Brownian motion ish

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Suppose I have a large particle of mass [itex]M[/itex] that is randomly emitting small particles. The magnitude of the momenta of the small particles is [itex]\delta p[/itex] (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 [itex]\delta t[/itex] 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 [itex]N[/itex] 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!
 

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  • #2
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Suppose I have a large particle of mass [itex]M[/itex] that is randomly emitting small particles. The magnitude of the momenta of the small particles is [itex]\delta p[/itex] (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 [itex]\delta t[/itex] 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 [itex]N[/itex] 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.
 
  • #3
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 [itex]\sim \sqrt{N} \delta p[/itex]. 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 [itex]\sum_{i} p(t_{i}) [/itex] where the sum is taken over time steps for the random walk.

My concern is that even though [itex] p [/itex] is expected to be [itex]\sim \sqrt{N} \delta p[/itex] at the end of the walk, I think that the sum may "accelerate" away from the origin because [itex] p [/itex] drifts from its origin.
 
  • #4
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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.
 
  • #5
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 [itex]\alpha = \frac{3}{2}[/itex] is the same thing I estimated with the following sketchy method:

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

Then [itex] \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} [/itex].

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

Thanks again.
 

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