Potential of Brownian particle

[Steve Zissou]

Hello physicians.
Consider the following Brownian motion particle:
\dot{x}(t)=\alpha(t)+\beta(t)\eta(t)
The kinetic energy of which would be
\frac{1}{2}v^2=\frac{1}{2}(\dot{x}(t))^2
(for some unit mass.)
The potential is...?

 

[sudu.ghonge]

Hello physicians.

In-case you're trying to refer to people who study physics, they're called 'physicists'. A 'Physician' is a professional who practices medicine.

Consider the following Brownian motion particle:
\dot{x}(t)=\alpha(t)+\beta(t)\eta(t)
The kinetic energy of which would be
\frac{1}{2}v^2=\frac{1}{2}(\dot{x}(t))^2
(for some unit mass.)
The potential is...?

Please be more elaborate. I have not studied brownian motion in depth but I can surely say that

\dot{x}(t)=\alpha(t)+\beta(t)\eta(t)

can also be written as
\dot{x}(t) = \gamma(t)
where \gamma(t) = \alpha(t)+\beta(t)\eta(t)
And brownian particles can be found in a variety of time dependent as well as independent potentials. And usually, the potential is a function of the position variables alone( In this case, 'x'. So, it might help if you could provide a relationship between, say, the force and position variables.

 

[Steve Zissou]

Sudu. Get a grip. I know what a physician is, and what a physicist is. It's what we call "humor."
Secondly, the way I've written it is the correct way to write it. There is a huge weight of convention and tradition here, and I'm hoping someone who understands that will help me out.

 

[sudu.ghonge]

It's what we call "humor."

No, we don't. And as far as I remember, this is not a forum where ideas of humor are discussed. You might very well have been a person who makes the common mistake. I don't need to "get a grip".
Peace out.

Secondly, the way I've written it is the correct way to write it. There is a huge weight of convention and tradition here, and I'm hoping someone who understands that will help me out.

Can you cite a source? I searched extensively and nowhere did I find this convention.

 

[Steve Zissou]

Ok I'll try again. Hopefully this will clarify.
Consider the well-known Langevin equation of a particle in Brownian motion.
Here is a reference:
http://www.mat.univie.ac.at/~esiprpr/esi2115.pdf
In the attached paper, the Langevin equation is written as:
\frac{\partial p_i}{\partial t}=-\gamma_0 p_i-\frac{\partial U}{\partial q_i}+\eta_i
.where the gamma is a friction term and the U is potential.
I would like to characterize a more general model. Let's say it like this:
\dot{x}(t)=\alpha(t)+\beta(t)\eta(t)
If we would like to, we are free to simply say
\dot{x}(t)=\alpha(t)+\eta(t).
Now, how would we characterize the potential here in my generalization? Thanks to all who would help.

 

[willem2]

You have an equation for the derivative of position (velocity), whereas the Langevin equationis for the derivative of momentum, which is equal to a force.

It's probably best to try to start with the langevin equation itself and try to vary that. The friction gamma could be a function of the velocity, for example, or the potential could vary with time (such as when you shake the fluid that the particle is in), and you'll have no problem with the potential, because you can just copy the term with U.

of course you could also use a force that is dependent of the position (or possibly constant) and do away with the potential.

 

[Steve Zissou]

Willem, thanks for the reply. Yes I am aware that there are important issues regarding the units of measure involved. I am aware that a force can be the grad of potential.

Let me try to clarify my question. There are many possible models for Brownian motion, not just Langevin. Also the Langevin equation need not be written in terms of momenta, it can be written in terms of distance formt he origin, &c. I am trying to get toward a generalization. Let's say we have an "equation of motion" for a particle under Brownian motion. We'll express it as:

whatever you like = a(t) + eta(t)

How would you characterize the potential acting on this particle?