Wednesday, 4 December 2019

statistical mechanics - How can I explicit the energy dependence of the Maxwell-Boltzmann distribution?


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 ?



Answer



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 $$ f_\text{MB}(E)\,\text{d}E = 2n\left(\frac{1}{k_BT}\right)^{3/2}\sqrt{\frac{E}{\pi}}e^{-E/k_BT}\,\text{d}E. $$


No comments:

Post a Comment

Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...