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{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
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{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
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{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
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some moment, their axes are coincident and that they zero their clocks 
at that momen.  If the velocity of   S_1 in  S_0 is u along the x axis 
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apply first a Lorentz from S_0 to S_1 and then a Transform from S_1 TO S_2.}
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{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
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tranform:}
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{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
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equate the terms of the second columns}
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{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard We note that these 
equations are independent of u and conclude that no solution is 
possible which will be valid for all u less than c. Therefore, we are 
unable to show that there is a Lorentz transform from S_0 to S_2.}
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that }
.EQN 0 22 275 0 0
({0:x_0}NAME)^(2)+({0:y_0}NAME)^(2)+({0:z_0}NAME)^(2)-({0:t_0}NAME)^(2)
.TXT 0 19 276 0 0
Cg a56.200000,56.200000,1
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard =}
.EQN 0 4 277 0 0
({0:x_2}NAME)^(2)+({0:y_2}NAME)^(2)+({0:z_2}NAME)^(2)-({0:t_2}NAME)^(2)
.TXT 0 18 278 0 0
Cg a20.200000,20.200000,13
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard is satisfied.}
.TXT 4 -63 279 0 0
Cg a83.200000,83.200000,153
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard This can be varified by 
doing the sums on a separate mathcad sheet. (It has a habbit of 
filling its memory and crashing if too much is done in one sheet)}