Re: How to measure one-way light speed.


"Gerald L. O'Barr" <globarr@xxxxxxxxx> wrote in message news:1164148909.709758.220900@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Martin Hogbin <goatREMOVETHIS...@xxxxxxxxxx> wrote:
Gerald L. O'Barr" <globarr...@xxxxxxxxx> wrote:
. . .
As far as I have studied it, the
rotating bar is able to maintain a close first
order affect of the SR relationship. But it is
not able to maintain a complete relationship. And
I do believe that this rotating bar approach is
the best way we have to break apart LET from SR.

Martin Hogbin <goatREMOVETHIS...@xxxxxxxxxx> wrote:
The rotating bar method of synchronising clocks is
a non-starter, for reasons which have been
explained to you many times.

Firstly from any practical perspective the method is a dead
loss. The speed of sound in steel is hundreds of thousands
of times less than the speed of light. How will you bring the
bar up to speed? How will you know if there are any
twists in it?

But suppose that by years of careful reasearch and
development you managed to overcome the practical
problems, what would you have achieved?

You would have produced one definition of
simultaneity in one frame of reference, that of the bar,
something you could have done far more easily
and accurately with a light beam or some slowly
transported clocks.

I know that some SR experts have just assumed
that a rotating bar will twist exactly the right
amount to maintain perfect SR sync.

In the (non-rotating) frame of the bar there is no need to
assume any twist..

Your calculations below assume the answer you are
trying to achieve.


For a simple diagram: Take triangle ABC:

A
|\
| \
| \
| \
|____\
B c


(If the diagram is not correct, save this post, bring it
up into any word processor that allows you to change
fonts, and pick a constant width font. This will bring
the diagram to the state in which it was made.)
Let AB be the direction and magnitude of velocity v
of the bar in the direction of the length, L, of the
bar.
Let BC be direction and magnitude of the velocity
of a point on the circumference of the bar, equal to
a velocity of rw, where r is the radius, and w is the
radians per second of the rate of spin.
In the ether approach, where we start at rest in
the ether, L, v, r and w, are all absolute values.
The maximum absolute velocity within the bar occurs
in those fibers that lie along AC, with an absolute
fiber maximum velocity of Vfm = SQRT(v*v + r*r*w*w).
The length of these fibers is equal to the same
ratio as their velocities, that is, triangle ABC is
the same similar triangle for both distances and for
velocities for those velocities that are in the same
directions as the distances.
Thus, Lf = L * AC/AB or = L * Vfm/V, etc.
The twist can now be considered. The physical
twist will be equal to the change in the maximum
length of the fibers, times the ratio of BC/AC.
And the clock sync (the change or difference in the
time being the change in the position or the angle of
the end) would thus be this change in distance
divided by rw.
Thus, what is the change in time between the two
ends of the bar? We can assume that the rotation
speed is very small compared to V. The standard
assumptions being rw < v << c. And thus, to
first order, Vfm = V. The approximate change in the
length of the fiber is thus a first order change with
a velocity of V, and a length of L, producing a total
change of:
L*(1/2)v^2/c^2.
The twist will be the above change of L*(1/2)v^2/c^2,
times rw/v.
This produces a value of L*(1/2)*V*r*w/c^2. This is
the amount of physical twist that occurs in the
circumference.

This physical distance produce a time difference of:
(1/2) L*V/c^2.

For correct SR sync, a full LV/c^2 would be expected.

Thus, as can be seen, we did get a twist, and the
twist was in the correct direction, and the twist was
about half of what would be desired. Such an act is
amazing, that even a rotating bar tries to hide the
fact that we have an ether. But again, it is not
perfect. And even when we use more accurate
approximations, we find that these rotating bars just
do not perfectly hide the ether effect. We must test
this approach!

Thanks again.
Gerald.

Now I just threw all this together without checking
any of my notes, but the results do end up the way I
remembered it. So it just might be right. I would
appreciate anyone's comments on this problem.



.

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