Estimating Displacement of Particle in Brownian Motion

[sjweinberg]

Suppose I have a large particle of mass $M$ that is randomly emitting small particles. The magnitude of the momenta of the small particles is $\delta p$ (and it is equal for all of them. Each particle is launched in a random direction (in 3 spatial dimensions--although we can work with 1 dimension if it's much easier). Assume also that these particles are emitted at a uniform rate with time $\delta t$ between emissions.

So here's my issue. It seems to me that this is a random walk in momentum space. What I would like to know is how to estimate the displacement of the particle after $N$ particles are pooped out. Thus, I need some way to "integrate the velocity".

However, I want to stress that I only care about an order of magnitude estimate of the displacement here. Has anyone dealt with this kind of a situation?

I appreciate any help greatly!

 

[ImaLooser]

Suppose I have a large particle of mass $M$ that is randomly emitting small particles. The magnitude of the momenta of the small particles is $\delta p$ (and it is equal for all of them. Each particle is launched in a random direction (in 3 spatial dimensions--although we can work with 1 dimension if it's much easier). Assume also that these particles are emitted at a uniform rate with time $\delta t$ between emissions.

So here's my issue. It seems to me that this is a random walk in momentum space. What I would like to know is how to estimate the displacement of the particle after $N$ particles are pooped out. Thus, I need some way to "integrate the velocity".

However, I want to stress that I only care about an order of magnitude estimate of the displacement here. Has anyone dealt with this kind of a situation?

I appreciate any help greatly!

We have the sum of N independent identically distributed random variables so this is going to converge to a Gaussian very quickly, that is with N>30 or so. The momentum will follow 3-D Gaussian with mean of zero, that has got to be available somewhere. (A 2D Gaussian is called a Rayleigh distribution.)

The 1D case will be a binomial distribution that converges to a Gaussian.

 

[sjweinberg]

We have the sum of N independent identically distributed random variables so this is going to converge to a Gaussian very quickly, that is with N>30 or so. The momentum will follow 3-D Gaussian with mean of zero, that has got to be available somewhere. (A 2D Gaussian is called a Rayleigh distribution.)

The 1D case will be a binomial distribution that converges to a Gaussian.

Thanks for your help.

I am aware that the momentum distribution will converge to a Gaussian of width $\sim \sqrt{N} \delta p$. However, do you know what this will mean for the position distribution? In other words, I am really interested in the distribution of the quantity $\sum_{i} p(t_{i})$ where the sum is taken over time steps for the random walk.

My concern is that even though $p$ is expected to be $\sim \sqrt{N} \delta p$ at the end of the walk, I think that the sum may "accelerate" away from the origin because $p$ drifts from its origin.

 

[mfb]

From a dimensional analysis: $\overline{|x|}=c~ \delta t~\delta v~ N^\alpha$
A quick simulation indicates $\alpha \approx 1.5$ and $c\approx 1/2$ in the 1-dimensional case. In 3 dimensions, c might be different, while alpha should stay the same.

 

[sjweinberg]

From a dimensional analysis: $\overline{|x|}=c~ \delta t~\delta v~ N^\alpha$
A quick simulation indicates $\alpha \approx 1.5$ and $c\approx 1/2$ in the 1-dimensional case. In 3 dimensions, c might be different, while alpha should stay the same.

Thanks for the help. In fact, your estimation of $\alpha = \frac{3}{2}$ is the same thing I estimated with the following sketchy method:

Let $n(t) = \frac{t}{\delta t}$ be the number of particles emitted after time $t$. Then, the speed of the large particle at time $t$ can be estimated as $\frac{\delta p \sqrt{n(t)}}{M} = \frac{\delta p }{M} \sqrt{\frac{t}{\delta t}}$.

Then $\left| x(t) \right| \sim \int_{0}^{t} \left| v(t) \right| dt \sim \delta t \delta v \left(\frac{t}{\delta t}\right)^{3/2}$.

I feel that this estimate is probably an overestimate which is where your $c \sim 1/2$ may come from.

Thanks again.