# Einstein's "On the effect on the motion of a liquid of a very small sphere suspended in it"

Entry by Zin Lin (AP 225, Sep 2012)

## Introduction

The success of a molecular kinetic theory of gases enabled scientists to actually make molecular measurements based on observed physical phenomena. However, formulation of a similar theory for liquids proved to be much less easy. This probably motivated Einstein to look at a slightly different problem, a liquid with particles dissolved in it where one might be able to separately treat the solvent as continuous, the solute discrete, and ultimately deduce useful properties for both. In "On the effect on the motion of a liquid of a very small sphere suspended in it", Einstein considered the change in heat dissipation of a liquid in the vicinity of solute particles.

## Hydrodynamic Equations

Consider a homogeneous incompressible liquid with viscocity $~k$. In the pure solvent (without solute), the velocity field in the neighborhood $~G$ of an arbitrary point $(x_0,~y_0,~z_0)$ is given (to linear order) by

$\displaystyle (u_0,~v_0,~w_0) = (A \xi,~ B \eta,~ C \zeta) \quad \text{ where } \quad \xi = x - x_0, ~\eta = y - y_0, ~\zeta = z - z_0,$

$\text{and} ~ A + B + C = 0 \quad \text{since the fluid is incompressible.}$

The motion of the liquid can be looked upon as a superposition of three kinds:

1. rigid translation (that is, without change of relative position between liquid particles)
2. rigid rotation
3. dilatation

Now consider a spherical rigid body with radius $~P$ placed at $(x_0,~y_0,~z_0)$ whose dimensions are much smaller compared to the domain $~G$. Einstein made the following assumptions:

1. the motion under consideration is so slow that the kinetic energy of the sphere as well as the liquid can be ignored. In other words, the velocities are so small and varying so slowly that the second order inertial terms are negligible.
2. the velocity of a surface element of the sphere is the same as the velocity of the liquid in the immediate neighborhood. In other words, the contact between the sphere's surface and the liquid exhibits everywhere a non-zero viscosity coefficient. (Otherwise, the sphere will slip through the liquid with a velocity independent of the liquid's velocity.)

Under such circumstances, one arrives at the hydrodynamic equations for the velocity field $(u, ~v,~w)$ and the pressure field $~p$:

$\displaystyle \frac{\partial p}{\partial \xi} = k \Delta u, \quad \frac{\partial p}{\partial \eta} = k \Delta v, \quad \frac{\partial p}{\partial \zeta} = k \Delta w, \quad \frac{\partial u}{\partial \xi} + \frac{\partial v}{\partial \eta} + \frac{\partial w}{\partial \zeta} = 0, \quad \quad\quad (1)$

where $\Delta$ stands for the operator $\frac{\partial^2}{\partial \xi^2} + \frac{\partial^2}{\partial \eta^2} + \frac{\partial^2}{\partial \zeta^2}$. It can be recognized that $~(1)$ is the familiar Navier-Stokes equations without the inertial terms.

From the symmetry of the motions of the liquid, the sphere can have neither translation nor rotation. Thus, one gets the boundary conditions:

$u=v=w=0 ~\text{ at}~ \rho = P ~\text{where} ~ \rho = \sqrt{\xi^2 + \eta^2 + \zeta^2}$

## Change in Heat Dissipation

Next, Einstein sought to calculate the amount of heat dissipated in the liquid lying within a sphere of radius $~R$ placed around $(x_0,~y_0,~z_0)$ and $~R \gg P$. Under the assumption that kinetic energy (as well as changes in kinetic energy) of the liquid is negligible, the heat dissipated is simply the mechanical work done on the liquid in the sphere $~R$.

The pressure forces on a vector surface element $~ds \hat{\rho}$ are

$\begin{pmatrix} X_n \\ Y_n \\ Z_n \end{pmatrix} = \frac{1}{\rho} \begin{pmatrix} X_{\xi} & X_{\eta} & X_{\zeta} \\ Y_{\xi} & Y_{\eta} & Y_{\zeta} \\ Z_{\xi} & Z_{\eta} & Z_{\zeta} \end{pmatrix} \begin{pmatrix} \xi \\ \eta \\ \zeta \end{pmatrix}$,

where the elements of the stress tensor are given by

\begin{align} X_{\xi} &= p - 2k \frac{\partial u}{\partial \xi}, \quad Y_{\zeta} = Z_{\eta} &= -k(\frac{\partial v}{\partial \zeta} +\frac{\partial w}{\partial \eta} ) \\ Y_{\eta} &= p - 2k \frac{\partial v}{\partial \eta}, \quad Z_{\xi} = X_{\zeta} &= -k(\frac{\partial w}{\partial \xi} +\frac{\partial u}{\partial \zeta} ) \\ Z_{\zeta} &= p - 2k \frac{\partial w}{\partial \zeta}, \quad X_{\eta} = Y_{\xi} &= -k(\frac{\partial u}{\partial \eta} +\frac{\partial v}{\partial \xi} ) \end{align}.

As given in Ref. [1], $~(u,~v,~w)$ and $~p$ are supplied by the solution to the hydrodynamic equations $(1)$ with appropriate boundary conditions. In this case, they are

\begin{align} u &= A \xi - {5 \over 2} P^3 \frac{\xi (A \xi^2 + B \eta^2 + C \zeta^2)}{\rho^5}\\ v &= B \eta - {5 \over 2} P^3 \frac{\eta (A \xi^2 + B \eta^2 + C \zeta^2)}{\rho^5}\\ w &= C \zeta - {5 \over 2} P^3 \frac{\zeta (A \xi^2 + B \eta^2 + C \zeta^2)}{\rho^5} \\ p &= - 5 k P^3 \frac{A \xi^2 + B \eta^2 + C \zeta^2}{\rho^5} + \text{const.} \end{align}

The mechanical work done or the heat dissipated is

$W = \int (X_n u + Y_n v + Z_n w) ds$

where the integral is performed over the surface of the sphere with radius $~R$, the result being [1]

$W = {8 \pi \over 3} (A^2 + B^2 + C^2) k ( R^3 + { P^3 \over 2})$.

In the absence of the rigid sphere $P = 0$, the work done is simply ${8 \pi \over 3} (A^2 + B^2 + C^2) k R^3$. Therefore, the presence of the sphere $~P$ modifies the amount of heat by ${4 \pi \over 3} (A^2 + B^2 + C^2) k P^3$