# 6.4: The Probability-density Function for Gas Velocities in One Dimension

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In Section 4.3, we find a differential equation in the function $${\rho }_x\left(v_x\right)$$. Unlike the velocity, which takes values from zero to infinity, the $$x$$-component, $$v_x$$, takes values from minus infinity to plus infinity. The probability density at an infinite velocity, in either direction, is necessarily zero. Therefore, we cannot evaluate the integral of $$d \rho_x\left(v_x\right)/ \rho_x\left(v_x\right)$$ from $$v_x=-\infty$$ to an arbitrary velocity, $$v_x$$. However, we know from Maxwell’s assumption that the probability density for $$v_x$$ must be independent of whether the molecule is traveling in the direction of the positive $$x$$-axis or the negative $$x$$-axis. That is, $$\rho_x\left(v_x\right)$$ must be an even function; the probability density function must be symmetric around $$v_x=0$$; $$\rho_x\left(v_x\right)=\rho_x\left(-v_x\right)$$. Hence, we can express $$\rho_x\left(v_x\right)$$ relative to its fixed value,$$\rho_x\left(0\right)$$, at $$v_x=0$$. We integrate $$d \rho_x\left(v_x\right)/ \rho_x\left(v_x\right)$$ from $$\rho_x\left(0\right)$$ to $$\rho_x\left(v_x\right)$$ as $$v_x$$ goes from zero to an arbitrary velocity, $$v_x$$, to find

$\int^{\rho_x\left(v_x\right)}_{\rho_x\left(0\right)} \frac{d \rho_x\left(v_x\right)}{\rho_x\left(v_x\right)}=-\lambda \int^{v_x}_0 v_xdv_x \nonumber$

or

$\rho_x\left(v_x\right)=\frac{df_x\left(v_x\right)}{dv_x}= \rho_x\left(0\right)\mathrm{exp}\left(\frac{-\lambda v^2_x}{2}\right) \nonumber$

The value of $$\rho_x\left(0\right)$$ must be such as to make the integral of $$\rho _x\left(v_x\right)$$ over all possible values of $$v_x$$, $$-\infty < v_x <\infty$$, equal to unity. That is, we must have

\begin{aligned} 1 & =\int^{\infty}_{-\infty} \rho_x\left(v_x\right) dv_x \\ ~ & =\int^{\infty}_{-\infty} \frac{df_x\left(v_x\right)}{dv_x}dv_x \\ ~ & = \rho_x \left(0\right)\int^{\infty}_{-\infty} \mathrm{exp} \left(\frac{-\lambda v^2_x}{2}\right) dv_x \\ ~ & =\rho_x\left(0\right)\sqrt{\frac{2\pi }{\lambda }} \end{aligned} \nonumber

where we use the definite integral $$\int^{\infty }_{-\infty } \mathrm{exp} \left(-ax^2\right) dx=\sqrt{ \pi /a}$$. (See Appendix D.) It follows that $$\rho_x\left(0\right)= \left( \lambda /2 \pi \right)^{1/2}$$. The one-dimensional probability-density function becomes

\begin{aligned} \rho_x\left(v_x\right) & =\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) \end{aligned} \nonumber

Note that this is the normal distribution with $$\mu =0$$ and $${\sigma }^2={\lambda }^{-1}$$. So $${\lambda }^{-1}$$ is the variance of the normal one-dimensional probability-density function. As noted above, in Section 4.6 we find that $$\lambda ={m}/{kT}$$.

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