Re: MMX proves Light anisotropic
- From: Peter Riedt <riedt1@xxxxxxxxxxx>
- Date: Fri, 10 Apr 2009 01:40:08 -0700 (PDT)
On Apr 10, 3:40 pm, "Androcles" <Headmas...@xxxxxxxxxxxxxxxx> wrote:
"Peter Riedt" <rie...@xxxxxxxxxxx> wrote in message
news:71efad29-fa30-48d6-b254-8d4fbaad09be@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
MMX proves Light anisotropic
Michelson&Morley write in the American Journal of Science 203/XXIV
November 1887 describing their interferometer experiment: ”The
distance travelled is 2D(1+vv/VV), and the length of the other path
is evidently 2D(1+vv/2VV)”. Using Michelson's formula 2D(1+vv/VV) we
get 22.00000022m for the total distance the light will travel to the
end of the parallel arm and back again and using 2D(1+vv/2VV) we get
22.00000011m for the total distance light will travel across the
perpendicular arm and back again.
To explain the absence of a fringe shift, Lorentz applied the formula
L’ = L*sqrt(1-vv/cc) to the parallel arm, reducing its length of 11m
to a contracted length of 10.999999945m. This equalised the length of
the light path of the parallel arm with the length of the light path
of the perpendicular arm and with equal light paths, there is no
fringe shift.
A similar gimmick can be applied if we vary the speed of light across
the parallel arm to c’ but leaving the speed of light across the
perpendicular arm at c just as Lorentz did in terms of distance.
If we apply the formula c’ = c*1/sqrt(1-vv/cc) to light, we get
300000001.5m/sec and with this greater light speed c’ the longer
parallel light path of 22.00000022m is covered in the same time as the
shorter perpendicular light path of 22.00000011 at c. Again, no fringe
shift.
Peter Riedt
===============================================
The distance travelled by the flight attendant as she travels
from the tail of the plane to the flight deck and back again is
2D(1+vv/VV) and the wingspan of the plane is evidently
2D(1+vv/2VV)
where D is the length of the plane, v is the flying speed of the plane
and V is the walking speed of the flight attendant.
Sanity check:
While the plane is taxiing...
Walking speed of flight attendant V: 3.16 mph
Taxi speed of plane v: 20 mph
D = 50 feet
2D(1+ 400/10) = 100*41 = 4100 feet.
Sounds reasonable, the flight attendant walks up and down
the plane as the plane travels from the gate to the runway.
Flying speed of plane: 500 mph
100 * (1 + 250000/10) = 2,500,100 feet = 473.5 miles.
Hmm... must have taken the flight attendant almost an hour
to walk the length of the plane and back. Ok, close enough
for government work.
Andy, your gedanken needs tidying up. You mix walking, taxiing, flying
speed and a few dimensional numbers. Try again.
Peter Riedt
.
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