Re: michelson morley experiment


"darkknight" <darkknight.21@xxxxxxxxx> wrote in message
news:00a642502jeogh7fkdipbsfpaopcphusaa@xxxxxxxxxx
On Mon, 17 Apr 2006 11:07:55 +1200, darkknight
<darkknight.21@xxxxxxxxx> wrote:

On Sun, 16 Apr 2006 21:38:49 GMT, Tom Roberts
<tjroberts137@xxxxxxxxxxxxx> wrote:



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.


I don't understand.

If the light path of the laser is at right angles to the 400 km per
second motion of the earth, then over a distance of 1 km, the laser
takes 1/300000 seconds to reach the target, during which time the
target has moved by (1/300000) * 400 km - which is .001333 km -
equals 130 cm. (or 1cm over 10 metres as you say).

If the light path of the laser is parallel to the direction of motion
(which it could be after 8 hours) then the target hasn't moved so a
difference of 1 cm would be seen on the "strike point" at the target.

I'm sure you're right coz I know next to nothing about physics but why
doesn't the "strike point" move?


Well, upon further thought, I guess the explanation could be that the
motion of the source affects the angle of the "laser beam" so that it
has a "sideways" component the equivalent of the 1 cm movement of the
target. This would allow the laser beam to leave the laser "straight
down its centre line" regardless of the movement of the laser.

Almost correct: the speed of light remains constant, so that the components
don't add up vectorially. But it's even more straightforward: the light
inside the laser follows a certain angle to start with (90 degrees in the
rocket's rest frame, but not in the stationary frame), and the light doesn't
change angle when it leaves the laser.

If this is true, this should mean that if a laser beam on a rocket
ship is fired sideways

You mean relative to the rocket

that the path of the laser beam is exactly
at right angles to the path of the rocket,

You mean relative to the CMBR frame I suppose...

if the rocket increases
speed by 400 km/second, the path of the laser beam will no longer be
exactly at 90 degrees to the path of the rocket, and will not be
parallel to the path of the laser beam before the rocket increased
speed.

That's correct - in the CMBR frame.

Also, the change in velocity of the earth from 340 km/sec to 400
km/sec should mean clocks run at different speeds at different times
of the year.

That's correct as well - in the CMBR frame. Of course, "in" the solar frame
they run (for all practical purposes) at constant speed.

Harald


.

Relevant Pages

  • Re: michelson morley experiment
    ... Because the telescope only sees light that travels down its centerline, ... (locations simultaneous in that frame). ... If the light path of the laser is at right angles to the 400 km per ... second motion of the earth, then over a distance of 1 km, the laser ...
    (sci.physics.relativity)
  • Re: michelson morley experiment
    ... the flight time of the light, the image at the target will remain ... rotational and orbital motion). ... This would allow the laser beam to leave the laser "straight ... strikes the target, the plot of the strikes on the target over the 24 ...
    (sci.physics.relativity)
  • Re: michelson morley experiment
    ... which the laser is positioned perpendicular to the essentially linear ... latitude and approximately 300 feet between source and target. ... frame as "stationary" frame. ... strikes the target, the plot of the strikes on the target over the 24 ...
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  • Re: michelson morley experiment
    ... >> In the frame of the earth surface at the location of the laser, ... >> light leaves the laser straight down its centerline. ... Because the telescope only sees light that travels down its centerline, and in the locally-inertial frame of the telescope the light does not travel directly along the line connecting source and telescope. ... > second motion of the earth, then over a distance of 1 km, the laser ...
    (sci.physics.relativity)
  • Re: michelson morley experiment
    ... > affect the path of the beam, there should be that "dancing around" as ... because the target does move in the time it takes the ... he didn't realize that that's only true in the CMBR frame, ...
    (sci.physics.relativity)
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