Androcles wrote:

In a reference frame in which the apparatus is stationary, x and y
don't
change as the apparatus is rotated (assuming x and y are the
arm-lengths).



Exactly. So you prove nothing. However, Einstein's postulate is
"light is always propagated in empty space with a definite velocity
c which is independent of the state of motion of the emitting body"
Reference :
http://www.fourmilab.ch/etexts/einstein/specrel/www/

The Earth is moving in the empty space, which is the reference frame.
Are you trying to deliberately obfuscate the problem?


Empty space doesn't have a particular reference frame in Einstein's
theory. One observer moving inertially in empty space may see the earth
moving at 0.8c, another may see it moving at 0.1c, another may see it at
rest (at least, for short amounts of time when the earth's path is not
visibly curved). Since there is no "Absolute Space" in relativity,
Einstein's statement would apply to any one of these reference frames.





No frequency shifts, since there is no relative motion, no
wavelengths shifts, so no change after rotation.

Do you dispute this?



Yes. Michelson and Maxwell thought there was aether and that
was indeed the rest frame. It is also Einstein's rest frame, he is
saying so by his second postulate and modifying the PoR so that
c = (c+v)/(1-v/c)

First, you wrote that equation incorrectly, it should be (c + v)/(1 + v/c)

Second, you continue to use that equation as if it was a "theory" of
Einstein's, when in fact it can be proved true by simple algebra:

(1 + v/c) = (1/c)*(c + v), by multiplying out the 1/c

so by substitution, (c + v)/(1 + v/c) = (c + v)/(1/c)*(c+v)

The (c+v) in the numerator and the denominator cancel out, leaving
1/(1/c), which of course is equal to c.

I would guess that the postulate you are thinking of is the formula for
addition of velocities, w = (u + v)/(1 + uv/c^2), explained on
http://math.ucr.edu/home/baez/physic.../velocity.html ...if
you plug in u=c, then this equation becomes (c + v)/(1 + v/c), and since
the algebra above shows that this is equal to c regardless of the value
of v, this shows that anything that's moving at c in one reference frame
will be moving at c in any other reference frame.

You challenged me in previous posts to derive the Lorentz equations from
(c + v)/(1 + v/c), but when I asked for clarification you didn't
respond...were you asking if the Lorentz equations can be derived from
the relativistic velocity addition formula, w = (u + v)/(1 + uv/c^2)?

You want to obfuscate the problem by declaring v = 0, satisfying
c+v = c+0 = c. If you declare that, then the problem is solved and
SR is irrelevant. The speed of Michelson's light is source dependent.
Without a |v| 0, SR vanishes without a trace.


SR will make exactly the same predictions about the result of the
experiment regardless of the value of v.






The speed of flight in both tubes is c (invariant),
and the tubes are of different lengths.
Therefore a count will occur, the x (up) count exceeding the count of
the
(y) down counter.


In a reference frame in which the MMX apparatus is moving, yes,
according
to SR, there is a length contraction. The length contraction applies
equally to the wavelength in the contracted arm, so the length as
measured
in wavelengths along the arm remains exactly the same.



That has nothing to do with the *speed* of light. Speed is defined as
distance/time. The distance along the contracted arm is less that
the distance along the transverse arm. How much you contract the
wavelength of the light to cram in the same count of wavelenghts along
the contracted arm is completely irrelevant. Time is the same in both
arms.


No, if one arm is contracted in my reference frame, then the time will
also be less, so light will travel at the same speed down both arms.



In the moving frame,



MOVING WITH RESPECT TO WHAT?
You can't have a moving frame unless you define what the rest frame is.



Presumably he meant moving with respect to the rest frame of the
interferometer in the MMX.




the arm along the motion is contracted
by 1/gamma, so we have x' = y'/gamma. The mirror moves at v, so the
light,
moving at c in the moving frame, moves at c-v towards the mirror, as
measured in the moving frame, and takes y'/(gamma (c-v)) to get there.
The
return journey is at c+v relative to the detector, as measured in the
moving frame, so the return journey takes y'/(gamma (c+v)). Therefore
the
total time along that arm for the round trip is

y'/gamma * (2c/(c^2 - v^2)).



There ya go.
You've just contradicted the invariance of the speed of light.
I'm done.
Androcles.


His math seems to be incorrect--it's better to analyze this problem
using the full Lorentz transformations. Suppose the two tubes have
length L. In the rest frame of the tubes, let tube A lie along the
x-axis and tube B lie along the y-axis. So if the two tubes intersect at
the origin, the coordinates of the end of tube A are x=L, y=0 and the
coordinates of the end of tube B are x=0, y=L. In the tube's rest frame,
if we send light pulses from the origin along both tubes at time t=0,
then it will hit the end of tube A at coordinates x=L, y=0 and t = L/c;
the other pulse will hit the end of tube B at coordinates x=0, y=L, t=L/c.

Now consider the perspective of an observer who is moving along the
x-axis at velocity v, and who defines his x' and y' axes to be parallel
to the x and y axis of the tube's frame, with the origin of his x'-y'
coordinate system coinciding with the origin of the x-y coordinate
system at time t=0 in the tube's frame and time t'=0 in his frame. We
can use the Lorentz transformations to find the space/time coordinates
of the light pulses hitting the ends of the tubes in his frame. For the
event of the pulse hitting the end of tube B, the Lorentz
transformations tell us:

x' = gamma*(x - vt), so plugging in x=0, t=L/c we get x' = gamma*(-vL/c)
y' = y, so y' = L
t' = gamma*(t - vx/c^2), so t' = gamma*(L/c)

If the pulse moved gamma*(-vL/c) along the x'-axis and L along the
y'-axis, then using the pythagorean theorem, the light travelled a total
distance of squareroot(gamma^2*(-vL/c)^2 + L^2) =
squareroot(v^2*L^2/c^2*(1 - v^2/c^2) + L^2) = squareroot(v^2*L^2/(c^2 -
v^2) + L^2*(c^2 - v^2)/(c^2 - v^2)) = squareroot(c^2*L^2/(c^2 - v^2)) =
squareroot(c^2*L^2/(c^2)*(1 - v^2/c^2)) = squareroot(L^2/(1 - v^2/c^2))
= L/squareroot(1 - v^2/c^2) = gamma*L

So it travelled a distance of gamma*L in a time of gamma*(L/c), meaning
the speed to travel down tube B was c in this frame.

For the event of the pulse hitting the end of tube A, we have:

x' = gamma*(x - vt), so plugging in x=L, t=L/c we get x' = gamma*(L -
vL/c) = gamma*(Lc/c - vL/c) = gamma*(1/c)*(Lc - Lv)
y' = y, so y' = 0
t' = gamma*(t - vx/c^2), so t' = gamma*(L/c - vL/c^2) = gamma*(Lc/c^2 -
vL/c^2) = gamma*(1/c^2)*(Lc - Lv)

So if you take distance/time to find the speed, the factors of gamma and
(Lc - Lv) cancel out, leaving you with (1/c)/(1/c^2), which is equal to
c. So, the speed to travel down tube A was c in this frame.

Jesse