The speed of a molecule in a uniform gas at equilibrium is a random
variable $v$ whose density function is given by ${f_V}\left( v \right) = a{v^2}{e^{ – b{v^2}}},{\rm{ }}v > 0$. Derive the distribution $W = {{m{v^2}} \over 2}$ of the kinetic energy and compute $E(W)$.
I believe I got the first part right, deriving the distribution. Found my inverse in terms of $W$, which is $v = \sqrt {{{2W} \over m}} $ and applied in the formula:
$${f_w} = {f_v}\left( {{g^{ – 1}}\left( w \right)} \right)\left| {{d \over {dw}}{g^{ – 1}}\left( w \right)} \right|$$
Which returned me, if I didn't mess up writing it,
$$\eqalign{
& {f_w} = a{\left( {\sqrt {{{2W} \over m}} } \right)^2}{e^{ – b{{\left( {\sqrt {{{2W} \over m}} } \right)}^2}}}{1 \over {\sqrt {2Wm} }} \cr
& {f_w} = {{\sqrt {2w} a} \over {{m^{{3 \over 2}}}}}{e^{{{ – 2bw} \over m}}} \cr} $$
Now for the E(W) things seem worse for me, I got stuck in this integral below and not sure how to do it.
$$\eqalign{
& E\left( W \right) = \int_0^\infty {w{f_w}dw} = \int_0^\infty {w{{\sqrt {2w} a} \over {{m^{{3 \over 2}}}}}{e^{{{ – 2bw} \over m}}}dw} = \int_0^\infty {{w^{{3 \over 2}}}{{\sqrt 2 a} \over {{m^{{3 \over 2}}}}}{e^{{{ – 2bw} \over m}}}dw} \cr
& E\left( W \right) = {{\sqrt 2 a} \over {{m^{{3 \over 2}}}}}\int_0^\infty {{w^{{3 \over 2}}}{e^{{{ – 2b} \over m}w}}dw} \cr} $$
As a side note, this is from class where people from different backgrounds get accepted, not everyone has their math sharp, so I'm thinking maybe there's a less cumbersome way of doing this I cannot see.
Best Answer
The claimed density for velocity $$f_V(v) = a v^2 e^{-b v^2}, \quad v > 0,$$ is only a density if the following relationship between the parameters holds: $$\color{red}{\boxed{a \sqrt{\pi} = 4 b^{3/2}}. \tag{1}}$$ Otherwise, the density does not integrate to $1$. So we will consider $a$ to be a function of the single parameter $b$ for this distribution.
Once you obtained the density of $W$, you should recognize that it follows a gamma distribution with shape parameter $\alpha = 3/2$ and rate $\beta = 2b/m$, since $$X \sim \operatorname{Gamma}(\alpha,\beta)$$ implies $$f_X(x) = \frac{\beta^\alpha x^{\alpha-1} e^{-\beta x}}{\Gamma(\alpha)}, \quad x > 0.$$ Noting our earlier restriction on $a$ in $(1)$, some straightforward algebra demonstrates this choice of parameters is correct.
Then, we can utilize the known properties of the gamma distribution to obtain moments of $W$; e.g., $$\operatorname{E}[W] = \frac{\alpha}{\beta} = \frac{3m}{2b}.$$
It is also worthwhile to note that we need not explicitly compute the distribution of the transformed variable $W$ to compute $\operatorname{E}[W]$, since $$\operatorname{E}[W] = \operatorname{E}[m V^2/2] = \frac{m}{2}\operatorname{E}[V^2],$$ which then allows us to use the density of $V$ instead: $$\operatorname{E}[V^2] = \int_{v=0}^\infty av^4 e^{-bv^2} \, dv,$$ and with a substitution of the form $u = bv^2$, $du = 2 b v \, dv$, $v^3 = (u/b)^{3/2}$, we obtain $$\operatorname{E}[V^2] = \frac{a}{2b} \int_{v=0}^\infty v^3 e^{-bv^2} \cdot (2bv) \, dv = \frac{a}{2 b^{5/2}} \int_{u=0}^\infty u^{3/2} e^{-u} \, du = \frac{ \Gamma(5/2)a }{2 b^{5/2}}.$$ Then again substitute $(1)$ to obtain the result.