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Topic: random motion on a triangle (Read 975 times) |
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JocK
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random motion on a triangle
« on: Jan 15th, 2005, 3:07am » |
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A particle performs a perfectly random Brownian motion inside a triangle until it hits one of the sides of the triangle. At what central point inside the triangle does the particle need to start such that it has equal likelihood of ending at any of the sides?
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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SWF
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Re: random motion on a triangle
« Reply #1 on: Jan 19th, 2005, 5:34pm » |
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Start with a unit circle centered on the origin in the w (u+i*v) complex plane. Divide into three equal sectors of 120 degrees each. Starting at the center has equal probabily of reaching the edges for each of the 3 sectors. Conformally map into a triangle and where the center of circle gets mapped is the point we are looking for.
The mapping w=(i-z)/(i+z) maps the circle into the upper half z (x+i*y) plane with the center of the circle at z=i. The point on the circle at w=-1 (taken as the boundaries of one of the 3 segments) maps to +/-infinity on the x axis. The other two boundaries map to z= +/-1/sqrt(3).
Map the upper half z plane into a triangle in the t complex plane with
dt/dz = (z- 1/sqrt(3))a/[pi]-1 * (z+1/sqrt(3))b/[pi]-1
where a and b are two of the interior angles of the triangle. Integrate z from z=1/sqrt(3) to z=i, to find the relative position of the point of interest from the vertex of the triange which has angle a. That integral does not look like fun so I will stop there.
Using a different approach- approximating long skinny isosceles triangles with apex angle [theta} and long side of L, by a sector of a circle, the point of interest is found to be approximately a distance L*exp(-[theta]/[pi] * sinh-1([sqrt]3) from the apex. For an equalaterial triangle the point must be the centriod (in this case the approximate formula is off by 12%: 60 degrees is not small enough an angle for it to work well).
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