Question

What is Brownian motion? Give three examples of such a motion. Using Einstein’s theory, obtain an expression for Einstein’s formula for mean square displacement of a Brownian particle.

Answer 1

Answer on Question 50676, Physics, Molecular Physics | Thermodynamics

Question:

What is Brownian motion? Give three examples of such a motion. Using Einstein's theory, obtain an expression for Einstein's formula for mean square displacement of a Brownian particle.

Answer:

Brownian motion is the random movement of microscopic particles suspended in a liquid or gas caused by their collision with the quick atoms or molecules in the surrounding medium. Three examples of such a motion: motion of pollen grains in water, diffusion of "holes" through a semiconductor, motion of smoke in a glass box.

Let us obtain an expression for Einstein's formula for mean square displacement of a Brownian particle. When Brownian particle is move the net force $F$ acts on it. Also on particle acts the friction force $f$ caused by the medium viscosity and directed opposite to the force $F$.

Let us suppose that the particle has a spherical shape of radius $a$. Then the friction force $f$ can be expressed by the Stokes' law:

$$f = 6 \pi \eta a v,$$

where, $\eta$ is the dynamic viscosity, $v$ is the velocity of the particle.

So, we can write the equation of motion of the particle:

$$m \ddot{r} = F - 6 \pi \eta a \dot{r},$$

where, $m$ is a mass of the particle, $r$ is the radius-vector of the particle relative to an arbitrary coordinate system, $\dot{r} = v$ is the velocity of the particle.

Let us consider the projection of the radius-vector on the axis $X$. Then the equation (1) looks like:

$$m \ddot{x} = F_{x} - 6 \pi \eta a \dot{x},$$

where, $F_{x}$ is the projection of the net force $F$ on the axis $X$.

We need to obtain the displacement of the Brownian particle $x$, which caused by the collisions with the molecules. Mean displacement of the particle $\bar{x}$ would be equal to

zero, because the displacements of the particle with equal probability can have both positive and negative values. But the mean square displacement $\overline{x^2}$ is not equal to zero, and we can rewrite the equation (2) so that it includes the value of $\overline{x^2}$ (we multiply both sides of the equation on $x$):

$$m x \ddot{x} = F_x - 6 \pi \eta a x \dot{x}.$$

Let us use the next identities:

$$x \ddot{x} = \frac{1}{2} \frac{d^2(x^2)}{dt^2} - \left(\frac{dx}{dt}\right)^2, \quad x \dot{x} = \frac{1}{2} \frac{d(x^2)}{dt}.$$

Substituting this identities into the equation (3) we obtain:

$$\frac{m}{2} \frac{d^2(x^2)}{dt^2} - m \left(\frac{dx}{dt}\right)^2 = -3 \pi \eta a \frac{d(x^2)}{dt} + x F_x.$$

This equality is valid for any particle and because of that it is also valid for mean values that includes in it, so we can write:

$$\frac{m}{2} \frac{d^2(\overline{x^2})}{dt^2} - m \overline{\left(\frac{dx}{dt}\right)^2} = -3 \pi \eta a \frac{d(\overline{x^2})}{dt} + \overline{x F_x},$$

where $\overline{x^2}$ is the mean square displacement of the particle, $\overline{\left(\frac{dx}{dt}\right)^2}$ is the mean square velocity of the particle, the mean value of $\overline{x F_x}$ is equal to zero because for a large number of particles $x$ and $F_x$ equally takes both positive and negative values. Therefore, the equation (2) takes the next form:

$$\frac{m}{2} \frac{d^2(\overline{x^2})}{dt^2} - m \overline{\left(\frac{dx}{dt}\right)^2} = -3 \pi \eta a \frac{d(\overline{x^2})}{dt}.$$

Because the motion of the particles is quite chaotic, then the mean squares velocity projections on all three coordinate axes must be equal to each other:

$$\overline{\left(\frac{dx}{dt}\right)^2} = \overline{\left(\frac{dy}{dt}\right)^2} = \overline{\left(\frac{dz}{dt}\right)^2}.$$

Also obvious, that the sum of this values must be equal to the mean square velocity of the particles $\overline{v^2}$:

$$\overline{\left(\frac{dx}{dt}\right)^2} + \overline{\left(\frac{dy}{dt}\right)^2} + \overline{\left(\frac{dz}{dt}\right)^2} = \overline{v^2}.$$

Therefore, we obtain:

$$\overline {{\left(\frac {d x}{d t}\right) ^ {2}}} = \frac {1}{3} \overline {{v ^ {2}}},$$ $$m \overline {{\left(\frac {d x}{d t}\right) ^ {2}}} = \frac {1}{3} m \overline {{v ^ {2}}} = \frac {2}{3} \frac {m \overline {{v ^ {2}}}}{2}.$$

Because the average kinetic energy of the Brownian particle must be equal to the average kinetic energy of the molecules of liquid (or gas), we can write:

$$\frac {m \overline {{v ^ {2}}}}{2} = \frac {3}{2} k T,$$ $$m \overline {{\left(\frac {d x}{d t}\right) ^ {2}}} = \frac {2}{3} \frac {m \overline {{v ^ {2}}}}{2} = k T.$$

Substituting the equation (5) into the equation (4) we obtain:

$$\frac {m}{2} \frac {d ^ {2} (\overline {{x ^ {2}}})}{d t ^ {2}} - k T = - 3 \pi \eta a \frac {d (\overline {{x ^ {2}}})}{d t}.$$

This equation can be easily integrated. Let us denote $\frac{d(\overline{x^2})}{dt} = Z$ and rewrite the equation:

$$\frac {m}{2} \frac {d Z}{d t} - k T = - 3 \pi \eta a Z.$$

After separation of variables we obtain:

$$\frac {d Z}{Z - \frac {k T}{3 \pi \eta a}} = - \frac {6 \pi \eta a}{m} d t.$$

Integrating the left-side of the equation within the limits from 0 to $Z$ and the right-side of the equation within the limits from 0 to $t$ we get:

$$\int_ {0} ^ {Z} \frac {d Z}{Z - \frac {k T}{3 \pi \eta a}} = - \int_ {0} ^ {t} \frac {6 \pi \eta a}{m} d t$$ $$\ln \left(Z - \frac {k T}{3 \pi \eta a}\right) - \ln \left(- \frac {k T}{3 \pi \eta a}\right) = - \frac {6 \pi \eta a}{m} t.$$

From this equation we can find $Z$ :

$$Z = \frac {k T}{3 \pi \eta a} \left(1 - e ^ {- \frac {6 \pi \eta a}{m} t}\right) = \frac {d (\overline {{x ^ {2}}})}{d t}.$$

The value of $e^{-\frac{6\pi\eta a}{m} t}$ is negligible, so we can write:

$$\frac {d}{d t} \left(\overline {{x ^ {2}}}\right) = \frac {k T}{3 \pi \eta a}. (6)$$

For the finite time intervals $\Delta t$ and appropriate displacements $\Delta \overline{x^2}$ the equation (6) takes form:

$$\frac {\Delta \overline {{x ^ {2}}}}{\Delta t} = \frac {k T}{3 \pi \eta a}.$$

Finally, we obtain an expression for Einstein's formula for mean square displacement of a Brownian particle:

$$\Delta \overline {{x ^ {2}}} = \frac {k T}{3 \pi \eta a} \Delta t.$$

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