# Langevin equations in translational and rotational direction

I want to describe the following system.

• A bead is connected with a tether.
• There is a force $\vec{F}_{up}=F_{up}\hat{y}$ that acts on the bead.
• The tether acts with a force on the bead, this force $F_{spring}$ is only dependent on the length/extension of the spring.
• A Brownian force acts in the rotational and translational directions. Just consider these as a random variable in x-y-theta direction.

Now I want to formulate the Langevin equation for the rotational motion.

I know that if there is no angular dependence, $\theta=0$ for all time, the Langevin equation are:

$\gamma_{y}\frac{dy}{dt}+F_{y}^{tether}(x,y)+\frac{dF_{y}^{tether}(x,y)}{dy}\hat{y}=F_{y}^{therm}+F_{mag}-mg$

$\gamma_{x}\frac{dx}{dt}+F_{x}^{tether}(x,y)+\frac{dF_{y}^{tether}(x,y)}{dx}\hat{x}=F_{x}^{therm}$

What I know:

The $\gamma_{theta}$ and $F_{them}^{\theta}$ are known from literature.

Should I construct a construct a Langevin equation in the same way, such as

$\gamma_{\theta} \frac{d\theta}{dt}+\frac{d\tau_{tether}}{d\theta}\theta+\tau_{tether}=F^{tether}_{\theta}$ ?

More specifically, should it include the term $\frac{d\tau_{tether}}{d\theta}\theta+\tau_{tether}$, which is completely analogous to the translational equations? Trivia: In the end I want to extend this into 3 dimensions and solve the system numerically, using Langevin dynamics. But I'm stuck on how to describe the force.

EDIT: I've edited the question into something more specific.

• That was a typo, I've changed it. The $F_{up}$ is in $y$-direction. – johnbaltis Nov 3 '14 at 19:47
• I'm not sure what the problem is, you will have two Langevin equations, one for translational and another for rotational motion. F_spring will naturally be in the translational equation and its associated torque will be in the rotational one. – Elvex Nov 3 '14 at 20:02
• That is what I try to establish. How will these Langevin equations look? – johnbaltis Nov 3 '14 at 20:24

All the forces acting on the bead will accelerate the center of mass as though the force is acting there - so the mass will accelerate with a magnitude and direction given by the vector sum of $$\frac{\vec{F_{up}} + \vec{F_{spring}}}{m}$$ .

The bead will further undergo a rotation. The torque is entirely due to the spring - but we are missing the angle of the spring to the vertical. Let's call that angle $\alpha$. Then the component of the force of the spring causing rotation is $F_{spring}\cos(\pi - \alpha - \theta - \pi / 2)=F_{spring}\sin(\alpha + \theta)$.

Diagram: From this it follows that the angle $\beta$ can be calculated from $x$, $y$ and $\theta$ (as I defined them in the drawing - I assume this is how you define them as well... not quite clear from your diagram).

Now you can see that

$$\tan\alpha = \frac{x + r\sin\theta}{y - r\cos\theta}$$

After which you can calculate $\beta$ from

$$\beta = \pi/2 - (\alpha + \theta)$$

And finally, the torque on the bead is

$$\Gamma = F_{spring} r \cos\beta$$

• Does this mean that the torque at the bead-tether attachment point is $\tau=R_{bead}F_{spring}\sin{(\alpha+\theta)}$? Can't $\alpha$ be expressed in $(x,y,\theta)$, I think so? – johnbaltis Nov 4 '14 at 1:52
• You did not define the position of the attachment point in terms of x and y. If you do we can eliminate $\alpha$... – Floris Nov 4 '14 at 2:02
• Well, the $(x,y)$ coordinates of the center-of-mass of the bead are known. – johnbaltis Nov 4 '14 at 15:25
• @johnbaltis - I have updated the answer accordingly – Floris Nov 4 '14 at 16:24
• Thank you so much! The geometry is solved. Do you perhaps something about the Langevin equation in rotation? – johnbaltis Nov 4 '14 at 23:06