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Looking on internet I came across the following formulation of the Maxwell-Boltzmann distribution for the kinetic energy (see the following link: https://www.chegg.com/homework-help/questions-and-answers/maxwell-boltzmann-distribution-function-f-v-describing-actual-distribution-molecular-speed-q22909713):

$\frac{8\pi}{m}\left(\frac{m}{2\pi kT}\right)^{\frac{3}{2}}\varepsilon e^{-\frac{\varepsilon}{kT}}$

Now, I know instead the following formulation:

$\frac{8\pi}{\sqrt{2}}\left(\frac{1}{2\pi kT}\right)^{\frac{3}{2}}\sqrt{\varepsilon}e^{-\frac{\varepsilon}{kT}}$

Which of these formulations is the correct one? Thank you for your time!

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2 Answers 2

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Adding onto the other correct answer, I think the best way to understand both formulas and to remember the difference is to start with the smallest surface area of knowledge possible. The most fundamental starting point for either formula is that the probability for a free particle to lie in a state with energy $\varepsilon$ is proportional to $e^{-\varepsilon / k_B T}$. This is known as a Boltzmann factor. With just this assumption, you can ask: what is the probability density for a particle to have velocity $\vec{v}$? This would be obtained by simply normalizing the above Boltzmann weight: $$ P(\vec{v}) = \frac{e^{-m v^2 / 2k_B T}}{\int d^3 v \, e^{-m v^2 / 2k_B T}} = \Big[ \frac{m}{2\pi k_B T} \Big]^{3/2} e^{-m v^2 / 2k_B T} $$ Multiplying by $d^3 v$, this is the probability for a particle to have velocity within a small cube centered on $\vec{v}$ with volume $d^3 v$. Now, you can ask: what is the probability density for a particle to have speed $v$? For a particle to have speed between $v$ and $v + dv$, its velocity must lie on a thin spherical shell of radius $v$ and thickness $dv$, with a total infinitesimal volume $4\pi v^2 dv$. The probability density for a particle to have speed $v$ is then simply $$ f(v) \, dv = P(v) \, d^3 v= P(\vec{v})4\pi v^2 dv = 4\pi \Big[ \frac{m}{2\pi k_B T} \Big]^{3/2} v^2 e^{-m v^2 / 2k_B T} dv $$ This recovers your first equation, and makes clear its meaning: it is the probability for the speed of a particle to lie between $v$ and $v + dv$ (divided by $dv$). Finally, you can ask: what is the probability density for a particle to have energy $\varepsilon$? To answer this, note that when the velocity $v$ lies within an interval $dv$, the energy lies within an interval $d\varepsilon = d\varepsilon / dv \, dv = mv\, dv$. Therefore, the probability for the energy to lie between $\varepsilon$ and $\varepsilon + d\varepsilon$ is $$ g(\varepsilon) d\varepsilon = f(v) dv = f(v) \frac{d\varepsilon}{mv} = 4\pi \Big[ \frac{1}{2\pi k_B T} \Big]^{3/2} v \sqrt{m} e^{-mv^2 / 2 k_B T} d\varepsilon = 4\pi \sqrt{2} \Big[ \frac{1}{2\pi k_B T} \Big]^{3/2} \sqrt{\varepsilon} e^{-\varepsilon / k_B T} d\varepsilon $$ This recovers your second equation, and again makes clear its meaning: it is the probability for the energy to lie between $\varepsilon$ and $\varepsilon + d\varepsilon$. It also makes clear why $g(\varepsilon)$ is different from $f(v)$: as $v$ becomes larger, a given interval $dv$ in the speed corresponds to a larger and larger energy interval $d\varepsilon$ since $\varepsilon$ is proportional to $v^2$.

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What's given on the site is $$f(v)=(\cdots)\epsilon(v) e^{-\beta\epsilon(v) }$$ What you have written is $$g(\epsilon)=(\cdots ) \sqrt{\epsilon}e^{-\beta \epsilon}$$

Now, what's the difference. In the first $f(v)dv$ gives the probability for the particle's velocity to be between $v$ and $v+dv$. In the second, $g(\epsilon)d\epsilon$ gives the probability for the particles' energy to be between $\epsilon$ and $\epsilon+d\epsilon$.

Both are correct! Though there isn't any meaning to write $f(v)$ as a function of $\epsilon(v)$ because it would remain the distribution of velocity. You can not regard $f(v(\epsilon))d\epsilon$ as probability for the particles' energy to be between $\epsilon$ and $\epsilon+d\epsilon$.

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