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Derivation of maxwell speed distribution

  1. Aug 24, 2010 #1
    Ok, so I'm reading Schroeder's Thermal Physics and I've reached the part where he 'derives' the Maxwell speed distribution. Most of what he writes makes perfect sense, however there is one bit that is rather confusing to me.

    Here's a quick breakdown of his derivation, up to the point where I become confused, just in case you don't have access to this book.

    He says that the probability of a molecule having a speed between v_1 and v_2 is equal to
    [tex]\int_{v_1}^{v_2} D(v)dv [/tex]
    where D(v) is the distribution function that is proportional to the product of the probability of a molecule having velocity v and the number of vectors v corresponding to the speed v.

    He then goes on to say the first factor mentioned above is proportional to the Boltzmann factor where the energy in the Boltzmann factors is just the molecule's kinetic energy.

    Everything up to this point makes sense to me.

    He then proceeds to talk about the second factor mentioned above, i.e., the number of vectors corresponding to the speed v. What follows is a quote from the book:

    "To evaluate this factor, imagine a three-dimensional "velocity space" in which each point represents a velocity vector. The set of velocity vectors corresponding to any given speed v lives on the surface of a sphere with radius v."

    Ok, this makes sense, although I wouldn't say that the set of velocity vectors corresponding to any given speed "lives on the surface a sphere", to me it looks like said set is just member of R^3. In fact, one may say that the set of velocity vectors corresponding to a given speed v could be expressed as
    [tex] \{(v_x,v_y,v_x) \in R^3 : v_x^2 + v_y^2 +v_z^2 = v^2\} [/tex].
    Anyway, he continues,

    "The larger v is, the bigger the sphere, and the more possible velocity vectors there are."

    The italics are mine; that is the statement that confuses me. Here is what I think he is saying: let [tex] v_1, v_2 [/tex] denote two speeds such that [tex] v_2 > v_1 [/tex]. Then
    [tex]card \{(v_x,v_y,v_x) \in R^3:v_x^2 + v_y^2 +v_z^2 = v_2^2\} > card \{(v_x,v_y,v_x) \in R^3:v_x^2 + v_y^2 +v_z^2 = v_1^2\} [/tex],
    where "card", of course, denotes the cardinality of the set. This is clearly a false statement; both sets are infinite.

    And he continues on with the derivation after that. It's really just that above point that is distressing.

    So... am I misinterpreting what he is trying to say? And if so, how should I read what he wrote? Or is this just sloppy language and/or logic? Or... help?!
    Last edited: Aug 24, 2010
  2. jcsd
  3. Aug 25, 2010 #2
    the right maths at the right time


    Assume A and B are two sets.
    Assume intersect(A,B) is empty (the intersection of the two sets).
    What would be wrong with such a statement:

    Card(Union(A,B)) >= Card(A) . . . . . . . . . . . . . . (statement)

    even when both sets A and B are infinite?
    It just means that to represesent the logic of infinite sets algebra, you need "cardinal numbers" instead of integers.

    I don't know anything about "cardinal numbers", I even don't know if that concept has been developped and if there something to read about that. However, the statement makes sense, and this sense can be explicited by many ways including by the simplest way: assuming v spans a discrete space instead of a continuous space and going to the limit. Another way out, is just to consider the volume of the sets (spheres) to be a measure of their cardinality. Probabilities might also offer another way to translate the statement.

    My flashback on my own process of learning physics is that I had put much too much hopes in mathematics. I believed, very naïvely, that learning more maths would made me better at physics. That's completely wrong: you need the right maths at the right time.

    For sure you understand that the problem of counting sets has really nothing to do with the more important question of "why the Maxwell distribution". Then, take your best Occam razor and cut that out of your analysis of the Maxwell distribution. Keep things as simple as possible, not more complicated than needed.
    You could become the world expert on counting inifinite sets, that you still might miss the answer to the main question: why the Maxwellian.
    And this answer was known for more than 130 years !!!
    To derive the Maxwell distribution, you might discover that very little maths are needed as well as very few assumptions.
    The main assumption is just the existence of this distribution.
    The symmetries of the problem then leave no other possibilities than the Maxwell distribution.
    Seen in this way, if your main interrest is in the maths, then I would suggest you to focus on the meaning of probabilities and on group theory, even though only basics will be needed. If you want mathematical insight on probabilities, then read Jaynes.

    To conclude, let me suggest you to read how Boltzmann and how Maxwell derived the distribution in their original work. You will find no trace of sophisticated set counting there, meaning that it is not useful for this purpose.

    Look further on this topic, it is really interresting.


    No sloppy language there, but wise language instead.
    It spares you a long and useless delay in understanding what really matters.
    Last edited: Aug 25, 2010
  4. Aug 25, 2010 #3
    Thanks for your reply lalbatros; however I think you missed my point. I will rephrase my question/objection.

    If v_1 and v_2 are two speeds and v_2 > v_1, why are there more velocity vectors associated with with v_2 than with v_1? (That's the author's claim, at least.) I don't think that it is true; it's kind of like saying that there are more points on the surface of a sphere of radius 2 than radius 1.

    By the way, I wasn't denying the correctness of the result, or questioning it's relevance or importance. I just don't understand this step in the derivation. Surely I'm not the first person in the last 130 years to stumble on this step.
  5. Aug 26, 2010 #4

    I have no access to your book as it is not available for preview on Amazon or on GoogleBooks.
    If you want more discussion, I would need at least a copy of the page where the argument appears.

    For my understanding the simplest derivation is based on:

    - assuming the existence of the Maxwell velocity distribution M
    - assuming the existence of distributions for the components of the velocities
    - assuming spherical symetry

    This leads to the constraint that M is a function of v² that must satisfy:

    M(vx²+vy²+vz²) = f(vx²) * f(vy²) * f(vz²)

    Only the exponential function satisfies this constraint and the Maxwell distribution results.

    As I said, I don't see why any other argument would be needed, unless the derivation starts from totally different assumptions.
    The assumptions above being very simple, I don't see why one should try another way, except maybe to have a deeper insight is statistical physics in general.

    For example, the Maxwell distribution can also be derived from the Maxwell-Boltzmann for energy distribution as a special case.
    This suggests that this MB distribution might also be derived from simple (symmetry) arguments, like the M distribution. Is that so?


  6. Aug 27, 2010 #5

    I agree with Lalbatros, if you get stacked with the way they derive the Maxwelll speed distribution formula, try another way -- there are several possible ways of deriving the formula, some of them are "more pedagogical" than the others.

    One very simple method is to use the classical canonical approach, starting from the partition function.

    My misfortune was that I for many years had to teach stat-mech courses, at undergrad and grad levels (misfortune - because I'm a lazy person, doing anything makes me unhappy). For deriving the Maxwell distribution function I used a recipe I had found in one book. It takes four or five lines. I don't have the details "at the top of my head" right now, but I can do that later today.

    "Truly classical" stat mech -- the one constructed by Maxwell, Botzmann, Gibbs, and a few other gents -- is a real *****. A "quantized" or "semi-quantum" approach is much better for a beginner, physics is not totally obscured by math in it! And essentially, all you need to switch from the "truly classical" approach to the "semi-quantum" one is to assume that there is not a continuum of states, but there are discrete states. And life becomes so much easier after that!
  7. Aug 27, 2010 #6
    The Wikipedia article on this subject is nearly identical to the discussion in Schroeder's book (indeed, the sole reference for the Wikipedia article on this subject IS Schroeder's book).

    Here is the link: http://en.wikipedia.org/wiki/Maxwell_speed_distribution" [Broken]

    I'm having trouble accepting the truth of the italicized statement (this is from the Wikipedia article, and is identical that what Schroeder says):

    "If we picture the particles with speed v in a 3-dimensional velocity space, we can see that these particles lie on the surface of a sphere with radius v. The larger v is, the bigger the sphere, and the more possible velocity vectors there are. So the number of possible velocity vectors for a given speed goes like the surface area of a sphere of radius v."

    I don't see how that is true, and I'd be very grateful if someone could prove that it is true.

    Anyway, I have since found a much more satisfactory discussion of this topic in Fundamentals of Statistical and Thermal Physics by Reif (chapter 7, section 10).
    Last edited by a moderator: May 4, 2017
  8. Aug 31, 2010 #7

    Sorry for the delay in answering your question. I have just come back home from my sabbatical year in Poland, and there are thousand of urgent things to do here at home...

    Well, the statement "on a larger sphere there are more velocity vectors" is indeed confusing. On every such sphere, no matter how big it is there is an infinite number of velocity vectors, right? When we deal with infinite sets, the meaning of "more" is quite tricky.

    So, my advice is that you switch to the "semi-quantum" approach to stat mech -- first, by considering the "standard statistical mechanical quantum model" of an ideal gas. In such approach, you consider a gas consisting of N non-interacting particles in a container of finite volume V And only after some additional consideration in the framework of this model, you may "go back" to the classical regime -- it is simple, you just expand the finite size of your gas to infinity. And then, I hope, the meaning of "more velocity vectors" will become clear to you.

    In the quantum ideal gas model, we begin with considering a single gas particle in a container of LxLxL dimensions. It's nothing else than the well-known "particle in the box" problem. A standard item in an introductory quantum mechanics problem is "a particle in a 1-D potential well", or "in a 1-D box". A transition to a 3-D box is just trivial.

    In a 1-D box of width L, as you surely remember, the particle wave-function are simply standing waves with nodes at x=0 and x=L. The lowest state (i.e., the ground state) is a standing wave of wavelength 2L, which means that in the "box" there is one-half of the basic wavelength. The next quantum state is a wave with three nodes, at x=0, x=1/2 L, and x=L, so the wavelength now is L; in the third state there are four nodes, so the wavelength is 2/3 L, and so on -- in general, for the n-th state the wavelength is (2/n)L.

    Don't worry, we don't need to go very far to get what we need -- however, there is one big problem form me. Namely, I don't know how to write math formulae in this editor! Someone told me that one can use LaTex. It's OK with me -- it even makes me quite happy -- because I know LaTex quite well, I have been using it for over 20 yars, and I can readily encode simple math formulae in LaTex without looking in the manual, because I have everything in my head. However, how should I use LaTex in this Forum?! Perhaps you can give me a hint, and then we can quickly get to where we need to.

    In short, what we want to do next: we want to start using the wavevector k instead of the wavelength -- and then describing the particle states in a 3-D box can bedone in term of a single 3-D vector, which we then "translate" to a particle velocity vector, thus obtaining the allowed particle velocity vectors in the box. It's really a simple task, but without a tool for writing elegant math equations I'm stuck.

    I'll tru to figure out by myself how to use LaTex here -- but, perhaps, if you give mi instructions, it may befaster.


  9. Aug 31, 2010 #8
    OK, I will try:
    This is for one dimension.

    And because the allowed values of wavelength are:

    where [tex]n=1,2,3,\ldots[/tex]

    we get:
    [tex]k_n=\frac{\pi n}{L}[/tex]

    These are the allowed values of the particle wavenumber ina a 1-D box.

    Now, let's see if it works!

    I checked using the "preview" tool -- it works fine!

    So, in my next posting we will write the formula for the allowed wavevectors in a 3-D box (a wavenumber is the same as a 1-D wavevector). And from that we will readily get the allowed particle velocity vectors.
    And then,after a few more really simple mathematical steps we will the result we need. But let me do that tomorrow.


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