# Simulation of the fission as a stochastic process

1. Jun 10, 2005

### Heimdall

Hello

I'm a french student, I'm actually not sure this is the good place to ask my question but as it deals with the nuclear fission I try here... don't hesitate to tell me if there is a better forum... thx..

well, I'm trying to solve numerically the Langevin equation, initially for the brownian motion, it is also used in nuclear physics as you probably know, to describe the fission of nuclei...

the Langevin equation is :

$$\ddot{x}+\beta \dot{x} + \frac{1}{m}\frac{\partial U}{\partial x} = \Gamma(t)$$

where $$\beta=\frac{\gamma}{m}$$ is a friction term (from a stocke's law) divided by the mass of the brownian particle. (so $$\beta$$ is $$s^{-1}$$)

$$U(x)$$ is an extern potential

and $$\Gamma(t)$$ is a stochastic force divided my the mass.

well, now I decide, as Kramers did, to tell that my potential U is represented by two parabolas as you can see on http://nicolas.aunai.free.fr/courb.htm [Broken]

I call B the height of the maximum, and $$x_b$$ the correspondant X-coordonnate (assuming that the minimum will be at x=0)

I decide that B will be my caracteristic energy for the problem, and $$x_b$$ my caracteristc length.

I want, in order to solve my equation numerically, to write it without dimension, so I introduce the following new variables :

$$Q = \frac{x}{x_b}$$

$$\Pi = \frac{U}{B}$$

$$t'=\omega t$$ where $$\omega$$ is the caracteristic pulsation of my parabolas

i.e. the equations of the parabolas are :

$$U_1(x)=\frac{1}{2}m\omega^2 x^2$$ for $$x<x_0$$

$$U_2(x)=-\frac{1}{2}m\omega^2(x-x_b)^2 + B$$ for $$x>x_0$$

assuming the continuity of the potential, the functions and the derivatives shoud be equal between them at $$x_0$$ the junction point. these conditions give us the height B which is equals to :

$$\frac{1}{4}m\omega^2 x_b^2$$

well now we have to re-write the langevin equation, replacing by the new variables... and here began my problems....

$$\ddot{x}+\beta \dot{x} + \frac{1}{m}\frac{\partial U}{\partial x} = \Gamma(t)$$

$$\dot{x}$$ means $$\frac{dx}{dt}$$, so since we have $$t'=\omega t$$, the derivation by t' gives us an $$\omega$$

$$\omega^2 x_b\ddot{Q}+\beta \omega x_b \dot{Q} + \frac{B}{m}\frac{\partial \Pi}{\partial Q} \frac{\partial Q}{\partial x} = \Gamma(t)$$

we know that $$\frac{\partial Q}{\partial x} = \frac{1}{x_b}$$

so we have :

$$\omega^2 x_b\ddot{Q}+\beta \omega x_b \dot{Q} + \frac{B}{mx_b}\frac{\partial \Pi}{\partial Q}= \Gamma(t)$$

and with the B value we have :

$$\omega^2 x_b\ddot{Q}+\beta \omega x_b \dot{Q} + \frac{\omega^2 x_b}{4}\frac{\partial \Pi}{\partial Q}= \Gamma(t)$$

finally, dividing by $$\omega^2 x_b$$ we obtain :

$$\ddot{Q}+\frac{\beta}{\omega}\dot{Q} + \frac{1}{4}\frac{\partial \Pi}{\partial Q}= \frac{1}{\omega^2 x_b}\Gamma(t)$$

which I think is almost the result that I'm looking for.. since the first term is without dimention, $$\frac{\beta}{\omega}$$ is without dimension too, and the third term too..

my problem is the stochastic term...

now I don't know how to do with it.... if you have an idea...

I now, that the autocorrelation function of the Langevin Force is :

$$<\Gamma(t)\Gamma(t')> = 2\beta T \delta(t_1-t_2)$$

since $$t_i = \frac{t_i'}{\omega}$$, can I write the following equation :

$$<\Gamma(t)\Gamma(t')> = 2\beta T \delta(\frac{t_1'}{\omega}-\frac{t_2'}{\omega})$$

and with $$\delta(a x)=\frac{1}{a}\delta(x)$$

I find :

$$<\Gamma(t)\Gamma(t')> = 2\omega \beta T \delta(t_1'-t_2')$$

so $$\Gamma(t)$$ should be proportionnal to $$\sqrt{(2\omega\beta T \delta(t_1'-t_2'))}$$

in the nuclear system, k (the Boltzmann constant) is equal to 1, so we can also introduce the new variable :

$$\Theta=\frac{T}{B}$$

so we can put it in our relation :

$$\Gamma(t)\ \ \ \alpha\ \ \ \sqrt{(2\omega\beta B\Theta\delta(t_1'-t_2'))}$$

well... I'm not sure where I'm going with this... can someone help me ?

thanks !

Last edited by a moderator: May 2, 2017