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### A Perfectly Reflecting Sphere

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Date: 09/09/2000 at 02:07:38
From: Jeff Ledbetter
Subject: Vector formula?

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
```

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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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Associated Topics:
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