_title:  Chart verified to be wrong 

message: 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)
thread_title:  Spelling in Picture 

message: 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)
thread_title:  Adding energy distribution 

message: 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)
thread_title:  Boltzman distribution 

message: 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)
replies: text: 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)
thread_title:  Blackbody radiation? 

message: 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)
thread_title:  PDF Box 

message: 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)
thread_title:  Order of sections? 

message: 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)
replies: text: 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)

text: 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)
thread_title: Graph etc of molecular speeds

message: 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:
*
* (the reviewer has some positive comments but concludes that Frieden's work is "misguided")
* eprint (highly favorable)
* 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:
#
##20 May 2005 confesses to being Roy Frieden in real life
##6 June 2006: adds cites of his papers to Extreme physical information
##23 May 2006 adds uncritical description of his own work in Lagrangian and uncritically cites his own controversial book
## 22 October 2004 attributes the uncertainty principle to the Cramer-Rao inequality, which is potentially misleading
##21 October 2004 adds uncritical mention of his controversial claim that the Maxwell-Boltzmann distribution can be obtained via his "method"
## 21 October 2004 adds uncritical mention of his controversial claim that the Klein-Gordon equation can be "derived" via his "method"
#
##9 September 2004 adds uncritical description of his work to Fisher information
## 8 September 2004 adds uncritical description of his highly dubious claim that EPI is a general approach to physics to Physical information
##16 August 2004 confesses IRL identity
##13 August 2004 creates uncritical account of his work in new article, Extreme physical information
##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)
thread_title: [[B. Roy Frieden]]'s anonymous POV-pushing edits

message: Can it be made clearer that f_EdE=f_p\frac{dp}{dE}dE is actually f_EdE=f_p\frac{dp_x}{dE}\frac{dp_y}{dE}\frac{dp_z}{dE}d^3E?
While on the subject, is the numerical factor correct? I keep getting 4 (from d^3E=4\pi dE instead of 2. Warrickball 11:14, 11 May 2007 (UTC)
thread_title: Equation 7 (Energy distibution)

message: 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)
thread_title:  boltzmann distribution is not energy distribution??? 

message: 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^2_{x,i}+p^2_{y,i}+p^2_{z,i} instead of p^2_{x}+p^2_{y}+p^2_{z} ??? 
Besides, what happened to g_i in front of (3). g_i should have been inherited from equation (1)?  —Preceding unsigned comment added by Resonance cascade (talk • contribs) 09:02, 6 February 2008 (UTC)
thread_title:  I have a big problem with equation (3) 

message: the paragraph above equation 4 states that f_p is proportional to N_i/N,and we can use ∫∫∫(c/z)exp〔..〕dp_x*dp_y*dp_z=1 to find the constant c.
My question is that would f_E or f_v be also proportional to N_i/N,if not so,why not?if so,why I get a wrong outcome by using the same technique,that is,I 
try toset f_v to be proportional to Ni/N,then f_v=(c/z)exp(...),but when I set ∫f_v*4πv^2*dv equal to 1 to solve c,I'll get a wrong answer.Is there a 
clearer explanation that why only f_p is proportional to Ni/N and f_v along with f_E is not?And how do we know f_p is proportional to Ni/N directly? f_p is 
a "probability density" function,while Ni/N is a "fraction" of the amount of  particles with energy ε_i to all of the molecules of the system. How can one 
visualize the connection between these two intuitively?  
—Preceding unsigned comment added by Yung.cheng (talk • contribs) 13:27, 4 June 2009 (UTC)
thread_title:  about equation4 

message: The original derivation by Maxwell assumed all three directions would behave in the same fashion, but a later derivation by Boltzmann dropped this assumption using kinetic theory.
I thought that the problematic assumption in Maxwell's derivation was that velocities in the three directions were independent. 72.75.67.226 (talk) 04:04, 8 October 2009 (UTC)
replies: text: Also, isn't it worth mentioning that the multivariate normal is unique in that it is isotropic and the three velocities are independent? 72.75.67.226 (talk) 04:12, 8 October 2009 (UTC)
thread_title:  Isotropy vs. Independence 

message: It says 600m/s for the most probable speed, which is about what you get if you plug in 300K and 14g/mol. However at room temperature nitrogen forms a molecule with two atoms, so you have to use 28g/mol which gives about 420m/s for the most probable speed. --84.56.21.137 (talk) 21:56, 2 January 2010 (UTC)
thread_title:  Typical speed of nitrogen at room temperature is wrong 

message: Upon checking this edit by Melcombe with my version of Maple 13, I found that the correction was indeed to the point. O.t.o.h. I also noticed that the infobox value for the kurtosis is different from what Maple says.
:Infobox says: \gamma_2=-4\frac{96-40\pi+3\pi^2}{(3 \pi - 8)^2}
:Maple says: \gamma_2=\frac{-192+16\pi+15\pi^2}{(3 \pi - 8)^2}
Do we have an online source somewhere? Preferably not Mathworld, as it has the skewness wrong as well. DVdm (talk) 17:48, 25 May 2010 (UTC)
replies: text: Maple's version is the kurtosis, the version in the infobox is the excess kurtosis, so is lower by 3. Both are correct. I must say that as the excess kurtosis is positive (0.1082) i think it would be clearer to rewrite the expression to avoid the minus sign out the front. Qwfp (talk) 20:13, 25 May 2010 (UTC)
replies: text: Ah yes. I had found the difference of 3 as well. I should have looked the Kurtosis article where the number 3 is sort of staring me in the face. Oops - and thanks. DVdm (talk) 21:07, 25 May 2010 (UTC)
thread_title:  Kurtosis correct? 

message: I have removed the paragraphs labelled as junk - they are indeed since they add nothing to the article (they were commented out anyway - but why keep them?). The intro has been re-structured to be more concise and to the point. Also made other various edits - such as:
* addition of sources, will slot into article later - very busy right now
* removed the tabs (proposed way back in 2009/10) - there are now sources
* add eqn navbox
* minor bits of clean up
* changed the leading titles - they are not helpful as they are
Hope the intro is now ok. --  F = q(E + v × B) 22:39, 22 February 2012 (UTC)
thread_title: "junk paragraphs"...

message: Maxwell–Boltzmann distribution
Boltzmann distribution
Maxwell–Boltzmann statistics
Someone needs to untangle this obvious mess.siNkarma86—Expert Sectioneer of Wikipedia86 = 19+9+14 + karma = 19+9+14 + talk 19:01, 8 May 2012 (UTC)
thread_title: Lack of unity of articles about the Maxwell-Boltzmann distribution

message: Could anyone please make a simple english version of this? Many thanks in advance!  — Preceding unsigned comment added by Mbb1056 (talk • contribs) 19:51, 7 April 2014 (UTC)
thread_title:  Simple English version 

message: I am not an expert, and to me it looks like there are two contradictory versions of the velocity distribution.
The first equation in the article reads 
: f(v) = \sqrt{\left(\frac{m}{2 \pi kT}\right)^3}\, 4\pi v^2 e^{- \frac{mv^2}{2kT}}, 
Later in the article, I see
:
f_\mathbf{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]
The difference I am pointing to is the addition of 4\pi v^2 in the first equation.
I am not insisting that there must be an error, but I think the difference may need to be commented/adressed, to lessen the confusion of new readers (like me).  
(EDIT: Oh, i guess the first one is for the absolute speed |v|, and has been integrated across a sphere surface. Perhaps this should be commented/adressed by someone more comfortable with the terminology. 
— Preceding unsigned comment added by Eaurs (talk • contribs) 09:54, 19 May 2016 (UTC)
thread_title:  Two different version of the velocity distribution. 

message: The validity should be discussed. The distribution for the velocities is a marginal distribution of the canonical distribution function. We obtain it via integrating out all position dependencies from the canonical distribution. the equation is therefore also valid for interacting systems. Compare https://books.google.de/books?id=Hp-TBQAAQBAJ&lpg=PA131&dq=maxwell%20boltzmann%20distribution%20marginal%20distribution&hl=de&pg=PA131#v=onepage&q=maxwell%20boltzmann%20distribution%20marginal%20distribution&f=false Biggerj1 (talk) 09:03, 6 December 2019 (UTC)
thread_title:  Validity 

message: We see that His name has been misspelt as "Boltzman" in a number of places on this talk page.  We will let these slip by.  However, be warned that The Boltzmann Police will quickly put to justice any criminal who attempts this sacrilege in a Wikipedia article. Boardhead (talk) 16:23, 14 February 2020 (UTC)
thread_title:  The Boltzmann Police 

message: For some reason the minus sign in the denonimator of the expression for the kurtosis is written in the source, but doesn't actually show up.  — Preceding unsigned comment added by TheGam3lerInDebt (talk • contribs) 22:05, 9 December 2020 (UTC)
thread_title:  Small mistake in the expression for the kurtosis 

message: discussing the equation below eq (9),
please discuss if the requested change of the formula below (9)  is valid 
and if it is possible to  give this equation a label (likewise : 9b)
It is about the distribution of a single degree of freedom using the equilibrium theorem
I certainly know, that \epsilon is not E/kT! 
But in combination  with the suggested change 8below) the dependency is indeed
f_\epsilon(\frac{\epsilon}{kT}). 
This makes sense, in order to apply a chi-square distribution. 
==== ARGUMENT  B.1: used citation [14] differs to equation on wikipedia ====
The cited [14] Laurendeau, Normand. p. 434 , (taken as reference for this equation),
states another dependency. 
I just bought the eBook. , It says:
f_\epsilon(\frac{\epsilon}{kT})= \frac{2}{\sqrt{\pi}}~\sqrt{\frac{\epsilon}{kT}}~\exp\left[-\frac{\epsilon}{kT}\right]
THIS FORMULA should be in the WIKIPEDIA page
The cited Laurendeau, Normand. ([14] ) also defines at (another page) p.294 in equation (15.24) the  expected:
\epsilon=\frac{1}{2}m v^2, 
==== ARGUMENT B.2: Probabilty Density Function f(...) must be unitless ====
so \epsilon has the UNIT of ENERGY.
Do we agree that f_\epsilon(...) is a unitless PDF ?
BUT the current factor \sqrt{\frac{1}{\epsilon~kT}} 
is  NOT unitless! It has the unit 1 by Energy. 
This supports argument B.1
==== ARGMENT B.3 :  parameter of f(....) ====
I believe we agree: 
f_\epsilon(...) depends on \epsilon and kT ?
If we have consent in ARGUMENTs B.1 and B.2 ,
the factor in the equation from citation [14] is:
\sqrt{\frac{\epsilon}{kT}}
So the overall parameter dependency is : \frac{\epsilon}{kT}
use as a Chi-squared distribution
If we have consent in ARGUMENTs B.1 and B.2:
In order to make use of a Chi-squared distribution an argument substitution 
the dependency must be :
f_\epsilon(\frac{\epsilon}{kT}) 
====  Summary of Arguments ====
* As f(...) must be a unitless probability the factor before exp() is unitless, too.
* When using citation [14], it has to be cited correctly 
* Certainly ONE can argue that kT is constant and may an be striped off.
If you agree the ARGUMENT B.1, and B.2 
* both parameter dependencies with and without the  fraction kT are valid options.
The formula should have a label e.g. (9b) and should be 
f_\epsilon(\frac{\epsilon}{kT}) = \frac{2}{\sqrt{\pi}}~\sqrt{\frac{\epsilon}{kT}}~\exp\left[-\frac{\epsilon}{kT}\right]
I know,
that this differs by a fraction of \frac{1}{\epsilon} from the formula of this wikipage.
Willyfoobar (talk) 16:09, 2 January 2022 (UTC)
replies: text: I can only repeat that "your" equation is absolutely valid, but it is a trivial rewrite of Eq.(9). The dimension of any PDF is that of its argument in the power of minus one. So when you integrate over it, the result is the dimensionless unity. Hence f(v), f(E), and f(E/kT) all are in different units; only f(E/kT) is dimensionless – because E/kT is. The original equation has a different meaning. It can be derived from the one-dimensional M–B distribution for velocity. Evgeny (talk) 21:55, 2 January 2022 (UTC)
thread_title:  Discussion part B 

message: * As f(...) must be a unitless probability the factor before exp() is unitless, too.
* When using citation [14], it has to be cited correctly 
* Certainly ONE can argue that kT is constant and may an be striped off.
If you agree the ARGUMENT B.1, and B.2 
* both parameter dependencies with and without the  fraction kT are valid options.
The formula should have a label e.g. (9b) and should be 
f_\epsilon(\frac{\epsilon}{kT}) = \frac{2}{\sqrt{\pi}}~\sqrt{\frac{\epsilon}{kT}}~\exp\left[-\frac{\epsilon}{kT}\right]
I know,
that this differs by a fraction of \frac{1}{\epsilon} from the formula of this wikipage.
Willyfoobar (talk) 16:09, 2 January 2022 (UTC)
replies: text: I can only repeat that "your" equation is absolutely valid, but it is a trivial rewrite of Eq.(9). The dimension of any PDF is that of its argument in the power of minus one. So when you integrate over it, the result is the dimensionless unity. Hence f(v), f(E), and f(E/kT) all are in different units; only f(E/kT) is dimensionless – because E/kT is. The original equation has a different meaning. It can be derived from the one-dimensional M–B distribution for velocity. Evgeny (talk) 21:55, 2 January 2022 (UTC)
thread_title:   Summary of Arguments 

message: It is certainly useful to understand the Maxwell-Boltzmann distribution in terms of all the physical constants that affect it.
But to appreciate what type of distribution this is among probability distributions, it would be extremely valuable if all the constants could be consolidated into as few as possible so one can view the form that the distribution takes in terms of velocity or in terms of speed.
I hope someone knowledgeable about this subject can add some summary lines that express these distributions as simply as possible. 2601:200:C000:1A0:3C4F:AE96:D771:5385 (talk) 00:50, 29 August 2022 (UTC)
thread_title:  Unfortunate domination by constants 

message: Near the beginning of the article under 'Distribution function' it reads
 f(v) ~d^3v = \biggl[\frac{m}{2 \pi kT}\biggr]^\frac{3}{2} \, \exp\left(-\frac{mv^2}{2kT}\right) ~ d^3v, 
i.e.
 f(v) = \biggl[\frac{m}{2 \pi kT}\biggr]^\frac{3}{2} \, \exp\left(-\frac{mv^2}{2kT}\right) ~  
whereas further down under 'Distribution for the Speed' ( https://en.wikipedia.org/wiki/Maxwell%E2%80%93Boltzmann_distribution#Distribution_for_the_speed ) it reads
f (v) =
\sqrt{\frac{2}{\pi}} \, \biggl[\frac{m}{kT}\biggr]^\frac{3}{2} v^2 \exp\left(-\frac{mv^2}{2kT}\right)
 140.228.40.161 (talk) 16:12, 21 April 2024 (UTC)
thread_title:  Inconsistent equations for distribution function of speeds