#14 How enzymes propel themselves

/*-------------Divider-------------*/
We have previously introduced the Einstein presentation of diffusion theory. We started from the conservation of probability and ended up with a well-known formula -- <x^2> = 2Dt. Today we are going to talk about the Langevin presentation of diffusion theory. Langevin tried to represent the thermal energy with a random force in Newton's second law, and solve it with statistical method. Macroscopically, the Langevin and Einstein representation would lead to the same prediction. However, Langevin's presentation is more appropriate if we are talking about only several particles.
/*-------------Divider-------------*/

Today we will try to imagine ourselves some cute enzyme molecules living in a 1 dimensional world. To know your final position, we must solve the equation of motion. Assume your mass m, and there is a viscous drag –ζv (ζ is called frictional drag coefficient). Assume thermal random force F(t). From Newton's second law, we could write down:

$m\frac{\textup{d}^2x}{\textup{d}t^2} = -\zeta \frac{\textup{d}x}{\textup{d}t}+F(t)$

in which F(t) follows

$\left \langle F(t) \right \rangle=0,\quad \left \langle F(t)F(t') \right \rangle=f\delta(t-t')$

To solve the equation, we have to average out the term of thermal random force, so we multiply the above equation with x, and we will get:

$m(\frac{\textup{d}}{\textup{d}t}(x\frac{\textup{d}x}{\textup{d}t})-(\frac{\textup{d}x}{\textup{d}t})^2)=-\zeta x\frac{\textup{d}x}{\textup{d}t}+F(t)x$

Note that

$\left \langle F(t)x \right \rangle=0,\quad\left \langle \frac{1}{2}m(\frac{\textup{d}x}{\textup{d}t})^2 \right \rangle=\frac{1}{2}kT$

Substitute it back to our equation, and we will get

$m\frac{\textup{d}}{\textup{d}t}\left \langle x\frac{\textup{d}x}{\textup{d}t} \right \rangle-kT=-\zeta\left \langle x\frac{\textup{d}x}{\textup{d}t} \right \rangle$

It's easy to solve this ODE (multiply the above equation with the integrating factor $e^{\frac{\zeta}{m}}$ ):

$\left \langle x^2 \right \rangle=\frac{2kT}{\zeta}(t-\frac{m}{\zeta}(1-e^{-\frac{\zeta t}{m}}))$

As t >> m/ζ, we will get

$\left \langle x^2 \right \rangle=\frac{2kT}{\zeta}t$

Compare it with what we derived from Einstein's presentation:

$\left \langle x^2 \right \rangle=2Dt$

We could then recognize an important relation called Einstein relation:

$D=\frac{kT}{\zeta}$

Noted that the origin of the fluctuation (the diffusion constant) force and the dissipation force (the drag coefficient) have the same origin -- they all come from the thermal motion of the solvent. The Einstein relation implies a more fundamental theorem called fluctuation-dissipation theorem.

We have to estimate the magnitude of m/ζ to know the time scale separating Einstein's theory and Langevin's theory. Take albumin as an example. The radius is about 3.5nm and its molecule weight is 66.5kDa. Assume the viscosity of water 1mPa-s. We could estimate m/ζ from Stokes relation (ζ=6πηr):

$\frac{m}{\zeta}=1.7\times 10^{-12}\textup{sec}$

It's a really short time scale. Therefore we shall arrive at the same conclusion with Einstein and Langevin's approach most of the time.

It's normally enough to stop here. However, we still have to derive a formula called "Kubo relation." Its derivation start from a simple integration:

$x(t) - x(0) = \int_{0}^{t}v(s)\textup{d}s$

Multiply the above equation by itself and take time average. For a stationary process, we will get:

$\\ \left \langle ((x(t) - x(0))^2 \right \ranglex= \int_{0}^{t}\textup{d}s_1 \int_{0}^{t}\textup{d}s_2 \left\langle v(s_1)v(s_2) \right\rangle\\ =\int_{0}^{t}\textup{d}s_2\int_{0}^{s_2}\textup{d}s_1\left\langle v(s_1)v(s_2) \right\rangle+\int_{0}^{t}\textup{d}s_2\int_{s_2}^{t}\textup{d}s_1\left\langle v(s_1)v(s_2) \right\rangle\\ =\int_{0}^{t}\textup{d}s_2\int_{0}^{s_2}\textup{d}s_1\left\langle v(s_2-s_1)v(0) \right\rangle+\int_{0}^{t}\textup{d}s_1\int_{0}^{s_1}\textup{d}s_2\left\langle v(s_1-s_2)v(0) \right\rangle\\ =2\int_{0}^{t}\textup{d}s\int_{0}^{s}\textup{d}u\left \langle v(u)v(0) \right \rangle\\ =2\int_{0}^{t}\textup{d}u\left \langle v(u)v(0) \right \rangle \int_{u}^{t}\textup{d}s\\ =2\int_{0}^{t}\textup{d}u(t-u)\left \langle v(u)v(0) \right \rangle$

As t approach ∞, we will arrive at

$\left \langle ((x(t) - x(0))^2 \right \rangle=2t\int_{0}^{\infty}\textup{d}u\left \langle v(u)v(0) \right \rangle$

That is to say:

$D=\int_{0}^{\infty}\textup{d}u\left \langle v(u)v(0) \right \rangle$

In 3 dimensional system, the equation becomes

$D=\frac{1}{3}\int_{0}^{\infty}\textup{d}u\left \langle v(u)v(0) \right \rangle$

To estimate the magnitude of <v(u)v(0)>, we could solve the Langevin equation in another way. Write down the Newton's second law in terms of speed:

$\frac{\textup{d}v}{\textup{d}t}=-\frac{\zeta}{m}v+\frac{F(t)}{m}$

The ODE could be solved easily:

$v(t) = v(0)e^{-\zeta t/m} + e^{-\zeta t/m}\int_{0}^{t}\frac{F(t')}{m}e^{\zeta t'/m} \textup{d}t'$

Substitute it back to our equation and we will get:

$D=\int_{0}^{\infty}\textup{d}u\left \langle v(u)v(0) \right \rangle = \int_{0}^{\infty}\textup{d}u\left \langle v(0)v(0) \right \rangle e^{-\zeta u/m} = \frac{m}{\zeta}<v(0)v(0)>$

This is a useful equation which allow us to estimate the diffusion constant with the expectation value of the kinetic energy.

#biophysics #3minBiophysics #生物物理 #三分鐘生物物理
#diffusion #Langevin #equation

Three Minute Biophysics

Search This Blog

Featured Post

#77 Brownian Carnot engine-I 微觀卡諾引擎-I

#14 How enzymes propel themselves - IV

Comments

Post a Comment