| Title: | Mathematics at DEC | 
| Moderator: | RUSURE::EDP | 
| Created: | Mon Feb 03 1986 | 
| Last Modified: | Fri Jun 06 1997 | 
| Last Successful Update: | Fri Jun 06 1997 | 
| Number of topics: | 2083 | 
| Total number of notes: | 14613 | 
    Given two functions fx and ft from R*R to R such that
    
    	for every u,x0, and t, fx(u*t+x0,t)=u'*ft(u*t+x0,t)+x0',
    
    find fx and ft.  Here, u' and x0' may be functions of u and x0 but not
    of t.  One may assume that fx and ft are continuous if necessary.  It
    would be nice to have a general solution, but initial conditions are:
    
    	fx(c*t+x0,t)=c*ft(c*t+x0,t)+x0',
    	fx(-c*t+x0,t)=-c*ft(-c*t+x0,t)+x0',
    	fx(x0,t)=-v*ft(x0,t)+x0',                                     
    	fx(0,0) = ft(0,0) = 0,
    
    where c and v are positive constants.
    
    These equations correspond to the postulates of special relativity; the
    first one above specifies that paths of uniform velocity translate to
    paths of uniform velocity; the first two of the initial conditions
    specify that the speed of light is constant in different frames; and
    the last two specify that the two frames are moving at v relative to
    each other and their origins coincide.  The solution is of course the
    familiar Lorentz transformation, but I need a rigorous derivation to
    show to a crackpot.
    
    I can show the Lorentz transformation satisfies each of the above, but
    that does not prove it is a unique solution.  Something's eluding me in
    maniuplating these equations into the necessary form -- any ideas?
    
    
    				-- edp
| T.R | Title | User | Personal Name | Date | Lines | 
|---|---|---|---|---|---|
| 1748.1 | RUSURE::EDP | Always mount a scratch monkey. | Wed May 05 1993 19:24 | 5 | |
|     Note that I expect the first equation can be shown to imply that
    fx(x,t)=a0*x+a1*t+a2 for some constants a0, a1, and a2.
    
    
    				-- edp
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| 1748.2 | will check | AUSSIE::GARSON | nouveau pauvre | Mon May 10 1993 03:48 | 6 | 
|     re .1
    
    I believe that I've seen this derivation in a text book. Unfortunately
    it was one that I just picked up off the shelf at the library, browsed
    through and put back on the shelf. It could take some time to
    re-locate if I am even remembering correctly.
 | |||||
| 1748.3 | AUSSIE::GARSON | nouveau pauvre | Fri May 14 1993 23:06 | 21 | |
|     .2 (cont.)
    
    "Introduction to Special Relativity", Robert Resnick (Wiley & Sons)
    ISBN 0 471 71725 8 (paperback)
    
    Section 2.2 pp56-61
    
    There you will find a derivation of
    
    x' = x'(x,y,z,t)
    y' = y'(x,y,z,t)
    z' = z'(x,y,z,t)
    t' = t'(x,y,z,t)
    
    i.e. the coordinate transformation from (x,y,z,t) to (x',y',z',t')
    where the two frames are inertial and in uniform relative motion.
    
    A key assumption is "isotropy" which the author uses to show implies
    that each function x'(), y'(), z'() and t'() is linear. The author also
    makes the assumption that the relative motion is parallel to the x-axis in
    order to simplify the algebra.
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