Re: michelson morley experiment

I believe you based your estimate on the non-inertial component, which
was not considered in the experiment because the value (as you show) is
not significant.  In Message No. 41 to Harry I outline what I see as
the experimenter's error.  He assumed that because it is established by
the CMBR that the Earth is moving essentially linearly (in the time it
takes for a laser beam to reach a target 300 feet away) through the
Cosmos at approx. 370 km/s that that motion must be considered in any
experiment on the Earth's surface.  In other words, he assumed that in
the transit time between the laser beam being fired and striking the
target, the target would have moved approx. four inches.  Apparently,
he didn't realize that that's only true in the CMBR frame, not in the
frame of the Earth's surface, as there is no relative motion between
the laser and the target in the frame of the Earth's surface.
Therefore, it appears that the whole basis for his assumption that
plotting strikes every hour would result in an elliptical pattern is
flawed.  In the frame of the CMBR, however, the target would move
approx. four inches in the transit time of the beam and an observer in
that frame, I guess, would notice that the beam would seem to take an
angled path to "catch up" to the target.

If the above analysis is wrong, I hope you'll correct me.

Vern

I just posted a response to darkknight that estimates how large the
"dancing" actually is for that 10 m path. Values are completely
unobservable. As surveyors around the world regularly observe.

I of course used SR in those estimates, not his model in which the CMBR
dipole=0 frame is special.

Yes. But for the above 10 meter path, the size of that ellipse is less
than an Angstrom, and is completely unobservable.

Tom Roberts    tjroberts@lucent.com

Yes, but according to your statement that the speed of the source laser
(anywhere from 340 to 400 km/s, depending on orientation) does not
affect the path of the beam, there should be that "dancing around" as
you put it, because the target does move in the time it takes the laser
to reach it.  How do you account for the fact that "This does not
happen."

Vern

In the frame of the earth surface at the location of the laser, the
light leaves the laser straight down its centerline.

Insofar as the non-inertial motions of the earth can be neglected during
the flight time of the light, the image at the target will remain
motionless, 24 hours a day 365 days a year (assuming true stability in
mounting and the optical path). Indeed, since the non-inertial motions
of the earth are so steady, they will be accounted for during setup, and
it's really the variation in them that matters (e.g. the beating between
rotational and orbital motion).

That 370 km/s is roughly 0.001 times the speed of light, and the angle
relative to the CMBR dipole=0 frame varies diurnally -- people would
notice if light danced around by that milliradian as the earth turned:
surveying over just 10 meters would be off by a cm! This does not happen.

Tom Roberts    tjroberts137@sbcglobal.net

[snip]

I believe this is an important issue and one that I have been
considering in relation to the laser experiments done by H. Webster
Kehr.  Tom, would you agree that if you have a stationary laser on the
surface of the Earth pointing at a stationary target on the surface of
the Earth, that irrespective of the motion of the Earth at
approximately 370 kilometers per second linearly towards the
constellation Leo, as evidenced by the CMBR, the laser beam would still
be pointed directly at the target in order to hit it and the laser beam
does not leave the laser at an angle?

Thanks,

Vern

I think you will not be able to measure the angle accurately enough.
Even for 60 MPH wind the angle is less than 5 degrees. And of course in
such a wind your friend would not hear you (or even be able to hold a
megaphone)....

Theoretically for this case and transverse wind, you (the speaker) must
point upwind by arcsin(windspeed/soundspeed); your friend (the receiver)
must point directly at the speaker, not upwind or downwind. This is also
the case for launching a motorboat across a river -- it must be
aimed/steered upriver to arrive transversely at a point opposite from
its starting point.

With light, we find no evidence of any motion that must be compensated
for, and a directional source always points directly at the receiver,
and vice versa (assuming source and detector are at rest in some
inertial frame).

"easy", no. But it has been done, for all known aether theories except
those that happen to be experimentally indistinguishable from SR. By
literally hundreds of different experiments [see the FAQ for references].

Tom Roberts    tjroberts@lucent.com

Do you think you need to aim a flashlight up stream to hit a target
perpendicular to your direction of motion?  Light doesn't work that
way, so your analogy is flawed.

Take a friend and a couple of megaphones out to a field on a windy day.
Use your megaphone to shout to your friend while he uses his to
determine the direction the sound comes from.  You will find that he
does not hear the sound come from up wind or down wind but straight
from you.  You will find it is best to aim straight at him, not up wind
or down wind.  So sound, which is carried by a medium, does not behave
the way you predict either.

It's not so easy to disprove the aether when you use what really
happens as opposed to poor analogies.

Bruce Richmond

So if I calculate the phase relationship between the two swimmers that
are described here
http://galileoandeinstein.physics.virginia.edu/lectures/michelson.html
will I find any change if I "rotate the apparatus" i.e. if the
direction of the current in the river is changed?

If I change the direction of the current in the river, the two
swimmers won't make it back to the original starting point unless I
also change the direction they swim, but if I change the direction
they swim, this is no longer the equivalent of the MM experiment, so
the fact that the two swimmers make it back to the starting point at
different times when the river is flowing parallel to the river bank,
proves nothing ??

Since the MM apparatus doesn't change the direction the "swimmers"
swim in, won't this produce a fringe shift as the apparatus is
rotated, according to the ether theory being tested?

Thanks

They originally set it up that way. Of course, one can not have
*exactly* same lengths but within error bars was the same length. Then
other similar setups were done with different arm lengths. See
Kennedy-Thorndyke.

It did not *rely* on that fact. I was setup to find what would happen
if rotated.

It (as was already known) showed that the TWLS was constant.
It did not disprove (modern) ehter theory because ether theory predicts
that there would be no fringe shift. But the 'earlier' ether theory was
badly interpreted and was thought that it it predicted a fringe shift.

---
If you want to be sure, then always doubt
}:-)

Hi

In the MM experiment, was the length of the two light paths set
exactly equal.  If so, how was this done?  If not, did the experiment
"rely" on the fact that there was no change in the interference
fringes when the apparatus was rotated?

Did the MM experiment demonstrate anything about the constancy of the
speed of light or did it merely "disprove" the aether theory?

Does this website have any credibility in mainstream physics?
http://home.iprimus.com.au/longhair1/page1.html

Thanks.