# Einstein relation (kinetic theory)

In physics (specifically, in kinetic theory) the Einstein relation (also known as Einstein–Smoluchowski relation) is a previously unexpected connection revealed independently by Albert Einstein in 1905 and by Marian Smoluchowski in 1906 in their papers on Brownian motion. The more general form of the equation is

$D=\mu \,k_{B}T$ where

This equation is an early example of a fluctuation-dissipation relation. 

Two frequently used important special forms of the relation are:

$D={{\mu _{q}\,k_{B}T} \over {q}}$ (Electrical mobility equation, for diffusion of charged particles)
$D={\frac {k_{B}T}{6\pi \,\eta \,r}}$ ("Stokes-Einstein equation", for diffusion of spherical particles through a liquid with low Reynolds number)

where

## Special cases

### Electrical mobility equation

For a particle with electrical charge q, its electrical mobility μq is related to its generalized mobility μ by the equation μ=μq/q. The parameter μq is the ratio of the particle's terminal drift velocity to an applied electric field. Hence, the equation in the case of a charged particle is given as

$D={\frac {\mu _{q}\,k_{B}T}{q}}$ ### Stokes-Einstein equation

In the limit of low Reynolds number, the mobility μ is the inverse of the drag coefficient $\zeta$ . A damping constant $\gamma =\zeta /m$ is frequently used for the momentum relaxation time (time needed for the inertia momentum to become negligible compared to the random momenta) of the diffusive object. For spherical particles of radius r, Stokes' law gives

$\zeta =6\pi \,\eta \,r,$ where $\eta$ is the viscosity of the medium. Thus the Einstein-Smoluchowski relation results into the Stokes-Einstein relation

$D={\frac {k_{\mathrm {B} }T}{6\pi \,\eta \,r}}$ $D_{\mathrm {r} }={\frac {k_{\mathrm {B} }T}{8\pi \,\eta \,r^{3}}}$ ### Semiconductor

In a semiconductor with an arbitrary density of states the Einstein relation is:

$D={\frac {\mu _{q}\,p}{q{\frac {dp}{d\eta }}}}=\mu _{p}p{\frac {\partial V}{\partial p}}=\mu _{p}{\frac {k_{B}T}{q}}$ where $\mu$ is the chemical potential, p the particle concentration, V the electrostatic potential (volt), T the temperature (K), $k_{B}$ Boltzmann constant, q the charge (C).

## Proof of general case

This is a proof in one dimension, but it is identical to a proof in two or three dimensions (just replace d/dx with $\nabla$ ). Essentially the same proof is found in many places, for example see Kubo.

Suppose some fixed, external potential energy U creates a force on the particle $F=-dU/dx$ (for example, an electric force). We assume that the particle would respond, other things being equal, by moving with velocity $v=\mu F$ . Now assume that there are a large number of such particles, with local concentration $\rho (x)$ as a function of position. After some time, equilibrium will be established: the particles will "pile up" around the areas with lowest U, but will still be spread out to some extent because of random diffusion. At this point, there is no net flow of particles: the tendency of particles to get pulled towards lower U (called the "drift current") is equal and opposite the tendency of particles to spread out due to diffusion (called the "diffusion current"). (See drift-diffusion equation.)

The net flow of particles due to the drift current alone is

$J_{\mathrm {drift} }(x)=\mu \,F(x)\,\rho (x)=-\rho (x)\mu {\frac {dU}{dx}}$ (i.e. the number of particles flowing past a point is the particle concentration times the average velocity).

The net flow of particles due to the diffusion current alone is, by Fick's laws

$J_{\mathrm {diffusion} }(x)=-D{\frac {d\rho }{dx}}$ (the minus sign means that particles flow from higher concentration to lower).

Equilibrium requires:

$0=J_{\mathrm {drift} }+J_{\mathrm {diffusion} }=-\rho (x)\mu {\frac {dU}{dx}}-D{\frac {d\rho }{dx}}$ On the other hand, in equilibrium, we can apply thermodynamics, in particular Boltzmann statistics, to infer that

$\rho (x)=Ae^{-U/(k_{B}T)}$ where A is some constant related to the total number of particles. Therefore, by the chain rule,

${\frac {d\rho }{dx}}=-{\frac {1}{k_{B}T}}{\frac {dU}{dx}}\rho (x).$ Finally, plugging this in:

$0=J_{\mathrm {drift} }+J_{\mathrm {diffusion} }=-\rho (x)\mu {\frac {dU}{dx}}+{\frac {D}{k_{B}T}}{\frac {dU}{dx}}\rho (x)=-\rho (x){\frac {dU}{dx}}\left(\mu -{\frac {D}{k_{B}T}}\right)$ Since this equation must hold everywhere,

$\mu ={\frac {D}{k_{B}T}}.$ 