Take limit of Maxwell-Boltzmann distribution

Question

Consider Maxwell-Boltzmann distribution:

$$f(v,T)=\left(\frac{m}{2\pi kT}\right)^{3/2} 4\pi \,v^2 \exp\left(-\frac{m}{2kT}v^2\right)$$ where $m$ is molecule's mass, $k$ is Boltzmann constant, $v$ is molecule's speed and $T$ is temperature. i want to calculate following limit: $$\lim_{T\rightarrow0} f(v,T)=?$$

I plotted the function $f$ in Matlab and the result for $T=0.0000000000000000000001$ is as follows. It seems that the limit goes to the Delta function at two points near zero. Can someone please help me to show that mathematically?

Gribouillis · Accepted Answer · 2017-08-28T06:19:07.360

It is true in general that if $f \in {L}^{1} \left(\mathbb{R}\right)$, then

$${\lim }_{{\epsilon} \rightarrow {0}^{+}} \frac{1}{{\epsilon}} f \left(\frac{{\nu}}{{\epsilon}}\right) = {{\delta}}_{0} \int_{\mathbb{R}}^{}f \left(x\right) d x$$

in the sense of distributions.

Proof: let ${\phi} \in {\mathscr{C}}_{0}^{\infty } \left(\mathbb{R}\right)$, then

$$\left\langle \frac{1}{{\epsilon}} f \left(\frac{\cdot }{{\epsilon}}\right) , {\phi}\right\rangle = \int_{\mathbb{R}}^{}\frac{1}{{\epsilon}} f \left(\frac{{\nu}}{{\epsilon}}\right) {\phi} {(\nu)} d {\nu} = \int_{\mathbb{R}}^{}f \left(x\right) {\phi} \left({\epsilon} x\right) d x$$

In the last integral, $f \left(x\right) {\phi} \left({\epsilon} x\right) \rightarrow f \left(x\right) {\phi} \left(0\right)$ almost everywhere and one has $$\left|f \left(x\right) {\phi} \left({\epsilon} x\right)\right| \leqslant \left(\max {|\phi|}\right) \left|f(x)\right| \in {L}^{1} (\mathbb{R})$$

The dominated convergence theorem implies that

$$\left\langle \frac{1}{{\epsilon}} f \left(\frac{\cdot }{{\epsilon}}\right), {\phi}\right\rangle \rightarrow {\phi} (0) \int_{\mathbb{R}}^{}f \left(x\right) d x$$

which proves our claim.

Now your result about Boltzmann follows with ${\epsilon} = \sqrt{T}$ and

$$f \left({\nu}\right) = {\left(\frac{m}{2 {\pi} k}\right)}^{3/2} 4 {\pi} {{\nu}}^{2} \exp \left({-\frac{m}{2 k}} {{\nu}}^{2}\right) {{\chi}}_{{\mathbb{R}}_{ > 0}}(\nu)$$

Edit: some more explanations with a FAQ

1) How does one prove that a family of function ${g}_{{\epsilon}}$ converges to the dirac function?

One proves that for any test functions ${\phi}$,

$$\int_{\mathbb{R}}^{}{g}_{{\epsilon}} \left(x\right) {\phi} \left(x\right) d x \mathop{\longrightarrow}\limits_{{\epsilon} \rightarrow 0} {\phi} \left(0\right)$$

one says that ${g}_{{\epsilon}}$ tends to the dirac measure in the sense of the distributions.

2) What is a test function ?

It is a function infinitely smooth (all its derivatives exist in $\mathbb{R}$) and its value is identically $0$ outside some interval $\left[a , b\right]$. The set of tests functions is denoted ${\mathscr{C}}_{0}^{\infty } \left(\mathbb{R}\right)$ or $\mathscr{D} \left(\mathbb{R}\right)$

3) What does the dominated convergence theorem say?

The DCT gives sufficient conditions to prove the following limit of integrals

$$\int_{\mathbb{R}}^{}{g}_{{\epsilon}} \left(x\right) d x \mathop{\longrightarrow}\limits_{{\epsilon} \rightarrow 0} \int_{\mathbb{R}}^{}g \left(x\right) d x$$

These conditions are that a) ${g}_{{\epsilon}} \left(x\right) \mathop{\longrightarrow}\limits g \left(x\right)$ for almost all $x \in \mathbb{R}$, that is to say for all $x$ but at most a set of measure $0$. b) The integrals $\int_{\mathbb{R}}^{}\left|{g}_{{\epsilon}} \left(x\right)\right| d x$ exist. and c) there exists some function $h \left(x\right)$ such that $\left|{g}_{{\epsilon}} \left(x\right)\right| \leqslant \left|h \left(x\right)\right|$ and $h \left(x\right)$ does not depend on ${\epsilon}$ and $\int_{\mathbb{R}}^{}\left|h \left(x\right)\right| d x$ exist.

I understood the whole answer of you, but I would appreciate it if you could explain the proof in a simpler way, or give me a reference. — Optimized Life, Aug 27 '17 at 07:58
What do you want me to explain in a simpler way? Do you know the dominated convergence theorem? Did you follow a course in distribution theory? — Gribouillis, Aug 27 '17 at 08:02
i dont know the dominated convergence theorem. I'm new to this area (distribution theory). — Optimized Life, Aug 27 '17 at 08:16
Thank you so much, i think i understand. Just another question, test function in my problem is $\delta_0$? — Optimized Life, Aug 27 '17 at 08:53
why $f \left(x\right) {\phi} \left({\epsilon} x\right) \rightarrow f \left(x\right) {\phi} \left(0\right)$? — Optimized Life, Aug 27 '17 at 09:53
Because $\phi$ is continuous, in particular it is continuous at $0$ which means that $\epsilon x \to 0 \Longrightarrow \phi(\epsilon x) \to \phi(0)$. A test function $\phi$ is any function that is continuous in $\mathbb{R}$, which all derivatives of any order $\phi^{(k)}$ exist and are continuous on $\mathbb{R}$ and such that $\phi(x) = 0$ when $|x|$ is large enough. — Gribouillis, Aug 27 '17 at 10:03

Victor Palea · Answer 2 · 2017-08-27T07:50:36.527

I'll make the constant parameters $=1$ for simplicity. The equation becomes

$$v^2\left(\frac{1}{T}\right)^{\frac{3}{2}}exp\left(-\frac{v^2}{T}\right)$$

Make the substitution $\frac{1}{T}=x$ so the relation becomes

$$v^2 \cdot x^{\frac{3}{2}}exp(-v^2 \cdot x) = v^2\frac{x^{\frac{3}{2}}}{exp(v^2 \cdot x)}$$

For $x\rightarrow \infty$ the relation goes to $0$.

Edit after discussion in comments.

I did some search and got to this answer. The function is not meant to work for $T=0$. Here is how I got to this.

From a physical point of view, we would expect that at $T=0$ the velocity distribution to be the delta function, such that all particles have $v=0$. Now, based on this distribution, that result is impossible because the function is made in such a way that $f(v=0,T)=0$ for $\forall T$.

Now, if we look at the position of the most probable velocity, that is

$$v_p=\sqrt{\frac{2kT}{m}}$$

we see $v_p\rightarrow 0$ as $T\rightarrow 0$. Also, the variance of the distribution is proportional to $T$. Those result are taken from wikipedia. So the distribution goes to $v=0$ and becomes narrower as it does so. But by it's definition, the value at $v=0$ will remain $0$ no matter the temperature.

That is, I assume, why when the limit of the distribution is taken, the result is a 'function' that is $0$ for $v>0$ and is already $0$ at $v=0$ because that's how it is defined.

So my conclusion is that the function is not meant to be used at $T=0$.

If you however are not satisfied with my answer, then have a look at this. Since one of your arguments is that the integral of the distribution should be constant for $T\ge 0$, and since intuition says that at $T=0$ the function is of type delta, if you want to check the integral of this, you will have to integrate over 'half' of the delta function, which does not look very friendly if you want to be rigorous.

Anyway, hope this help somehow.

here $v$ is not constant, the main problem is with $v^2$ before $exp$. If we remove this $v^2$, then for $T \rightarrow 0$, $f$ goes to the delta function at $v=0$. — Optimized Life, Aug 27 '17 at 06:38
I know it is not a constant, but it does not matter. I'll edit the answer. — Victor Palea, Aug 27 '17 at 06:41
As shown in the figure, the limit is not zero. Since in that case the distribution function does not have the necessary conditions for the distribution function — Optimized Life, Aug 27 '17 at 06:45
This distribution function is a well-known distribution function in physics — Optimized Life, Aug 27 '17 at 06:46
I included the velocity. Even in this case, the limit will give 0. And it makes sense from a physical point of view. — Victor Palea, Aug 27 '17 at 06:46
if $f$ is a distribution function, then
$$\sum_{v}f(v,T)=1$$ — Optimized Life, Aug 27 '17 at 06:48

score 0 · Answer 3 · answered Aug 27 '17 at 07:22

First, I'd like to point out that your graph is incorrect. The Maxwell-Boltzmann distribution is a speed distribution; it is only defined for nonnegative $v$.

Anyways, let's write this in a simpler form using the thermal speed $v_T = \sqrt{T/m}$. $$ f(v,v_T) = \sqrt{\frac{2}{\pi}}\frac{1}{v_T} \left(\frac{v}{v_T}\right)^2\exp\left[-\frac{1}{2}\left(\frac{v}{v_T}\right)^2\right] $$ So how does this function behave? We have \begin{eqnarray} \forall v , \lim_{v_T\rightarrow 0}f(v,v_T) &=& 0\\ \forall v_T , \int_0^\infty f(v,v_T)dv &=& 1\\ \max_{v\in[0,\infty)}f(v,v_T) = f(\sqrt{2}v_T, v_T) &=& \frac{2\sqrt{2}}{e\sqrt{\pi}}\frac{1}{v_T} \end{eqnarray} There's clear parallels to the delta function here: vanishes almost everywhere, has finite integral, and appears to be the limit of an increasingly peaky function. However, the limit is actually zero everywhere, whereas most limit expressions of delta functions at least have one point where the limit diverges. So in that way its limit isn't a delta function. Pointwise, it converges to the zero function.

On the other hand, the delta function isn't a function at all. Its real defining factor is the integral sifting property: $ \int_a^b g(x)\delta(x)dx = f(0)$ if $a < 0 < b$. So if you could show $$ \lim_{v_T\rightarrow 0}\int_0^\infty g(v)f(v,v_T)dv = g(0) $$ for all $g$, that would in a sense show its limit is (sort of) a delta function. This can probably be done using function value estimates, but I'll leave that as an exercise for the reader.

Take limit of Maxwell-Boltzmann distribution

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