Maxwell-Boltzmann Distribution of Velocities

For a collection of particles that do not interact, we recall that the ideal gas description is sufficient to predict properties of fluids at certain limiting conditions. Within such a description, the particles, of certain masses, are considered to be moving with velocities that are distributed in a certain fashion. We now consider this distribution of velocities, starting with the distribution of velocity in one-dimension. The generalization to three dimensions follows.

We are concered with finding a velocity distribution function. This mathematical description provides the probability of a particle having the Cartesian components of velocity in the range $v_x + dv_x$, $v_y + dv_y$, and $v_z + dv_z$.

We let the distribution function be:

$$\Omega(v_x, v_y, v_z) = f(v_x) f(v_y) f(v_z)$$

With the definition of the probability density distribution above, we can determine the probability of a particle having the components of velocity $v_x + dv_x$, $v_y + dv_y$, and $v_z + dv_z$ as:

$$\Omega(v_x, v_y, v_z) dv_x dv_y dv_z = [ f(v_x) dv_x ] [ f(v_y) dv_y ] [ f(v_z) dv_z ]$$

$$\Omega(v_x, v_y, v_z) dv_x dv_y dv_z = [f(v_x) f(v_y) f(v_z)] dv_x dv_y dv_z$$

An assumption regarding the nature of this distribution: the gas is isotropic such that the direction of particle movement does not affect the properties of the fluid. In this sense, the velocity distribution is effectively dependent on the magnitude of the velocity. The magnitude of the velocity is:

$$\mid {\bf\nu} \mid = \sqrt{ v_x^2 + v_y^2 + v_z^2 }$$

We now consider the natural logarithm of the distribution function:

$$ln \Omega(v_x, v_y, v_z) = ln f(v_x) + ln f(v_y) + ln f(v_z)$$

$$ln \Omega(\nu) = ln f(v_x) + ln f(v_y) + ln f(v_z)$$

with

$$\nu^2 = v_x^2 + v_y^2 + v_z^2$$

To determine $f(v_x)$, we consider the derivative of the probability function $\Omega(\nu)$ with respect to the variable $v_x$.

• Effectively, we are interested in finding that distribution of velocities (along the x-direction) that distributes the total kinetic energy of the system among the available degrees of freedom in the collection of particles (recall that the internal energy of an ideal gas is only dependent on the temperature and amount of fluid) under the constraint that the total number of particles is constant, the volume we consider remains constant, and the temperature is constant and thus known.

$$\left ( \frac{ \partial ln \Omega(\nu)}{ \partial v_x} \right)_{v_y, v_z} = \frac{d ln f(v_x)}{d v_x}$$

$$\left ( \frac{ d ln \Omega(\nu)}{ d \nu} \right) \left ( \frac{\partial \nu}{\partial v_x}\right)_{v_y, v_z} = \frac{d ln f(v_x)}{d v_x}$$

One can show that $\left( \frac{\partial \nu}{\partial v_x}\right)_{v_y, v_z} = \frac{v_x}{\nu}$.

$$\left( \frac{\partial \nu}{\partial v_x}\right)_{v_y, v_z} ... ...{\partial v_x} \left ( v_x^2 + v_y^2 + v_z^2 \right )^{1/2} \right )_{v_y, v_z}$$

$$= \frac{1}{2} (2v_x) \left ( v_x^2 + v_y^2 + v_z^2 \right )^{-1/2}$$

$$= \frac{v_x}{\nu}$$

Using this last relation, we obtain:

$$\left ( \frac{d ln \Omega(\nu)}{d \nu} \right ) \left ( \frac... ... \nu}{\partial v_x} \right)_{v_y, v_z} = \frac{d ln f(v_x)}{d v_x}$$

$$\left ( \frac{d ln \Omega(\nu)}{d \nu} \right ) \left ( \frac{v_x}{v} \right ) = \frac{d ln f(v_x)}{d v_x}$$

$$\left ( \frac{d ln \Omega(\nu)}{\nu d \nu} \right ) = \frac{d ln f(v_x)}{v_x d v_x}$$

Since each direction is equivalent (under our assumption of isotropy of the medium), we can write for the $v_y$ and $v_z$ differentials (and you can show yourself independently):

$$\left ( \frac{d ln \Omega(\nu)}{\nu d \nu} \right ) = \frac{d ln f(v_y)}{v_y d v_y}$$

$$\left ( \frac{d ln \Omega(\nu)}{\nu d \nu} \right ) = \frac{d ln f(v_z)}{v_z d v_z}$$

Thus, the following equality is clear:

$$\frac{d ln f(v_y)}{v_y d v_y} = \frac{d ln f(v_x)}{v_x d v_x} = \frac{d ln f(v_z)}{v_z d v_z}$$

Since the individual functions of x, y, and z are equal to one another for all space, we can justifiably argue that each is a constant (equal in all cases); in general we can write:

$$\frac{d ln f(v_j)}{v_j d v_j} = \frac{d f(v_j)}{v_j f(v_j) dv_j} = - \gamma for j = x,y,z$$

We use the negative of a positive $\gamma$ in order to obtain a probability function that is well-behaved as $v_j$ approaches $\infty$. The solutions for the previous equation are:

$$\displaystyle\int \frac{d f(v_j) }{f(v_j)} = - \displaystyle\int \gamma v_j dv_j$$

$$ln f(v_j) = -\frac{1}{2} \gamma v_j^2$$

$$f(v_j) = A e^{- \gamma v_j^2 / 2 }$$

This last equation is the general expression of our solution; it applies to the individual probability distributions in each direction (x,y,z) (recall we decomposed our total distribution into a product of the three individual distributions).

We still need to consider 2 missing elements:

• A
• $\gamma$

To determine an expression for $A$, we need to consider that the distribution is normalized over the domain for which it is defined. What does this mean operationally? Since this expression represents a probability, the sum of the individual probabilities for each infinitesimal region of space (or in this case, each window in `velocity" space) must sum to 1. Thus we have:

$$\displaystyle\int_{- \infty}^{\infty} f(v_j) dv_j = 1 =... ...isplaystyle\int_{- \infty}^{\infty} A e^{- \gamma v_j^2 / 2 } dv_j$$

$$1 = 2 A \displaystyle\int_{0}^{\infty} A e^{- \gamma v_j^2 / 2 } dv_j$$

$$1 = A \sqrt{ \frac{2 \pi}{\gamma}}$$

$$A = \sqrt{ \frac{\gamma}{2 \pi}}$$

We used the property of even integrands to change the lower limit of integration. Thus, the distribution becomes:

$$f(v_j) = \sqrt{ \frac{\gamma}{2 \pi}} e^{- \gamma v_j^2 / 2 }$$

To evaluate $\gamma$, we introduce from statistical mechanics (which we just state here without proof, which is left for another course or to the individual student):

$$\left < v_x^2 \right > = \frac{k_B T}{m}$$

where $m$ is the mass of the particles. Thus,

$$\left < v_x^2 \right > = \frac{k_B T}{m} = \displaystyle\int_{-\infty}^{\infty} v_x^2 f(v_x) dv_x$$

$$ = \sqrt{ \frac{\gamma}{2 \pi}} \displaystyle\int_{-\infty}^{\infty} v_x^2 e^{- \gamma v_x^2 / 2 } dv_x$$

$$ = \sqrt{ \frac{\gamma}{2 \pi}} \left ( \frac{1}{\gamma} \sqrt{ \frac{2 \pi}{\gamma} } \right )$$

$$ = \frac{1}{\gamma}$$

$$\frac{m}{k_b T} = \gamma$$

We have thus arrived at the famous Maxwell-Boltzmann velocity distribution in one dimension:

$$f(v_j) = \left ( \frac{m}{2 \pi k_B T} \right)^{1/2} ... ... \left ( \frac{M}{2 \pi R T} \right)^{1/2} e^{-M v_j^2 /2 R T)}$$

where mass, $m$, is in units of $kg$ and molar mass, $M$, is in units of $\frac{kg}{mol}$ (we use the relation $R = N_A k_B$ for the conversion between the forms using $m$ and $M$.

In three dimensions (Cartesian represenatation) the total distribution becomes:

$$\Omega(v_x, v_y, v_z) = \displaystyle\prod_{j=1,3} f(v_j)$$

$$\Omega(v_x, v_y, v_z) = \displaystyle\prod_{j=1,3} \left ( \frac{m}{2 \pi k_B T} \right)^{1/2} e^{-m v_j^2 /2 k_B T)}$$

$$\Omega(\nu) = \displaystyle\prod_{j=1,3} \left ( \frac{m}{2 \pi k_B T} \right)^{1/2} e^{-m v_j^2 /2 k_B T)}$$

Example Problem 33.1 (Engel and Reid)

• Compute average velocity, $\left < v_x \right >$.

  
  

  $$\left < v_x \right > = \displaystyle\int_{-\infty}^{\infty} v_x f(v_x) dv_x$$

  $$ = \displaystyle\int_{-\infty}^{\infty} v_x \left ( \frac{m}{2 \pi k_B T} \right)^{1/2} e^{-m v_x^2 /2 k_B T)} dv_x$$

  $$ = \left ( \frac{m}{2 \pi k_B T} \right)^{1/2} \displaystyle\int_{-\infty}^{\infty} v_x e^{-m v_x^2 /2 k_B T)} dv_x$$

  $$ = 0$$

  
  

The result reflects the vectorial nature of velocity.