Re: Test SPF4

On Jan 14, 9:10 pm, "Sue..." <suzysewns...@xxxxxxxxxxxx> wrote:
On Jan 14, 11:42 pm, "Ken S. Tucker" <dynam...@xxxxxxxxxxxx> wrote:



Hi Peter.
On Jan 14, 1:09 pm, Peter <end...@xxxxxxxxxxx> wrote:

On 14 Jan., 19:39, "Ken S. Tucker" <dynam...@xxxxxxxxxxxx> wrote:

Based on c being constant, (g=1 - v^2/c^2), E=mc^2,

E' = E /sqrt(g) and t' = t*sqrt(g),

results, therefore,

E' * t' = E * t = N*ergs * seconds = h = invariant,

( N~ 6.626*10^-27 is a scalar, therefore an invariant).

this is obviously wrong

Check this out...
Example: Ken and Peter have identical unit masses
and unit second, at rest. Then, Ken (K') moves at 8/10 c
relative to Peter (K) where

sqrt(g) = sqrt(1 - 64/100) = sqrt(36/100) = 6/10.

A unit mass "m" in Peter's FoR is measured to be 10/6
of Ken's "m' "otherwise equal mass, using m' = m/sqrt(g),
so Peter's unit mass is heavier relative to Ken.

A phenomena (the unit second) requiring 1 second in
Peter's FoR takes only 6/10 of a second to occur in Ken's
FoR, relatively to Ken, using, t' = t sqrt(g), so the rate of
events in Peter's FoR transpire more slowly measured by
Ken.
How's that?

<< Thus, we can account for the ever decreasing
acceleration of a particle subject to a constant
force [see Eq. (1542)] by supposing that the
inertial mass of the particle increases with
its velocity according to the rule (1546).
Henceforth, $m_0$ is termed the rest mass, and
$m$ the inertial mass. >>http://farside.ph.utexas.edu/teaching/em/lectures/node126.html
Sue...

Eq.(1552) must be wrong, the m_0 is a rest mass,
while an invariant is reqired.
Thanks
Ken S. Tucker
.

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