# Average kinetic energy in 1 dimension according to Maxwell-Boltzmann Distribution

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?

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.

• 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?
– Phy
Commented Jul 28, 2019 at 22:58
• You multiply first as in your first equation: $2A \exp{[-E/(k_B T)]}$, then normalize to obtain $A$. It follows that $A=1/(2k_B T)$.
– Lith
Commented Jul 29, 2019 at 22:12
• 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)
– Phy
Commented Jul 29, 2019 at 22:20
• 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 $\frac{1}{k_BT} \cdot e^{-\frac{E}{k_BT}}$.
– Phy
Commented Jul 30, 2019 at 0:03
• 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!
– Lith
Commented Jul 30, 2019 at 8:53