# 6.5: Combining the One-dimensional Probability Density Functions

In Section 4.4, we derive the probability density function for one Cartesian component of the velocity of a gas molecule. The probability density functions for the other two Cartesian components are the same function. For $$\mathop{v}\limits^{\rightharpoonup}=\left(v_x,v_y,v_z\right)$$, we have $$v^2=v^2_x+v^2_y+v^2_z$$, and

\begin{aligned} \frac{df_x\left(v_x\right)}{dv_x}= \left(\frac{\lambda }{2\pi }\right)^{1/2}\mathrm{exp}\left(\frac{-\lambda v^2_x}{2}\right) \\ \frac{df_y\left(v_y\right)}{dv_y}= \left(\frac{\lambda }{2\pi} \right)^{1/2}\mathrm{exp}\left(\frac{-\lambda v^2_y}{2}\right) \\ \frac{df_z\left(v_z\right)}{dv_z}= \left(\frac{\lambda }{2\pi }\right)^{1/2}\mathrm{exp}\left(\frac{-\lambda v^2_z}{2}\right) \end{aligned}

We now want to derive the three-dimensional probability density function from these relationships. Given these probability density functions for the Cartesian components of $$\mathop{v}\limits^{\rightharpoonup}$$, we can find the probability density function in spherical coordinates

$\begin{array}{l} \left(\frac{df_x\left(v_x\right)}{dv_x}\right)\left(\frac{df_y\left(v_y\right)}{dv_y}\right)\left(\frac{df_z\left(v_z\right)}{dv_z}\right) \\ = \left(\frac{\lambda }{2\pi }\right)^{3/2}\mathrm{exp}\left(\frac{-\lambda v^2_x}{2}\right)exp\left(\frac{-\lambda v^2_y}{2}\right)exp\left(\frac{-\lambda v^2_z}{2}\right) \\ = \left(\frac{\lambda }{2\pi }\right)^{3/2}\mathrm{exp}\left(\frac{-\lambda v^2}{2}\right) \\ =\rho \left(v,\theta ,\varphi \right) \end{array}$

Since the differential volume element in spherical coordinates is $$v^2 \mathrm{sin} \theta ~ dvd\theta d\varphi$$, the probability that a molecule has a a velocity vector whose magnitude lies between $$v$$ and $$v+dv$$, while its $$\theta$$-component lies between $$\theta$$ and$$\ \theta +d\theta$$, and its $$\varphi$$-component lies between $$\varphi$$ and $$\varphi +d\varphi$$ becomes

$\begin{array}{l} \left(\frac{df_v\left(v\right)}{dv}\right)\left(\frac{df_{\theta }\left(\theta \right)}{d\theta }\right)\left(\frac{df_{\varphi }\left(\varphi \right)}{d\varphi }\right)dvd\theta d\varphi \\ ~~ =\rho \left(v,\theta ,\varphi \right)v^2 \mathrm{sin} \theta dvd\theta d\varphi \\ ~~ =\left(\frac{\lambda }{2\pi }\right)^{3/2}v^2\mathrm{exp}\left(\frac{-\lambda v^2}{2}\right) \mathrm{sin} \theta dvd\theta d\varphi \end{array}$

(We found the same result in Section 4.3, of course.) We can find the probability-density function for the scalar velocity by eliminating the dependence on the angular components. To do this, we need only sum up, at a given value of $$v$$, the contributions from all possible values of $$\theta$$ and $$\varphi$$, recalling that $$0\le \theta <\pi$$ and $$0\le \varphi <2\pi$$. This sum is just

\begin{aligned} \frac{df_v\left(v\right)}{dv}\int^{\pi }_{\theta =0} \left(\frac{df_{\theta }\left(\theta \right)}{d\theta }\right) d\theta \int^{2\pi }_{\varphi =0} \left(\frac{df_{\varphi }\left(\varphi \right)}{d\varphi }\right)d\varphi = \\ =\left(\frac{\lambda }{2\pi }\right)^{3/2}v^2exp\left(\frac{-\lambda v^2}{2}\right)\int^{\pi }_{\theta =0} \mathrm{sin} \theta d\theta \int^{2\pi }_{\varphi =0} d\varphi \end{aligned}

Since $$\int^{\pi }_{\theta =0}{\left(\frac{df_{\theta }\left(\theta \right)}{d\theta }\right)}d\theta =\int^{2\pi }_{\varphi =0}{\left(\frac{{df}_{\varphi }\left(\varphi \right)}{d\varphi }\right)d\varphi }=1$$, $$\int^{\pi }_0 \mathrm{sin} \theta d\theta =2$$, and $$\int^{2\pi }_0 d\varphi =2\pi$$, we again obtain the Maxwell-Boltzmann probability-density function for the scalar velocity:

$\frac{df_v\left(v\right)}{dv}=4\pi \left(\frac{\lambda }{2\pi }\right)^{3/2}v^2exp\left(\frac{-\lambda v^2}{2}\right)$

Unlike the distribution function for the Cartesian components of velocity, the Maxwell-Boltzmann distribution for scalar velocities is not a normal distribution. Possible speeds lie in the interval $$0\le v<\infty$$. Because of the $$v^2$$ term, the Maxwell-Boltzmann equation is asymmetric; it has a pronounced tail at high velocities.