Re: The velocity of light going pass a moving train.

Subject: Re: The velocity of light going pass a
moving train.

Sue... <suzysewns...@xxxxxxxxxxxx> wrote:
BZ <WQAHBGMXS...@xxxxxxxxxxxxx> wrote:
Dono <s...@xxxxxxxxxxx> wrote:
. . . .

<many deletes by O'Barr>


Sue... <suzysewns...@xxxxxxxxxxxx> wrote:
Now! I nominate you to explain why we can't see
light on its journey from the sun to the moon and
why raindrops viewed from a moving car don't take
longer to hit the ground in spite of the diagonal
path. >:-)

O'Barr comments:
Why can't we see light moving from the sun to the
moon? We cannot see light moving from the sun to
the moon, when we are not in the direct line of
motion, because in this specified region exists pure
ether, which by theory is a perfect carrier of light.
Thus, if any portion of this light were being
deflected all along its path, so that we could in a
sense observe it, then the ether would not be what it
is by theory, and light would follow a much greater
rate of fall-off than just 1/r^2. So Sue, what else
could it be, than exactly as it is?
And why would rain change its rate of vertical
fall? It is just like the old game where you shoot a
bullet out of a gun in a horizontal direction, and
drop a bullet at the same time, which bullet hits the
ground first? The greater horizontal velocity of a
bullet does not in principle interfere with the
vertical relationships that are present. And also,
any real observer will note that there is a wind that
is causing the rain to move horizontally, and the
force of gravity will still be there to keep the
vertical fall the same.
So, Sue, do you want to add any second order
effects to any of this?

Additional O'Barr comments:
When we on this earth, in going around the sun, we
find that between summer and winter, the location of
stars shift due to the fact that our speed was first
in one direction, and then in another direction.
This angle is slightly different between classical
and SR. And one reason why there is a difference
is because the tools being used do change with
changes in our motions. But in the train example,
during the problem, the tools being used by those in
the train remain constant, and the tools being used
in the frame of the tracks remain constant. Thus, we
should be sure that we are understanding the
differences here as to what change in angle is
actually being measured.
To repeat: The angle that is measured in the
frame of the track is exactly classical, and it does
not matter how fast the train is moving. The correct
vector analyses is the simple absolute velocity of
light, c, and the absolute velocity of the train, v,
and what these vectors produce is what will be
measured in the track frame. The only problem is in
knowing the exact angle of the light, and this
certainly is not hard to know, as it has been
correctly stated.

Thanks for reading.
Gerald L. O'Barr <globarr...@xxxxxxxxx>
Remove ... for e-mail.

.

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