The format of the 3 dimensional MB distribution is $A \cdot e^{-\frac{E}{k_BT}} \cdot g(E)$ in which $A$ can be derived using normalization (integration up to $\infty$ must be 1) and $g(E)$ being the degeneracy according to $g(E)=\frac{V\pi \cdot 2^{2.5}m^{1.5}}{h^3}$

The 3 dimensional average kinetic energy $\bar E$ of a particles system can then be calculated by multiplying this MB distribution with $E$ and integrating it over infinity, which yields: $$\bar E = \int_0^{\infty} \frac{2}{\sqrt \pi} \cdot (\frac{1}{k_BT})^{\frac{3}{2}} \cdot e^{-\frac{E}{k_BT}} \cdot \sqrt{E} \cdot E \cdot dE = \frac{3}{2}k_BT$$

The format for the 1 dimensional MB distribution (e.g. the x-coordinate) is $A\cdot e^{-\frac{E_x}{k_BT}}$ where $A$ is derived by normalizing the integration to 1, which gives $A= \frac{1}{k_BT}$ When calculating the 1 dimensional average energy $\bar E_x$, this MB distribution should also be multiplied by the energy $E_x$ and integrated up to $\infty$ which gives: $$\bar E_x=\int^{\infty}_0 \frac{1}{k_BT}\cdot e^{-\frac{E_x}{k_BT}}\cdot E_x\cdot dE = k_BT$$ But this should be $\frac{1}{2}k_BT$ instead. The peculiar thing is that when writing $E_x$ in terms of $\frac{1}{2}mv_x^2$ within the formula $A\cdot e^{-\frac{E_x}{k_BT}}$, normalizing $A$ to that, multiplying the formula with $\frac{1}{2}mv_x^2$ and integrating it up to $\infty$, then one would indeed get $\frac{1}{2}k_BT$. $$\int^{\infty}_0\frac{\sqrt{2m}}{\sqrt{\pi k_BT}}\cdot e^{-\frac{mv_x^2}{2k_BT}}\cdot \frac{1}{2}mv_x^2 \cdot dv=\frac{1}{2}k_BT$$ But it wasn't necessary for the 3 dimensional MB distribution to write the format down in terms of $v$ to get the correct average kinetic energy.

Why does the 1 dimensional MB distribution in terms of $E_x$ give an incorrect average energy and how would one realise that this is the wrong way to do it?


2 Answers 2


I think you forgot to take into account $g(E)$ in the calculation for the $1\mathrm{D}$ case. Actually, $g(E) \propto 1/\sqrt{E}$ in that case, which will lead to a different (and hopefully correct) answer.

The reason why you get a different answer by using $v_x$ as your integration variable is because the integrand $\propto \exp \left(-\frac{1mv^2}{2kT} \right) \frac{1}{2}mv^2 $, when changing the variable to $E \propto v^2$, has to be multiplied by $\frac{dv}{dE} \propto 1/\sqrt{E}$ which gives you back the degeneracy factor $g(E)$.


If I'm not mistaken, the degeneracy factor for the one-dimensional case is $g(E)=2$, since two free particles moving with the same speed but opposite directions ($-\hat{\mathbf{x}}$ and $\hat{\mathbf{x}}$) have the same kinetic energy.
In that case, the normalizing factor $A$ would be $1/(2k_B T)$ and you would obtain your desired result.

  • $\begingroup$ I don't understand why the factor 2 should be in the denominator instead of the numerator. If there's a degeneracy of 2, shouldn't the formula be multiplied by 2 instead of a half? $\endgroup$
    – Phy
    Commented Jul 28, 2019 at 22:58
  • $\begingroup$ You multiply first as in your first equation: $2A \exp{[-E/(k_B T)]}$, then normalize to obtain $A$. It follows that $A=1/(2k_B T)$. $\endgroup$
    – Lith
    Commented Jul 29, 2019 at 22:12
  • $\begingroup$ Thanks. This makes me think though, why isn't it necessary to take into account the different directions for the 3 dimensional formula $A \cdot e^{-\frac{E}{k_BT}} \cdot g(E)$? From what I understand, the degeneracy $g(E)$ only takes into account the different combination values of $E_x$, $E_y$ and $E_z$ that give the same sum, but not the $2$ different possible directions of each of them, which should be $8$ as a total (2 for each of the 3 degrees of freedom) $\endgroup$
    – Phy
    Commented Jul 29, 2019 at 22:20
  • $\begingroup$ Also, I noticed something regarding the multiplication with $2$ because of the two opposite directions. If there are two directions, then the 1D distribution should be $2A \cdot e^{-\frac{E}{k_BT}}$ and normalizing would give indeed $A= 1/(2k_BT)$. The formula would then be $2 \cdot \frac{1}{2k_BT} \cdot e^{-\frac{E}{k_BT}}$. But this simplifies again to $\frac{1}{k_BT} \cdot e^{-\frac{E}{k_BT}}$. $\endgroup$
    – Phy
    Commented Jul 30, 2019 at 0:03
  • $\begingroup$ Yeah, I noticed yesterday my answer is probably wrong (sorry about that). Now I'm at work but I will try to study this later today! $\endgroup$
    – Lith
    Commented Jul 30, 2019 at 8:53

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