Maxwell Velocity Distribution

Consider a molecule of mass $m$ in a gas that is sufficiently dilute for the intermolecular forces to be negligible (i.e., an ideal gas). The energy of the molecule is written

$$\epsilon = \frac{{\bf p}^{ 2}}{2 m} + \epsilon^{ \rm int},$$ (7.202)

where $\bf p$ is its momentum vector, and $\epsilon^{ \rm int}$ is its internal (i.e., non-translational) energy. The latter energy is due to molecular rotation, vibration, et cetera. Translational degrees of freedom can be treated classically to an excellent approximation, whereas internal degrees of freedom usually require a quantum-mechanical approach. Classically, the probability of finding the molecule in a given internal state with a position vector in the range $\bf r$ to ${\bf r} + d{\bf r}$ , and a momentum vector in the range $\bf p$ to ${\bf p} + d{\bf p}$ , is proportional to the number of cells (of ``volume'' $h_0$ ) contained in the corresponding region of phase-space, weighted by the Boltzmann factor. In fact, because classical phase-space is divided up into uniform cells, the number of cells is just proportional to the ``volume'' of the region under consideration. This ``volume'' is written $d^{ 3}{\bf r} d^{ 3}{\bf p}$ . Thus, the probability of finding the molecule in a given internal state $s$ is

$$P_s({\bf r}, {\bf p}) d^{ 3}{\bf r} d^{ 3}{\bf p} \propto \e... ...\right) \exp(-\beta \epsilon^{ \rm int}_s) d^{ 3}{\bf r} d^{ 3}{\bf p},$$ (7.203)

where $P_s$ is a probability density defined in the usual manner. The probability $P({\bf r}, {\bf p}) d^{ 3}{\bf r} d^{ 3}{\bf p}$ of finding the molecule in any internal state with position and momentum vectors in the specified range is obtained by summing the previous expression over all possible internal states. The sum over $\exp(-\beta \epsilon^{ \rm int}_s)$ just contributes a constant of proportionality (because the internal states do not depend on ${\bf r}$ or ${\bf p}$ ), so

$$P({\bf r}, {\bf p}) d^{ 3}{\bf r} d^{ 3}{\bf p} \propto \exp\left(-\frac{\beta p^{ 2}}{2 m}\right) d^{ 3}{\bf r} d^{ 3}{\bf p}.$$ (7.204)

Of course, we can multiply this probability by the total number of molecules, $N$ , in order to obtain the mean number of molecules with position and momentum vectors in the specified range.

Suppose that we now wish to determine $f({\bf r}, {\bf v}) d^{ 3}{\bf r} d^{ 3}{\bf v}$ : that is, the mean number of molecules with positions between ${\bf r}$ and ${\bf r} + d{\bf r}$ , and velocities in the range ${\bf v}$ and ${\bf v}+d{\bf v}$ . Because ${\bf v} = {\bf p}/m$ , it is easily seen that

$$f({\bf r}, {\bf v}) d^{ 3}{\bf r} d^{ 3}{\bf v} = C \exp\left(-\frac{\beta m v^{ 2}}{2}\right) d^{ 3}{\bf r} d^{ 3}{\bf v},$$ (7.205)

where $C$ is a constant of proportionality. This constant can be determined by the condition

$$\int_{({\bf r})} \int_{({\bf v})} f({\bf r}, {\bf v}) d^{ 3}{\bf r} d^{ 3}{\bf v} = N.$$ (7.206)

In other word, the sum over molecules with all possible positions and velocities gives the total number of molecules, $N$ . The integral over the molecular position coordinates just gives the volume, $V$ , of the gas, because the Boltzmann factor is independent of position. The integration over the velocity coordinates can be reduced to the product of three identical integrals (one for $v_x$ , one for $v_y$ , and one for $v_z$ ), so we have

$$C V \left[\int_{-\infty}^{\infty} \exp\left(-\frac{\beta m v_z^{ 2}}{2}\right) dv_z\right]^{ 3} = N.$$ (7.207)

Now,

$$\int_{-\infty}^{\infty} \exp\left(-\frac{\beta m v_z^{ 2}}{2}... ...{-\infty}^{\infty} \exp\left(-y^{ 2}\right)dy = \sqrt{\frac{2\pi}{\beta m}},$$ (7.208)

so $C =(N/V) (\beta m / 2\pi)^{3/2}$ . (See Exercise 2.) Thus, the properly normalized distribution function for molecular velocities is written

$$f({\bf v}) d^{ 3}{\bf r} d^{ 3}{\bf v} = n \left(\frac{m}{2\... ..., \exp\left(-\frac{m v^{ 2}}{ 2 k T}\right)d^{ 3}{\bf r} d^{ 3}{\bf v}.$$ (7.209)

Here, $n=N/V$ is the number density of the molecules. We have omitted the variable ${\bf r}$ in the argument of $f$ , because $f$ clearly does not depend on position. In other words, the distribution of molecular velocities is uniform in space. This is hardly surprising, because there is nothing to distinguish one region of space from another in our calculation. The previous distribution is called the Maxwell velocity distribution, because it was discovered by James Clark Maxwell in the middle of the nineteenth century. The average number of molecules per unit volume with velocities in the range ${\bf v}$ to ${\bf v}+d{\bf v}$ is obviously $f({\bf v}) d^{ 3}{\bf v}$ .

Let us consider the distribution of a given component of velocity: the $z$ -component (say). Suppose that $g(v_z) dv_z$ is the average number of molecules per unit volume with the $z$ -component of velocity in the range $v_z$ to $v_z+dv_z$ , irrespective of the values of their other velocity components. It is fairly obvious that this distribution is obtained from the Maxwell distribution by summing (integrating actually) over all possible values of $v_x$ and $v_y$ , with $v_z$ in the specified range. Thus,

$$g(v_z) dv_z = \int_{(v_x)} \int_{(v_y)} f({\bf v}) d^{ 3}{\bf v}.$$ (7.210)

This gives

$$g(v_z) dv_z = n \left(\frac{m}{2\pi k T}\right)^{ 3/2} \int_{(v_x)} \int_... ...rac{m}{2 k T}\right)(v_x^{ 2}+v_y^{ 2}+v_z^{ 2})\right] dv_x dv_y dv_z$$

$$=n \left(\frac{m}{2\pi k T}\right)^{ 3/2} \exp\left(-\frac{m\... ...{-\infty}^{\infty} \exp\left(-\frac{m v_x^{ 2}}{ 2 k T}\right)\right]^{ 2}$$

$$=n \left(\frac{m}{2\pi k T}\right)^{ 3/2}\exp\left(-\frac{m v_z^{ 2}}{2 k T}\right) \left(\sqrt{\frac{2\pi k T}{m}}\right)^{ 2},$$ (7.211)

or

$$g(v_z) dv_z = n \left(\frac{m}{2\pi k T} \right)^{ 1/2} \exp\left(-\frac{m v_z^{ 2}}{2 k T}\right) dv_z.$$ (7.212)

Of course, this expression is properly normalized, so that

$$\int_{-\infty}^{\infty} g(v_z) dv_z = n.$$ (7.213)

It is clear that each component (because there is nothing special about the $z$ -component) of the velocity is distributed with a Gaussian probability distribution (see Section 2.9), centered on a mean value

$$\overline{v_z} = 0,$$ (7.214)

with variance

$$\overline{v_z^{ 2}} = \frac{k T}{m}.$$ (7.215)

Equation (7.214) implies that each molecule is just as likely to be moving in the plus $z$ -direction as in the minus $z$ -direction. Equation (7.215) can be rearranged to give

$$\overline{\frac{1}{2} m v_z^{ 2}} = \frac{1}{2} k T,$$ (7.216)

in accordance with the equipartition theorem.

Note that Equation (7.209) can be rewritten

$$\frac{ f({\bf v}) d^{ 3}{\bf v}}{n} =\left[\frac{g(v_x) dv_x... ...\right] \left[\frac{g(v_y) dv_y}{n}\right]\left[\frac{g(v_z) dv_z}{n}\right],$$ (7.217)

where $g(v_x)$ and $g(v_y)$ are defined in an analogous way to $g(v_z)$ . Thus, the probability that the velocity lies in the range ${\bf v}$ to ${\bf v}+d{\bf v}$ is just equal to the product of the probabilities that the velocity components lie in their respective ranges. In other words, the individual velocity components act like statistically-independent variables.

Suppose that we now wish to calculate $F(v) dv$ : that is, the average number of molecules per unit volume with a speed $v=\vert{\bf v}\vert$ in the range $v$ to $v+dv$ . It is obvious that we can obtain this quantity by summing over all molecules with speeds in this range, irrespective of the direction of their velocities. Thus,

$$F(v) dv = \int f({\bf v}) d^{ 3}{\bf v},$$ (7.218)

where the integral extends over all velocities satisfying

$$v < \vert{\bf v}\vert < v + dv.$$ (7.219)

This inequality is satisfied by a spherical shell of radius $v$ and thickness $dv$ in velocity space. Because $f({\bf v})$ only depends on $\vert v\vert$ , so $f({\bf v})\equiv f(v)$ , the previous integral is just $f(v)$ multiplied by the volume of the spherical shell in velocity space. So,

$$F(v) dv = f(v) 4\pi v^{ 2} dv,$$ (7.220)

which gives

$$F(v) dv = 4\pi n\left(\frac{m}{2\pi k T}\right)^{ 3/2} v^{ 2} \exp\left(-\frac{m v^{ 2}}{2 k T}\right)dv.$$ (7.221)

This is the famous Maxwell distribution of molecular speeds. Of course, it is properly normalized, so that

$$\int_0^\infty F(v) dv = n.$$ (7.222)

Note that the Maxwell distribution exhibits a maximum at some non-zero value of $v$ . The reason for this is quite simple. As $v$ increases, the Boltzmann factor decreases, but the volume of phase-space available to the molecule (which is proportional to $v^{ 2}$ ) increases: the net result is a distribution with a non-zero maximum.

Figure: The Maxwell velocity distribution as a function of molecular speed, in units of the most probable speed ($\tilde{v}$ ). The dashed, dash-dotted, and dotted lines indicates the most probable speed, the mean speed, and the root-mean-square speed, respectively.

The mean molecular speed is given by

$$\overline{v} = \frac{1}{n} \int_0^\infty F(v) v dv.$$ (7.223)

Thus, we obtain

$$\overline{v} = 4\pi \left(\frac{m}{2\pi k T}\right)^{ 3/2} \int_0^\infty v^{ 3} \exp \left(-\frac{m v^{ 2}}{2 k T}\right)dv,$$ (7.224)

or

$$\overline{v} = 4\pi \left(\frac{m}{2\pi k T}\right)^{ 3/2} \l... ...c{2 k T}{m}\right)^{ 2} \int_0^\infty y^{ 3} \exp\left(-y^{ 2}\right) dy.$$ (7.225)

Now

$$\int_0^\infty y^{ 3} \exp\left(-y^{ 2}\right)dy = \frac{1}{2}$$ (7.226)

(see Exercise 2), so

$$\overline{v} = \sqrt{\frac{8}{\pi} \frac{k T}{m}}.$$ (7.227)

A similar calculation gives

$$v_{\rm rms} = \left[\overline{v^{ 2}}\right]^{ 1/2} = \sqrt{\frac{3 k T}{m}}.$$ (7.228)

(See Exercise 14.) However, this result can also be obtained from the equipartition theorem. Because

$$\overline{\frac{1}{2} m v^{ 2}} = \overline{\frac{1}{2} m (v_x^{ 2}+v_y^{ 2}+v_z^{ 2})} = 3 \left(\frac{1}{2} k T\right),$$ (7.229)

then Equation (7.228) follows immediately. It is easily demonstrated that the most probable molecular speed (i.e., the maximum of the Maxwell distribution function) is

$$\tilde{v} = \sqrt{\frac{2 k T}{m}}.$$ (7.230)

The speed of sound in an ideal gas is given by

$$c_s = \sqrt{\frac{\gamma p}{\rho}},$$ (7.231)

where $\gamma$ is the ratio of specific heats. This can also be written

$$c_s=\sqrt{\frac{\gamma k T}{m}},$$ (7.232)

because $p = n k T$ and $\rho = n m$ . It is clear that the various average speeds that we have just calculated are all of order the sound speed (i.e., a few hundred meters per second at room temperature). In ordinary air ( $\gamma=1.4$ ) the sound speed is about 84% of the most probable molecular speed, and about 74% of the mean molecular speed. Because sound waves ultimately propagate via molecular motion, it makes sense that they travel at slightly less than the most probable and mean molecular speeds.

Figure 7.7 shows the Maxwell velocity distribution as a function of molecular speed in units of the most probable speed. Also shown are the mean speed and the root-mean-square speed.