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In terms of energy, the 3 dimensional MB Distribution is giving the probability for a particle to have an energy $E \geq E + dE$ is: $$f(E) = \frac{2}{\sqrt \pi} \cdot \bigg(\frac{1}{k_BT}\bigg)^{\frac{3}{2}} \cdot e^{-\frac{E}{k_BT}} \cdot \sqrt{E} \cdot dE$$ It is said that the 1 dimensional MB Distribution, giving the probability for a particle to have a certain energy in 1 degree of freedom, say $E_x \geq E_x + dE_x$, is: $$f(E_x) = \sqrt{\frac{1}{\pi E_x k_B T}} \cdot e^{-\frac{E_x}{k_BT}} \cdot dE_x$$ How is this 1D MB-Distribution derived from scratch and from the 3D MB Distribution?

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The MB distribution for the one degree of freedom in terms of momentum is: $$ w(p_x) dp_x = \frac1{\sqrt{2\pi m k_BT}} e^{-\frac{p_x^2}{2mk_BT}} dp_x $$ The relation between $p_x$ and $E$ is $$ E = \frac{p_x^2}{2m}\ \longleftrightarrow \ p_x(E) = \pm\sqrt{2mE}. $$ Probability "conservation" requires $$ f(E) dE = 2 w(p_x(E)) dp_x(E), $$ where $p_x = \sqrt{2mE}$ is chosen and $2$ factor accounts for two possibilities. From the last equation follows $$ f(E) = 2\ w(p_x(E))\ |p_x'(E)|. $$ This formula leads to the needed expression.

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  • $\begingroup$ Thanks. I got 2 questions. 1. Is the $2$ multiplier from the fact that each momentum value can have to directions? 2. Is it possible to derive the 1 dimension MB Distribution from scratch instead of just rewriting another existing formula in terms of energy? $\endgroup$
    – Phy
    Jul 29, 2019 at 21:39
  • $\begingroup$ @JohnnyGui My answers are 1. Yes, 2. I don't know a method to derive MBD in 1D from scratch. $\endgroup$
    – Gec
    Jul 29, 2019 at 22:00
  • $\begingroup$ Doesn't the $w(p_x)dp_x$ already take the 2 possible directions into account? If not, how would $w(p_x)dp_x$ then give correct probabilities in the first place? $\endgroup$
    – Phy
    Jul 29, 2019 at 22:03
  • $\begingroup$ I've been trying to write an answer satisfying the "from scratch" criterion but it's turning out to be very difficult without some experience with statistical mechanics; do you have any? There isn't a lot of prior knowledge involved, but there's quite a few integrals over phase space involved and I don't want to waste a lot of effort on something no-one will read. $\endgroup$ Jul 29, 2019 at 22:34
  • $\begingroup$ @Heatherfield This is exactly what I was asking for and what I'm trying to understand. I don't know how this phase space is derived and used to derive the 1D MB Distribution. The format for the 3D MB Distribution is $A \cdot e^{-\frac{E}{k_BT}} \cdot g(E)$ where $A$ is a constant that can be gotten through integration and normalizing it to 1 and $g(E)$ being the degeneracy. I'm doubting that the 1D format is merely this formula without $g(E)$. $\endgroup$
    – Phy
    Jul 29, 2019 at 23:28
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Consider a canonical ensemble of a single particle in a 1D container of length $L$. According to the (classical) Boltzmann probability distribution, the probability density function for the state of the system being the microstate with position $x'$ and momentum $p'$ is $$ f_{x,p}(x',p') = \frac{e^{-\beta H(x', p')}}{\int\limits_0^L dx'' \int\limits_{-\infty}^\infty dp'' e^{-\beta H(x'',p'')}}$$ where $ H(x, p) = p^2/2m $ is the energy of the particle with position $x$ and momentum $p$, and $\beta = 1/k_BT$. The marginal probablilty density function for the momentum is $$ f_p(p') = \int\limits_0^L dx' f_{x,p}(x',p') = \frac{\int\limits_0^L dx'e^{-\beta p'^2/2m}}{\int\limits_0^L dx'' \int\limits_{-\infty}^\infty dp'' e^{-\beta p''^2/2m}} = \frac{L e^{-\beta p'^2/2m}}{L\int\limits_{-\infty}^\infty dp'' e^{-\beta p''^2/2m}} = \sqrt{\frac{\beta}{2\pi m}} e^{-\beta p'^2/2m}. $$

To calculate the probability distribution function for the energy, note that the probability that the energy is less than $E'$ is given by

$$ \int\limits_0^E' dE'' f_E(E'') = \int\limits_{-\sqrt{2mE'}}^\sqrt{2mE'} dp' f_p(p'') = 2\int\limits_0^\sqrt{2mE'} dp' f_p(p'') . $$ Differentiating both sides and using the fundamental theorem of calculus, $$f_E(E') = 2 \sqrt{\frac{m}{2E'}} f_p(\sqrt{2mE'}) = \sqrt{\frac{2m}{E'}} \sqrt{\frac{\beta}{2\pi m}} e^{-\beta E'} = \sqrt{\frac{\beta}{\pi E'}} e^{-\beta E'}. $$

If you don't know where the Boltzmann probability distribution that I started with comes from, it is a standard result of statistical mechanics that should be found in any statistical mechanics textbook. If you want to ask about that, I think that probably deserves its own question.

If you wish to start with a particle free to move in a 3D container of volume $V$,

$$ f_{\vec{x},\vec{p}}(\vec{x}',\vec{p}') = \frac{e^{-\beta H(\vec{x}', \vec{p}')}}{\int\limits_V d^3\vec{x}'' \int\limits_{-\infty}^\infty d^3\vec{p}'' e^{-\beta H(\vec{x}'',\vec{p}'')}}$$

$$ f_\vec{p}(\vec{p}') = \int\limits_V d^3\vec{x}' f_{\vec{x},\vec{p}}(\vec{x}',\vec{p}') = \frac{\int\limits_V d^3\vec{x}'e^{-\beta p'^2/2m}}{\int\limits_V d^3\vec{x}'' \int\limits_{-\infty}^\infty d^3\vec{p}'' e^{-\beta p''^2/2m}} = \frac{V e^{-\beta p'^2/2m}}{V\int\limits_{-\infty}^\infty d^3\vec{p}'' e^{-\beta p''^2/2m}} = \frac{e^{-\beta p'^2/2m}}{\left( \int\limits_{-\infty}^\infty dp_x'' e^{-\beta p_x''^2/2m} \right) \left( \int\limits_{-\infty}^\infty dp_y'' e^{-\beta p_y''^2/2m} \right) \left( \int\limits_{-\infty}^\infty dp_z'' e^{-\beta p_z''^2/2m} \right)} = \left( \frac{\beta}{2\pi m} \right)^{3/2} e^{-\beta p'^2/2m} = \left( \frac{\beta}{2\pi m} \right)^{3/2} e^{-\beta (p_x'^2 + p_y'^2 + p_z'^2)^2/2m}. $$

This is the joint probability density function for all three components of the momentum. If you want the marginal probability density function for the $x$-component only, $$ f_{p_x}(p_x') = \int\limits_{-\infty}^\infty dp_y' \int\limits_{-\infty}^\infty dp_z' f_\vec{p}(\vec{p}') = \left( \frac{\beta}{2\pi m} \right)^{3/2} e^{-\beta p_x'^2/2m} \int\limits_{-\infty}^\infty dp_y' e^{-\beta p_y'^2/2m} \int\limits_{-\infty}^\infty dp_z' e^{-\beta p_z'^2/2m} = \sqrt{\frac{\beta}{2\pi m}} e^{-\beta p_x'^2/2m} $$ so you end up with the same distribution as the 1D case. If you want the distribution of the energy due to the motion along the $x$ direction (i.e. $E_x = p_x^2/2m$), the math to derive it is identical to the 1D case above, and you end up with $$f_{E_x}(E_x') = \sqrt{\frac{\beta}{\pi E_x'}} e^{-\beta E_x'}. $$

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  • $\begingroup$ Thanks a lot for this. Should it also be possible to write the 3D MB distribution in terms of velocity and from that deriving the one dimensional velocity term $v_x$ (using the formula for a volume $dv$ on a velocity sphere), and then writing it back in terms of energy, which would give your derived 1 dimensional MB distribution? $\endgroup$
    – Phy
    Jul 30, 2019 at 14:37
  • $\begingroup$ Please see my updated answer. $\endgroup$
    – Puk
    Jul 30, 2019 at 18:50
  • $\begingroup$ Thanks so much for your effort. One last question, the link here physics.byu.edu/faculty/christensen/Physics%20427/FTI/… also derived the 1D MB Distribution in terms of $v_x$. When rewriting this formula in terms of $E_x$, there is an extra factor of $\frac{1}{2}$ that comes up compared to your derivation. I'm amazed that, when written in terms of velocity, one does not have to take the possibility of 2 directions into account, but when it is rewritten in $E_x$, then it must be multiplied by $2$ to make sense. $\endgroup$
    – Phy
    Jul 30, 2019 at 23:10
  • $\begingroup$ I'm not sure what the question is. Are you saying that that the distribution of $E_x$ that follows from the link (which I don't see by the way) differs from the expression I gave by a factor of 2, or are you saying they explicitly multiply something by 2 while I don't? Note that this factor of 2 also appears in the second equation I gave for the probability of the energy being less than $E'$. It arises because $f_{p}{(p')$ is an even function being integrated over a symmetric interval. Physically, it is a consequence of the fact that there are two momenta corresponding to each positive energy. $\endgroup$
    – Puk
    Jul 30, 2019 at 23:53
  • $\begingroup$ You mean you don't see the link? I have rewritten the link's formula (which was in terms of $v_x$) in terms of $E_x$ myself and an extra factor of $\frac{1}{2}$ came up compared to your derivation. Is it because, in terms of $v_x$, the formula calculates the probability of that velocity only in 1 direction, while when converting it into $E_x$, you should also take the opposite possible direction into consideration and therefore the probability in terms of $E_x$ should be multiplied by a factor of $2$? $\endgroup$
    – Phy
    Jul 31, 2019 at 22:50
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Simply we can derive constant from one dimensional Maxwell-Boltzmann distribution as follows: Here, Maxwell-Boltzmann distribution is $$f(u)=Ae^{(-mu^2/2KT)}$$ let, $a = m/KT$

we know $\int_{-\infty}^{\infty} e^{(-au^2/2)}du = \sqrt{\frac{2\pi}{a}}=\sqrt{\frac{2\pi m KT}{m}}$

So, the normalized one dimensional Maxwellian distribution, $$\int_{-\infty}^{\infty}f(u)du= 1 $$ $$ A = \sqrt{\frac{m}{2\pi m KT}}$$

  • for $3$D cases, $A = (\frac{m}{2\pi m KT})^{3/2}$

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