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    Re: Geometry of SNO-T
    From: Herbert Prinz
    Date: 2004 Oct 14, 08:48 -0400

    Alexandre
    
    Thank you for posting your proof and doing the tedious work of typing it
    in. Unless you find a flaw in the following, we are in 100% agreement
    now.
    
    I have been trying to give a purely geometrical explanation, but could
    not see how to do it without a drawing. Your proof has shown the way.
    Most list members will be able to follow your proof up to the point
    where you introduce the general formula for reflection. This and the dot
    product might scare a few people away. So, if you don't mind, here is
    your proof again, after a small cosmetic change in the form of a
    coordinate transformation. I will still use cartesian coordinates,
    because I can't draw, but the argument is entirely based on good old
    proportions, Greek style, avoids complicated vector algebra and can
    therefore be drawn on paper if the reader cares to.
    
    Let O = (0,0,0) be the front edge of the mirror at which we reflect. Let
    the mirror intersect the x-y-plane in the y-axis. Let M be a point on
    the y-axis.
    
    Let A be the left visor that we see directly. Let B = (p,q,0) be the
    right visor that we see in the mirror. Because angles AOM = MOB, the
    right triangles that form the coordinates of points A and B are similar
    and symmetric, and the coordinates are proportional. So A can be written
    as k*(p,-q,0) or
    
        A = (k*p,  k*-q ,  0)
    
    Let B' be the reflection of B, and therefore
    
        B' = (p, -q, 0)
    
    The rest goes like you say: After the observer lifts his eye, the rays
    go through O' = (0,0, h).
    
       O'A = (k*p,  k*-q ,  -h)
       O'B = (p, -q, -h)
    
    These vectors can only be parallel if h = 0 or k = 1.
    
    If h = 0, then k is arbitrary, meaning that the mirror does not have to
    be in the symmetry plane of the two visors. But since the pivotal axis
    has to be in it, we know that the mirror may be outside of the pivotal
    axis. Therefore the test is always valid if we look in the x-y-plane.
    
    If h > 0, then k = 1. Inserting this into the definition of  A, above,
    we get
    
       A = (p,  -q ,  0)
    
    This means that A and B' can be seen in the same direction from a point
    O' outside the x-y-plane only if A and B are symmetric with respect to
    the mirror. In other words the mirror is the symmetry plane. But the
    pivotal axis is always in the symmetry plane, and thus in the mirror.
    
    ***
    
    (Since line AB is perpendicular to the mirror if and only if k = 1, this
    is also criterion for whether the visors will coincide for every h. We
    don't need this here, I just mention it for completeness, since I
    brought it up earlier.)
    
    Herbert Prinz
    
    
    

       
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