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Reading: Gardiner Sec. 1.2
Homework 2 due Tuesday, March 17 at 2 PM.
The friction constant depends on the dynamic viscosity of the
fluid and the size and shape of the particle.
How shall we model the force representing the thermal
fluctuations?
The thermal fluctuation force represents the effective
randomness of the solvent molecules. For large solute particles,
there would be large numbers of collisions with the solvent
molecules, and by LLN/CLT the random force would become small
relative to their average effect (which is encoded in the friction
term). But for particles m, the number of effective collisions with
the solvent molecules fails to be large enough for the LLN to be in
force, and so the random component should be explicitly
incorporated.
The time scale on which the solute particle changes momentum is,
depending on its size, s to ; we'll show how to compute this
later.
•
The typical time scale over which a water molecule moves between
collisions (with other water
molecules) is
•
The time scale over which a water molecule interacts with the
solute particle (when they do) is
•
Let's think about what properties the thermally fluctuating
force should have. To this end, it's relevant to take into account
some more physical time scales
In summary, the time scale on which the momentum of the solute
particle fluctuates is much longer (by at least a factor of ) than
the time scale on which the force it feels from the collision with
solvent molecules is correlated.
Langevin Equation Model for Brownian MotionFriday, March 13,
20152:04 PM
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(We are also assuming here that the water molecules act roughly
independently of each other, which is a reasonably good
approximation with caveats from low Reynolds number
hydrodynamics.)
The time scale separation also tells us that we can model as
Gaussian because CLT resulting from many solvent particle
collisions per resolved time scale.
Therefore, if we agree to only ask questions about the behavior
of the Brownian particle over time
scales , then the thermal fluctuation force can be modeled as
continuous white noise, i.e., with no memory between different
moments of time. The way this can be encoded in the physicist-style
language is to say the force is delta-correlated in time:
Now, what is the nature of the matrix
We expect the thermal forces to be statistically stationary
(time-homogenous), so the correlation function should only depend
on the time difference , not the times separately.
Statistical isotropy of the thermal forces:
Actually we need to be a little careful about statistical
isotropy, because the current
velocity of the Brownian particle actually breaks isotropy. In
fact, the
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velocity of the Brownian particle actually breaks isotropy. In
fact, the deterministic friction force is just the mean of the
force of collision with the solvent particles. Implicitly, we
imagine that this mean effect has been subtracted out, and that the
remaining randomness in the solvent particle motion is decoupled
from the motion of the large Brownian solute particle.
Therefore, we arrive at the physicist's version of the Langevin
equation:
mean •
correlation function •
We will first treat this model in physicist-mode (see Risken
Sections 1.1, 3.1), calculate the solution for the velocity and
some basic properties, and then later show how these calculations
are translated into the language of SDEs.
where is a Gaussian random function with:
Almost always, the thermal force is modeled as independent of
.
Let's solve for the random velocity . Just same way as for
deterministic force; use integrating factor:
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What are the properties of the resulting random function ?
Most basic property is the mean:
Note that expectation commutes with integrals over deterministic
domain by linearity (and Fubini's theorem).
This is exactly expressing the loss of momentum due to friction,
and is the same result as for a purely deterministic setting.
The effects of the randomness will be seen through the next
order description of the random velocity function, namely the
correlation function.
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Note that different dummy variables are used in each argument of
the covariance; this will facilitate the next step, where we deploy
the bilinearity of covariance:
Now, we replace:
We use the property that if is a continuous function, then
What if we integrate delta functions over finite intervals?
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What if we integrate delta functions over finite intervals?
The upshot is that the integrand is continuous everywhere except
at so you can show the delta-function integration rule still works
provided that . But if the delta function is located at or , the
answer is not well-defined. Calculation continued next time...
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