Brownian Particle bound by a Spring / internal Energy

1. Mar 25, 2013

Abigale

Hi,
i regard a Brownian Particle connectet to a Spring and there is a heat-reservoir.
The distribution of the x-coordinate of the particle follows the Diffusion-Equation (Fokker-Planck-Equation):

$\partial_{t}P(x,t)=\frac{D}{2} \partial_{x}^{2}P(x,t)- \Gamma\partial_{x}[f(x)P(x,t)]$

A deterministic Force is given by $f(x)=-\frac{d}{dx}U(x)$.
Whereby
$U(x)= \frac{1}{2}(x-x_{0})^{2}$ is a Potential.

Also i know that the equilibrium-equation is a Gaussian-Function.
$P_{eq}=(\frac{\beta}{2\pi})^{1/2}\exp[{-\frac{\beta}{2}(x-x_{0})^2}]$

I want to determine the Expected Value of the (potential) internal Energy.
But i don't know how I can get it. ;-(

thank you a lot!!!
Bye Abigale

2. Mar 25, 2013

klawlor419

For any function say f(x), to find its average value you take the product of it with the probability of finding a value x (your probability distribution in this case) and sum it up over all possible values of x which is in your case the displacement of the particle.

3. Mar 25, 2013

Abigale

Expected Value of the Brownian Particle/ continued

Hey thx,
i continued calculating the Expected Value for the internal Energy of this System.

Might this way be okay?
I am not sure.

$\bar{U}=\int^{+\infty}_{-\infty}~U(x)P_{eq}dx \\ ~~~=(\frac{\beta}{2\pi})^{1/2}\int^{+\infty}_{-\infty}~\frac{1}{2}(x-x_{0})^2 \exp[{-\frac{\beta}{2}(x-x_{0})^2}]~~dx$

Than i used a trick:

$\lbrace I=\int dx \exp{-\alpha x^2}\\ ~~\Rightarrow \frac{dI}{d\alpha}=\frac{d}{d\alpha}\int dx \exp{-\alpha x^2}\\ ~~~~~~~~~~~~=\int dx \frac{d}{d\alpha} \exp{-\alpha x^2}\\ ~~~~~~~~~~~~=\int dx~-x^2 \exp{-\alpha x^2} \rbrace$

So it ensues:
$\bar{U}=-(\frac{\beta}{2\pi})^{1/2}\int^{+\infty}_{-\infty}~-\frac{1}{2}(x-x_{0})^2 \exp[{-\frac{\beta}{2}(x-x_{0})^2}]~~dx\\ ~~~~=-(\frac{\beta}{2\pi})^{1/2}\int^{+\infty}_{-\infty}~\frac{d}{d\beta} \exp[{-\frac{\beta}{2}(x-x_{0})^2}]~~dx \\ ~~~~=-(\frac{\beta}{2\pi})^{1/2}\frac{d}{d\beta}\int^{+\infty}_{-\infty}~ \exp[{-\frac{\beta}{2}(x-x_{0})^2}]~~dx\\ \\ \\$

And with the Gaussian Function:

$\\ \bar{U}=-(\frac{\beta}{2\pi})^{1/2}\frac{d}{d\beta}\sqrt{\frac{2\pi}{\beta}}\\ ~~~=\frac{1}{2}\frac{1}{\beta}$

$\beta$ is the inverse temperature.
So the Expected Value of the internal Energy should be proportional to the Temperature, this seems good.
But i am not sure if the result/callculations i have done are correct.
Could someone help me?

Thanks a lot.
Bye Abigale

4. Mar 25, 2013

klawlor419

That looks good and makes sense. In fact, its easily confirmed by the equipartition theorem which states that any quadratic degree of freedom will contribute one-half kT to the average energy.

The next thing I think you need to consider is that the probability distribution may get slightly altered by the oscillator potential. I recognize your P(eq) as the equilibrium probability distribution for purely an oscillator potential. If you notice though it doesn't satisfy the diffusion equation with time present.

Point being that there is a more general solution to this adjusted Fokker-Planck equation, for which you can extract some time-dependence on the average potential energy. Of course for the large-t limit it will go to the equilibrium distribution, but the behavior in the mean time is interesting

5. Mar 25, 2013

klawlor419

I cannot see the solution to the equation yet.. will let you know if I get it.

6. Mar 25, 2013

klawlor419

7. Mar 26, 2013

Abigale

Hallo physicist ,

Let us consider a change of the rest-position of the spring from $x_0$ to $x_{o}^{'}$.
This happened within the short time-intervall [$t_0,t_1$].

The change of the rest-position is very fast.
So the assumption can be done, that the particle-coordinate $x$ and the distribution $P(x,t)$ is not changed.

Now i want to callculate the Expexted Value of the Internal-Energy immediately after the change, so $\bar U (t_1)$.

I am thinking for many hours and got maybe an idea, but i am not sure if it is right.
Could you help me to find what is wrong, or do you have a better idea?

First way, but seems to easy (What is wrong?):

$\begin{split} \bar{U} &=\int\limits_{-\infty}^{\infty} \frac{1}{2} (x-x_{0}^{'})^2 P(x,t_{0}) ~dx\\ \end{split}$

So here i think $P(x,t_{0})$ must then be the $P_{eq}$ from my first post.

$\Rightarrow\begin{split} \bar{U} &=\int\limits_{-\infty}^{\infty} \frac{1}{2} (x-x_{0}^{'})^2 \cdot (\frac{\beta}{2\pi})^{1/2}\exp{[-\frac{\beta}{2}(x-x_{0})^{2}]} ~dx\\ \end{split}$

Second way (maybe even wrong?):

I regard the "Diffusion-Equation" and include the Potential, with the changed rest-position $x_{0}^{'}$ :

$U(x)=\frac{1}{2}(x-x_{0}^{'})^{2}$

So i get the equation:

$\partial_{t} P(x,t_{0}) = \frac{D}{2} \partial_{x}^{2} P(x,t_{0}) - \Gamma \partial_{x} [ -\frac{d}{dx}(\frac{1}{2}(x-x_{0}^{'})^{2}) \cdot P(x,t_{0}) ]$

After some partial derrivations i get:

$\begin{split} \partial_{t} P(x,t_{0})& = \underbrace{~\sqrt{\frac{\beta}{2}} \exp{[-\frac{\beta}{2}(x-x_{0})^{2}]}} ~~~ \underbrace{ \lbrace~ ~ \frac{D}{2}(-\beta+(\beta(x-x_{0}))^{2}) + \Gamma (1-\beta(x-x_{0})(x-x_{0}^{'}))~~ \rbrace}\\ &=~~~~~~P(x,t_{0})~~~~~~~~~~~~~~~~~~~~~~~\cdot~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~f \end{split}$

Now, i regard the small time intervall $[t_{0},t_{1}]=\delta t$.
So i claim that :

$\begin{split} P(x,t_{1})&=P(x,t_{0}+\delta t)\overset{\text{taylor}}=P(x,t_{0})+\delta t \partial_{t}P(x,t_{0})\\ &=P(x,t_{0})\lbrace 1+ \delta t f \rbrace\\ &\overset{\exp{\delta t f \approx 1 + \delta t f}}=P(x,t_{0})\exp{[\delta t~ f]} \end{split}$

And after that i do the integration:

$\begin{split} \bar{U} &=\int\limits_{-\infty}^{\infty} \frac{1}{2} (x-x_{0}^{'})^2 \cdot P(x,t_{1}) dx\\ &=\int\limits_{-\infty}^{\infty} \frac{1}{2} (x-x_{0}^{'})^2 \cdot P(x,t_{0})\exp{[\delta t~ f]} ~~dx\\ \end{split}$

and thx for the help.

Also i am interessted in the work i have to do for the change of the rest-position.
But this later guys.

Bye Abby

8. Mar 26, 2013

klawlor419

Abigale

I wrote up a LaTeX file and will attach it. I tried to address your questions and I asked a few as well. Still have not figured out how to use LaTeX on the forum.

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9. Apr 6, 2013

klawlor419

Any further luck with this problem?

10. Apr 6, 2013

Abigale

Not yet, I regard an similar problem with other boundary conditions for better understanding the fpe. But I slowly understand it. I study physics and I hear the first time statistical mechanics. I will write you later more.
P.s.
Latex can be used by writing two times the dollar-sign:
dollardollar
/xxx
dollardollar