[플라즈마 물리]Normalization Constant & Average Kinetic Energy

Problem The 3D Maxwellian distribution is given by
$$f(\vec{v}) = A_3 e^{\left( -\frac{m(v_x^2 + v_y^2 + v_z^2)}{2KT} \right)}$$
with $$n = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(\vec{v})dv_x dv_y dv_z$$
(a) Show that the normalization constant is given by $$A_3 = n \left( \frac{m}{2\pi KT} \right)^{\frac{3}{2}}$$
(b) Show that the average kinetic energy is $$E_{av} = \frac{3}{2}KT$$
Answer (a)

Notice that
$$\begin{aligned} \label{eq_gauss} \int_{-\infty}^{\infty} e^{-ax^2}dx = \sqrt{\frac{\pi}{a}}\end{aligned}$$
$$\begin{aligned} n &= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(\vec{v})dv_x dv_y dv_z \nonumber \\ &= A_3 \int_{-\infty}^{\infty} e^{ -\frac{m(v_x^2)}{2KT}}dv_x \int_{-\infty}^{\infty} e^{ -\frac{m(v_y^2)}{2KT}}dv_y \int_{-\infty}^{\infty} e^{ -\frac{m(v_z^2)}{2KT}}dv_z \nonumber \\ &= A_3 \left( \sqrt{\frac{ 2 K T \pi}{m}} \right)^{3}\end{aligned}$$
Hence,
$$A_3 = n \left( \frac{m}{2\pi KT} \right)^{\frac{3}{2}}$$
Answer (b)

Average energy can be calculated as
$$\begin{aligned} E_{av} &= \frac{\int_{-\infty}^{\infty} \frac{1}{2}mv^2 f(\vec{v})dv}{\int_{-\infty}^{\infty}f(\vec{v})dv} \nonumber \\ &=\frac{m}{2n}\int_{-\infty}^{\infty} (v_x^2 + v_y^2 + v_z^2) A_3 e^{ -\frac{m(v_x^2 + v_y^2 + v_z^2)}{2KT} }dv\end{aligned}$$
The denominator is $n$ as we calculated in previous problem. Notice that, $$\int x^2 e^{-ax^2}dx = \frac{1}{2}\sqrt{\frac{\pi}{a^3}}$$
If we calculate the first term of equation,
$$\begin{aligned} E_{av_x} &= \frac{m}{2n} \int_{-\infty}^{\infty} A_3(v_x^2 ) e^{ -\frac{m(v_x^2 + v_y^2 + v_z^2)}{2KT} }dv_x \nonumber \\ &=\frac{m}{2n}n \left( \frac{m}{2\pi KT} \right)^{\frac{3}{2}} \frac{1}{2} \left( \frac{2 KT}{m} \right)^{\frac{3}{2}}\sqrt{\pi}\frac{2KT \pi}{m} \nonumber \\ &=\frac{1}{2}KT\end{aligned}$$
Hence, if we calculate $v_y$ and $v_z$ components, final result becomes $$E_{av} = \frac{3}{2}KT$$
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