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Consider the most probable and RMS speeds for a kinetic gas: $$ v_{mps} = \sqrt{\frac{2RT}{M}}$$

$$ v_{rms} = \sqrt{ \frac{3RT}{M}}$$

Now consider the average speed:

$$ v_{avg} = \sqrt{\frac{8RT}{\pi M}}$$

Why does a factor of $\pi$ pop up in the above equation? I know the mathematical reason is due to integrations of gaussian integrals but usually $\pi$ we associate with circles/spheres. So, is there a physical way to give meaning as to why we should expect a $\pi$ in the formula here?

Related

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    $\begingroup$ Related: math.stackexchange.com/q/28558 and math.stackexchange.com/q/3239012 $\endgroup$ Aug 28, 2021 at 17:18
  • $\begingroup$ What does mps stand for? $\endgroup$
    – J.G.
    Aug 29, 2021 at 7:36
  • $\begingroup$ Most probable speed @J.G. $\endgroup$
    – Babu
    Aug 29, 2021 at 7:37
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    $\begingroup$ @Buraian I was typing this question and then saw yours .. +1.. $\endgroup$
    – Ankit
    Aug 29, 2021 at 10:09
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    $\begingroup$ Greats minds think alike @Ankit😂 $\endgroup$
    – Babu
    Aug 29, 2021 at 10:10

3 Answers 3

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Based on the $\pi$ term in the given $f_V(v)$, the question is not so much where the $\pi$ comes from in $v_{avg}$, but rather, “Where does the $\pi$ go in the others?”

First, note the $\pi$ term in our velocity pdf: $$f_p(v)=4 \pi v^2 (\frac{1}{\pi} \frac{m}{2k_bT})^{1.5} \exp(\frac{-mv^2}{2k_bT}) $$ $$ f_p(v)= j \frac{v^2}{\sqrt{\pi}} \exp(-kv^2)$$

$$\text{with }~k \equiv \frac{m}{2k_bT} ~,~ j\equiv 4k^{1.5}$$

First, it’s disappearing from $v_{mp}$ is maximization:


1.$v_{mp}$: The most probable speed is determined by setting the derivative of the pdf equal to zero (We are finding the maximum density point):

$$ \frac{\partial f_p}{\partial v} = \frac{j}{\sqrt{\pi}} (2v e^{-kv^2} -2kv^3 e^{-kv^2})=0$$

$$\implies \frac{2 j v e^{-kv^2}}{\sqrt{\pi}} (1-kv^2)=0$$

The resulting $v_{mp}=\operatorname{argmax}[f_p(\cdot)]=\frac{1}{\sqrt{k}}= \sqrt{\frac{2k_bT}{m}}$ does not include a $\pi$.


2.Now for $v_{avg}$, where the $\pi$, perhaps unsurprisingly, ends up as a radical in the denominator, as it is in the distribution. $$v_{avg} = \int vf_p(v)dv$$

With a very minor notation abuse.

$$ v_{avg}= \frac{j}{\sqrt{\pi}} \int v^3 \exp(-kv^2)dv$$

Based in this link from Vincent Thacker in the comments https://math.stackexchange.com/q/28558 , we see a $\pi$ will not appear with an exponent of $3$, hence the original $f_V(v)$’s $\frac{1}{\sqrt{\pi}}$ will go through to the $v_{avg}$.

Specifically:

$$ v_{avg}= \frac{j}{\sqrt{\pi}} \int v^3 ~ e^{-kv^2}dv$$

$$v_{avg}= \frac{-j}{2k^2\sqrt{\pi}}(kv^2+1) e^{-kv^2} \Big|_0^{\infty}$$

$$= 0 - \frac{-4k^{1.5}}{2k^2\sqrt{\pi}} ~ \cdot 1=\frac{2}{\sqrt{k \pi}} =\sqrt{\frac{8k_bT}{\pi m}}$$


3.Consider the $f_p(v)$ to $f_p(v^2)$ general transition with $x,y$: Note that $$F_Y(y)=P(Y\leq y)=P(X^2\leq y)=P(X\leq\sqrt{y})=F_X(\sqrt{y})=\int_0^\sqrt{y}3t^2\,dt=y^{3/2}$$ hence $$f_Y(y)=\frac{3}{2}\sqrt{y}\implies v^4$$

Based again in this link from Vincent Thacker in the comments https://math.stackexchange.com/q/28558 , we see a $\sqrt{\pi}$ will appear in the numerator with an exponent of $4$, hence the original $f_V(v)$’s $\frac{1}{\sqrt{\pi}}$ will not be in the expressions for $v_{rms}^2$ and then $v_{rms}$.

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  • $\begingroup$ Nice.. but this is again mathematical $\endgroup$
    – Babu
    Aug 29, 2021 at 10:08
  • $\begingroup$ Theres lots of words in there saying what the source of the pi is. $\endgroup$
    – Al Brown
    Aug 29, 2021 at 12:46
  • $\begingroup$ I have upvoted both answers, but I'll wait a bit more and see if I other answers. However, I do understand your point that the question should be why the pi doesn't appear there. Maybe there is something fundamental being missed. $\endgroup$
    – Babu
    Aug 30, 2021 at 8:13
  • $\begingroup$ Yes makes sense. I did think more about why intuitively pi is there. Did you see my last comment about not a circle but rather normal curve kicking out a pi, under other answer? Also, for $v_{mp}$, there is the reminder that pi never is part of the mean. If x is normal, I think E(x^3) and E(x^5) have a pi? If time I might write something on that in answer. Question actually does make lotta sense and I at least upvoted. Not even sure ours is guassian but gotta roll.. $\endgroup$
    – Al Brown
    Aug 30, 2021 at 18:45
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OP is asking for prose description of my other answer saying where the $\pi$ came from. So here that is - the prose part of the other answer expanded, plus a couple summary sentences at the end (jump to that if you wish):


Based on the $\pi$ term in the given $f_p(v)$, the question is not so much where the $\pi$ comes from in $v_{avg}$, but rather, “Where does the $\pi$ go in the others?”

First, note the $\pi$ term in our velocity pdf: $$f_p(v)=4 \pi v^2 (\frac{1}{\pi} \frac{m}{2k_bT})^{1.5} \exp(\frac{-mv^2}{2k_bT}) $$ If we plug in constants for the unchanging stuff and simplify, the $\pi$ becomes a radical in the denominator and the functional form is easier to see: $$ f_p(v)= (\tfrac{j}{\sqrt{\pi}}) ~ v^2 e^{-kv^2}$$

Where did this $\frac{1}{\sqrt{\pi}}$ go in two of our answers? After all, $\frac{1}{\sqrt{\pi}}$ appears exactly in $v_{avg}$. Why not in $v_{mp}$ and $v_{rms}$?

First, it’s disappearing from $v_{mp}$ is due to maximization.


1.$v_{mp}$: The most probable speed is determined by setting the derivative of the pdf equal to zero (We are finding the maximum density point):

The resulting $v_{mp}=\operatorname{argmax}$ does not include a $\pi$.

Generally when you maximize, the constants out front dont matter. They don’t affect the maximizing value $x^*$ that you pick. Whatever maximizes some $f(x)$ also maximizes $k ~f(x)$. So it’s unsurprising to not see a $k$ in the result.

That’s true in our maximization. Meaning $k$ is $\frac{1}{\sqrt{\pi}}$ here. So that explains why the $\pi$ term disappeared from the given pdf when finding $v_{mp}$. We didn’t actually integrate any $f_p$ form like with the other two, just picked a $v$ to maximize it.

2.Now for $v_{avg}$, where the $\pi$, perhaps unsurprisingly, ends up as a radical in the denominator, as it is in the distribution function $f_p$, because: $$v_{avg} = \int vf_p(v)dv$$

Based on this link from Vincent Thacker in the comments https://math.stackexchange.com/q/28558 for how to integrate that, we see a $\pi$ will not appear just from integrating with an exponent of $3$ (the $\int v^3 e^{-kv} dv$ form), hence the original $f_p(v)$’s $\frac{1}{\sqrt{\pi}}$ will go through to the $v_{avg}$. Because integrating did not create a $\pi$ to cancel it out.

3.Based again on the link for how to integrate that, we see a $\sqrt{\pi}$ will appear in the numerator with an exponent of $4$, (the $\int v^4 e^{-kv} dv$ form), and that is available to cancel out, hence the original $f_p(v)$’s $\frac{1}{\sqrt{\pi}}$ will not be in the expressions for $v_{rms}^2$ and then $v_{rms}$.

In Summary, the given pdf for the velocity had a $\sqrt{\pi}$ in the denominator, so it was initially surprising to not see it in all three results. For $v_{mp}$, the $\frac{1}{\sqrt{\pi}}$ was a leading coefficient which did not affect the selection of the pdf’s maximizing $v$ (where the slope is zero). For $v_{rms}$, it disappeared because integration of a form $\int v^4 e^{-kv} dv$ creates a $\sqrt{\pi}$ as a result, canceling the $\frac{1}{\sqrt{\pi}}$ in $f_p$.

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  • $\begingroup$ I accept this , but, your answer is not from the angle I wanted the question to be answered in $\endgroup$
    – Babu
    Aug 29, 2021 at 12:48
  • $\begingroup$ It’s the only possible answer. Integrating a $\int v^4 e^{-kv}$ form makes a $\pi$ appear as a mathematical result, which cancels the given $\pi$. Thats why it left $v_{rms}$. Again, it leaving the given pdf is what should be the confusion. Not the fact that youre given a pdf with it there and it ends up in your answer. $\endgroup$
    – Al Brown
    Aug 29, 2021 at 13:12
  • $\begingroup$ I would say that the pi is not about circles but is about the normal curve for $f_p(v)$. Have a good week $\endgroup$
    – Al Brown
    Aug 29, 2021 at 21:52
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To evaluate $I=\int_{-\infty}^\infty dx e^{-x^2}$ the trick is often to evaluate $$ I^2=\int_{-\infty}^\infty dx dy e^{-x^2}e^{-y^2}= 2\pi \int_{0}^\infty \rho e^{-\rho^2} = \pi $$ after the change to plane polar coordinates, so that $I=\sqrt{\pi}$. The $\pi$ factor comes from integrating over the polar angles.

A lot of non-zero Gaussian integrals contain $\sqrt{\pi}$ factor since they can be obtained by parametric differentiation w/r to $\lambda$ of the basic result \begin{align} \int_{-\infty}^\infty e^{-\lambda x^2}=\sqrt{\frac{\pi}{\lambda}}\, , \end{align} v.g. \begin{align} \frac{d}{d\lambda}\int_{-\infty}^\infty e^{-\lambda x^2}= -\int_{-\infty}^\infty x^2 e^{-\lambda x^2} =-\frac{\sqrt{\pi}}{2\lambda^{3/2}}\, , \end{align}

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    $\begingroup$ The OP has already told us that he knows how the expression is derived. $\endgroup$
    – garyp
    Aug 28, 2021 at 17:36
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    $\begingroup$ @garyp well I'm not sure that stating this come from Gaussian integrals is same as knowing, and if the OP does know then it hardly makes sense to ask the question in the first place... $\endgroup$ Aug 28, 2021 at 18:23
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    $\begingroup$ I guess it's how you read the post. My interpretation of the question is: what does average speed have to do with circles? (And why not the others.) The linked answers in the comment to the OP go there. Whether or not they get there can be debated. $\endgroup$
    – garyp
    Aug 28, 2021 at 21:13
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    $\begingroup$ @garyp it seems you were likely right… $\endgroup$ Aug 29, 2021 at 6:52

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