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Top Document: Einstein (1905) Absurdities Previous Document: 3. The light direction absurdity. Next Document: 5. The amazing transverse gamma absurdity. See reader questions & answers on this topic! - Help others by sharing your knowledge
Perhaps the most marvelous thing about Einstein's Special
Relativity derivation is the math he used to get from his
tau function in t and x' to his tau=f(t,x) transform.
[We let his a=phi(v)=1, as he concludes later.]
[1] tau = (t-vx'/(cc-vv)).
[2] tau = (t-vx/cc)/sqrt(1-(v/c)^2).
First of all, to get to [2], we certainly have
to rid [1] of x'. x'=x-vt.
[3] tau = (t-v(x-vt)/(cc-vv))
= (tcc-tvv-vx-vvt)/(cc-vv)
= (tcc - vx)/(cc-vv)
Now, divide numerator and denominator on the right
by cc:
[4] tau = (t-vx/cc)/(1-vv/cc).
There's only one way to get [2] from [4]. Let
tau<>tau, a logical absurdity in this situation;
Einstein has proceeded far beyond tau the unknown
function. The only unknown is a, which he later
says is phi(v)=1.
And if it is legal to get [2] by multiplying only one side
by sqrt(1-vv/cc), then it is also correct to multiply
only one side by (1-vv/cc), and get the galilean transform.
Or to multiply one side by pi and get "t and -vx/cc
are really circle diameters" transforms. [You know,
the circumference of a circle is Pi*diameter?]
But in all cases - both the absurd Einsteinian and Pi
transforms - it is not legal to treat only one side of
an equation in a non-identity fashion. The left side of
the tau function would not be tau, but gamma*tau or Pi*tau.
The appearance of gamma is just as magically marvelous
in the X' transform (we used X' for the moving system
x value coordinate, remember?):
X' = ccx'/(cc-vv).
= (ccx-ccvt)/(cc-vv)
= (x-vt)/(1-vv/cc).
Not X' = (x-vt)/sqrt(1-vv/cc).
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Last Update March 27 2014 @ 02:12 PM
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