There is no length contraction
Marcel Luttgens
______________________________
In the article "The Lorentz transformation (LT) are false"
(see http://perso.wanadoo.fr/mluttgens/LTfalse.htm ),
I have demonstrated that the correct "Lorentz transformation" are
X' = V't', where V' = (V-v)/(1-Vv/c^2), and
t' = t*sqrt(1-v^2/c^2).
Those relations have been obtained
1) by considering two frames of reference, S and S', each in uniform
translatory motion relative to the other, the velocity of S' relative
to S being v.
2) by counting time from the instant at which the origins of the frames,
O and O', momentarily coincide. At that instant, it is assumed that
t = t' = 0.
3) by considering an object starting from the coincident origins at
t = t' = 0, with a constant velocity V.
According to the frame S, x=vt is the position of the origin of S' after
a time interval t, and X=Vt the position of the object after the same
interval t.
But after such time interval, the clock of S' will read t', and
the object will be at X' = V't' from the origin of S'.
The existence of length contraction has been put forward by both
Fitzgerald and Lorenz in order to explain the negative result
of the Michelson and Morley experiment. They supposed that motion
through the ether caused the material composing the interferometer
to shorten by the factor sqrt(1-v^2/c^2) in a direction parallel
to the motion. This contraction would equalize the light paths
and prevent a displacement of the fringes.
Indeed, let's call S the ether frame, and S' the interferometer frame.
Both arms of the interferometer have the same length l.
The arm of the interferometer, which is parallel to the x-axis,
is limited by a mirror M2, whereas the arm parallel to the y-axis
is limited by a mirror M1. As the interferometer is supposed to
move at v relatively to the frame S, the light wave moving along the
x-axis has a velocity c-v on the outgoing trip, and c+v on the return
trip. Hence, the total time required by this wave is
t2 = l/(c-v) + l/(c+v) = (2l/c) * 1/(1-v^2/c^2).
According to the frame S, the wave moving transversely travels
a distance ct' along the hypotenuse of a right-angled triangle,
whereas in the same time, the mirror M1 advances a distance vt'.
Hence, (ct')^2 = (vt')^2 + l^2, and t' = (l/c) * 1/sqrt(1-v^2/c^2).
This wave returns to the origin of the frame S' after a total time
t1 = (2l/c) * 1/sqrt(1-v^2/c^2).
If the length of the arm moving along the x-axis is contacted
by sqrt(1-v^2/c^2), one has have of course t2 = t1.
But this demonstration doesn't take the phenomenon of "time dilation"
into account:
If the interferometer is at rest in the "ether", i.e., if the frame S'
is at rest relatively to the frame S, the light wave sent at t=t'=0
will be, according to S or S', at X=ct, and at Y=ct after the time
interval t. Let's note that ct corresponds to the arms'length l.
But if the frame S' moves at v relatively to the frame S, one gets
X' = V't' = ct' for the wave following the x'-axis (which slides
along the x-axis), and Y' = ct' for the wave following the y'-axis.
As t' = t * sqrt(1-v^2/c^2), one obtains
X' = ct * sqrt(1-v^2/c^2)
Y' = ct * sqrt(1-v^2/c^2)
on the outgoing trip, and also, for reasons of symmetry, on the
return trip.
The lengths of the two paths X' and Y' being equal, no fringe
displacement will be observed.
As X=ct, one could represent X' by X * sqrt(1-v^2/c^2), and
claim that the length of the arm moving along the x-axis is contacted
by sqrt(1-v^2/c^2). But this would be false, as the apparent
contraction is entirely due to the time "dilation".
N.B.: One could instead use the Einsteinian derivation of the Lorentz
transformation:
"Let's suppose that a light signal, starting from the coincident
origins of frames S and S' at t = t' = 0, travels toward positive x.
After a time t, it will be at x = ct, and also at x' = ct',
since the speed of light is the same in all frames.
If the signal travels toward negative x, x = -ct and x' = -ct'
Now, a light signal follow the y' axis. Relatively to S,
it travels obliquely, for, while the signal goes a distance ct,
the y'-axis advances a distance x = vt.
Thus c^2t^2 = v^2t^2 + y^2, whence y = sqrt(c^2 - v^2) * t.
But, also, y' = ct'."
Iow, also according to Einstein, x' = ct' and y' = ct', meaning that
the two paths x' and y' are equal.
However, Einstein used his position LT x' = g(x - vt) to "prove"
the existence of length contraction:
Assuming that the length of an object at rest in S' corresponds to
the difference Lo between the coordinates x2' and x1' of its ends,
he substituted in Lo = x2' - x1' the values of x2' and x1' calculated
from his position LT for a given value of t, thus obtaining
Lo = g(x2 - x1) = gL, where L is the length of the object in S.
He then concluded
1) that any body measures shorter in terms of a frame
relative to which it is moving with speed v than it does as measured
in a frame relative to which it is at rest, the ratio of shortening
being sqrt(1-v^2/c^2).
2) that, relative to a single frame, any physical body set into
motion with speed v shortens in the direction of its motion, as was
postulated by Fitzgerald and Lorentz, in the same ratio sqrt(1-v^2/c^2).
As shown above, such apparent contraction is entirely due to the
phenomenon of time "dilation". All what Einstein "proved" is the
falseness of his transformations and his lack of logical thinking.
Marcel Luttgens
Jamieson Christie
>
> As shown above, such apparent contraction is entirely due to the
> phenomenon of time "dilation". All what Einstein "proved" is the
> falseness of his transformations and his lack of logical thinking.
Length contraction and time dilation are two aspects of the same
phenomenon. This is already known. What's your point?
Jamieson Christie
Dirk Van de moortel
"Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message news:b45b8808.03051...@posting.google.com...
> There is no length contraction
> ______________________________
>
> In the article "The Lorentz transformation (LT) are false"
> (see http://perso.wanadoo.fr/mluttgens/LTfalse.htm ),
> I have demonstrated that the correct "Lorentz transformation" are
>
> X' = V't', where V' = (V-v)/(1-Vv/c^2), and
> t' = t*sqrt(1-v^2/c^2).
>
> Those relations have been obtained
By not understanding what a transformation is in the first place.
Let's see...
>
> 1) by considering two frames of reference, S and S', each in uniform
> translatory motion relative to the other, the velocity of S' relative
> to S being v.
>
> 2) by counting time from the instant at which the origins of the frames,
> O and O', momentarily coincide. At that instant, it is assumed that
> t = t' = 0.
>
> 3) by considering an object starting from the coincident origins at
> t = t' = 0, with a constant velocity V.
>
> According to the frame S, x=vt is the position of the origin of S' after
> a time interval t, and X=Vt the position of the object after the same
> interval t.
> But after such time interval, the clock of S' will read t', and
> the object will be at X' = V't' from the origin of S'.
"After such a time interval" does not make any sense and
makes you go down again.
Events have coordinates:
x and t according to S
x' and t' according to S'
x" and t" according to the object itself.
We have been through this a thousand times before.
Download the picture at
http://users.pandora.be/vdmoortel/dirk/Stuff/Marcel2.gif
edit it and use it to tell us what you don't understand.
If you don't do that, we cannot help.
Dirk Vdm
Stephen Speicher
> There is no length contraction
You're confused. There is length contraction, but there is no
Marcel Luttgens.
--
Stephen
s...@speicher.com
Ignorance is just a placeholder for knowledge.
Printed using 100% recycled electrons.
-----------------------------------------------------------
Marcel Luttgens
> [snip]
> >
> > As shown above, such apparent contraction is entirely due to the
> > phenomenon of time "dilation". All what Einstein "proved" is the
> > falseness of his transformations and his lack of logical thinking.
>
> Length contraction and time dilation are two aspects of the same
> phenomenon.
According to *your* relativity theory!
>This is already known. What's your point?
>
> Jamieson Christie
Marcel Luttgens
Dirk Van de moortel
"Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message news:b45b8808.03051...@posting.google.com...
> > [snip]
> > >
> > > As shown above, such apparent contraction is entirely due to the
> > > phenomenon of time "dilation". All what Einstein "proved" is the
> > > falseness of his transformations and his lack of logical thinking.
> >
> > Length contraction and time dilation are two aspects of the same
> > phenomenon.
>
> According to *your* relativity theory!
Not according to *yours* of course:
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/IfOnlyIf.html
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/SRSymbols.html
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/CorrectRelations.html
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Forget.html
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/SRLuttgens.html
Biology is wrong, the Luttgens way:
| When biologists say that birds can fly, they are wrong.
| Here is the proof: suppose a bird is flying from city A to city B.
| Now, when the bird arrives in city B, shoot it and pick it up.
| If it is dead, hold it before your eyes and let it go. It will
| immediately fall right in front of your feet. This proves that the
| bird is heavier than air and that it can *not* fly.
| Are all biologists so dense?
Dirk Vdm
Marcel Luttgens
> On 14 May 2003, Marcel Luttgens wrote:
>
> > There is no length contraction
>
> You're confused. There is length contraction, but there is no
> Marcel Luttgens.
This is an addendum for the SR crackpots like you, Dirk Van de moortel,
and most of the SRists, who are unable or unwilling to realize Einstein's
logical mistake:
1) According to Einstein's own theory, the length l of the interferometer's
arm, which is parallel to the xx'-axis, corresponds, after a time
interval t, to x=ct when the interferometer is at rest "in the ether"
(Iow, when S' is at rest wrt S). Let's remember that x is the distance
travelled, after a time interval t measured in S, or S'(because S' is
at rest wrt S), by a light signal sent at t=t'=0.
But when the interferometer moves at v wrt the ether (Iow, when S' moves
at v wrt S), x' = ct'.
Otoh, the length l of the arm, which is parallel to the yy'-axis, is
given by y=ct when the interferometer is at rest "in the ether".
But when the interferometer moves at v wrt the ether (Iow, when S'
moves at v wrt S), y' = ct'.
The two paths x' and y' being equal, no fringe displacement can be
observed.
And indeed, the MMX showed no fringe displacemment.
Let's note that, according to Einstein's transformation (the LT),
x' = g(x-vt) and t' = g(t-xv/c^2). In the present case, x = ct, hence,
x' = gt(c-v) and t' = gt(1-v/c), and x' is indeed equal to ct'.
2) Otoh, Einstein used his position transform x' = g(x-vt) to demonstrate
that the length of the arm moving along the xx'-axis (called hereafter the
parallel arm) is contracted by sqrt(1-v^2/c^2).
He literally claimed that "any body measures shorter in terms of a frame
relative to which it is moving with speed v than it does as measured
in a frame relative to which it is at rest, the ratio of shortening
being sqrt(1-v^2/c^2)."
3) As, according to Einstein, the parallel arm is at rest in S', its length
l = ct, measured in the "ether" frame S becomes x = ct * sqrt(1-v^2/c^2),
because the body (i.e., the arm) is moving at v wrt S. Hence, x' = g(x-vt)
becomes x' = g(ct * sqrt(1-v^2/c^2) - vt). As t' = gt(1-v/c), x' is no more
equal to ct', and y' is now different from x', meaning that the two light paths
are no more equal. Conclusively, the MMX should show a fringe displacement.
As it doesn't, everybody sane should conclude that Einstein's transforms are
false, and that "length contraction" is, at best, the result of a logical
mistake, or, at worst, the product of pure sophistry.
Marcel Luttgens
Jeff Krimmel
news:<Pine.LNX.4.33.03051...@localhost.localdomain>...
> > On 14 May 2003, Marcel Luttgens wrote:
> >
> > > There is no length contraction
> >
> > You're confused. There is length contraction, but there is no
> > Marcel Luttgens.
>
> This is an addendum for the SR crackpots like you, Dirk Van de moortel,
> and most of the SRists, who are unable or unwilling to realize Einstein's
Rational person: "Marcel, you're stupid."
Marcel: "Oh yeah? If you thought _that_ was stupid, then take a look at
this..."
Marcel, we all know you are as dumb as dirt. When we poke fun at you, you
really shouldn't feel the need to prove this fact to us any more. We already
know.
Jeff
Dirk Van de moortel
"Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message news:b45b8808.03051...@posting.google.com...
> > On 14 May 2003, Marcel Luttgens wrote:
> >
> > > There is no length contraction
> >
> > You're confused. There is length contraction, but there is no
> > Marcel Luttgens.
>
> This is an addendum for the SR crackpots like you, Dirk Van de moortel,
> and most of the SRists, who are unable or unwilling to realize Einstein's
> logical mistake:
>
> 1) According to Einstein's own theory, the length l of the interferometer's
> arm, which is parallel to the xx'-axis, corresponds, after a time
> interval t, to x=ct when the interferometer is at rest "in the ether"
> (Iow, when S' is at rest wrt S). Let's remember that x is the distance
> travelled, after a time interval t measured in S, or S'(because S' is
> at rest wrt S), by a light signal sent at t=t'=0.
> But when the interferometer moves at v wrt the ether (Iow, when S' moves
> at v wrt S), x' = ct'.
> Otoh, the length l of the arm, which is parallel to the yy'-axis, is
> given by y=ct when the interferometer is at rest "in the ether".
> But when the interferometer moves at v wrt the ether (Iow, when S'
> moves at v wrt S), y' = ct'.
> The two paths x' and y' being equal, no fringe displacement can be
> observed.
> And indeed, the MMX showed no fringe displacemment.
Title: "Arms growing at lightspeed!"
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ArmsGrow.html
Dirk Vdm
Richard
Huh? Where did you study math? When the arm is moving at v wrt the
ether then the two way trip time is given by:
t' = t (gamma)^2
t being the trip time when the arm is at rest, t' the trip time when the
arm is in motion at v wrt the ether.
For the lateral arm the Galilean transform gives us:
t' = t gamma
This is strictly according to the 'Galilean' transform, v = speed of arm
wrt ether. (According to Einstein the ether is always at rest wrt the
observer, but we know better:) Nevertheless, even according to LET:
Assuming no length change of the lateral arm, we get for the trip time
in the lateral arm
t'= x/c
Substituting
t gamma = x/c or
x = c t gamma
For the longitudinal arm the Galilean transform gives:
x' = c t gamma^2
Hence in order to generate equal trip times:
x' = x gamma
From this can easily be derived that the longitudinal arm L' contracts
by a factor gamma wrt the lateral arm L, or
L' = L gamma
It should be noted that this is only assuming no contraction or
elongation of the 'lateral' arm. The proposition that no such
contraction occurs to the lateral arm is baseless, and as such, the
entire analysis is void. It is a simple matter to provide any number of
arbitrary changes to the lateral arm if an adjustment is made to the
longitudinal arm accordingly. I don't believe that the contraction takes
place according to the above, but rather, that the interferometer was
virtually at rest wrt the absolute 'local' ether of which the Earth is
the source. If a contraction 'does' occur, then it is a physical effect
and it is doubtful that it would be independent of arm material, and
moreover from the standpoint of electrodynamic contraction, the lateral
arm would necessarily also contract, though by a different factor, since
the photons mediating the interactions along that direction are also
delayed.
Only further experiments can decide the complex details about length
contraction.
--
Richard Perry
http://www.cswnet.com/~rper
Richard
> Huh? Where did you study math? When the [longitudinal] arm is moving at v wrt the
Bilge
>______________________________
>
>In the article "The Lorentz transformation (LT) are false"
>(see http://perso.wanadoo.fr/mluttgens/LTfalse.htm ),
>I have demonstrated that the correct "Lorentz transformation" are
>
>X' = V't', where V' = (V-v)/(1-Vv/c^2), and
>t' = t*sqrt(1-v^2/c^2).
>
>Those relations have been obtained
Where is your similar refutation of lorentz transforms known
as rotations?
Stephen Speicher
The thing is, it is not as if nobody tried to help Marcel
understand. In my opinion, Dirk deserves a medal for the
continued patience he exhibited while attempting to teach Marcel
over a long period of time, despite the response of a mind which
chooses not to see or grasp. With Marcel it is not simply lack of
knowledge, but the refusal to let go of his ignorance. Sad,
because I do not really think that Marcel is dumb, but rather
that he prefers terminal ignorance.
Marcel Luttgens
You have simply reproduced the classical analysis of the MMX, which
I have also given in my article "There is no length contraction".
You should take into account Einstein's postulate, according to which
the speed of light is the same in all frames.
Then, automatically, you get x' = ct', which is the length of the
"longitudinal" arm when the apparatus is moving in the "ether".
But, also, according to special relativity, y' = ct', which is the
corresponding length of the "lateral" arm.
The two lengths being equal, no fringe displacement can be observed.
You should note that this result is obtained without assuming
arm contraction. If the longitudinal arm were contracted, x' would not
be equal to y' any more, and a fringe displacement should be observed.
As this is not the case, one has to conclude, assuming that Einstein's
postulate is correct, that no contraction occurs.
Otoh, Einstein used his transformation x' = g(x-vt) to demonstrate
that "any body measures shorter in terms of a frame relative to which
it is moving with speed v than it does as measured in a frame relative
to which it is at rest, the ratio of shortening being sqrt(1-v^2/c^2)."
But this contradicts the negative result of the MMX. So, if you accept
the premises of special relativity, you have to conclude that Einstein's
trasformation are false. I have shown in
http://perso.wanadoo.fr/mluttgens/LTfalse.htm
how Einstein mistakenly obtained his transforms, and given the correct
formulae.
Marcel Luttgens
Dirk Van de moortel
"Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message news:b45b8808.03051...@posting.google.com...
[snip]
>
> You have simply reproduced the classical analysis of the MMX, which
> I have also given in my article "There is no length contraction".
>
> You should take into account Einstein's postulate, according to which
> the speed of light is the same in all frames.
> Then, automatically, you get x' = ct', which is the length of the
> "longitudinal" arm when the apparatus is moving in the "ether".
> But, also, according to special relativity, y' = ct', which is the
> corresponding length of the "lateral" arm.
Don't you even notice what an utterly stupid thing you say here?
Title: "And the arms never stop growing - hence the null result!"
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ArmsGrow2.html
Trying to beat Ken Seto or something?
Dirk Vdm
Richard
No it isn't.
> But, also, according to special relativity, y' = ct', which is the
> corresponding length of the "lateral" arm.
> The two lengths being equal, no fringe displacement can be observed.
The problem is that x is not equal to L, nor is x' equal to L'.
x and x' are the displacement of the beams, not the lengths of the
arms.
>
> You should note that this result is obtained without assuming
> arm contraction. If the longitudinal arm were contracted, x' would not
> be equal to y' any more, and a fringe displacement should be observed.
> As this is not the case, one has to conclude, assuming that Einstein's
> postulate is correct, that no contraction occurs.
>
> Otoh, Einstein used his transformation x' = g(x-vt) to demonstrate
> that "any body measures shorter in terms of a frame relative to which
> it is moving with speed v than it does as measured in a frame relative
> to which it is at rest, the ratio of shortening being sqrt(1-v^2/c^2)."
> But this contradicts the negative result of the MMX. So, if you accept
> the premises of special relativity, you have to conclude that Einstein's
> trasformation are false. I have shown in
> http://perso.wanadoo.fr/mluttgens/LTfalse.htm
> how Einstein mistakenly obtained his transforms, and given the correct
> formulae.
>
> Marcel Luttgens
--
Richard Perry
http://www.cswnet.com/~rper
Richard
Marcel Luttgens wrote:
<snipped previous
The 'correct' formulae are as follows:
Give arms of equal 'rest' length, oriented at 90deg wrt each other, and
given one of the arms moving longitudinally wrt the medium (whatever you
chose to interpret the 'medium' as).
We will assume rest frame to be Earth frame, and empirically light
propagates at c through vacuum in the near Earth environment. Since the
MM interferometer was at rest wrt Earth then we can assume the medium to
be at rest wrt Earth as a condition of the experiment, which agrees with
LET and SR, as well as the various so-called Modified Ether Theories.
So regardless of which theory is the most accurate, the same
'mathematical' argument applies to MM regardless of which of these
theories that we choose to apply to the events. However, whether that
math actually applies is another matter to be addressed afterward.
The one-way trip time along the longitudinal arm, according to the
Galilean transform, is derived as follows:
L + vt_1 = ct_1
L = ct_1 - vt_1
L = t_1(c - v)
t_1 = L/(c - v)
The one-way trip time in the other direction (return trip) is given by:
L - vt_2 = ct_2
L = ct_2 + vt_2
L = t_2(c + v)
t_2 = L/(c + v)
The two-way trip time along the longitudinal arm is:
t' = t_1 + t_2
Or
t' = [L/(c - v)] + [L/(c + v)]
Which reduces to:
t' = 2L/c(1 - (v/c)^2)
But the time required for the two-way trip when the arm is at rest wrt
the medium is just
t = 2L/c
Thus by substitution:
t = t/(1 - (v/c)^2)
Or
t' = t/gamma^2 |gamma = sqrt(1 - (v/c)^2)
__________________
Keep in mind that at no time will the Galilean transform be departed
from in this argument; these are all Galilean values and equations. t'
is the Galilean time required for the two-way trip when the arm is in
motion wrt the medium, and t is the Galilean time required for the
two-way trip when the same arm is at rest wrt the medium. (The argument
will apply equally as well to sound propagation when c is alternately
interpreted to be the speed of sound wrt a non-moving isotropic medium)
Pay close attention, and acquaint yourself with the definitions of the
terms used, because with them the Galilean derivation of the length
contraction of the longitudinal arm will be derived as a special case of
the argument, and this should be no surprise since the argument is just
Lorentz's 'textbook' take on MM.
__________________
For the lateral arm we simply apply the Pythagorean Theorem to find the
component of velocity of the beam in the lateral direction:
c_y = sqrt(c^2 - v^2)
The one-way trip time is
L'/c_y
Hence the two-way trip time is
t'' = 2L'/c_y
Or
t'' = 2L'/sqrt(c^2 - v^2)
But
t = 2L'/c
Hence
t'' = tc/sqrt(c^2 - v^2)
Or
t'' = t/(sqrt((c^2 - v^2)/c^2))
And
t'' = t/(sqrt(1 - (v/c)^2)
Or
t'' = t/gamma
If the trip times along the two arms are equal then
t' = t''
Thus
t/gamma^2 = t/gamma
And thus it is required by the premises that
gamma = 1
Which is to say, the arms are at rest wrt the medium.
Which is, unfortunately, the stance assumed by Einstein, i.e. that the
interferometer frame is always at rest wrt the medium.
If, OTOH, we assert that
gamma =/= 1, then we necessarily have assumed that at least one premise
above was incorrect. There are only two such premises to call into
question
1) that light doesn't propagate at c wrt a medium
2) that the lengths of the arms are equal when the interferometer is in
motion wrt the medium.
Oddly, both LET and SR assume case 2, i.e. that the arms are not equal
in length when in motion wrt the medium.
SR, though derives a contradiction, since on the one hand it postulates
that the arms are always at rest wrt the medium, but on the other, that
when in motion wrt the medium the arms differ in length. LET doesn't
suffer this paradoxical stance in that it doesn't assert that the arms
are always at rest wrt the medium, but rather only that the arms differ
in length when in motion wrt the medium. The assumption was made by
Lorentz that the longitudinal arm contracted along its length by a
factor gamma. And indeed, upon substitution of L*gamma for L in the
above equations, we 'do' in fact arrive at the conclusion:
t' = t''
independently of the degree of motion of the arms wrt the medium.
However, as noted previously, this assumes no contraction of the lateral
arm, and thus cannot be correct in that the contraction is supposed to
be due to an alteration in the electromagnetic potentials in the arm
material along the longitudinal direction, which in turn is caused by
the two-way lag in the mediating forces. A similar lag occurs in the
lateral direction also, thus negating Lorentz's conclusion absolutely.
The only remaining non-contradictory possibilities are that the
interferometer wasn't moving wrt the medium, (or was 'barely' moving),
or that the lateral arm was likewise contracted. It would remain to be
experimentally determined what that degree of contraction would be,
since such would be beyond the means of logic to provide this answer
without empirical data upon which to rest it.
As for your argument that a contraction contradicts SR, I said no less
above. (If you were paying attention you will remember the argument)
However, its adherents provide for themselves a solution, that at least
satisfies them, i.e. they apply LET logic and then claim obliviously
that the solution derives from SR. Since it doesn't contradiction the
Lorentz transformation, then to them it is an 'SR' solution. Not so,
however, since SR predicts reciprocity of the contractions and
dilations, which is an absurd posture, if not downright stooopid, as has
also been proven beyond doubt to them infinite times since the
conception of SR. Not very bright those SRists:)
Richard Perry
http://www.cswnet.com/~rper
Marcel Luttgens
As this is a Galilean transform, it would be better not to use
SR symbols like primed variables or gamma.
For instance, t1 = t/(1 - (v/c)^2) is preferable to t' = t/gamma^2.
I would prefer the form t2 = t/(sqrt(1 - (v/c)^2).
>
> If the trip times along the two arms are equal then
>
> t' = t''
>
> Thus
>
> t/gamma^2 = t/gamma
>
> And thus it is required by the premises that
>
> gamma = 1
>
> Which is to say, the arms are at rest wrt the medium.
>
> Which is, unfortunately, the stance assumed by Einstein, i.e. that the
> interferometer frame is always at rest wrt the medium.
Not exactly, because, according to Einstein, x' = ct' and y' = ct'.
When the apparatus is at rest wrt the medium, x = x' = ct and y = y' = ct,
where ct corresponds to the rest length of the arms.
But when the apparatus is moving at v wrt the medium, the time t' measured
by the interferometer's clock is a function of the time t measured by the
medium's clock, and of v.
According to Einstein, t' = gt(1-v/c), thus
x' = y' = ct * (1-v/c)/sqrt(1-v^2/c^2)
= ct * sqrt[(1-v/c)/(1+v/c)],
but, in fact, t' = t * sqrt(1-v^2/c^2),
hence x' = y' = ct * sqrt(1-v^2/c^2).
Anyhow, the arms are not contracted, but time is "dilated".
>
> If, OTOH, we assert that
>
> gamma =/= 1, then we necessarily have assumed that at least one premise
> above was incorrect. There are only two such premises to call into
> question
>
> 1) that light doesn't propagate at c wrt a medium
>
> 2) that the lengths of the arms are equal when the interferometer is in
> motion wrt the medium.
>
> Oddly, both LET and SR assume case 2, i.e. that the arms are not equal
> in length when in motion wrt the medium.
SR is self-contradictory, because on one hand, its formulae x'=ct' and
y'=ct' indicate that the arms are not contracted, but on the other hand,
the Einstein's transformation x' = g(x-vt) can and has be used to demonstrate
that a contraction of the longitudinal arm must occur.
>
> SR, though derives a contradiction, since on the one hand it postulates
> that the arms are always at rest wrt the medium,
More precisely, the arms keep their rest length, but time is affected
by the motion of the apparatus.
>but on the other, that
> when in motion wrt the medium the arms differ in length. LET doesn't
> suffer this paradoxical stance in that it doesn't assert that the arms
> are always at rest wrt the medium, but rather only that the arms differ
> in length when in motion wrt the medium. The assumption was made by
> Lorentz that the longitudinal arm contracted along its length by a
> factor gamma. And indeed, upon substitution of L*gamma for L in the
> above equations, we 'do' in fact arrive at the conclusion:
>
> t' = t''
>
> independently of the degree of motion of the arms wrt the medium.
> However, as noted previously, this assumes no contraction of the lateral
> arm, and thus cannot be correct in that the contraction is supposed to
> be due to an alteration in the electromagnetic potentials in the arm
> material along the longitudinal direction, which in turn is caused by
> the two-way lag in the mediating forces. A similar lag occurs in the
> lateral direction also, thus negating Lorentz's conclusion absolutely.
This is indeed the big problem for etherists as well as for Einsteinian
relativists. You could go to http://perso.wanadoo.fr/mluttgens/sapere.htm
and read the chapter 2.1. Tower of Babel. Enjoy!
>
> The only remaining non-contradictory possibilities are that the
> interferometer wasn't moving wrt the medium, (or was 'barely' moving),
> or that the lateral arm was likewise contracted. It would remain to be
> experimentally determined what that degree of contraction would be,
> since such would be beyond the means of logic to provide this answer
> without empirical data upon which to rest it.
The third possibility is that there is no length contraction at all.
>
>
> As for your argument that a contraction contradicts SR, I said no less
> above. (If you were paying attention you will remember the argument)
> However, its adherents provide for themselves a solution, that at least
> satisfies them, i.e. they apply LET logic and then claim obliviously
> that the solution derives from SR. Since it doesn't contradiction the
> Lorentz transformation, then to them it is an 'SR' solution. Not so,
> however, since SR predicts reciprocity of the contractions and
> dilations, which is an absurd posture, if not downright stooopid, as has
> also been proven beyond doubt to them infinite times since the
> conception of SR. Not very bright those SRists:)
>
I can only agree with you. Nobody can convince ayatollahs with logical
arguments.
>
>
> Richard Perry
> http://www.cswnet.com/~rper
Marcel Luttgens
Dirk Van de moortel
"Richard" <no_mail...@yahoo.com> wrote in message news:3EC868C4...@yahoo.com...
[snip]
> t' = t/gamma^2 |gamma = sqrt(1 - (v/c)^2)
So gamma = sqrt(1 - (v/c)^2) < 1
[snip]
> And
>
> t'' = t/(sqrt(1 - (v/c)^2)
>
> Or
>
> t'' = t/gamma
So again gamma = sqrt(1 - (v/c)^2) < 1
[snip]
> As for your argument that a contraction contradicts SR, I said no less
> above. (If you were paying attention you will remember the argument)
> However, its adherents provide for themselves a solution, that at least
> satisfies them, i.e. they apply LET logic and then claim obliviously
> that the solution derives from SR. Since it doesn't contradiction the
> Lorentz transformation, then to them it is an 'SR' solution. Not so,
> however, since SR predicts reciprocity of the contractions and
> dilations, which is an absurd posture, if not downright stooopid, as has
> also been proven beyond doubt to them infinite times since the
> conception of SR. Not very bright those SRists:)
Indeed, all these silly SRists (and LETters) have always been
thinking that gamma > 1.
Luckily, after more than a century of being so utterly wrong,
they have the World Famous Immortal Richard Perry, in an
intimate communication with The Even More So World Famous
Immortal Marcel Luttgens to put them on the right track again.
From the cessperrypool of sci.physics.relativity:
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/HowdyDoo.html
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/LorentzPerry.html
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/SRValid.html
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/AAB.html
From the cessluttgenspool of sci.physics.relativity:
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ArmsGrow2.html
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ArmsGrow.html
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/IfOnlyIf.html
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/SRSymbols.html
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/CorrectRelations.html
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Forget.html
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/SRLuttgens.html
Dirk Vdm
Richard
What in the 'hell' is wrong with your brain? If ever there was an
immortal fumble you just posted it. If you'll look at the terms (as they
are defined) and the use of gamma (as it is defined), the equations
posted set the moving light-clocks' ticking rates slower than the
stationary light-clock's ticking rate. What are you objecting to?
I think it's time to post a 'Dirk's Dorky Dalliances' page.
Marcel Luttgens
Instead of constantly referring to your mostly irrelevant
pages, you could at least try to refute without quibbling
the following proof of the falseness of the LT, based on the MMX.
Let be L the rest length of the arms.
When the apparatus is at rest in the ether, ct = L.
When the apparatus is moving at v wrt the ether,
according to the Galilean analysis of the MMX,
the two-way trip time of light along the longitudinal arm is
t(x) = 2L'/c(1 - v^2/c^2), where L' is the length of the arm
moving parallel to the x-axis, whereas, for the lateral arm,
one gets, by applying the Pythagorean theorem,
t(y) = 2L/c*sqrt(1 - v^2/c^2).
As no fringe shift is observed, one logically assume that t(x) = t(y).
This implies that L' = L * sqrt(1-v^2/c^2), iow, that the length
of the longitudinal arm is contracted by sqrt(1-v^2/c^2).
Now let's call S the ether frame and S' the interferometer's frame.
S' moves at v wrt S.
Note that the longitudial arm moves parallel to the x,x'-axis,
whereas the lateral arm corresponds to the y'-axis.
According to SR, y' = ct'. If you disagree, please tell us why.
As the result of the MMX is negative, the light paths must be
equal, hence x' = y' = ct' = L'.
Do you agree?
Let's consider that t' = t * sqrt(1-v^2/c^2).
Then, x' = ct * sqrt(1-v^2/c^2), to which corresponds
L' = L * sqrt(1-v^2/c^2). Note that this last relation doesn't
imply a length contraction, but only a time "dilation".
Can you follow?
By replacing L' by L * sqrt(1-v^2/c^2) in the Galilean relation
t(x) = 2L'/c(1 - v^2/c^2), one gets
t(x) = 2L/c*sqrt(1 - v^2/c^2) = t(y).
The travel times of the light signal along the two arms being
equal, the MMX must be negative.
Do you disagree?
Now, instead of t' = t * sqrt(1-v^2/c^2), let's use the Einsteinian
"time" transform t' = g(t-xv/c^2).
Since x = ct, t' = t * sqrt[(1-v/c)/(1+v/c)], and
x' = ct' = ct * sqrt[(1-v/c)/(1+v/c)].
The equivalent form being L' = L * sqrt[(1-v/c)/(1+v/c)],
do you think that t(x) is still equal to t(y) when using
this value of L'?
If not, don't you have to conclude that the Einsteinian
"time" LT is false, and that the correct relation is
t' = t * sqrt(1-v^2/c^2), as shown above?
Do you realize that, as the "time" LT is false, so is the
"position" LT x' = g(x-vt)?
Marcel Luttgens
Dirk Van de moortel
"Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message news:b45b8808.03052...@posting.google.com...
*Your* pages, Marcel.
Dirk Vdm
Marcel Luttgens
> > Instead of constantly referring to your mostly irrelevant
> > pages,
>
> *Your* pages, Marcel.
>
> Dirk Vdm
Iow, you have nothing to say.
Marcel Luttgens
Dirk Van de moortel
"Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message news:b45b8808.03052...@posting.google.com...
At this momenet, nothing that hasn't been said before.
Dirk Vdm
Marcel Luttgens
the absence of length contraction and of the falseness of the LT,
based on the MMX.
This is golden opportunity to show the crackpottery of SR
opponents.
1) Let L be the rest length of the arms.
When the apparatus is at rest in the ether, ct = L.
When the apparatus is moving at v wrt the ether,
according to the Galilean analysis of the MMX,
the two-way trip time of light along the longitudinal arm is
t(x) = 2L'/c(1 - v^2/c^2), where L' is the length of the arm
moving parallel to the x-axis, whereas, for the lateral arm,
one gets, by applying the Pythagorean theorem,
t(y) = 2L/c*sqrt(1 - v^2/c^2).
As no fringe shift is observed, one logically assumes that
t(x) = t(y).
This implies that L' = L * sqrt(1-v^2/c^2), iow, that the length
of the longitudinal arm is contracted by sqrt(1-v^2/c^2).
2) Now let's call S the ether frame and S' the interferometer's
frame.
S' moves at v wrt S.
The longitudial arm moves parallel to the x,x'-axis,
whereas the lateral arm is situated along the y'-axis.
3) According to SR, y' = ct'.
As the result of the MMX is negative, the light paths must be
equal, hence x' = y' = ct' = L'.
4) Let's consider that t' = t * sqrt(1-v^2/c^2).
Then, x' = ct * sqrt(1-v^2/c^2), to which corresponds
the relation L' = L * sqrt(1-v^2/c^2).
This relation doesn't imply a length contraction of the arm.
This can look paradoxical, but let's consider the following
scenario:
"Two cars run side by side on parallel lanes from A to B
at a velocity V = 50 miles/h.
According to the map, the distance AB = 100 miles.
Arrived at B, one of the driver looked at his watch, and
claimed that, since he has traveled during t = 2 hours
at 50 miles/h, the distance AB = Vt is indeed 100 miles.
The other driver claimed that, according to his own watch,
he drived during t' = 1 hour 1/4, adding that, since t' < t,
the distance AB is smaller than 100 miles, because he travelled
only AB = Vt'= 87.5 miles.
The first driver said: "You are wrong, because your
watch must be 3/4 hours slow, compared to mine." The other
drive replied: "No, I am right, because my lane contracted
by the factor t'/t = 7/8."
THe first driver then asked: "Isn't your last name Einstein?"
5) By replacing L' by L * sqrt(1-v^2/c^2) in the Galilean relation
t(x) = 2L'/c(1 - v^2/c^2), one gets
t(x) = 2L/c*sqrt(1 - v^2/c^2) = t(y).
The travel times of the light signal along the two arms being
equal, the MMX must be negative.
6) Now, instead of t' = t * sqrt(1-v^2/c^2), let's use the
Einsteinian "time" transform t' = g(t-xv/c^2).
Since x = ct, t' = t * sqrt[(1-v/c)/(1+v/c)], and
x' = ct' = ct * sqrt[(1-v/c)/(1+v/c)].
The equivalent form being L' = L * sqrt[(1-v/c)/(1+v/c)],
t(x) now different from t(y), in contradiction with the
negative result of the MMX.
7) One has to conclude
- that there is no length contraction
- that the Einsteinian "time" LT is false, the correct relation
being t' = t * sqrt(1-v^2/c^2), as shown above.
8) As the "time" LT is false, so is the "position" LT x' = g(x-vt),
because the two transforms are related by x' = ct'.
Marcel Luttgens
Dirk Van de moortel
"Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message news:b45b8808.0305...@posting.google.com...
Mixing vanilla icecream and mustard and complaining that
something horrible comes out.
This equation is not a transform.
It is merely one part of a transformation.
The other part is
x' = g(x-vt).
The fact that you call part of a transformation a "transform",
is already enough evidence that you don't know what you
are talking about.
But let's move on and look at the content rather than at the
form...
Both equations
x' = g(x-vt)
t' = g(t-xv/c^2)
are valid for all spacetime events.
They allow us to calculate an event's x' and t' if we know
its x and t and vice versa. This is something you *still* have
not grasped yet.
> Since x = ct,
Ha, here we have our light signal sent out at time t=0
in the positive x-direction, and an equation telling us that
at time t, the signal will have reached a distance ct from
the origin.
> t' = t * sqrt[(1-v/c)/(1+v/c)], and
> x' = ct' = ct * sqrt[(1-v/c)/(1+v/c)].
Absolutely: all these equations are perfectly valid for the
coordinates (x,t) and (x',t') of that single light signal sent
out at time t=0 in the positive x-direction.
Well done, Marcel. Good job.
> The equivalent form being L' = L * sqrt[(1-v/c)/(1+v/c)],
> t(x) now different from t(y), in contradiction with the
> negative result of the MMX.
Where did you get *that* silly idea?
You are talking about the endpoint of an arm with proper
length L here. This arm is fixed to frame S', moving at
speed *v* to frame S, while the light signal for which that
beautiful equation
x' = x * sqrt[(1-v/c)/(1+v/c)]
is valid, is moving at speed *c* to frame S and S'.
As always and always and always you are mixing vanilla ice
and mustard again.
Naughty troll. Again.
Caught with your pants down. Again.
>
> 7) One has to conclude
>
- that you are a dumbfuck with *extremely* bad taste. Again.
But if you really need help, all you have to do, is to say that
you are a bit retarded and that you need some help with
elementary analytic geometry and linear algebra concepts
like coordinates and transformations.
Say the word, we'll be glad to help.
Dirk Vdm
Marcel Luttgens
Do you deny that y' = ct', and the lengths of the light paths
must be equal along the two axes? If not, you have to accept
that x' = y'.
Try to realize that this conclusion is based on x' = y'.
> > 6) Now, instead of t' = t * sqrt(1-v^2/c^2), let's use the
> > Einsteinian "time" transform t' = g(t-xv/c^2).
>
> This equation is not a transform.
> It is merely one part of a transformation.
> The other part is
> x' = g(x-vt).
You are quibbling again. Call t' = g(t-xv/c^2) a SR relation if
you prefer.
> The fact that you call part of a transformation a "transform",
> is already enough evidence that you don't know what you
> are talking about.
Blah, blah, blah...
> But let's move on and look at the content rather than at the
> form...
> Both equations
> x' = g(x-vt)
> t' = g(t-xv/c^2)
> are valid for all spacetime events.
> They allow us to calculate an event's x' and t' if we know
> its x and t and vice versa. This is something you *still* have
> not grasped yet.
This is trivial.
>
> > Since x = ct,
>
> Ha, here we have our light signal sent out at time t=0
> in the positive x-direction, and an equation telling us that
> at time t, the signal will have reached a distance ct from
> the origin.
>
And that the distance ct corresponds the the length of the arm.
> > t' = t * sqrt[(1-v/c)/(1+v/c)], and
> > x' = ct' = ct * sqrt[(1-v/c)/(1+v/c)].
>
> Absolutely: all these equations are perfectly valid for the
> coordinates (x,t) and (x',t') of that single light signal sent
> out at time t=0 in the positive x-direction.
> Well done, Marcel. Good job.
>
So, you accept now that x' = y' = ct'.
> > The equivalent form being L' = L * sqrt[(1-v/c)/(1+v/c)],
> > t(x) now different from t(y), in contradiction with the
> > negative result of the MMX.
>
> Where did you get *that* silly idea?
> You are talking about the endpoint of an arm with proper
> length L here. This arm is fixed to frame S', moving at
> speed *v* to frame S, while the light signal for which that
> beautiful equation
> x' = x * sqrt[(1-v/c)/(1+v/c)]
> is valid, is moving at speed *c* to frame S and S'.
You are once more reproducing extracts of SR textbooks,
without realizing that x = ct is the rest length L of the arm,
because the light path in S is limited by the end mirror
of the arm.
So, x' = L * sqrt[(1-v/c)/(1+v/c)] implies x' = L'.
Don't evade the facts
1) that the relation t' = t * sqrt(1-v^2/c^2).
leads to x' = ct * sqrt(1-v^2/c^2), to which corresponds
the relation L' = L * sqrt(1-v^2/c^2),
2) that this relation explains the negative result of the MMX.
3) that the SR relation x' = ct' = ct * sqrt[(1-v/c)/(1+v/c)]
can mutatis mutandis be written L' = L * sqrt[(1-v/c)/(1+v/c)].
4) that this last relation doesn't explain the negative result
of the MMX.
5) that the SR relation t' = g(t-xv/c^2), being experimentally
refuted, is false.
6) that, automatically, the whole Einsteinian transformation
is false.
>
> As always and always and always you are mixing vanilla ice
> and mustard again.
> Naughty troll. Again.
> Caught with your pants down. Again.
>
> >
> > 7) One has to conclude
> >
>
> - that you are a dumbfuck with *extremely* bad taste. Again.
Wheras you are a true SR gentleman.
>
> But if you really need help, all you have to do, is to say that
> you are a bit retarded and that you need some help with
> elementary analytic geometry and linear algebra concepts
> like coordinates and transformations.
> Say the word, we'll be glad to help.
The one who badly needs help, especially in elementary logic, is you.
When I wrote "This is golden opportunity to show the crackpottery
of SR opponents", I made a Freundian slip. In fact, I meant
"This is golden opportunity to show the crackpottery of SR
proponents". You didn't miss it!
>
> Dirk Vdm
Marcel Luttgens
Dirk Van de moortel
"Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message news:b45b8808.0305...@posting.google.com...
http://users.pandora.be/vdmoortel/dirk/Stuff/MarcelAtSchool.gif
Dirk Vdm
Marcel Luttgens
> "Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message news:b45b8808.0305...@posting.google.com...
>
> http://users.pandora.be/vdmoortel/dirk/Stuff/MarcelAtSchool.gif
>
> Dirk Vdm
Escapist crackpot!
Marcel Luttgens
Paul B. Andersen
"Marcel Luttgens" <mutt...@wanadoo.fr> skrev i melding
news:b45b8808.0305...@posting.google.com...
> This is golden opportunity to show the crackpottery of SR
> opponents.
Nobody can do that better than you, Marcel.
Look:
A real convincing demonstration of crackpottery, isn't it?
BTW, you didn't really expect anyone to take this nonsense seriously?
Paul
Marcel Luttgens
Then refute it.
Marcel Luttgens
Marcel Luttgens
"Naughty troll,
You are a dumbfuck with *extremely* bad taste,
Crackpot!, etc... etc...",
no so-called SR experts have been able to refute my
demonstration of the non-existence of physical length
contraction, and of the falseness of the Lorentz
transformation.
The whole demonstration can be found at
http://perso.wanadoo.fr/mluttgens/mmx.htm
Marcel Luttgens
Dirk Van de moortel
"Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message news:b45b8808.03052...@posting.google.com...
> "Naughty troll,
> You are a dumbfuck with *extremely* bad taste,
> Crackpot!, etc... etc...",
> no so-called SR experts have been able to refute my
> demonstration of the non-existence of physical length
> contraction, and of the falseness of the Lorentz
> transformation.
Your so-called "demonstration" was a cheat, just like all the
"demonstrations" you ever made and you ever are going to
make.
Each time you attempt to say something about relativity and/or the
Lorentz transformation, you use these silly phrases like
"at the same moment", or
"after such a time interval", or
"where T is the time of the observer",
and on each occasion you are sneaking in some notion of absolute
time, which of course will "prove" the Lorentz transformation to
be "wrong".
People have been pointing this out to you since so many years.
The fact that you still continue doing this, is clear proof that you
either do it deliberatly, which exposes you as a troll, or that you
just can't help it, which exposes you as a very stupid person.
But even if you *are* a troll, you must be very stupid as well,
since you don't even seem to realize how utterly transparent you
are.
But, I must admit, you *do* provide some entertainment.
Please, do keep it that way.
Dirk Vdm
HenriWilson
My proof of that is trivial. See www.users.bigpond.com/hewn/contractions.exe
>
>Marcel Luttgens
Henri Wilson.
The BIG BANG Theory = The creationists' attempt to hijack science!
And it nearly worked!!!!!
See my newly UPGRADED animations at:
http://www.users.bigpond.com/HeWn/index.htm
YBM
[...]
> My proof of that is trivial. See www.users.bigpond.com/hewn/contractions.exe
Sorry dear, this is not a proof, this is a buggy ridiculous bloatware
ms-windows executable supposed to run on a late XXth century
architecture called IA32.
What about physics, crackpot ?
YBM
You shouldn't say that. I've read all of you recent contributions and it
is indeed the best abstract of your thoughts.
Vous ne devriez pas dire ça, j'ai lu toutes vos contributions récentes,
et ce dessin est clairement le meilleur résumé qu'on puisse en faire.
Marcel Luttgens
> <snipped previous
>
> Or
>
>
> t'' = t/(sqrt(1 - (v/c)^2)
>
> Or
>
> t'' = t/gamma
>
>
> t' = t''
>
> Thus
>
> t/gamma^2 = t/gamma
>
> And thus it is required by the premises that
>
> gamma = 1
>
> Which is to say, the arms are at rest wrt the medium.
>
> Which is, unfortunately, the stance assumed by Einstein, i.e. that the
> interferometer frame is always at rest wrt the medium.
>
>
> gamma =/= 1, then we necessarily have assumed that at least one premise
> above was incorrect. There are only two such premises to call into
> question
>
> 1) that light doesn't propagate at c wrt a medium
>
> 2) that the lengths of the arms are equal when the interferometer is in
> motion wrt the medium.
>
> Oddly, both LET and SR assume case 2, i.e. that the arms are not equal
> in length when in motion wrt the medium.
>
> when in motion wrt the medium the arms differ in length. LET doesn't
> suffer this paradoxical stance in that it doesn't assert that the arms
> are always at rest wrt the medium, but rather only that the arms differ
> in length when in motion wrt the medium. The assumption was made by
> Lorentz that the longitudinal arm contracted along its length by a
> factor gamma. And indeed, upon substitution of L*gamma for L in the
> above equations, we 'do' in fact arrive at the conclusion:
>
> t' = t''
>
> independently of the degree of motion of the arms wrt the medium.
> However, as noted previously, this assumes no contraction of the lateral
> arm, and thus cannot be correct in that the contraction is supposed to
> be due to an alteration in the electromagnetic potentials in the arm
> material along the longitudinal direction, which in turn is caused by
> the two-way lag in the mediating forces. A similar lag occurs in the
> lateral direction also, thus negating Lorentz's conclusion absolutely.
>
> interferometer wasn't moving wrt the medium, (or was 'barely' moving),
> or that the lateral arm was likewise contracted. It would remain to be
> experimentally determined what that degree of contraction would be,
> since such would be beyond the means of logic to provide this answer
> without empirical data upon which to rest it.
>
>
> above. (If you were paying attention you will remember the argument)
> However, its adherents provide for themselves a solution, that at least
> satisfies them, i.e. they apply LET logic and then claim obliviously
> that the solution derives from SR. Since it doesn't contradiction the
> Lorentz transformation, then to them it is an 'SR' solution. Not so,
> however, since SR predicts reciprocity of the contractions and
> dilations, which is an absurd posture, if not downright stooopid, as has
> also been proven beyond doubt to them infinite times since the
> conception of SR. Not very bright those SRists:)
>
>
> http://www.cswnet.com/~rper
You wrote:
"The one-way trip time of the beam along the longitudinal arm,
according to the Galilean transform, is derived as follows:
(demonstration snipped)
The two-way trip time along the longitudinal arm is:
t' = 2L/c(1 - (v/c)^2)
For the lateral arm we simply apply the Pythagorean Theorem to find the
component of velocity of the beam in the lateral direction:
(demonstration snipped)
Hence the two-way trip time is
t'' = 2L'/sqrt(c^2 - v^2)"
Of course, I agree with your demonstration, but I would prefer to
express its results as follows:
The two-way trip time of the beam along the longitudinal arm is
t(x) = 2(L'/c) * 1/(1 - v^/c^2) (1)
For the lateral arm, the two-way trip time of the beam is
t(y) = 2(L/c) * 1/sqrt(1 - v^2/c^2) (2)
You also wrote (in adapted form):
"Upon substitution of L*gamma for L' in the above equations,
we 'do' in fact arrive at the conclusion:
t(x) = t(y)
The assumption was made by Lorentz that the longitudinal arm contracted
along its length.
Einstein also claimed that when in motion wrt the medium the arms
differ in length."
Neither Lorentz nor Einstein realized that L'/c and L/c represent times
in above relations (1) and (2), which can be written
t(x) = 2t' * 1/(1 - v^/c^2) (1')
t(y) = 2t * 1/sqrt(1 - v^2/c^2) (2')
If t' = t * 1/(1 - v^/c^2), the relation (1') becomes
t(x) = 2t * 1/sqrt(1 - v^/c^2), which is identical to the relation (2').
Thus, if clocks are slowed down by a factor 1/(1 - v^/c^2),
as observed from the moving frame of reference, the two-way trip time
of the beam is identical for the two arms, and there is no need to
assume a contraction of the longitudinal arm. Moreover, such assumption
would be false.
Let's by the way note that, according to the so-called
"Lorentz" transformation,
t' = t * sqrt[(1-v/c)/(1+v/c)], and
x' = ct * sqrt[(1-v/c)/(1+v/c)].
As the negative result of the MMX is wholly explained by a time "dilation"
according to the relation t' = t * 1/(1 - v^/c^2), one has to conclude
that the "Lorentz" transformation are false. Then, of course, SR itself
is false.
In fact, the "Lorentz" transformation reduce to
t' = t * 1/(1 - v^/c^2)
x' = V't', where V' = (V-v)/(1-Vv/c^2).
In this case, V = c, hence, x' = ct'.
Marcel Luttgens
Dirk Van de moortel
"Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message news:b45b8808.03053...@posting.google.com...
Since the first post (16 May 2003 14:55:30 GMT) where Marcel
Luttgens declared that the lengths of the arms are given by L = ct,
and this post (31 May 2003 08:59:46 GMT), exactly 1274656
seconds have passed.
Therefore by now the arms of the interferometer have reached a
length of 382132255344448 meters.
And growing.
Dirk Vdm
Marcel Luttgens
Sorry, the last part of my message should read:
Neither Lorentz nor Einstein realized that L'/c and L/c represent times
in above relations (1) and (2), which can be written
t(x) = 2t' * 1/(1 - v^/c^2) (1')
t(y) = 2t * 1/sqrt(1 - v^2/c^2) (2')
If t' = t * sqrt(1 - v^/c^2), the relation (1') becomes
t(x) = 2t * 1/sqrt(1 - v^/c^2), which is identical to the relation (2').
Thus, if clocks are slowed down by a factor sqrt(1 - v^/c^2),
as observed from the moving frame of reference, the two-way trip time
of the beam is identical for the two arms, and there is no need to
assume a contraction of the longitudinal arm. Moreover, such assumption
would be false.
Let's by the way note that, according to the so-called
"Lorentz" transformation,
t' = t * sqrt[(1-v/c)/(1+v/c)], and
x' = ct * sqrt[(1-v/c)/(1+v/c)].
As the negative result of the MMX is wholly explained by a time "dilation"
according to the relation t' = t * sqrt(1 - v^/c^2), one has to conclude
that the "Lorentz" transformation are false. Then, of course, SR itself
is false.
In fact, the "Lorentz" transformation reduce to
t' = t * sqrt(1 - v^/c^2)
Marcel Luttgens
Try to refute what I wrote, not what you misunderstood:
Neither Lorentz nor Einstein realized that L'/c and L/c represent times
in above relations (1) and (2), which can be written
t(x) = 2t' * 1/(1 - v^/c^2) (1')
t(y) = 2t * 1/sqrt(1 - v^2/c^2) (2')
If t' = t * sqrt(1 - v^/c^2), the relation (1') becomes
t(x) = 2t * 1/sqrt(1 - v^/c^2), which is identical to the relation (2').
Thus, if clocks are slowed down by a factor sqrt(1 - v^/c^2),
as observed from the moving frame of reference, the two-way trip time
of the beam is identical for the two arms, and there is no need to
assume a contraction of the longitudinal arm. Moreover, such assumption
would be false.
Let's by the way note that, according to the so-called
"Lorentz" transformation,
t' = t * sqrt[(1-v/c)/(1+v/c)], and
x' = ct * sqrt[(1-v/c)/(1+v/c)].
As the negative result of the MMX is wholly explained by a time "dilation"
according to the relation t' = t * sqrt(1 - v^/c^2), one has to conclude
that the "Lorentz" transformation are false. Then, of course, SR itself
is false.
In fact, the "Lorentz" transformation reduce to
t' = t * sqrt(1 - v^/c^2)
Dirk Van de moortel
"Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message news:b45b8808.03060...@posting.google.com...
> "Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote in message
news:<kK3Ca.30375$1u5....@afrodite.telenet-ops.be>...
> > "Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message news:b45b8808.03053...@posting.google.com...
> >
> > Since the first post (16 May 2003 14:55:30 GMT) where Marcel
> > Luttgens declared that the lengths of the arms are given by L = ct,
> > and this post (31 May 2003 08:59:46 GMT), exactly 1274656
> > seconds have passed.
> > Therefore by now the arms of the interferometer have reached a
> > length of 382132255344448 meters.
> > And growing.
> >
> > Dirk Vdm
>
>
> Try to refute what I wrote,
Done, but as usual you didn't want to understand that
your argument was a cheat.
> not what you misunderstood:
Apart from you, we all have perfectly understood that
http://w0rli.home.att.net/youare.swf
Dirk Vdm
Marcel Luttgens
Nobody has understood a refutation that you never clearly presented
on this NG.
Be fair, try again to refute the following analysis, clearly this time.
Dirk Van de moortel
which is only valid for events with
x = vt
x' = 0
, the relation (1') becomes
> t(x) = 2t * 1/sqrt(1 - v^/c^2), which is identical to the relation (2').
>
> Thus, if clocks are slowed down by a factor sqrt(1 - v^/c^2),
> as observed from the moving frame of reference, the two-way trip time
> of the beam is identical for the two arms, and there is no need to
> assume a contraction of the longitudinal arm. Moreover, such assumption
> would be false.
>
> Let's by the way note that, according to the so-called
> "Lorentz" transformation,
>
> t' = t * sqrt[(1-v/c)/(1+v/c)]
which is only valid for events with
x = ct
x' = ct'
>, and
> x' = ct * sqrt[(1-v/c)/(1+v/c)].
>
> As the negative result of the MMX is wholly explained by a time "dilation"
> according to the relation t' = t * sqrt(1 - v^/c^2), one has to conclude
> that the "Lorentz" transformation are false. Then, of course, SR itself
> is false.
>
> In fact, the "Lorentz" transformation reduce to
>
> t' = t * sqrt(1 - v^/c^2)
> x' = V't', where V' = (V-v)/(1-Vv/c^2).
>
> In this case, V = c, hence, x' = ct'."
>
>
> Marcel Luttgens
Repeating the last paragraph of:
http://groups.google.com/groups?&as_umsgid=MlMza.17596$1u5....@afrodite.telenet-ops.be
But if you really need help, all you have to do, is to say that
you are a bit retarded and that you need some help with
elementary analytic geometry and linear algebra concepts
like coordinates and transformations.
Say the word, we'll be glad to help.
Dirk Vdm
Marcel Luttgens
But explains the null result of the MMX!
>
> , the relation (1') becomes
> > t(x) = 2t * 1/sqrt(1 - v^/c^2), which is identical to the relation (2').
> >
> > Thus, if clocks are slowed down by a factor sqrt(1 - v^/c^2),
> > as observed from the moving frame of reference, the two-way trip time
> > of the beam is identical for the two arms, and there is no need to
> > assume a contraction of the longitudinal arm. Moreover, such assumption
> > would be false.
> >
> > Let's by the way note that, according to the so-called
> > "Lorentz" transformation,
> >
> > t' = t * sqrt[(1-v/c)/(1+v/c)]
>
> which is only valid for events with
> x = ct
> x' = ct'
But doesn't explain the negative result of the MMX!
>
> >, and
> > x' = ct * sqrt[(1-v/c)/(1+v/c)].
> >
> > As the negative result of the MMX is wholly explained by a time "dilation"
> > according to the relation t' = t * sqrt(1 - v^/c^2), one has to conclude
> > that the "Lorentz" transformation are false. Then, of course, SR itself
> > is false.
> >
> > In fact, the "Lorentz" transformation reduce to
> >
> > t' = t * sqrt(1 - v^/c^2)
> > x' = V't', where V' = (V-v)/(1-Vv/c^2).
> >
> > In this case, V = c, hence, x' = ct'."
> >
> >
> > Marcel Luttgens
>
> Repeating the last paragraph of:
> http://groups.google.com/groups?&as_umsgid=MlMza.17596$1u5....@afrodite.telenet-ops.be
> But if you really need help, all you have to do, is to say that
> you are a bit retarded and that you need some help with
> elementary analytic geometry and linear algebra concepts
> like coordinates and transformations.
> Say the word, we'll be glad to help.
>
> Dirk Vdm
As long as you, the self-appointed SR champion, can't logically refute
the following proof of SR falseness, one has to conclude that
the idiot is you (cf. http://w0rli.home.att.net/youare.swf ):
Neither Lorentz nor Einstein realized that L'/c and L/c represent times
in relations (1) and (2):
The two-way trip time of the beam along the longitudinal arm is
t(x) = 2(L'/c) * 1/(1 - v^/c^2) (1)
For the lateral arm, the two-way trip time of the beam is
t(y) = 2(L/c) * 1/sqrt(1 - v^2/c^2) (2),
relations which can be written
t(x) = 2t' * 1/(1 - v^/c^2) (1')
t(y) = 2t * 1/sqrt(1 - v^2/c^2) (2')
If t' = t * sqrt(1 - v^/c^2), the relation (1') becomes
t(x) = 2t * 1/sqrt(1 - v^/c^2), which is identical to the relation (2').
Thus, if clocks are slowed down by a factor sqrt(1 - v^/c^2),
as observed from the moving frame of reference, the two-way trip time
of the beam is identical for the two arms, and there is no need to
assume a contraction of the longitudinal arm. Moreover, such assumption
would be false (cf. Hafele & Keating experiment, which confirms the
slowing down of moving clocks).
Otoh, according to the so-called "Lorentz" transformation,
t' = t * sqrt[(1-v/c)/(1+v/c)], and
x' = ct * sqrt[(1-v/c)/(1+v/c)].
Since the negative result of the MMX is wholly explained by a time
"dilation" according to the relation t' = t * sqrt(1 - v^/c^2), one
has to conclude that the "Lorentz" transformation are false. Then,
of course, SR itself is false. Iow, SR = BS, and SRists are crackpots,
unless they can invent a new logic.
In fact, the "Lorentz" transformation reduce to
t' = t * sqrt(1 - v^/c^2)
x' = V't', where V' = (V-v)/(1-Vv/c^2).
In this case, V = c, hence, x' = ct'.
(See http://perso.wanadoo.fr/mluttgens/LTfalse.htm ).
Marcel Luttgens
Dirk Van de moortel
Chocolate....
>
> >
> > , the relation (1') becomes
> > > t(x) = 2t * 1/sqrt(1 - v^/c^2), which is identical to the relation (2').
> > >
> > > Thus, if clocks are slowed down by a factor sqrt(1 - v^/c^2),
> > > as observed from the moving frame of reference, the two-way trip time
> > > of the beam is identical for the two arms, and there is no need to
> > > assume a contraction of the longitudinal arm. Moreover, such assumption
> > > would be false.
> > >
> > > Let's by the way note that, according to the so-called
> > > "Lorentz" transformation,
> > >
> > > t' = t * sqrt[(1-v/c)/(1+v/c)]
> >
> > which is only valid for events with
> > x = ct
> > x' = ct'
>
> But doesn't explain the negative result of the MMX!
... filled with mayonnaise.
Bad taste.
Dirk Vdm
Paul B. Andersen
> Be fair, try again to refute the following analysis, clearly this time.
Re inserting "above relations":
t(x) = 2(L'/c) * 1/(1 - v^/c^2) (1)
t(y) = 2(L/c) * 1/sqrt(1 - v^2/c^2) (2)
> "Neither Lorentz nor Einstein realized that L'/c and L/c represent times
> in above relations (1) and (2),
Of course they are constants with dimension "time".
That is blatantly obvious since the other factor is dimensionless.
So why did you think anyone could miss that?
> which can be written
>
> t(x) = 2t' * 1/(1 - v^/c^2) (1')
> t(y) = 2t * 1/sqrt(1 - v^2/c^2) (2')
Where t' = L'/c and t = L/c
Note that neither t' nor t are the temporal coordinates
of any specific event, or the reading of any clock.
They are constants with dimension time.
Nothing else.
One can of course wonder about the wisdom in selecting
the names t' and t on these constants. T1 and T2 or
something like that would be much better.
The motivation for this ambiguous naming is however
made clear below.
> If t' = t * sqrt(1 - v^/c^2), the relation (1') becomes
And remembering that the constants t' and t in this equation are
nothing but other names for: L'/c and L/c respectively, we get:
L'/c = (L/c)*sqrt(1 - v^2/c^2),
L' = L*sqrt(1-v^2/c^2)
> t(x) = 2t * 1/sqrt(1 - v^2/c^2), which is identical to the relation (2').
Quite.
If we assume the parallel arm is shortened by sqrt(1-v^2/c^2),
the round trip times become equal.
> Thus, if clocks are slowed down by a factor sqrt(1 - v^/c^2),
> as observed from the moving frame of reference, the two-way trip time
> of the beam is identical for the two arms, and there is no need to
> assume a contraction of the longitudinal arm. Moreover, such assumption
> would be false.
This is obvious nonsense!
You have assumed nothing about clock slowing.
t' is another name on the constant L'/c, which is not a time on a clock.
t is another name on the constant L/c, which is not a time on a clock.
Thus the equation: t' = t * sqrt(1 - v^2/c^2)
is an assumption about the length of the rods,
not about the rate of clocks.
You have shown that if the parallel arm is shortened by 1/gamma,
then the MMX will predict null fringe shifts.
Marcel, do you really manage to fool yourself by these silly
substitutions, where you deliberately choose ambiguous names for
constants in order to change their meaning to something entirely different
that what they are?
I hardly think that can be possible.
So why do you do it?
You can't seriously think you can prove anything but your
own stupidity this way.
Or can you?
Paul
Marcel Luttgens
> "Marcel Luttgens" <mutt...@wanadoo.fr> skrev i melding
> news:b45b8808.03060...@posting.google.com...
>
> > Be fair, try again to refute the following analysis, clearly this time.
>
> Re inserting "above relations":
> t(x) = 2(L'/c) * 1/(1 - v^/c^2) (1)
> t(y) = 2(L/c) * 1/sqrt(1 - v^2/c^2) (2)
>
> > "Neither Lorentz nor Einstein realized that L'/c and L/c represent times
> > in above relations (1) and (2),
>
> Of course they are constants with dimension "time".
> That is blatantly obvious since the other factor is dimensionless.
> So why did you think anyone could miss that?
>
> > which can be written
> >
> > t(x) = 2t' * 1/(1 - v^/c^2) (1')
> > t(y) = 2t * 1/sqrt(1 - v^2/c^2) (2')
>
> Where t' = L'/c and t = L/c
>
> Note that neither t' nor t are the temporal coordinates
> of any specific event, or the reading of any clock.
> They are constants with dimension time.
> Nothing else.
Nice try!
A length divided by a velocity is not a time anymore?
Doesn't t(y) represents twice the time taken by the light
beam to travel the hypotenuse of a right-angled triangle,
whose other sides are ct and vt(y)?
Any fool, except perhaps the donkeys themselves, can solve
the "pons asinorum", obtain t = [t(y)/2] * sqrt(1 - v^2/c^2),
and realize that t is a time.
Consider that t can be calculated when the hypotenuse
has reached a certain length, for instance H.
Otoh, t(x) = t(y) when t' = t * sqrt(1 - v^/c^2).
How can you sanely claim that t' is a constant with dimension
time, whereas it is a function of t, *and* of v/c ?
>
> One can of course wonder about the wisdom in selecting
> the names t' and t on these constants. T1 and T2 or
> something like that would be much better.
>
> The motivation for this ambiguous naming is however
> made clear below.
>
> > If t' = t * sqrt(1 - v^/c^2), the relation (1') becomes
>
> And remembering that the constants t' and t in this equation are
> nothing but other names for: L'/c and L/c respectively, we get:
> L'/c = (L/c)*sqrt(1 - v^2/c^2),
> L' = L*sqrt(1-v^2/c^2)
Ignoring time "dilation".
>
> > t(x) = 2t * 1/sqrt(1 - v^2/c^2), which is identical to the relation (2').
>
> Quite.
> If we assume the parallel arm is shortened by sqrt(1-v^2/c^2),
> the round trip times become equal.
>
> > Thus, if clocks are slowed down by a factor sqrt(1 - v^/c^2),
> > as observed from the moving frame of reference, the two-way trip time
> > of the beam is identical for the two arms, and there is no need to
> > assume a contraction of the longitudinal arm. Moreover, such assumption
> > would be false.
>
> This is obvious nonsense!
> You have assumed nothing about clock slowing.
No, but SR rightly does, even if its relation for clock slowing is false.
> t' is another name on the constant L'/c, which is not a time on a clock.
A very peculiar constant!
> t is another name on the constant L/c, which is not a time on a clock.
t is the time needed by light to travel the length L of the arm when
the apparatus is at rest in the "ether".
t(y) is a direct function of t, t(x) becomes a function of t
when t' is replaced by t * sqrt(1 - v^2/c^2).
>
> Thus the equation: t' = t * sqrt(1 - v^2/c^2)
> is an assumption about the length of the rods,
> not about the rate of clocks.
>
> You have shown that if the parallel arm is shortened by 1/gamma,
> then the MMX will predict null fringe shifts.
>
As a SRist, you shouldn't forget clock slowing. But you can't have
simultaneously length shortening *and* clock slowing. Try to be
coherent.
By the way, do you get length contraction or time slowing when
using SR, or both? Can you show the SRists's analysis of the MMX?
Do you get x = ct = L, and x' = ct'?
What is the SR relation for t'?
What do you get for y and y'?
>
> Paul
Marcel Luttgens
Marcel Luttgens
This is an addendum to my last message:
Paul B. Andersen doesn't know what a variable or a constant
represents in an equation.
In mathematics or physics, a variable is a part of a mathematical
expression that may assume any value in a specific, related
set of values, or a symbol for such a part, whereas a constant is
a quantity that always has the same value.
In physics, as a practical definition, a constant would be a variable,
whose value cannot be modified in a particular experimental design.
For instance,
- An apple falling from a branch situated at a distance d
from the ground will reach the ground after a time
t = sqrt(2d/a), where a is the acceleration of gravity.
In this relation, d is a variable, because its value can be
chosen at will by the experimenter, whereas a is a constant,
because its value cannot be modified in the tree environment.
The fact that some branch is situated at 10 meters from the ground
doesn't make d a constant, because the experimenter can in
principle chose another branch.
- Mutatis mutandi, in the equation
t(y) = 2(L/c) * 1/sqrt(1 - v^2/c^2),
L is a variable, not a constant. Indeed, the experimenter can
modify at will the length of the interferometer's arm.
If L/c is replaced by t in the equation, t becomes the variable,
because c is a constant.
Claiming, like Paul B. Andersen, that t is another name on the
constant L/c, reveals serious gaps in mathematics.
Marcel Luttgens
Jeff Krimmel
"Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message
news:b45b8808.0306...@posting.google.com...
> mutt...@wanadoo.fr (Marcel Luttgens) wrote in message
news:<b45b8808.03060...@posting.google.com>...
>
> Paul B. Andersen doesn't know what a variable or a constant
> represents in an equation.
I'm willing to bet it is indeed Marcel Luttgens that "doesn't know what a
variable or a constant represents in an equation." Usually I would follow
this with a few excerpts of some of your more inane "mathematical arguments"
(though that prescribes some sort of legitimacy to these "arguments" which
does not in fact exist), but anyone that has read just one of your posts is
already painfully aware.
Jeff
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Marcel Luttgens
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>
> "Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message
> news:b45b8808.0306...@posting.google.com...
> > mutt...@wanadoo.fr (Marcel Luttgens) wrote in message
> news:<b45b8808.03060...@posting.google.com>...
> >
> > Paul B. Andersen doesn't know what a variable or a constant
> > represents in an equation.
>
> I'm willing to bet it is indeed Marcel Luttgens that "doesn't know what a
> variable or a constant represents in an equation." Usually I would follow
> this with a few excerpts of some of your more inane "mathematical arguments"
> (though that prescribes some sort of legitimacy to these "arguments" which
> does not in fact exist), but anyone that has read just one of your posts is
> already painfully aware.
>
> Jeff
>
Don't bet! Explain that L is not a variable in the equation
t(y) = 2(L/c) * 1/sqrt(1 - v^2/c^2).
If you don't, or can't, yoy are just another SR crackpot.
Maecel Luttgens
Paul B. Andersen
"Marcel Luttgens" <mutt...@wanadoo.fr> skrev i melding news:b45b8808.03060...@posting.google.com...
> "Paul B. Andersen" <paul.b....@hia.no> wrote in message news:<bbib4m$cu3$1...@dolly.uninett.no>...
> > "Marcel Luttgens" <mutt...@wanadoo.fr> skrev i melding
> > news:b45b8808.03060...@posting.google.com...
> >
> > > Be fair, try again to refute the following analysis, clearly this time.
> >
> > Re inserting "above relations":
> > t(x) = 2(L'/c) * 1/(1 - v^/c^2) (1)
> > t(y) = 2(L/c) * 1/sqrt(1 - v^2/c^2) (2)
> >
> > > "Neither Lorentz nor Einstein realized that L'/c and L/c represent times
> > > in above relations (1) and (2),
> >
> > Of course they are constants with dimension "time".
> > That is blatantly obvious since the other factor is dimensionless.
> > So why did you think anyone could miss that?
> >
> > > which can be written
> > >
> > > t(x) = 2t' * 1/(1 - v^/c^2) (1')
> > > t(y) = 2t * 1/sqrt(1 - v^2/c^2) (2')
> >
> > Where t' = L'/c and t = L/c
> >
> > Note that neither t' nor t are the temporal coordinates
> > of any specific event, or the reading of any clock.
> > They are constants with dimension time.
> > Nothing else.
>
> Nice try!
> A length divided by a velocity is not a time anymore?
Sure it is a time.
It is a time not depending on the co-ordinates (including the temporal one)
"Constant" means in this context "not depending on the co-ordinates".
Thus it is a constant time - or a constant with dimention time.
> Doesn't t(y) represents twice the time taken by the light
> beam to travel the hypotenuse of a right-angled triangle,
> whose other sides are ct and vt(y)?
> Any fool, except perhaps the donkeys themselves, can solve
> the "pons asinorum", obtain t = [t(y)/2] * sqrt(1 - v^2/c^2),
> and realize that t is a time.
Sure.
It the "time" t = L/c, a measure of the length of the arm.
t is NOT a temporal co-ordinate, and do not depend on
on the co-ordinates.
> Consider that t can be calculated when the hypotenuse
> has reached a certain length, for instance H.
"t can be calculated WHEN"?
t is not a temporal co-ordinate dependeing on "when".
t = L/c, a measure of the length of the arm.
Nothing else.
> Otoh, t(x) = t(y) when t' = t * sqrt(1 - v^/c^2).
> How can you sanely claim that t' is a constant with dimension
> time, whereas it is a function of t, *and* of v/c ?
As t = L/c, a measure of the length of the parallel rod,
and t' = L'/c, a measure of the length of the transverse arm
the equation t' = t*sqrt(1-v^2/c^2) relates the lengths of the arms
and is identical to: L' = L*sqrt(1-v^2/c^2)
Sure L' is a function of L, *and* v/c, if you consider
L and v to be variables.
I used the word "constant" as "not depending on the co-ordinates",
and not as "having the same value everywhere in all eternity",
like in "constant of nature".
The mathematical term for the latter is "absolute constant".
L' , L and v are all constants in the sense that none of
them are co-ordinate dependent.
> > One can of course wonder about the wisdom in selecting
> > the names t' and t on these constants. T1 and T2 or
> > something like that would be much better.
> >
> > The motivation for this ambiguous naming is however
> > made clear below.
> >
> > > If t' = t * sqrt(1 - v^/c^2), the relation (1') becomes
> >
> > And remembering that the constants t' and t in this equation are
> > nothing but other names for: L'/c and L/c respectively, we get:
> > L'/c = (L/c)*sqrt(1 - v^2/c^2),
> > L' = L*sqrt(1-v^2/c^2)
>
> Ignoring time "dilation".
It's YOUR calculation.
YOU have related the lengths of the arms.
YOU have said nothing about time dilation.
In fact, you have not said anything about the time of anything.
Your t and t' are not temporal coordinates, they are alternative measures
of the lengths of the arms.
> > > t(x) = 2t * 1/sqrt(1 - v^2/c^2), which is identical to the relation (2').
> >
> > Quite.
> > If we assume the parallel arm is shortened by sqrt(1-v^2/c^2),
> > the round trip times become equal.
> >
> > > Thus, if clocks are slowed down by a factor sqrt(1 - v^/c^2),
> > > as observed from the moving frame of reference, the two-way trip time
> > > of the beam is identical for the two arms, and there is no need to
> > > assume a contraction of the longitudinal arm. Moreover, such assumption
> > > would be false.
> >
> > This is obvious nonsense!
> > You have assumed nothing about clock slowing.
>
> No, but SR rightly does, even if its relation for clock slowing is false.
>
> > t' is another name on the constant L'/c, which is not a time on a clock.
>
> A very peculiar constant!
As peculiar as the constants a and b in the equation
y = ax + b
where x and y are coordinates.
> > t is another name on the constant L/c, which is not a time on a clock.
>
> t is the time needed by light to travel the length L of the arm when
> the apparatus is at rest in the "ether".
In other words, t is a measure of the length of the arm.
> t(y) is a direct function of t, t(x) becomes a function of t
> when t' is replaced by t * sqrt(1 - v^2/c^2).
Or - since t = L/c and t' = L'/c - in other words:
t(y) is a direct function of L, t(x) becomes a function of L
when L' is replaced by L*sqrt(1 - v^2/c^2)
Not very surprising that the round trip time depend on
the lengths of the arms, is it?
> > Thus the equation: t' = t * sqrt(1 - v^2/c^2)
> > is an assumption about the length of the rods,
> > not about the rate of clocks.
> >
> > You have shown that if the parallel arm is shortened by 1/gamma,
> > then the MMX will predict null fringe shifts.
> >
>
> As a SRist, you shouldn't forget clock slowing. But you can't have
> simultaneously length shortening *and* clock slowing. Try to be
> coherent.
It is YOUR calculation, Marcel.
I have calculated nothing, and forgotten nothing.
YOU haven't said anything about clock slowing.
All YOU have done is in a very awkward and convoluted way
to point out that if we in the equations:
t(x) = 2(L'/c) * 1/(1 - v^2/c^2) (1)
t(y) = 2(L/c) * 1/sqrt(1 - v^2/c^2) (2)
assume that the length L' of the parallel arm is shortened
according to the equation L' = L*sqrt(1 -v^2/c^2)
then t(x) = t(y), that is, the round trip times become equal.
What your point with your stupid substitutions was, is hard to see.
If it was to confuse, you obviously succeeded.
You are indeed very confused about what you were doing.
> By the way, do you get length contraction or time slowing when
> using SR, or both? Can you show the SRists's analysis of the MMX?
> Do you get x = ct = L, and x' = ct'?
> What is the SR relation for t'?
> What do you get for y and y'?
You know of course the answers, Marcel.
I have shown you the Lorentz transform applied on
a moving interferometer numerous times a long time ago.
And so have others.
It is simple and straight forward. No contradictions.
So what is your problem?
Are you anable to apply the LT yourself?
(Stupid question, of course.)
Paul
Paul B. Andersen
What the hell are you babbling about?
It is obviously a matter of definition what are variables and
what are constants.
The scenario in question is:
Given two ortogonal arms moving with the speed v in
the "rest frame" (or "ether frame" or whatever you wish
to call it.) The length of the arm parallel to the motion
is L', and the length of the transverse arm is L, both
measured in the stationary frame.
What is the round trip time of light for the two arms
measured in the stationary frame?
In this scenario L', L and v are all given.
They are constants by bloody definition.
The answer to the question above is:
t(x) = 2*(L'/c)/(1 - v^2/c^2)
t(y) = 2*(L/c)/(1 - v^2/c^2)
Thiseare a solution, and everything are constants.
And note. Everything is referred to the stationary frame.
You may of course ask the question:
How do the round trip time dependon the length of the arm?
THEN L is a variable by bloody definition.
But that is a different scenario!
It is not the question asked!
It is utterly irrelevant!
You may of course substitute - say L'/c = Z1 and
L/c = Z2 and write the equations thus:
t(x) = 2*Z1/(1 - v^2/c^2)
t(y) = 2*Z2/sqrt(1 - v^2/c^2)
Z1 and Z2 are still measures of the lengths of the respective
arms, measured by the time i takes light to transverse
the _stationary_ lengths L' and L..
Both Z1 and Z2 are referred to the stationary frame!
----------------------------------------------------
It is of course utterly ridiculous to claim that one of these
are "time delated" relative to the other because they have dimension "time".
How the hell did you invent anything this stupid just
because you selected names where one of them
had a prime and the other not?
Paul
Dirk Van de moortel
"Paul B. Andersen" <paul.b....@hia.no> wrote in message news:bbquji$p0r$1...@dolly.uninett.no...
Apparently since that remarkable thread
http://groups.google.com/groups?&threadm=20010921091535...@nso-mn.aol.com
our silly troll has learned exactly nothing.
Dirk Vdm
Marcel Luttgens
It is useless to discuss ad nauseum about the validity of SR.
Only the results of a well chosen experiment could convince
the participants on this NG.
' A very simple and cheap experiment using an interferometer
' could confirm or disprove the Lorentz/Einstein hypothesis
' of length contraction in the ratio sqrt(1-v^2/c^2).
' Instead of travelling through air, the light beam would
' get through a medium with higher refractive index, for
' instance window glass with n = 1.5. Then, light would
' travel at V = 2/3 c along the inteferometer's arms, instead
' of at c.
' Then, the wave moving longitudinally has a velocity V-v
' on the outgoing trip, and V+v on the return trip, where
' v is the velocity of the interferometer through the "ether".
' The corresponding round-trip time is
' t2 = L'/(V-v) + L'/(V+v) = (2L'/V) * 1/(1-v^2/V2), where
' L' is the length of the parallel arm.
' Applying the Pythagoras theorem, it is easy to calculate
' that the wave moving transversely will need
' t1 = (2L/V) * sqrt(1-v^2/V^2) seconds to get back to the
' half-silvered mirror (L is the length of the perpendicular
' arm, and is equal to L' when the apparatus is at rest
' in the "ether").
' According to Lorentz/Einstein, L' = L * sqrt(1-v^2/c^2),
' hence t2 should be given by the formula
' t2 = (2L/V) * sqrt(1-v^2/c^2) / (1-v^2/V2).
' As t2 is different from t1, one should observe a fringe shift.
'
' For instance, L = 100 cm, V = 2c/3, v = 10e-4 c, and
' Wl = 6e-5 cm (the wavelength of the used light).
' Then, N = 4.17 hundredths of a fringe. By the way, Michelson
' and Morley claimed that they could detect a shift of 1 hundredth
' of a fringe.
'
' If such experiment is negative, one has to conclude that
' the parallel arm underwent a contraction in the ratio
' sqrt(1-v^2/V^2), which disproves the Lorentz/Einstein
' hypothesis of a contraction by sqrt(1-v^2/c^2).
' If the experiment is positive, in other words, if a fringe shift
' is observed, one has to conclude that motion through ether
' can be detected, which falsifies SR.
Marcel Luttgens
Bill Hobba
Marcel Luttgens wrote:
> It is useless to discuss ad nauseum about the validity of SR.
> Only the results of a well chosen experiment could convince
> the participants on this NG.
Very true. Science is based on experiment. That is the very reason length
contraction exists. When applied to a current carrying wire it predicts a
magnetic field exactly as found by experiment. Explain that without length
contraction or do magnetic fields exist a priori in your scheme of things
and so no explanation need to be sought.
Thanks
Bill
Dirk Van de moortel
"Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message news:b45b8808.03061...@posting.google.com...
> "Paul B. Andersen" <paul.b....@hia.no> wrote in message news:<bbq7mq$ir8$1...@dolly.uninett.no>...
>
> It is useless to discuss ad nauseum about the validity of SR.
> Only the results of a well chosen experiment could convince
> the participants on this NG.
Well said.
For all the participants on this NG who, until now, have
not found the time to be convinced by this kind of well
chosen experiments, here's a short overview:
http://math.ucr.edu/home/baez/physics/Relativity/SR/experiments.html
Dirk Vdm
Paul B. Andersen
"Marcel Luttgens" <mutt...@wanadoo.fr> skrev i melding news:b45b8808.03061...@posting.google.com...
> "Paul B. Andersen" <paul.b....@hia.no> wrote in message news:<bbq7mq$ir8$1...@dolly.uninett.no>...
No, I didn't write anything of the following.
> It is useless to discuss ad nauseum about the validity of SR.
> Only the results of a well chosen experiment could convince
> the participants on this NG.
Well said.
> ' A very simple and cheap experiment using an interferometer
> ' could confirm or disprove the Lorentz/Einstein hypothesis
> ' of length contraction in the ratio sqrt(1-v^2/c^2).
> ' Instead of travelling through air, the light beam would
> ' get through a medium with higher refractive index, for
> ' instance window glass with n = 1.5. Then, light would
> ' travel at V = 2/3 c along the inteferometer's arms, instead
> ' of at c.
[Snip Marcel's wrong derivation of what SR predicts
for the experiment. SR obviously predicts null fringe shifts.
Marcel's error is so obvious that I don't find it necessary
to point it out. Anybody but Marcel will see it, and Marcel
isn't amendable to reason anyway.]
But the following is a real gem!
> ' If such experiment is negative, one has to conclude that
> ' the parallel arm underwent a contraction in the ratio
> ' sqrt(1-v^2/V^2), which disproves the Lorentz/Einstein
> ' hypothesis of a contraction by sqrt(1-v^2/c^2).
> ' If the experiment is positive, in other words, if a fringe shift
> ' is observed, one has to conclude that motion through ether
> ' can be detected, which falsifies SR.
The experiment will falsify SR regardless of its results!
Congratulation, Marcel.
You did it again! :-)
Paul
Marcel Luttgens
The experiment I proposed has seemingly never been done.
Note that:
"If such experiment is negative, one has to conclude that
the parallel arm underwent a contraction in the ratio
sqrt(1-v^2/V^2), which disproves the Lorentz/Einstein
hypothesis of a contraction by sqrt(1-v^2/c^2).
If the experiment is positive, in other words, if a fringe shift
is observed, one has to conclude that motion through ether
can be detected, which falsifies SR."
Even as a thought experiment, it shows that SR is false from A to Z.
In particular, the LT cannot explain its results, whether negative or
positive.
If I understand you well, you are claiming that the ratio of length
contraction is a function of the speed c of light in vacuum.
Realize that a negative result of such experiment would imply that
the contraction is a function of the speed of light in a
specific medium, for instance glass. Change the medium, and you get
another contraction ratio.
But I agree with you, "Science is based on experiment", so why not
perform the experiment? NASA could easily do it.
Marcel Luttgens
Marcel Luttgens
> "Marcel Luttgens" <mutt...@wanadoo.fr> skrev i melding news:b45b8808.03061...@posting.google.com...
> > "Paul B. Andersen" <paul.b....@hia.no> wrote in message news:<bbq7mq$ir8$1...@dolly.uninett.no>...
>
> No, I didn't write anything of the following.
>
> > It is useless to discuss ad nauseum about the validity of SR.
> > Only the results of a well chosen experiment could convince
> > the participants on this NG.
>
> Well said.
>
> > ' A very simple and cheap experiment using an interferometer
> > ' could confirm or disprove the Lorentz/Einstein hypothesis
> > ' of length contraction in the ratio sqrt(1-v^2/c^2).
> > ' Instead of travelling through air, the light beam would
> > ' get through a medium with higher refractive index, for
> > ' instance window glass with n = 1.5. Then, light would
> > ' travel at V = 2/3 c along the inteferometer's arms, instead
> > ' of at c.
>
> [Snip Marcel's wrong derivation of what SR predicts
> for the experiment. SR obviously predicts null fringe shifts.
> Marcel's error is so obvious that I don't find it necessary
> to point it out. Anybody but Marcel will see it, and Marcel
> isn't amendable to reason anyway.]
Where did I derive what SR predicts?
I only used a Galilean analysis!
>
> But the following is a real gem!
>
> > ' If such experiment is negative, one has to conclude that
> > ' the parallel arm underwent a contraction in the ratio
> > ' sqrt(1-v^2/V^2), which disproves the Lorentz/Einstein
> > ' hypothesis of a contraction by sqrt(1-v^2/c^2).
> > ' If the experiment is positive, in other words, if a fringe shift
> > ' is observed, one has to conclude that motion through ether
> > ' can be detected, which falsifies SR.
>
> The experiment will falsify SR regardless of its results!
Exactly, SR is just a big hoax!
>
> Congratulation, Marcel.
> You did it again! :-)
>
> Paul
Marcel Luttgens
Dirk Van de moortel
"Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message news:b45b8808.0306...@posting.google.com...
> "Paul B. Andersen" <paul.b....@hia.no> wrote in message news:<bcrtho$fl5$1...@dolly.uninett.no>...
> > "Marcel Luttgens" <mutt...@wanadoo.fr> skrev i melding news:b45b8808.03061...@posting.google.com...
> > > "Paul B. Andersen" <paul.b....@hia.no> wrote in message news:<bbq7mq$ir8$1...@dolly.uninett.no>...
> >
> > No, I didn't write anything of the following.
> >
> > > It is useless to discuss ad nauseum about the validity of SR.
> > > Only the results of a well chosen experiment could convince
> > > the participants on this NG.
> >
> > Well said.
> >
> > > ' A very simple and cheap experiment using an interferometer
> > > ' could confirm or disprove the Lorentz/Einstein hypothesis
> > > ' of length contraction in the ratio sqrt(1-v^2/c^2).
> > > ' Instead of travelling through air, the light beam would
> > > ' get through a medium with higher refractive index, for
> > > ' instance window glass with n = 1.5. Then, light would
> > > ' travel at V = 2/3 c along the inteferometer's arms, instead
> > > ' of at c.
> >
> > [Snip Marcel's wrong derivation of what SR predicts
> > for the experiment. SR obviously predicts null fringe shifts.
> > Marcel's error is so obvious that I don't find it necessary
> > to point it out. Anybody but Marcel will see it, and Marcel
> > isn't amendable to reason anyway.]
>
> Where did I derive what SR predicts?
> I only used a Galilean analysis!
And then - like always - prove that SR s wrong.
Where have we seen that before?
>
> >
> > But the following is a real gem!
> >
> > > ' If such experiment is negative, one has to conclude that
> > > ' the parallel arm underwent a contraction in the ratio
> > > ' sqrt(1-v^2/V^2), which disproves the Lorentz/Einstein
> > > ' hypothesis of a contraction by sqrt(1-v^2/c^2).
> > > ' If the experiment is positive, in other words, if a fringe shift
> > > ' is observed, one has to conclude that motion through ether
> > > ' can be detected, which falsifies SR.
> >
> > The experiment will falsify SR regardless of its results!
>
> Exactly, SR is just a big hoax!
Yes, when you are allowed to use Galilean Relativity,
you can prove that SR is false.
Congratulations indeed. You are so frighteningly smart.
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/OnlyGalilean.html
Title "I only used a Galilean analysis!"
Dirk Vdm
Mikko Levanto
> It is useless to discuss ad nauseum about the validity of SR.
> Only the results of a well chosen experiment could convince
> the participants on this NG.
Maybe some of them. Many have are demonstrably immune to
any experimental evidence.
Mikko
Peter K.
Of course there is lenght contraction.. .and it's in line with the
direction of motion (say horizontal thus vertical remains same
length).
Maxwell's Permittivity forces act upon each particle and force a
fission of matter at the speed of light. (Same as saying infinite
Energy needed for mass to attain speed of light ...just a different
angle of perception (copernicus))
None the less I disagree with Einstein on Time. Time remains constant
and is controlled by Maxwell's Permittivity which dictates the maximum
light's velocity and can be used as a frame of reference(Relativity)
for all our Universe.
Maxwell's Permittivity remains constant no matter what frame of
reference (in our Universe's Big Bang) similar to light's speed.
(I believe Maxwell's Permittivity Constant controls the rate of
orbiting time for all planets and molecules and it's value was set
from the total Energy emitted at the Big Bang).
Paul B. Andersen
"Marcel Luttgens" <mutt...@wanadoo.fr> skrev i melding news:b45b8808.0306...@posting.google.com...
> "Paul B. Andersen" <paul.b....@hia.no> wrote in message news:<bcrtho$fl5$1...@dolly.uninett.no>...
> > "Marcel Luttgens" <mutt...@wanadoo.fr> skrev i melding news:b45b8808.03061...@posting.google.com...
> >
> > > Only the results of a well chosen experiment could convince
> > > the participants on this NG.
> >
> > Well said.
And the way to test the validity of a theory is to derive what
it predicts for an experiment, do the experiment, and then see
if the predictions are correct.
In other words, you must derive whether or not SR predicts
fringe shifts in this "well chosen" variant of the MMX.
But what do you do?
This:
> Where did I derive what SR predicts?
> I only used a Galilean analysis!
So you will test the valitity of SR by deriving
the predictions of Galilean relativity.
(Actually in a Michelson type ether)
> > But the following is a real gem!
> >
> > > ' If such experiment is negative, one has to conclude that
> > > ' the parallel arm underwent a contraction in the ratio
> > > ' sqrt(1-v^2/V^2), which disproves the Lorentz/Einstein
> > > ' hypothesis of a contraction by sqrt(1-v^2/c^2).
So if there are no fringe shifts, which prove the Galilean
prediction for the time along the parallell arm wrong,
this proves that the contraction predicted by SR is must be wrong
since it is different from the contraction not predicted by the Galilean
analysis.
> > Congratulation, Marcel.
> > You did it again! :-)
And you did it even better than I initially thought possible! :-)
This is the greatest display of faulty logic I ever saw.
You are the very best in this department!
Congratulations again.
Paul
Marcel Luttgens
According to the Galilean analysis of the MMX, where light
travels through air (whose refractive index n is about 1),
the two-way trip time of the beam along the longitudinal arm is
t(x) = 2(L'/c) * 1/(1 - v^/c^2). (1).
For the lateral arm, the overall trip time of the beam is
t(y) = 2(L/c) * 1/sqrt(1 - v^2/c^2). (2)
So, if L' = L * sqrt(1-v^2/c^2), iow, if the parallel arm
is contracted by sqrt(1-v^2/c^2), the two transit times of light
are equal, and the experiment must be negative.
Let us note that the wave moving longitudinally has a velocity c-v
on the outgoing trip, hence it needs t = L'/(c-v) seconds to reach
the mirror. As L' is given by L * sqrt(1-v^2/c^2), one gets
t = L * sqrt(1-v^2/c^2) / (c-v) = (L/c) * sqrt(1-v^2/c^2) / (1-v/c).
According to Einstein's formula t' = g(t-vx/c^2), one gets
t' = t(1 - v/c) / sqrt(1-v^2/c^2) , when x = ct.
Conversely, t = t' * sqrt(1-v^2/c^2) / (1-v/c), which is equivalent
to the Galilean result if t' = L/c. And indeed, according to
Einstein's derivation of his above formula, y' = ct', and y' = y = L,
thus L = ct' and t' = L/c.
So, the Galilean analysis and the use of Einstein's formula give
identical results when light travels through air. There is nothing
surprising about that, since Einstein manipulated accordingly the
equations he used in his derivation of the LT.
Iow, even SRists can safely use Galilean analysis.
But when light travels through a medium with n > 1, its velocity
is V < c, and the Galilean relations (1) and (2) become
t(x) = L'/(V-v) + L'/(V+v)
= 2(L'/V) * 1/(1-v^2/V2)
t(y) = 2(L/V) * sqrt(1-v^2/V^2).
If L' = L * sqrt(1-v^2/V^2), the two transit times are identical,
and an interferometric experiment where air is replaced for instance
by glass would be negative.
However, most scientists consider that the material composing the
interferometer always shortens by sqrt(1-v^2/c^2) in a direction
parallel to the motion. If they are right, the times t(x) and t(y)
are different when n > 1, and a fringe shift should be observed.
The only way to prove them right is to perform such experiment.
By the way, the Lorentz transformation cannot be applied when
the speed of light is V < c. Iow, SR is not a general theory.
Marcel Luttgens
Dirk Van de moortel
"Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message news:b45b8808.0306...@posting.google.com...
[snip previously debunked stuff]
.
>
> By the way, the Lorentz transformation cannot be applied when
> the speed of light is V < c. Iow, SR is not a general theory.
"When the speed of light is V < c."
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/SpeedV.html
Four fumbles in one thread, that must be a record.
Dirk Vdm
Bill Hobba
> > Very true. Science is based on experiment. That is the very reason
length
> > contraction exists. When applied to a current carrying wire it predicts
a
> > magnetic field exactly as found by experiment. Explain that without
length
> > contraction or do magnetic fields exist a priori in your scheme of
things
> > and so no explanation need to be sought.
> >
Marcel Luttgens replied:
>
> The experiment I proposed has seemingly never been done.
> Note that:
You are free to propose any experiment you see fit. All I am saying is that
length contraction when applied to a current carrying wire leads to the
existence of magnetic fields. If length contraction did not exist then
magnetic fields would not exists in contradiction to experiment. Thus if
your experiment showed no length contraction it would represent a crisis
similar to the discovery of a perpetual motion machine - basic physical
principles would be violated. Thus similar to perpetual motion machines
many people, including myself, have no interest in analyzing schemes that
violate such. But each to their won.
Thanks
Bill
Marcel Luttgens
Crackpot! Show how SR can be applied in this specific case.
Marcel Luttgens
Marcel Luttgens
Here is a much more efficient interferometric experiment, that
should detect motion through "ether".
One would use a hybrid interferometer, i.e.
the light beam would travel through air along one arm
(the "air" arm), and through glass along the other arm
(the "glass" arm). The rest length of both arms is L.
Let's suppose that the "air" arm is parallel to the motion of
the apparatus through "ether". As the refractive index of air
is about 1, let's assume that the light speed is c.
Then, the wave moving longitudinally has a velocity c-v
on the outgoing trip, and c+v on the return trip, where
v is the velocity of the interferometer through the "ether".
The corresponding round-trip time is
t(air) = L'/(c-v) + L'/(c+v) = (2L'/c) * 1/(1-v^2/c2), where
L' is the length of the parallel arm.
If, as generally assumed, the parallel arm is contracted
in the ratio sqrt(1-v^2/c^2), one gets
t(air) = (2L/c) * 1/sqrt(1-v^2/c^2).
Let's suppose that the refractive index of the glass is n.
The speed of light through glass is of course V = c/n.
As the "glass" arm is perpendicular to the velocity vector v,
one can apply the Pythagoras theorem, and calculate that the
wave moving transversely will need
t(glass) = (2L/V) * sqrt(1-v^2/V^2) seconds
to get back to the half-silvered mirror.
The two transit times of light t(air) and t(glass) being
greatly different, one should observe an important shift
in the fringe pattern.
I leave to the SRists the pleasure to calculate such shift,
by using of course SR.
Marcel Luttgens
Marcel Luttgens
should detect motion through "ether".
One would use a hybrid interferometer, i.e.
the light beam would travel through air along one arm
(the "air" arm), and through glass along the other arm
(the "glass" arm). The rest length of both arms is L.
Let's suppose that the "air" arm is parallel to the motion of
the apparatus through "ether". As the refractive index of air
is about 1, let's assume that the light speed is c.
Then, the wave moving longitudinally has a velocity c-v
on the outgoing trip, and c+v on the return trip, where
v is the velocity of the interferometer through the "ether".
The corresponding round-trip time is
t(air) = L'/(c-v) + L'/(c+v) = (2L'/c) * 1/(1-v^2/c2), where
L' is the length of the parallel arm.
If, as generally assumed, the parallel arm is contracted
in the ratio sqrt(1-v^2/c^2), one gets
t(air) = (2L/c) * 1/sqrt(1-v^2/c^2).
Let's suppose that the refractive index of the glass is n.
The speed of light through glass is of course V = c/n.
As the "glass" arm is perpendicular to the velocity vector v,
one can apply the Pythagoras theorem, and calculate that the
wave moving transversely will need
t(glass) = (2L/V) * sqrt(1-v^2/V^2) seconds
to get back to the half-silvered mirror.
The two transit times of light t(air) and t(glass) being
greatly different, one should observe an important shift
in the fringe pattern.
I leave to the SRists the pleasure to calculate such shift,
by using of course SR.
Marcel Luttgens
Addendum:
Of course, what counts is not the instantaneous shift, but
the shift variations in time, due to the Earth rotation, its
orbital motion, the motion of the solar system through the
"ether", etc...
Marcel Luttgens