A Perfectly Reflecting SphereDate: 09/09/2000 at 02:07:38 From: Jeff Ledbetter Subject: Vector formula? Please help! I am trying to find some assistance in developing a mathematical representation of a series of events for the purposes of helping my sons with a science project they are undertaking. The series of events is as follows: A laser is fired from one of the "poles" of a perfect sphere into the interior of the sphere. The interior surface of the sphere is a perfectly mirrored surface, so that the laser reflects an infinite number of times. I think that the equation I am trying to develop would be a vector representation of the angle of reflection at each intersection of the laser beam with the interior mirrored surface, even though velocity does not really affect the outcome. In other words, what I am looking for is a way to know the coordinates of each intersection of the laser beam in the interior of the sphere, an infinite number of times. Although I have a minor in math and probably should be able to develop the equation myself, it has been a long while since I gave a great deal of thought to vector mathematics. My sons and I would very much appreciate any assistance or direction you could provide. Thank you. Jeff, Shane, and Tajjen Ledbetter Date: 09/09/2000 at 08:41:21 From: Doctor Jerry Subject: Re: Vector formula? Hi Jeff, Here are my thoughts on your question. Suppose we have reached the point p on the sphere and know the direction u (a unit vector) the light will take as it leaves from p. If p happened to be the point from which the laser is fired, then I'm assuming that you would know the direction in which the laser is fired. An essential part of the problem is, given p and u, to figure out the next point, call it q, that the laser hits and the direction v (a unit vector) the light will take as it leaves q. I'll first determine the point q. Note that when I say point q I'm also saying that q is a position vector of this point. I'll assume that we're on a sphere of radius a and center at the origin. The light is traveling on the line with vector equation r = p+t*u, where t >= 0 is a parameter. We want to determine t so that r.r = a^2 (this is r dot r). r.r = (p+t*u).(p+t*u) = p.p + 2t*u.p + t^2 u.u Now, r.r = p.p = a^2 and u.u = 1. So, 0 = 2t*u.p + t^2 and so (ignoring t = 0), t = -2u.p So, q = p + t*u = p + (-2u.p)u This determines q. As to the direction of the next bounce, I'm imagining the point q with the head of the vector u just touching q. Draw the interior unit normal n to the sphere at the point q. I believe it to be true that the plane P determined by the vectors u and n will contain the bounced ray. So, in the plane P, I have a point q, a vector u whose head just touches q, and the unit vector v leaving q, with the lines of u and w symmetrically placed about n. This is angle of incidence equal to the angle of reflection. I think that we can write -u + w = s*n, where s is a scalar to be determined. If I dot both sides of the above equation with n, -u.n + w.n = s Let z be the acute angle between -u and n; this is also the angle between w and n. So, cos(z) + cos(z) = s s = 2cos(z) So, w = u + s*n = u + (2cos(z)) So w is almost determined. The only thing we need is z, the angle between -u and the unit normal n at q. Because we are on an origin-centered sphere, n = -q. So, (-u).n = 1*1*cos(z) and (-u).(-q) = cos(z) cos(z) = u.q Summarizing and hoping I haven't made any dumb mistakes, given p and u, q = p + (-2u.p)u and w = u + 2u.q Whether if you string these together you can get any decent formulas, I don't know. - Doctor Jerry, The Math Forum http://mathforum.org/dr.math/ |
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