# Which contraction factor?

3 views

### MLuttgens

Aug 4, 1999, 3:00:00 AM8/4/99
to
Let's consider an interferometer's arm of length L, that makes
an angle a with the velocity vector.

According to SR, the arm will contract by
f=sqrt(1-v^2) when a=0Â°, and by f=1 when a=90Â°.
Iow, L'=L*sqrt(1-v^2) when a=0Â° and L'=L when a=90Â°.
Hence, for any angle a other than 0Â° or 90Â°, the contraction factor f
must be situated between sqrt(1-v^2) and 1.

Is the general formula giving f,

1) f = sqrt(1-v^2) * cos(a),
2) f = sqrt(1-(v*cos(a))^2), or
3) another formula?

Marcel Luttgens

### Cees Roos

Aug 4, 1999, 3:00:00 AM8/4/99
to

MLuttgens <mlut...@aol.com> wrote in message
news:19990804055504...@ngol08.aol.com...
f = sqrt((1-v^2)/(1-(v*sin(a))^2))

> Marcel Luttgens
--
Regards, Cees Roos
I think it's much more interesting to live not knowing than
to have answers which might be wrong. R. Feynman 1981

### MLuttgens

Aug 5, 1999, 3:00:00 AM8/5/99
to
In article <7o9s7n\$e9k\$1...@news1.xs4all.nl>, "Cees Roos" <cr...@xs4all.nl> wrote
:

>Date : Wed, 4 Aug 1999 18:40:20 +0200

>
>
>MLuttgens <mlut...@aol.com> wrote in message
>news:19990804055504...@ngol08.aol.com...
>> Let's consider an interferometer's arm of length L, that makes
>> an angle a with the velocity vector.
>>
>> According to SR, the arm will contract by
>> f=sqrt(1-v^2) when a=0Â°, and by f=1 when a=90Â°.
>> Iow, L'=L*sqrt(1-v^2) when a=0Â° and L'=L when a=90Â°.
>> Hence, for any angle a other than 0Â° or 90Â°, the contraction factor f
>> must be situated between sqrt(1-v^2) and 1.
>>
>> Is the general formula giving f,
>>
>> 1) f = sqrt(1-v^2) * cos(a),
>> 2) f = sqrt(1-(v*cos(a))^2), or
>> 3) another formula?
>>
>
>f = sqrt((1-v^2)/(1-(v*sin(a))^2))
>

Thank you,

In "Re: To Dennis, Keto and all the Etherists", on Jul 20, 1999,
I already claimed that f=sqrt(1-v^2)/sqrt(1-(v*sin(a))^2) is
the simplest factor that allows to transform T(a) =
(2L/(1-v^2))*sqrt(1-(v*sin(a))^2) to T(a)' = 2L/sqrt(1-v^2),
which is independent of all angles a.

But perhaps did you find in the meantime a direct demonstration
of that factor? It would be much better than reverse
engineering!

Anyhow, there are now 2 formulae matching SR's prediction for
a=0Â° or a=90Â°:
1) f = sqrt(1-(v*cos(a))^2)
2) f = sqrt((1-v^2)/(1-(v*sin(a))^2))

Which one would Wayne Throop, Tom Roberts, Steve Carlip,
and other SR experts chose? Or another formula?

>Cees Roos

Marcel Luttgens

### Cees Roos

Aug 5, 1999, 3:00:00 AM8/5/99
to

MLuttgens <mlut...@aol.com> wrote in message
news:19990805050822...@ngol08.aol.com...

> In article <7o9s7n\$e9k\$1...@news1.xs4all.nl>, "Cees Roos" <cr...@xs4all.nl>
wrote
[snip]

> >f = sqrt((1-v^2)/(1-(v*sin(a))^2))
> >
>
> Thank you,
>
> In "Re: To Dennis, Keto and all the Etherists", on Jul 20, 1999,
> I already claimed that f=sqrt(1-v^2)/sqrt(1-(v*sin(a))^2) is
> the simplest factor that allows to transform T(a) =
> (2L/(1-v^2))*sqrt(1-(v*sin(a))^2) to T(a)' = 2L/sqrt(1-v^2),
> which is independent of all angles a.
>
> But perhaps did you find in the meantime a direct demonstration
> of that factor? It would be much better than reverse
> engineering!
>
Consider two diagrams:
1: Barlength l' (contracted length), angle a.

/|
/ |
/ |
/ |
l' / | l' * sin(a)
/ |
/ |
/ |
/_a________|
l' * cos(a)

2: Barlength l (uncontracted length), angle a'.

/|
/ |
/ |
/ |
l / | l * sin(a') ( == l' * sin(a) )
/ |
/ |
/ |
/_a'_______|
l * cos(a')

Contraction in x-direction only, so y-sizes equal.
(Excuse me for the diagrams, I'm a lousy ASCII artist)

l'^2 = ( l' * cos(a) )^2 + (l' * sin(a) )^2 =
l'^2 = ( l * cos(a') / gamma(v)) ^2 + (l * sin(a') )^2
= l^2 - ( v * l * cos(a') )^2 (1)

( l * cos(a') )^2 = l^2 * ( 1 - (sin(a') )^2 ) =
= l^2 - ( l * sin(a') )^2
= l^2 - ( l' * sin(a) )^2 (2)

Substitute (2) in (1):
l'^2 = l^2 - v^2 * ( l^2 - ( l' * sin(a) )^2 )
= l^2 - ( v * l )^2 + (v * l' * sin(a) )^2

l'^2 * ( 1 - ( v * sin(a) )^2 ) = l^2 * ( 1 - v^2 )

l'^2 = l^2 * ( 1 - v^2 ) / ( 1 - ( v * sin(a) )^2 )

l' = l * sqrt( ( 1 - v^2 ) / ( 1 - ( v * sin(a) )^2 ) )

> Anyhow, there are now 2 formulae matching SR's prediction for
> a=0Â° or a=90Â°:
>

Three.

> 1) f = sqrt(1-(v*cos(a))^2)
> 2) f = sqrt((1-v^2)/(1-(v*sin(a))^2))
>
> Which one would Wayne Throop, Tom Roberts, Steve Carlip,
> and other SR experts chose? Or another formula?
>

### DE WITTE Roland

Aug 5, 1999, 3:00:00 AM8/5/99
to

MLuttgens a Ã©crit dans le message
<19990804055504...@ngol08.aol.com>...

>Let's consider an interferometer's arm of length L, that makes
>an angle a with the velocity vector.
>
>According to SR, the arm will contract by
>f=sqrt(1-v^2) when a=0Â°, and by f=1 when a=90Â°.
>Iow, L'=L*sqrt(1-v^2) when a=0Â° and L'=L when a=90Â°.
>Hence, for any angle a other than 0Â° or 90Â°, the contraction factor f
>must be situated between sqrt(1-v^2) and 1.
>
>Is the general formula giving f,
>
>1) f = sqrt(1-v^2) * cos(a),
>2) f = sqrt(1-(v*cos(a))^2), or
>3) another formula?

It is an other formula.
see http://www.ping.be/electron/mmx.htm

DE WITTE Roland
http://www.ping.be/electron
Under construction.

>
>Marcel Luttgens

### MLuttgens

Aug 6, 1999, 3:00:00 AM8/6/99
to
In article <7ocges\$r55\$1...@news1.xs4all.nl>, "Cees Roos" <cr...@xs4all.nl> wrote
:

>Date : Thu, 5 Aug 1999 19:08:18 +0200

Your diagrams and demonstration are OK!

>
>l'^2 = ( l' * cos(a) )^2 + (l' * sin(a) )^2 =
>l'^2 = ( l * cos(a') / gamma(v)) ^2 + (l * sin(a') )^2
> = l^2 - ( v * l * cos(a') )^2 (1)
>
>( l * cos(a') )^2 = l^2 * ( 1 - (sin(a') )^2 ) =
> = l^2 - ( l * sin(a') )^2
> = l^2 - ( l' * sin(a) )^2 (2)
>
>Substitute (2) in (1):
>l'^2 = l^2 - v^2 * ( l^2 - ( l' * sin(a) )^2 )
> = l^2 - ( v * l )^2 + (v * l' * sin(a) )^2
>
>l'^2 * ( 1 - ( v * sin(a) )^2 ) = l^2 * ( 1 - v^2 )
>
>l'^2 = l^2 * ( 1 - v^2 ) / ( 1 - ( v * sin(a) )^2 )
>
>l' = l * sqrt( ( 1 - v^2 ) / ( 1 - ( v * sin(a) )^2 ) )
>
>
>
>> Anyhow, there are now 2 formulae matching SR's prediction for
>> a=0Â° or a=90Â°:
>>
>Three.
>

Which is the third one?

>> 1) f = sqrt(1-(v*cos(a))^2)
>> 2) f = sqrt((1-v^2)/(1-(v*sin(a))^2))
>>
>> Which one would Wayne Throop, Tom Roberts, Steve Carlip,
>> and other SR experts chose? Or another formula?
>>
>> Marcel Luttgens
>--
>Regards, Cees Roos

I will not comment now on your derivation, because I don't
want to influence the other SR experts.
In view of the theoretical significance of such formula, their
opinion is really needed and welcome.

Marcel Luttgens

### MLuttgens

Aug 6, 1999, 3:00:00 AM8/6/99
to
In article <7oct83\$hvr\$1...@news.planetinternet.be>, "DE WITTE Roland"
<roland....@ping.be> wrote :

>Date : Thu, 5 Aug 1999 17:40:02 +0100

>
>
>MLuttgens a Ã©crit dans le message
><19990804055504...@ngol08.aol.com>...
>
>>Let's consider an interferometer's arm of length L, that makes
>>an angle a with the velocity vector.
>>
>>According to SR, the arm will contract by
>>f=sqrt(1-v^2) when a=0Â°, and by f=1 when a=90Â°.
>>Iow, L'=L*sqrt(1-v^2) when a=0Â° and L'=L when a=90Â°.
>>Hence, for any angle a other than 0Â° or 90Â°, the contraction factor f
>>must be situated between sqrt(1-v^2) and 1.
>>
>>Is the general formula giving f,
>>
>>1) f = sqrt(1-v^2) * cos(a),
>>2) f = sqrt(1-(v*cos(a))^2), or
>>3) another formula?
>
>It is an other formula.
>see http://www.ping.be/electron/mmx.htm
>

to present your derivation in the NG.
Anyhow, I don't see any difference between your formula
and that of Cees Roos.

### Cees Roos

Aug 6, 1999, 3:00:00 AM8/6/99
to

MLuttgens <mlut...@aol.com> wrote in message
news:19990806050850...@ngol02.aol.com...

> In article <7ocges\$r55\$1...@news1.xs4all.nl>, "Cees Roos" <cr...@xs4all.nl>
wrote
[snip]

> >l' = l * sqrt( ( 1 - v^2 ) / ( 1 - ( v * sin(a) )^2 ) )
> >
> >
> >
> >> Anyhow, there are now 2 formulae matching SR's prediction for
> >> a=0Â° or a=90Â°:
> >>
> >Three.
> >
>
> Which is the third one?
>
The one I gave you !-)

> >> 1) f = sqrt(1-(v*cos(a))^2)
> >> 2) f = sqrt((1-v^2)/(1-(v*sin(a))^2))
> >>
> >> Which one would Wayne Throop, Tom Roberts, Steve Carlip,
> >> and other SR experts chose? Or another formula?
> >>
> >> Marcel Luttgens
> >--
> >Regards, Cees Roos
>
> I will not comment now on your derivation, because I don't
> want to influence the other SR experts.
> In view of the theoretical significance of such formula, their
> opinion is really needed and welcome.
>
> Marcel Luttgens
--
Regards, Cees Roos

### DE WITTE Roland

Aug 6, 1999, 3:00:00 AM8/6/99
to

MLuttgens a Ã©crit dans le message
<19990806050851...@ngol02.aol.com>...

>>>Is the general formula giving f,
>>>
>>>1) f = sqrt(1-v^2) * cos(a),
>>>2) f = sqrt(1-(v*cos(a))^2), or
>>>3) another formula?
>>
>>It is an other formula.
>>see http://www.ping.be/electron/mmx.htm
>>
>
>to present your derivation in the NG.

It is difficult without geometric figures. I think that at present everybody
has Internet explorer 4 with the possibility to clic on the web-site page

>Anyhow, I don't see any difference between your formula
>and that of Cees Roos.

Yes, may be, I don't read all the messages, but often only the headers of
the threads. That because my ideas are nearly always different than the
other peoples in this newsgroup.

### Wayne Throop

Aug 9, 1999, 3:00:00 AM8/9/99
to
: mlut...@aol.com (MLuttgens)
: Anyhow, there are now 2 formulae matching SR's prediction for
: a=0Â° or a=90Â°:
: 1) f = sqrt(1-(v*cos(a))^2)

: 2) f = sqrt((1-v^2)/(1-(v*sin(a))^2))
:
: Which one would Wayne Throop, Tom Roberts, Steve Carlip,
: and other SR experts chose? Or another formula?

MLuttgens is going around pretending there's some difficulty with the
correct expression of the length of a moving rod when oriented at an
angle to motion. Note: MLuttgens has actually also pretended that
deriving this distance in terms of a rod was invalid, since in an
interferometer, there neen not be a literal rod. But the discussion is
about two endpoints and a distance between them. Whether that distance
is occupied by a rod is completely irrelevant; I state the problem in
terms of "a rod" simply for illustrative purposes. MLuttgens'
"there might not be a rod" objection is ludicrous.

In SR or LET, the x-extent of the rod is foreshortened in the
coordinate system in which the rod moves, so there is no difficulty at all.
The length is

L' = L*sqrt(cos(a)^2*(1-v^2)+sin(a)^2)
= sqrt(cos(a)^2+sin(a)^2-v^2cos(a)^2))
= sqrt(1-(v*cos(a))^2)

To say this "matches SR's prediction for 0 or 90 degrees"
is a ridiculous understatement. That is SR's predicted arm length
for any angle "a".

Note: the angle "a" is the angle in the comoving coordinate system (or
in LET, measured with a comoving protractor). If you want it as a
function of an angle "b" in the non-comoving coordinate system (or in
LET, measured with a protractor "at rest"), you substitute
a=atan(tan(b)/sqrt(1-v^2)).

Now, since we find that for all angles "a" (measured in the comoving
frame), the light bounce time along that arm (in the non-comoving frame)
is 2*L/sqrt(1-v^2). So, if we want the form in terms of "b", so
everything is from the viewpoint of the non-comoving frame, we just
substitute for all atan(tan(b)/sqrt(1-v^2)) for all "a"s in that
formula. Go ahead, MLuttgens; do that. Be sure to replace
every single "a" in 2*L/sqrt(1-v^2). What do you get?

Now... what controversy or puzzle is supposed to be lurking here?
The whole thing is simple, straightforward, and unambiguous.
MLuttgens himself did the derivation of 2*L/sqrt(1-v^2) given "a".
It's a bit tricky as such things go, but not really all that
difficult; nothing past high-school trig.

So why does he pretend there's a problem here? I dunno.
I first thought, it's idiocy. Then I thought, it's an intentional
lie, and hence malice. Now I don't know. There's just no explaining
MLuttgens' behavior logically at all.

Wayne Throop thr...@sheol.org http://sheol.org/throopw

### MLuttgens

Aug 9, 1999, 3:00:00 AM8/9/99
to
In article <9341...@sheol.org>, thr...@sheol.org (Wayne Throop) wrote :

>Date : Mon, 09 Aug 1999 00:03:29 GMT

>
>: mlut...@aol.com (MLuttgens)
>: Anyhow, there are now 2 formulae matching SR's prediction for
>: a=0Â° or a=90Â°:
>: 1) f = sqrt(1-(v*cos(a))^2)
>: 2) f = sqrt((1-v^2)/(1-(v*sin(a))^2))
>:
>: Which one would Wayne Throop, Tom Roberts, Steve Carlip,
>: and other SR experts chose? Or another formula?
>
>MLuttgens is going around pretending there's some difficulty with the
>correct expression of the length of a moving rod when oriented at an
>angle to motion. Note: MLuttgens has actually also pretended that
>deriving this distance in terms of a rod was invalid, since in an
>interferometer, there neen not be a literal rod. But the discussion is
>about two endpoints and a distance between them. Whether that distance
>is occupied by a rod is completely irrelevant; I state the problem in
>terms of "a rod" simply for illustrative purposes. MLuttgens'
>"there might not be a rod" objection is ludicrous.
>
>In SR or LET, the x-extent of the rod is foreshortened in the
>coordinate system in which the rod moves, so there is no
>difficulty at all.

So you keep claiming that the contracted projection of the
arm has a physical meaning, and can be used to calculate
the contracted length of a moving rod when oriented at an
angle to motion.
Imagine now that the arm is situated on the x-axis,
and the velocity vector makes an angle a with the x-axis?
What would be your derivation of the contraction factor f
in such case?

Thank you. About the "Now, I don't know", which is true,
just wait and see.

To recap, the relevant formulae are

1) f = sqrt(1-(v*cos(a))^2)
According to WayneThroop (see above),
"L'=L*sqrt(1-(v*cos(a))^2), which is SR's predicted
arm length for any angle "a", "a" being the angle
measured in the comoving coordinate system.

2) f = sqrt((1-v^2)/(1-(v*sin(a))^2))
(Cees Roos and De Witte)

The "whole thing" is far from being "simple, straightforward, and unambiguous".

What is the opinion of Tom Roberts, Steve Carlip and
other SR experts?

Marcel Luttgens

### Wayne Throop

Aug 9, 1999, 3:00:00 AM8/9/99
to
: mlut...@aol.com (MLuttgens)
: So you keep claiming that the contracted projection of the arm has a

: physical meaning, and can be used to calculate the contracted length
: of a moving rod when oriented at an angle to motion.

Liar. You know very well that that's not what I said,
nor is it implied by anything I've said. I can't "keep claiming"
things you mistake me for saying or implying.

The arms directional contraction has a physical meaning. The projection
onto a coordinate axis is conventional. And MLuttgens knew that,
because he'd brought up the exact same bogus objection before.

: The "whole thing" is far from being "simple, straightforward, and
: unambiguous".

Only in what passes for MLuttgens' "mind".

### Wayne Throop

Aug 10, 1999, 3:00:00 AM8/10/99
to
: mlut...@aol.com (MLuttgens)
: 1) f = sqrt(1-(v*cos(a))^2)

: According to WayneThroop (see above),
: "L'=L*sqrt(1-(v*cos(a))^2), which is SR's predicted
: arm length for any angle "a", "a" being the angle
: measured in the comoving coordinate system.
:
: 2) f = sqrt((1-v^2)/(1-(v*sin(a))^2))
: (Cees Roos and De Witte)

By the way, what makes you think those are two distinct cases?

Hint: in (1), the angle is measured in the comoving frame.
but in (2), the angle is measured in the non-comoving frame.

An issue which has been explained to MLuttgens several times,
to no noticeable effect.

: The "whole thing" is far from being "simple, straightforward,
: and unambiguous".

Sorry. Still simple. Still straightforward. Still unambiguous.

### MLuttgens

Aug 10, 1999, 3:00:00 AM8/10/99
to
NB:
WT = Wayne Throop, ML = MLuttgens

WT:

In SR or LET, the x-extent of the rod is foreshortened in the
coordinate system in which the rod moves, so there is no
difficulty at all.

ML:

So you keep claiming that the contracted projection of the
arm has a physical meaning, and can be used to calculate
the contracted length of a moving rod when oriented at an
angle to motion.

WT:

Liar. You know very well that that's not what I said,
nor is it implied by anything I've said. I can't "keep claiming"
things you mistake me for saying or implying.

The arms directional contraction has a physical meaning.
The projection onto a coordinate axis is conventional.
And MLuttgens knew that, because he'd brought up the exact
same bogus objection before.

ML(new):
It is perhaps not what you said, but it is implied by your
derivation.
Btw, does "conventional" means: "depending on or conforming
to formal or accepted standards or rules rather than nature"?
Could you elaborate a little further?

ML:

Imagine now that the arm is situated on the x-axis,
and the velocity vector makes an angle a with the x-axis?
What would be your derivation of the contraction factor f
in such case?

ML(new):
No reaction from WT to that scenario.
Does it mean that WT is unable to derive f in that case?

ML:

To recap, the relevant formulae are

1) f = sqrt(1-(v*cos(a))^2)

According to WayneThroop (see above),
"L'=L*sqrt(1-(v*cos(a))^2), which is SR's predicted
arm length for any angle "a", "a" being the angle
measured in the comoving coordinate system.

2) f = sqrt((1-v^2)/(1-(v*sin(a))^2))
(Cees Roos and De Witte)

WT:

By the way, what makes you think those are two distinct cases?
Hint: in (1), the angle is measured in the comoving frame.
but in (2), the angle is measured in the non-comoving frame.

ML(new):
The angles in the two formulae are related according to
the relation sin(a') = sin(a) / sqrt(1-(v^cos(a))^2), hence (2)

Marcel Luttgens

### Wayne Throop

Aug 10, 1999, 3:00:00 AM8/10/99
to
::: mlut...@aol.com (MLuttgens)
::: So you keep claiming that the contracted projection of the arm has a

::: physical meaning, and can be used to calculate the contracted length
::: of a moving rod when oriented at an angle to motion.

:: thr...@sheol.org (Wayne Throop)
:: You know very well that that's not what I said, nor is it implied by

:: anything I've said. I can't "keep claiming" things you mistake me
:: for saying or imply

: mlut...@aol.com (MLuttgens)
: It is perhaps not what you said, but it is implied by your derivation.

Why MLuttgens bothers lying about things like this is beyond me.
A projection is not physical, I never said it was, and I never said
anything that implies that it is.

The derivation uses coordinates; in particular, distance measures
from a standard origin. Coordinates are not themselves physical.
Therefore, the simple use of length contraction on standard
coordinates has no implication whatsoever that a projection is physical.

So what *is* physical? The contraction of physical objects in the
direction of motion. If you lay out coordinates using such physical objects,
you get a coordinate system foreshortened in the direction of travel.

And this has all been explained to MLuttgens several times already.

:: By the way, what makes you think those are two distinct cases? Hint:

:: in (1), the angle is measured in the comoving frame. but in (2), the
:: angle is measured in the non-comoving frame.

: The angles in the two formulae are related according to the relation

: sin(a') = sin(a) / sqrt(1-(v^cos(a))^2), hence (2) should read f =
: sqrt((1-v^2)/(1-(v*sin(a'))^2)).

So, you expect people to use a terminology you've just newly invented ?
And why is (a') the angle in "stationary" coordinates, while (a) is the
angle in "moving" coordinates, opposite of the usual conventions?

I explicitly said what I meant by "a"; you have no excuse to now pretend
I was being unclear in any way whatsoever, or that my terminology is
suddenly "wrong", when you knew exactly what I meant all along.

Slimy weasel.

The fact remains: the two forms are identical, once you realize (as is
obvious) that one is expressed as a function of the angle in comoving
coordinates, and the other ie expressed as a function of the angle in
non-comoving coordinates. No ammount of MLuttgens' squirming and
wriggling and thrashing about will change this simple fact.

### Tom Roberts

Aug 12, 1999, 3:00:00 AM8/12/99
to
MLuttgens wrote:
> Anyhow, there are now 2 formulae matching SR's prediction for
> a=0Â° or a=90Â°:
> 1) f = sqrt(1-(v*cos(a))^2)
> 2) f = sqrt((1-v^2)/(1-(v*sin(a))^2))

In SR it is impossible to come up with a truly meaningful formula for
this -- this is not a purely spatial situation in SR when viewed from
another frame.

Other posters have provided the SR formula for this, as asked.
But I think the entire question and underlying approach are
flawed.

SR is more than length contraction, and you are attempting to consider
just the length contraction of a rod (or distance between two points
at rest in the frame). When viewed from another frame there is an
admixture of time difference between the points, as well as a different
spatial difference. Your entire approach ignores this basic fact.

That is (viewing from another frame), as a varies not only
does the spatial distance between the endpoints vary but also
the time difference varies (for events at the endpoints of the
rod which are simultaneous in the rest frame of the rod).
Attempting to ignore this variation in time difference
frequently/usually leads you to wrong conclusions.

In general it is invalid to "take a shortcut" and just consider
length contraction; you must use the full Lorentz transform, and
you must consider specific events, not merely endpoints of a rod --
the latter includes too much ambiguity in other frames. Ditto for
time dilation.

This _does_ depends upon how you use the result. In some specific
cases it may be valid to use the results given by other posters.
But in general it is easy to get into trouble with shortcuts like
this....

If, say, you want to analyze the MMX in SR, then the best
approach is to use the invariance of c, and avoid any
transforms at all! Such invariance principles are _MUCH_
more powerful than slogging through the details of the
coordinate transforms. Note that length contraction alone
cannot give that invariance, nor can length contraction
plus time dilation; it takes the full Lorentz transform
to demonstrate the invariance of c.

Exercise: compare and contrast the use of invariance
principles like this vs coordinate transforms to the use
of energy conservation vs F=ma in Newtonian mechanics.
Advanced exercise: compare and contrast to Lagrangian vs
Newtonian mechanics, and Hamiltonian vs Newtonian mechanics.

Tom Roberts tjro...@lucent.com

### MLuttgens

Aug 13, 1999, 3:00:00 AM8/13/99
to
In article <37B2DEFF...@lucent.com>, Tom Roberts <tjro...@lucent.com>
wrote :

>of a rod --the latter includes too much ambiguity in other

>frames. Ditto for time dilation.
>
>This _does_ depends upon how you use the result.
>In some specific cases it may be valid to use the results
>given by other posters.
>But in general it is easy to get into trouble with shortcuts like
>this....
>
> If, say, you want to analyze the MMX in SR, then the best
> approach is to use the invariance of c, and avoid any
> transforms at all! Such invariance principles are _MUCH_
> more powerful than slogging through the details of the
> coordinate transforms. Note that length contraction alone
> cannot give that invariance, nor can length contraction
> plus time dilation; it takes the full Lorentz transform
> to demonstrate the invariance of c.
>
> Exercise: compare and contrast the use of invariance
> principles like this vs coordinate transforms to the use
> of energy conservation vs F=ma in Newtonian mechanics.
> Advanced exercise: compare and contrast to Lagrangian vs
> Newtonian mechanics, and Hamiltonian vs Newtonian
> mechanics.
>
>
>Tom Roberts

Thank you for this interesting analysis of the problem.

I cannot pretend that I am surprised at the impossibility, in SR,
to come up with a truly meaningful formula, essentially, Imo,
because the derivation of the Lorentz transform was based
on two frames of reference in uniform relative translatory
motion and the consideration of light signals sent along
the x' and y'-axes, and certainly not of signals making some
angle with the coordinate axes.
So, I am inclined to thinking that the Lorentz transform itself
is moot, and that a more general transform, not limited to
angles of 0Â° and 90Â°, should be derived. In any case, such
an approach can only be fruitful, much more than going round
in circles for months.
The new transform would probably show that the
interferometer's arms contract according to sqrt(1-(v*cos(a))^2),
and that the angle a between arm and velocity vector
is frame independent. Let's note that a is measured in the
interferometer frame.
Otoh, in the second formula f = sqrt((1-v^2)/(1-(v*sin(a))^2)),
a is measured in the non-comoving frame (the frame at rest
in the ether). Using the present Lorentz transform, and
calling b that new angle a, one finds that cos(a)^2 =
(1-(sin(b))^2)/(1-(v*sin(b))^2). Replacing cos(a)^2 by that
value in f=sqrt(1-(v*cos(a))^2), the second formula is readily
obtained. But if a is frame independent, the second formula
is of course false.

As an approach to derive a general Lorentz transform,
one could first consider an arm of length L making an angle
of 45Â° with the velocity vector v situated on the x_axis. By
following the same steps as Lorentz in his demonstration,
one would obtain
gamma=1/sqrt(1-(v*cos(45))^2)
x'=gamma(x-v*cos(45)*t) y'=y
t'=gamma(t-v*cos(45)*x)
But one could as well imagine that the velocity vector is situated
on the y-axis, and obtain a perfect symmetrical result:
gamma=1/sqrt(1-(v*sin(45))^2)
y'=gamma(y-v*sin(45)*t) x'=x
t'=gamma(t-v*sin(45)*y)
Hence, in vue of that symmetry, it is impossible for the angle a
to change, because its anti-clockwise increase (velocity vector
along x) is exactly compensated by its clockwise decrease
(vector along y).

Marcel Luttgens

### Paul B. Andersen

Aug 13, 1999, 3:00:00 AM8/13/99
to MLuttgens
MLuttgens wrote:
>
> To recap, the relevant formulae are
>
> 1) f = sqrt(1-(v*cos(a))^2)
> According to WayneThroop (see above),
> "L'=L*sqrt(1-(v*cos(a))^2), which is SR's predicted
> arm length for any angle "a", "a" being the angle
> measured in the comoving coordinate system.

Correct.

> 2) f = sqrt((1-v^2)/(1-(v*sin(a))^2))
> (Cees Roos and De Witte)

Which is also correct, but the angle a in this equation
is the angle of the _moving_ rod in the "ether frame",
e.g. in the frame where the rod is moving with the speed v.

If we rename this angle to a', we will have the following
relationship between these angles:

sin(a') = sin(a)/sqrt(1-(v*cos(a))^2)
a = angle in the rod frame
a'= angle in the "ether frame"

if you insert this in 2) you will get 1) above.

> The "whole thing" is far from being "simple, straightforward,
> and unambiguous".

However, this length is not very significant regarding the MMX.

In the frame where the rod is moving, the important question
is the length the light have to go forth and back the rod,
and as the rod is moving, it should be quite obvious that
this length is not twice the length of the rod.

In fact, it is according to SR: LP = 2*L/sqrt(1-v^2)

(I am to lazy to show the calculations right now,
but I probably will if provoked. :-) )

Since it is independent of the angle of the rod, SR predicts
a null result for the MMX for any angle between the two rods,
not only pi/2.

Paul

### Paul B. Andersen

Aug 13, 1999, 3:00:00 AM8/13/99
to

Utter nonsense. :-)
Of course it does not matter in which direction the light
or anything else moves.

> So, I am inclined to thinking that the Lorentz transform itself
> is moot, and that a more general transform, not limited to
> angles of 0Â° and 90Â°, should be derived. In any case, such
> an approach can only be fruitful, much more than going round
> in circles for months.

You are right about the circles.
You see problems where none are.
From where have you got the strange idea that there
is a problem with the LT if the light goes in other
directions than along the axes?

> The new transform would probably show that the
> interferometer's arms contract according to sqrt(1-(v*cos(a))^2),
> and that the angle a between arm and velocity vector
> is frame independent. Let's note that a is measured in the
> interferometer frame.

But why would you want this angle to be frame independent?

> Otoh, in the second formula f = sqrt((1-v^2)/(1-(v*sin(a))^2)),
> a is measured in the non-comoving frame (the frame at rest
> in the ether).

Right.

> Using the present Lorentz transform, and
> calling b that new angle a, one finds that cos(a)^2 =
> (1-(sin(b))^2)/(1-(v*sin(b))^2).

Right.
cos(a) = cos(b)/sqrt(1-(v*sin(b))^2)
and
sin(b) = sin(a)/sqrt(1-(v*sin(a))^2)

> Replacing cos(a)^2 by that
> value in f=sqrt(1-(v*cos(a))^2), the second formula is readily
> obtained.

Right.

> But if a is frame independent, the second formula
> is of course false.

But the angle is frame dependent and the second
formula is right as is the first.
So what's the problem?

> As an approach to derive a general Lorentz transform,
> one could first consider an arm of length L making an angle
> of 45Â° with the velocity vector v situated on the x_axis. By
> following the same steps as Lorentz in his demonstration,
> one would obtain
> gamma=1/sqrt(1-(v*cos(45))^2)
> x'=gamma(x-v*cos(45)*t) y'=y
> t'=gamma(t-v*cos(45)*x)
> But one could as well imagine that the velocity vector is situated
> on the y-axis, and obtain a perfect symmetrical result:
> gamma=1/sqrt(1-(v*sin(45))^2)
> y'=gamma(y-v*sin(45)*t) x'=x
> t'=gamma(t-v*sin(45)*y)
> Hence, in vue of that symmetry, it is impossible for the angle a
> to change, because its anti-clockwise increase (velocity vector
> along x) is exactly compensated by its clockwise decrease
> (vector along y).
>
> Marcel Luttgens

Tell me, which problem are you seeing with the Lorentz transform
since you think a new should be derived?

L' = L*sqrt(1-(v*cos(a))^2) - a angle in rod frame
L' = L*sqrt((1-v^2)/(1-(v*sin(a))^2)) - a angle in "ether frame"

The answers are identical. So what's the problem?

Is it that you have not realized that the rod length
Ã¬n the ether frame is not very significant for the MMX?

Paul

### MLuttgens

Aug 14, 1999, 3:00:00 AM8/14/99
to
In article <37B4635F...@hia.no>, "Paul B. Andersen"
<paul.b....@hia.no> wrote :

It does matter for the MMX analysis.

>> So, I am inclined to thinking that the Lorentz transform itself
>> is moot, and that a more general transform, not limited to
>> angles of 0Â° and 90Â°, should be derived. In any case, such
>> an approach can only be fruitful, much more than going round
>> in circles for months.
>
>You are right about the circles.
>You see problems where none are.
>From where have you got the strange idea that there
>is a problem with the LT if the light goes in other
>directions than along the axes?
>

Why don't you try to derive the general LT first?

>> The new transform would probably show that the
>> interferometer's arms contract according to sqrt(1-(v*cos(a))^2),
>> and that the angle a between arm and velocity vector
>> is frame independent. Let's note that a is measured in the
>> interferometer frame.
>
>But why would you want this angle to be frame independent?
>

It is fundamental for the correct interpretation of the MMX.

>> Otoh, in the second formula f = sqrt((1-v^2)/(1-(v*sin(a))^2)),
>> a is measured in the non-comoving frame (the frame at rest
>> in the ether).
>
>Right.
>
>> Using the present Lorentz transform, and
>> calling b that new angle a, one finds that cos(a)^2 =
>> (1-(sin(b))^2)/(1-(v*sin(b))^2).
>
>Right.
> cos(a) = cos(b)/sqrt(1-(v*sin(b))^2)
>and
> sin(b) = sin(a)/sqrt(1-(v*sin(a))^2)
>
>> Replacing cos(a)^2 by that
>> value in f=sqrt(1-(v*cos(a))^2), the second formula is readily
>> obtained.
>
>Right.
>
>> But if a is frame independent, the second formula
>> is of course false.
>
>But the angle is frame dependent and the second
>formula is right as is the first.
>So what's the problem?
>

How would you derive the LT if the arm is on the x-axis,
and the velocity vector made an angle of 45Â° with the arm?
Would you still find that a is frame dependent?
You are just asserting without really knowing.

>> As an approach to derive a general Lorentz transform,
>> one could first consider an arm of length L making an angle
>> of 45Â° with the velocity vector v situated on the x_axis. By
>> following the same steps as Lorentz in his demonstration,
>> one would obtain
>> gamma=1/sqrt(1-(v*cos(45))^2)
>> x'=gamma(x-v*cos(45)*t) y'=y
>> t'=gamma(t-v*cos(45)*x)
>> But one could as well imagine that the velocity vector is situated
>> on the y-axis, and obtain a perfect symmetrical result:
>> gamma=1/sqrt(1-(v*sin(45))^2)
>> y'=gamma(y-v*sin(45)*t) x'=x
>> t'=gamma(t-v*sin(45)*y)
>> Hence, in vue of that symmetry, it is impossible for the angle a
>> to change, because its anti-clockwise increase (velocity vector
>> along x) is exactly compensated by its clockwise decrease
>> (vector along y).
>>
>> Marcel Luttgens
>
>Tell me, which problem are you seeing with the Lorentz transform
>since you think a new should be derived?
>

The problem is that the present LT is not general

>L' = L*sqrt(1-(v*cos(a))^2) - a angle in rod frame
>L' = L*sqrt((1-v^2)/(1-(v*sin(a))^2)) - a angle in "ether frame"
>
>The answers are identical. So what's the problem?
>

They are identical only if one assumes length contraction.
Or such contraction has never been experimentally observed.
Without length contraction, the correct formula is the first one,
where f=sqrt(1-(v*cos(a))^2) is a time slowing factor.

>Is it that you have not realized that the rod length
>Ã¬n the ether frame is not very significant for the MMX?
>
>Paul

Lucky are those who blindly believe!

Marcel Luttgens

### Wayne Throop

Aug 15, 1999, 3:00:00 AM8/15/99
to
: "Paul B. Andersen" <paul.b....@hia.no>
: However, this length is not very significant regarding the MMX.

:
: In the frame where the rod is moving, the important question
: is the length the light have to go forth and back the rod,
: and as the rod is moving, it should be quite obvious that
: this length is not twice the length of the rod.
:
: In fact, it is according to SR: LP = 2*L/sqrt(1-v^2)
:
: (I am to lazy to show the calculations right now,
: but I probably will if provoked. :-) )

It's been done. With no detectable effect on MLuttgens' strange claims.
No, wait: there *was* one effect: he then claimed that length contraction
changed both x and y displacements between the endpoints; implicitly,
in such a way as to keep the angle a=a' for all orientations. Sigh.

### Wayne Throop

Aug 15, 1999, 3:00:00 AM8/15/99
to
: Tom Roberts
: If, say, you want to analyze the MMX in SR, then the best approach is

: to use the invariance of c, and avoid any transforms at all! Such
: invariance principles are _MUCH_ more powerful than slogging through
: the details of the coordinate transforms. Note that length
: contraction alone cannot give that invariance, nor can length
: contraction plus time dilation; it takes the full Lorentz transform to
: demonstrate the invariance of c.

Of course; I agree fully. Use of SR's invariants is just a
good-old-fashioned-simpler way of looking at things.

But the point here is not to analyze MM. It's to convince MLuttgens
that the invariants actually work correctly; that it really does not
matter which frame you do the calculation in, you get the same answer.
Therefore, we know, immediately, that there are exactly zero fringe
shifts in the gadget's rest frame. The task is then to show MLuttgens
the tedious-but-basically-simple fact of the matter: you also get zero
frings shifts in ALL OTHER frames as well.

So how did "length contraction alone" get into it? Well, that's because
MLuttgens claims that "time dilation alone" can account for MM (as opposed
to other experiments), which is clearly incorrect. It's a historical
artifact of the discussion path. Neither one gives invariance, but
1: time dilation cannot ex plain MM, and 2: length contraction can.

And finally, MLuttgens has some strange notion of length contraction
which is not LET nor SR length contraction: the angle "a" between a
coordinate axis and a line containing two points is invariant across
a boost in MLuttgens' bizzaro-world version.

### Wayne Throop

Aug 15, 1999, 3:00:00 AM8/15/99
to
:: Of course it does not matter in which direction the light
:: or anything else moves.

: mlut...@aol.com (MLuttgens)
: It does matter for the MMX analysis.

Wrong, of course; demonstrably so.
The round trip times are equal no matter the direction.

: I cannot pretend that I am surprised at the impossibility, in SR, to

: come up with a truly meaningful formula,

The SR formula is perfectly meaningful.
Just because MLuttgens pretends it has no meaning
is not a flaw in SR's predicted results.

### Paul B. Andersen

Aug 16, 1999, 3:00:00 AM8/16/99
to
MLuttgens wrote:
>
> In article <37B4635F...@hia.no>, "Paul B. Andersen"
> <paul.b....@hia.no> wrote :
>
> >
> >MLuttgens wrote:
> >>
> >> I cannot pretend that I am surprised at the impossibility, in SR,
> >> to come up with a truly meaningful formula, essentially, Imo,
> >> because the derivation of the Lorentz transform was based
> >> on two frames of reference in uniform relative translatory
> >> motion and the consideration of light signals sent along
> >> the x' and y'-axes, and certainly not of signals making some
> >> angle with the coordinate axes.
> >
> >Utter nonsense. :-)
> >Of course it does not matter in which direction the light
> >or anything else moves.
> >
>
> It does matter for the MMX analysis.

The point is that the LT obviously can handle light
going in any direction.

> >> So, I am inclined to thinking that the Lorentz transform itself
> >> is moot, and that a more general transform, not limited to
> >> angles of 0Â° and 90Â°, should be derived. In any case, such
> >> an approach can only be fruitful, much more than going round
> >> in circles for months.
> >
> >You are right about the circles.
> >You see problems where none are.
> >From where have you got the strange idea that there
> >is a problem with the LT if the light goes in other
> >directions than along the axes?
> >
>
> Why don't you try to derive the general LT first?

What do you mean by that?
Do you mean with the boost in a general direction
not along the x-axis?
Of course you can express the LT with the boost in a general
direction. I don't have to derive it. It's done.

> >> The new transform would probably show that the
> >> interferometer's arms contract according to sqrt(1-(v*cos(a))^2),
> >> and that the angle a between arm and velocity vector
> >> is frame independent. Let's note that a is measured in the
> >> interferometer frame.
> >
> >But why would you want this angle to be frame independent?
> >
>
> It is fundamental for the correct interpretation of the MMX.

Indeed.
The frame dependence of this angle is due to the velocity dependent
shortening of the rod which is necessary to explain the MMX.

> >> Otoh, in the second formula f = sqrt((1-v^2)/(1-(v*sin(a))^2)),
> >> a is measured in the non-comoving frame (the frame at rest
> >> in the ether).
> >
> >Right.
> >
> >> Using the present Lorentz transform, and
> >> calling b that new angle a, one finds that cos(a)^2 =
> >> (1-(sin(b))^2)/(1-(v*sin(b))^2).
> >
> >Right.
> > cos(a) = cos(b)/sqrt(1-(v*sin(b))^2)
> >and
> > sin(b) = sin(a)/sqrt(1-(v*sin(a))^2)
> >
> >> Replacing cos(a)^2 by that
> >> value in f=sqrt(1-(v*cos(a))^2), the second formula is readily
> >> obtained.
> >
> >Right.
> >
> >> But if a is frame independent, the second formula
> >> is of course false.
> >
> >But the angle is frame dependent and the second
> >formula is right as is the first.
> >So what's the problem?
> >
>
> How would you derive the LT if the arm is on the x-axis,
> and the velocity vector made an angle of 45Â° with the arm?

Use the LT with the boost in that direction, of course.

> Would you still find that a is frame dependent?

Yes, the angle of the rod is frame dependent. Period.
However, the angle will be the same for all frames which
are moving in such a way that the angle between the rod and
the velocity vector happens to be 0 or pi/2.
So what?

> You are just asserting without really knowing.

Asserting what without knowing?

> >> As an approach to derive a general Lorentz transform,
> >> one could first consider an arm of length L making an angle
> >> of 45Â° with the velocity vector v situated on the x_axis. By
> >> following the same steps as Lorentz in his demonstration,
> >> one would obtain
> >> gamma=1/sqrt(1-(v*cos(45))^2)
> >> x'=gamma(x-v*cos(45)*t) y'=y
> >> t'=gamma(t-v*cos(45)*x)
> >> But one could as well imagine that the velocity vector is situated
> >> on the y-axis, and obtain a perfect symmetrical result:
> >> gamma=1/sqrt(1-(v*sin(45))^2)
> >> y'=gamma(y-v*sin(45)*t) x'=x
> >> t'=gamma(t-v*sin(45)*y)
> >> Hence, in vue of that symmetry, it is impossible for the angle a
> >> to change, because its anti-clockwise increase (velocity vector
> >> along x) is exactly compensated by its clockwise decrease
> >> (vector along y).
> >>
> >> Marcel Luttgens
> >
> >Tell me, which problem are you seeing with the Lorentz transform
> >since you think a new should be derived?
> >
>

> The problem is that the present LT is not general.

If you by "the present LT" mean the expression for the LT written
in it's most common form with the boost in the x-direction,
so no - _that form_ is not general. Obviously.
But the LT as such is.

> >L' = L*sqrt(1-(v*cos(a))^2) - a angle in rod frame
> >L' = L*sqrt((1-v^2)/(1-(v*sin(a))^2)) - a angle in "ether frame"
> >
> >The answers are identical. So what's the problem?
> >
>
> They are identical only if one assumes length contraction.
> Or such contraction has never been experimentally observed.

But it's consequences are.

> Without length contraction, the correct formula is the first one,
> where f=sqrt(1-(v*cos(a))^2) is a time slowing factor.

Ah. That nonsense again.
There is only in a fantasy world that coinciding events are
not coinciding in every frame of reference.
That is quite impossible in the real world.

> >Is it that you have not realized that the rod length
> >Ã¬n the ether frame is not very significant for the MMX?
> >
> >Paul
>
> Lucky are those who blindly believe!

Indeed.
Blindly believing that the MMX somehow can be explained
by a _direction dependent_ time dilation is -
well - a blind belief.
It would vaporize the moment you open your eyes.

BTW, I cannot understand where you are heading with all
You do understand that the LT explains the MMX just fine.
Or don't you?

Paul

### MLuttgens

Aug 17, 1999, 3:00:00 AM8/17/99
to
In article <37B7BC24...@hia.no>, "Paul B. Andersen"
<paul.b....@hia.no> wrote :

MLuttgens:
The angle a between arm and velocity vector, measured in the
interferometer frame, is frame independent.
P.B. Andersen:
The angle of the rod is frame dependent. Period.

However, the angle will be the same for all frames which
are moving in such a way that the angle between the rod and
the velocity vector happens to be 0 or pi/2.

L' = L*sqrt(1-(v*cos(a))^2) - a angle in rod frame
L' = L*sqrt((1-v^2)/(1-(v*sin(a))^2)) - a angle in "ether frame"
The answers are identical. So what's the problem?

MLuttgens:

They are identical only if one assumes length contraction.
Or such contraction has never been experimentally observed.

P.B. Andersen:

But it's consequences are.

MLuttgens:

Without length contraction, the correct formula is the first one,
where f=sqrt(1-(v*cos(a))^2) is a time slowing factor.

P.B. Andersen

Ah. That nonsense again.
There is only in a fantasy world that coinciding events are
not coinciding in every frame of reference.
That is quite impossible in the real world.

BTW, I cannot understand where you are heading with all
You do understand that the LT explains the MMX just fine.
Or don't you?

MLuttgens (new):

The LT does not explain the MMX fine.

I claim

- that the negative result of the MMX can only mean
that light took the same time to travel (round-trip) the two legs,

- that a simple geometrical analysis shows that
t(perpendicular) = (2L/c) / sqrt(1 - (v/c)^2), and
t(parallel) = (2L/c) / (1 - (v/c)^2),

- that those two round-trip times can only be equal if the time
along the parallel leg is slowed by a factor sqrt(1 - (v/c)^2),
and becomes
t(parallel) = [(2L/c) / (1 - (v/c)^2)] * sqrt(1 - (v/c)^2), thus
t(parallel) = (2L/c) / sqrt(1 - (v/c)^2),
= t(perpendicular),
meaning that clock slowing in the direction parallel to motion
suffices to explain the "null" result of the MMX,

- that clock slowing is a well attested phenomenon (GPS,
Haefele and Keating, etc...), whereas physical length contraction
has never been experimentally observed,

- that the fact that the formula giving t(parallel) can be written
t(parallel) = (2L*sqrt(1-(v/c)^2) / c[1 - (v/c)^2]) is not a proof of
length contraction, but must be considered as a mere
mathematical "artefact".

- that, consequently, the correct LT is

x' = x y' = y
t' = gamma * (t - vx / c^2), with gamma = 1/sqrt(1 - (v/c)^2),

and the "general" LT is, assuming that a is the angle
between arm and velocity vector, measured in the interferometer
frame, and v*cos(a) is the relevant velocity:

x' = x y' = y
t' = gamma * (t - v*cos(a)*x / c^2),
with gamma = 1/sqrt(1 - (v*cos(a))^2)/c^2)

Indeed, as there is no physical length contraction, the angle a
is constant ( = frame independent), and neither y nor x are
modified.

- that a general geometrical analysis of the MMX leads to the
formula t = (2L/c*(1-v^2)) * sqrt(1 - ((v/c)*sin(a))^2), giving
the round-trip travel time of light when time slowing is not
taken into consideration, and to the formula
t = (2L/c*(1-(v/c)^2)) * sqrt(1 - ((v/c)*sin(a))^2) *
sqrt(1 - ((v/c)*cos(a))^2)
when the clock slowing factor
sqrt(1 - ((v/c)*cos(a))^2) is applied.

Note that *without* time slowing, we get
for the perpendicular arm (sin(a)=1):
t = (2L/c*(1-(v/c)^2)) * sqrt(1 - (v/c)^2)
= (2L/c) / sqrt(1 - (v/c)^2), which is the same formula
as above,
and for the parallel arm (sin(a)=0):
t = (2L/c) / (1-(v/c)^2), again the same formula as above.

But *with* time slowing, by multiplying those times by the
factor sqrt(1 - ((v/c)*cos(a))^2), we obtain
for the perpendicular arm (cos(a)=0):
t = (2L/c) / sqrt(1 - (v/c)^2),
and for the parallel arm (cos(a)=1):
t = ((2L/c) / (1-(v/c)^2)) * sqrt(1-(v/c)^2)
= (2L/c) / sqrt(1 - (v/c)^2),
thus identical times for both arms.

Btw, I dont find any pertinence to your remark

"There is only in a fantasy world that coinciding events are
not coinciding in every frame of reference".

Marcel Luttgens

### Paul B. Andersen

Aug 17, 1999, 3:00:00 AM8/17/99
to
MLuttgens wrote:
>
> The LT does not explain the MMX fine.
>
> I claim
>
> - that the negative result of the MMX can only mean
> that light took the same time to travel (round-trip) the two legs,

Right.

> - that a simple geometrical analysis shows that
> t(perpendicular) = (2L/c) / sqrt(1 - (v/c)^2), and
> t(parallel) = (2L/c) / (1 - (v/c)^2),

According to the Galilean transformation, right.

> - that those two round-trip times can only be equal if the time
> along the parallel leg is slowed by a factor sqrt(1 - (v/c)^2),
> and becomes
> t(parallel) = [(2L/c) / (1 - (v/c)^2)] * sqrt(1 - (v/c)^2), thus
> t(parallel) = (2L/c) / sqrt(1 - (v/c)^2),
> = t(perpendicular),
> meaning that clock slowing in the direction parallel to motion
> suffices to explain the "null" result of the MMX,

Still nonsense.
This inevitably leads to that coinciding events are
not coinciding in all frames.
I find it impossible to understand why it is not
obvious to you that this is impossible.

> - that clock slowing is a well attested phenomenon (GPS,

> Haefele and Keating, etc...), whereas physical length contraction
> has never been experimentally observed,

The consequences of length contraction is observed.

> - that the fact that the formula giving t(parallel) can be written
> t(parallel) = (2L*sqrt(1-(v/c)^2) / c[1 - (v/c)^2]) is not a proof of
> length contraction, but must be considered as a mere
> mathematical "artefact".

But that's what it is.
You cannot explain it without length contraction.

> - that, consequently, the correct LT is
>
> x' = x y' = y
> t' = gamma * (t - vx / c^2), with gamma = 1/sqrt(1 - (v/c)^2),

Funny claim.
That transformation predicts fringe shifts in the MMX.
Was it not the MMX you wanted to explain?

> and the "general" LT is, assuming that a is the angle
> between arm and velocity vector, measured in the interferometer
> frame, and v*cos(a) is the relevant velocity:
>
> x' = x y' = y
> t' = gamma * (t - v*cos(a)*x / c^2),
> with gamma = 1/sqrt(1 - (v*cos(a))^2)/c^2)

Even funnier. :-)
How can a co-ordinate transformation depend on
of the orientation of some rod?
The speed of light in a frame will depend on
Funny place, that dream world of yours. :-)

> Indeed, as there is no physical length contraction, the angle a
> is constant ( = frame independent), and neither y nor x are
> modified.

Right.
But your transformation will predict a number of things
which is not observed.
Like non invariant speed of light.

> - that a general geometrical analysis of the MMX leads to the
> formula t = (2L/c*(1-v^2)) * sqrt(1 - ((v/c)*sin(a))^2), giving
> the round-trip travel time of light when time slowing is not
> taken into consideration, and to the formula
> t = (2L/c*(1-(v/c)^2)) * sqrt(1 - ((v/c)*sin(a))^2) *
> sqrt(1 - ((v/c)*cos(a))^2)
> when the clock slowing factor
> sqrt(1 - ((v/c)*cos(a))^2) is applied.
>
> Note that *without* time slowing, we get
> for the perpendicular arm (sin(a)=1):
> t = (2L/c*(1-(v/c)^2)) * sqrt(1 - (v/c)^2)
> = (2L/c) / sqrt(1 - (v/c)^2), which is the same formula
> as above,
> and for the parallel arm (sin(a)=0):
> t = (2L/c) / (1-(v/c)^2), again the same formula as above.
>
> But *with* time slowing, by multiplying those times by the
> factor sqrt(1 - ((v/c)*cos(a))^2), we obtain
> for the perpendicular arm (cos(a)=0):
> t = (2L/c) / sqrt(1 - (v/c)^2),
> and for the parallel arm (cos(a)=1):
> t = ((2L/c) / (1-(v/c)^2)) * sqrt(1-(v/c)^2)
> = (2L/c) / sqrt(1 - (v/c)^2),
> thus identical times for both arms.

But it is impossible nonsense.

> Btw, I dont find any pertinence to your remark

> "There is only in a fantasy world that coinciding events are
> not coinciding in every frame of reference".

Don't you?
I think the pertinence of that remark should be blatantly
obvious. I have explained it before.
In the rod frame, the light paths are equal, and the two light
beams which are emitted as coinciding events, will hit the end
of the rods as coinciding events.
But without length contraction, the light paths in the "ether frame"
will be different. Thus the light will not hit the other end
of the rod as coinciding events. This is self contradictory nonsense,
which cannot be patched by demanding that the two light beams should
be timed by clocks running at different rates.
Which in any case is impossible nonsense.
Both light beams can in principle be timed by one single
clock at the intersection of the arms. The two light beams will
not hit this clock as coinciding events. Now you demand that
this clock shall run at two different rates - one rate for each
beam - and thus find that the beams use the same time.

I find it hard to understand how anybody can claim something
so utterly impossible and self contradicting.

I think I now am fed up with stating the obvious.

Paul

### MLuttgens

Aug 18, 1999, 3:00:00 AM8/18/99
to
In article <37B9DB97...@hia.no>, "Paul B. Andersen"
<paul.b....@hia.no> wrote :

>

>MLuttgens wrote:
>>
>> The LT does not explain the MMX fine.
>>
>> I claim
>>
>> - that the negative result of the MMX can only mean
>> that light took the same time to travel (round-trip) the two legs,
>
>Right.
>
>> - that a simple geometrical analysis shows that
>> t(perpendicular) = (2L/c) / sqrt(1 - (v/c)^2), and
>> t(parallel) = (2L/c) / (1 - (v/c)^2),
>
>According to the Galilean transformation, right.
>
>> - that those two round-trip times can only be equal if the time
>> along the parallel leg is slowed by a factor sqrt(1 - (v/c)^2),
>> and becomes
>> t(parallel) = [(2L/c) / (1 - (v/c)^2)] * sqrt(1 - (v/c)^2), thus
>> t(parallel) = (2L/c) / sqrt(1 - (v/c)^2),
>> = t(perpendicular),
>> meaning that clock slowing in the direction parallel to motion
>> suffices to explain the "null" result of the MMX,
>
>Still nonsense.
>This inevitably leads to that coinciding events are
>not coinciding in all frames.
>I find it impossible to understand why it is not
>obvious to you that this is impossible.
>

You are the one who is spouting nonsense.
Don't you realize that
1) t(parallel) = [(2L/c) / (1 - (v/c)^2)] * sqrt(1 - (v/c)^2),
which explains the negative result of the MMX is terms
of time slowing, is strictly equivalent to
2) t(parallel) = [(2L * sqrt(1 - (v/c)^2)] / c) / (1 - (v/c)^2),
which explains the null result by a contraction of the arm's
length L by the factor sqrt(1 - (v/c)^2).

Or you accept the form 2), but reject its mathematically
equivalent form 1), because, according to you, it "leads to that

coinciding events are not coinciding in all frames".

If your objection were pertinent, it *should* apply to
both forms.
Moreover, as I said many times, time slowing has been
experimentally demonstrated, but never length contraction.
As you are obviously impervious to logical thinking, I take
no further interest in such discussion.

Marcel Luttgens

### Paul B. Andersen

Aug 18, 1999, 3:00:00 AM8/18/99
to
MLuttgens wrote:
>
> In article <37B9DB97...@hia.no>, "Paul B. Andersen"
> <paul.b....@hia.no> wrote :
>
> >
> >MLuttgens wrote:
> >>
> >> The LT does not explain the MMX fine.
> >>
> >> I claim
> >>
> >> - that the negative result of the MMX can only mean
> >> that light took the same time to travel (round-trip) the two legs,
> >
> >Right.
> >
> >> - that a simple geometrical analysis shows that
> >> t(perpendicular) = (2L/c) / sqrt(1 - (v/c)^2), and
> >> t(parallel) = (2L/c) / (1 - (v/c)^2),
> >
> >According to the Galilean transformation, right.
> >
> >> - that those two round-trip times can only be equal if the time
> >> along the parallel leg is slowed by a factor sqrt(1 - (v/c)^2),
> >> and becomes
> >> t(parallel) = [(2L/c) / (1 - (v/c)^2)] * sqrt(1 - (v/c)^2), thus
> >> t(parallel) = (2L/c) / sqrt(1 - (v/c)^2),
> >> = t(perpendicular),
> >> meaning that clock slowing in the direction parallel to motion
> >> suffices to explain the "null" result of the MMX,
> >
> >Still nonsense.
> >This inevitably leads to that coinciding events are
> >not coinciding in all frames.
> >I find it impossible to understand why it is not
> >obvious to you that this is impossible.
> >
>
> You are the one who is spouting nonsense.
> Don't you realize that
> 1) t(parallel) = [(2L/c) / (1 - (v/c)^2)] * sqrt(1 - (v/c)^2),
> which explains the negative result of the MMX is terms
> of time slowing, is strictly equivalent to
> 2) t(parallel) = [(2L * sqrt(1 - (v/c)^2)] / c) / (1 - (v/c)^2),
> which explains the null result by a contraction of the arm's
> length L by the factor sqrt(1 - (v/c)^2).
>
> Or you accept the form 2), but reject its mathematically
> equivalent form 1), because, according to you, it "leads to that
> coinciding events are not coinciding in all frames".
> If your objection were pertinent, it *should* apply to
> both forms.

Of course the two forms are mathematically equivalent.
That is not the issue.

> Moreover, as I said many times, time slowing has been
> experimentally demonstrated, but never length contraction.
> As you are obviously impervious to logical thinking, I take
> no further interest in such discussion.
>
> Marcel Luttgens

To state what should be blatantly obvious to anybody
not impervious to logical thinking yet again:

Consider this:
According to you (with no rod shortening), the path lengths
in the ether frame for the light going along the two beams are:
Parallel arm: LPp = L/(1 - v^2/c^2)
Transverse arm: LPt = L/sqrt(1 - v^2/c^2)
that means that if the light fronts start at ether time t = 0,
they will get back to the intersections of the arms at the
ether times:
Parallel arm: tp = 2*(L/c)/(1 - v^2/c^2)
Transverse arm: tt = 2*(L/c)/sqrt(1 - v^2/c^2)
that means that the events are not coinciding.
When the instrument is rotated 90 degrees, the times
will be interchanged.
Hence this theory predicts fringe shifts for the MMX
when the calculations are carried out in the ether frame.

So let's time them with a clock moving along with the interferometer.
This can obviously be done with a single clock at the intersections
of the arms, where all the relevant events take place.
This clock show t' = 0 when the lightfronts starts.
Now, according to you, this single clock must run at
the rate (1 - (v/c)^2) relative to ether time to
yield the time along the parallel arm:
tp' = [(2L/c) / (1 - (v/c)^2)] * (1 - (v/c)^2) = 2L/c
while it must run at the rate sqrt(1 - (v/c)^2) to yield
the time along the transverse arm:
tp' = [(2L/c)/sqrt(1 - (v/c)^2)] * sqrt(1 - (v/c)^2) = 2L/c

How can you make this single clock run at two different rates
at the same time?

I have asked you this question a couple of times before,
but you have snipped it without comment every time.

Will you face it this time, or will you yet again snip
it without comment because I am "obviously impervious to
logical thinking"?

I expect the latter.
Prove me wrong!

Paul

### MLuttgens

Aug 18, 1999, 3:00:00 AM8/18/99
to
In article <37BAB5FE...@hia.no>, "Paul B. Andersen"
<paul.b....@hia.no> wrote :

>Date : Wed, 18 Aug 1999 14:32:46 +0100

>
>MLuttgens wrote:
>
> In article <37B9DB97...@hia.no>, "Paul B. Andersen"
> <paul.b....@hia.no> wrote :
>

[snip]

>So let's time them with a clock moving along with the interferometer.
>This can obviously be done with a single clock at the intersections
>of the arms, where all the relevant events take place.
>This clock show t' = 0 when the lightfronts starts.
>Now, according to you, this single clock must run at
>the rate (1 - (v/c)^2) relative to ether time to
>yield the time along the parallel arm:
> tp' = [(2L/c) / (1 - (v/c)^2)] * (1 - (v/c)^2) = 2L/c
>while it must run at the rate sqrt(1 - (v/c)^2) to yield
>the time along the transverse arm:
> tp' = [(2L/c)/sqrt(1 - (v/c)^2)] * sqrt(1 - (v/c)^2) = 2L/c
>
>How can you make this single clock run at two different rates
>at the same time?
>
>I have asked you this question a couple of times before,
>but you have snipped it without comment every time.
>
>Will you face it this time, or will you yet again snip
>it without comment because I am "obviously impervious to
>logical thinking"?
>
>I expect the latter.
>Prove me wrong!
>

How could the interferometer's clock run at two different rates
at the same time?
According to everybody, it slows down by sqrt(1 - (v/c)^2)
wrt a clock at rest.

If, as you are claiming, the time slowing explanation of the
MMX's negative result is invalid, so must be the length
contraction one, because it relies on the same equation.
So, how do you explain, with your moving clock, that the rod

Anyhow, suppose that 2L/c = 1 and that v = 0.6 c.
The observer at rest in the ether will see that the lightfront
travelling along the transverse arm took, according to his
clock, tt = 2L/c / sqrt(1 - (v/c)^2) = 1.25 s to come back
at the arms' intersection, against tp = 2L/c / (1 - (v/c)^2) =
1.5625 s for the the "parallel" lightfront. It is important to
note that this time difference has absolutely nothing to do
with the physical reality of the meeting of the lightfronts
at the intersection. This is precisely the origin of your
persistent error: a confusion between observation and reality.
Knowing that such time difference can only be explained by
the motion of the interferometer through the ether, the clever
observer will even calculate v from the expression sqrt (1-v^2) =
1.25 / 1.5625 = .8, and find that v = 0.6.

>Paul

Marcel Luttgens

### Paul B. Andersen

Aug 18, 1999, 3:00:00 AM8/18/99
to
MLuttgens wrote:
>
> In article <37BAB5FE...@hia.no>, "Paul B. Andersen"
> <paul.b....@hia.no> wrote :
>
> >Date : Wed, 18 Aug 1999 14:32:46 +0100
> >
> >MLuttgens wrote:
> >
> > In article <37B9DB97...@hia.no>, "Paul B. Andersen"
> > <paul.b....@hia.no> wrote :
> >
>
> [snip]
>
> >So let's time them with a clock moving along with the interferometer.
> >This can obviously be done with a single clock at the intersections
> >of the arms, where all the relevant events take place.
> >This clock show t' = 0 when the lightfronts starts.
> >Now, according to you, this single clock must run at
> >the rate (1 - (v/c)^2) relative to ether time to
> >yield the time along the parallel arm:
> > tp' = [(2L/c) / (1 - (v/c)^2)] * (1 - (v/c)^2) = 2L/c
> >while it must run at the rate sqrt(1 - (v/c)^2) to yield
> >the time along the transverse arm:
> > tt' = [(2L/c)/sqrt(1 - (v/c)^2)] * sqrt(1 - (v/c)^2) = 2L/c

> >
> >How can you make this single clock run at two different rates
> >at the same time?
> >
> >I have asked you this question a couple of times before,
> >but you have snipped it without comment every time.
> >
> >Will you face it this time, or will you yet again snip
> >it without comment because I am "obviously impervious to
> >logical thinking"?
> >
> >I expect the latter.
> >Prove me wrong!
> >
>
> How could the interferometer's clock run at two different rates
> at the same time?
> According to everybody, it slows down by sqrt(1 - (v/c)^2)
> wrt a clock at rest.

So you have changed your mind? The clock run at only one rate?
In that case the predictions of your theory are:
The light fronts hit the end of the rods at the times
in the rod frame:
tp' = [(2L/c)/(1-(v/c)^2)]*sqrt(1-(v/c)^2) = (2L/c)/sqrt(1-(v/c)^2)
tt' = [(2L/c)/sqrt(1 - (v/c)^2)] * sqrt(1 - (v/c)^2) = 2L/c
They does not hit the end of the rod as coinciding events.
So your theory predict fringe shifts.
Clock slowing alone cannot predict a null-result.

> If, as you are claiming, the time slowing explanation of the
> MMX's negative result is invalid, so must be the length
> contraction one, because it relies on the same equation.

SR has both time slowing and rod contraction.
I never said time slowing was impossible.
I said that two different slowings of a single clock
at the same time is impossible.

But it is the rod shortening that explains the MMX.
A theory with no time slowing, like Lorentz first theory
could explain it just fine.
A theory with time slowing alone can not,
as you so thoroughly if involuntarily have demonstrated.

> So, how do you explain, with your moving clock, that the rod
> shrinking analysis is contradiction free?

The simple point is that two rods which are differently
oriented relative to the velocity vector can shrink
differently, but you cannot make a single clock run
at two different rates at the same time.

If you do not see the difference between those,
you must be blind.

> Anyhow, suppose that 2L/c = 1 and that v = 0.6 c.
> The observer at rest in the ether will see that the lightfront
> travelling along the transverse arm took, according to his
> clock, tt = 2L/c / sqrt(1 - (v/c)^2) = 1.25 s to come back
> at the arms' intersection, against tp = 2L/c / (1 - (v/c)^2) =
> 1.5625 s for the the "parallel" lightfront. It is important to
> note that this time difference has absolutely nothing to do
> with the physical reality of the meeting of the lightfronts
> at the intersection.

So what are you talking about? A dream world?
I am talking about the real world.

> This is precisely the origin of your
> persistent error: a confusion between observation and reality.

If the light fronts are observed not to meet in coinciding
events, then they really don't do so.
You cannot make that an error by persistently claiming
the impossible.

> Knowing that such time difference can only be explained by
> the motion of the interferometer through the ether, the clever
> observer will even calculate v from the expression sqrt (1-v^2) =
> 1.25 / 1.5625 = .8, and find that v = 0.6.

Sure. But that does not make the events co-incide.

But all this evasive talk is rather mute now, isn't it?

You admitted above that the the moving clock and thus the time
in the interferometer frame can only run at one rate
relative to the ether time, and not at one rate for each
differently oriented rod you might have in that frame.

So it should be settled that time dilation alone cannot
predict a null result of the MMX.
Right?

Paul

### MLuttgens

Aug 19, 1999, 3:00:00 AM8/19/99
to
In article <37BB2888...@hia.no>, "Paul B. Andersen"
<paul.b....@hia.no> wrote :

>Date : Wed, 18 Aug 1999 22:41:28 +0100

I never claimed that a clock can simultaneously run at two
different rates!

>In that case the predictions of your theory are:
>The light fronts hit the end of the rods at the times
>in the rod frame:
> tp' = [(2L/c)/(1-(v/c)^2)]*sqrt(1-(v/c)^2) = (2L/c)/sqrt(1-(v/c)^2)
> tt' = [(2L/c)/sqrt(1 - (v/c)^2)] * sqrt(1 - (v/c)^2) = 2L/c
>They does not hit the end of the rod as coinciding events.
>So your theory predict fringe shifts.
>Clock slowing alone cannot predict a null-result.
>

No, my theory (?) predicts

tp' = (2L/c)/(1-(v/c)^2) *sqrt(1-(v/c)^2) = (2L/c)/sqrt(1-(v/c)^2)

tt' = (2L/c)/sqrt(1 - (v/c)^2) * 1 = (2L/c)/sqrt(1-(v/c)^2)
But I admit that the clock at the arms' intersection poses problem.

And you predict
tp' = (2L*sqrt(1-(v/c)^2) / c) / (1-(v/c)^2) = (2L/c) / sqrt(1-(v/c)^2)
tt' = (2L * 1 / c) / sqrt(1 - (v/c)^2) = (2L/c) / sqrt(1-(v/c)^2)
How simple!
But herunder, you accept that "SR has both time slowing and
rod contraction".
How do you integrate time slowing in your above formulae?
Do you just multiply tp' and tt' by sqrt(1-(v/c)^2), and obtain
2L/c in both cases?
But then, you would imply that time slowing affects the transverse
arm (supposed to be oriented at 90Â° wrt the velocity vector),
in contradiction with SR itself. And if you consider that
time slowing is limited to the parallel arm, you also face the
problem of a single clock running at two different rates at the
same time.

>> If, as you are claiming, the time slowing explanation of the

>> MMX's negative result is invalid, so must be the length
>> contraction one, because it relies on the same equation.
>
>SR has both time slowing and rod contraction.

Then SR is flawed, because both cannot exist at the same time.
You have to chose your solution, either time slowing or length
contraction.

>I never said time slowing was impossible.
>I said that two different slowings of a single clock
>at the same time is impossible.
>
>But it is the rod shortening that explains the MMX.
>A theory with no time slowing, like Lorentz first theory
>could explain it just fine.

Lorentz ignored that there *is* time slowing on rods oriented
at an angle other than 90Â° wrt the velocity vector. *You* cannot
neglect that fact.

>A theory with time slowing alone can not,
>as you so thoroughly if involuntarily have demonstrated.
>
>> So, how do you explain, with your moving clock, that the rod
>> shrinking analysis is contradiction free?
>
>The simple point is that two rods which are differently
>oriented relative to the velocity vector can shrink
>differently, but you cannot make a single clock run
>at two different rates at the same time.
>

They can shrink differently, but as "SR has both time slowing
and then, you have also the problem of the single clock.
If you consider only rod shrinking, you are at variance with SR,
and one can as well claim that you use in fact time slowing
in disguise.

[snip]

>Paul

Marcel Luttgens

### Paul B. Andersen

Aug 20, 1999, 3:00:00 AM8/20/99
to

Yes, you implicitly did.
And repeat it below.

> >In that case the predictions of your theory are:
> >The light fronts hit the end of the rods at the times
> >in the rod frame:
> > tp' = [(2L/c)/(1-(v/c)^2)]*sqrt(1-(v/c)^2) = (2L/c)/sqrt(1-(v/c)^2)
> > tt' = [(2L/c)/sqrt(1 - (v/c)^2)] * sqrt(1 - (v/c)^2) = 2L/c
> >They does not hit the end of the rod as coinciding events.
> >So your theory predict fringe shifts.
> >Clock slowing alone cannot predict a null-result.
> >
>
> No, my theory (?) predicts
> tp' = (2L/c)/(1-(v/c)^2) *sqrt(1-(v/c)^2) = (2L/c)/sqrt(1-(v/c)^2)
> tt' = (2L/c)/sqrt(1 - (v/c)^2) * 1 = (2L/c)/sqrt(1-(v/c)^2)

See?
You say the clock at the same time must run at the rate sqrt(1-(v/c)^2)
and 1 relative to ether time.

> But I admit that the clock at the arms' intersection poses problem.

So you admit that your idea is impossible, but choose to ignore it?

> And you predict
> tp' = (2L*sqrt(1-(v/c)^2) / c) / (1-(v/c)^2) = (2L/c) / sqrt(1-(v/c)^2)
> tt' = (2L * 1 / c) / sqrt(1 - (v/c)^2) = (2L/c) / sqrt(1-(v/c)^2)

Yes, but these times are in the ether frame. (tp and tt).
In the interferometer frame the times will be:
tp' = 2L/c, tt' = 2L/c

> How simple!
> But herunder, you accept that "SR has both time slowing and
> rod contraction".
> How do you integrate time slowing in your above formulae?
> Do you just multiply tp' and tt' by sqrt(1-(v/c)^2), and obtain
> 2L/c in both cases?

Of course.

> But then, you would imply that time slowing affects the transverse
> arm (supposed to be oriented at 90Â° wrt the velocity vector),
> in contradiction with SR itself.

Of course it affects both events equally.
The point is that the two events "light from transverse arm
hit end of arm" and "light from rarallel arm hit end of arm"
are coinciding. They have equal temporal co-ordinate in all frames,
but what that co-ordinate is, depend on the frame.
The temporal co-ordinate is for both events (2L/c)/sqrt(1-(v/c)^2)
in the ether frame, and 2L/c in the interferometer frame.

> And if you consider that
> time slowing is limited to the parallel arm, you also face the
> problem of a single clock running at two different rates at the
> same time.

Indeed.
That's why time slowing is not "limited to one arm".

> >> If, as you are claiming, the time slowing explanation of the
> >> MMX's negative result is invalid, so must be the length
> >> contraction one, because it relies on the same equation.
> >
> >SR has both time slowing and rod contraction.
>
> Then SR is flawed, because both cannot exist at the same time.
> You have to chose your solution, either time slowing or length
> contraction.

And why cannot both exist at the same time, pray tell?

> >I never said time slowing was impossible.
> >I said that two different slowings of a single clock
> >at the same time is impossible.
> >
> >But it is the rod shortening that explains the MMX.
> >A theory with no time slowing, like Lorentz first theory
> >could explain it just fine.
>
> Lorentz ignored that there *is* time slowing on rods oriented
> at an angle other than 90Â° wrt the velocity vector. *You* cannot
> neglect that fact.

What the heck are you talking about?
In his first "contraction only" theory, Lorentz ignored
time slowing completely. That theory (or rather hypothesis) was
devised with the sole purpose of explaining the MMX, and
time slowing is irrelevant for that purpose.
In his 1904 theory, he had time slowing as well as contraction
because that was necessary to make Maxwell's equations invariant
when transformed.
But he never had the crazy idea that this time slowing
in any way was dependent on the orientation of the arms.

> >A theory with time slowing alone can not,
> >as you so thoroughly if involuntarily have demonstrated.
> >
> >> So, how do you explain, with your moving clock, that the rod
> >> shrinking analysis is contradiction free?
> >
> >The simple point is that two rods which are differently
> >oriented relative to the velocity vector can shrink
> >differently, but you cannot make a single clock run
> >at two different rates at the same time.
> >
>
> They can shrink differently, but as "SR has both time slowing
> and rod contraction", you must add time slowing to your analysis,
> and then, you have also the problem of the single clock.

Why is it a problem that the clock has to run at single rate?

> If you consider only rod shrinking, you are at variance with SR,
> and one can as well claim that you use in fact time slowing
> in disguise.

Right.
So that's why we don't do that. We have both.

Honestly, you seem very confused.
I find it very hard to figure out exactly what your
misconceptions really are.

Paul

### MLuttgens

Aug 21, 1999, 3:00:00 AM8/21/99
to
In article <37BD13C9...@hia.no>, "Paul B. Andersen"
<paul.b....@hia.no> wrote :

>Date : Fri, 20 Aug 1999 09:37:29 +0100

The factor 1 only means that there is no time slowing along the
transverse arm.

>> But I admit that the clock at the arms' intersection poses problem.
>
>So you admit that your idea is impossible, but choose to ignore it?
>

I have a simple solution, i.e. the intersection clock is
irrelevant for the transverse case. What matters is not the arm,
but the light path.
When the light front travels some distance, the intersection
with its clock move along the x-axis, away from the light path.
Hence, what the clock still measures has no signification for
the determination of the transverse time tt'.

>> And you predict
>> tp' = (2L*sqrt(1-(v/c)^2) / c) / (1-(v/c)^2) = (2L/c) / sqrt(1-(v/c)^2)
>> tt' = (2L * 1 / c) / sqrt(1 - (v/c)^2) = (2L/c) / sqrt(1-(v/c)^2)
>
>Yes, but these times are in the ether frame. (tp and tt).

The times tp' are given by two different forms of the same
equation, and both forms give the same result, i.e.
(2L/c) / sqrt(1-(v/c)^2). Claiming that using one form justify length
contraction is no more that a semantical trick.

>In the interferometer frame the times will be:
> tp' = 2L/c, tt' = 2L/c
>

Yes, by applying the factor sqrt(1-(v/c)^2) twice !

>> How simple!
>> But herunder, you accept that "SR has both time slowing and
>> rod contraction".
>> How do you integrate time slowing in your above formulae?
>> Do you just multiply tp' and tt' by sqrt(1-(v/c)^2), and obtain
>> 2L/c in both cases?
>
>Of course.
>
>> But then, you would imply that time slowing affects the transverse
>> arm (supposed to be oriented at 90Â° wrt the velocity vector),
>> in contradiction with SR itself.
>
>Of course it affects both events equally.
>The point is that the two events "light from transverse arm

>hit end of arm" and "light from pararallel arm hit end of arm"

>are coinciding. They have equal temporal co-ordinate in all frames,
>but what that co-ordinate is, depend on the frame.
>The temporal co-ordinate is for both events (2L/c)/sqrt(1-(v/c)^2)
>in the ether frame, and 2L/c in the interferometer frame.
>
>> And if you consider that
>> time slowing is limited to the parallel arm, you also face the
>> problem of a single clock running at two different rates at the
>> same time.
>
>Indeed.
>That's why time slowing is not "limited to one arm".
>

>> >> If, as you are claiming, the time slowing explanation of the
>> >> MMX's negative result is invalid, so must be the length
>> >> contraction one, because it relies on the same equation.
>> >
>> >SR has both time slowing and rod contraction.
>>
>> Then SR is flawed, because both cannot exist at the same time.
>> You have to chose your solution, either time slowing or length
>> contraction.
>
>And why cannot both exist at the same time, pray tell?
>
>> >I never said time slowing was impossible.
>> >I said that two different slowings of a single clock
>> >at the same time is impossible.
>> >
>> >But it is the rod shortening that explains the MMX.
>> >A theory with no time slowing, like Lorentz first theory
>> >could explain it just fine.
>>
>> Lorentz ignored that there *is* time slowing on rods oriented
>> at an angle other than 90Â° wrt the velocity vector. *You* cannot
>> neglect that fact.
>
>What the heck are you talking about?
>In his first "contraction only" theory, Lorentz ignored
>time slowing completely.

That's what I meant.

>That theory (or rather hypothesis) was
>devised with the sole purpose of explaining the MMX, and
>time slowing is irrelevant for that purpose.

No, time slowing perfectly explains the MMX, you don't need
length contraction at all. As I just demonstrated above, the
intersection clock is a red herring.

>In his 1904 theory, he had time slowing as well as contraction
>because that was necessary to make Maxwell's equations invariant
>when transformed.
>But he never had the crazy idea that this time slowing
>in any way was dependent on the orientation of the arms.
>
>> >A theory with time slowing alone can not,
>> >as you so thoroughly if involuntarily have demonstrated.
>> >
>> >> So, how do you explain, with your moving clock, that the rod
>> >> shrinking analysis is contradiction free?
>> >
>> >The simple point is that two rods which are differently
>> >oriented relative to the velocity vector can shrink
>> >differently, but you cannot make a single clock run
>> >at two different rates at the same time.
>> >
>>
>> They can shrink differently, but as "SR has both time slowing
>> and rod contraction", you must add time slowing to your analysis,
>> and then, you have also the problem of the single clock.
>
>Why is it a problem that the clock has to run at single rate?
>
>> If you consider only rod shrinking, you are at variance with SR,
>> and one can as well claim that you use in fact time slowing
>> in disguise.
>
>Right.
>So that's why we don't do that. We have both.
>

Which is a physical nonsense.

>Honestly, you seem very confused.
>I find it very hard to figure out exactly what your
>misconceptions really are.
>
>Paul

Paul, your theory is false, I'm trying to convince you.

Consider an apparatus composed of two sources of
monochromatic light equidistant from a detector of interference
fringes. The length of arms 1 and 2 is L.

In the first case, the apparatus is perpendicular to the velocity
vector v. When the fronts arrive at the detector, t1 = t2 =
(L/c) / sqrt(1-(v/c)^2). Note that this result is obtained with
the help of the time slowing factor sqrt(1-(v/c)^2), that you
call arbitrarily a length contraction factor.

To get the times in the interferometer frame, you, as a smart
relativist, myltiply t1 and t2 by the time slowing factor sqrt(1-(v/c)^2,
and obtain t1' = t2' = L/c, a correct result according to SR.

In the second case, the apparatus is comoving with the velocity
vector, so, according to you, t1 = (L/(c-v)) * sqrt(1-(v/c)^2),
because of length contraction, and, similarly,
t2 = (L/(c+v)) * sqrt(1-(v/c)^2).

Now, again to get the interferometer times t1' and t2', you
multiply t1 and t2 a second time by sqrt(1-(v/c)^2), which you
now consider as a time slowing factor, and you obtain
t1' = (L/(c-v)) * (1-(v/c)^2) = (L/c) * (1+v/c), and
t2' = (L/(c+v)) * (1-(v/c)^2) = (L/c) * (1-v/c).

How do you explain that t1' and t2' are not only different from
L/c, but also that you get now interference fringes?
Do you still find SR a coherent and correct theory?

Marcel Luttgens

### Paul B. Andersen

Aug 21, 1999, 3:00:00 AM8/21/99
to

Which is ridiculous.

> >> But I admit that the clock at the arms' intersection poses problem.
> >
> >So you admit that your idea is impossible, but choose to ignore it?
> >
>
> I have a simple solution, i.e. the intersection clock is
> irrelevant for the transverse case. What matters is not the arm,
> but the light path.
> When the light front travels some distance, the intersection
> with its clock move along the x-axis, away from the light path.
> Hence, what the clock still measures has no signification for
> the determination of the transverse time tt'.

:-)
Some proposals defies logic. This is one of them.

> >> And you predict
> >> tp' = (2L*sqrt(1-(v/c)^2) / c) / (1-(v/c)^2) = (2L/c) / sqrt(1-(v/c)^2)
> >> tt' = (2L * 1 / c) / sqrt(1 - (v/c)^2) = (2L/c) / sqrt(1-(v/c)^2)
> >
> >Yes, but these times are in the ether frame. (tp and tt).
>
> The times tp' are given by two different forms of the same
> equation, and both forms give the same result, i.e.
> (2L/c) / sqrt(1-(v/c)^2). Claiming that using one form justify length
> contraction is no more that a semantical trick.

I said this is the times in the ether frame.
How can that be a semantic trick?

> >In the interferometer frame the times will be:
> > tp' = 2L/c, tt' = 2L/c
> >
>
> Yes, by applying the factor sqrt(1-(v/c)^2) twice !

Sure.
But only once as time dilation.

> >> How simple!
> >> But herunder, you accept that "SR has both time slowing and
> >> rod contraction".
> >> How do you integrate time slowing in your above formulae?
> >> Do you just multiply tp' and tt' by sqrt(1-(v/c)^2), and obtain
> >> 2L/c in both cases?
> >
> >Of course.
> >
> >> But then, you would imply that time slowing affects the transverse
> >> arm (supposed to be oriented at 90Â° wrt the velocity vector),
> >> in contradiction with SR itself.
> >
> >Of course it affects both events equally.
> >The point is that the two events "light from transverse arm
> >hit end of arm" and "light from pararallel arm hit end of arm"
> >are coinciding. They have equal temporal co-ordinate in all frames,
> >but what that co-ordinate is, depend on the frame.
> >The temporal co-ordinate is for both events (2L/c)/sqrt(1-(v/c)^2)
> >in the ether frame, and 2L/c in the interferometer frame.
> >
> >> And if you consider that
> >> time slowing is limited to the parallel arm, you also face the
> >> problem of a single clock running at two different rates at the
> >> same time.
> >
> >Indeed.
> >That's why time slowing is not "limited to one arm".

> In contradiction with SR !

I hope you are joking, but fear you are confused.

I see you keep asserting.

> >In his 1904 theory, he had time slowing as well as contraction
> >because that was necessary to make Maxwell's equations invariant
> >when transformed.
> >But he never had the crazy idea that this time slowing
> >in any way was dependent on the orientation of the arms.
> >
> >> >A theory with time slowing alone can not,
> >> >as you so thoroughly if involuntarily have demonstrated.
> >> >
> >> >> So, how do you explain, with your moving clock, that the rod
> >> >> shrinking analysis is contradiction free?
> >> >
> >> >The simple point is that two rods which are differently
> >> >oriented relative to the velocity vector can shrink
> >> >differently, but you cannot make a single clock run
> >> >at two different rates at the same time.
> >> >
> >>
> >> They can shrink differently, but as "SR has both time slowing
> >> and rod contraction", you must add time slowing to your analysis,
> >> and then, you have also the problem of the single clock.
> >
> >Why is it a problem that the clock has to run at single rate?
> >
> >> If you consider only rod shrinking, you are at variance with SR,
> >> and one can as well claim that you use in fact time slowing
> >> in disguise.
> >
> >Right.
> >So that's why we don't do that. We have both.
> >
>
> Which is a physical nonsense.

Why?

> >Honestly, you seem very confused.
> >I find it very hard to figure out exactly what your
> >misconceptions really are.
> >
> >Paul
>
> Paul, your theory is false, I'm trying to convince you.

You mean SR is false.

> Consider an apparatus composed of two sources of
> monochromatic light equidistant from a detector of interference
> fringes. The length of arms 1 and 2 is L.

I suppose you mean like this:

S -> light D light <- S
|-----------|-----------|
L L

> In the first case, the apparatus is perpendicular to the velocity
> vector v. When the fronts arrive at the detector, t1 = t2 =
> (L/c) / sqrt(1-(v/c)^2). Note that this result is obtained with
> the help of the time slowing factor sqrt(1-(v/c)^2), that you
> call arbitrarily a length contraction factor.

You do not specify your examples wery well.
So I will have to guess.
The apparatus is moving perpendicular in the ether frame.
Your times t1 and t2 for the light to go from the sourse to
the detector are measured ** in the ether frame **.

There are neither any contraction nor time dilation in this
case of course. The length of the light pathes are
simply L/sqrt(1-(v/c)^2), and since the light is going with c,
the times must be as you say.

> To get the times in the interferometer frame, you, as a smart
> relativist, myltiply t1 and t2 by the time slowing factor sqrt(1-(v/c)^2,
> and obtain t1' = t2' = L/c, a correct result according to SR.

Right.

> In the second case, the apparatus is comoving with the velocity
> vector, so, according to you, t1 = (L/(c-v)) * sqrt(1-(v/c)^2),
> because of length contraction, and, similarly,
> t2 = (L/(c+v)) * sqrt(1-(v/c)^2).

I suppose you mean that the length axis of the apparatus is
parallel with the velocity, and that the light fronts are
both emitted at ether time t = 0,
e.g. *** sumultaneously in the ether frame. ***

In that case, your equations are correct.

Note that the times are different, e.g the two events
"hit detector" are not coinciding.
Rather obvious, since the detector moves.

In this scenario the two events are not coinciding.
===================================================

> Now, again to get the interferometer times t1' and t2', you
> multiply t1 and t2 a second time by sqrt(1-(v/c)^2), which you
> now consider as a time slowing factor, and you obtain
> t1' = (L/(c-v)) * (1-(v/c)^2) = (L/c) * (1+v/c), and
> t2' = (L/(c+v)) * (1-(v/c)^2) = (L/c) * (1-v/c).

Right.

> How do you explain that t1' and t2' are not only different from
> L/c, but also that you get now interference fringes?

Because it is correct.
In the "apparatus frame", the light fronts from the sources
are emitted at the _different_ times t01' and t02', where
t01' = Lv/c^2 and t02' = -Lv/c^2. Since the pulses both
use the time L/c on the journey, they will hit the detector
at the times:
t1' = Lv/c^2 + L/c = (L/c)*(1+v/c)
t2' = -Lv/c^2 + L/c = (L/c)*(1-v/c)

Of course the events are not coinciding.
SR would have been a pretty crazy theory if the events
which are not coinciding as described in your scenario
had been coinciding when transformed to the other frame.
Right?

Events are either coinciding or they are not.
In this case, they are not.
In any frame.

> Do you still find SR a coherent and correct theory?

Indeed.

But you have demonstrated your confusion yet again.

You are ascribing your failure to understand SR to
problems with SR.
Not uncommon, but the lack of self criticism that
displays always puzzles me.

Paul

### MLuttgens

Aug 23, 1999, 3:00:00 AM8/23/99
to
In article <37BEBBF1...@hia.no>, "Paul B. Andersen"
<paul.b....@hia.no> wrote :

>Date : Sat, 21 Aug 1999 15:47:13 +0100
>
>MLuttgens wrote:

[snip]

>> I have a simple solution, etc...

>Some proposals defies logic. This is one of them.
>

Another solution that doesn't defy logic is that Michelson and
Morley were unable to detect fringe shifts, simply because they
are too small. Indeed, for arms of 11 m, a wavelength of
5.9E-5 cm, and a velocity v/c of 1E-4, we find

FRINGE SHIFTS:

Alpha= 30 9.322033689285912D-02
Alpha= 33.75 7.134776073470397D-02
Alpha= 37.5 4.825439608104603D-02
Alpha= 41.25 2.433539524511251D-02
Alpha= 45 0
Alpha= 48.75 -2.433539524511251D-02
Alpha= 52.5 -4.825439608104603D-02
Alpha= 56.25 -7.134776073470397D-02
Alpha= 60 -9.322033689285912D-02

N.B. : Alpha is the angle between one arm of the MM interferometer
and the velocity vector
The fringes are so small for angles between 30Â° and
60Â°, that it is doubtful that Michelson and Morley could
detect them.
The obvious conclusion is that length contraction is not
needed to explain their negative result.

[snip]

>> Consider an apparatus composed of two sources of
>> monochromatic light equidistant from a detector of interference
>> fringes. The length of arms 1 and 2 is L.
>
>I suppose you mean like this:
>
> S -> light D light <- S
> |-----------|-----------|
> L L
>

Yes

>> In the first case, the apparatus is perpendicular to the velocity
>> vector v. When the fronts arrive at the detector, t1 = t2 =
>> (L/c) / sqrt(1-(v/c)^2). Note that this result is obtained with
>> the help of the time slowing factor sqrt(1-(v/c)^2), that you
>> call arbitrarily a length contraction factor.
>
>You do not specify your examples wery well.
>So I will have to guess.
>The apparatus is moving perpendicular in the ether frame.
>Your times t1 and t2 for the light to go from the sourse to
>the detector are measured ** in the ether frame **.
>

You guessed correctly.

>There are neither any contraction nor time dilation in this
>case of course. The length of the light pathes are
>simply L/sqrt(1-(v/c)^2), and since the light is going with c,
>the times must be as you say.
>

In the interferometer frame, the length of the pathes are
of course L, hence L(ether frame) = L(interferometre frame) /
sqrt(1-(v/c)^2), or L(interferometer frame) = L(ether frame) *
sqrt(1-(v/c)^2), meaning that L, measured in the ether frame,
must be contracted by sqrt(1-(v/c)^2) to get the length of the
arm in the interferometer frame.

Congratulation, with the above thought experiment, you
have demonstrated that the detection of a preferred frame, i.e.
the ether, is possible with the help of a one-way interferometer.

[snip]
>Paul

Marcel Luttgens

### Paul B. Andersen

Aug 23, 1999, 3:00:00 AM8/23/99
to
MLuttgens wrote:
>
> In article <37BEBBF1...@hia.no>, "Paul B. Andersen"
> <paul.b....@hia.no> wrote :
>
> >Date : Sat, 21 Aug 1999 15:47:13 +0100
> >
> >MLuttgens wrote:
>
> [snip]
>
> >> I have a simple solution, etc...

> >Some proposals defies logic. This is one of them.
> >
>
> Another solution that doesn't defy logic is that Michelson and
> Morley were unable to detect fringe shifts, simply because they
> are too small. Indeed, for arms of 11 m, a wavelength of
> 5.9E-5 cm, and a velocity v/c of 1E-4, we find
>
> FRINGE SHIFTS:
>
> Alpha= 30 9.322033689285912D-02
> Alpha= 33.75 7.134776073470397D-02
> Alpha= 37.5 4.825439608104603D-02
> Alpha= 41.25 2.433539524511251D-02
> Alpha= 45 0
> Alpha= 48.75 -2.433539524511251D-02
> Alpha= 52.5 -4.825439608104603D-02
> Alpha= 56.25 -7.134776073470397D-02
> Alpha= 60 -9.322033689285912D-02
>
> N.B. : Alpha is the angle between one arm of the MM interferometer
> and the velocity vector
> The fringes are so small for angles between 30Â° and
> 60Â°, that it is doubtful that Michelson and Morley could
> detect them.
> The obvious conclusion is that length contraction is not
> needed to explain their negative result.

They rotated the interferometer all the way around and looked for
the greatest shifts in the position of the fringes during the rotation.
So what are those angles above supposed to mean?

Again I find it amazing that you seem to believe that the great
experimentalists Michelson and Morley were plain stupid, and
did an elementary error which nobody but you have detected
during more than a century.

>
> [snip]

>
> >> Consider an apparatus composed of two sources of
> >> monochromatic light equidistant from a detector of interference
> >> fringes. The length of arms 1 and 2 is L.
> >
> >I suppose you mean like this:
> >
> > S -> light D light <- S
> > |-----------|-----------|
> > L L
> >
>

> Yes

>
> >> In the first case, the apparatus is perpendicular to the velocity
> >> vector v. When the fronts arrive at the detector, t1 = t2 =
> >> (L/c) / sqrt(1-(v/c)^2). Note that this result is obtained with
> >> the help of the time slowing factor sqrt(1-(v/c)^2), that you
> >> call arbitrarily a length contraction factor.
> >
> >You do not specify your examples wery well.
> >So I will have to guess.
> >The apparatus is moving perpendicular in the ether frame.
> >Your times t1 and t2 for the light to go from the sourse to
> >the detector are measured ** in the ether frame **.
> >
>

> You guessed correctly.

>
> >There are neither any contraction nor time dilation in this
> >case of course. The length of the light pathes are
> >simply L/sqrt(1-(v/c)^2), and since the light is going with c,
> >the times must be as you say.
> >
>

> In the interferometer frame, the length of the pathes are
> of course L, hence L(ether frame) = L(interferometre frame) /
> sqrt(1-(v/c)^2), or L(interferometer frame) = L(ether frame) *
> sqrt(1-(v/c)^2), meaning that L, measured in the ether frame,
> must be contracted by sqrt(1-(v/c)^2) to get the length of the
> arm in the interferometer frame.

No. No. No.
There is no contraction of a transversely moving arm.
The length of the arm remains L in the ether frame.
But since the arm is moving, the path length of the light
in the ether frame is of course longer than the arm,
it is L/sqrt(1 - v^2/c^2).

Isn't it about time you get this right, after so long time
and so many postings?

> Congratulation, with the above thought experiment, you
> have demonstrated that the detection of a preferred frame, i.e.
> the ether, is possible with the help of a one-way interferometer.

So you think so?
You mean you pick an arbitrary frame, call it the "ether frame",
emit the pulses _simultaneously_ in this frame, and you can
in principle measure the speed of the apparatus relative to
that "ether frame". Sure you can.

You can detect the frame **because you emitted the pulses
simultaneously in this frame.**

Paul

### MLuttgens

Aug 24, 1999, 3:00:00 AM8/24/99
to
In article <37C1C20C...@hia.no>, "Paul B. Andersen"
<paul.b....@hia.no> wrote :

>Date : Mon, 23 Aug 1999 22:50:04 +0100

>
>MLuttgens wrote:
>>
>> In article <37BEBBF1...@hia.no>, "Paul B. Andersen"
>> <paul.b....@hia.no> wrote :
>>
>> >Date : Sat, 21 Aug 1999 15:47:13 +0100
>> >
>> >MLuttgens wrote:

[snip]

>> Another solution that doesn't defy logic is that Michelson and

>> Morley were unable to detect fringe shifts, simply because they
>> are too small.

[snip]

>They rotated the interferometer all the way around and looked for
>the greatest shifts in the position of the fringes during the rotation.
>So what are those angles above supposed to mean?
>
>Again I find it amazing that you seem to believe that the great
>experimentalists Michelson and Morley were plain stupid, and
>did an elementary error which nobody but you have detected
>during more than a century.

You are right, they should possibly have detected the shifts
if the Earth's velocity in the ether was, as they assumed,
30 km/s. But we know to-day that the Earth is moving at
about 370 km/s wrt the CMBR. Using that velocity, one calculates
the following FRINGE SHIFTS:

Arms' length: 10 cm 11 m
__________ _____ ______

Alpha= 0 .2578 28.3597
Alpha= 7.5 .2490 27.3934
Alpha= 15 .2232 24.5602
Alpha= 22.5 .1823 20.0533
Alpha= 30 .1289 14.1798
Alpha= 37.5 .0667 7.3400
Alpha= 45 0 0
Alpha= 52.5 -.0667 - 7.3400
Alpha= 60 -.1289 -14.1798
Alpha= 67.5 -.1823 -20.0533
Alpha= 75 -.2232 -24.5602
Alpha= 82.5 -.2490 -27.3934
Alpha= 90 -.2578 -28.3597

N.B. : Alpha is the angle between one arm of the MM interferometer

and the velocity vector v (=370 km/s).

It is clear that Michelson & Morley couldn't detect the shifts with
arms of 11 m. So, their "null" result has no physical meaning,
and one cannot conclude that it justifies length contraction.
If they had used much shorter arms, e.g. of 10 cm, they
normally would have observed fringe shifts, hence obtained a
positive result.

Otoh, the following thought experiment using a "one-way"
interferometer with opposite arms demonstrates that the
detection of a preferred frame, i.e. the ether, is possible, even
according to SR, as you showed yourself below.

Now you are claiming:

"You mean you pick an arbitrary frame, call it the "ether frame",
emit the pulses _simultaneously_ in this frame, and you can
in principle measure the speed of the apparatus relative to
that "ether frame". Sure you can.
You can detect the frame **because you emitted the pulses

simultaneously in this frame.** ".

Your objection could as well be used against the MMX. In fact,
it is nonsensical, because the "one-way" interferometer is
of course situated on Earth. You are forgetting that the
Earth is moving in the ether, so how could the pulses be emitted
in the ether frame? They are obviously emitted in the Earth frame,
it is silly to claim the contrary.
I repeat, if your objection made sense, it should apply to the
MMX as well, and one could indeed find "amazing that you

seem to believe that the great experimentalists Michelson

and Morley were plain stupid".

Marcel Luttgens

### Paul B. Andersen

Aug 24, 1999, 3:00:00 AM8/24/99
to
MLuttgens wrote:
>
> In article <37C1C20C...@hia.no>, "Paul B. Andersen"
> <paul.b....@hia.no> wrote :
>
> >Date : Mon, 23 Aug 1999 22:50:04 +0100
> >
> >MLuttgens wrote:
> [snip]
> >> Another solution that doesn't defy logic is that Michelson and
> >> Morley were unable to detect fringe shifts, simply because they
> >> are too small.
>
> [snip]
>
> >What are you talking about?
> >They rotated the interferometer all the way around and looked for
> >the greatest shifts in the position of the fringes during the rotation.
> >So what are those angles above supposed to mean?
> >
> >Again I find it amazing that you seem to believe that the great
> >experimentalists Michelson and Morley were plain stupid, and
> >did an elementary error which nobody but you have detected
> >during more than a century.
>
> You are right, they should possibly have detected the shifts
> if the Earth's velocity in the ether was, as they assumed,
> 30 km/s. But we know to-day that the Earth is moving at
> about 370 km/s wrt the CMBR.

In which case the fringe shifts should be much easier to
detect, of course.
That is, if the ether with properties as expected by Michelson

Oh, my dear.
Another bright guy teaching the world about the MMX,
when it is obvious that he does not even know how
a Michelson interferometer works.

I suppose the numbers which you call fringe shifts above
are the difference between the effective lengths of the arms
measured in number of wavelengths.
If your numbers are right, the difference between the maximum
and minimum is ca. 57 wavelengths. That means that as the
interferometer is rotated, 57 dark/bright fringes will move past
a point on the screen (or a point on the grid in the scope).
This is _very_ much easier to detect than a shift a mere fraction
of the interfringe distance, as expected by Michelson.
And you say it could not be detected?
From whence did you get that funny idea?

> Otoh, the following thought experiment using a "one-way"
> interferometer with opposite arms demonstrates that the
> detection of a preferred frame, i.e. the ether, is possible, even
> according to SR, as you showed yourself below.

How do you manage to read the below that way? :-)

> Now you are claiming:
> "You mean you pick an arbitrary frame, call it the "ether frame",
> emit the pulses _simultaneously_ in this frame, and you can
> in principle measure the speed of the apparatus relative to
> that "ether frame". Sure you can.
> You can detect the frame **because you emitted the pulses
> simultaneously in this frame.** ".

From which it should be clear that detection of the "true ether
frame" is impossible. You can only detect the frame in which you
have _selected_ to synchronize the emission of the pulses.
I thought it to be rather obvious that you have no way of
knowing if this arbitrarily selected frame really is
the "ether frame".

BTW, remember that the scenario where the pulses were emitted
simultaneously in what you called the "ether frame" was
specified by _you_:

|Marcel Luttgens:

|> In the second case, the apparatus is comoving with the velocity
|> vector, so, according to you, t1 = (L/(c-v)) * sqrt(1-(v/c)^2),
|> because of length contraction, and, similarly,
|> t2 = (L/(c+v)) * sqrt(1-(v/c)^2).
|

|Paul B. Andersen:

| I suppose you mean that the length axis of the apparatus is
| parallel with the velocity, and that the light fronts are
| both emitted at ether time t = 0,

| e.g. *** simultaneously in the ether frame. ***

|
| In that case, your equations are correct.

> Your objection could as well be used against the MMX. In fact,

> it is nonsensical, because the "one-way" interferometer is
> of course situated on Earth. You are forgetting that the
> Earth is moving in the ether, so how could the pulses be emitted
> in the ether frame? They are obviously emitted in the Earth frame,
> it is silly to claim the contrary.

My objection to what?
But you are quite right.
Your "one way interferometer" where the light fronts according
to your description are emitted simultaneously in the "ether frame"
is indeed not very relevant to the MMX experiment.

But you sure are a funny guy! :-)
You specify a thought experiment to which I respond,
and then you tell me how silly it is to claim that
the pulses in a Michelson interferometer are emitted
simultaneously in the "ether frame", as if anybody

But we sure can agree on it's silliness! :-)

> I repeat, if your objection made sense, it should apply to the
> MMX as well, and one could indeed find "amazing that you
> seem to believe that the great experimentalists Michelson
> and Morley were plain stupid".

I don't think I understand which "objection" you are referring to,
but we sure can agree that SR predicts null fringe shifts.

So if Michelson had believed that SR predicted fringe shifts
in the MMX experiment, he would indeed have been stupid.

But do you think he believed that?
Or do you think I believe he belived that? :-)

Paul

### MLuttgens

Aug 24, 1999, 3:00:00 AM8/24/99
to
In article <37C2A951...@hia.no>, "Paul B. Andersen"
<paul.b....@hia.no> wrote :

>Date : Tue, 24 Aug 1999 15:16:49 +0100

You could interpret the shifts in terms of arm lengths, implying
length contraction, but Michelson & Morley simply expected
that the two waves would interfere in the microscope with a
phase difference, and that the fringe pattern would be shifted
at most through 0.37 fringe.

>That means that as the
>interferometer is rotated, 57 dark/bright fringes will move past
>a point on the screen (or a point on the grid in the scope).
>This is _very_ much easier to detect than a shift a mere fraction
>of the interfringe distance, as expected by Michelson.
>And you say it could not be detected?
>From whence did you get that funny idea?
>

M&M used a microscope in order to detect the fringe
shifts, because they were expecting a mere fraction of a fringe.
They even claimed that they could detect shifts of a hundredth
of a fringe! Or the wavelength they used was about 5.9E-5 cm,
thus, to 57 fringes, corresponds a distance of about 0.03 mm,
against a distance of about 6E-6 mm for a hundredth of a
fringe. You should know that a microscope capable of detecting
0.000006 mm cannot observe 0.03 mm.

>> Otoh, the following thought experiment using a "one-way"
>> interferometer with opposite arms demonstrates that the
>> detection of a preferred frame, i.e. the ether, is possible, even
>> according to SR, as you showed yourself below.
>
>How do you manage to read the below that way? :-)
>
>> Now you are claiming:
>> "You mean you pick an arbitrary frame, call it the "ether frame",
>> emit the pulses _simultaneously_ in this frame, and you can
>> in principle measure the speed of the apparatus relative to
>> that "ether frame". Sure you can.
>> You can detect the frame **because you emitted the pulses
>> simultaneously in this frame.** ".

In fact, no pulses are emitted. I expressly said:
"Consider an apparatus composed of two sources of
monochromatic light equidistant from a detector of interference

fringes". You don't need synchronization. Just observe the
shifting of the fringe pattern when the apparatus is rotated.

>From which it should be clear that detection of the "true ether
>frame" is impossible. You can only detect the frame in which you
>have _selected_ to synchronize the emission of the pulses.
>I thought it to be rather obvious that you have no way of
>knowing if this arbitrarily selected frame really is
>the "ether frame".
>

If pulses were emitted, as you seem to prefer, they could be
synchronized in the interferometer frame, not in the ether frame
or any other "arbitrarily selected" frame, because that's
physically impossible. You are on Earth and moving at 370 km/s
in the ether. In order to synchronize the pulses in the ether frame,
you would need first to find a place at rest in the ether.

Why do you claim such a thing for the "one-way" interferometer?

>But we sure can agree on it's silliness! :-)
>
>> I repeat, if your objection made sense, it should apply to the
>> MMX as well, and one could indeed find "amazing that you
>> seem to believe that the great experimentalists Michelson
>> and Morley were plain stupid".
>
>I don't think I understand which "objection" you are referring to,
>but we sure can agree that SR predicts null fringe shifts.
>

Of course, because SR is precisely based on the negative
result of the MMX.

>So if Michelson had believed that SR predicted fringe shifts
>in the MMX experiment, he would indeed have been stupid.
>
>But do you think he believed that?

>Or do you think I believe he believed that? :-)

Only an experiment can settle the problem.
The easiest way is to repeat the MMX, but with arms no
more than one meter long.

>
>Paul

Marcel Luttgens

### Wayne Throop

Aug 24, 1999, 3:00:00 AM8/24/99
to
:: MLuttgens
:: Congratulation, with the above thought experiment, you have

:: demonstrated that the detection of a preferred frame, i.e. the
:: ether, is possible with the help of a one-way interferometer.

: "Paul B. Andersen" <paul.b....@hia.no>
: So you think so?

: You mean you pick an arbitrary frame, call it the "ether frame",
: emit the pulses _simultaneously_ in this frame, and you can
: in principle measure the speed of the apparatus relative to
: that "ether frame". Sure you can.
:
: You can detect the frame **because you emitted the pulses
: simultaneously in this frame.**

Indeed, essentially ALL of MLuttgens's threads start out with
Yet Another Way of making this exact same mistake: presuming some
pair of distant events is simultaneous in what-he-calls-the-ether-frame,
and then deducing a velocity relative to that frame.

Well, DUH! It appears to be impossible to explain his error to him.

### MLuttgens

Aug 25, 1999, 3:00:00 AM8/25/99
to
In article <9355...@sheol.org>, thr...@sheol.org (Wayne Throop) wrote :

>Date : Tue, 24 Aug 1999 23:19:12 GMT

We were speaking of to experiments:

1) The MMX, who has been initially analysed by its authors
in terms of round-trips of light along the two arms of their
apparatus. In the case of the arm parallel to the velocity
vector v assumed to be 30 km/s, they stated that the wave
moving outwards has a velocity c-v relative to the apparatus,
and that on the return trip, the wave has a velocity c+v,
hence that the round-trip time required by that wave is
t2=(2L/c)/(1-(v/c)^2).
Otoh, the wave moving transversely, according to M&M,
travels along the hypotenuse of a right-angled triangle
having a side of length L, hence the corresponding
round-trip time t1=(2L/c)/sqrt(1-(v/c)^2).
From the time difference t2-t1, and knowing L, v (assumed)
and the wavelength of the used light, they calculated a
maximum shift of about one third of a fringe.
Nowhere in their analysis has the issue of the simultaneity
of emitted pulses been raised, and nobody on this NG ever
objected to their omission.

2) A proposed thought experiment with a "one-way"
interferometer, using two opposite arms instead of
perpendicular ones like in the MMX, two sources of
monochromatic light (easy to obtain with appropriate
filters) and a detector of interferences situated at equal
distances from the light sources.
The analysis of that experiment exactly follows that of
M&M, and demonstrates that the velocity of the apparatus
in the ether can be detected, even when assuming length
contraction of the arms when they are parallel to the velocity
vector.
Why would somebody want to introduce the notion of
simultaneity in this analysis, identical in its principle to
that of Michelson and Morley, if not for obfuscating
the issue?

It would be easy to repeat the MMX, but with arms no more
than one meter long, because the Earth velocity in the ether
is not 30 km/s, but about 370 km/s, and to perform an
experiment with a "one-way" interferometer, also with very
short arms, so that everybody could satisfy himself of the
validity of SR. That would be much better than endlessly
quibbling.

Marcel Luttgens

### Paul B. Andersen

Aug 25, 1999, 3:00:00 AM8/25/99
to

When I said "effective length", I meant the length measured
in number of wavelength, as is the normal meaning of the expression.
If the length remains the some but the wavelength is shortened,
then the "effective length" would be longer.

> >That means that as the
> >interferometer is rotated, 57 dark/bright fringes will move past
> >a point on the screen (or a point on the grid in the scope).
> >This is _very_ much easier to detect than a shift a mere fraction
> >of the interfringe distance, as expected by Michelson.
> >And you say it could not be detected?
> >From whence did you get that funny idea?
> >
>
> M&M used a microscope in order to detect the fringe
> shifts, because they were expecting a mere fraction of a fringe.
> They even claimed that they could detect shifts of a hundredth
> of a fringe! Or the wavelength they used was about 5.9E-5 cm,
> thus, to 57 fringes, corresponds a distance of about 0.03 mm,
> against a distance of about 6E-6 mm for a hundredth of a
> fringe. You should know that a microscope capable of detecting
> 0.000006 mm cannot observe 0.03 mm.

Which illustrates that you do not know how an interferometer
works. The point with an interferometer, and what has given
it it's name, is that two light beams combine in an interference
pattern. This pattern can be projected onto a screen, or viewed
directly by placing the eye in the position of the screen.
In the latter case a telescope or microscope (same thing, both
names are used in this case) is commonly used. The exact shape
of the interference pattern depend on the exact geometry of the
interferometer, but it is often a "bull's eye" pattern of dark
and bright fringes. If the phase of one of the light beams
changes relative to the other, the fringes will move.
If the phase changes by pi, equivalent to a change
of half a wavelength of the "effective length" of one of
light paths, the dark (say) dot in the middle will be bright,
the fist bright fringe will be dark, etc. - all the fringes
will move outwards (or inwards) by one half "interfringe distance".
The point is that the distance between two dark/bright fringes
is much bigger than the wavelength - it can be in the order of mm,
dependent on the exact geometry of the interferometer.
That is the point with the interferometer, a very small change
in in the light path of one of the beams (< 1um) will give a much
bigger change in the position of the fringes.

I can assure you that if the 57 fringes had rolled outwards,
passing by a point on the grid in their telescope as the
interferometer was rotated, then it would have been very
apparent and impossible to miss.

Saying that M&M could not detect the fringe shifts because
they had turned up the magnification in their scope to
such a degree that they did not see the picture is
plain nonsense.

> >> Otoh, the following thought experiment using a "one-way"
> >> interferometer with opposite arms demonstrates that the
> >> detection of a preferred frame, i.e. the ether, is possible, even
> >> according to SR, as you showed yourself below.
> >
> >How do you manage to read the below that way? :-)
> >
> >> Now you are claiming:
> >> "You mean you pick an arbitrary frame, call it the "ether frame",
> >> emit the pulses _simultaneously_ in this frame, and you can
> >> in principle measure the speed of the apparatus relative to
> >> that "ether frame". Sure you can.
> >> You can detect the frame **because you emitted the pulses
> >> simultaneously in this frame.** ".
>
> In fact, no pulses are emitted. I expressly said:
> "Consider an apparatus composed of two sources of
> monochromatic light equidistant from a detector of interference
> fringes". You don't need synchronization. Just observe the
> shifting of the fringe pattern when the apparatus is rotated.

Yes, you do indeed need synchronisation ** in the thought
experiment you presented **, which is _not_ the MMX.
You have to synchronize the phase of the two sources.

> >From which it should be clear that detection of the "true ether
> >frame" is impossible. You can only detect the frame in which you
> >have _selected_ to synchronize the emission of the pulses.
> >I thought it to be rather obvious that you have no way of
> >knowing if this arbitrarily selected frame really is
> >the "ether frame".
> >
>
> If pulses were emitted, as you seem to prefer, they could be
> synchronized in the interferometer frame, not in the ether frame
> or any other "arbitrarily selected" frame, because that's
> physically impossible. You are on Earth and moving at 370 km/s
> in the ether. In order to synchronize the pulses in the ether frame,
> you would need first to find a place at rest in the ether.

Sure it is practically impossible, but it is not in principle
physically impossible to synchronize the phase of the two
sources to be equal in an arbitrary selected frame in which
the interferometer is moving.
It is _your_ thought experiment we are talking about. Remember?

Because it is correct - in principle.

Listen:
_You_ presented a thought experiment. In that thought
experiment _you_ calculated that two light fronts would
hit the detector at different times in what _you_ called
the "ether frame". I pointed out that the implication of
_your_ calculations were that the light fronts would have to
be emitted simultaneously in _your_ ether frame for _your_
calculations to be valid. Translated to continuos waves,
this means that the sources would have to be phase
synchronized in _your_ ether frame, and that the phases
would not be equal at the detector.
SR would explain this phase difference (or time difference
- same thing) by that the sources are _not_ phase synchronized
in the interferometer frame (not emitting the wave fronts
simultaneously - again the same thing).
By measuring the phase difference (or time difference),
you can calculate the state of motion of the frame
of reference in which the sources would be synchronous.

This experiment is hardly feasible with light. But with
lower frequencies, it is. Remember that a light source
is a clock - imagine the phase of the light as a spinning
hand. If we have two sources - like atomic clocks - emitting
much lower frequencies then we could do it.
Say we have two atomic clocks some distance apart, and we
know that these clocks are synchronized in the ECI-frame
(that _is_ how we synchronize clocks on the Earth).
If we have a detector midway between these clocks,
comparing the phase of a periodic signal emitted from
these clock, then we can measure the speed of the ECI-frame
relative to the surface of the Earth - or the other
way around if you prefer.

> >But we sure can agree on it's silliness! :-)
> >
> >> I repeat, if your objection made sense, it should apply to the
> >> MMX as well, and one could indeed find "amazing that you
> >> seem to believe that the great experimentalists Michelson
> >> and Morley were plain stupid".
> >
> >I don't think I understand which "objection" you are referring to,
> >but we sure can agree that SR predicts null fringe shifts.
> >
>
> Of course, because SR is precisely based on the negative
> result of the MMX.
>
> >So if Michelson had believed that SR predicted fringe shifts
> >in the MMX experiment, he would indeed have been stupid.
> >
> >But do you think he believed that?
> >Or do you think I believe he believed that? :-)
>
> Only an experiment can settle the problem.
> The easiest way is to repeat the MMX, but with arms no
> more than one meter long.

Oh, come on. Don't be such an idiot.
Or rather - don't think that the experimentalists are idiots.
The longer the arms - the easier the detection will be.
BTW, Michelson's first experiment (1881) had much shorter arms,
and he detected nothing. So he made the arms much longer
in the 1886 experiment - still detecting nothing.
The experiment has since been repeated numerous times with
various lengths of the arms - with the same result.

And if you repeat once more that the reason why nothing is
detected is that the effect is much bigger than expected
- then I will scream in anguish! :-)

Paul

### Tom Roberts

Aug 25, 1999, 3:00:00 AM8/25/99
to
MLuttgens wrote:
> 2) A proposed thought experiment with a "one-way"
> interferometer, using two opposite arms instead of
> perpendicular ones like in the MMX, two sources of

> monochromatic light (easy to obtain with appropriate
> filters) and a detector of interferences situated at equal
> distances from the light sources.

Cialdea did just that. Null result.

Cialdea, Lett. Nuovo Cimento 4#16, p821 (1972)

Note that such an experiment does not really measure any "one way"
speed of light, it merely reflects your convention for synchronizing
clocks.

I go into detail on this in my forthcoming post, "Subject:
Torr and Kolen's Experiment Cannot See the Ether". Look
for it in a day or two.

> The analysis of that experiment exactly follows that of
> M&M, and demonstrates that the velocity of the apparatus
> in the ether can be detected, even when assuming length
> contraction of the arms when they are parallel to the velocity
> vector.

No such "detection" of the ether is possible, for _any_ reasonable
ether theory.

By "reasonable ether theory" I mean one which has not already
been refuted by other experiments. That implies that in any
such theory the round-trip speed of light is isotropically c
(regardless of what it predicts for 1-way speed). That's
essentially all that is required to show that the ether is
undetectable (because the 1-way speed is inextricably
entangled with the way one synchronizes clocks).

> It would be easy to repeat the MMX, but with arms no more
> than one meter long, because the Earth velocity in the ether
> is not 30 km/s, but about 370 km/s,

Longer arms are more sensitive, not less. Your notion that dozens
of fringe shifts would be missed is ludicrous.

> and to perform an
> experiment with a "one-way" interferometer, also with very
> short arms, so that everybody could satisfy himself of the
> validity of SR.

It's been done. Cialdea used a distance of about 2 meters between his
lasers. That's a one-way path.

Tom Roberts tjro...@lucent.com

### Wayne Throop

Aug 25, 1999, 3:00:00 AM8/25/99
to
:: Indeed, essentially ALL of MLuttgens's threads start out with Yet

:: Another Way of making this exact same mistake: presuming some pair of
:: distant events is simultaneous in what-he-calls-the-ether-frame, and
:: then deducing a velocity relative to that frame.

: mlut...@aol.com (MLuttgens)
: A proposed thought experiment with a "one-way" interferometer, using

: two opposite arms instead of perpendicular ones like in the MMX, two
: sources of monochromatic light (easy to obtain with appropriate
: filters) and a detector of interferences situated at equal distances

: from the light sources. The analysis of that experiment exactly

: follows that of M&M, and demonstrates that the velocity of the
: apparatus in the ether can be detected, even when assuming length
: contraction of the arms when they are parallel to the velocity vector.

Obviously, the velocity of the ether can be detected if there is
length contraction and not time dilation. Again, "well, DUH!".

: Why would somebody want to introduce the notion of simultaneity in
: this analysis,

They would want to introduce the notion of simultaneity in order to
do the analysis correctly. When the light leaves the source is a
critical factor in when it reaches the target: small timing changes
in when the light reaches the target is what an interferometer
is senstivie to.

And, of course, in his various analyses of all the various cases
MLuttgens has brought up for the last few years, he does not "leave out"
simultaneity; he simply assumes simultaneity in the ether frame of two
remote events. He has many varients on this one mistake. Many, many variants.

### MLuttgens

Aug 26, 1999, 3:00:00 AM8/26/99
to
In article <37C3BECB...@hia.no>, "Paul B. Andersen"
<paul.b....@hia.no> wrote :

>Date : Wed, 25 Aug 1999 11:00:43 +0100

O.K.

>> >That means that as the
>> >interferometer is rotated, 57 dark/bright fringes will move past
>> >a point on the screen (or a point on the grid in the scope).
>> >This is _very_ much easier to detect than a shift a mere fraction
>> >of the interfringe distance, as expected by Michelson.
>> >And you say it could not be detected?
>> >From whence did you get that funny idea?
>> >
>>
>> M&M used a microscope in order to detect the fringe
>> shifts, because they were expecting a mere fraction of a fringe.
>> They even claimed that they could detect shifts of a hundredth
>> of a fringe! Or the wavelength they used was about 5.9E-5 cm,
>> thus, to 57 fringes, corresponds a distance of about 0.03 mm,
>> against a distance of about 6E-6 mm for a hundredth of a
>> fringe. You should know that a microscope capable of detecting
>> 0.000006 mm cannot observe 0.03 mm.
>
>Which illustrates that you do not know how an interferometer
>works.

I just assumed that they used a microscope with a
great magnification, because they expected a fringe shift
of about 0.2 microns, not one of 30 microns.

I don't see why. Do you think that after division of the beam
by the "half-silvered glass plate", and reflection by the mirrors,
the two partial waves obtained by M&M were still in phase?
Do you believe that the arms' lengths of their interferometer
were identical?
Anyhow, synchronization is NOT needed to *qualitatively*
detect the ether. Do you claim the contrary?

>> >From which it should be clear that detection of the "true ether
>> >frame" is impossible. You can only detect the frame in which you
>> >have _selected_ to synchronize the emission of the pulses.
>> >I thought it to be rather obvious that you have no way of
>> >knowing if this arbitrarily selected frame really is
>> >the "ether frame".
>> >
>>
>> If pulses were emitted, as you seem to prefer, they could be
>> synchronized in the interferometer frame, not in the ether frame
>> or any other "arbitrarily selected" frame, because that's
>> physically impossible. You are on Earth and moving at 370 km/s
>> in the ether. In order to synchronize the pulses in the ether frame,
>> you would need first to find a place at rest in the ether.
>
>Sure it is practically impossible, but it is not in principle
>physically impossible to synchronize the phase of the two
>sources to be equal in an arbitrary selected frame in which
>the interferometer is moving.
>It is _your_ thought experiment we are talking about. Remember?
>

Yes, but it is not in principle different from the MMX.

>calculations to be valid. Translated to continuous waves,

>this means that the sources would have to be phase
>synchronized in _your_ ether frame, and that the phases
>would not be equal at the detector.

Do you deny that a modification of the fringe pattern will be
seen when rotating the apparatus?
If you claim that the rotation of the apparatus will not be
accompanied by a change of pattern, you don't believe in SR.

>SR would explain this phase difference (or time difference
>- same thing) by that the sources are _not_ phase synchronized
>in the interferometer frame (not emitting the wave fronts
>simultaneously - again the same thing).
>By measuring the phase difference (or time difference),
>you can calculate the state of motion of the frame
>of reference in which the sources would be synchronous.
>

Sure.

>This experiment is hardly feasible with light.

Let's try it. Synchro is not necessary, so the experiment
is perfectly feasible.

>But with
>lower frequencies, it is. Remember that a light source
>is a clock - imagine the phase of the light as a spinning
>hand. If we have two sources - like atomic clocks - emitting
>much lower frequencies then we could do it.
>Say we have two atomic clocks some distance apart, and we
>know that these clocks are synchronized in the ECI-frame
>(that _is_ how we synchronize clocks on the Earth).
>If we have a detector midway between these clocks,
>comparing the phase of a periodic signal emitted from
>these clock, then we can measure the speed of the ECI-frame
>relative to the surface of the Earth - or the other
>way around if you prefer.
>

I prefer the other way. But I don't see why another check of
the Earth's angular velocity would be needed. Haefele and
No, let's perform the experiment with the "one-way"
interferometer using two sources of monochromatic light of
same wavelength (filters can be used), and let's observe
the fringe shifts. Do you fear that it could be positive?

Marcel Luttgens

### MLuttgens

Aug 26, 1999, 3:00:00 AM8/26/99
to
In article <9356...@sheol.org>, thr...@sheol.org (Wayne Throop) wrote :

>Date : Wed, 25 Aug 1999 21:49:22 GMT

>
>:: Indeed, essentially ALL of MLuttgens's threads start out with Yet
>:: Another Way of making this exact same mistake: presuming some pair of
>:: distant events is simultaneous in what-he-calls-the-ether-frame, and
>:: then deducing a velocity relative to that frame.
>
>: mlut...@aol.com (MLuttgens)
>: A proposed thought experiment with a "one-way" interferometer, using
>: two opposite arms instead of perpendicular ones like in the MMX, two
>: sources of monochromatic light (easy to obtain with appropriate
>: filters) and a detector of interferences situated at equal distances
>: from the light sources. The analysis of that experiment exactly
>: follows that of M&M, and demonstrates that the velocity of the
>: apparatus in the ether can be detected, even when assuming length
>: contraction of the arms when they are parallel to the velocity vector.
>
>Obviously, the velocity of the ether can be detected if there is
>length contraction and not time dilation. Again, "well, DUH!".
>

A first step of Wayne Throop in the right direction.
He just needs to grasp that time dilation doesn't modifiy the
outcome of the experiment.

>: Why would somebody want to introduce the notion of simultaneity in
>: this analysis,
>
>They would want to introduce the notion of simultaneity in order to
>do the analysis correctly. When the light leaves the source is a
>critical factor in when it reaches the target: small timing changes
>in when the light reaches the target is what an interferometer
>is senstivie to.
>

You still have not grasped that synchronization is not necessary
to detect the mere existence of the ether. The theoretically
expected modification of the fringe pattern is enough for that.
If you want to calculate the velocity of the apparatus in the
ether, then, in principle, you need simultaneity.

>And, of course, in his various analyses of all the various cases
>MLuttgens has brought up for the last few years, he does not "leave out"
>simultaneity; he simply assumes simultaneity in the ether frame of two
>remote events. He has many varients on this one mistake. Many, many
>variants.
>
>Wayne Throop

Marcel Luttgens

### MLuttgens

Aug 27, 1999, 3:00:00 AM8/27/99
to
In article <37C42AA0...@lucent.com>, Tom Roberts <tjro...@lucent.com>
wrote :

>Date : Wed, 25 Aug 1999 12:40:48 -0500
>
>MLuttgens wrote:
>> 2) A proposed thought experiment with a "one-way"

>> interferometer, using two opposite arms instead of
>> perpendicular ones like in the MMX, two sources of
>> monochromatic light (easy to obtain with appropriate
>> filters) and a detector of interferences situated at equal
>> distances from the light sources.
>

>Cialdea did just that. Null result.
>
> Cialdea, Lett. Nuovo Cimento 4#16, p821 (1972)
>

Beware, "Testis unus, testis nullus".

>Note that such an experiment does not really measure any "one way"
>speed of light, it merely reflects your convention for synchronizing
>clocks.
>
> I go into detail on this in my forthcoming post, "Subject:
> Torr and Kolen's Experiment Cannot See the Ether". Look
> for it in a day or two.
>

I read it, it is a very interesting article, dense and serious, so I need
some time to form an opinion.

>
>> The analysis of that experiment exactly follows that of
>> M&M, and demonstrates that the velocity of the apparatus
>> in the ether can be detected, even when assuming length
>> contraction of the arms when they are parallel to the velocity
>> vector.
>

>No such "detection" of the ether is possible, for _any_ reasonable
>ether theory.
>
> By "reasonable ether theory" I mean one which has not >
> been refuted by other experiments.

I understand, but Imo, only one "one-way" experiment is not enough.

> That implies that in any
> such theory the round-trip speed of light is isotropically c
> (regardless of what it predicts for 1-way speed). That's
> essentially all that is required to show that the ether is
> undetectable (because the 1-way speed is inextricably
> entangled with the way one synchronizes clocks).
>
>
>> It would be easy to repeat the MMX, but with arms no more
>> than one meter long, because the Earth velocity in the ether
>> is not 30 km/s, but about 370 km/s,
>
>Longer arms are more sensitive, not less. Your notion that dozens
>of fringe shifts would be missed is ludicrous.
>

Yes, my hypothesis is probably not very strong.

>
>> and to perform an
>> experiment with a "one-way" interferometer, also with very
>> short arms, so that everybody could satisfy himself of the
>> validity of SR.
>
>It's been done. Cialdea used a distance of about 2 meters
>between his lasers. That's a one-way path.
>

Some more experiments are badly needed:

1) Let's consider the Lorentz analysis of the MMX, supposing that
the velocity vector v is oriented East-West, and L is the length of
the arms.

"Ether" time for the transverse arm:
Tt(e) = (2L/c) * 1/sqrt(1-(v/c)^2)
Corresponding "interferometer" time, assuming time slowing,
but no length contraction:
Tt(i) = Tt(e) * sqrt(1-(v/c)^2) = 2L/c.
"Ether" time for the parallel arm:
Tp(e) = (2L/c) * 1/(1-(v/c)^2)
Corresponding "interferometer" time, assuming time slowing,
and length contraction:
Tp(i) = Tp(e) * sqrt(1-(v/c)^2) * sqrt(1-(v/c)^2) = 2L/c.
As Tt(i) = Tp(i), the phase difference is zero, and no fringe shift
is expected.

2) Let's perform a similar analysis with the "one-way"
interferometer of the thought experiment.

Case 1: The opposite arms are oriented North-South
"Ether" time for the Northern arm:
Tn(e) = (L/c) * 1/sqrt(1-(v/c)^2)
Corresponding "interferometer" time, assuming time slowing,
but no length contraction:
Tn(i) = Tn(e) * sqrt(1-(v/c)^2) = L/c.
The times for the Southern arm are of course identical, and
the time difference is zero.

Case 2: The opposite arms are oriented East-West, Iow,

they are parallel to the velocity vector.

"Ether" time for the Eastern arm:
Te(e) = L/(c-v) = (L/c) * 1/(1-(v/c))
Corresponding "interferometer" time, assuming time slowing,
and length contraction:
Te(i) = Te(e) * sqrt(1-(v/c)^2) * sqrt(1-(v/c)^2)
= Te(e) * (1-(v/c)^2)
= (L/c) * (1+(v/c))
"Ether" time for the Western arm:
Tw(e) = L/(c+v) = (L/c) * 1/(1+(v/c))
Corresponding "interferometer" time, assuming time slowing,
and length contraction:
Tw(i) = Tw(e) * sqrt(1-(v/c)^2) * sqrt(1-(v/c)^2)
= Tw(e) * (1-(v/c)^2)
= (L/c) * (1-(v/c))
In this case, the time difference is not zero. Indeed,
Te(i) - Tw(i) = (2L/c) * (v/c), hence, when rotating the apparatus,
a modification of the fringes pattern is expected.

So, SR allows to predict a negative result of the MMX, but
a positive one for the thought experiment.
It is important to note that the positive result obtained with
one-way light paths implies a detection of the "ether wind".

Otoh, if Cialdea really observed no fringe shift, his experiment
would contradict SR. Indeed, to transform Te(e) to Te(i) = L/c,
we need to apply a contraction factor, which is different
from sqrt (1-(v/c)^2):
f(e) = sqrt [(1-(v/c)) / (1+(v/c))], but not sqrt (1-(v/c)^2).
Similarly, to obtain L/c from Tw(e), the needed contraction
factor is f(w) = sqrt [(1+(v/c)) / (1-(v/c))].
For instance, Tw(i) then becomes Tw(e) * f(w) * sqrt(1-(v/c)^2)
= (L/c) * 1/(1+(v/c)) * sqrt [(1+(v/c)) / (1-(v/c)) * sqrt(1-(v/c)^2)
= L/c.

What a mess!

>
>Tom Roberts

Marcel Luttgens