Which contraction factor?
MLuttgens
an angle a with the velocity vector.
According to SR, the arm will contract by
f=sqrt(1-v^2) when a=0°, and by f=1 when a=90°.
Iow, L'=L*sqrt(1-v^2) when a=0° and L'=L when a=90°.
Hence, for any angle a other than 0° or 90°, 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
MLuttgens <mlut...@aol.com> wrote in message
news:19990804055504...@ngol08.aol.com...
> 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
:
>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=0°, and by f=1 when a=90°.
>> Iow, L'=L*sqrt(1-v^2) when a=0° and L'=L when a=90°.
>> Hence, for any angle a other than 0° or 90°, 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=0° or a=90°:
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
> >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!
>
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=0° or a=90°:
>
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
MLuttgens a écrit 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=0°, and by f=1 when a=90°.
>Iow, L'=L*sqrt(1-v^2) when a=0° and L'=L when a=90°.
>Hence, for any angle a other than 0° or 90°, 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
:
>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=0° or a=90°:
>>
>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
<roland....@ping.be> wrote :
>Date : Thu, 5 Aug 1999 17:40:02 +0100
>
>
>MLuttgens a écrit 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=0°, and by f=1 when a=90°.
>>Iow, L'=L*sqrt(1-v^2) when a=0° and L'=L when a=90°.
>>Hence, for any angle a other than 0° or 90°, 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
>
Thank you, I downloaded you page, but it would be better
to present your derivation in the NG.
Anyhow, I don't see any difference between your formula
and that of Cees Roos.
Cees Roos
> >l' = l * sqrt( ( 1 - v^2 ) / ( 1 - ( v * sin(a) )^2 ) )
> >
> >
> >
> >> Anyhow, there are now 2 formulae matching SR's prediction for
> >> a=0° or a=90°:
> >>
> >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
--
Regards, Cees Roos
DE WITTE Roland
MLuttgens a écrit dans le message
>>>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
>>
>
>Thank you, I downloaded you page, but it would be better
>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
address.
>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
: Anyhow, there are now 2 formulae matching SR's prediction for
: a=0° or a=90°:
: 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
>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=0° or a=90°:
>: 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
: 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
: 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
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)
should read f = sqrt((1-v^2)/(1-(v*sin(a'))^2)).
Marcel Luttgens
Wayne Throop
::: 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
> Anyhow, there are now 2 formulae matching SR's prediction for
> a=0° or a=90°:
> 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
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 0° and 90°, 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 45° 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
>
> 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".
What exactly do you find so difficult about this?
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
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 0° and 90°, 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 45° 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?
You have got your answers:
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
ìn the ether frame is not very significant for the MMX?
Paul
MLuttgens
<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 0° and 90°, 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 45° 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 45° 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
>You have got your answers:
>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
>ìn the ether frame is not very significant for the MMX?
>
>Paul
Lucky are those who blindly believe!
Marcel Luttgens
Wayne Throop
: 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
: 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
:: 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
>
> 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 0° and 90°, 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 45° 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 45° 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.
> >You have got your answers:
> >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
> >ìn 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
this nonsense about the LT.
You do understand that the LT explains the MMX just fine.
Or don't you?
Paul
MLuttgens
<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.
You have got your answers:
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
this nonsense about the LT.
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
>
> 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
the orientation of your rod.
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
<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
>
> In article <37B9DB97...@hia.no>, "Paul B. Andersen"
> <paul.b....@hia.no> 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.
> >
>
> 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.
The point is that your theory is self contradictory.
> 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
<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
shrinking analysis is contradiction free?
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
>
> 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:
> >
> >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
<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 90° 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 90° 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 rod contraction", you must add time slowing to your analysis,
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
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 90° 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 90° 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
<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 90° 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 !
>> >> 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 90° 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
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 90° 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
<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 30° and
60°, 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
>
> In article <37BEBBF1...@hia.no>, "Paul B. Andersen"
> <paul.b....@hia.no> wrote :
>
> >
> >MLuttgens wrote:
>
> [snip]
>
> >Some proposals defies logic. This is one of them.
> >
>
> 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 30° and
> 60°, 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.
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.
>
> [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
<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]
>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. 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
>
> 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
had existed.
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
had claimed something like that.
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
<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
:: 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
>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
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
> 2) A proposed thought experiment with a "one-way"
> interferometer, using two opposite arms instead 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
:: 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
<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
Keating already did it.
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
>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
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 >
already
> 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
Paul B. Andersen
>
[snip discussion of Michelson's interferometer]
> Paul B. Andersen wrote :
> >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.
Am I wrong when assuming that you must have realized
this by now?
[snip]
Still talking about the "one-way interferometer":
> Do you deny that a modification of the fringe pattern will be
> seen when rotating the apparatus?
Yes.
Both SR and LET predict no changes in the fringe pattern
when the apparatus is rotated.
> If you claim that the rotation of the apparatus will not be
> accompanied by a change of pattern, you don't believe in SR.
You can't be serious. You surely must know better.
To get back to your initial calculations:
When the "one-way interferometer" were aligned
perpendicular to the velocity, you calculated
that there would be no phase difference (or time difference -
same thing) between the light beams at the detector.
The unstated assumption behind this calculation is that
the light sources are synchronized - e.g. simultaneously
emitting the same phase in **the ether frame**.
This is correct.
================
When the "one-way interferometer" were aligned
parallel to the velocity, you calculated
that there would be a phase difference (or time difference -
same thing) between the light beams at the detector.
The unstated assumption behind this calculation is that
the light sources are synchronized - e.g. simultaneously
emitting the same phase in **the ether frame**.
This is correct.
================
But your error is from the above two separate cases
to conclude that there would be a change in the relative
phase between the beams at the detector when the apparatus
is rotated.
The point is that you do not get from the first case
to the second by rotating the apparatus.
You are ignoring what happens to synchronization of the
sources when the apparatus is rotated.
In "ether speak" the explanation goes like this:
Remember that the monochromatic sources are clocks, and behave
accordingly. If these "clocks" are synchronized in the ether
frame when the apparatus is transverse, rotating the apparatus
means that one clock is moving faster through the ether than
the other. This will mean that the "clocks" are _not_ synchronous
in the ether frame when the apparatus is parallel to the velocity;
they will be offset relative to each other, and this offset is
exactly such that the relative phase of the lighbeams will be
unchanged at the detector.
BTW, it does not matter _how_ the sources are initially synchronized.
They point is that the synchronization observed in the ether frame
will change in such a way that the fringe pattern stays unchanged.
[..]
> 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?
Not at all. :-)
But I am not quite sure if it is feasible. The two lasers
would have to be of the exact same frequency, not drifting
relative to each others. Can you get that stable lasers?
Paul
Wayne Throop
:: Obviously, the velocity of the ether can be detected if there is
:: length contraction and not time dilation. Again, "well, DUH!".
: mlut...@aol.com (MLuttgens)
: A first step of Wayne Throop in the right direction.
It cannot possibly be a "step", since this has my position,
and obvious, from the very start of my participation in relativity discussions.
: He just needs to grasp that time dilation doesn't modifiy the outcome
: of the experiment.
It was MLuttgens who must "grasp" that. MLuttgens is the one who claims
that time dilation dilation without length contraction yields a null result.
: You still have not grasped that synchronization is not necessary to
: detect the mere existence of the ether.
Synchronization, schmynchronization.
MLuttgens has still not grasped that it is impossible to detect
the rest state of a Lorentzian ether.
Arlin Brown
Wayne writes: MLuttgens has still not grasped that it is impossible to
detect the rest state of a Lorentzian ether.
Arlin Jean: How about the AET ether? AET is based on absolute sync and
absolute sync does not invalidate SR math, except for the untested 1ppm
energy anomaly which some day could help determine whether the ether
exists. The ether can certainly be detected and probably has been.
Relative SR sync can not exist unless the real one-way light speed
varies from frame to frame. If it did not vary, it would not be
necessary for an observer to re-set her separated clocks upon changing
frames. When separated clocks in two different frames are set
(erroneously called "sync") the same way, e.g., having been set in the
same frame initially, the OWLS will be measured to be different, in
general, because it IS different. But WHY is it necessary to re-set
clocks when arriving in a different frame, except to conform with the
unproven SR theory? There's nothing at all wrong with NOT re-setting
clocks when changing frames. IF all frames are equally valid, what's
wrong with all observers synching their clocks in the earth frame and
then go on their merry way to other frames. SR could then be used
against itself and would be forced to allow all of my absolute-sync AET
equations to kick in and give all the correct measurement values and
would form a consistent set of valid equations and a new paradigm with
no paradoxes of any kind, because all observers would agree about
everything.
---
Take care,
Jeannie
MLuttgens
wrote :
>Date : Fri, 27 Aug 1999 16:50:36 -0500
>
>MLuttgens wrote:
>> Imo, only one "one-way" experiment is not enough.
>
>You did not understand what I said. There are _NO_ "one-way"
>experiments. NONE. All there are is tests of the isotropy of the
>round-trip speed of light. Some of these use what appear to be
>one-way paths in a simplistic analysis; all turn out to actually
>be round-trip paths in one way or another.
>
O.K., there are _NO_ "one-way" experiments. A clever way to
evade the embarrassing problems. I wonder if Steve Carlip
would agree with you.
So, in order to avoid simplistic analysis of non-existent one-way
experiments, I will limit myself to the one century old MMX:
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.
I presume that you will agree that the parallel arm is contracted
by the factor sqrt(1-(v/c)^2).
But Tp(e) = (2L/c) * 1/(1-(v/c)^2), i.e. the "ether" round-trip time
of light along the parallel arm, has been calculated from the
outward time
To(e) = L/(c-v),
and the backward time
Tb(e) = L/(c+v).
Indeed, Tp(e) = To(e) + Tb(e) = (2L/c) * 1/(1-(v/c)^2).
According to Tom Roberts (and all relativists?), the corresponding
"interferometer"times are To(i) = Tb(i) = L/c.
So, we will calculate the contraction factors fo and fb leading
from To(e) to To(i) and from Tb(e) to Tb(i).
It is easy to find that
fo = sqrt [(1-(v/c)) / (1+(v/c))], and
fb = sqrt [(1+(v/c)) / (1-(v/c))].
Indeed,
To(i) = L/(c-v) * fo * the time contraction factor sqrt(1-(v/c)^2)
= (L/c) * 1/sqrt(1-(v/c)^2) * sqrt(1-(v/c)^2) = L/c
Tb(i) = L/(c+v) * fb * the time contraction factor sqrt(1-(v/c)^2)
= (L/c) * 1/sqrt(1-(v/c)^2) * sqrt(1-(v/c)^2) = L/c
Let's note that the length contraction factor
fo = sqrt [(1-(v/c)) / (1+(v/c))] is smaller than one (or 1 if v=0),
meaning that the arm will contract if the direction of the
light beam which travels along it is the same as that of the
velocity vector, and that the length contraction factor
fb = sqrt [(1+(v/c)) / (1-(v/c))] is greater than one (or 1 if v=0),
hence that the arm will *dilate* when the direction of the beam
is opposite to that of the velocity vector !
Iow, the arm of the interferometer will contract or dilate according
to the direction of the light beam which travels along it!
Btw, how would you label any theory (other than SR of course)
leading to such a contradictory result?
Marcel Luttgens
Tom Roberts
> O.K., there are _NO_ "one-way" experiments. A clever way to
> evade the embarrassing problems.
This is not a "clever way" to avoid anything, this is merely the way mathematical
theories work. The constraint that the round-trip speed of light be isotropically
c in avery inertial frame is a powerful restriction. But it is also powerfully-
well established experimentally (to an accuracy of a few parts in 10^15 for
earthbound labs -- that is a limit on the anisotropy of the round-trip speed of
light which is less than a micron per second).
Tom Roberts tjro...@lucent.com
Paul Stowe
writes:
>MLuttgens wrote:
>> O.K., there are _NO_ "one-way" experiments. A clever way to
>> evade the embarrassing problems.
>
>This is not a "clever way" to avoid anything, this is merely the
>way mathematical theories work. The constraint that the round-trip
>speed of light be isotropically c in avery inertial frame
So what, the issue isn't the round trip it's the 'one-way'. So the
question remains, why, in almost one hundred years, hasn't this aspect
been put to the arbiter of science, the good old experiment. It isn't
like it really hard to do.
>is a powerful restriction. But it is also powerfully-well established
>experimentally (to an accuracy of a few parts in 10^15 for earthbound
>labs -- that is a limit on the anisotropy of the round-trip speed of
>light which is less than a micron per second).
Again, get it straight, the issue IS NOT the round trip experimenta!
Paul Stowe
Frank Wappler
> [...] a limit on the anisotropy of the round-trip speed of light
> [... is] established experimentally (to an accuracy of
> a few parts in 10^15 for earthbound labs [...)]
Which experiment would have unambiguously established that?
Obviously not the Brillet-Hall experiment (PRL42(9), 549, 1979)
which only obtained a limit on the variation of a _frequency_.
Instead, "isotropy of the round-trip speed of light" is already
implied by the experimental prescriptions about measured distances
(such as "cavity length of a Fabry-Perot to remain stable",
independent of spatial direction), together with the reproducible
measurement procedure which defines pairwise "distance" in terms of
light signal round-trips in the first place.
Apparently Brillet and Hall realized their experimental prescription
indeed quite accurately.
Regards, Frank W ~@) R
Tom Roberts
> Tom Roberts wrote:
> > [...] a limit on the anisotropy of the round-trip speed of light
> > [... is] established experimentally (to an accuracy of
> > a few parts in 10^15 for earthbound labs [...)]
> Which experiment would have unambiguously established that?
> Obviously not the Brillet-Hall experiment (PRL42(9), 549, 1979)
> which only obtained a limit on the variation of a _frequency_.
Yes, Brillet and Hall. They did indeed measure the frequency directly. But
for this frequency to remain a constant as they measured it is necessary for
the round-trip speed of light to also remain constant to that same relative
accuracy.
One cannot interpret an experiment without a theory, and any theory used to
analyze this experiment whill have that same conclusion.
Tom Roberts tjro...@lucent.com
Tom Roberts
> So what, the issue isn't the round trip it's the 'one-way'.
I know that, and was directly addressing it. I was continuing remarks I made
earlier in the thread, and this didn't fully appear in this one article.
> So the
> question remains, why, in almost one hundred years, hasn't this aspect
> been put to the arbiter of science, the good old experiment. It isn't
> like it really hard to do.
It is _IMPOSSIBLE_ to do, in the following sense:
Consider the class of theories I keep talking about, all theories in which
the round-trip speed of light is isotropically c in any inertial frame;
they differ from each other in what the one-way speed of light is in
different frames. Arbitrarily select one of them and claim: "_THIS_ is the
way the world _REALLY_ works". Now perform an experiment, _ANY_ experiment.
Analyze it with the selected theory and obtain agreement between theory and
experiment. Now select _ANY_OTHER_ theory of that class and _ALSO_ obtain
agreement between theory and experiment. There is _NO_POSSIBLE_EXPERIMENT_
which can distinguish among the theories of this class, because of their
mathematical structure and interrelationships.
Note that theories outside this class, with an anisotropic round-
trip speed of light, are uninteresting because they have already
been soundly refuted by many experiments.
The reason for this is quite simple: the only difference among these theories
is the way multiple clocks are synchronized, and that is an _ARBITRARY_ and a
_HUMAN_ choice -- it can have no possible physical consequences (but it does
affect the numerical measurements those clocks report).
For a detailed discussion of this, see Zhang, _Special_Relativity_and_
_its_Experimental_Foundations_ (1997).
I have recently posted a series of articles applying this to
specific experiments which some people around here falsely claim
have observed the ether:
Subject: Sivertooth's Experiment Cannot See the Ether
Subject: Sivertooth's Experiment Cannot See the Ether - II
Subject: Torr and Kolen's Experiment Cannot See the Ether
In preparation:
Subject: DeWitte's Experiments Cannot See the Ether
Tom Roberts tjro...@lucent.com
Frank Wappler
> [...] the issue isn't the round trip it's the 'one-way'.
> So the question remains, why, in almost one hundred years,
> hasn't this aspect been put to the arbiter of science,
> the good old experiment.
> It isn't like it [is] really hard to do.
Then how would you measure "one way speed of light",
at least in principle?
Frank Wappler
> Frank Wappler wrote:
> > [...] the Brillet-Hall experiment (PRL42(9), 549, 1979) [...]
> > obtained a limit on the variation of a _frequency_.
> But for this frequency to remain a constant as they measured
> it is necessary for the round-trip speed of light
> to also remain constant to that same relative accuracy.
Not at all - there's a gauge freedom:
It is sufficient for the _ratio_ "c/lambda" to remain constant,
which is of course guaranteed by the conventional measurement
procedure to determine "lambda" in the first place,
via time interval (or frequency) measurements,
and using the symbol "c/2", (or "c").
> One cannot interpret an experiment without a theory
Rather: without a preselected _measurement procedure_;
since "theory" is usually reserved for what _can be_
experimentally tested, falsified or corroborated:
a prediction about values in any one individual trial,
relative to the experimental result obtained in that trial;
or an algorithm which summarizes several experimental result
of past trials
and which would make predictions about values in "future" trials.
> and any theory used to analyze this experiment will have
> that same conclusion.
No, different measurement procedures, e.g. for deriving "values"
of "c" and "lambda" from the obtained frequency values,
can obtain _different_ results about "isotropy" (of either one);
and the theories/algorihms that summerize the corresponding
values would have to be different as well.
Of course, in contrast to the SR procedure (which implies isotropy),
other measurement procedures might not be reproducible,
but referring to arbitrarily chosen directions or trials,
or depending on individual artefacts.
Paul Stowe
writes:
>
>Paul Stowe wrote:
>> So what, the issue isn't the round trip it's the 'one-way'.
> I know that, and was directly addressing it. I was continuing remarks
> I made earlier in the thread, and this didn't fully appear in this
> one article.
>
>> this aspect been put to the arbiter of science, the good old
>
> It is IMPOSSIBLE to do, in the following sense:
>
> Consider the class of theories I keep talking about, all theories in
> inertial frame;
Let's be clear, that the round trip speed is perceived as c in ever
inertial FOR, OK.
> they differ from each other in what the one-way speed of light is in
> different frames.
Here again, you've lost me. The vector math on this is clear, not
arbitrary. It's c - v in the direction of motion and c + v against it.
This need NOT be based upon an assumption of an aether, only upon
Einstein's first postulate (that the speed of light is globally
invariant).
Thus the 'round trip' time is 2L/cgamma^2 in standard Galilean
Transforms and 2L/cgamma in SR's LP group. As demonstrated by the MMX
class of experiments, the difference is exactly gamma in the direction
of motion. So, any 'one-way' anisotropicy in a single inertial FOR,
where multiple clocks can be synchronized and then remain so, will by
definition refute the assumption of symmetry of speed of c in a the
fore/aft paths along the line of motion. This would not however,
invalid the round trip results or the perception based upon this 'round
trip' data of c being locally invariant.
> Arbitrarily select one of them and claim: "THIS is the way the world
> REALLY works". Now perform an experiment, ANY experiment. Analyze
> it with the selected theory and obtain agreement between theory and
> experiment. Now select ANY OTHER theory of that class and ALSO obtain
> agreement between theory and experiment. There is NO POSSIBLE
> EXPERIMENT which can distinguish among the theories of this class,
> because of their mathematical structure and interrelationships.
>
> Note that theories outside this class, with an anisotropic round-
> trip speed of light, are uninteresting because they have already
> been soundly refuted by many experiments.
>
> The reason for this is quite simple: the only difference among these
> theories is the way multiple clocks are synchronized, and that is an
> ARBITRARY and a HUMAN choice -- it can have no possible physical
> consequences (but it does affect the numerical measurements those
> clocks report).
Again, going through a step by logical step specific example to explain
this would (at least for me) help in understanding what you're trying
to say.
> For a detailed discussion of this, see Zhang,
> Special Relativity and its Experimental Foundations (1997).
Paul Stowe
Paul Stowe
Wappler) writes:
>
>Paul Stowe wrote:
>
>> [...] the issue isn't the round trip it's the 'one-way'.
>> hasn't this aspect been put to the arbiter of science,
>> the good old experiment.
>
>Then how would you measure "one way speed of light",
>at least in principle?
>
OK here's a very simple example, we have three atomic clocks, one tied
into a transmitter (b) and two into receivers (a & c). They are ALL
synchronized to within one nano-second of each other and then placed
100 meters apart on level ground as shown below:
a b c
Now b is programmed to emit a 10 nano-second burst of radio waves every
hour on the hour (precisely) and the receivers simply record
(precisely) the time they see the pulses. Now, since they are ALL in
the very same inertial frame (no relative motion between them) the
superimposed times of reception for a & c should NEVER be different.
If there is ANY difference in reception (arrival) times, we surely
cannot say it is due to the clocks being out of synch! If however,
there is a sidereal sinusoidal pattern, and if performed for a year, a
annual sinusoidal pattern in the asynchronous reception times, we most
certainly CAN NOT hand wave this away saying 'we need to esynch' the
clocks.
Paul Stowe
Frank Wappler
> Frank Wappler [wrote:]
> > Paul Stowe wrote:
> > > It isn't like it [is] really hard to do.
> > > Then how would you measure "one way speed of light",
> > at least in principle?
> OK here's a very simple example, we have three atomic clocks,
> one tied into a transmitter (b) and two into receivers (a & c).
> They are ALL synchronized to within one nano-second of each other
What's the procedures to do that?
How would you check _that_ those three are and remain "synchronized"
as specified, after and while they are separated?
> and then placed 100 meters apart on level ground as shown below:
> a b c
How would you measure and check those distances throughout the
experiment?
What do you mean by "meter" -
the unit of "distance" (pairwise, i.e. of two wrt. each other)
according to the conventional SI definition as
"1 meter == c/2 2_seconds_light_signal_roundtrip_interval;
if the beginning and end states of that interval, and the state
of reflecting the light signal, and the time interval unit `second'
have been calibrated between those two through Einstein's
calibration procedure"
?
Or what else?
How would you determine whether or not those three have been
"placed on level ground", or at least "on a straight line"
if not via measurements of pairwise distance,
requiring for instance that
distance( a, c ) == distance( a, b ) + distance( b, c )
(for some selected definition of the operator "+") ?
The rest of your proposal is not hard to do/understand:
> Now b is programmed to emit a 10 nano-second burst of radio waves
> every hour on the hour (precisely) and the receivers simply record
> (precisely) the time they see the pulses.
Fine - that's part of what they had to do in order to detemine
their "synchronization" and their pairwise distances in units
of "meter" anyways.
> Now, since they are ALL in the very same inertial frame
> (no relative motion between them)
... at least as far as that's implied in your prescription that
each of their pairwise distances should have a specific value
throughout the experiment ...
> the superimposed times of reception for a & c should NEVER be different.
Right, at least that's guaranteed if "synchronization" and
measurements of pairwise distance are obtained through
Einstein's calibration procedure and the associated distance definition.
(Or would you suggest any other measurement procedures?)
> If there is ANY difference in reception (arrival) times, we
> surely cannot say it is due to the clocks being out of synch!
That's right _if_ we have (or had?) an _independent_ procedure
to make sure of this "synch relation"; about which I'm asking above.
OTOH, using the Einstein procedures, the exchange of light signals
that you prescribe would be _part of_ how "synchronization" and
"distances" are being determined and checked throughout the
experiment in the first place.
ANY difference would indicate that the prescribed specifications
were not satisfied, but that instead the clocks had "accelerated"
wrt. each other.
> If however, there is a sidereal sinusoidal pattern, and if
> performed for a year, a annual sinusoidal pattern in the
> asynchronous reception times, we most certainly CAN NOT
> hand wave this away saying 'we need to esynch' the clocks.
Well - while accelerations of the clocks wrt. each other are clearly
detrimental to the supposed "measurement of one way speed of light",
measured accelerations were surely interesting results in themselves:
Would they allow to derive "the most probable potential"
of the region containing the three clocks?,
and even to determine the "mass" of the "center/singularity"
of that potential, (if there is one)?
Best regards, Frank W ~@) R
p.s.
Seems you've sent a reply to Tom Roberts as follow-up to my
preceding post ...
Paul Stowe
Wappler) writes:
>
>Paul Stowe wrote:
>
>> Frank Wappler [wrote:]
>
>> > Paul Stowe wrote:
>> > > It isn't like it [is] really hard to do.
>
>> > > Then how would you measure "one way speed of light",
>> > at least in principle?
>
>> OK here's a very simple example, we have three atomic clocks,
>> one tied into a transmitter (b) and two into receivers (a & c).
>
>> They are ALL synchronized to within one nano-second of each other
>
>What's the procedures to do that?
The clocks would be placed side by side & synchronized.
>How would you check _that_ those three are and remain "synchronized"
>as specified, after and while they are separated?
There are several methods available, the easiest would be to bring them
back to b on an interval that assured random orientation relative to
the experiment.
>> and then placed 100 meters apart on level ground as shown below:
>
>> a b c
>
>How would you measure and check those distances throughout the
>experiment?
Again several methods are available, but NONE involving light!
>What do you mean by "meter" -
Hmm, one meter, or 39.37007 inches. Again NOT involving the light
definition!
>How would you determine whether or not those three have been
>"placed on level ground", or at least "on a straight line"
>if not via measurements of pairwise distance,
>requiring for instance that
Using standard surveying methods. This is NOT a difficult task.
>distance( a, c ) == distance( a, b ) + distance( b, c )
>
>(for some selected definition of the operator "+") ?
>
>
>The rest of your proposal is not hard to do/understand:
>
>> Now b is programmed to emit a 10 nano-second burst of radio waves
>> every hour on the hour (precisely) and the receivers simply record
>> (precisely) the time they see the pulses.
>
>Fine - that's part of what they had to do in order to detemine
>their "synchronization" and their pairwise distances in units
>of "meter" anyways.
No the distancing is done the old fashion way. You know you can use a
string tied to a stake...
>> Now, since they are ALL in the very same inertial frame
>> (no relative motion between them)
>
>... at least as far as that's implied in your prescription that
>each of their pairwise distances should have a specific value
>throughout the experiment ...
>
>> the superimposed times of reception for a & c should NEVER be
different.
>
>Right, at least that's guaranteed if "synchronization" and
>measurements of pairwise distance are obtained through
>Einstein's calibration procedure and the associated distance
> definition.
What calibration? The only calibration is done in synchonizing the
clocks when side by side.
>(Or would you suggest any other measurement procedures?)
See above ...
>> If there is ANY difference in reception (arrival) times, we
>> surely cannot say it is due to the clocks being out of synch!
>
>That's right _if_ we have (or had?) an _independent_ procedure
>to make sure of this "synch relation"; about which I'm asking above.
OK forget the radio and recievers, once synchronized and moved into
place, should these clocks drift out of sync?
>OTOH, using the Einstein procedures, the exchange of light signals
>that you prescribe would be _part of_ how "synchronization" and
>"distances" are being determined and checked throughout the
>experiment in the first place.
No, again I suggest a more direct, old fashion approach.
>ANY difference would indicate that the prescribed specifications
>were not satisfied, but that instead the clocks had "accelerated"
>wrt. each other.
That would depend on the resulting data pattern, if any.
>> If however, there is a sidereal sinusoidal pattern, and if
>> performed for a year, a annual sinusoidal pattern in the
>> asynchronous reception times, we most certainly CAN NOT
>> hand wave this away saying 'we need to esynch' the clocks.
>
>Well - while accelerations of the clocks wrt. each other are clearly
>detrimental to the supposed "measurement of one way speed of light",
>measured accelerations were surely interesting results in themselves:
>
>Would they allow to derive "the most probable potential"
>of the region containing the three clocks?,
>and even to determine the "mass" of the "center/singularity"
>of that potential, (if there is one)?
I attempted to think this out as throughly as possible, for both
applicablity and ease of execution. It would, I believe, making a
verrrry interesting experiment.
Paul Stowe
MLuttgens
wrote :
>Date : Sat, 28 Aug 1999 13:55:24 -0500
>
>MLuttgens wrote:
>> O.K., there are _NO_ "one-way" experiments. A clever way to
>> evade the embarrassing problems.
>
>This is not a "clever way" to avoid anything, this is merely
>the way mathematical theories work. The constraint that the
>round-trip speed of light be isotropically c in avery inertial
>frame is a powerful restriction. But it is also powerfully-
>well established experimentally (to an accuracy of a few parts
>in 10^15 for earthbound labs -- that is a limit on the anisotropy
>of the round-trip speed of light which is less than a micron per
>second).
>
>
>Tom Roberts
You are certainly right about the round-trip anisotropy, even
if higher order v/c effects exist, but are too small to be
detected with present devices.
But what about the one-way speed of light?
Are you sure that in the interferometer frame, the one-way
travel time of light is L/c for the parallel arm?
In such a case, the arm would contract or *dilate* according
to the light direction, which seems rather absurd (see the
demonstration in my last post).
Otoh, you claimed that "one-way" experiments are in fact
tests of the isotropy of the round-trip speed of light.
Does that means that no modification of the fringe pattern
could be observed when rotating a "one-way" interferometer?
Marcel Luttgens
Tom Roberts
> > which the round-trip speed of light is isotropically c in any
> > inertial frame;
> Let's be clear, that the round trip speed is perceived as c in ever
> inertial FOR, OK.
> > they differ from each other in what the one-way speed of light is in
> > different frames.
> Here again, you've lost me. The vector math on this is clear, not
> arbitrary. It's c - v in the direction of motion and c + v against it.
No!!! c-v and c+v do NOT give an isotropic round-trip speed of light.
The theory you are describing does not belong to my class, and is
clearly refuted by many experiments (MMX, Brillet-Hall, etc.).
> Thus the 'round trip' time is 2L/cgamma^2 in standard Galilean
> Transforms and 2L/cgamma in SR's LP group.
Both of your formulas are wrong, and neither one is a correct calculation
of the round-trip speed of light. For one-way speeds c+v and c-v, the
round-trip speed is given by:
2/c_rt = 1/(c+v) + 1/(c-v)
c_rt = c (1 - v^2/c^2)
This is NOT c. And it won't be isotropic because perpendicular to v the
round-trip speed will be c. The "round-trip time" you compute for an
out-and-back path of length 2L is just 2L/c_rt, and is not isotropic
either.
Tom Roberts tjro...@lucent.com
Tom Roberts
> Wayne writes: MLuttgens has still not grasped that it is impossible to
> detect the rest state of a Lorentzian ether.
> Arlin Jean: How about the AET ether?
Nope, your "AET Ether frame" cannot be detected either. Because the
one-way speed of light is inextricably intertwined with the
synchronization of clocks, and clock synchronization is _conventional_
(i.e. there is no unique way to do it, and one must make an arbitrary
choice -- that choice can have no physical consequences).
See my recent articles:
Subject: Silvertooth's Experiment Cannot See the Ether.
Subject: Silvertooth's Experiment Cannot See the Ether - II.
Subject: Torr and Kolen's Experiment Cannot See the Ether.
Your "AET" is a member of the class of theories I discuss.
Tom Roberts tjro...@lucent.com
Tom Roberts
> It is sufficient for the _ratio_ "c/lambda" to remain constant,
No, it is sufficient for the quantity c/(2L/n) to remain constant,
where L is the length of the Fabry-Perot cavity, n is the number of
half-wavelengths in L, and c is the round-trip speed of light. A change
in n is clearly ruled out (n must be an integer).
One can assume, as did Brillet and Hall, that c is invariant, and then
deduce that L did not vary. One can also assume, as I do, that L is
constant, and then deduce that c did not vary. The experimenters went
to great effort to ensure that L does not vary.
Tom Roberts tjro...@lucent.com
Paul Stowe
writes:
>Paul Stowe wrote:
>> In <37C88A99...@lucent.com> Tom Roberts <tjro...@lucent.com>
>
>> > Consider the class of theories I keep talking about, all
>> > theories in which the round-trip speed of light is isotropically
>> > c in any inertial frame;
>>
>> Let's be clear, that the round trip speed is perceived as c in ever
>> inertial FOR, OK.
>>
>> > they differ from each other in what the one-way speed of light
>> > is in different frames.
>>
>> Here again, you've lost me. The vector math on this is clear, not
>> arbitrary. It's c - v in the direction of motion and c + v against
>> it.
>
>No!!! c-v and c+v do NOT give an isotropic round-trip speed of
>light. The theory you are describing does not belong to my class,
>and is clearly refuted by many experiments (MMX, Brillet-Hall, etc.).
>
>> Thus the 'round trip' time is 2L/cgamma^2 in standard Galilean
>> Transforms and 2L/cgamma in SR's LP group.
>
>Both of your formulas are wrong, and neither one is a correct
>
> 2/c_rt = 1/(c+v) + 1/(c-v)
> c_rt = c(1 - v^2/c^2)
Nope it not, but you CAN define to be, but doing just what Einstein
did, say it WILL appear to be c because if we have a 'distortion' of
gamma in the direction of travel, which takes your equation (which I
will translate back into times by inserting path length L) we get:
t_rt = Lgamma/(c - v) + Lgamma/(c + v) <Eq.1>
or as you say;
t_rt = 2Lgamma/cgamma^2 = [2L/c]/gamma <Eq.2>
Or, the classic time dilation...
HOWEVER, Equation 1 can be resolved to: [Let B = v/c]
t_rt = (2L/c)[Sqrt(1 + B/1 - B) + Sqrt(1 - B/1 + B)]
We note here the symmetry I was referring to, this form is
mathematically equal to [2L/c]/gamma or Einstein's time dilation
equation. But, as you say and can see, the one ways clearly aren't
equal. This difference is testable, and while the round trips are
demonstrably equal to SR, the one ways are different AND NOT AT ALL
ARBITRARY.
>This is NOT c. And it won't be isotropic because perpendicular to v
>the round-trip speed will be c. The "round-trip time" you compute for
>an out-and-back path of length 2L is just 2L/c_rt, and is not
>isotropic either.
The key here is Lorentz's original premise of 1895, which he makes in
his 1904 paper also, that matter and energy processes are distorted by
gamma in the direction of travel. This I see as a testable difference
between LET and SR.
Paul Stowe
Wayne Throop
: How about the AET ether?
How about it?
: AET is based on absolute sync and absolute sync does not invalidate SR
: math, except for the untested 1ppm energy anomaly which some day could
: help determine whether the ether exists. The ether can certainly be
: detected and probably has been.
Shrug. I can't help it that Jeannie clings to this error.
In an ether "based on absolute sync", in which time dilation and length
contraction happen to every experimental apparatus and process relative
to their absolute durations and lengths, the rest state of the ether
cannot be discovered by any such apparatus.
That's simply a mathematical fact, same as "2+2=4",
or "there is no way to trisect an anble with compas and straightedge".
Tom Roberts
> In <37C98CB6...@lucent.com> Tom Roberts <tjro...@lucent.com>
> writes:
> >No!!! c-v and c+v do NOT give an isotropic round-trip speed of
> >light.
> did, say it WILL appear to be c because if we have a 'distortion' of
> gamma in the direction of travel, which takes your equation (which I
> will translate back into times by inserting path length L) we get:
> t_rt = Lgamma/(c - v) + Lgamma/(c + v) <Eq.1>
Where do you get that "gamma" from??
Here we are in a single inertial frame, with the 1-way speed of light
c+v in one direction, and c-v in the other, and the round-trip speed
of light is this frame is c(1-v^2/c^2) for an out-and-back path in
that direction. That's not isotropic. And there is no gamma -- this
is all in a single inertial frame.
If you want to assume that "gammas" appear, then you need to be more
precise in describing what you are doing -- they sure don't appear
when working in a single frame, which is what I was doing, and what
I thought you meant....
Let me make the following assumptions:
1) there is a unique ether frame.
2) in the ether frame, light propagates with a 1-way speed which is
isotropically c.
3) the coordinate transform from the ether frame to any inertial frame
is linear, and is characterized by the velocity of the inertial
frame as measured in the ether frame.
4) in any inertial frame, the round-trip speed of light is
isotropically c, as measured in that inertial frame.
Note (1) and (2) are essentially what it means to have an ether;
(3) can easily be deduced from the properties of inertial frames,
and (4) is required by numerous experiments (MMX, Brillet-Hall,
etc.).
Implicitly I am restricting the discussion to inertial frames; non-
inertial frames can be dealt with later. Yes, this implicitly assumes
that the ether frame is inertial -- if you think that is unreasonable,
say so.
If necessary, just consider the small region of a tabletop
experiment for the duration of light signals to traverse it.
If you buy into those 4 assumptions, then it can be shown that the
one-way speed of light is unobservable (in the sense I used earlier --
these assumptions determine a class of theories which are experimentally
indistinguishable; this class includes SR and all reasonable ether
theories). Note that (4) is a quite limiting assumption, but the
experiments essentially force it upon any reasonable theory.
Tom Roberts tjro...@lucent.com
Paul Stowe
writes:
>Paul Stowe wrote:
>> In <37C98CB6...@lucent.com> Tom Roberts <tjro...@lucent.com>
>> writes:
>> >No!!! c-v and c+v do NOT give an isotropic round-trip speed of
>> >light.
>>
>> 'distortion' of gamma in the direction of travel, which takes
>> your equation (which I will translate back into times by
>> inserting path length L) we get:
>>
>> t_rt = Lgamma/(c - v) + Lgamma/(c + v) <Eq.1>
>
> Where do you get that "gamma" from??
It's always there, just when 'in' a local FOR where relative v = 0, we
simplify this to:
t_rt = L[Sqrt(1 - 0)]/(c - 0) + LSqrt(1 - 0)]/(c + 0)
which of course becomes simply:
t_rt = L/c + L/c = 2L/c
> Here we are in a single inertial frame, with the 1-way speed of
> light c + v in one direction, and c - v in the other, and the
> round-trip speed of light is this frame is c(1-v^2/c^2) for an
> out-and-back path in that direction. That's not isotropic. And
> there is no gamma -- this is all in a single inertial frame.
And as mentioned above, v relative = 0. Ah-ha, I see where presumption
that c 'is' isotropic comes from. Not postulate #1 (c is globally
invariant), but postulate #2, the local speed of light IS c, NOT 'will
appear' to be c in a round trip. For with postulate #1, as Lorentz
also presumed, any motion relative to this 'global' framework will/must
have a v associated with it. That velocity would affect the one way
measures but NOT the round trip, given an actual physical distortion of
gamma in the direction of said motion.
> If you want to assume that "gammas" appear, then you need to be
> more precise in describing what you are doing -- they sure don't
> appear when working in a single frame, which is what I was doing,
> and what I thought you meant....
Oh they're always there, they just factor out as demonstrated above.
>Let me make the following assumptions:
>
> 1) there is a unique ether frame.
> 2) in the ether frame, light propagates with a 1-way speed which
> is isotropically c.
> 3) the coordinate transform from the ether frame to any inertial
> frame is linear, and is characterized by the velocity of the
> inertial frame as measured in the ether frame.
> 4) in any inertial frame, the round-trip speed of light is
> isotropically c, as measured in that inertial frame.
>
> Note (1) and (2) are essentially what it means to have an ether;
> (3) can easily be deduced from the properties of inertial
> frames, and (4) is required by numerous experiments (MMX,
> Brillet-Hall, etc.).
However, your #3 becomes invalid if a distortion factor of gamma is
applicable to said motion. That is to say Sqrt[1 - (v/c)^2] certainly
is not linear! This gamma would apply to so-called wave-like phenomena
of said ether, as demonstrated in the known properties of moving
acoustical 'sources'.
> Implicitly I am restricting the discussion to inertial frames;
> non->inertial frames can be dealt with later. Yes, this
> implicitly assumes that the ether frame is inertial -- if
> you think that is unreasonable, say so.
>
> If necessary, just consider the small region of a tabletop
> experiment for the duration of light signals to traverse it.
>
> If you buy into those 4 assumptions, then it can be shown that
> the one-way speed of light is unobservable (in the sense I used
> earlier -- these assumptions determine a class of theories which
> are experimentally indistinguishable; this class includes SR and
> all reasonable ether theories). Note that (4) is a quite limiting
> assumption, but the experiments essentially force it upon any
> reasonable theory.
As I said before, I think this difference has the potential to
discriminate between LET and basic SR. A.k.a, a means of falsifing
LET, or conversely, some aspects of SR. But experiments would have to
tell us which.
Paul Stowe
Arlin Brown
Wayne writes: In an ether "based on absolute sync", in which time
dilation and length contraction happen to every experimental apparatus
and process relative to their absolute durations and lengths, the rest
state of the ether cannot be discovered by any such apparatus.
Arlin Jean: Where is your proof? You made a false claim. Now back it up
and show conclusively that it is true. You can't do it. All you can do
is to make gratuitous assertions.
Wayne: That's simply a mathematical fact, same as "2+2=4", or "there is
no way to trisect an anble with compas and straightedge".
Arlin Jean: A mathematical fact? OK. Now back it up with a mathematical
proof or forever hold your your peace. BTW, Archimedes trisected any
given angle with a straight edge and compass. If you don't believe it,
state the problem properly and I'll take it from there. You will have no
third chance to re-state the problem after it is solved. If you think
you stated the problem properly above, then you lose. I'll give you a
second chance to state it properly.
---
Take care,
Jeannie
Frank Wappler
> [...] once synchronized and moved into place,
> should these clocks drift out of sync?
Obviously any two clocks _could_ do that.
The question is how to find out unambiguously whether or not
they actually _did_, in each trial, throughout the experiment.
If your experimental procedure prescribes that they "shall not"
then you should discard the trials in which they _were found_
"drifting", and only consider the trials in which they
_were found_ not to.
> Frank Wappler [wrote:]
> > How would you check _that_ those three are and remain "synchronized"
> > as specified, after and while they are separated?
> easiest would be to bring them back to b on an interval that
> assured random orientation relative to the experiment.
What about the check _while_ they are separated ("100 meters", as you
prescribed) - aren't those the trials for which it really matters?
> [...] Using standard surveying methods.
The standard surveying methods, as well as the SI definition of
"meter" _are based on_ the exchange of light signals;
implementing the Einstein procedures more or less completely.
> Again several methods are available, but NONE involving light! [...]
> Hmm, one meter, or 39.37007 inches. [...]
> You know you can use a string tied to a stake [...]
Then what do you mean by "string"?
How would you measure and compare the distance from _one end_ of the
"string" (i.e. "from the stake") to the _other end_ of the "string"?;
How should anyone (reading about your experiment) reproduce what you
happen to mean by "inch", in trial or the other?
How could you be sure yourself?
> > "synchronization" and measurements of pairwise
> > distance are obtained through Einstein's calibration procedure
> > and the associated distance definition.
> What calibration? The only calibration is done in synchonizing the
> clocks when side by side.
At least, if any measurements are to be derived from
"difference in reception (arrival) times of clock a vs. clock c"
one must determine the relation or map between the individual
times/readings of clock a and the individual times/readings of clock c
in the first place.
Conventionally that's determined through Einstein's calibration
procedure:
"Pairs of events/states/clock_readings/proper_times of two observers
correspond to each other (they are "simultaneous") if they contain
observations of the same light signals from "the middle between"
those two observers."
where "the middle between" clock a and clock c may be identified
trial by trial as the observer satisfying the following requirements:
- "the middle" must find the light signal roundtrip interval
to clock a (and back) same as to clock c (and back);
- clock a must find two roundtrips to "the middle" same as one to clock c;
- clock c must find two roundtrips to "the middle" same as one to clock a;
(One could formulate further requirements, involving additional
auxiliary observers/surveyers.)
An associated definition of pairwise "distance" can be based
on the results of such calibration procedures:
"distance of a pair of observers/surveyers wrt. each other is
c/2 * light_signal_roundtrip_interval;
if the beginning and end states of that interval, and the state
of reflecting the light signal, have been calibrated between
those two through Einstein's calibration procedure"
(The unit "1 meter" being defined correspndinglt through a
light_signal_roundtrip_interval of "2 seconds".)
Note that obviously one cannot in turn use pairwise distances
to identify "the middle between" any pair of observers/surveyers
in the first place.
So - do you have any suggestions to determine pairwise distances
that can be generally and unambiguosuly reproduced and understood,
other than "using some string"?
> > measured accelerations were surely interesting results in themselves:
> > Would they allow to derive "the most probable potential"
> > of the region containing the three clocks?,
> > and even to determine the "mass" of the "center/singularity"
> > of that potential, (if there is one)?
> I attempted to think this out as throughly as possible, for both
> applicablity and ease of execution.
The most important criterion would be _reproducibility_ of a
measurement procedure, i.e. that anyone involved can agree
on the result obtained, and that anyone else (e.g. reading about it)
can understand it unambiguously.
Of course one would have to apply it, therefore the procedure must
refer to observations (e.g. whethe or not an observer/surveyer sees
two signals together, as one event/state/clock_reading/proper_time).
The simpler a reproducible measurement procedure can be conducted,
and the fewer observers/surveyors have to contribute to it,
the easier it may succeed and obtain results, trial by trial.
> It would, I believe, making a verrrry interesting experiment.
Right, determinations of "most probable potentials" and "masses"
are obviously more complex measurements than determinations of
pairwise coordinates, such as "distances" or "angles".
Frank Wappler
> Frank Wappler wrote:
> > It is sufficient for the _ratio_ "c/lambda" to remain constant,
> No, it is sufficient for the quantity c/(2L/n) to remain constant,
Can you please explain the experimental difference between those
two constraints?
> where L is the length of the Fabry-Perot cavity, n is the number of
> half-wavelengths in L, and c is the round-trip speed of light.
> A change in n is clearly ruled out (n must be an integer).
> One can assume, as did Brillet and Hall, that c is invariant
Isn't that an assumption about "spatial isotropy"
for the "propagation of light"?
> One can also assume, as I do, that L is constant
Isn't that an assumption about "spatial isotropy"?
> The experimenters went to great effort to ensure that L does not vary.
How did they detemine whether or not their great effort had the
desired consequence; IOW, how did they measure and compare "L",
trial by trial?
Thanks again, Frank W ~@) R
Paul Stowe
Well, if you are suggesting using only two clocks a & b and one
transmitter (a) & receiver (b), that should also work under the same
procedure. As shown below:
a<------ 100 m ------>b
The distance IS arbitrarily set large enough to assure that the signal
delay is ten times greater that the clocks accuracy. Again 'a'
transmits at a predetermined interval and 'b' records the reception
time. As long as the cycle spans at least twelve hours, and perferably
183 days... it should still test the one way symmetry.
Paul Stowe
MLuttgens
Stowe)
wrote :
>Date : 28 Aug 1999 18:07:45 GMT
>
>In <19990828044322...@ngol02.aol.com> mlut...@aol.com
>(MLuttgens) writes:
>>
>>In article <7q53et$i...@dfw-ixnews10.ix.netcom.com>,
>pst...@ix.netcom.com(Paul
>>Stowe) wrote :
>>
>>> The test can be performed, and any necessity to re-synchronize
>>> clocks in a 'single' inertial FOR, would, IMO bring into
>>> question the validity of using gamma for one way light travel
>>> measurements.
>>>
>>> Paul Stowe
>>
>> Perhaps my reply to Tom Roberts to his article
>> <37C7082C...@lucent.com>
>> (thread: "Re: Which contraction factor?") will interest you:
>>
>> OK., there are _NO_ "one-way" experiments. A clever way to
>> evade the embarrassing problems. I wonder if Steve Carlip
>> would agree with you. So, in order to avoid simplistic analysis
>> of non-existent one-way experiments, I will limit myself to the
>> one century old MMX:
>>
>> 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) <Eq.1>
>>
>> Corresponding "interferometer" time, assuming time slowing,
>> but no length contraction:
>>
>> Tt(i) = Tt(e) * sqrt(1-(v/c)^2) = 2L/c. <Eq.2>
>
>OK your Tt(i) is the 'apparent' time along the y-z plane in the 'lab'
>or frame moving at v along an arbitrary x axis. Light speed c remains
>unchanged (globally [as in all frames] invariant), and the apparent
>speed [c'] is cSqrt[1 - (v/c)^2] which is shown by your equation 1
>above. Thus, to assume or 'say' that transverse speed in the lab frame
>is unchanged, we must 'assume' that the lab time has 'slowed down' to
>allow c' to appear to be the same as c. This IS the SR logical
>paradox, that is to say, if c IS globally invariant (meaning that it
>really isn't affected by any motion), then it truly cannot be the same
>in frames with different relative motion. To "make it so', you MUST
>apply this 'assumption'.
>
An interesting analysis!
>Now this has been OK in SR because, and only because, there is a
>contraction by gamma in the direction of travel which allows symmetry
>along all three axis, even when moving.
>
>Purist will claim that NO contraction occurs, it's just a coordinate
>transformation, and THAT is a mathematically valid (but physically
>flawed) way of approaching this issue.
>
I claim that NO contraction occur, because physical length
contraction has never be observed.
More significatively, there should be length dilation, not contraction,
in one-way situations where the light direction is opposite to that
of the velocity vector (see below).
>> "Ether" time for the parallel arm:
>>
>> Tp(e) = (2L/c) * 1/(1-(v/c)^2) <Eq.3>
>>
>> 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. <Eq.4>
>>
>> As Tt(i) = Tp(i), the phase difference is zero, and no fringe
>> shift is expected.
>>
>> I presume that you will agree that the parallel arm is contracted
>> by the factor sqrt(1-(v/c)^2).
>
>Yes, and is due to the motion v, as can be tied to, and explained by,
>the mechanism presented in my thread "Acoustical Relativity"...
>
>> But Tp(e) = (2L/c) * 1/(1-(v/c)^2), i.e. the "ether"
>> round-trip time of light along the parallel arm, has been
>> calculated from the outward time
>>
>> To(e) = L/(c-v) <Eq.5>
>>
>> and the backward time
>>
>> Tb(e) = L/(c+v) <Eq.6>
>>
>> Indeed, Tp(e) = To(e) + Tb(e) = (2L/c) * 1/(1-(v/c)^2).
>>
>> According to Tom Roberts (and all relativists?), the
>> corresponding "interferometer"times are To(i) = Tb(i) = L/c.
>
>I know, this IS the standard 'interpretation.
>
>> So, we will calculate the contraction factors fo and fb leading
>> from To(e) to To(i) and from Tb(e) to Tb(i). It is easy to find
>> that:
>>
>> fo = sqrt [(1-(v/c)) / (1+(v/c))] <Eq.7>
>>
>> and
>>
>> fb = sqrt [(1+(v/c)) / (1-(v/c))]. <Eq.8>
>
>OK this jives with:
>
> t_1 = (L/c)Sqrt[(1 + v/c)/(1 - v/c)]
>
>and
>
> t_2 = (L/c)Sqrt[(1 - v/c)/(1 + v/c)]
>
>Where my t_1 is your To <Eq.5>, and my t_2 is your Tb <Eq.6> BUT
>includes the gamma contraction factor that yours, at that point, do
>not.
>
Yes, including the "round-trip" contraction factor sqrt(1-(v/c)^2),
t_1 = (L/(c-v)) * sqrt(1-(v/c)^2) = (L/c)Sqrt[(1 + v/c)/(1 - v/c)]
t_2 = (L/(c+v)) * sqrt(1-(v/c)^2) = (L/c)Sqrt[(1 - v/c)/(1 + v/c)], and
the round-trip time
t = t_1 + t_2 = (2L/c) * 1/sqrt(1-(v/c)^2)
By multiplying t by the time slowing factor sqrt(1-(v/c)^2), you get
t(i) = 2L/c, which is the expected SR result for the interferometer
frame.
But logically, the time slowing factor should also apply to the
one-way times t_1 and t_2, and lead to 2L/c, but that's not the
case, so there must be a flaw somewhere in your analysis.
Imo, it comes from the application of the "round-trip" contraction
factor to the one-way "ether" times L/(c-v) and L/(c+v) !
In fact, the only way to obtain 2L/c from those "ether" times
To(e) and Tb(e) is to apply specific "one-way" contraction
factors, i.e. fo = sqrt [(1-(v/c)) / (1+(v/c))] and
fb = sqrt [(1+(v/c)) / (1-(v/c))].
Then, you would obtain
t_1 = To(e) * fo = (L/(c-v)) * sqrt [(1-(v/c)) / (1+(v/c))]
= (L/c) * 1/sqrt(1-(v/c)^2), and
t_2 = Tb(e) * fb = (L/(c+v)) * sqrt [(1+(v/c)) / (1-(v/c))]
= (L/c) * 1/sqrt(1-(v/c)^2).
And indeed, by multiplying t_1 and t_2 by the time slowing
factor sqrt(1-(v/c)^2), one get in each case L/c, and of course
a total time t(i) = 2L/c.
An obvious consequence of the factor fb is that the interferometer
"arm" will *dilate* when the direction of the light beam is opposite
to that of the velocity vector.
This, and the mere existence of three different "contraction"
factors is a strong argument for their non-physical existence.
Another possibility is that only the "round-trip" contraction
factor makes sense, but then, the interferometer times would be
To(i) = t_1 * sqrt(1-(v/c)^2
= (L/c)Sqrt[(1 + v/c)/(1 - v/c)] * sqrt(1-(v/c)^2
= (L/c) * (1+(v/c)), and
Tb(i) = t_2 * sqrt(1-(v/c)^2
= (L/c)Sqrt[(1 - v/c)/(1 + v/c)] * sqrt(1-(v/c)^2
= (L/c) * (1-(v/c)),
thus a time difference
To(i) - Tb(i) = (L/c) * (1+(v/c)) - (L/c) * (1-v/c))
= 2Lv/c^2
would be observed in one-way experiments.
Such experiments are badly needed !
For instance, your proposed experiment would be rather cheap,
and easy, to conduct. Imo, at least 5 such experiments should be
simultaneously performed, in view to allow a statistical
analysis.
Your experiment:
"Here's a very simple example, we have three atomic clocks,
one tied into a transmitter (b) and two into receivers (a & c).
They are ALL synchronized to within one nano-second of
each other and then placed 100 meters apart on level ground
as shown below:
a b c
Now b is programmed to emit a 10 nano-second burst of
radio waves every hour on the hour (precisely) and the receivers
simply record (precisely) the time they see the pulses."
Note that the maximum expected time difference 2Lv/c^2
(when a-b-c is parallel to the velocity vector) is 6.67E-11 s
if v=30 km/s, and 8.22E-10 s if v=370 km/s, so a 10 nanosecond
burst is probably too long. A picosecond would be better (is it
possible?, I don't know).
>> Indeed,
>>
>> To(i) = L/(c-v) * fo * sqrt(1-(v/c)^2) <Eq.9>
>>
>> Which is:
>>
>> (L/c) * 1/sqrt(1-(v/c)^2) * sqrt(1-(v/c)^2) <Eq.10>
>>
>> And also:
>>
>> Tb(i) = L/(c+v) * fb * sqrt(1-(v/c)^2) <Eq.11>
>>
>> Finally:
>>
>> (L/c) * 1/sqrt(1-(v/c)^2) * sqrt(1-(v/c)^2) = L/c <Eq.12>
>
>OK, I'm lost, but I think you've factored in gamma one too many times.
>
>> Let's note that the length contraction factor
>> fo = sqrt [(1-(v/c)) / (1+(v/c))] is smaller than one (or 1 if
>> v=0), meaning that the arm will contract if the direction of
>> the light beam which travels along it is the same as that of
>> the velocity vector, and that the length contraction factor
>> fb = sqrt [(1+(v/c)) / (1-(v/c))] is greater than one (or 1 if
>> v=0), hence that the arm will *dilate* when the direction of the
>> beam is opposite to that of the velocity vector !
>
>I disagree, the contraction is a factor of gamma, period. The outbound
>trip time is my t_1, and back is my t_2. This leads to a round trip
>time of:
>
> 2L
> t = ------------------
> cSqrt[1 - (v/c)^2]
>
>which of course is:
>
> 2Lgamma
> t = --------------
> c[1 - (v/c)^2]
>
>
>Explicitly bringing in the gamma contraction factor along x.
>
>> Iow, the arm of the interferometer will contract or dilate
>> according to the direction of the light beam which travels
>> along it!
>
>No, the arm is simply contracted by gamma, due to its velocity v.
>
>> Btw, how would you label any theory (other than SR of course)
>> leading to such a contradictory result?
>
>Let's NOT go there, suffice to say ad hoc.
>
>Paul Stowe
>
P.S. : I also post this article in the thread " Re: Which contraction
factor? "
Marcel Luttgens
Frank Wappler
> Well, if you are suggesting using only two clocks a & b
I was sketching the conventional procedure to measure
distance of any one pair wrt. each other, in any one trial.
Applied to your experiment stated earlier, which involved
clocks a, b, and c, one should measure the three distances
(a, b), (a, c), and (b, c).
(That's assuming the procedure guarantees the same distance
values both ways; other, even reproducible definitions of
pairwise distance might not guarantee that, and then one
should have measured the distance values (a to b), (b to a),
(a to c), (c to a), (b to c) and (c to b).)
But let's consider your modified experiment:
> one transmitter (a) & receiver (b), that should also work
> under the same procedure. As shown below:
> a<------ 100 m ------>b
What do you mean by "m"?
> The distance IS arbitrarily set large enough to assure that
> the signal delay is ten times greater that the clocks accuracy.
How is that distance value _measured_, in order "to assure"?
What do you mean by/how would you measure "clocks accuracy"
to begin with? Is this a quantity derived from past trials?
How would you measure its value _while_ the experiment is in progress?
> Again 'a' transmits at a predetermined interval
> and 'b' records the reception time.
Are the emissions events predetermined and specified
in terms of states/readings/proper_times of "a"?
How would you determine their relation to the
reception events/states/readings/proper_times of "b"
such as:
"I, b, have seen a light signal emitted by a.",
"After that, I, b, have seen anoth light signal emitted by a.",
"After that, I, b, have seen anoth light signal emitted by a." ...
> As long as the cycle spans at least twelve hours, and perferably
> 183 days... it should still test the one way symmetry.
There's basic asymmetry in your prescription already:
clock a is _distinct_ and even "at a nonzero distance from" clock b.
What sort of "symmetry" are you attempting to "test" there anyways?
Tom Roberts
> Tom Roberts wrote:
> > Frank Wappler wrote:
> > > It is sufficient for the _ratio_ "c/lambda" to remain constant,
> > No, it is sufficient for the quantity c/(2L/n) to remain constant,
> Can you please explain the experimental difference between those
> two constraints?
The experimenter has no control over lambda, except for the imposition
of the Fabry-Perot boundary conditions. He has direct control over the
length of the Fabry-Perot cavity.
> > One can assume, as did Brillet and Hall, that c is invariant
> Isn't that an assumption about "spatial isotropy"
> for the "propagation of light"?
It is an assumption about the isotropy of the speed of light. They
chose to assume that and to derive isotropy of space.
> > One can also assume, as I do, that L is constant
> Isn't that an assumption about "spatial isotropy"?
No, it's a description of their experimental technique. They went to
great effort to ensure that L did not vary.
Note I use an operational definition of length: measure it
with a "perfect" ruler. So your inquiry about "spatial
isotropy" is irrelevant -- any "change" in L would be matched
by a "change" in the ruler. This is, of course, the usual
meaning of "length" in modern physical theories.
> > The experimenters went to great effort to ensure that L does not vary.
> How did they detemine whether or not their great effort had the
> desired consequence; IOW, how did they measure and compare "L",
> trial by trial?
By observing that nobody came along and sawed off a part of it, or
stretched or compressed it, etc. And by ensuring that no known physical
effect could do so either.
Tom Roberts tjro...@lucent.com
Tom Roberts
> In <37C9AFF0...@lucent.com> Tom Roberts <tjro...@lucent.com>
> writes:
> > Where do you get that "gamma" from??
> It's always there,
NO!!! it is _NEVER_ there when we work in a single frame, as I
explicitly said I was doing.
Where is "gamma" on your kitchen clock? Or even on an atomic
clock used at NIST? Where is it on your rulers?
gamma is a function of v, and for v=0, gamma=1, and is in any case
irrelevant because all clocks are at rest in the frame I am working in.
> Ah-ha, I see where presumption
> that c 'is' isotropic comes from. Not postulate #1 (c is globally
> invariant), but postulate #2, the local speed of light IS c, NOT 'will
> appear' to be c in a round trip.
That is in SR. I was discussing the entire class of theories, and
_NEITHER_ postulate of SR is valid for any members of the class except
SR. Yes, the second postulate of SR directly _assumes_ the isotropy
of the 1-way speed of light. But the other theories in this class
do not.
> > If you want to assume that "gammas" appear, then you need to be
> > more precise in describing what you are doing -- they sure don't
> > appear when working in a single frame, which is what I was doing,
> > and what I thought you meant....
> Oh they're always there, they just factor out as demonstrated above.
You need to describe _specifically_ what you have in mind. There _IS_
no gamma in what I have been describing.
> >Let me make the following assumptions:
> > [...]
> > 3) the coordinate transform from the ether frame to any inertial
> > frame is linear, and is characterized by the velocity of the
> > inertial frame as measured in the ether frame.
> > [...]
> However, your #3 becomes invalid if a distortion factor of gamma is
> applicable to said motion. That is to say Sqrt[1 - (v/c)^2] certainly
> is not linear!
I said the _COORDINATE_TRANSFORM_ is linear. That means that x' is
a linear function of x amd t, and t' is a linear function of x and t.
This is inherently required of coordinate transforms among Cartesian
coordinates in inertial frames. The functional dependence of gamma on
v is irrelevant, because neither is a coordinate -- for a given
transform both are constants.
> As I said before, I think this difference has the potential to
> discriminate between LET and basic SR. A.k.a, a means of falsifing
> LET, or conversely, some aspects of SR. But experiments would have to
> tell us which.
Then you are seriously lacking in understanding the relationship
between LET and SR. THEY ARE THE SAME MATHEMATICAL THEORY, but with
different physical interpretations of the quantities appearing in the
equations. But the relationship between theory and experiment is
independent of such interpretations, and SR and LET are experimentally
indistinguishable. _Provably_ so.
In fact, the situation is more dismal than that (as far as observing
the ether is concerned). Not only LET, but _every_ "reasonable" ether
theory (i.e. every member of the class of theories) is experimentally
indistinguishable from SR.
"reasonable" means "not already refuted by other experiments",
so the round-trip speed of light is isotropically c in every
intertial frame, for every theory belonging to this class.
Tom Roberts tjro...@lucent.com
Tom Roberts
> Well, if you are suggesting using only two clocks a & b and one
> transmitter (a) & receiver (b), that should also work under the same
> procedure. As shown below:
> a<------ 100 m ------>b
> The distance IS arbitrarily set large enough to assure that the signal
> delay is ten times greater that the clocks accuracy. Again 'a'
> transmits at a predetermined interval and 'b' records the reception
> time. As long as the cycle spans at least twelve hours, and perferably
> 183 days... it should still test the one way symmetry.
Nope. Every "reasonable" ether theory predicts a null result for this
(i.e. the phase relationship between signals received from a by b and the
local clock b remains constant, independent of orientation). You have
ignored the way slow clock transport affects the synchronization of your
clocks.
This is essentially Torr and Kolen's experiment. For a detailed discussion
see:
Subject: Torr and Kolen's Experiment Cannot See the Ether.
Tom Roberts tjro...@lucent.com
Paul Alan Cardinale
Arlin Brown wrote:
>
<snip>
> BTW, Archimedes trisected any
> given angle with a straight edge and compass.
>
Most likely, arlin brown is referring to the method of trisecting an
angle
that requires putting a mark on the straight edge. However, in putting
that mark on the tool, one has changed it from being merely a straight
edge
into a ruler. Thus the method is disqualified from being considered
using
ONLY a straight edge and a compass.
Paul Cardinale
Arlin Brown
paul writes: Arlin Brown wrote: BTW, Archimedes trisected any given
angle with a straight edge and compass.
paul: Most likely, arlin brown is referring to the method of trisecting
an angle that requires putting a mark on the straight edge.
Arlin Jean: Sorry, Archimedes did not make any marks on the straight
edge.
---
Best,
Jeannie
Frank Wappler
> > Tom Roberts wrote:
> > > [Brillet and Hall ...] went to great effort to ensure
> > > to ensure that L does not vary.
> > How did they detemine whether or not their great effort had the
> > desired consequence; IOW, how did they measure and compare "L",
> > trial by trial?
> They went to great effort to ensure that L did not vary.
Quite possibly - but how _did_ they measure and compare "L"?
> By observing that nobody came along and sawed off a part of it,
> or stretched or compressed it, etc.
How does one/how would they measure _which_ effect any of those
actions (or the lack thereof) had on "L";
before as well as during their experiment?
If they were (or if someone else was) able to build and align their
"Fabry Perot interferometer" of a specific cavity length "L"
in the first place, then could they not conduct that same procedure
and check on the value of "L" throughout the experiment,
at least in principle?
> And by ensuring that no known physical effect could do so either.
Why should attention be restricted to "known physical effects"
anyways - weren't they (at least potentially) attempting to
"measure a new effect on L"
(correlated perhaps with other measurements obtained in the same trial)?
Correlations of "known physical effects" to values of "length", or
pairwise distance, must have been determined somehow in the first place;
those (conventional) procedures can be reproduced, at least in principle.
> Note I use an operational definition of length:
> measure it with a "perfect" ruler.
> So your inquiry about "spatial isotropy" is irrelevant --
> any "change" in L would be matched by a "change" in the ruler.
Why should that be? - what do you mean by "perfect ruler"?
Note that Brillet and Hall didn't make any reference to a "perfect ruler",
and that I don't understand what you mean by by that term a priori.
However, as much as anyone else, I can understand Einstein's calibration
procedure and the associated distance defintion, which allow to obtain
reproducible "length" measurements in the first place, independent of
any other measurements about "physical effects" in the same trial.
However, those conventional measurement procedures imply
"spatial isotropy of the roundtrip speed of light" necessarily;
thereby ruling out any "experimental test" of this quantity in turn
if those conventional measurement procedures were used to set up
and check such experiments in the first place.
Paul Stowe
Wappler) writes:
>
>Paul Stowe wrote:
>
>> Well, if you are suggesting using only two clocks a & b
>
>distance of any one pair wrt. each other, in any one trial.
>Applied to your experiment stated earlier, which involved
>clocks a, b, and c, one should measure the three distances
>(a, b), (a, c), and (b, c).
>(That's assuming the procedure guarantees the same distance
>values both ways; other, even reproducible definitions of
>pairwise distance might not guarantee that, and then one
>should have measured the distance values (a to b), (b to a),
>(a to c), (c to a), (b to c) and (c to b).)
>
>But let's consider your modified experiment:
>
>> under the same procedure. As shown below:
>
>> a<------ 100 m ------>b
>
I thought that m was a standard denotation of meter as in m/sec.
>> The distance IS arbitrarily set large enough to assure that
>> the signal delay is ten times greater that the clocks accuracy.
>
>How is that distance value _measured_, in order "to assure"?
>What do you mean by/how would you measure "clocks accuracy"
>to begin with? Is this a quantity derived from past trials?
>How would you measure its value _while_ the experiment is in progress?
This of course depends upon the smallest interval a given clock can
resolve or output, you know the 'ticks'. I think atomic clocks can
resolve down to about one nanosecond. Thus at a distance (L) of 100
meters, L/c = 333.56 nanoseconds.
>> Again 'a' transmits at a predetermined interval
>> and 'b' records the reception time.
>
>Are the emissions events predetermined and specified
>in terms of states/readings/proper_times of "a"?
The suggested procedure is as follows:
o Synchronized the clocks to within the precision tolerance
o Move them 100 meters apart and allow them to remain
untouched for 12 hours
o Bring them back together and compare current time values for
variations.
o Repeat this as necessary until one can gain a two sigma
confidence band on any drift (I doubt there would be any
noticable)
o Now place the clocks and turn on the transmitter/reciever and
start the pulse/recording cycle. Note the timing and duration of
these IS arbirary. The more the merrier.
o Analyse and plot the recorded data of delta time, that is the
time of actual transmission minus the times of actual reception.
By SR standards, this delta t should not vary, period!
>How would you determine their relation to the
>reception events/states/readings/proper_times of "b"
>such as:
>
>"I, b, have seen a light signal emitted by a.",
>"After that, I, b, have seen anoth light signal emitted by a.",
>"After that, I, b, have seen anoth light signal emitted by a." ...
>
>> As long as the cycle spans at least twelve hours, and perferably
>> 183 days... it should still test the one way symmetry.
>
>There's basic asymmetry in your prescription already:
>clock a is _distinct_ and even "at a nonzero distance from" clock b.
>
>What sort of "symmetry" are you attempting to "test" there anyways?
This test is discern which of the following equations are valid.
dt = L/c
NO VARIANCE, Or dt varies from:
(L/c)Sqrt[1 + v/c/1 - v/c]
to:
(L/c)Sqrt[1 - v/c/1 + v/c]
where v is to be determined by the magnitude of the variation in dt (IF
ANY). Prediction would be v ~ 350 Km/sec.
Key to this IS, this data (variations in dt) should NOT be random, but
sinusoidal in a 12 hour & 183 day period. This is suggested by
Lorentz's base frame and the first postulate of SR.
Paul Stowe
Paul Stowe
writes:
>Paul Stowe wrote:
>> Well, if you are suggesting using only two clocks a & b and one
>> transmitter (a) & receiver (b), that should also work under the same
>> procedure. As shown below:
>> a<------ 100 m ------>b
>> The distance IS arbitrarily set large enough to assure that the
signal
>> delay is ten times greater that the clocks accuracy. Again 'a'
>> transmits at a predetermined interval and 'b' records the reception
>> time. As long as the cycle spans at least twelve hours, and
perferably
>> 183 days... it should still test the one way symmetry.
>
>(i.e. the phase relationship between signals received from a by b and
>the local clock b remains constant, independent of orientation). You
>have ignored the way slow clock transport affects the synchronization
>of your clocks.
>
>This is essentially Torr and Kolen's experiment. For a detailed
discussion
>see:
>
> Subject: Torr and Kolen's Experiment Cannot See the Ether.
But Tom, didn't Torr and Kolen Experiment have not null results? Last
I checked, experiments are the ultimate arbiters of science, and just
because a theory 'predicts' something, measurements CAN override these
predictions.
Paul Stowe
Jim Carr
}
} [...] a limit on the anisotropy of the round-trip speed of light
} [... is] established experimentally (to an accuracy of
} a few parts in 10^15 for earthbound labs [...)]
In article <7q9llh$h...@mary.csc.albany.edu>
fw7...@csc.albany.edu (Frank Wappler) writes:
>
>Which experiment would have unambiguously established that?
Read the fairly recent paper by Clifford Will (on xxx.lanl.gov,
searchable if you lost the ID for it that has been posted here
a few times in the recent past) that is a follow on to his book.
>Obviously not the Brillet-Hall experiment (PRL42(9), 549, 1979)
>which only obtained a limit on the variation of a _frequency_.
And which is *twenty* years old.
--
James A. Carr <j...@scri.fsu.edu> | Commercial e-mail is _NOT_
http://www.scri.fsu.edu/~jac/ | desired to this or any address
Supercomputer Computations Res. Inst. | that resolves to my account
Florida State, Tallahassee FL 32306 | for any reason at any time.
Frank Wappler
> The suggested procedure is as follows:
> o Synchronized the clocks to within the precision tolerance
That's supposedly while those two are "side by side", as you wrote
earlier - IIUYC this means that light signal roundtrip intervals
between those two are supposed to be zero, and that then
for instance clock a's state
"a_k: when a sees clock b in state b_2:
when b sees clock a in state a_k: ..."
would be called "synchronized" to clock b's state
"b_2: when b sees clock a in state a_k:
when a sees clock b in state b_2: ..." .
> o Move them 100 meters apart and allow them to remain
> untouched for 12 hours
Besides the question how you would _determine whether or not_
they individually "remained untouched for 12 hours", why would this
be relevant to the light signals which those two clocks exchange?
> o Bring them back together and compare current time values for
> variations.
Yes, the states of those two clocks will be different; let's say
"a_q: when a sees clock b in state b_5:
when b sees clock a in state a_q: ..."
(which, as at least clock by a's assertion , is _after_ "a_k"), and
"b_7: when b sees clock a in state a_q:
when a sees clock b in state b_7: ..."
(which, as at least clock by b's assertion , is _after_ "b_2").
Since thereby "a_q" and "b_7" are "synchronized" to each other, and
"a_k" and "b_2" are "synchronized" to each other (as found above),
clock b will be able to agree that "a_q" was _after_ "a_k";
and in turn clock a will be able to agree that "b_7" was _after_ "b_2".
> o Repeat this as necessary until one can gain a two sigma
> confidence band on any drift (I doubt there would be any
> noticable)
Can you please explain this in more detail, let's say given
four pairs of states synchronized as above:
("a_f" and "b_0"), ("a_k" and "b_2"), ("a_q" and "b_7"), ("a_u" and "b_91")
(obviously one could use more than one letter to denote distinct states
of clock a, and more than one or two numbers to denote distinct states
of clock b).
> o Now place the clocks and turn on the transmitter/receiver and
> start the pulse/recording cycle. Note the timing and duration of
> these IS arbitrary. The more the merrier.
Yes, those would be for instance clock a's state
"a_j: when a sees clock b in state b_113:
when b sees clock a in state a_gzz:
when a sees clock b in state b_1004: ..."
etc. (The more the merrier.)
(The notation means that at least by clock b's assertion state
"b_2" comes after "b_113" comes after "b_1004" comes after "b_1";
and at least by clock a's assertion state
"a_h" comes after "a_gzz" comes after "a_g", ...)
> o Analyse and plot the recorded data of delta time, that is the
> time of actual transmission minus the times of actual reception.
But how should one calculate this difference between some state of
clock a, and some other state of clock b?
That's easy only for the few pairs that have been found "synchronized"
above: the difference/distance between "a_k" and "b_2" should be zero,
as both will agree since clock b's "b_2" corresponds to clock a's "a_k",
and the difference "between" one and the same state, "a_k" and "a_k"
is evidently zero.
But in order to express a difference between other pairs,
for instance between "a_gzz" and "b_1004", they would first of all
have to establish which state of clock b corresponds to a's "a_gzz";
and/or which state of clock a corresponds to b's "b_1004".
How do you suggest they accomplish this,
if not using Einstein's calibration procedure?
(Of course, both must then also be calibrated in terms of a time interval
unit "s". Again - how, if not via Einstein's calibration procedure?)
(Of course someone might say: just let them both realize the unit "s"
independently, each referencing a "Cs133 atom, in the ground state,
undisturbed by external fields". But how should clock a and clock b
agree _that_ their individual atoms are at least in the same
"groundstate" and are "undisturbed by external fields" at least to
similar effect, if not via Einstein's calibration procedure?)
> By SR standards, this delta t should not vary, period!
> > > a<------ 100 m ------>b
> > What do you mean by "m"?
> I thought that m was a standard denotation of meter as in m/sec.
So did I. Do you know that "meter" is a unit of distance which
is defined and to be realized via light signals roundtrips?
Obviously, if this definition of "m" is used to set up your experiment,
and the exchange of light signals is used to calibrate the states
of clock a and clock b along with that, following Einstein's procedure,
then the "delta t" should be the same throughout the experiment:
> L/c = 333.56 nanoseconds
I get 333.564 nanoseconds and a little - well, let's just consider
the experiment on the level of about ten picoseconds accuracy.
> This test is discern which of the following equations are valid.
> dt = L/c NO VARIANCE,
... that's guaranteed by the procedure above ...
> Or dt varies from:
> (L/c)Sqrt[1 + v/c/1 - v/c]
> to:
> (L/c)Sqrt[1 - v/c/1 + v/c]
Sure, if there were any variation being measured, then one might as well
parametrize it this way; whatever "v" (of dimension "speed") might mean.
The problem illustrated above is _how to measure_ "dt" in the first place,
such that at least clock a and clock b can agree on the result.
> Prediction would be v ~ 350 Km/sec.
"m" ... again!?
Paul Stowe
You're playing semantical games, its trival and wasting bandwith.
>> o Analyse and plot the recorded data of delta time, that is the
>> time of actual transmission minus the times of actual
>> reception.
>
>But how should one calculate this difference between some state of
>clock a, and some other state of clock b?
Your whole clock argument is an adaptation of Zeno's Paradox. Each
clock is an independent unit, each 'ticks' at a rate set in the case of
'atomic' clocks by the vibrational state of the Cesium 133 atom.
'manually' setting the times in synch in NO WAY alters this, just like
synchronizing two quarks wristwatches.
>That's easy only for the few pairs that have been found "synchronized"
>above: the difference/distance between "a_k" and "b_2" should be zero,
>as both will agree since clock b's "b_2" corresponds to clock a's
>"a_k", and the difference "between" one and the same state, "a_k" and
>"a_k" is evidently zero.
>
>But in order to express a difference between other pairs,
>for instance between "a_gzz" and "b_1004", they would first of all
>have to establish which state of clock b corresponds to a's "a_gzz";
>and/or which state of clock a corresponds to b's "b_1004".
>
>How do you suggest they accomplish this,
>if not using Einstein's calibration procedure?
>
>(Of course, both must then also be calibrated in terms of a time
>interval unit "s". Again - how, if not via Einstein's calibration
>procedure?)
>
>(Of course someone might say: just let them both realize the unit "s"
>independently, each referencing a "Cs133 atom, in the ground state,
>undisturbed by external fields". But how should clock a and clock b
>agree _that_ their individual atoms are at least in the same
>"groundstate" and are "undisturbed by external fields" at least to
>similar effect, if not via Einstein's calibration procedure?)
The answer is, they don't have to. They only have to 'tick' at the
same rate AND initially be set in agreement. Now you can of course
attempt to 'define' the issue away, but that is the behavior of what
most around here would call a 'crank'.
>> By SR standards, this delta t should not vary, period!
>
>> > > a<------ 100 m ------>b
>
>> > What do you mean by "m"?
>
>> I thought that m was a standard denotation of meter as in m/sec.
>
>So did I. Do you know that "meter" is a unit of distance which
>is defined and to be realized via light signals roundtrips?
>
>Obviously, if this definition of "m" is used to set up your
>experiment, and the exchange of light signals is used to calibrate
>the states of clock a and clock b along with that, following
>Einstein's procedure, then the "delta t" should be the same throughout
>the experiment:
>
>> L/c = 333.56 nanoseconds
>
>I get 333.564 nanoseconds and a little - well, let's just consider
>the experiment on the level of about ten picoseconds accuracy.
>
>> This test is discern which of the following equations are valid.
>
>> dt = L/c NO VARIANCE,
>
>... that's guaranteed by the procedure above ...
Good, then the experiment WOULD show this...
>> Or dt varies from:
>> (L/c)Sqrt[1 + v/c/1 - v/c]
>> to:
>> (L/c)Sqrt[1 - v/c/1 + v/c]
>
>Sure, if there were any variation being measured, then one might as
>well parametrize it this way; whatever "v" (of dimension "speed")
>might mean.
This is not an arbitrary variation, it has a very distinctive pattern.
>The problem illustrated above is _how to measure_ "dt" in the first
>place, such that at least clock a and clock b can agree on the result.
Oh please (eye rolling) you would fit right in with Zeno and his
fellows.
>> Prediction would be v ~ 350 Km/sec.
>
>"m" ... again!?
I hate to break the news to you but the meter definition and length
pre-dates its 'modern' light redefinition by hundreds of years.
Paul Stowe
Tom Roberts
> But Tom, didn't Torr and Kolen Experiment have not null results?
Apparently so.
There experiment is directly comparable to both Krishner et al and
Cialdea, both of which obtained null results.
I have been unable to obtain copies of Torr and Kolen's paper in which
they present results. By "reverse engineering" things said in this
newsgroup (always dangerous), I estimate they observed a daily variation
of about 5 ns in the phase difference they measure. I also estimate that
their Rhubidium clocks are nowhere near that accurate. I have this
sneaking suspicion that they actually reported something like 5+-20 ns
for their signal, and the reports of a "non-null" result are greatly
exaggerated, by people who have clearly demonstrated their lack of
experience and understanding. In particular, _NOBODY_ who has ever
mentioned their experiment as supporting an ether had _EVER_ quoted
their actual results or the error-bars on them; that to me is suspicious.
But in any case any claim that they "observed the ether" is _WRONG_,
because there is no reasonable ether theory which predicts a non-null
result for this experiment. In other words, any ether theorists are
as mystified as are SR theorists about their non-null result.
> Last
> I checked, experiments are the ultimate arbiters of science, and just
> because a theory 'predicts' something, measurements CAN override these
> predictions.
No experiment can "override" the predictions of a theory. A theory is
what it is. But a consistent and reproducible difference between
experiment and a theory indicates the theory is wrong, and a new theory
is needed. Note that Torr and Kolen do not have a _reproducible_
discrepancy with theory (Krishner et al and Cialdea did not confirm
their non-null result, and I believe that both had much better
accuracy).
Tom Roberts tjro...@lucent.com
Tom Roberts
> >Otoh, you claimed that "one-way" experiments are in fact
> >tests of the isotropy of the round-trip speed of light.
Essentially, yes. The one-way speed of light is unobservable because
it is inextricably intertwined with the synchronization convention one
chooses to use.
> >Does that means that no modification of the fringe pattern
> >could be observed when rotating a "one-way" interferometer?
I don't know what a "one-way" interferometer is. By its nature, an
interferometer looks at the difference between two rays propagated
from a single source, and that makes it inherently round-trip (two-way).
But look at Cialdea's experiment -- that is about as close to a
"one-way interferometer" as I can imagine. They observed no "fringe
shifts" (really a stable interference pattern), and no reasonable ether
theory predicts any.
Cialdea, Lett. Nuovo Cimento 4#16, p821 (1972)
Cialdea mounted two He-Ne lasers 2 meters apart and parallel
to each other on a rotatable table. The output of one hits a
45-degree mirror to direct it toward the other, where the two
are combined with a 45-degree beam-splitter/combiner, and the
combined beam goes into a fast photodiode whose output is
displayed on an oscilloscope (the lasers are prependicular to
the light path). Because of the multiple modes of the lasers,
the interference pattern on the oscilloscope is a complicated
function of time. As long as the phase relationships among the
modes remain stable, the pattern won't change. As the table is
rotated the pattern does not change, and he deduced a limit of
0.9 m/sec on any anisotropy in the speed of light.
While this may look like a one-way experiment, it is not sensitive to
the difference in the one-way speed of light among all the different
theories in the class of "reasonable" theories I discuss -- _ALL_
reasonable ether theories predict a null result, as does SR.
Tom Roberts tjro...@lucent.com
Frank Wappler
> Frank Wappler [wrote]:
> > Which experiment would have unambiguously established [...
> > a limit on the "anisotropy of of the round-trip speed of light"]?
> Read the fairly recent paper by Clifford Will
Apparently gr-qc/9811036, Clifford M. Will,
The Confrontation between General Relativity and Experiment:
A 1998 Update
which contained for instance a reference to
T. P. Kirshner, et. al. (incl. C. M. Will),
Test of the isotropy of the one-way speed of light using
hydrogen-maser frequency standards, PRD42 (2), 731, 1990.
This experiment observed relative phase relations of two
"hydrogen-maser frequency standards" wrt. each other.
Kirshner, et. al. parametrize phase as "phi == 2Pi nu L / c".
In order to derive a statement about "isotropy of c" from their
observations, how did they determine the relation between the
individual frequencies "nu" of the two masers; and how was determined
whether or not the "path lengths L" were equal in both directions?
> > Obviously not the Brillet-Hall experiment (PRL42(9), 549, 1979)
> > which only obtained a limit on the variation of a _frequency_.
> And which is *twenty* years old.
Well - that's the beauty about reproducible measurement procedures
and their results: once they have been formulated and _if_ they have
been obtained accordingly, then they _remain_ understandable and valid.
Regards, Frank W ~@) R
p.s.
A question about the so-called "equivalence principle"
that was mentioned in Will's paper:
An observer sits on one side of a platform and
"continues to feel constantly accelerated towards"
that platform, with "acceleration a".
How long does it take for the observer to conclude
that the platform didn't continue to operate as a rocket
and to "eject fuel" (even in the most efficient way),
but that it (simply) continued to have a certain "mass"?
MLuttgens
wrote :
>Date : Tue, 31 Aug 1999 12:47:33 -0500
Thanks.
But I keep claiming that a single "one-way" experiment is
not enough.
Why are the relativists seemingly reluctant to perform confirmatory
experiments, if not because they are unsure of the validity of
their theory?
Marcel Luttgens
Tom Roberts
> But I keep claiming that a single "one-way" experiment is
> not enough.
There are others. The most relevant is Krishner et al. Also look at
the two-photon experiment by Riis et al -- it's sort-of one-way (but
uses both directions simultaneously).
But given the solidity of the experimental observation that the
round-trip speed of light is isotropically c, there's little
incentive to do "1-way" experiments, as it is well known that the
1-way speed of light is unobservable (i.e. no experiment can
distinguish among different theories differing in the 1-way speed).
> Why are the relativists seemingly reluctant to perform confirmatory
> experiments, if not because they are unsure of the validity of
> their theory?
Because experiments are expensive, and one-way experiments require
two (or more) stable timing references; atomic clocks are very expensive.
Lasers are much less so, but that experiment has been done.
And experiments are also expensive in terms of one's career -- nobody
wants to waste time on an experiment which is highly unlikely to
observe anything new.
Tom Roberts tjro...@lucnet.com
DJMenCk
>> pre-dates its 'modern' light redefinition by hundreds of years.
Dennis: Paul, is this your first run in with Wappler?
Wappler: >Really? - what specificly and unambiguously would "meter" have meant
>before its 1983 "redefinition"?
>
>How would you determine and reproduce "circumfence of the earth",
>and communicate the result to anyone (for instance in order to
>explain the notion of "earth" in the first place)?
>How does the "length" of one stick relate to the "length" of some
>other stick; or to the "length" of the same stick in another trial?
Dennis: Again, I repeat, there are certain fundamental assumptions about
reality that most sane
people share and which make experiments and communication possible. If you
want to challenge all fundamental assumptions--like what one means by "string"
or "meter" or "next to each other" --it just becomes silly.
Save it for the dorm-rooms, the bong-fests, and the Dead concerts. Most of us
here have trivialized all such "deep" arguments when they were 17--and haven't
bothered with them since.
--D
Paul Stowe
(DJMenCk) writes:
>
>Stowe: >> I hate to break the news to you but the meter definition and
>length pre-dates its 'modern' light redefinition by hundreds of years.
>
>Dennis: Paul, is this your first run in with Wappler?
Umm, yeah...
Paul Stowe
Tom Roberts
> Further, do a search on "Atomic Clocks", I think
> you will find that one can obtain platforms (essentially radio coupled
> units to the NIST unit) for very reasonable costs.
Those are useless in looking for an ether -- the propagation of the radio waves
makes them not be independent clocks.
> But contrary to your position, I have given at least one derivation,
>[...]
I have no idea what you are talking about. Please describe the physical
situation and the derivation in one place.
But look first at my article Subject: Torr and Kolen's Experiment
Cannot See the Ether, and check whether or not your derivation
is consistent with both the approach presented there, and the
application of that approach to Torr and Kolen's experiment.
I suspect that you are not properly taking into account how slow
clock transport affects the synchronization of your clocks. Remember
that a slowly-transported clock retains its synchronization difference
with the coordinate clocks it is successively collocated with, and
that the coordinate clocks were synchronized using 1-way light signals.
> Last but BY NO MEANS least, I would think that if one were confident in
> the validity of Einstein's 'assumption', that one-way velocities were
> 'truly' isotropic, then they would say, "go for it, do all the
> tests/experiments you want". A null result would further solidify SR's
> premise, maybe even help in elevating it to a 'law' status. Why argue
> against running experiments that can actually discriminate between
> these two possibilities?
I'm not arguing against running experiments, I'm merely pointing out that
this entire class of theories makes identical predictions for them, so there
is no point in attempting to distinguish among the different theories of
this class (i.e. there's no point in trying to measure the 1-way speed of
light -- to do that you need to synchronize two clocks, and however you choose
to do that will directly affect the answer you get). All "reasonable" ether
theories belong to this class, as well as SR.
Tom Roberts tjro...@lucent.com
Paul Stowe
writes:
General Comments:
You have said one-way tests are theoretically impossible, since 'any'
answer one wishes to obtain is possible. However, above you say "
...one-way experiments require two (or more) stable timing references;
atomic clocks are very expensive" which, if taken at face value would
suggest otherwise. Further, do a search on "Atomic Clocks", I think
you will find that one can obtain platforms (essentially radio coupled
units to the NIST unit) for very reasonable costs.
But contrary to your position, I have given at least one derivation,
which gives the transit times [t] as:
_______________
L / 1 + vCos[a]/c
t =- \ / -------------
c v 1 - vCos[a]/c
Where angle [a] is the angle of the line between the
transmitter/receiver [T & R] and the velocity vector [v] references
Lorentz's primal and Einstein's global frame [of postulate #1], as
illustrated below:
T
/
/ a
---------------->v
a /
/
R
And L is the distance between the transmitter and receiver.
Now, as you know, when the above equation is applied to the
'round-trip' it results in the classical form of:
2L
t = ------------------
cSqrt[1 - (v/c)^2]
Giving us the proper time dilation formula.
Last but BY NO MEANS least, I would think that if one were confident in
the validity of Einstein's 'assumption', that one-way velocities were
'truly' isotropic, then they would say, "go for it, do all the
tests/experiments you want". A null result would further solidify SR's
premise, maybe even help in elevating it to a 'law' status. Why argue
against running experiments that can actually discriminate between
these two possibilities?
Paul Stowe
Frank Wappler
Marcel Luttgens wrote:
> Why are the relativists seemingly reluctant to perform
> confirmatory experiments, if not because they are unsure
> of the validity of their theory?
Trying to "test, validate, confirm or falsify experimentally"
the very measurement procedures by which any experimental results
are derived in first place _is circular_.
Such attempts are irrational, whether they concern
"experimental tests" of the Einstein procedures/SR,
or any other measurement procedures.
Instead, measurement procedures must be selected a priori
in order to conduct and analyze experiments, and obtain results.
Einstein's calibration procedure and the associated distance
definition are selected a priori and validated because
(and only insofar as) they are unambiguously understandable and
reproducible by all observers (who can count, at least in principle).
Frank Wappler
> Again, I repeat, there are certain fundamental assumptions about
> reality that most sane people share and which make experiments
> and communication possible.
As in our previous correspondence, I repeat that I agree with you,
and that those certain fundamental assumptions about reality
are well known:
sane observers share the ability to compare, to count, to do math.
> If you want to challenge all fundamental assumptions
... no, certainly not the one stated above ...
> --like what one means by "string" or "meter" or "next to each other"
> --it just becomes silly.
On the contrary, if you tell someone nothing more but
"to take a piece of string", then you'll be amazed at how diverse and
still _legitimate_ "pieces of strings" would be taken and satisfy
your (loose) prescription.
"Meter" is much better reproducible, at least since 1983, _because_
it doesn't merely refer to some incidental pair of rod ends or the other,
or some unique and irreproducible planetary body, but it comes with a
_procedure_ which pretty much anyone who reads it can understand.
"Next to each other" is indeed a primitive of that procedure, too;
but that doesn't mean its totally unambiguous either,
lest you thought that SR/Einstein's procedures were in any sense perfect.
Regards, Frank W ~@) R
p.s.
> Save it for the dorm-rooms, the bong-fests, and the Dead concerts.
> Most of us here have trivialized all such "deep" arguments
> when they were 17--and haven't bothered with them since.
Most of us here don't do a Physics Ph.D. either. Go figure ...
Frank Wappler
Frank Wappler wrote:
> > Frank Wappler [wrote]: [...]
The preceding post was sent in response to a post by Jim Carr.
That attribution had gotten lost - sorry. Frank W ~@) R
MLuttgens
wrote :
>Date : Wed, 01 Sep 1999 12:42:43 -0500
>
>MLuttgens wrote:
>> But I keep claiming that a single "one-way" experiment is
>> not enough.
>
>There are others. The most relevant is Krishner et al. Also look at
>the two-photon experiment by Riis et al -- it's sort-of one-way (but
>uses both directions simultaneously).
>
>But given the solidity of the experimental observation that the
>round-trip speed of light is isotropically c, there's little
>incentive to do "1-way" experiments, as it is well known that the
>1-way speed of light is unobservable (i.e. no experiment can
>distinguish among different theories differing in the 1-way speed).
>
Are you sure that the 1-way speed of light is always
inobservable? Think of experiments which don't use mirrors.
Btw, what is the purpose of distinguishing among different theories,
if some experiment can detect the "ether" wind?
>
>> Why are the relativists seemingly reluctant to perform confirmatory
>> experiments, if not because they are unsure of the validity of
>> their theory?
>
>Because experiments are expensive, and one-way
>experiments require two (or more) stable timing references;
No, some one-way experiments *does not* require such references.
>atomic clocks are very expensive.
>Lasers are much less so, but that experiment has been done.
>
>And experiments are also expensive in terms of one's
>career -- nobody wants to waste time on an experiment which
>is highly unlikely to observe anything new.
>
Till now, according to you, only three one-way experiments
have been performed (Cialdea, Krishner et al. and perhaps
Riis et al.). That's not enough, in view of the theoretical
implications of the existence of an absolute frame. SR and
GR would have to be adapted if such a frame is detected,
and the GUT could become reality much sooner.
Highly unlikely doesn't mean zero probability, and the
discoverer of the "ether" would probably be rewarded with
a significative career advancement, if not with a Nobel.
>
>Tom Roberts
Marcel Luttgens