Random walk in spherical coordinates

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Modeling receptors on a cell surface in spherical coordinates simplifies the representation of their movement while ensuring they remain within the membrane. The random walk can be achieved by treating the angles theta and phi similarly to Cartesian coordinates, allowing for a 2D random walk approach. For uniform distribution on the sphere's surface, theta should be uniformly chosen between 0 and 2π, while phi should be selected using sinφ uniformly between -1 and 1. This method effectively maintains the receptors' confinement to the cell membrane while facilitating their movement modeling. Adopting this spherical coordinate approach enhances the accuracy of receptor interaction simulations within cellular environments.
arandall
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Hi,

I'm modeling receptors moving along a cell surface that interact with proteins inside of a cell. I figured it would be easier to model the receptors in spherical coordinates, however I'm unsure of how to model a random walk. In cartesian coordinates, I basically model a step as:

x = x + sqrt(6*D*timeStep)*randn
y = y + sqrt(6*D*timeStep)*randn
z = z + sqrt(6*D*timeStep)*randn

Where D is my diffusion constant. How can I do this just using theta and phi? Modeling random walk in spherical coordinates will be really nice, because I can fix r such that the receptors can't leave the membrane of the cell, and just focus on how it moves in 2D with respect to the membrane.
 
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Because the receptors are so much smaller than the size of the cell, it should be fine if you treat theta and phi just like x-y; i.e. pretend its a 2D random walk in Cartesian coordinates.
 
To choose points on the surface of a sphere uniformly, the two angles should be chosen as follows (I'll call them latitude and longitude):

Longitude (θ) - θ uniform between 0 and 2π.
Latitude (φ) - sinφ uniform between -1 and 1.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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