I'm having a bit of a problem figuring out the energy dependent Maxwell-Boltzmann distribution.

According to my book (Ashcroft & Mermin) they write the velocity dependent distribution as:

$${{f}_{MB}}\left( \mathbf{v} \right)=n{{\left( \frac{m}{2\pi {{k}_{B}}T} \right)}^{3/2}}{{e}^{-m{{v}^{2}}/2{{k}_{B}}T}},$$

where $n = N/V.$

But how do I change the variables so it will become energy ($\epsilon$) dependent? The term in the exponential, $-\frac{mv^{2}}{2k_{B}T}$, I should be able to make the switch $\epsilon = \frac{mv^{2}}{2}$ so that I will get $e^{-\frac{\epsilon}{k_{B}T}}$, but I'm pretty sure that is not the only thing I need to do to make it energy dependent $(f_{MB}(\epsilon))$, or am I wrong ?

  • $\begingroup$ I think you are right. The only energy dependent factor in your expression is the exponential. Therefore, that alone changes to $\exp(-\epsilon/k_BT)$. $\endgroup$
    – Amey Joshi
    May 16, 2013 at 13:22

2 Answers 2


You have to take into account the differentials. The actual equation is $$ f_\text{MB}(\mathbf{v})\,\text{d}v_x\text{d}v_y\text{d}v_z = n\left(\frac{m}{2\pi k_BT}\right)^{3/2}e^{-mv^2/2k_BT}\,\text{d}v_x\text{d}v_y\text{d}v_z. $$ Changing to spherical coordinates, we get $$ \text{d}v_x\text{d}v_y\text{d}v_z = v^2\sin\theta\,\text{d}\theta\,\text{d}\varphi\,\text{d}v. $$ Integrating $\theta$ and $\varphi$, this becomes $$ v^2\,\text{d}v\int_0^{2\pi}\text{d}\varphi\int_0^\pi\sin\theta\,\text{d}\theta = 4\pi v^2\,\text{d}v, $$ so we have $$ f_\text{MB}(v)\,\text{d}v = 4\pi n\left(\frac{m}{2\pi k_BT}\right)^{3/2}v^2e^{-mv^2/2k_BT}\,\text{d}v. $$ Now you can change $v$ to $E$. Using $$ \text{d}E = mv\,\text{d}v = \sqrt{2mE}\,\text{d}v, $$ you eventually obtain $$ f_\text{MB}(E)\,\text{d}E = 2n\left(\frac{1}{k_BT}\right)^{3/2}\sqrt{\frac{E}{\pi}}e^{-E/k_BT}\,\text{d}E. $$

  • 1
    $\begingroup$ Why the hell was my answer downvoted? $\endgroup$
    – Pulsar
    May 16, 2013 at 13:44
  • $\begingroup$ I didn't do it... But it looks good. I'm just going it through atm. $\endgroup$ May 16, 2013 at 13:49

The distribution function allows to find some kind of probability by summing of $f_vdv$ if it is the distribution over velocities or $f_EdE$ if it concerns energies. One can be transformed to other as:

$$f_v(v) dv=f_v(E) v dE,$$

since $dv=vdE$ if $E$ is a function of $v$

Taking into account $E=\frac{mv^2}{2}$, one gets:

$$f_v(v)dv=f_v(E) \sqrt{\frac{2E}{m}} dE \Rightarrow f_E (E)=f_v(E) \sqrt{\frac{2E}{m}},$$


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