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Old October 19th 07, 06:50 PM posted to sci.physics.relativity,sci.physics,sci.astro
T.M. Sommers
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Posts: 35
Default Improved Relativity Theory and Doppler Theory of Gravity

kenseto wrote:
A Paper entitled "Improved Relativity Theory (IRT) and Doppler Theory
of Gravity (DTG)" is available in the following link:
http://www.geocities.com/kn_seto/2007IRT.pdf


Your "theory" still suffers from problems that were pointed out
long ago. Take, for instance, you definition of relative
velocity on p. 4:

v = \lambda (f_aa - f_ab),

along with your definitions of the speed of light,

c = f_aa \lambda

and \gamma,

1 / \gamma = F_ab / F_aa.

Now, f_aa is the "[i]nstantaneous frequency measurement of A's
standard light source as measured by A", and F_aa is the
"[f]requency of a standard light source in A's frame as measured
by A." There is no reason at all not to equate these two
arbitrary "standard" light sources, so that f_aa = F_aa. Then,
unless something is very wrong, f_ab = F_ab. Then you have,
assuming your \gamma is the same as everyone else's \gamma:

f_ab / f_ab = \sqrt ( 1 - v^2 / c^2 )
= \sqrt ( 1 - (\lambda (f_aa - f_ab))^2 / (f_aa \lambda)^2 ).

Square both sides, multiply by f_aa^2, and expand the squa

f_ab^2 = f_aa^2 - f_aa^2 + 2 f_aa f_ab - f_ab^2,

or

2 f_ab^2 = 2 f_aa f_ab

which means that f_aa must equal f_ab. In other words, just as
before, your equation for v only works if v = 0.


As another example of a fundamental problem, look at Eq. 11 on p.
6, which purportedly gives the energy of a single particle:

E = M_0 \lambda^2 f_aa^2.

This means that you can change a particle's energy just by
choosing a different "standard" light source. That is absurd.

--
Thomas M. Sommers -- -- AB2SB