User:Willyfoobar/sandbox

= Energy Distribution of an Ideal-Gas with n-degrees of Freedom =

Used Functions
$$\Gamma(x)$$ and $$\gamma(x)$$, are the upper and lower incomplete gamma function.

With the value $$\Gamma(\frac{3}{2})=\frac{\sqrt{\pi}}{2}$$

The PDF and the CDF of the Chi-squared distribution are:

$$\chi^2_{PDF}(x)=\frac{1}{ {2}^{\frac{k}{2}}~\Gamma( \frac{k}{2} ) }~{x}^{\frac{k}{2}-1}~\exp(-\frac{x}{2})$$ $$\chi^2_{CDF}(x)=\frac{1}{ \Gamma( \frac{k}{2} ) }~\gamma(\frac{k}{2}, x)$$

The distribution for a velocity in a single direction (Degrees of Freedom n=1)
$$f(V_i) = \sqrt{\frac{m}{2~\pi~kT}}~\exp\left[-\frac{m~V^2_i}{2~kT}\right]$$

The distribution for absolute velocity (Degrees of Freedom n=3)
$$f(v) = {(\frac{m}{2~\pi~kT})}^{\frac{3}{2}}~4~\pi~v^2 \exp\left[-\frac{m~v^2}{2~kT}\right]$$

The Energy Distribution
using: $$\epsilon = \frac{1}{2}~m~v^2=\frac{1}{2}~m~(V^2_x+V^2_y+V^2_y)$$ $$k=3$$ and a scale parameter $$a=\sqrt{\frac{kT}{m}}$$

$$f(v) = {(\frac{1}{2~\pi})}^{\frac{1}{2}} ~2~\frac{v^2}{a^3} ~\exp\left[-\frac{1}{2}~{(\frac{v}{a})}^2\right]$$

CALCULUS TRY 2:
$$f_\epsilon(\frac{\epsilon}{kT}) = \frac{2}{\sqrt{\pi}}~\sqrt{\frac{\epsilon}{kT}}~\exp\left[-\frac{\epsilon}{kT}\right]$$

Rewriting the Energy Distribution as Maxwell-Boltzmann Distribution
Using $$k=3$$ and $$x = 2 ~ \frac{\epsilon}{kT}$$.

Equation (x) can be rewritten as

$$f_\epsilon(\frac{x}{2}) = \frac{1}{\Gamma{\frac{k}{2} } }~{x}^{\frac{k}{2}-1}~\exp(\frac{x}{2})= \chi^2_{PDF}(x)$$

So the Energy Cumulative Distribution Function (CDF) of three degree of freedom $$n=3$$ is:

$$ F_\epsilon(\frac{x}{2}) = \chi^2_{CDF}(x) = \frac{1}{\Gamma{\frac{k}{2}}}~\gamma(\frac{k}{2}, x)$$

$$ F_\epsilon(\frac{\epsilon}{kT}) = \frac{1}{\Gamma{\frac{3}{2}}}~\gamma(\frac{3}{2}, 2 ~ \frac{\epsilon}{kT})$$

Energy Distribution of n-Degrees of Freedom
With the energy of a single degree $$E_{n=1}$$ $$\epsilon=3 ~ E_{n=1}$$ We have $$E_{n=1} = \epsilon \times \frac{1}{3} ~ $$

I believe the following is NOT VALID?
the energy in a diatomic gas with n=5 is $$E_{n=5}=5 \times E_{n=1} = \frac{5}{3} \times \epsilon$$

$$\epsilon=E_{n=5} \times \frac{3}{5}  $$

$$ F_\epsilon(\frac{\epsilon}{kT}) = \frac{1}{\Gamma{\frac{3}{2}}} \gamma(\frac{3}{2}, 2\frac{3}{5} ~ \frac{E_{n=5}}{kT})$$