Re: Contraction has been abolished by Special Relativity



rbwinn wrote:

On Oct 9, 7:29�am, doug <x...@xxxxxx> wrote:

rbwinn wrote:

On Oct 8, 9:26 pm, doug <x...@xxxxxx> wrote:


show how any experiment you listed shows these equations wrong.
x'=x-vt
y'=y
z'=z
t'=t

w=velocity of light
x=wt
x'=wn'

x'=x-vt
wn'=wt-vt
n'=t(1-v/w)

n' is the time it takes a photon to travel a distance of x' in the
moving frame of reference.

You keep saying this as though we had forgotten what it is you have
been confused about.

OK, doug, show us how any of the experiments you listed show this to
be wrong.

Pretty clearly you did not look at them. See if you can find one that
supports your view.

The Michelson-Morley experiment supports my view. �According to
scientists, it shows light traveling at 186,000 miles per second in
two different frames of reference.
Robert B. Winn

No, it does not support the GT. Do the math and show how you think
it does.


Light was directed down the arm of an interferometer where it was
reflected by a mirror at the end of the arm. Now with regard to the
Galilean transformation equations this is not anywhere near as
scientists try to make it. The Equation for time in the Galilean
transformation equations is t'=t. That means there is a common
reference for time between the two frames of reference. That common
reference for time is time as measured by rotation of the earth, which
is the same in both frames of reference. But frequency and speed of
light is not measured in both frames of reference by rotation of the
earth. It is measured by transitions of a cesium isotope molecule in
both frames of reference, which is called scientific time. The
transitions of a cesium isotope molecule in the moving frame of
reference are not the same as transitions of a cesium isotope molecule
in the frame of reference at rest. This is called relativity of
time. So to be brief, n' in the foregoing equations is time according
to transitions of a cesium isotope molecule in the moving frame of
reference.
When the light is traveling down the arm of the interferomether
it has a velocity of +c relative to both frames of reference. Suppose
that the light starts at x1=0 and is reflected by the mirror at x2=x.
Then in the frame of reference at rest, the light travels x2-x1=x in a
time of t or x=ct. In the moving frame of reference, x'1=0 and x'2=
x-vt.
x'2-x'1=x-vt. While this is happening, the earth rotates on its
axis. This is shown by t'=t.
An observer timing the speed of light with a cesium clock in the
moving frame of reference gets a velocity of light of c=186,000 miles
per second for the velocity of the light the same as an observer in
the frame of reference at rest.

w= velocity of light
x=wt
x'=wn'
x'=x-vt
wn'=wt-vt
n'=t(1-v/w)

Until the light is reflected by the mirror, it has a velocity of c,
so the time it takes the light to travel to the mirror in the moving
frame of reference according to the cesium clock in that frame of
reference is n'=t(1-v/c). It takes less time for the light to reach
the mirror in the moving frame of reference.
Then the light is reflected by the mirror. Its velocity changes
from +c to -c relative to the two frames of reference as measured by
scientific time. In the frame of reference at rest, the light travels
at this velocity from x1 at the mirror to
x2, which is the origin of x. This will be the negative of the x
where the light was traveling at a positive velocity, or
x(reflected) = -ct or -x.
Notice that t remains positive and is the same for light going
either direction in the frame of reference at rest.
In the moving frame of reference, the light reaches the mirror at
the end of the arm at a time of n' according to the cesium clock in
the moving frame of reference, and travels at a velocity of -c
relative to both frames of reference, reaching the origin of x' at a
time of n'=t(1-v/c) according to the cesium clock in the moving frame
of reference, continuing on to the origin of x in the frame of
reference at rest, which it reaches in a time of (-x-vt)/-c = (x+vt)/
c, which is a longer time than t.
The time it takes for a photon to travel from the origin of the frame
of reference at rest and back after being reflected by the mirror is
t'=t. The time it takes a photon to travel from the origin of the
moving frame of reference to the mirror and back to the origin of the
moving frame of reference is the time shown by a cesium clock in the
moving frame of reference and is a shorter time than t'=t.
So we have relativity of time without a distance contraction and
without relativity of simultaneity. The light has a speed of c in
both frames of reference, and it has the same frequency and wave
length in both frames of reference if scientific time, or identical
cesium clocks are used in both frames of reference.
This interpretation of the Galilean transformation equations
gives identical answers to several decimal places for lower velocities
as the Lorentz equations, such as 30 miles per second, the velocity of
the planet Mercury. As the speed of light is approached, the answers
given by this interpretation of the Galilean transformation equations
are increasingly less than Lorentz equation values for x' and
scientific time.
I hope that this is a sufficiently clear explanation for
scientists to understand.
Robert B. Winn
You have gone through a contorted "explanation" of what you think
is happening. There are lots of problems here. First of all, you
even confuse yourself in you notation of the frames. Second there
is no physics content here. To analyze this, you need to express
it mathematically and show what you predict. Then you can compare
that with the experimental results. You have made a word salad but
that does not help show you ideas.
Scientists understand the situation just fine. So far you seem to
be very confused.

.

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