If I'm not completely confused by all possible changes in space,
time, simultaneity and even velocity, resulting from the Lorentz
transformation, then I've found a very simple way to determine
our velocity wrt the ether in LET.
I suppose that this reasoning also entails that Special Relativity
is empirically refuted. Anyway, I'm sure that SR is fundamentally
wrong (despite of being a very elegant and astonishingly well
Let as assume that the earth moves wrt the ether in such a way
that point C moves at v = 0.001 c (300 km/s) wrt the ether in
direction of point R. The distances from C to L and to R are both
10'000 nano light seconds (3 km).
L C R v wrt ether
surface of the earth
The gamma factor of 1.000'000'5 can be ignored, because the
following reasonings are based on a first order effect proportional
From C short laser pulses are sent to L and R at a frequency of
10^8 (10 ns delay between two successive emissions). By comparing
the number of pulses on the two paths (from C to L and from C to R)
it becomes possible to determine the velocity wrt the ether.
At the same moving-frame time, there are only 999 pulses between L
and C, but 1001 between C and L.
This becomes obvious if we take into consideration that the same
pulses can be used to synchronize the clocks at L and R from C. As
observed from the ether frame, it needs 9'990 ns to synchronize
L and 10'010 ns to synchronize R from C.
During these 10'010 ns 1001 pulses have been sent in both
directions. When the first of these pulses reaches R, then the
first two pulses in the other direction have already surpassed L
and it is the third pulse at L which is declared to be moving-frame
simultanous with the first pulse at R. Therefore we have 999
pulses in the one and 1001 in the other direction.
Just after having written the last sentence, I recognized the big
CONTRADICTION I have committed. It is of course in both cases the
first pulse which is declared moving-frame simultanous. If the
first pulse is emitted in both directions from the center at ether
time t = 0 and at moving-frame time t' = 0 then at t' = 10'000 ns
we have this situation:
Locus t' t Pulse
L 10'000 ns 9'990 ns 1
C 10'000 ns 10'000 ns 1000
R 10'000 ns 10'010 ns 1
The moving-frame time t' = 10'000 ns as observed from the ether frame
appears at first at L (exactly when pulse 1 arrives), then moves
continuously to C where it arrives 10 ns later (exactly after pulse
1000 has been emitted) and arrives another 10 ns later at R (again
when pulse 1 arrives). So we have complete symmetry between the
pulses in the moving frame. Really astonishing!
But I cannot believe that the basic asymmetry in the ether has no
effect at all. Isn't there a problem with the intensity of light?
Intensity is inversely proportional to the distance square. If the
ether is the medium of light, then light intensity decreases with
ether distance, doesn't it?
In the above case (300 km/s wrt ether) the result is a difference
in light intensity corresponding to a factor of 1.004, certainly
enough to be measurable.
Wolfgang Gottfried G. (0:022.65)
My previous post of this thread:
This is not really a question for beliefs, it is a question for
_COMPUTATIONS_. If one generalizes the clear experimental result
that the round-trip speed of light is isotropic in all inertial
frames occupied by the earth to all inertial frames whatsoever,
then computations show that any ether theory consistent with that
is experimentally indistinguishable from SR. Need I reference
my trio of articles which discuss this again?
> Isn't there a problem with the intensity of light?
Not for any theory in the equivalence class of theories I discuss.
> Intensity is inversely proportional to the distance square. If the
> ether is the medium of light, then light intensity decreases with
> ether distance, doesn't it?
Yes, when measured in the ether frame. But intensity is also
proportional to the solid angle subtended by the detector, and
_that_ is not invariant. Do the computation and you will find
that competing effects cancel out exactly.
Tom Roberts tjro...@lucent.com
:: But I cannot believe that the basic asymmetry in the ether has no
:: effect at all. Isn't there a problem with the intensity of light?
:: Intensity is inversely proportional to the distance square. If the
:: ether is the medium of light, then light intensity decreases with
:: ether distance, doesn't it?
: Yes, when measured in the ether frame. But intensity is also
: proportional to the solid angle subtended by the detector, and
: _that_ is not invariant. Do the computation and you will find
: that competing effects cancel out exactly.
| - - - - - - - o - - - - - - - |
detector source of detector
The sound intensity registered by the two detectors, unlike the
measured frequency, is not independent of the speed of the medium
(the air) wrt the system.
A spherical light wave is Lorentz invariant:
x^2 + y^2 + z^2 = (ct)^2 --> x'^2 + y^2 + z^2 = (ct')^2
But I don't think that all segments corresponding to given solid
angles are Lorentz invariant in the same way. We can for instance
divide the original spherical wave into two parts separated by
the circularly propagating wave
x = 0, y^2 + z^2 = (ct)^2
Thinking about stellar aberration is enough to recognize that
this circular wave cannot be Lorentz-transformed to
x' = 0, y^2 + z^2 = (ct')^2
This means that if the number of photons per solid angle is
uniform in the system at rest then it cannot be uniform in the
moving system and vice versa.
In SR it is reasonable to assume that it is always the frame
of the source where the number of photons per solid angle is
uniform. I doubt however that this is a reasonable assumption
in an ether theory.
| - - - - - - - o - - - - - - - | -->
detector spherical detector v
Let us assume that this system moves at v = sqrt(0.99)c wrt the
ether and that the distance between the source and each dectector
is compressed from 10 m (restlength) to 1 m in the ether.
Whereas light to the right detector has to travel over 199.5 m in
the ether, the light path to the opposite detector is only 0.5 m.
1 m / (1 - sqrt(0.99) = 199.5 m
1 m / (1 + sqrt(0.99) = 0.5 m
The sound analogy would result in an intensity difference
factor of (199.5 m / 0.5 m)^2. So the intensity registered by
the left dectector would be (at least) 159 thousand times
stronger than by the right detector.
Wolfgang Gottfried G. (0:032.6)
I think you are correct.
And the experimental confirmation of this agrees with SR's prediction.
( I'm refering to the "searchlight" effect in emissions from
moving sources, mentioned fairly often by Tom Roberts, IIRC. )
What you are still missing is any way of determining an ether rest frame.