The Wilson Chainsaw Demonstration Redrafted.
David Evens
Henry Wilson wrote:
>
> My apologies for omitting vital information in the original
> announcement of this world shattering experiment. At 4.am, on the
> coldest May day on record, here, and with a mild hangover as well as a
> nearby girlfriend humming the 'Internet Song', it is not surprising
> that the whole thing was a minor stuffup.
>
> The demonstration is seriously intended to show how a pair of
> perpendicular chainsaws can provide a slow motion simulation of the
> MMX apparatus.
> For a better analogy, I have modified the configuration so that the
> two chains are actually driven by the one common central cog. Refer to
> diagram, which is the best I can do.(use fixed pitch)
>
> (imaginary mirror)
> ___
> O
> | |
> | |
> up| |down
> | |
> | |____>c__
> O__________O| (imaginary mirror)
> <-c
>
> In the original message, I failed to explain that the chain velocity
> is denoted by the symbol 'c' for the purpose of the simulation only.
> The chain does not travel at the real velocity of light but at a
> normal one.
>
> The crux of the experiment is to regard the motions of the two chains
> as simulating the forward and reflected rays of each split light beam
> of an MM interferometer. Driving them both by a common cog, ensures
> that the chains will always maintain equal velocities, thus simulating
> the principle of equal light speed in their own frame of reference.
> The chain teeth represent individual wave crests.
>
> On inspection, it is obvious that, no matter how the whole apparatus
> moves horizontally, the vertical blade always remains exactly
> vertical.That is not what standard MMX diagrams show.
> The two chains always remain in perfect synch, establishing that the
> travel times in both arms is always equal. This explains why a null
> result should be expected in the MMX.
> Each TOOTH on the VERTICAL saw follows a diagonal path, the velocity
> along which is equal to sqrt(c^2+v^2). It is NOT c!! The chain itself,
> remains vertical!
And, of course, the chain does not resemble the light in the MMX as a
result.
> Now that you have a better understanding of the arrangement, I invite
> you to study the original article again. and please, no more insults,
> this is serious stuff.
> I repeat. Wasn't it a pity Einstein didn't own a chainsaw!
WHy? He might have made the same brain-dead error you keep demanding is
somehow consistent with reality.
EL
[EL]
I still say that you have one ***"NEW"*** beautiful point in
your post.
Only one, do not break your arm now. :-)
Is a light beam a photonic concatenation?
If the MMX would cause a frequency shift, how much would that be?
Would their equipment be good enough to measure it.
Ok. you made the very first point (concatenation).
:-)
EL Hemetis [ http://www.freeyellow.com/members8/hemetis/ ]
*
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Henry Wilson
Now that you have a better understanding of the arrangement, I invite
Paul B. Andersen
>
> [snip ]
Henry, please understand:
We all know that the MMX can be explained by source dependency of light.
It never was disputed.
It is so obvious that anybody will realize it right away.
So it is unnecessary to invent bad analogies to convince us.
Were you really not aware of that this is common knowledge?
Paul
EL
[EL]
What the hell is this very STUPID question from one who claims
humungous knowledge?
Rewind and Repeat:
What the hell is this very STUPID question from one who claims
humungous knowledge?
Is it not so much more obvious that he did not know that that
was common knowledge and that is why he made the analogy?
I would rather encourage him to produce more healthy thinking
than the generally common diarrhea we read around here.
Thusly, you make me sick.
Your mere showing off is a very strong evidence for any
psychologist of problems inherent in your character.
You could have kindly gave him a reference of a published work
that demonstrates the same proof, but you did not.
Oh! Mr. Know-it-all, Oh! no, I should say Mr. MMX, like you
sniff around for the letters.
Aether in your ear.
Aether in your eyes.
Aether in your brain.
Aether in yours.
I do not like you.
Go away.
You are slime, very bad person, very bad. Anderson
EL Kurt Hemetis. :-)
EL
[Wislon]
>EL the crucial point I am making is that there never is a
diagonal
>beam of light traveling at c. An incorrect claim that there is
one,
>formed the basis of SR.
>Individual wave crests move diagonal in the rest frame and
approach
>the cross mirror at sqrt(v^2+c^2). The chainsaw demo
illustrates that
>point very neatly. The teeth of the chain represent individual
crests.
>
>Has no-one else the courage to point out this massive error in
SR
>which has caused physics to be sidetracked for 100 years?
>
>
[EL]
I once pointed out the error in the construction of the MMX to
the so named P. Under Son. :-) And he accused me of everything
in the black book of the "Aetherist in opposition to dogma".
I gave you the most critical analogy clue (concatenation).
If photonic waves were not concatenated then all our frequency,
lambda, c and the shifts and physics and god and me and you are
hospitable beings, mindless brainless and nonsensical.
The only hope remaining for a diagonal is to stretch the ray
shifting it in the red direction.
What is the expected shift?
Is it measurable?
How did they prepare the triangulations?
Your analogy is crude but very demonstrative.
It was never in any common knowledge base as alleged by the liar.
All what he could do, is to pressure me to find a mistake in the
mathematical solution of an originally wrongly constructed
experiment.
When I explained that the mere mathematical construct is at
error to use such a small scalar to get any significant phase
shift at a short distance he dared me to prove it.
Now, he says that it is common knowledge base (So pathetic).
Did I provoke your evils enough Mr. Anderson??? ;-)
Defend Einstein and the MMX blindly. No need to think.
Why.
It is in the Journals.
It is experimentally verified.
It is awarded a Nobel prize.
It is much better than the similar but older dogma that lived
for centuries as a flat earth.
Aether, say Aether, good say Aether again. Aether.
Do you understand? Aether.
David Evens
Henry Wilson wrote:
> On Wed, 31 May 2000 23:38:44 -0400, David Evens
> <dev...@technologist.com> wrote:
> >result.
> >
> cross beam? Try it! You will see that the light beam itself always
> remains vertical, to any observer, just as the saw blade does.
Then why is it that the plot shows that the wave crests move DIAGONALLY
in any frame in which the apparatus is moving, just as the teeth of the
saw blade are?
> >> Now that you have a better understanding of the arrangement, I invite
> >> you to study the original article again. and please, no more insults,
> >> this is serious stuff.
> >> I repeat. Wasn't it a pity Einstein didn't own a chainsaw!
> >
> >WHy? He might have made the same brain-dead error you keep demanding is
> >somehow consistent with reality.
> No. He just jumped on a bandwagon and everyone else blindly followed.
How can the FIRST person to come up with an idea jump on a bandwagon?
EL
The light in the MMX will be seen to move diagonally in any
frame in which the apparatus is moving, since the light must
move between a source and a mirror and back again while the
source and mirror are moving along. The light OBVIOUSLY has to
move diagonally, since the mirror is not going to be at the
place it was when the light reaches its distance as it was when
the light was emitted. It takes TIME for the light to move from
source to mirror, and in that time the mirror has MOVED. In
EXACTLY the same way, incidentally, the PIECES of the saw blade
ALWAYS move diagonally in the frames in which the saw is not at
overall rest, since they have to move along in order to REMAIN
in the saw blade.
(then)
You ignored the fact that your own analogy shows your conclusion
to be wrong: For the pieces of the blade to remain in the blade
as it moves along, they MUST move diagonally, just as the wave
crests of light in the MMX MUST move diagonally to remain in the
apparatus as it moves along.
[EL]
You seem to be absolutely correct. That is by following one
photon from from the photons frame of reference (that is neither
source nor destination or even background), which is a dynamic
frame of reference. But it is true that a single wave would
trace a diagonal on a hypothetically moving background (Aether).
The motion inline ray is agreed to be compresses and expanded
equally and nulls out the effect.
We are all expecting the resultant vector inherent in the
temporal triangle formed by the orthogonal spatial ray to show a
phase shift as if it was *REALLY* moving with or against the
Aether.
Here is exactly the error. Inherent logic does not work on
physical nature.
The orthogonal ray shall not show any net result in any
direction because it also nulls itself **Physically*.
Whatever is added going up MUST be lost coming down under the
same physical conditions.
Wilson's analogy is beautiful, and you can use a band and a
motor as illustrated and put it in the wind. You should expect
the band to curve if it was standing still. But it is under the
tension dynamically. Would you think that light beams shall look
like they were curved by Aether flow?
Yet regardless of the effects they must return to motor shaft as
light must return to mirror on the spot. The MMX experiment was
not in error "IT IS an error".
Henry Wilson
<paul.b....@hia.no> wrote:
>Henry Wilson wrote:
>>
>> [snip ]
>
>Henry, please understand:
>We all know that the MMX can be explained by source dependency of light.
>It never was disputed.
>It is so obvious that anybody will realize it right away.
>So it is unnecessary to invent bad analogies to convince us.
>
>Were you really not aware of that this is common knowledge?
>
>Paul
I see that you fail to comment on the main point of my discovery,
which is that no diagonal beam of light ever exists. The time taken
for the vertical beam to return is always just 2L/c. This is where
Einstein was completely wrong! Pythagoras doesn't come into the
picture anywhere.
As for the forward beam, my analysis is no different from the standard
one.
Henry Wilson
<hemetis...@lilac.ocn.ne.jp.invalid> wrote:
>
>
>[EL]
>
>I still say that you have one ***"NEW"*** beautiful point in
>your post.
>
>Only one, do not break your arm now. :-)
>
>Is a light beam a photonic concatenation?
>
>If the MMX would cause a frequency shift, how much would that be?
>
>Would their equipment be good enough to measure it.
>
>Ok. you made the very first point (concatenation).
>
>:-)
Henry Wilson
Henry Wilson
>
>Henry Wilson wrote:
Would you care to enlarge on that.
>> Now that you have a better understanding of the arrangement, I invite
>> you to study the original article again. and please, no more insults,
>> this is serious stuff.
>> I repeat. Wasn't it a pity Einstein didn't own a chainsaw!
>
>WHy? He might have made the same brain-dead error you keep demanding is
>somehow consistent with reality.
Would you please enlarge on why my discovery is incorrect.
David Evens
The light in the MMX will be seen to move diagonally in any frame in
which the apparatus is moving, since the light must move between a
source and a mirror and back again while the source and mirror are
moving along. The light OBVIOUSLY has to move diagonally, since the
mirror is not going to be at the place it was when the light reaches its
distance as it was when the light was emitted. It takes TIME for the
light to move from source to mirror, and in that time the mirror has
MOVED. In EXACTLY the same way, incidentally, the PIECES of the saw
blade ALWAYS move diagonally in the frames in which the saw is not at
overall rest, since they have to move along in order to REMAIN in the
saw blade.
> >> Now that you have a better understanding of the arrangement, I invite
> >> you to study the original article again. and please, no more insults,
> >> this is serious stuff.
> >> I repeat. Wasn't it a pity Einstein didn't own a chainsaw!
> >
> >WHy? He might have made the same brain-dead error you keep demanding is
> >somehow consistent with reality.
> Would you please enlarge on why my discovery is incorrect.
YOu ignored the fact that your own analogy shows your conclusion to be
David Evens
Then why can you not SUPPORT your claim that light behaves in a manner
entirely unlike the manner in which it is observed to behave?
> As for the forward beam, my analysis is no different from the standard
> one.
So you admit that light behaves in the manner it is observed to behave
provided that it agrees with your randomly selected prejudices.
Bilge
>
>The crux of the experiment is to regard the motions of the two chains
>as simulating the forward and reflected rays of each split light beam
>of an MM interferometer. Driving them both by a common cog, ensures
>that the chains will always maintain equal velocities, thus simulating
>the principle of equal light speed in their own frame of reference.
>The chain teeth represent individual wave crests.
>
Good. I'll start with the common cog as my frame of reference.
>On inspection, it is obvious that, no matter how the whole apparatus
>moves horizontally, the vertical blade always remains exactly
>vertical.That is not what standard MMX diagrams show.
It is if the frame of reference is common cog. The common cog
is not moving. If it is moving, then the vertical does not remain
vertical. This is predicted and in real life appears as predicted
in the transverse doppler shift.
>The two chains always remain in perfect synch, establishing that the
>travel times in both arms is always equal. This explains why a null
>result should be expected in the MMX.
>Each TOOTH on the VERTICAL saw follows a diagonal path, the velocity
If each tooth follows a diagonal path, then it should be obvious
the arm isn't vertical. You can't define it to be two different
things ang then pick and choose between contradictory definitions
to illustrate a point.
>along which is equal to sqrt(c^2+v^2). It is NOT c!! The chain itself,
>remains vertical!
>
The chain only remains vertical in the rest frame of the cog.
An observer watching the apparatus pass in the direction of
the horizontal arm by will NEVER see the vertical arm as
vertical. The only reaason it appears vertical to the observer
in the cog restframe is that the observer is in that frame
at every measurement and can observe the same light sent by
his own test apparatus. An observer that sends two light pulses
one to be refelected from the near end and one from the far end
of the vertical arm will only see one of them return if both
are sent at the same time. The observer can NEVER be in the
same co-moving frame as both ends of the vertical arm unless
their relative velocities are all 0.
Relativity states that the rod will be seen as defined by the
simultaneous arrival of light from it.
d=vt d^2 = x^2 + x^2 - 2x^2 cos(\theta)
.\ x=ct
d. \ Points at O that are simultaneous correspond to light
.____ O sent at different times from the near and far ends or
x=ct in other words the rod will always appear tilted. The
observer will see the rod slowly rotate as it advances.
>Now that you have a better understanding of the arrangement, I invite
>you to study the original article again. and please, no more insults,
>this is serious stuff.
>I repeat. Wasn't it a pity Einstein didn't own a chainsaw!
You just don't understand your own apparatus and how relativity
describes it.
Paul B. Andersen
>
> On Thu, 01 Jun 2000 22:48:17 +0200, "Paul B. Andersen"
> <paul.b....@hia.no> wrote:
>
> >Henry Wilson wrote:
> >>
> >> [snip ]
> >
> >Henry, please understand:
> >We all know that the MMX can be explained by source dependency of light.
> >It never was disputed.
> >It is so obvious that anybody will realize it right away.
> >So it is unnecessary to invent bad analogies to convince us.
> >
> >Were you really not aware of that this is common knowledge?
> >
> >Paul
> I see that you fail to comment on the main point of my discovery,
> which is that no diagonal beam of light ever exists. The time taken
> for the vertical beam to return is always just 2L/c.
I think the releavant part of your thought experiment is:
| On inspection, it is obvious that, no matter how the whole apparatus
| moves horizontally, the vertical blade always remains exactly
| vertical.That is not what standard MMX diagrams show.
| The two chains always remain in perfect synch, establishing that the
| travel times in both arms is always equal. This explains why a null
| result should be expected in the MMX.
| Each TOOTH on the VERTICAL saw follows a diagonal path, the velocity
| along which is equal to sqrt(c^2+v^2). It is NOT c!! The chain itself,
| remains vertical!
It is hardly a "new discovery" that the interferometer arm (or blade)
which is vertical in the rest frame of the interferometer (or saws),
also is vertical in a frame which is moving relative to the interferometer
(your ground frame). It very obviously is.
But the light path is "diagonal" in the latter frame.
The "light path" is analogous to the "tooth path" in your thought
experiment, which you quite correctly say is a diagonal path.
It is this light/tooth path that is drawn diagonally in
the "standard MMX diagrams". The arm itself is always vertical.
Your thought experiment is an analogue to the ballistic light theory.
And deprived of all the Billy Halfdollar noise, it's even a good
analogy. (Why do you clutter up what might have been a good thought
experiment with all that nonsense?)
Your analogy quite correctly show that the ballistic light (or
source dependent light) theory will predict zero fringe shift
when the interferometer is rotated.
=================================================================
| But the point is that this is old news. |
| It has been known since day one in the history of the MMX. |
| Nobody has ever disputed that source dependent speed of light |
| is consistent with the MMX. |
=================================================================
Both SR and "ballistic light" predict a null result of the MMX.
MMX did show a null result.
The conclusion you can draw from this is that the MMX falsifies
neither of the theories.
And it proves neither of them.
The ether theory in which Michelson believed ("Galilean ether")
did however predict a fringe shift.
This particular ether theory was thus falsified.
Which is the historic importance of the MMX.
When "ballistic light" was never considered an option, it was
because other experiments had falsified the theory.
At the time the MMX was done "ballistic light" was considered
dead because it was falsified by other experiments.
That's why Michelson and Lorentz et al never considered it
to be an option, even if they were well aware that it could
explain the null result of the MMX.
At the present date source dependence of the speed of light
is thoroughly falsified by a number of experiments.
Not accepting that will inevitably give you a high score on
the crackpot index. :-)
Paul
EL
<paul.b....@hia.no> wrote:
>
>Have you learned it by now?
>
>Paul
>
>
[EL]
I can see that you are a very well educated person.
But I can also see that you are the same ass hole until now. :-)
I pity your students if you have any.
If you do not have any students, please do not commit that crime
and try to teach.
A good teacher must be equally good at listening and
understanding what is being said. You so not have that talent.
ryker1
<hemetis...@lilac.ocn.ne.jp.invalid> wrote:
>
>[Wislon]
>>EL the crucial point I am making is that there never is a
>diagonal
>>beam of light traveling at c. An incorrect claim that there is
>one,
>>formed the basis of SR.
>>Individual wave crests move diagonal in the rest frame and
>approach
>>the cross mirror at sqrt(v^2+c^2). The chainsaw demo
>illustrates that
>>point very neatly. The teeth of the chain represent individual
>crests.
>>
>>Has no-one else the courage to point out this massive error in
>SR
>>which has caused physics to be sidetracked for 100 years?
>>
>>
>
>EL Hemetis [ http://www.freeyellow.com/members8/hemetis/
]
Rod: If we sent pulses of light from the emitter that
traveled at v , then the light wouldn't , couldn't
comove with the emitter right ? No chain , no pain ! :)
Rod Ryker...
It is reasoning and faith that bind truth .
EL
[EL]
Drop your ego from your glasses and read carefully because I do
see value in your words.
Up to this moment, you were the most ordered mentality in
response to my posts.
This reflects two things, one is a powerful control and focus on
problem at hand.
But the second one is a rigid personality devoid of humor.
I was in error to expect you to be a perfection (which no one
is).
I shall give it a try and crystallize my character into a
hardest diamond to mach your rigidity.
Let me see, no jokes, no pun, no fun, no humor, no wit, no
analogy, no metaphor and ...what is left?
Ah! We were debating the MMX.
MMX setup:
1- Monochromatic light source.
2- Half silvered mirror. (critical point of interaction "O")
3- Full reflection mirror (M1 )
4- Full reflection mirror (M2)
Construction:
M1_O = M2_O = L
Light source is inline with the M2_O axis, which is the
direction of Aether current at maximum postulated effect (Major
axis of experiment).
M1_O is perpendicular to the major axis and inline with the
observer.
We should call the major axis the "Reference axis".
We should call the perpendicular axis the "Observation axis".
The original analogy on which the MMX was designed is the
famous "boat in the river" analogy. (Sorry this one was
obligatory).
The diagonal is designed such that :
V' = V (1-((v^2)/(V^2)))^(0.5)
During the time in which the physical boat is advancing to
finish the course "L" (even though the heading is designed as V)
by analyzing the progress path we shall see the boat moving in a
straight line along the path "L".
Thusly the vector V is only designed to compensate v and result
in a net vector V' to make the boat move ORTHOGONAL to the
current.
If we did not control the heading at the vector V the boat shall
land at a point on the other side that makes no perpendiculars
to the current.
Then we lose "L" and our calculations are in "Construction
error", (messing up the vector triangle).
In the MMX the reference ray is always perpendicular to the
effect ray.
Again according to the analogy, we know that the reference ray
should by faster with stream and slower against stream in an
equal measure.
The resulting reference is hence unaffected from a global point
of view.
Since the MMX did not include any mechanism for controlling the
perpendicular resultant vector on its way to the mirror, the ray
should land at an offset from the perfect point on the
perpendicular following a curved path and not a diagonal.
If each photon was a boat, each photon certainly did not have a
rudder to control the vector along the path.
Thusly it should interact in a passive manner to the flow and
produce a curvature rather than a steady vector.
By tilting the mirror we can compensate for such an offset to
return the ray to its original point of reflection.
Pay attention here now.
If we rotate the whole device 90 degrees the reference beam
shall be offset in such a way that compensates for the
perpendicular offset ray.
If you rotate the whole device 45 degrees the offsets shall
always compensate.
That is the real reason of the null result.
To cause a phase shift we need a delay in one of the beams. In
simple words, if the number of photons per ray never changed and
are in phase and the length of the path is the same, from where
can the phase shift start and under what influence? Once the
rays leave the monochromatic source we can only split the ray
into two in phase rays (as long as the length of the path is
invariable).
How did M&M expect Aether to change the length of the path? All
what we can get is a negligible displacement, which is still in
phase.
This should cause no significant change in the interference
fringes, because the rays are still in phase or better said they
are still at the original phase relative positions.
Please notice that a sideways displacement makes no phase shift,
it is the "in direction of propagation" displacement that causes
the shift.
The boat in the original analogy shall never be longer or
shorter, and the number of "CONCATENATED" boats within the
perpendicular length is invariable.
Thusly, the whole MMX construction is a fraudulent setup.
There is no mathematical error it is absolutely a conceptual
error if it was not on purpose.
So be my guest and show me how does this setup disprove the
existence of Aether.
EL Hemetis [ http://www.freeyellow.com/members8/hemetis/ ]
*
David Evens
Henry Wilson wrote:
> On Fri, 02 Jun 2000 00:02:37 -0400, David Evens
> <dev...@technologist.com> wrote:
> >Henry Wilson wrote:
> >> On Wed, 31 May 2000 23:38:44 -0400, David Evens
> >> <dev...@technologist.com> wrote:
> >> >Henry Wilson wrote:
> >>
> >> >> For a better analogy, I have modified the configuration so that the
> >> >> two chains are actually driven by the one common central cog. Refer to
> >> >> diagram, which is the best I can do.(use fixed pitch)
> >> >>
> >> >> (imaginary mirror)
> >> >> ___
> >> >> O
> >> >> | |
> >> >> | |
> >> >> up| |down
> >> >> | |
> >> >> | |____>c__
> >> >> O__________O| (imaginary mirror)
> >> >> <-c
> >> >>
> >> >> In the original message, I failed to explain that the chain velocity
> >> >> is denoted by the symbol 'c' for the purpose of the simulation only.
> >> >> The chain does not travel at the real velocity of light but at a
> >> >> normal one.
> >> >>
> >> >> as simulating the forward and reflected rays of each split light beam
> >> >> of an MM interferometer. Driving them both by a common cog, ensures
> >> >> that the chains will always maintain equal velocities, thus simulating
> >> >> the principle of equal light speed in their own frame of reference.
> >> >> The chain teeth represent individual wave crests.
> >> >>
> >> >> moves horizontally, the vertical blade always remains exactly
> >> >> vertical.That is not what standard MMX diagrams show.
> >> >> The two chains always remain in perfect synch, establishing that the
> >> >> travel times in both arms is always equal. This explains why a null
> >> >> result should be expected in the MMX.
> >> >> Each TOOTH on the VERTICAL saw follows a diagonal path, the velocity
> >> >> along which is equal to sqrt(c^2+v^2). It is NOT c!! The chain itself,
> >> >> remains vertical!
> >> >
> >> >result.
> >>
> >> Would you care to enlarge on that.
> >
> >The light in the MMX will be seen to move diagonally in any frame in
> >which the apparatus is moving, since the light must move between a
> >source and a mirror and back again while the source and mirror are
> >moving along. The light OBVIOUSLY has to move diagonally, since the
> >mirror is not going to be at the place it was when the light reaches its
> >distance as it was when the light was emitted. It takes TIME for the
> >light to move from source to mirror, and in that time the mirror has
> >MOVED. In EXACTLY the same way, incidentally, the PIECES of the saw
> >blade ALWAYS move diagonally in the frames in which the saw is not at
> >overall rest, since they have to move along in order to REMAIN in the
> >saw blade.
> >
> >> >> you to study the original article again. and please, no more insults,
> >> >> this is serious stuff.
> >> >> I repeat. Wasn't it a pity Einstein didn't own a chainsaw!
> >> >
> >> >somehow consistent with reality.
> >> Would you please enlarge on why my discovery is incorrect.
> >
> >YOu ignored the fact that your own analogy shows your conclusion to be
> >wrong: For the pieces of the blade to remain in the blade as it moves
> >along, they MUST move diagonally, just as the wave crests of light in
> >the MMX MUST move diagonally to remain in the apparatus as it moves
> >along.
> now.
You mean that you WANTED to prove yourself wrong?
> The main consequence of this is that only the chain teeth (ie, the
> light wave crests) move diagonally. The chain blade ( ie, the beam of
> light) still always appears vertical and to move at c, to all
> observers.
This, of course, cannot be relevant, since the overall blade does not
carry the information any more than the overall light beam.
> The time taken for each tooth (each wavecrest) to travel across each
> arm is just L/c, to all observers.
Of course, the speed c is not the speed of the elements of the saw blade
in the rest frame of a general observer, but only the component of the
velocity that is along the saw's length. This does not correspond to
the speed of the light in the MMX, which MUST be diagonal.
> Einstein and his deciples claimed that the light beam itself turned
> over diagonally and moved at c. That is not what happens, as the
> chainsaw example clearly illustrates.
Had you BOTHERED to read my post, you would have been unable to avoid
noticing that the PIECES of the saw blade ALL MOVE DIAGONALLY, just like
ALL THE WAVE CRESTS OF LIGHT MOVE DIAGONALLY.
> Each tooth (wave crest) closes in on the mirror at a speed of
> sqrt(c^2+v^2) wrt the rest frame. That does not violate any law about
> exceeding light speed.
And, of course, this is irrelevant to the fact that in ALL frames, the
SPEED OF LIGHT is ALWAYS measured to be the same. What you have
discovered is that the components of this velocity in different
direction differ for different observers. This is not interesting.
This is, in fact, one of the things that explains why time MUST appear
to run slower in all other frames, regardless of the observer doing the
measuring.
Paul B. Andersen
>
> I once pointed out the error in the construction of the MMX to
> the so named P. Under Son. :-) And he accused me of everything
> in the black book of the "Aetherist in opposition to dogma".
So are you an etherist? I was no aware of that.
I accused you of one thing:
Paul.B.Andersen wrote on Dec 11 1999 in thread "Fraud exposition"
> EL Hemetis wrote:
> > [about the M&M interferometer]
> > They used 500.0 nm wave length monochromatic light (Sodium lamp).
> > They used in their best enhancement 11 meters long arms.
> > The velocity of light is approximately 3.00 x 10^8 m/s
> > Time consideration is in the order of 4.00 x 10^-7 s
> > They calculated 0.400 fringes as a fringe shift expectation.
> > If you apply a dimensional analysis you shall discover that the
> > expected fringe shift should be in the order of nano-meters' fractions
> > because the wave interference is fractional to a wave length per
> > fringe, which we are expecting to be less than its half, yet they were
> > looking for it in a telescope with a hair line which was impossible to
> > calibrate for Angstroms but they were still looking for a shift, which
> > is less than the marking hairline width.
> > How brilliant, what a waste of time?
>
> So you do not even know what an interferometer is. :-)
Paul B. Andersen
>
> *
> * Sent from RemarQ http://www.remarq.com The Internet's Discussion Network *
> The fastest and easiest way to search and participate in Usenet - Free!
I understand that you are still sore because I several months ago
challenged you to substantiate your claim that "Michelson was a fraud."
Your "arguments" was much the same then as in this posting of yours.
Paul
Henry Wilson
The main consequence of this is that only the chain teeth (ie, the
light wave crests) move diagonally. The chain blade ( ie, the beam of
light) still always appears vertical and to move at c, to all
observers.
The time taken for each tooth (each wavecrest) to travel across each
arm is just L/c, to all observers.
Einstein and his deciples claimed that the light beam itself turned
over diagonally and moved at c. That is not what happens, as the
chainsaw example clearly illustrates.
Henry Wilson
<paul.b....@hia.no> wrote:
>Henry Wilson wrote:
>>
>
>> I see that you fail to comment on the main point of my discovery,
>> which is that no diagonal beam of light ever exists. The time taken
>> for the vertical beam to return is always just 2L/c.
>
>I think the releavant part of your thought experiment is:
>| On inspection, it is obvious that, no matter how the whole apparatus
>| moves horizontally, the vertical blade always remains exactly
>| vertical.That is not what standard MMX diagrams show.
>| The two chains always remain in perfect synch, establishing that the
>| travel times in both arms is always equal. This explains why a null
>| result should be expected in the MMX.
>| Each TOOTH on the VERTICAL saw follows a diagonal path, the velocity
>| along which is equal to sqrt(c^2+v^2). It is NOT c!! The chain itself,
>| remains vertical!
>
>It is hardly a "new discovery" that the interferometer arm (or blade)
>which is vertical in the rest frame of the interferometer (or saws),
>also is vertical in a frame which is moving relative to the interferometer
>(your ground frame). It very obviously is.
>But the light path is "diagonal" in the latter frame.
>The "light path" is analogous to the "tooth path" in your thought
>experiment, which you quite correctly say is a diagonal path.
>It is this light/tooth path that is drawn diagonally in
>the "standard MMX diagrams". The arm itself is always vertical.
>
>Your thought experiment is an analogue to the ballistic light theory.
>And deprived of all the Billy Halfdollar noise, it's even a good
>analogy. (Why do you clutter up what might have been a good thought
>experiment with all that nonsense?)
>
>Your analogy quite correctly show that the ballistic light (or
>source dependent light) theory will predict zero fringe shift
>when the interferometer is rotated.
This is not ballistic theory, It's pure wave stuff. You are missing a
vital point. Each wave crest, (chain tooth) moves along a different
diagonal. The number of wave crests (teeth) along the cross beam
vertical doesn't alter with velocity.
I'll try to draw what happens. Use fixed pitch fonts.
M
|
|
|x
/|
/ |
/ |y
/ / |
/ / |
/ / /|z
/ / / |
_________/____/____/____|_____
A B C -->v
The apparatus is moving at v, wrt a rest frame. A wave crest is
emitted at A, the next wave crests are emitted at B, C, etc,,.
No matter what the value of v is, the spacing x,y,z, between the wave
fronts along the vertical remains constant. For a fixed light
frequency, the spacing between points A,B and C increases as the
diagonal becomes less steep.
An observer in the apparatus frame sees the same number of wavelengths
in the cross beam, irrespective of speed
The chainsaw demonstration shows that clearly.
>
>=================================================================
>| But the point is that this is old news. |
>| It has been known since day one in the history of the MMX. |
>| Nobody has ever disputed that source dependent speed of light |
>| is consistent with the MMX. |
>=================================================================
>
>
>Both SR and "ballistic light" predict a null result of the MMX.
>MMX did show a null result.
>The conclusion you can draw from this is that the MMX falsifies
>neither of the theories.
>And it proves neither of them.
>
>The ether theory in which Michelson believed ("Galilean ether")
>did however predict a fringe shift.
>This particular ether theory was thus falsified.
>Which is the historic importance of the MMX.
The ether theory insisted that light would follow a diagonal beam and
move at c along that diagonal. After the null result, Einstein still
assumed the ether prediction but then invented the Lorentz contraction
to explain the null.
The point is, there should have been a null,anyway, just on geometry,
and Einstein had no justification for the contraction idea.
Henry Wilson
(Bilge) wrote:
>Henry Wilson said some stuff about
>
> >
> >The crux of the experiment is to regard the motions of the two chains
> >as simulating the forward and reflected rays of each split light beam
> >of an MM interferometer. Driving them both by a common cog, ensures
> >that the chains will always maintain equal velocities, thus simulating
> >the principle of equal light speed in their own frame of reference.
> >The chain teeth represent individual wave crests.
> >
> Good. I'll start with the common cog as my frame of reference.
>
> >On inspection, it is obvious that, no matter how the whole apparatus
> >moves horizontally, the vertical blade always remains exactly
> >vertical.That is not what standard MMX diagrams show.
>
> It is if the frame of reference is common cog. The common cog
> is not moving. If it is moving, then the vertical does not remain
> vertical. This is predicted and in real life appears as predicted
> in the transverse doppler shift.
Even Paul Anderson agrees with me on this. Sorry, the light beam
remains vertical to all observers.
>
> >The two chains always remain in perfect synch, establishing that the
> >travel times in both arms is always equal. This explains why a null
> >result should be expected in the MMX.
>
> >Each TOOTH on the VERTICAL saw follows a diagonal path, the velocity
>
> If each tooth follows a diagonal path, then it should be obvious
> the arm isn't vertical. You can't define it to be two different
> things ang then pick and choose between contradictory definitions
> to illustrate a point.
See my diagram in my next message.
>
> >along which is equal to sqrt(c^2+v^2). It is NOT c!! The chain itself,
> >remains vertical!
> >
> The chain only remains vertical in the rest frame of the cog.
> An observer watching the apparatus pass in the direction of
> the horizontal arm by will NEVER see the vertical arm as
> vertical.
You have to compensate for the time taken for the info to reach him.
Then at always appears vertical.
>The only reaason it appears vertical to the observer
> in the cog restframe is that the observer is in that frame
> at every measurement and can observe the same light sent by
> his own test apparatus. An observer that sends two light pulses
> one to be refelected from the near end and one from the far end
> of the vertical arm will only see one of them return if both
> are sent at the same time. The observer can NEVER be in the
> same co-moving frame as both ends of the vertical arm unless
> their relative velocities are all 0.
This is why I used distant 'instantaneous observers in my chainsaw
demo. But, like I said, whether the observer's eyes followed the saws
(thus moving their FOR with the car) or remained fixed makes no
difference to the vertical orientation of the blade.
>
> Relativity states that the rod will be seen as defined by the
> simultaneous arrival of light from it.
>
>
>d=vt d^2 = x^2 + x^2 - 2x^2 cos(\theta)
> .\ x=ct
>d. \ Points at O that are simultaneous correspond to light
>.____ O sent at different times from the near and far ends or
> x=ct in other words the rod will always appear tilted. The
> observer will see the rod slowly rotate as it advances.
this is just an observational effect. I don't disagree but it is not
really relevant here.
>
> >Now that you have a better understanding of the arrangement, I invite
> >you to study the original article again. and please, no more insults,
> >this is serious stuff.
> >I repeat. Wasn't it a pity Einstein didn't own a chainsaw!
>
> You just don't understand your own apparatus and how relativity
> describes it.
>
Bilge, point a laser at a distant wall. Spin the laser until the spot
moves much faster than c. The spot, if it could be analysed close up,
would be found to consists of a series of spots representing
individual wave crests. Thus these crests move wrt the wall at greater
than c, even though the laser beam itself moves at c.
This is exactly the same as the point I am making about the MMX.
Bilge
>Even Paul Anderson agrees with me on this. Sorry, the light beam
>remains vertical to all observers.
velocity. I'm sorry if you don't like it, but that's the way
it is. Relativity does say what you'd like it to
say.
>This is why I used distant 'instantaneous observers in my chainsaw
>demo. But, like I said, whether the observer's eyes followed the saws
>(thus moving their FOR with the car) or remained fixed makes no
>difference to the vertical orientation of the blade.
There is no such thing as "distant instantaneous observers".
If you are at any non-zero velocity relative to something,
You can only be co-moving with one point for one instant.
Every other point is connected by the time it takes the light
to cross the distance to that point. If you want to consider
the relativistic picture, You can never be co-moving with
the near and far end at the same instant. at time t, the
far end is t+d/v away. Only when d/v is insignificant can
you ignore it. Which is exactly what differes between classical
and relativistic mechanics.
>this is just an observational effect. I don't disagree but it is not
>really relevant here.
>
It is relavent. That's what defines simultaneous.
>Bilge, point a laser at a distant wall. Spin the laser until the spot
>moves much faster than c. The spot, if it could be analysed close up,
>would be found to consists of a series of spots representing
Since you cannot rotate the laser that fast, this is meaningless.
Anyone can violate physical laws and prove whatever they wish if
they first break the physical laws as part of the premise. You
cannot say relativity is incorrect based upon the hypothesis
that it is incorrect and expect to be taken seriously. Any
idiot can prove a tautology.
Paul B. Andersen
But the heading of the vector will be controled by the
"aiming device" of the light beam.
I will address this point below.
> Then we lose "L" and our calculations are in "Construction
> error", (messing up the vector triangle).
>
> In the MMX the reference ray is always perpendicular to the
> effect ray.
>
> Again according to the analogy, we know that the reference ray
> should by faster with stream and slower against stream in an
> equal measure.
>
> The resulting reference is hence unaffected from a global point
> of view.
Ooops.
If you travel a distance of 1m with the speed 10m/s + 5m/s = 15m/s
forth, and the speed 10m/s - 5m/s = 5m/s back, is then
the average speed 10 m/s?
No. It is 7.5 m/s.
> Since the MMX did not include any mechanism for controlling the
> perpendicular resultant vector on its way to the mirror, the ray
> should land at an offset from the perfect point on the
> perpendicular following a curved path and not a diagonal.
See below.
> If each photon was a boat, each photon certainly did not have a
> rudder to control the vector along the path.
>
> Thusly it should interact in a passive manner to the flow and
> produce a curvature rather than a steady vector.
>
> By tilting the mirror we can compensate for such an offset to
> return the ray to its original point of reflection.
No tilting of the mirrors would be necessary.
See below.
Right. Will do.
Note that the following analysis is based on the assumption
that there is an ether of the kind expected by Michelson.
The question is: what is the prediction of the MMX if such
an ether exists?
First point: ("heading of the boats")
-------------------------------------
What will happen to the light beam from an optical aiming
device if there is an ether cross wind? Will it be
"blown off course"?
The important point to realize is that the ether will
blow right through the aiming device.
An elementary aiming device is a screen with an aperture:
| uniform ether wind
V with speed v
\|/ | |
--*------------------------------------->|
/|\ | |
spherical screen target
light source with aperture screen
It should be immediately obvious that the light beam will
always hit at the same point on the target screen regardless
of the speed of the ether wind. The simple point is that
only the photons with a vertical speed components opposite
to the speed of the ether will go through the aperture.
You can refine the aiming device with parabolic mirrors
and/or lenses, but the above will remain true.
The light beam will always follow the optical axis of
the aiming device, and will not be "blown" off this
axis regardless of the velocity of the ether wind.
However, the length of the light path through the ether
(that is the length in the ether frame) will however depend
on the ether wind.
Thus when the interferometer has been collimated, the light beams
will always be along the arms and no adjusting of the mirrors
will be necessary when the interferometer is rotated.
So let us analyse the interferometer:
-------------------------------------
Let us assume that the length of each arm is d.
First we will calculate the transit-time of the light in each arm:
Arm parallel to movement:
The transit time t1 = time forth + time back
t1 = d/(c+v) + d/(c-v) = 2*d*c/(c^2 - v^2)
t1 = 2*d/(c*(1 - v^2/c^2))
A first order approximation will be:
t1 = 2*(d/c)*(1 + v^2/c^2) assuming that v << c.
Arm perpendicular to movement:
v
*----> * -
\ / |
\ / |
c\ / d
\ / |
\/ |
* -
Since the arm is moving through the ether, the length of the
light path in the ether will be: 2*d/sqrt(1 - v^2/c^2)
The transit time t2 = path-length/c
t2 = 2*d/(c*sqrt(1 - v^2/c^2))
A first order approximation will be:
t2 = 2*(d/c)*(1 + (1/2)*v^2/c^2) assuming that v << c.
The difference in the effective lengths of the light paths will
be: dl = (t1 - t2)*c wich to a first order approximation will be:
dl = (d/c)*(2 + 2*v^2/c^2 - 2 - 2*(1/2)*v^2/c^2)*c = d*v^2/c^2
When the apparatus is rotated 90 degrees, then the change in the
effective length of the arms will be opposite, so we get this
difference twice.
Change in effective length = 2*d*v^2/c^2
which was Michelson's prediction.
Paul
Ken H. Seto
<paul.b....@hia.no> wrote:
>Henry Wilson wrote:
>>
>> [snip ]
>
>Henry, please understand:
>We all know that the MMX can be explained by source dependency of light.
This is false.The MMX can be explained by the source _independency_ of
the speed of light.
Ken Seto.
EL
[EL]
This is precisely what I am objecting. "Aiming" is not
sufficient to guide the photons in a straight line, and by
tilting mirrors we end up with a game of racket and a ball
bouncing in a curved path. That is if we took gravity to
represent the force of Aether current,the rackets for the
mirrors and the ball for each photon.
>> Then we lose "L" and our calculations are in "Construction
>> error", (messing up the vector triangle).
>>
>> In the MMX the reference ray is always perpendicular to the
>> effect ray.
>>
>> Again according to the analogy, we know that the reference ray
>> should by faster with stream and slower against stream in an
>> equal measure.
>>
>> The resulting reference is hence unaffected from a global
point
>> of view.
>
>Ooops.
>If you travel a distance of 1m with the speed 10m/s + 5m/s =
15m/s
>forth, and the speed 10m/s - 5m/s = 5m/s back, is then
>the average speed 10 m/s?
>No. It is 7.5 m/s.
>
[EL]
That was very funny, because you just demonstrated in numbers
what I said in words:
For the equal measure you put 5m/s.
Faster upstream became 15m/s.
Slower downstream became 5m/s.
Then you calculated an average speed, which is a mental
convenience only and has nothing to do with REAL events taking
place.
The reference landing point shall not be displaced by any inline
operations.
Only the phase can be affected due to blue/ red shifting, the
calculation of which has nothing to do with average speed but we
must calculate each leg and arm separately to visualize the
compression/ expansion of the wave. After you finish
demonstrating wave compression/ expansion you shall be surprized
to find out that the wave frequency was unaffected due to an
equal shift-rate.
The monochromatic green light shall not become blue or yellow in
the laboratory frame by changing the mirror's variable
parameters.
This should make my point quite clear.
So, what was that oops for? :-)
You didn't really think I missed the trivial fact that the
average of any two velocities is a resultant vector, or did you?
Or is it that you want it to look like it is?
Bad Paul. :-)
>> Since the MMX did not include any mechanism for controlling
the
>> perpendicular resultant vector on its way to the mirror, the
ray
>> should land at an offset from the perfect point on the
>> perpendicular following a curved path and not a diagonal.
>
>See below.
>
[EL]
See above.
>> If each photon was a boat, each photon certainly did not have
a
>> rudder to control the vector along the path.
>>
>> Thusly it should interact in a passive manner to the flow and
>> produce a curvature rather than a steady vector.
>>
>> By tilting the mirror we can compensate for such an offset to
>> return the ray to its original point of reflection.
>
>No tilting of the mirrors would be necessary.
>See below.
>
[EL]
See above.
<snip published and very well known unnecessary arithmetic>
[EL]
Paul please, you can not shove copies of school arithmetic at me
as an acceptable debate technique. This is history.
I know the M&M calculations very well, we have no boats here, so
please pay attention to what is being disputed.
The M&M setup is making a wrong assumption from start, which is
the assumption that the speed of light shall change in different
Aether speeds. Based on that wrong assumption they treated waves
in a field as a higher order floating particulation, which is
not the case.
You shall do your best to pull me back to velocity vector
calculations where you must win the debate and where I have no
objection about, which makes no point at all. I am not in the
ring in which you are ready for boxing.
Electromagnetic radiation as was understood by Maxwell are waves
in a field, once we start the particle babble we must hit the
wall.
EMR is not a boat in a canal but waves in the canal.
According to Fresnel and Fizeau, the "phase" velocity of light
in a medium shall change from an "absolute" value by a quantity
determined by v(1-1/n^2), but it is Aether, which we take as
reference for an index n=1. This should show that phase velocity
in Aether is an absolute arbitration. It is the reference we
started with. Light phase velocity shall always be a constant in
all Aether velocities.
Since phase velocity is v = c/n = c/1 = c, as calculated by
Maxwell, and shall never change, there is no point in treating
Aether as any other medium because you have no refraction index
at all to start with (no change in medium before the ray hits
the eye), (optical glassware is compensated for).
M&M used a Newtonian V = c + v and V' = c - v, which is a very
naive approach (or fraudulent from a master perfectly versed in
optical wizardry)
Apart from all that, if phase velocity is not the issue and
phase shift from a fixed setup is an substantiated, we are left
with nothing else than red/ blue wave frequency shifting which
is very doubtful because the source and the destination are both
in the same time frame moving together.
Conclusion: MMX must give a null In all Aether conditions and
prove nothing.
David Evens
Henry Wilson wrote:
> On Fri, 02 Jun 2000 23:49:02 -0400, David Evens
> <dev...@technologist.com> wrote:
> >Henry Wilson wrote:
> >> On Fri, 02 Jun 2000 00:02:37 -0400, David Evens
> >> <dev...@technologist.com> wrote:
> >> >Henry Wilson wrote:
> >> >> On Wed, 31 May 2000 23:38:44 -0400, David Evens
>
> >> light wave crests) move diagonally. The chain blade ( ie, the beam of
> >> light) still always appears vertical and to move at c, to all
> >> observers.
> >
> >This, of course, cannot be relevant, since the overall blade does not
> >carry the information any more than the overall light beam.
> chain and demonstrating that it remains vertical.
Given the demonstration of the total irrelevance of this, it cannot
matter. Had you BOTHERED to understand the diagrams you mindlessly
attack, you would not be able to avoid noticing that the LIGHT BEAMS are
ALWAYS vertical, despite the fact that NO LIGHT TRAVELS VERTICALLY.
> >> The time taken for each tooth (each wavecrest) to travel across each
> >> arm is just L/c, to all observers.
> >
> >Of course, the speed c is not the speed of the elements of the saw blade
> >in the rest frame of a general observer, but only the component of the
> >velocity that is along the saw's length. This does not correspond to
> >the speed of the light in the MMX, which MUST be diagonal.
> No
No, what? No, the elements of the saw blade are NOT moving at some
speed other than in the general observer's frame?
> >> Einstein and his deciples claimed that the light beam itself turned
> >> over diagonally and moved at c. That is not what happens, as the
> >> chainsaw example clearly illustrates.
> >
> >Had you BOTHERED to read my post, you would have been unable to avoid
> >noticing that the PIECES of the saw blade ALL MOVE DIAGONALLY, just like
> >ALL THE WAVE CRESTS OF LIGHT MOVE DIAGONALLY.
> That's correct. But each one moves on a different diagonal.
Just as the diagrams you mindlessly attack say.
> At any time, all the wave crests are aligned vertically, irrespective
> of speed just like the teeth on the saw.
> Draw it!
> You will soon see that there is no diagonal beam of light.
What beam of light are you pretending is diagonal in the MMX diagrams
that you keep mindlessly attacking?
> >> Each tooth (wave crest) closes in on the mirror at a speed of
> >> sqrt(c^2+v^2) wrt the rest frame. That does not violate any law about
> >> exceeding light speed.
> >
> >And, of course, this is irrelevant to the fact that in ALL frames, the
> >SPEED OF LIGHT is ALWAYS measured to be the same.
> That's an assumption.
In reality, that is an OBSERVATION.
> >What you have
> >discovered is that the components of this velocity in different
> >direction differ for different observers.
> and apparent velocity can exceed c
Only for a non-inertial observer. Analysis of the scenario ALWAYS
results in inertial observers always seeing light raveling at c, and all
material bodies traveling slower.
> >This is not interesting.
> >This is, in fact, one of the things that explains why time MUST appear
> >to run slower in all other frames, regardless of the observer doing the
> >measuring.
> that's a circular argument.
Where did you imagine a circular argument appearing?
> However. Nothing I have said rules out the possibility that the ratio
> of the standards defining 'l/t' remains constant in all moving frames.
In reality, of course, REALITY prohibits this, since the light would
have to move at different speeds in different frame, which does not
occur.
David Evens
DO start to PAY ATTENTION, Ken. The fact is, that the MMX does NOT rule
out source dependence. It DOES limit the possible combinations of
conditions that can exist.
Henry Wilson
<dev...@technologist.com> wrote:
>
>Henry Wilson wrote:
>> On Fri, 02 Jun 2000 00:02:37 -0400, David Evens
>> <dev...@technologist.com> wrote:
>> >Henry Wilson wrote:
>> >> On Wed, 31 May 2000 23:38:44 -0400, David Evens
>> The main consequence of this is that only the chain teeth (ie, the
>> light wave crests) move diagonally. The chain blade ( ie, the beam of
>> light) still always appears vertical and to move at c, to all
>> observers.
>
>This, of course, cannot be relevant, since the overall blade does not
>carry the information any more than the overall light beam.
The blade is only useful wrt respect to the fact that it confines the
chain and demonstrating that it remains vertical.
>
>> The time taken for each tooth (each wavecrest) to travel across each
>> arm is just L/c, to all observers.
>
>Of course, the speed c is not the speed of the elements of the saw blade
>in the rest frame of a general observer, but only the component of the
>velocity that is along the saw's length. This does not correspond to
>the speed of the light in the MMX, which MUST be diagonal.
No
>
>> Einstein and his deciples claimed that the light beam itself turned
>> over diagonally and moved at c. That is not what happens, as the
>> chainsaw example clearly illustrates.
>
>Had you BOTHERED to read my post, you would have been unable to avoid
>noticing that the PIECES of the saw blade ALL MOVE DIAGONALLY, just like
>ALL THE WAVE CRESTS OF LIGHT MOVE DIAGONALLY.
That's correct. But each one moves on a different diagonal.
At any time, all the wave crests are aligned vertically, irrespective
of speed just like the teeth on the saw.
Draw it!
You will soon see that there is no diagonal beam of light.
>
>> Each tooth (wave crest) closes in on the mirror at a speed of
>> sqrt(c^2+v^2) wrt the rest frame. That does not violate any law about
>> exceeding light speed.
>
>And, of course, this is irrelevant to the fact that in ALL frames, the
>SPEED OF LIGHT is ALWAYS measured to be the same.
That's an assumption.
>What you have
>discovered is that the components of this velocity in different
>direction differ for different observers.
and apparent velocity can exceed c
>This is not interesting.
>This is, in fact, one of the things that explains why time MUST appear
>to run slower in all other frames, regardless of the observer doing the
>measuring.
that's a circular argument.
Paul B. Andersen
>
> <snip published and very well known unnecessary arithmetic>
>
> [EL]
> Paul please, you can not shove copies of school arithmetic at me
> as an acceptable debate technique. This is history.
And you have found no errors in this historic arithmetic?
> I know the M&M calculations very well, we have no boats here, so
> please pay attention to what is being disputed.
> The M&M setup is making a wrong assumption from start, which is
> the assumption that the speed of light shall change in different
> Aether speeds.
Of course Michelson did this wrong assumption.
That assumption was part of the particular ether theory
he believed to be valid.
Naive or not, Galilean relativity was part of the particular
ether theory Michelson belived to be valid.
> Apart from all that, if phase velocity is not the issue and
> phase shift from a fixed setup is an substantiated, we are left
> with nothing else than red/ blue wave frequency shifting which
> is very doubtful because the source and the destination are both
> in the same time frame moving together.
>
> Conclusion: MMX must give a null In all Aether conditions and
> prove nothing.
The MMX would indeed have given a fringe shift if the particular
ether theory Michelson believed to be valid had been so.
Thus the MMX falsified this theory.
If you are trying to say that Michelson's ether theory
was no consistent theory and therefore false independent of
experimental evidence, then I think you will have to prove
that in a more coherent manner than the above.
Paul
Paul B. Andersen
>
> On Fri, 02 Jun 2000 23:55:20 +0200, "Paul B. Andersen"
> >> which is that no diagonal beam of light ever exists. The time taken
> >> for the vertical beam to return is always just 2L/c.
> >
> >I think the releavant part of your thought experiment is:
> >| moves horizontally, the vertical blade always remains exactly
> >| vertical.That is not what standard MMX diagrams show.
> >| travel times in both arms is always equal. This explains why a null
> >| result should be expected in the MMX.
> >| Each TOOTH on the VERTICAL saw follows a diagonal path, the velocity
> >| remains vertical!
> >
> >which is vertical in the rest frame of the interferometer (or saws),
> >also is vertical in a frame which is moving relative to the interferometer
> >(your ground frame). It very obviously is.
> >But the light path is "diagonal" in the latter frame.
> >The "light path" is analogous to the "tooth path" in your thought
> >experiment, which you quite correctly say is a diagonal path.
> >It is this light/tooth path that is drawn diagonally in
> >the "standard MMX diagrams". The arm itself is always vertical.
> >
> >Your thought experiment is an analogue to the ballistic light theory.
> >And deprived of all the Billy Halfdollar noise, it's even a good
> >analogy. (Why do you clutter up what might have been a good thought
> >experiment with all that nonsense?)
> >
> >Your analogy quite correctly show that the ballistic light (or
> >source dependent light) theory will predict zero fringe shift
> This is not ballistic theory, It's pure wave stuff.
Particle or wave does not matter.
What does matter is the assumption that the speed of light
is c only in the rest frame of the source.
Indeed it does.
The point you seems to miss is that this conclusion follows
from you assumption: source dependent speed of light.
It would _not_ have been correct in "Galilean ether" (by which
I mean the particular ether theory in which Michelson believed),
where this assumption is not valid.
> >=================================================================
> >| But the point is that this is old news. |
> >| It has been known since day one in the history of the MMX. |
> >| Nobody has ever disputed that source dependent speed of light |
> >| is consistent with the MMX. |
> >=================================================================
> >
> >
> >Both SR and "ballistic light" predict a null result of the MMX.
> >MMX did show a null result.
> >The conclusion you can draw from this is that the MMX falsifies
> >neither of the theories.
> >And it proves neither of them.
> >
> >The ether theory in which Michelson believed ("Galilean ether")
> >did however predict a fringe shift.
> >This particular ether theory was thus falsified.
> >Which is the historic importance of the MMX.
> The ether theory insisted that light would follow a diagonal beam and
> move at c along that diagonal. After the null result, Einstein still
> assumed the ether prediction but then invented the Lorentz contraction
> to explain the null.
I think you confuse the names.
It was Lorentz and Fitzgerald who independently invented this
contraction to save the ether theory.
Hence its name: Lorentz-Fitzgerald contraction.
Einstein did not say that the arms of the interferometer contracts.
> The point is, there should have been a null,anyway, just on geometry,
> and Einstein had no justification for the contraction idea.
Nonsense.
It would definitely not have been null in "Galilean ether".
Paul
EL
<paul.b....@hia.no> wrote:
>EL wrote:
>>
>> <snip published and very well known unnecessary arithmetic>
>>
>> [EL]
>> Paul please, you can not shove copies of school arithmetic at
me
>> as an acceptable debate technique. This is history.
>
>
[EL]
The arithmetic itself is high school arithmetic, how can such
scientists make an error that would be published?
The error is in the setup of the experiment and concept
combined, they were looking for a phase shift interference,
which would have been acceptable if the moving currents were of
any other medium else than Aether, because it is Aether that was
taken as a reference for a refraction index. Thusly we shall
always yield a constant answer "UNITY". Thusly the change would
be "NULL". Nevertheless, it is very possible to observe a
sideways displacement effect (not phase shift interference), but
it would be extremely trivial or within the size of the
calibration hair itself. Since the arithmetic is not only
equations and procedures but also numerical reflection of the
experimental data, I have resented the limitations of the
design, which did not allow for any surprise to be found. The
target was 0.4 fringe shifts while the sideway displacement
could have been (consequently), less than 0.00004 fringe to
fringe lengths.
>> I know the M&M calculations very well, we have no boats here,
so
>> please pay attention to what is being disputed.
>> The M&M setup is making a wrong assumption from start, which
is
>> the assumption that the speed of light shall change in
different
>> Aether speeds.
>
>Of course Michelson did this wrong assumption.
>That assumption was part of the particular ether theory
>he believed to be valid.
>
>Naive or not, Galilean relativity was part of the particular
>ether theory Michelson belived to be valid.
>
[EL]
That is not a good enough excuse to use the Fizeau-Fresnel
interferometry in a Newtonean Frame, which was an obvious error
even at that period of time.
And neglecting that was reason not to realize that Aether to
Aether have no refraction to start with for any dynamic medium
velocity determination as in the case of the Fizeau experiment.
The whole ancient idea was based on the discovery that the
refraction index (with respect to Aether) would change if the
medium was moving with or against the direction of light.
In order to observe that change quantitatively, Fizeau used the
Fresnel well established interferometry by that time.
Thusly, what is the meaning of an attempt to produce a phase
shift interference without a refraction index? But of course
they might have tried but it was an obvious NULL from start, so
they dropped the whole orthodox method and adopted the Newtonean
calculations to disprove Aether. Was it a conspiracy of some
kind or what?
>> Apart from all that, if phase velocity is not the issue and
>> phase shift from a fixed setup is an substantiated, we are
left
>> with nothing else than red/ blue wave frequency shifting which
>> is very doubtful because the source and the destination are
both
>> in the same time frame moving together.
>>
>> Conclusion: MMX must give a null In all Aether conditions and
>> prove nothing.
>
>The MMX would indeed have given a fringe shift if the particular
>ether theory Michelson believed to be valid had been so.
>Thus the MMX falsified this theory.
>
>If you are trying to say that Michelson's ether theory
>was no consistent theory and therefore false independent of
>experimental evidence, then I think you will have to prove
>that in a more coherent manner than the above.
>
>Paul
>
[EL]
I am tired of proving over and over Paul, my time is really
limited and I have to type very fast most of the time.
I need no glories of any proof. I am humbly raising an issue for
investigation by those who are more concerned with it.
My major question is, based on what did M&M design the
interferometer to measure the speed of Aether?
Regards.
Henry Wilson
<dev...@technologist.com> wrote:
>
>Henry Wilson wrote:
>> On Fri, 02 Jun 2000 23:49:02 -0400, David Evens
>> <dev...@technologist.com> wrote:
>> >Henry Wilson wrote:
>> >> On Fri, 02 Jun 2000 00:02:37 -0400, David Evens
>> >> <dev...@technologist.com> wrote:
>> >> >Henry Wilson wrote:
>> >> >> On Wed, 31 May 2000 23:38:44 -0400, David Evens
>>
>> >> The main consequence of this is that only the chain teeth (ie, the
>> >> light wave crests) move diagonally. The chain blade ( ie, the beam of
>> >> light) still always appears vertical and to move at c, to all
>> >> observers.
>> >
>> >This, of course, cannot be relevant, since the overall blade does not
>> >carry the information any more than the overall light beam.
>> The blade is only useful wrt respect to the fact that it confines the
>> chain and demonstrating that it remains vertical.
>
>Given the demonstration of the total irrelevance of this, it cannot
>matter. Had you BOTHERED to understand the diagrams you mindlessly
>attack, you would not be able to avoid noticing that the LIGHT BEAMS are
>ALWAYS vertical, despite the fact that NO LIGHT TRAVELS VERTICALLY.
You obviously know nothing about the subject.
>
>> >> The time taken for each tooth (each wavecrest) to travel across each
>> >> arm is just L/c, to all observers.
>> >
>> >Of course, the speed c is not the speed of the elements of the saw blade
>> >in the rest frame of a general observer, but only the component of the
>> >velocity that is along the saw's length. This does not correspond to
>> >the speed of the light in the MMX, which MUST be diagonal.
>> No
>
>No, what? No, the elements of the saw blade are NOT moving at some
>speed other than in the general observer's frame?
>
>> >> Einstein and his deciples claimed that the light beam itself turned
>> >> over diagonally and moved at c. That is not what happens, as the
>> >> chainsaw example clearly illustrates.
>> >
>> >Had you BOTHERED to read my post, you would have been unable to avoid
>> >noticing that the PIECES of the saw blade ALL MOVE DIAGONALLY, just like
>> >ALL THE WAVE CRESTS OF LIGHT MOVE DIAGONALLY.
>> That's correct. But each one moves on a different diagonal.
>
>Just as the diagrams you mindlessly attack say.
Tell me about one.
Every MMX diagram I have seen states that the cross beam of light
moves diagonally and then goes on to insist that it must travel at c
along this path. That is false. As the chainsaw shows, the entire
lightbeam moves horizontally, just like a rigid rod, while the light
moves up it at c. The number of wavelengths in the cross beam does not
change because of its longitudinal motion. This is the crucial factor
that Einstein et al. refused to accept.
>
>> At any time, all the wave crests are aligned vertically, irrespective
>> of speed just like the teeth on the saw.
>> Draw it!
>> You will soon see that there is no diagonal beam of light.
>
>What beam of light are you pretending is diagonal in the MMX diagrams
>that you keep mindlessly attacking?
Get hold of any book on the subject and you will see a diagram showing
a cross beam drawn at an oblique angle. It will then state that the
beam travels at c up that diagonal. That is plain BULLSHIT!
>
>> >> Each tooth (wave crest) closes in on the mirror at a speed of
>> >> sqrt(c^2+v^2) wrt the rest frame. That does not violate any law about
>> >> exceeding light speed.
>> >
>> >And, of course, this is irrelevant to the fact that in ALL frames, the
>> >SPEED OF LIGHT is ALWAYS measured to be the same.
>> That's an assumption.
>
>In reality, that is an OBSERVATION.
Nobody has ever measured the one way velocity of light from a moving
source.
>
>> >What you have
>> >discovered is that the components of this velocity in different
>> >direction differ for different observers.
>> and apparent velocity can exceed c
>
>Only for a non-inertial observer. Analysis of the scenario ALWAYS
>results in inertial observers always seeing light raveling at c, and all
>material bodies traveling slower.
>
>> >This is not interesting.
>> >This is, in fact, one of the things that explains why time MUST appear
>> >to run slower in all other frames, regardless of the observer doing the
>> >measuring.
>> that's a circular argument.
>
>Where did you imagine a circular argument appearing?
>
>> However. Nothing I have said rules out the possibility that the ratio
>> of the standards defining 'l/t' remains constant in all moving frames.
>
>In reality, of course, REALITY prohibits this, since the light would
>have to move at different speeds in different frame, which does not
>occur.
SR says that light always travels at c in any frame. That implies that
the ratio of reference scales for velocity units, L/T, remains
constant in all frames. Using the incorrect assumption that the
velocity of the diagonal beam in the MMX travels at c, the constancy
of L/T is assured because of the same contraction in L and T.
Henry Wilson
<paul.b....@hia.no> wrote:
>EL wrote:
>>
>> <snip published and very well known unnecessary arithmetic>
>> Conclusion: MMX must give a null In all Aether conditions and
>> prove nothing.
>
>The MMX would indeed have given a fringe shift if the particular
>ether theory Michelson believed to be valid had been so.
>Thus the MMX falsified this theory.
>
>Paul
But instead of looking for the error an accepting the obvious
explanation, Einstein stuck to the aether notion that the transverse
light beam moved diagonally at c. It does not!
He then wrongly invented the absolutely ridiculous idea of a length
contraction to avoid accepting the officially unpopular notion that,
in certain configurations, light speed might appear to be source
dependent.
Henry Wilson
<paul.b....@hia.no> wrote:
At least we agree on something.
>
>> >=================================================================
>> >| But the point is that this is old news. |
>> >| It has been known since day one in the history of the MMX. |
>> >| Nobody has ever disputed that source dependent speed of light |
>> >| is consistent with the MMX. |
>> >=================================================================
>
>> The ether theory insisted that light would follow a diagonal beam and
>> move at c along that diagonal. After the null result, Einstein still
>> assumed the ether prediction but then invented the Lorentz contraction
>> to explain the null.
>
>I think you confuse the names.
>It was Lorentz and Fitzgerald who independently invented this
>contraction to save the ether theory.
>Hence its name: Lorentz-Fitzgerald contraction.
>
>Einstein did not say that the arms of the interferometer contracts.
>
>> The point is, there should have been a null,anyway, just on geometry,
>> and Einstein had no justification for the contraction idea.
>
>Nonsense.
>It would definitely not have been null in "Galilean ether".
>
I agree. I am not an etherist, even though I think some kind of
absolutivity is a very attractive theory.
>Paul
Paul B. Andersen
>
> In article <393A9BF0...@hia.no>, "Paul B. Andersen"
> <paul.b....@hia.no> wrote:
> >EL wrote:
> >>
> >> <snip published and very well known unnecessary arithmetic>
> >>
> >> [EL]
> >> Paul please, you can not shove copies of school arithmetic at
> me
> >> as an acceptable debate technique. This is history.
> >
> >And you have found no errors in this historic arithmetic?
> >
>
> [EL]
> The arithmetic itself is high school arithmetic, how can such
> scientists make an error that would be published?
Well. Michelson did such an error in his 1881 paper.
He corrected it in his 1886 paper, though.
> The error is in the setup of the experiment and concept
> combined, they were looking for a phase shift interference,
> which would have been acceptable if the moving currents were of
> any other medium else than Aether, because it is Aether that was
> taken as a reference for a refraction index. Thusly we shall
> always yield a constant answer "UNITY". Thusly the change would
> be "NULL". Nevertheless, it is very possible to observe a
> sideways displacement effect (not phase shift interference), but
> it would be extremely trivial or within the size of the
> calibration hair itself. Since the arithmetic is not only
> equations and procedures but also numerical reflection of the
> experimental data, I have resented the limitations of the
> design, which did not allow for any surprise to be found. The
> target was 0.4 fringe shifts while the sideway displacement
> could have been (consequently), less than 0.00004 fringe to
> fringe lengths.
Could you please rephrase this in an intelligible manner?
Michelson assumed the existence of a rigid ether as the medium
of EM-waves. Maxwell's equation was assumed to be valid only
in the stationary frame of this ether. The Galilean transformation
was assumed to apply for transformation from the "ether frame"
to other frames like the interferometer frame. (This was taken
for granted at the time.)
Are you saying that this theory was inconsistent?
In that case, why is it inconsistent?
What exactly are you trying to say?
Fizeau's experiment was at the time considered to confirm Fresnel's
theory that optical media dragged the ether to an extent dependent
on their index of refraction.
Are you saying that it was an error of Michelson not to
consider the drag of the ether due to the air?
According to Fresnel's theory, this drag would be so small that
it could be neglected with no big error.
If this is not what you are saying - what _are_ you saying?
> >> Apart from all that, if phase velocity is not the issue and
> >> phase shift from a fixed setup is an substantiated, we are
> left
> >> with nothing else than red/ blue wave frequency shifting which
> >> is very doubtful because the source and the destination are
> both
> >> in the same time frame moving together.
> >>
> >> Conclusion: MMX must give a null In all Aether conditions and
> >> prove nothing.
> >
> >The MMX would indeed have given a fringe shift if the particular
> >ether theory Michelson believed to be valid had been so.
> >Thus the MMX falsified this theory.
> >
> >If you are trying to say that Michelson's ether theory
> >was no consistent theory and therefore false independent of
> >experimental evidence, then I think you will have to prove
> >that in a more coherent manner than the above.
> >
> >Paul
> >
> [EL]
> I am tired of proving over and over Paul, my time is really
> limited and I have to type very fast most of the time.
> I need no glories of any proof. I am humbly raising an issue for
> investigation by those who are more concerned with it.
Proving over and over?
I am unable to see you have proven anything once.
I am not even able to see which issue you so humbly are raising.
> My major question is, based on what did M&M design the
> interferometer to measure the speed of Aether?
Isn't that obvious? You find the answer in the arithmetic
you do not question. It shows quite clearly that a fringe
shift should be expected if the assumption on which the
arithmetic is based were valid. Namely the existence of
an ether with particular properties.
Paul
Paul B. Andersen
>
> On Thu, 01 Jun 2000 22:48:17 +0200, "Paul B. Andersen"
> <paul.b....@hia.no> wrote:
> >Henry, please understand:
> >We all know that the MMX can be explained by source dependency of light.
>
> This is false.
So my statement was wrong. The correct statement is:
We all but Ken Seto understand that the MMX can be explained
by source dependency of light.
Paul
JeffMo
>Henry Wilson said some stuff about
> >Bilge, point a laser at a distant wall. Spin the laser until the spot
> >moves much faster than c. The spot, if it could be analysed close up,
> >would be found to consists of a series of spots representing
>
> Since you cannot rotate the laser that fast, this is meaningless.
I doubt Henry's theories as much as anyone, but this objection is
wrong. It is certainly physically possible to make a laser spot move
faster than c. The spot is a sort of phantom object, though, like a
shadow. It is not the SAME photons moving superluminally, but
different photons being reflected from a surface. There is no
superluminal motion of any given mass, energy, or information.
JeffMo
"[...] any effort at safe sex is totally, utterly immoral from top to bottom."
-- Rev. James Reuter, Office of Mass Media, Catholic Church of the Philippines
David Evens
Henry Wilson wrote:
> On Sat, 03 Jun 2000 23:54:49 -0400, David Evens> >> On Fri, 02 Jun 2000 23:49:02 -0400, David Evens
> >> <dev...@technologist.com> wrote:
> >> >Henry Wilson wrote:
> >> >> On Fri, 02 Jun 2000 00:02:37 -0400, David Evens
> >> >> <dev...@technologist.com> wrote:
> >> >> >Henry Wilson wrote:
> >> >> >> On Wed, 31 May 2000 23:38:44 -0400, David Evens
> >>
> >> >> light wave crests) move diagonally. The chain blade ( ie, the beam of
> >> >> light) still always appears vertical and to move at c, to all
> >> >> observers.
> >> >
> >> >This, of course, cannot be relevant, since the overall blade does not
> >> >carry the information any more than the overall light beam.
> >> chain and demonstrating that it remains vertical.
> >
> >Given the demonstration of the total irrelevance of this, it cannot
> >matter. Had you BOTHERED to understand the diagrams you mindlessly
> >attack, you would not be able to avoid noticing that the LIGHT BEAMS are
> >ALWAYS vertical, despite the fact that NO LIGHT TRAVELS VERTICALLY.
> You obviously know nothing about the subject.
Your grammar is becoming as incompetent as your physics. You REALLY
messed up your tenses.
> >> >> The time taken for each tooth (each wavecrest) to travel across each
> >> >> arm is just L/c, to all observers.
> >> >
> >> >Of course, the speed c is not the speed of the elements of the saw blade
> >> >in the rest frame of a general observer, but only the component of the
> >> >velocity that is along the saw's length. This does not correspond to
> >> >the speed of the light in the MMX, which MUST be diagonal.
> >> No
> >
> >No, what? No, the elements of the saw blade are NOT moving at some
> >speed other than in the general observer's frame?
> >
> >> >> Einstein and his deciples claimed that the light beam itself turned
> >> >> over diagonally and moved at c. That is not what happens, as the
> >> >> chainsaw example clearly illustrates.
> >> >
> >> >Had you BOTHERED to read my post, you would have been unable to avoid
> >> >noticing that the PIECES of the saw blade ALL MOVE DIAGONALLY, just like
> >> >ALL THE WAVE CRESTS OF LIGHT MOVE DIAGONALLY.
> >> That's correct. But each one moves on a different diagonal.
> >
> >Just as the diagrams you mindlessly attack say.
> Tell me about one.
> Every MMX diagram I have seen states that the cross beam of light
> moves diagonally and then goes on to insist that it must travel at c
> along this path. That is false. As the chainsaw shows, the entire
> lightbeam moves horizontally, just like a rigid rod, while the light
> moves up it at c.
How can the light move up the beam at c while it moves ACROSS on the
diagonals at c, as the physical laws of the universe require?
>The number of wavelengths in the cross beam does not
> change because of its longitudinal motion. This is the crucial factor
> that Einstein et al. refused to accept.
Where DID you get this delusion? The fact that the NUMBER of
wavelengths in the beam doesn't change with its horizontal velocity
PROVES that time appears to be running more slowly in the frame of the
apparatus than it does in the rest frame of the observer. This, quite
simply, PROVES that SR is predicting things correctly.
> >> At any time, all the wave crests are aligned vertically, irrespective
> >> of speed just like the teeth on the saw.
> >> Draw it!
> >> You will soon see that there is no diagonal beam of light.
> >
> >What beam of light are you pretending is diagonal in the MMX diagrams
> >that you keep mindlessly attacking?
> Get hold of any book on the subject and you will see a diagram showing
> a cross beam drawn at an oblique angle. It will then state that the
> beam travels at c up that diagonal. That is plain BULLSHIT!
Indeed it is, since NO SUCH BEAM IS SHOWN. Had you EVER BOTHERED to
examine what is being diagrammed, you could not avoid noticing that it
is AN INDIVIDUAL PHOTON PATH that is diagonal. Since the photons MUST
travel along the diagonal in rest frames in which the apparatus is not
at rest, then OBVIOUSLY they cannot travel at c vertically, as this
would result in them traveling superluminally in the rest frame of the
observer.
> >> >> Each tooth (wave crest) closes in on the mirror at a speed of
> >> >> sqrt(c^2+v^2) wrt the rest frame. That does not violate any law about
> >> >> exceeding light speed.
> >> >
> >> >And, of course, this is irrelevant to the fact that in ALL frames, the
> >> >SPEED OF LIGHT is ALWAYS measured to be the same.
> >> That's an assumption.
> >
> >In reality, that is an OBSERVATION.
> Nobody has ever measured the one way velocity of light from a moving
> source.
Ah, yes, THIS random assumption again. Looks like we have another of
the deluded idiots who assume that the universe cannot POSSIBLY operate
the way it is observed to operate.
> >> >What you have
> >> >discovered is that the components of this velocity in different
> >> >direction differ for different observers.
> >> and apparent velocity can exceed c
> >
> >Only for a non-inertial observer. Analysis of the scenario ALWAYS
> >results in inertial observers always seeing light raveling at c, and all
> >material bodies traveling slower.
> >
> >> >This is not interesting.
> >> >This is, in fact, one of the things that explains why time MUST appear
> >> >to run slower in all other frames, regardless of the observer doing the
> >> >measuring.
> >> that's a circular argument.
> >
> >Where did you imagine a circular argument appearing?
> >
> >> However. Nothing I have said rules out the possibility that the ratio
> >> of the standards defining 'l/t' remains constant in all moving frames.
> >
> >In reality, of course, REALITY prohibits this, since the light would
> >have to move at different speeds in different frame, which does not
> >occur.
> SR says that light always travels at c in any frame. That implies that
> the ratio of reference scales for velocity units, L/T, remains
> constant in all frames. Using the incorrect assumption that the
> velocity of the diagonal beam in the MMX travels at c, the constancy
> of L/T is assured because of the same contraction in L and T.
Where did you get the delusion that L is contracting in the MMX? After
all, L is NORMAL to the velocity, and SR says that the contraction
occurs in the direction of motion.
David Evens
Henry Wilson wrote:
> On Sun, 04 Jun 2000 20:12:00 +0200, "Paul B. Andersen"
> <paul.b....@hia.no> wrote:
> >EL wrote:
> >>
> >> <snip published and very well known unnecessary arithmetic>
>
> >> prove nothing.
> >
> >ether theory Michelson believed to be valid had been so.
> >Thus the MMX falsified this theory.
> >
> But instead of looking for the error an accepting the obvious
> explanation, Einstein stuck to the aether notion that the transverse
> light beam moved diagonally at c. It does not!
Where did you get these two delusions? Light is OBSERVED to travel at
the same speed in al directions in all inertial frames.
> He then wrongly invented the absolutely ridiculous idea of a length
> contraction to avoid accepting the officially unpopular notion that,
> in certain configurations, light speed might appear to be source
> dependent.
That is to say, to model the universe in such a way that the predictions
match the observations, instead of some ignorant fool's prejudices about
the universe.
PA Carr
> ro...@radioactivex.lebesque-al.net (Bilge) wrote:
> >Henry Wilson said some stuff about
>
> > >Bilge, point a laser at a distant wall. Spin the laser until the spot
> > >moves much faster than c. The spot, if it could be analysed close up,
> > >would be found to consists of a series of spots representing
> >
> > Since you cannot rotate the laser that fast, this is meaningless.
>
> I doubt Henry's theories as much as anyone, but this objection is
> wrong. It is certainly physically possible to make a laser spot move
> faster than c. The spot is a sort of phantom object, though, like a
> shadow. It is not the SAME photons moving superluminally, but
> different photons being reflected from a surface. There is no
> superluminal motion of any given mass, energy, or information.
>
> JeffMo
I've got to disagree here, JeffMo. The paper that smart cited a while
ago, which he 'expanded the data' of does record superluminal effects. I
haven't got the paper with me now, but I can find the reference if you're
interested. If memory serves (and it probably doens't ;-) ), the first
observation of this phenomenon was in the 80s.
Regards,
Paul.
Paul B. Andersen
>
> On Sun, 04 Jun 2000 20:55:52 +0200, "Paul B. Andersen"
> At least we agree on something.
What is then your point?
You said: "the point is, there should have been a null,anyway,
just on geometry"
implying that the MMX didn't falsify Michelson's ether.
Now you admit that if Michelson's assumption about the ether
were right, there would have been a fringe shift,
and thus that the MMX do indeed falsify Michelson's ether.
> >> >=================================================================
> >> >| But the point is that this is old news. |
> >> >| It has been known since day one in the history of the MMX. |
> >> >| Nobody has ever disputed that source dependent speed of light |
> >> >| is consistent with the MMX. |
> >> >=================================================================
>
> >
> >> The ether theory insisted that light would follow a diagonal beam and
> >> move at c along that diagonal. After the null result, Einstein still
> >> assumed the ether prediction but then invented the Lorentz contraction
> >> to explain the null.
> >
> >I think you confuse the names.
> >It was Lorentz and Fitzgerald who independently invented this
> >contraction to save the ether theory.
> >Hence its name: Lorentz-Fitzgerald contraction.
> >
> >Einstein did not say that the arms of the interferometer contracts.
> >
> >> The point is, there should have been a null,anyway, just on geometry,
> >> and Einstein had no justification for the contraction idea.
> >
> >Nonsense.
> >It would definitely not have been null in "Galilean ether".
> >
> I agree. I am not an etherist, even though I think some kind of
> absolutivity is a very attractive theory.
So does that mean that you finally have realized that your
thought experiment only show what we all knew all the time,
namely that the MMX null result is consistent with
"source dependent speed of light"?
Paul
EL
[EL]
Paul,
I noticed something while reading your post.
You confirmed my admiration. :-)
It is obvious now that our dispute was based on a completely
different conception of the characteristics of what Aether is.
Yes, it is a pervading medium but what?
So, I see now that MMX disproved a specific set of parameters,
which defined a classical Aether and not all and any definition
of Aether.
I was under the impression due to the main stream relativists'
talk, and who believe blindly in the refutation of Aether in an
absolute essence, that the MMX was behind such belief. So I
never really found a definition of Aether as was in M&M 's
conception, and probably there is not.
What you did is pointing out the logical expectations that
reflected the only meaning they could have believed, which is
not what I had in mind and thusly concluded a fraudulent fake
experiment.
Naturally, it is very plausible that they had a different
conception than our modern one.
Your reaction to my detailed explanation gave me two choices:
1- That you could be incompetent and that my explanation just
went over your head (which is obviously not the case). :-)
2- That you are trapped in the same concepts as M&M due to long
debates on the very same subject. (which is most probable).
I only care to emphasize on the fact that interferometry is a
very precise measuring tool, when it concerns distance traveled
by light.
Once you set the coherent split wave in synchrony, any change in
phase shall correspond to a fringe change.
My dispute was on the "Why did they expect Aether to induce any
change at all?". Movement? Of Aether? How?
The Fizeau experiment used a moving liquid (such as water) that
was run once in every direction.
This affected the refraction index from the liquid to air. If we
were pumping air in the Fizeau experiment, what reading do we
expect to read?
Kind regards.
JeffMo
>On Mon, 5 Jun 2000, JeffMo wrote:
>
>> ro...@radioactivex.lebesque-al.net (Bilge) wrote:
>> >Henry Wilson said some stuff about
>>
>> > >Bilge, point a laser at a distant wall. Spin the laser until the spot
>> > >moves much faster than c. The spot, if it could be analysed close up,
>> > >would be found to consists of a series of spots representing
>> >
>> > Since you cannot rotate the laser that fast, this is meaningless.
>>
>> I doubt Henry's theories as much as anyone, but this objection is
>> wrong. It is certainly physically possible to make a laser spot move
>> faster than c. The spot is a sort of phantom object, though, like a
>> shadow. It is not the SAME photons moving superluminally, but
>> different photons being reflected from a surface. There is no
>> superluminal motion of any given mass, energy, or information.
>
>I've got to disagree here, JeffMo. The paper that smart cited a while
>ago, which he 'expanded the data' of does record superluminal effects. I
>haven't got the paper with me now, but I can find the reference if you're
>interested. If memory serves (and it probably doens't ;-) ), the first
>observation of this phenomenon was in the 80s.
This misses the point of what I was trying to say, and I think I can
clarify this for you: Add the phrase "in such a 'moving laser spot'
demonstration" to the end of my paragraph above.
I do not have the inclination nor the evidence to dispute the recent
discoveries which appear to indicate some superluminal phenomena. All
I was doing was pointing out that rotating a laser fast enough to make
its distant spot on some surface appear to move faster than c is
physically possible, and does not necessitate any single energy- or
information-carrying object to move superluminally in order for it to
occur.
I believe this is covered in the FAQ.
JeffMo
>[EL]
>Paul,
>I noticed something while reading your post.
>You confirmed my admiration. :-)
>It is obvious now that our dispute was based on a completely
>different conception of the characteristics of what Aether is.
>Yes, it is a pervading medium but what?
>
>So, I see now that MMX disproved a specific set of parameters,
>which defined a classical Aether and not all and any definition
>of Aether.
This is correct. In fact, it would be hard to conceive of an
experiment which could serve as a counterexample to "all and any
definion[s]" of anything! For with perverse opponents, you could say
that your experiment has ruled out the possible existence of any
planets in the solar system bigger than Jupiter, and then they can
inform you that their personal definition includes stars as a subset
of planets!
This is why it is good to agree on relevant definitions BEFORE the
debate gets hot and earnest.
Bilge
>
>I doubt Henry's theories as much as anyone, but this objection is
>wrong. It is certainly physically possible to make a laser spot move
>faster than c. The spot is a sort of phantom object, though, like a
>shadow. It is not the SAME photons moving superluminally, but
>different photons being reflected from a surface. There is no
>superluminal motion of any given mass, energy, or information.
>
It doesn't count if you have to violate relativity to call it
the same spot. All you are doing is playing a word game then. There
is no phantom object any more than a stream of bullets gives me
phantom bullets from a rotating gun. An observer which views successive
spots in a proper interval will either the distance between the two
get smaller. In the rest frame of the system containing the vertical
axis of the laser's rotation in the plane of the laser, two spots are
simultaneous iff ct1 - ct2 = 0, but to be simultaneous in any reference
frame, those times have to be transformed.
No matter how you slice it, no non-zero value of the product of w
and r produce what the premise suggests - a moving spot which every frame
will agree is simultaneous. At best, you can define a spot which is app-
roximately simultaneous between two frames with no relative velocity.
Everyone else will see the distance between the same two points to be
different than theobserver at the origin obtains from:
d^2 = 2r^ - 2r^2cos(\theta), \theta = w(t2-t1).
I'm not about to play semantic word games with henry and have to
later backpeddle when he either intentionally or through his misunder-
standing of what is implied from context, misquotes me so that he can
claim I contradict myself. It's hard enough to guess what won't be
misconstrued by an honest misunderstanding made by someone with a
genuine interest in the correct answer. Henry's (and any number of
other's that do the same thing, but usually less egregiously) only
interest is to find two or more statements that seem to be contra-
dictory. If the contradiction is not an honest mistake, then the
option to take the statements out of context and invent a contra-
diction is the next best thing. While it's hard to avoid assuming
things that seem perfectly understood as assumptions, I don't intend
to provide any additional opportunities if I can help it. Strictly
speaking, you cannot rotate the laser fast enough to give him the
definition he seeks, since phase two will be to obscure this fact
and create an impossible super-luminal scenario that would be allowed
if the premises were true. There is simply no justification for refering
to a spot at fixed r as the the same spot without a condition placed
on what you mean by the "same spot". Just like with the sagnac
effect, this provides a great opportunity to abuse the math and
declare sr incorrect.
He apparantly has a greater capacity to search and cross-correlate
posts than he does for getting out a pen and paper and doing the arith-
metic himself. Of course, that would nip the argument in the bud, too.
PA Carr
> PA Carr <pac...@york.ac.uk> wrote:
>
> >On Mon, 5 Jun 2000, JeffMo wrote:
> >
> >> ro...@radioactivex.lebesque-al.net (Bilge) wrote:
> >> >Henry Wilson said some stuff about
> >>
> >> > >Bilge, point a laser at a distant wall. Spin the laser until the spot
> >> > >moves much faster than c. The spot, if it could be analysed close up,
> >> > >would be found to consists of a series of spots representing
> >> >
> >> > Since you cannot rotate the laser that fast, this is meaningless.
> >>
> >> I doubt Henry's theories as much as anyone, but this objection is
> >> wrong. It is certainly physically possible to make a laser spot move
> >> faster than c. The spot is a sort of phantom object, though, like a
> >> shadow. It is not the SAME photons moving superluminally, but
> >> different photons being reflected from a surface. There is no
> >> superluminal motion of any given mass, energy, or information.
> >
> >I've got to disagree here, JeffMo. The paper that smart cited a while
> >ago, which he 'expanded the data' of does record superluminal effects. I
> >haven't got the paper with me now, but I can find the reference if you're
> >interested. If memory serves (and it probably doens't ;-) ), the first
> >observation of this phenomenon was in the 80s.
>
> This misses the point of what I was trying to say, and I think I can
> clarify this for you: Add the phrase "in such a 'moving laser spot'
> demonstration" to the end of my paragraph above.
Sorry ;-). I assumed that your final statement was general,
because of your inclusion of 'mass, energy and information'.
Regards,
Paul.
Bilge
>I do not have the inclination nor the evidence to dispute the recent
>discoveries which appear to indicate some superluminal phenomena. All
>I was doing was pointing out that rotating a laser fast enough to make
>its distant spot on some surface appear to move faster than c is
>physically possible, and does not necessitate any single energy- or
>information-carrying object to move superluminally in order for it to
>occur.
Only if you take for granted as part of the assumption that something
is already moving faster than c to set up the conditions necessary to make
that observation, in which case, you've guaranteed the outcome a priori.
All you need to do inorder to prove that is to pulse the laser at a frequency
of w/n and trace the line through points that are simultaneous.
JeffMo
I'm sorry, I don't follow what you're saying. I'm claiming basically
the same as what is in the FAQ....that a distant laser "spot" can
"move" at speeds greater than c, without any REAL thing actually
moving faster than c.
JeffMo
> >wrong. It is certainly physically possible to make a laser spot move
> >faster than c. The spot is a sort of phantom object, though, like a
> >shadow. It is not the SAME photons moving superluminally, but
> >different photons being reflected from a surface. There is no
> >superluminal motion of any given mass, energy, or information.
> >
>
>the same spot.
We both agree. It's basically a trick of human perception that we
tend to label it the same spot. It's not a thing going faster than c,
even though it would look like it was, to people who didn't think it
through, and realize that it's really the reflection of a
continuously-flowing STREAM, which is different photons in each
"instant."
>There is simply no justification for refering
>to a spot at fixed r as the the same spot without a condition placed
>on what you mean by the "same spot". Just like with the sagnac
>effect, this provides a great opportunity to abuse the math and
>declare sr incorrect.
I agree with you wholeheartedly, sir.
JeffMo
No problem. I *welcome* such clarification. I was unclear, and I
bore the burden of clarification. No apology necessary.
ryker1
>> <paul.b....@hia.no> wrote:
>> >EL wrote:
>> >>
>> >> <snip published and very well known unnecessary arithmetic>
>>
>> >> Conclusion: MMX must give a null In all Aether conditions
and
>> >> prove nothing.
>> >
>> >The MMX would indeed have given a fringe shift if the
particular
>> >ether theory Michelson believed to be valid had been so.
>> >Thus the MMX falsified this theory.
>> >
>> >Paul
>> But instead of looking for the error an accepting the obvious
>> explanation, Einstein stuck to the aether notion that the
transverse
>> light beam moved diagonally at c. It does not!
>
>Where did you get these two delusions? Light is OBSERVED to
travel at
>the same speed in al directions in all inertial frames.
Rod: The speed of light does travel at c in all frames .
However , it is the direction wrt the frames direction
mathmatically that creates argument .
If the light is moving vetically up wrt a comoving frame ,
the light does not travel at c vertically in that frame .
But does travel at c at angle (m) wrt another frame .
GEO MET RIC GAMES . :)
Rod Ryker...
It is reasoning and faith that bind truth .
http://homestead.deja.com/user.ryker1/index.html
Henry Wilson
<dev...@technologist.com> wrote:
>
>Henry Wilson wrote:
>> On Sat, 03 Jun 2000 23:54:49 -0400, David Evens
>> >
>> >Given the demonstration of the total irrelevance of this, it cannot
>> >matter. Had you BOTHERED to understand the diagrams you mindlessly
>> >attack, you would not be able to avoid noticing that the LIGHT BEAMS are
>> >ALWAYS vertical, despite the fact that NO LIGHT TRAVELS VERTICALLY.
>> You obviously know nothing about the subject.
>
>Your grammar is becoming as incompetent as your physics. You REALLY
>messed up your tenses.
Hahahahaha! you wrote that yourself!
hahahahahah!
>
>> >> >Had you BOTHERED to read my post, you would have been unable to avoid
>> >> >noticing that the PIECES of the saw blade ALL MOVE DIAGONALLY, just like
>> >> >ALL THE WAVE CRESTS OF LIGHT MOVE DIAGONALLY.
>> >> That's correct. But each one moves on a different diagonal.
>> >
>> >Just as the diagrams you mindlessly attack say.
>> Tell me about one.
>> Every MMX diagram I have seen states that the cross beam of light
>> moves diagonally and then goes on to insist that it must travel at c
>> along this path. That is false. As the chainsaw shows, the entire
>> lightbeam moves horizontally, just like a rigid rod, while the light
>> moves up it at c.
>
>How can the light move up the beam at c while it moves ACROSS on the
>diagonals at c, as the physical laws of the universe require?
No light beam moves along a diagonal, that's why. Infinitesimally
small elements of the vertical light beam each follow a separate
diagonal path to the top mirror. Why would anyone in their right mind
want to claim that they move at c along this path. It obviously moves
vrtiacally at c. The important factor is that the number of
wavelengths in the cross beam doesn't depend on velocity of the
apparatus.
This is where Einstein and his disciples are clearly wrong.
Anyway, why don't you concentrate on the parallel beam. It is much
more interesting!
>
>>The number of wavelengths in the cross beam does not
>> change because of its longitudinal motion. This is the crucial factor
>> that Einstein et al. refused to accept.
>
>Where DID you get this delusion? The fact that the NUMBER of
>wavelengths in the beam doesn't change with its horizontal velocity
>PROVES that time appears to be running more slowly in the frame of the
>apparatus than it does in the rest frame of the observer. This, quite
>simply, PROVES that SR is predicting things correctly.
You obviously know nothing about the mmx.
>
>> >> At any time, all the wave crests are aligned vertically, irrespective
>> >> of speed just like the teeth on the saw.
>> >> Draw it!
>> >> You will soon see that there is no diagonal beam of light.
>> >
>> >What beam of light are you pretending is diagonal in the MMX diagrams
>> >that you keep mindlessly attacking?
>> Get hold of any book on the subject and you will see a diagram showing
>> a cross beam drawn at an oblique angle. It will then state that the
>> beam travels at c up that diagonal. That is plain BULLSHIT!
>
>Indeed it is, since NO SUCH BEAM IS SHOWN. Had you EVER BOTHERED to
>examine what is being diagrammed, you could not avoid noticing that it
>is AN INDIVIDUAL PHOTON PATH that is diagonal. Since the photons MUST
>travel along the diagonal in rest frames in which the apparatus is not
>at rest, then OBVIOUSLY they cannot travel at c vertically, as this
>would result in them traveling superluminally in the rest frame of the
>observer.
Sorry, It is not even an individual photon that moves along a single
diagonal path. It is as infinitesimally small element of a photon that
does.
>
>> >
>> >In reality, that is an OBSERVATION.
>> Nobody has ever measured the one way velocity of light from a moving
>> source.
>
>Ah, yes, THIS random assumption again. Looks like we have another of
>the deluded idiots who assume that the universe cannot POSSIBLY operate
>the way it is observed to operate.
You mean the way srists want it to operate. It isn't hard to
manipulate the interpretation of experimental data to fit your
favourite theory.
L is the length of both arms of the MMX. You obviously know nothing
about the experiment or Einsteins explanation of the null result.
Henry Wilson
<dev...@technologist.com> wrote:
>
>Henry Wilson wrote:
>> On Sun, 04 Jun 2000 20:12:00 +0200, "Paul B. Andersen"
>> <paul.b....@hia.no> wrote:
>> >EL wrote:
>> >>
>> >> <snip published and very well known unnecessary arithmetic>
>>
>> >> Conclusion: MMX must give a null In all Aether conditions and
>> >> prove nothing.
>> >
>> >The MMX would indeed have given a fringe shift if the particular
>> >ether theory Michelson believed to be valid had been so.
>> >Thus the MMX falsified this theory.
>> >
>> >Paul
>> But instead of looking for the error an accepting the obvious
>> explanation, Einstein stuck to the aether notion that the transverse
>> light beam moved diagonally at c. It does not!
>
>Where did you get these two delusions? Light is OBSERVED to travel at
>the same speed in al directions in all inertial frames.
Two points here.
1. There is no diagonal light beam, as I have stated many times.
2.The fact that light speed is always MEASURED to have the same value
c in any frame, doesn't really tell us much about the relevance of
velocity as it applies to light.
What is the velocity of a free photon, smartarse?
>
>> He then wrongly invented the absolutely ridiculous idea of a length
>> contraction to avoid accepting the officially unpopular notion that,
>> in certain configurations, light speed might appear to be source
>> dependent.
>
>That is to say, to model the universe in such a way that the predictions
>match the observations, instead of some ignorant fool's prejudices about
>the universe.
As I said before. It isn't hard to manipulate experimental results to
provide the answers that you want, particularly when you know that
hardly anyone will be game enough to challenge them.
Henry Wilson
wrote:
>ro...@radioactivex.lebesque-al.net (Bilge) wrote:
>>JeffMo said some stuff about
>>
>> >I doubt Henry's theories as much as anyone, but this objection is
>> >wrong. It is certainly physically possible to make a laser spot move
>> >faster than c. The spot is a sort of phantom object, though, like a
>> >shadow. It is not the SAME photons moving superluminally, but
>> >different photons being reflected from a surface. There is no
>> >superluminal motion of any given mass, energy, or information.
>> >
>>
>> It doesn't count if you have to violate relativity to call it
>>the same spot.
>
>We both agree. It's basically a trick of human perception that we
>tend to label it the same spot. It's not a thing going faster than c,
>even though it would look like it was, to people who didn't think it
>through, and realize that it's really the reflection of a
>continuously-flowing STREAM, which is different photons in each
>"instant."
>
>>There is simply no justification for refering
>>to a spot at fixed r as the the same spot without a condition placed
>>on what you mean by the "same spot". Just like with the sagnac
>>effect, this provides a great opportunity to abuse the math and
>>declare sr incorrect.
>
>I agree with you wholeheartedly, sir.
>
>JeffMo
And so do I. Hooray, we all agree!!!!This backs up the point I have
been making, all along.
WHAT MOVES DIAGONALLY IN THE MMX IS NOT A LIGHT BEAM MOVING AT C BUT A
SERIES OF INFINITESIMALLY SMALL ELEMENTS THAT APPEAR TO MOVE AT
SUPERLUMINAL SPEEDS.
Bilge
>I'm sorry, I don't follow what you're saying. I'm claiming basically
>the same as what is in the FAQ....that a distant laser "spot" can
>"move" at speeds greater than c, without any REAL thing actually
>moving faster than c.
>
Suppose and observer has a detector that subtends some small angle.
A spot of a finite size has both a leading and trailing edge. Assume
that when stationary the spot's leading and trailing edge coincide
with the detector edges. You would then define the speed of the spot
as the time it take an edge to sweep across the detector and it would
be equal to t = r/w where r = distance to the source, w = angular
velocity. What the observer sees is something different. The width of
the beam is contracted and he sees a pulse that is smaller in width
with high frequency components that correspond to the narrowing of
the pulse. The pulse interval no longer covers the entire width
of the detector and in fact, is will appear and disappear well within
the detector geometry. The observer no longer sees the same angle
it would with the laser slowly rotating. So, observers will not
see a spot move faster than c. They'll see a shortened pulse with
fourier components due to the pulse shape caused by the doppler
shifted laser light.
JeffMo
>And so do I. Hooray, we all agree!!!!This backs up the point I have
>been making, all along.
>WHAT MOVES DIAGONALLY IN THE MMX IS NOT A LIGHT BEAM MOVING AT C BUT A
>SERIES OF INFINITESIMALLY SMALL ELEMENTS THAT APPEAR TO MOVE AT
>SUPERLUMINAL SPEEDS.
I thought that what moves in the MMX are photons, that appear to move
at c.
JeffMo
David Evens
Henry Wilson wrote:
> On Tue, 06 Jun 2000 22:24:32 GMT, jef...@dipstick.cfw.com (JeffMo)
> wrote:
> >ro...@radioactivex.lebesque-al.net (Bilge) wrote:
> >>
> >> >wrong. It is certainly physically possible to make a laser spot move
> >> >faster than c. The spot is a sort of phantom object, though, like a
> >> >shadow. It is not the SAME photons moving superluminally, but
> >> >different photons being reflected from a surface. There is no
> >> >superluminal motion of any given mass, energy, or information.
> >> >
> >>
> >> It doesn't count if you have to violate relativity to call it
> >>the same spot.
> >
> >We both agree. It's basically a trick of human perception that we
> >tend to label it the same spot. It's not a thing going faster than c,
> >even though it would look like it was, to people who didn't think it
> >through, and realize that it's really the reflection of a
> >continuously-flowing STREAM, which is different photons in each
> >"instant."
> >
> >>There is simply no justification for refering
> >>to a spot at fixed r as the the same spot without a condition placed
> >>on what you mean by the "same spot". Just like with the sagnac
> >>effect, this provides a great opportunity to abuse the math and
> >>declare sr incorrect.
> >
> >I agree with you wholeheartedly, sir.
> >
> >JeffMo
> been making, all along.
> WHAT MOVES DIAGONALLY IN THE MMX IS NOT A LIGHT BEAM MOVING AT C BUT A
> SERIES OF INFINITESIMALLY SMALL ELEMENTS THAT APPEAR TO MOVE AT
> SUPERLUMINAL SPEEDS.
Then why is there nothing that appears to move at superluminal speeds
ANYWHERE in the MMX?
David Evens
Henry Wilson wrote:
> On Mon, 05 Jun 2000 21:55:48 -0400, David Evens
> <dev...@technologist.com> wrote:
> >Henry Wilson wrote:
> >> On Sun, 04 Jun 2000 20:12:00 +0200, "Paul B. Andersen"
> >> <paul.b....@hia.no> wrote:
> >> >EL wrote:
> >> >>
> >> >> <snip published and very well known unnecessary arithmetic>
> >>
> >> >> Conclusion: MMX must give a null In all Aether conditions and
> >> >> prove nothing.
> >> >
> >> >The MMX would indeed have given a fringe shift if the particular
> >> >ether theory Michelson believed to be valid had been so.
> >> >Thus the MMX falsified this theory.
> >> >
> >> >Paul
> >> But instead of looking for the error an accepting the obvious
> >> explanation, Einstein stuck to the aether notion that the transverse
> >> light beam moved diagonally at c. It does not!
> >
> >Where did you get these two delusions? Light is OBSERVED to travel at
> >the same speed in al directions in all inertial frames.
> Two points here.
> 1. There is no diagonal light beam, as I have stated many times.
Yes, we KNOW that you think that the light does not travel up the
transverse arm and back. You have not, however, been able to explain
how this is possible, since the light clearly IS in the apparatus.
> 2.The fact that light speed is always MEASURED to have the same value
> c in any frame, doesn't really tell us much about the relevance of
> velocity as it applies to light.
Other than the fact that it shows that your assumptions about light are
stupidly wrong.
> What is the velocity of a free photon, smartarse?
All photons conduct information at a limiting speed of c. The question
of what the speed of an actual photon is another matter. It is also
irrelevant, since it is the speed of information transit that the MMX
actually measures.
> >> He then wrongly invented the absolutely ridiculous idea of a length
> >> contraction to avoid accepting the officially unpopular notion that,
> >> in certain configurations, light speed might appear to be source
> >> dependent.
> >
> >That is to say, to model the universe in such a way that the predictions
> >match the observations, instead of some ignorant fool's prejudices about
> >the universe.
> As I said before. It isn't hard to manipulate experimental results to
> provide the answers that you want, particularly when you know that
> hardly anyone will be game enough to challenge them.
Then why are you incapable of doing so?
David Evens
Henry Wilson wrote:
> On Mon, 05 Jun 2000 21:04:54 -0400, David Evens
> <dev...@technologist.com> wrote:
> >Henry Wilson wrote:
> >> On Sat, 03 Jun 2000 23:54:49 -0400, David Evens
>
> >> >
> >> >Given the demonstration of the total irrelevance of this, it cannot
> >> >matter. Had you BOTHERED to understand the diagrams you mindlessly
> >> >attack, you would not be able to avoid noticing that the LIGHT BEAMS are
> >> >ALWAYS vertical, despite the fact that NO LIGHT TRAVELS VERTICALLY.
> >> You obviously know nothing about the subject.
> >
> >Your grammar is becoming as incompetent as your physics. You REALLY
> >messed up your tenses.
> Hahahahaha! you wrote that yourself!
> hahahahahah!
How would you know? You never read anything I write anyway.
> >> >> >Had you BOTHERED to read my post, you would have been unable to avoid
> >> >> >noticing that the PIECES of the saw blade ALL MOVE DIAGONALLY, just like
> >> >> >ALL THE WAVE CRESTS OF LIGHT MOVE DIAGONALLY.
> >> >> That's correct. But each one moves on a different diagonal.
> >> >
> >> >Just as the diagrams you mindlessly attack say.
> >> Tell me about one.
> >> Every MMX diagram I have seen states that the cross beam of light
> >> moves diagonally and then goes on to insist that it must travel at c
> >> along this path. That is false. As the chainsaw shows, the entire
> >> lightbeam moves horizontally, just like a rigid rod, while the light
> >> moves up it at c.
> >
> >How can the light move up the beam at c while it moves ACROSS on the
> >diagonals at c, as the physical laws of the universe require?
> No light beam moves along a diagonal, that's why. Infinitesimally
> small elements of the vertical light beam each follow a separate
> diagonal path to the top mirror.
That is to say, all the light moves on the diagonals, at c.
> Why would anyone in their right mind
> want to claim that they move at c along this path.
Because that is the path it must follow to stay in the apparatus.
> It obviously moves
> vrtiacally at c.
Then how does it stay in the apparatus, which isn't at rest?
> The important factor is that the number of
> wavelengths in the cross beam doesn't depend on velocity of the
> apparatus.
> This is where Einstein and his disciples are clearly wrong.
Then you are claiming that time dilation does not occur, as it would
have to if the light is going to move as it is observed to move, which
is at c along the diagonal (which the diagrams always deal with only as
individual photons, which OBVIOUSLY move ONLY on the diagonals).
> Anyway, why don't you concentrate on the parallel beam. It is much
> more interesting!
You're making stupidly wrong claims about the transverse beams, which is
much more interesting to ridicule you on. Of course, if you want to
start making stupid claims about the parallel, I can start making you
look stupid(er) with them, too.
> >>The number of wavelengths in the cross beam does not
> >> change because of its longitudinal motion. This is the crucial factor
> >> that Einstein et al. refused to accept.
> >
> >Where DID you get this delusion? The fact that the NUMBER of
> >wavelengths in the beam doesn't change with its horizontal velocity
> >PROVES that time appears to be running more slowly in the frame of the
> >apparatus than it does in the rest frame of the observer. This, quite
> >simply, PROVES that SR is predicting things correctly.
> You obviously know nothing about the mmx.
I see that you are unable to support your delusions about the MMX, since
you did nothing but claim that the observations about it are wrong.
> >> >> At any time, all the wave crests are aligned vertically, irrespective
> >> >> of speed just like the teeth on the saw.
> >> >> Draw it!
> >> >> You will soon see that there is no diagonal beam of light.
> >> >
> >> >What beam of light are you pretending is diagonal in the MMX diagrams
> >> >that you keep mindlessly attacking?
> >> Get hold of any book on the subject and you will see a diagram showing
> >> a cross beam drawn at an oblique angle. It will then state that the
> >> beam travels at c up that diagonal. That is plain BULLSHIT!
> >
> >Indeed it is, since NO SUCH BEAM IS SHOWN. Had you EVER BOTHERED to
> >examine what is being diagrammed, you could not avoid noticing that it
> >is AN INDIVIDUAL PHOTON PATH that is diagonal. Since the photons MUST
> >travel along the diagonal in rest frames in which the apparatus is not
> >at rest, then OBVIOUSLY they cannot travel at c vertically, as this
> >would result in them traveling superluminally in the rest frame of the
> >observer.
> Sorry, It is not even an individual photon that moves along a single
> diagonal path. It is as infinitesimally small element of a photon that
> does.
I see you know nothing about light IN ANY WAY, since it is not possible
to divide a particle in this manner.
> >> >In reality, that is an OBSERVATION.
> >> Nobody has ever measured the one way velocity of light from a moving
> >> source.
> >
> >Ah, yes, THIS random assumption again. Looks like we have another of
> >the deluded idiots who assume that the universe cannot POSSIBLY operate
> >the way it is observed to operate.
> You mean the way srists want it to operate. It isn't hard to
> manipulate the interpretation of experimental data to fit your
> favourite theory.
Then why are you incapable of doing so?
> >> >> >What you have
L is the length of both arms in the REST FRAME OF THE APPARATUS. In any
OTHER frame, there is a length contraction of the arm in the direction
of motion, but the arm normal to the direction of motion does NOT
contract. This eliminates your fantasy that the transverse arm
contracts.
Henry Wilson
(Bilge) wrote:
>JeffMo said some stuff about
>
Not so. You are assuming that each minute element of the beam acquires
a rotary component from the rotating laser. You should know that, once
each photon leaves the laser, it travels in a straight line.
Depending on the rotation speed, what appears on the detector, is a
linear wavelike intensity pattern, corresponding to the wave crests
and troughs that are featured in the light beam. This would result in
a line, of width equal to the beam size, appearing as a series of
bright spots separated by dark bands in between, as the speed was
increased.
Of course, no detector or human eye could ever see such an event
unless the rotation speed was set to exactly synch with 2 pi.r, if you
know what I mean.
Henry Wilson
wrote:
>He...@the.edge(Henry Wilson) wrote:
>
>>And so do I. Hooray, we all agree!!!!This backs up the point I have
>>been making, all along.
>>WHAT MOVES DIAGONALLY IN THE MMX IS NOT A LIGHT BEAM MOVING AT C BUT A
>>SERIES OF INFINITESIMALLY SMALL ELEMENTS THAT APPEAR TO MOVE AT
>>SUPERLUMINAL SPEEDS.
>
>I thought that what moves in the MMX are photons, that appear to move
>at c.
>
>JeffMo
They do but always along the line between the two mirrors. No complete
photon ever moves along a single diagonal path.
Consequently the number of wavelengths in the cross beam remains
constant no matter what the speed of the apparatus.
This realization eliminates the sqrt factor in conventional gamma.
Bilge
>Not so. You are assuming that each minute element of the beam acquires
>a rotary component from the rotating laser. You should know that, once
>each photon leaves the laser, it travels in a straight line.
I assumed it travels a straight line. It's not possible for any
light leaving the laser to not have a rotational component affect
it. The source cannot have a emittance of 0 mm-mrad unless the beam
size is 0mm or the divergence at every point is 0 rad without violating
liouvilles theorem. Quantum mecanics alone garauntees this isn't
possible, but even that isn't required. Construct a ray from any
real physical process. The only light that can reach the observer
is emmitted by a charge with a velocity in the direction of the
obvserver and must be doppler shifted before ever exiting the laser.
But even this isn't necessary to show you're wrong. Any light
created between the rotation point and the exit can either be
considered to have its position defined by the point inbetween or
the point on a physical common exit window. No light reaaching
the exit widow can leave parallel to any other light created
inside the laser in the case you use an exit window. All of the
light will undergo some refraction and exit with a non-zero
emittance since it entered with a non-zero emittance as you
have defined the source. If you don't want a physical window then
you still lose unless your source is a single point at the origin,
in which case, it cant be rotating. Any finite extent source will
require light that emerges at from the end to consist of light
emitted at different points in the rotation corresponding to
different distance to the aperature. So, no ray will ever be
a line perpendicular to the source and observer except the fiducial
one which represents nothing but a mathematical construct from which
to reference the real rays. In this case, there isn't even any
light produced at the point containing the reference ray, unless
you have an infinitely thin, infinitely narrow source. In that
case, the direction of the emitted light is totally undefined
and can't reach the observer either.
JeffMo
>On Wed, 07 Jun 2000 18:34:09 GMT, jef...@dipstick.cfw.com (JeffMo)
>wrote:
>
>>He...@the.edge(Henry Wilson) wrote:
>>
>>>And so do I. Hooray, we all agree!!!!This backs up the point I have
>>>been making, all along.
>>>WHAT MOVES DIAGONALLY IN THE MMX IS NOT A LIGHT BEAM MOVING AT C BUT A
>>>SERIES OF INFINITESIMALLY SMALL ELEMENTS THAT APPEAR TO MOVE AT
>>>SUPERLUMINAL SPEEDS.
>>
>>I thought that what moves in the MMX are photons, that appear to move
>>at c.
>They do but always along the line between the two mirrors.
Don't the mirrors move during the time the photons are traveling?
Doesn't "the line between the two mirrors" move during the time the
photons are traveling? Wouldn't that mean that the photons travel a
"diagonal path," even while remaining "along the line?"
> No complete
>photon ever moves along a single diagonal path.
Does something less than a "complete photon" move? Or does a
"complete photon" move along multiple paths?
I'm really not following what you are claiming here. I apologize if
it's just my stupidity driving this subthread. Ignore me if you feel
I'm holding you back from a more substantive discussion, in all
seriousness.
Henry Wilson
<dev...@technologist.com> wrote:
>
>Henry Wilson wrote:
>> >
>> >How can the light move up the beam at c while it moves ACROSS on the
>> >diagonals at c, as the physical laws of the universe require?
>> No light beam moves along a diagonal, that's why. Infinitesimally
>> small elements of the vertical light beam each follow a separate
>> diagonal path to the top mirror.
>
>That is to say, all the light moves on the diagonals, at c.
NO! You cannot give a name to an infinitesimal point. It is not a
photon. A photon has size and volume otherwise it couldn't hold
intrinsic information about its source.
>
>> Why would anyone in their right mind
>> want to claim that they move at c along this path.
>
>Because that is the path it must follow to stay in the apparatus.
why should it move at c? that was my question. Why shouldn't it move
at sqrt(c^2+v^2) like the chainsaw teeth?
>
>> It obviously moves
>> vrtiacally at c.
>
>Then how does it stay in the apparatus, which isn't at rest?
the vertical component of its velocity is c. like the velocity of
falling raindrops going past your car window. they always fall
vertically at the same rate even though they appear to move diagonally
as you travel.
>
>> The important factor is that the number of
>> wavelengths in the cross beam doesn't depend on velocity of the
>> apparatus.
>> This is where Einstein and his disciples are clearly wrong.
>
>Then you are claiming that time dilation does not occur, as it would
>have to if the light is going to move as it is observed to move, which
>is at c along the diagonal (which the diagrams always deal with only as
>individual photons, which OBVIOUSLY move ONLY on the diagonals).
you can call them individual photons if you like.
The critical point is that the direction of the axis of the E and M
wave components that lie along the 'length' of a photon always remains
vertical even though the axis, as a whole, moves diagonally.
_
/
/ this is wrong.
/ The wave axis does not tilt.
/
_
|
| this is what happens
| as all sections of the light beam
| move along different diagonals.
I suggest you plot the paths individual wave crests and you might be
able to understand the significance of my discovery.
>
>> Anyway, why don't you concentrate on the parallel beam. It is much
>> more interesting!
>
>You're making stupidly wrong claims about the transverse beams, which is
>much more interesting to ridicule you on. Of course, if you want to
>start making stupid claims about the parallel, I can start making you
>look stupid(er) with them, too.
I wish I could include a .jpg here so that I could get the message
across. I will try to draw it using fixed pitch fonts. I did this
before but you obviously didn't follow.
M
__q
/\ /|
/ / \|r,p
/ / / |s
/ / / /| ^
/ / / / |t |c
/ / / / / |u
/ / / / / /|
/ / / / / / |v
/ / / / / / / |w
__________________/__/__/__/__/__/__/__/|________________| M -->V
a b c d e f g h
The diagonal lines represent the paths taken by individual sections of
the cross beam. a,b,c....., represent points in the rest frame where
successive wavecrests leave the splitting mirror. p,q,r,s... show
where these wavecrests lie when b reaches the top mirror. p is a point
on the return beam representing the position of a. Simple geometry
shows that the vertical separation of these points (wavelength)
remains constant irrespective of velocity V.
At any time, all sections are aligned vertically under the cross
mirror. The wave axes of each beam element are never tilted
diagonally, however. They remain vertical like so:
|
|
|
|
|
This is the crucial point of my revelation. The wave axis of each
element moves diagonally but remains upright, like an infinitesimal
rigid vertical rod. This is why the concept of the light beam moving
diagonally at c has no credibility.
>> Sorry, It is not even an individual photon that moves along a single
>> diagonal path. It is as infinitesimally small element of a photon that
>> does.
>
>I see you know nothing about light IN ANY WAY, since it is not possible
>to divide a particle in this manner.
>
>> >> >In reality, that is an OBSERVATION.
>> >> Nobody has ever measured the one way velocity of light from a moving
>> >> source.
>> >
>> >Ah, yes, THIS random assumption again. Looks like we have another of
>> >occurs in the direction of motion.
>> L is the length of both arms of the MMX. You obviously know nothing
>> about the experiment or Einsteins explanation of the null result.
>
>L is the length of both arms in the REST FRAME OF THE APPARATUS. In any
>OTHER frame, there is a length contraction of the arm in the direction
>of motion, but the arm normal to the direction of motion does NOT
>contract. This eliminates your fantasy that the transverse arm
>contracts.
I didn't say it did.
I said the standard treatment claims that the number of wavelengths in
the cross beam varies with horizontal velocity - in proportion to the
length of the diagonal, to be exact. This is not true!
I hope you can follow the diagram and now understand the point I have
been trying to get across.
EL
[EL]
Dear JeffMo
The moving mirrors are only a convenient dynamic way to
reconstruct the sides of two infinite parallel reflecting walls.
The beam would be moving diagonally in a stationary Aether (zero
velocity) or absolutely perpendicular in a moving current of
Aether.
Relativistically we need to define the reference frame. Is it
the moving light or is it the moving Aether.
While they could be (and certainly they are) both moving, we
need to understand that they have no external third reference.
If the device is moving with respect to Aether we need to
imagine Aether being static.
If Aether is moving through the device we need to see NO
diagonals and we can not play it both ways simultaneously.
That is why this thread is confused.
We MUST be very clear about the reference anchor. What is it?
I have been arguing that the MMX construction was wrong from
scratch because of this particular point plus some other minor
considerations.
Calculating for a diagonal path infers a static Aether with zero
velocity and a null result.
Calculating for a dynamic Aether infers a perpendicular path (A
static beam axis) with no length variation and thusly no phase
shifting and the result is also null.
For that MMX to succeed they needed static mirrors as a
reference while the Aether drives the beam of the perpendicular
and shift the phase and create change in the interference
pattern.
There was no way to adjust an absolute reference from which a
deviation could be stimulated.
Changing the direction of the arms shall never change the
relative velocity between Aether and the light source and the
observer.
What should have been done is to compare two images, one direct
and one reflected, like a mirror in your car.
Light coming through the wind shield is compressed while that in
the mirror is stretched and can cause a phase shift.
If the car was not moving the mirror shall be just an integrated
part of the scenery, while on moving we can tell the difference.
MMX is watching the same image (red shifted) equally from two
points of view.
No design was made for capturing different images for comparison.
Henry's chainsaw is just a demonstration of how WRONG the MMX
was (is) but they are identical.
The chainsaw tooth-synchronicity is identical to photonic phase
synchronicity and there is nothing affecting them.
The MMX should have used TWO opposing monochromatic light
sources interfering at a unification prism.
After adjusting the phases to coincide we rotate the device 90
degrees and watch the change, only then there would be any
meaning.
Using one and the same source demands a bidirectional path in
which we stretch and compress, losing gain, and returning to
synchronous state. Relativity can be very tricky without very
accurate frame of reference definition.
Kind regards.
EL Hemetis [ http://www.freeyellow.com/members8/hemetis/ ]
*
David Evens
Henry Wilson wrote:
> On Wed, 07 Jun 2000 23:23:04 -0400, David Evens
> <dev...@technologist.com> wrote:
> >Henry Wilson wrote:
> >> >
> >> >How can the light move up the beam at c while it moves ACROSS on the
> >> >diagonals at c, as the physical laws of the universe require?
> >> No light beam moves along a diagonal, that's why. Infinitesimally
> >> small elements of the vertical light beam each follow a separate
> >> diagonal path to the top mirror.
> >
> >That is to say, all the light moves on the diagonals, at c.
> NO! You cannot give a name to an infinitesimal point. It is not a
> photon. A photon has size and volume otherwise it couldn't hold
> intrinsic information about its source.
Given the fact that a photon has zero size, it is fortunate that photons
carry no hidden information about their sources.
> >> Why would anyone in their right mind
> >> want to claim that they move at c along this path.
> >
> >Because that is the path it must follow to stay in the apparatus.
> why should it move at c? that was my question. Why shouldn't it move
> at sqrt(c^2+v^2) like the chainsaw teeth?
Other than the fact that the laws of physics require light to travel at
the same speed in all inertial frames?
> >> It obviously moves
> >> vrtiacally at c.
> >
> >Then how does it stay in the apparatus, which isn't at rest?
> the vertical component of its velocity is c. like the velocity of
> falling raindrops going past your car window. they always fall
> vertically at the same rate even though they appear to move diagonally
> as you travel.
I see that your grotesque ignorance of physics doesn't stop you from
making these bizarre claims.
> >> The important factor is that the number of
> >> wavelengths in the cross beam doesn't depend on velocity of the
> >> apparatus.
> >> This is where Einstein and his disciples are clearly wrong.
> >
> >Then you are claiming that time dilation does not occur, as it would
> >have to if the light is going to move as it is observed to move, which
> >is at c along the diagonal (which the diagrams always deal with only as
> >individual photons, which OBVIOUSLY move ONLY on the diagonals).
> you can call them individual photons if you like.
That is to say, they are only intelligibly treated as photons, since
they are units of light.
> The critical point is that the direction of the axis of the E and M
> wave components that lie along the 'length' of a photon always remains
> vertical even though the axis, as a whole, moves diagonally.
Where did you get the delusion that there was a length to photons?
Where did you get the delusion that photons have field axes, when the
field is clearly connected to the WAVES?
> _
> /
> / this is wrong.
> / The wave axis does not tilt.
> /
> _
> |
> | this is what happens
> | as all sections of the light beam
> | move along different diagonals.
And the light stays inside the apparatus without moving with because...?
> I suggest you plot the paths individual wave crests and you might be
> able to understand the significance of my discovery.
You REALLY should do this, since it could not avoid demonstrating that
you are stupidly wrong.
> >> Anyway, why don't you concentrate on the parallel beam. It is much
> >> more interesting!
> >
> >You're making stupidly wrong claims about the transverse beams, which is
> >much more interesting to ridicule you on. Of course, if you want to
> >start making stupid claims about the parallel, I can start making you
> >look stupid(er) with them, too.
> I wish I could include a .jpg here so that I could get the message
> across.
Why am I not surprised that you don't know how to do that?
> I will try to draw it using fixed pitch fonts. I did this
> before but you obviously didn't follow.
Just because your ignorance of how to use your newsreader matches your
ignorance of physics does not mean that you will be able to support your
delusions.
> M
> __q
> /\ /|
> / / \|r,p
> / / / |s
> / / / /| ^
> / / / / |t |c
> / / / / / |u
> / / / / / /|
> / / / / / / |v
> / / / / / / / |w
> __________________/__/__/__/__/__/__/__/|________________| M -->V
> a b c d e f g h
>
> The diagonal lines represent the paths taken by individual sections of
> the cross beam. a,b,c....., represent points in the rest frame where
> successive wavecrests leave the splitting mirror. p,q,r,s... show
> where these wavecrests lie when b reaches the top mirror. p is a point
> on the return beam representing the position of a. Simple geometry
> shows that the vertical separation of these points (wavelength)
> remains constant irrespective of velocity V.
> At any time, all sections are aligned vertically under the cross
> mirror. The wave axes of each beam element are never tilted
> diagonally, however. They remain vertical like so:
>
> |
> |
> |
> |
> |
So you are claiming that the laws of reflection are wrong, too.
> This is the crucial point of my revelation. The wave axis of each
> element moves diagonally but remains upright, like an infinitesimal
> rigid vertical rod. This is why the concept of the light beam moving
> diagonally at c has no credibility.
And the OBSERVATION that the beam moves diagonally at c is explained
by...?
> >> Sorry, It is not even an individual photon that moves along a single
> >> diagonal path. It is as infinitesimally small element of a photon that
> >> does.
> >
> >I see you know nothing about light IN ANY WAY, since it is not possible
> >to divide a particle in this manner.
> >
> >> >> >In reality, that is an OBSERVATION.
> >> >> Nobody has ever measured the one way velocity of light from a moving
> >> >> source.
> >> >
> >> >Ah, yes, THIS random assumption again. Looks like we have another of
>
> >> >occurs in the direction of motion.
> >> L is the length of both arms of the MMX. You obviously know nothing
> >> about the experiment or Einsteins explanation of the null result.
> >
> >L is the length of both arms in the REST FRAME OF THE APPARATUS. In any
> >OTHER frame, there is a length contraction of the arm in the direction
> >of motion, but the arm normal to the direction of motion does NOT
> >contract. This eliminates your fantasy that the transverse arm
> >contracts.
> I didn't say it did.
So, you explain your claim that there is a contraction involved by...?
> I said the standard treatment claims that the number of wavelengths in
> the cross beam varies with horizontal velocity - in proportion to the
> length of the diagonal, to be exact. This is not true!
Indeed your claim is not true, since it is the CONSTANCY of the number
of wavelength in the transverse beam that inevitably leads to the
conclusion that clocks in other frames must appear to be running slow.
> I hope you can follow the diagram and now understand the point I have
> been trying to get across.
I have followed your diagrams, and understand perfectly well what your
mistakes are. You have IGNORED the fact that you ARE making mistakes.
EL
*
Given the fact that a photon has zero size, it is fortunate that
photons carry no hidden information about their sources.
Other than the fact that the laws of physics require light to
travel at the same speed in all inertial frames?
I see that your grotesque ignorance of physics doesn't stop you
from making these bizarre claims.
That is to say, they are only intelligibly treated as photons,
since they are units of light.
Where did you get the delusion that there was a length to
photons?
Where did you get the delusion that photons have field axes,
when the field is clearly connected to the WAVES?
And the light stays inside the apparatus without moving with
because...?
You REALLY should do this, since it could not avoid
demonstrating that you are stupidly wrong.
Why am I not surprised that you don't know how to do that?
Just because your ignorance of how to use your newsreader
matches your ignorance of physics does not mean that you will be
able to support your delusions.
So you are claiming that the laws of reflection are wrong, too.
And the OBSERVATION that the beam moves diagonally at c is
explained by...?
So, you explain your claim that there is a contraction involved
by...?
Indeed your claim is not true, since it is the CONSTANCY of the
number of wavelength in the transverse beam that inevitably
leads to the conclusion that clocks in other frames must appear
to be running slow.
I have followed your diagrams, and understand perfectly well
what your mistakes are. You have IGNORED the fact that you ARE
making mistakes.
[EL]
What is this garbage?
I could not find one coherent sentence in your post else than
rude insults and most ignorant conception of physics.
Who gave that fucken *FACT* that a photon has "Zero"
size??????????
Is not a photon the Quantification of the particle of light
which happens to have a variable wavelength? Why is it a zero?
If you are referring to the photons of enlightenment of your
mind, I must agree with you. ;-)
Please study better than that dogma and be objective to
objective people.
Naturally you are absolutely allowed to toy with kooks, but I do
not think that Wilson is one of them.
If you have knowledge, spit it out and educate him.
If you have non, just SHUT the fuck UP.
Henry Wilson
<dev...@technologist.com> wrote:
>
>Henry Wilson wrote:
>> >That is to say, all the light moves on the diagonals, at c.
>> NO! You cannot give a name to an infinitesimal point. It is not a
>> photon. A photon has size and volume otherwise it couldn't hold
>> intrinsic information about its source.
>
>Given the fact that a photon has zero size, it is fortunate that photons
>carry no hidden information about their sources.
You really are stubborn, aren't you. You obviously know you are wrong
but you will never admit it. I wouldn't even be surprised if all my
messages suddenly appeared in the Journal of Physics under your name.
>
>> >> Why would anyone in their right mind
>> >> want to claim that they move at c along this path.
>> >
>> >Because that is the path it must follow to stay in the apparatus.
bullshit. look at the diagram below. It stays vertical - which is why
it always remains focussed on the centre of the top mirror.
>> why should it move at c? that was my question. Why shouldn't it move
>> at sqrt(c^2+v^2) like the chainsaw teeth?
>
>Other than the fact that the laws of physics require light to travel at
>the same speed in all inertial frames?
What is moving at that speed is not a light beam. It is an
infinitesimal element. There are plenty of instance in physics of that
kind of thing happening.
>
>> >> It obviously moves
>> >> vrtiacally at c.
>> >
>> >Then how does it stay in the apparatus, which isn't at rest?
>> the vertical component of its velocity is c. like the velocity of
>> falling raindrops going past your car window. they always fall
>> vertically at the same rate even though they appear to move diagonally
>> as you travel.
>
>I see that your grotesque ignorance of physics doesn't stop you from
>making these bizarre claims.
I suppose they would sound bizarre to anyone as brainwashed and
gullible as you.
>
>> >> The important factor is that the number of
>> >> wavelengths in the cross beam doesn't depend on velocity of the
>> >> apparatus.
>> >> This is where Einstein and his disciples are clearly wrong.
>> >
>> >Then you are claiming that time dilation does not occur, as it would
>> >have to if the light is going to move as it is observed to move, which
>> >is at c along the diagonal (which the diagrams always deal with only as
>> >individual photons, which OBVIOUSLY move ONLY on the diagonals).
>> you can call them individual photons if you like.
>
>That is to say, they are only intelligibly treated as photons, since
>they are units of light.
>
>> The critical point is that the direction of the axis of the E and M
>> wave components that lie along the 'length' of a photon always remains
>> vertical even though the axis, as a whole, moves diagonally.
>
>Where did you get the delusion that there was a length to photons?
>
>Where did you get the delusion that photons have field axes, when the
>field is clearly connected to the WAVES?
I don't think you are making much sense, here. It's a bit hard to
imagine a photon having E and an M perpendicular wave components,
without having length.
>
>> _
>> /
>> / this is wrong.
>> / The wave axis does not tilt.
>> /
>> _
>> |
>> | this is what happens
>> | as all sections of the light beam
>> | move along different diagonals.
>
>And the light stays inside the apparatus without moving with because...?
Look, pick up a pencil, hold it vertically, then move it sideways.
Does it tip over diagonally?
>
>> I suggest you plot the paths individual wave crests and you might be
>> able to understand the significance of my discovery.
>
>You REALLY should do this, since it could not avoid demonstrating that
>you are stupidly wrong.
>
>> >> Anyway, why don't you concentrate on the parallel beam. It is much
>> >> more interesting!
>> >
>
>> I will try to draw it using fixed pitch fonts. I did this
>> before but you obviously didn't follow.
>
>Just because your ignorance of how to use your newsreader matches your
>ignorance of physics does not mean that you will be able to support your
>delusions.
Do you actually know how to switch to fixed pitch fonts to dsplay
diagrams correctly on all newsreaders. I suppose you are using crappy
old MS software.
>
>> M
>> __q
>> /\ /|
>> / / \|r,p
>> / / / |s
>> / / / /| ^
>> / / / / |t |c
>> / / / / / |u
>> / / / / / /|
>> / / / / / / |v
>> / / / / / / / |w
>> __________________/__/__/__/__/__/__/__/|________________| M -->V
>> a b c d e f g h
>>
>> The diagonal lines represent the paths taken by individual sections of
>> the cross beam. a,b,c....., represent points in the rest frame where
>> successive wavecrests leave the splitting mirror. p,q,r,s... show
>> where these wavecrests lie when b reaches the top mirror. p is a point
>> on the return beam representing the position of a. Simple geometry
>> shows that the vertical separation of these points (wavelength)
>> remains constant irrespective of velocity V.
>> At any time, all sections are aligned vertically under the cross
>> mirror. The wave axes of each beam element are never tilted
>> diagonally, however. They remain vertical like so:
>>
>> |
>> |
>> |
>> |
>> |
>
>So you are claiming that the laws of reflection are wrong, too.
Where is the reflection here. What I have drawn here is the way a
short section of upward moving beam look like. The mirror is always
situated directly above each infinitesimally small vertical element.
>
>> This is the crucial point of my revelation. The wave axis of each
>> element moves diagonally but remains upright, like an infinitesimal
>> rigid vertical rod. This is why the concept of the light beam moving
>> diagonally at c has no credibility.
>
>And the OBSERVATION that the beam moves diagonally at c is explained
>by...?
Ignorance of basic geometry and lack of spatial ability on the part of
those who make such incorrect claims.
>
>> >L is the length of both arms in the REST FRAME OF THE APPARATUS. In any
>> >OTHER frame, there is a length contraction of the arm in the direction
>> >of motion, but the arm normal to the direction of motion does NOT
>> >contract. This eliminates your fantasy that the transverse arm
>> >contracts.
>> I didn't say it did.
>
>So, you explain your claim that there is a contraction involved by...?
Will you please read up on the standard analysis of the MMX before you
post any more embarassing nonsense.
>
>> I said the standard treatment claims that the number of wavelengths in
>> the cross beam varies with horizontal velocity - in proportion to the
>> length of the diagonal, to be exact. This is not true!
>
>Indeed your claim is not true, since it is the CONSTANCY of the number
>of wavelength in the transverse beam that inevitably leads to the
>conclusion that clocks in other frames must appear to be running slow.
Oh really???
>
>> I hope you can follow the diagram and now understand the point I have
>> been trying to get across.
>
>I have followed your diagrams, and understand perfectly well what your
>mistakes are. You have IGNORED the fact that you ARE making mistakes.
I have corrected one basic mistake in the MMX, that being the fact
that no diagonal beam of light ever moves at velocity c.
Now that I have cleaned up the interpretive errors of the cross beam,
lets talk about the parallel one because that is much more
controversial.
Henry Wilson
<hemetis...@lilac.ocn.ne.jp.invalid> wrote:
>[David Evens]
>*
>Given the fact that a photon has zero size, it is fortunate that
>photons carry no hidden information about their sources.
>*
>Other than the fact that the laws of physics require light to
>travel at the same speed in all inertial frames?
>*
>I see that your grotesque ignorance of physics doesn't stop you
>from making these bizarre claims.
>*
>That is to say, they are only intelligibly treated as photons,
>since they are units of light.
>*
>Where did you get the delusion that there was a length to
>photons?
>*
>Where did you get the delusion that photons have field axes,
>when the field is clearly connected to the WAVES?
>*
>And the light stays inside the apparatus without moving with
>because...?
>*
>You REALLY should do this, since it could not avoid
>demonstrating that you are stupidly wrong.
>*
>Why am I not surprised that you don't know how to do that?
>*
>Just because your ignorance of how to use your newsreader
>matches your ignorance of physics does not mean that you will be
>able to support your delusions.
>*
>So you are claiming that the laws of reflection are wrong, too.
>*
>And the OBSERVATION that the beam moves diagonally at c is
>explained by...?
>*
>So, you explain your claim that there is a contraction involved
>by...?
>*
>Indeed your claim is not true, since it is the CONSTANCY of the
>number of wavelength in the transverse beam that inevitably
>leads to the conclusion that clocks in other frames must appear
>to be running slow.
>*
>I have followed your diagrams, and understand perfectly well
>what your mistakes are. You have IGNORED the fact that you ARE
>making mistakes.
>*
>
>[EL]
>
>What is this garbage?
>I could not find one coherent sentence in your post else than
>rude insults and most ignorant conception of physics.
>Who gave that fucken *FACT* that a photon has "Zero"
>size??????????
>Is not a photon the Quantification of the particle of light
>which happens to have a variable wavelength? Why is it a zero?
>If you are referring to the photons of enlightenment of your
>mind, I must agree with you. ;-)
>
>Please study better than that dogma and be objective to
>objective people.
>Naturally you are absolutely allowed to toy with kooks, but I do
>not think that Wilson is one of them.
>If you have knowledge, spit it out and educate him.
>If you have non, just SHUT the fuck UP.
>
Thank you EL. I'm trying to maintain my composure with this clown. I
think he's almost converted!
Henry Wilson
Bilgey, old chap, you don't actually spin the laser, You spin a tiny
mirror onto which the laser beam is directed. At 60,000 rpm, or 1000
rps, the spot would appear to move at the speed of light at a radius
of about 45 kms. A bit tricky but not impossible!
It might be possible to use a number of rotating mirrors to
effectively increase the spot rotation speed and reduce the radius.
Henry Wilson
wrote:
>He...@the.edge(Henry Wilson) wrote:
>
>
>Don't the mirrors move during the time the photons are traveling?
>Doesn't "the line between the two mirrors" move during the time the
>photons are traveling? Wouldn't that mean that the photons travel a
>"diagonal path," even while remaining "along the line?"
Infinitesimal elements do but always on different diagonals, assuming
of course, that photons have some kind of wavelike pattern along a
longitudinal axis. The axes stay vertical, just like segments of the
chainsaw. The teeth always point vertically, in all frames, even
though the entire saw moves sideways and each tooth moves diagonally.
I know that sounds strange but the chainsaw clearly demonstrates the
point.
>> No complete
>>photon ever moves along a single diagonal path.
>
>Does something less than a "complete photon" move? Or does a
>"complete photon" move along multiple paths?
A complete photon moves along a broad diagonal line but its axis
remains upright.
>
>I'm really not following what you are claiming here. I apologize if
>it's just my stupidity driving this subthread. Ignore me if you feel
>I'm holding you back from a more substantive discussion, in all
>seriousness.
No, quite the contrary. The picture is becoming progressively clearer
to me, because of all this discussion. Keep it up Jeffmo.
I might even write a paper on this soon.
Henry Wilson
<hemetis...@lilac.ocn.ne.jp.invalid> wrote:
>
>[EL]
>Dear JeffMo
>
>The moving mirrors are only a convenient dynamic way to
>reconstruct the sides of two infinite parallel reflecting walls.
>The beam would be moving diagonally in a stationary Aether (zero
>velocity) or absolutely perpendicular in a moving current of
>Aether.
>Relativistically we need to define the reference frame. Is it
>the moving light or is it the moving Aether.
>While they could be (and certainly they are) both moving, we
>need to understand that they have no external third reference.
>If the device is moving with respect to Aether we need to
>imagine Aether being static.
>If Aether is moving through the device we need to see NO
>diagonals and we can not play it both ways simultaneously.
Yes that's a good way to lok at it.
>That is why this thread is confused.
>
>We MUST be very clear about the reference anchor. What is it?
>
>I have been arguing that the MMX construction was wrong from
>scratch because of this particular point plus some other minor
>considerations.
>
>Calculating for a diagonal path infers a static Aether with zero
>velocity and a null result.
>
>Calculating for a dynamic Aether infers a perpendicular path (A
>static beam axis) with no length variation and thusly no phase
>shifting and the result is also null.
More importantly, the time taken for the beam to reach the mirror is
unaffected by your aether velocity.
>
>For that MMX to succeed they needed static mirrors as a
>reference while the Aether drives the beam of the perpendicular
>and shift the phase and create change in the interference
>pattern.
>
>There was no way to adjust an absolute reference from which a
>deviation could be stimulated.
>
>Changing the direction of the arms shall never change the
>relative velocity between Aether and the light source and the
>observer.
>
>The MMX should have used TWO opposing monochromatic light
>sources interfering at a unification prism.
Excellent thinking El.
>
>After adjusting the phases to coincide we rotate the device 90
>degrees and watch the change, only then there would be any
>meaning.
Yes!!!!
Henry Wilson
<dev...@technologist.com> wrote:
>
>Henry Wilson wrote:
>> On Tue, 06 Jun 2000 22:24:32 GMT, jef...@dipstick.cfw.com (JeffMo)
>> wrote:
>> >ro...@radioactivex.lebesque-al.net (Bilge) wrote:
>> >>JeffMo said some stuff about
>> >>
>> >> >I doubt Henry's theories as much as anyone, but this objection is
>> >> >wrong. It is certainly physically possible to make a laser spot move
>> >> >faster than c. The spot is a sort of phantom object, though, like a
>> >> >shadow. It is not the SAME photons moving superluminally, but
>> >> >different photons being reflected from a surface. There is no
>> >> >superluminal motion of any given mass, energy, or information.
>> >> >
>> >>
>> >> It doesn't count if you have to violate relativity to call it
>> >>the same spot.
>> >
>> >We both agree. It's basically a trick of human perception that we
>> >tend to label it the same spot. It's not a thing going faster than c,
>> >even though it would look like it was, to people who didn't think it
>> >through, and realize that it's really the reflection of a
>> >continuously-flowing STREAM, which is different photons in each
>> >"instant."
>> >
>> >>There is simply no justification for refering
>> >>to a spot at fixed r as the the same spot without a condition placed
>> >>on what you mean by the "same spot". Just like with the sagnac
>> >>effect, this provides a great opportunity to abuse the math and
>> >>declare sr incorrect.
>> >
>> >I agree with you wholeheartedly, sir.
>> >
>> >JeffMo
>> And so do I. Hooray, we all agree!!!!This backs up the point I have
>> been making, all along.
>> WHAT MOVES DIAGONALLY IN THE MMX IS NOT A LIGHT BEAM MOVING AT C BUT A
>> SERIES OF INFINITESIMALLY SMALL ELEMENTS THAT APPEAR TO MOVE AT
>> SUPERLUMINAL SPEEDS.
>
>Then why is there nothing that appears to move at superluminal speeds
>ANYWHERE in the MMX?
They only do it on paper. They are infinitesimally small, which is why
I used the chainsaw to demonstrate the principle involved. This slows
everything down and greatly magnifies and assimilates the movement of
each minute photon length element.
PS. How's the paper going? Please give me an acknowledgement, will
you!
Bilge
>They only do it on paper. They are infinitesimally small, which is why
>I used the chainsaw to demonstrate the principle involved. This slows
>everything down and greatly magnifies and assimilates the movement of
>each minute photon length element.
>
Sorry char^H^H^H^Hhenry, but you cant allow your interval to shrink
if there are values that have to remain constant in any limiting process.
In this case, your assumption violates liouville's theorem. For a source
with a finite emittance, that value can never be made smaller. So,
regardless of what size laser beam you use, shrinking the same laser
to make it thinner cannot decreease the emittance without making it a
different laser. If you do that, you limiting process isn't legitimate,
since you are taking something that represents the average of the larger
laser. In short, your narrow beam cannot make both the momentum spread
and the angular divergence of the laser smaller at the same time. To
make the momentum spread small or the divergence small the other must
increase to keep the phase space constant.
Henry, you can't simply declare by fiat that all of your variables
must shrink just because you get a result you like. First, you have to
see how many physical laws are violated by ignoring the conditions
imposed by physics and in this case, ancient classical physics - Going
back to the 18th century. You violated physics before anyone had a chance
to argue using a phase space that's lorentz invariant. You violated
it in your own rest frame. It's a cinch you wont recover it with a
lorentz boost.
Bilge
>>
>Bilgey, old chap, you don't actually spin the laser, You spin a tiny
>mirror onto which the laser beam is directed. At 60,000 rpm, or 1000
>rps, the spot would appear to move at the speed of light at a radius
>of about 45 kms. A bit tricky but not impossible!
>It might be possible to use a number of rotating mirrors to
>effectively increase the spot rotation speed and reduce the radius.
Harry old chap, if you could bridge the gap, you would notice it
applies just as well to a spinning mirror. However if that were true,
this conversation wouldn't be taking place, either. I believe this
configuration is even worse though. If the laser is a finite width,
it still has a finite emittance.
<- v
/ <----------- y r
/| <----------- | /
+|| <----------- | /
/||| |/
/ ||| +--------- x
v -> ||| r*cos(\theta)
However, this doesnt show the light that leave the laser and
arrives at the destination simultaneously. That occurs for differnt
points in the rotation:
reflection is
symmetric about
the normal at this
point in the rotation.
.
.<-------------
. both leave the source at the same time.
.
. <-------------
.
| ------- |
\Delta\theta = w\delta t (do you really want the transformed
coordinates?)
. l
.
.
.
. This angle of reflection is smaller
. <-------------<
direction of first ray
/
/
/ | direction of second ray.
/ |
X
---------- plane perpendicular to the observer at X. He sees the second
ray and not the first. Both rays are also doppler shifted, but
by different amounts. Consequently, the rays that the observer
sees in all will be ones with different frequencies that depend upon the
angular velocity of the mirror.
Of course, moving toward the center minimizes this, BECAUSE the
center has a lower linear velocity. So. What else would you expect?
It's easy to get rid show there are no relativistic effects when you
simplify the problem to one with no relativistic movement. It hardly
matters what the velocity at a distance r from the center of rotation
if you don't shine any light there to illustrate the effect of the
velocity that matters.
This is exactly the same reason your chainsaw abomination is wrong.
You assumed that the light remained vertical because no one would see
the width of the lenses and mirrors contract because you don't in the
restframe of the apparatus. That's wrong. The mirrors and lenses have
different angles as the rest of the world sees it.
David Evens
Henry Wilson wrote:
> On Thu, 08 Jun 2000 19:43:40 -0400, David Evens
> <dev...@technologist.com> wrote:
> >Henry Wilson wrote:
> >> >That is to say, all the light moves on the diagonals, at c.
> >> NO! You cannot give a name to an infinitesimal point. It is not a
> >> photon. A photon has size and volume otherwise it couldn't hold
> >> intrinsic information about its source.
> >
> >Given the fact that a photon has zero size, it is fortunate that photons
> >carry no hidden information about their sources.
> You really are stubborn, aren't you. You obviously know you are wrong
> but you will never admit it. I wouldn't even be surprised if all my
> messages suddenly appeared in the Journal of Physics under your name.
Why are you suddenly talking to yourself?
> >> >> Why would anyone in their right mind
> >> >> want to claim that they move at c along this path.
> >> >
> >> >Because that is the path it must follow to stay in the apparatus.
> bullshit. look at the diagram below. It stays vertical - which is why
> it always remains focussed on the centre of the top mirror.
> >> why should it move at c? that was my question. Why shouldn't it move
> >> at sqrt(c^2+v^2) like the chainsaw teeth?
> >
> >Other than the fact that the laws of physics require light to travel at
> >the same speed in all inertial frames?
> What is moving at that speed is not a light beam. It is an
> infinitesimal element. There are plenty of instance in physics of that
> kind of thing happening.
To bad (for you) that light can only be divided into finitely small
elements. These finitely small elements are called 'photons,' and are
not subject to further subdivision.
> >> >> It obviously moves
> >> >> vrtiacally at c.
> >> >
> >> >Then how does it stay in the apparatus, which isn't at rest?
> >> the vertical component of its velocity is c. like the velocity of
> >> falling raindrops going past your car window. they always fall
> >> vertically at the same rate even though they appear to move diagonally
> >> as you travel.
> >
> >I see that your grotesque ignorance of physics doesn't stop you from
> >making these bizarre claims.
> I suppose they would sound bizarre to anyone as brainwashed and
> gullible as you.
Why are you talking to yourself?
> >> >> The important factor is that the number of
> >> >> wavelengths in the cross beam doesn't depend on velocity of the
> >> >> apparatus.
> >> >> This is where Einstein and his disciples are clearly wrong.
> >> >
> >> >Then you are claiming that time dilation does not occur, as it would
> >> >have to if the light is going to move as it is observed to move, which
> >> >is at c along the diagonal (which the diagrams always deal with only as
> >> >individual photons, which OBVIOUSLY move ONLY on the diagonals).
> >> you can call them individual photons if you like.
> >
> >That is to say, they are only intelligibly treated as photons, since
> >they are units of light.
> >
> >> The critical point is that the direction of the axis of the E and M
> >> wave components that lie along the 'length' of a photon always remains
> >> vertical even though the axis, as a whole, moves diagonally.
> >
> >Where did you get the delusion that there was a length to photons?
> >
> >Where did you get the delusion that photons have field axes, when the
> >field is clearly connected to the WAVES?
> I don't think you are making much sense, here. It's a bit hard to
> imagine a photon having E and an M perpendicular wave components,
> without having length.
I see you didn't bother reading what you were replying to again.
Had you bothered to read it, you would not have been able to avoid
noticing that I pointed out that the FIELD is associate with the WAVES,
not the PHOTONS.
> >> _
> >> /
> >> / this is wrong.
> >> / The wave axis does not tilt.
> >> /
> >> _
> >> |
> >> | this is what happens
> >> | as all sections of the light beam
> >> | move along different diagonals.
> >
> >And the light stays inside the apparatus without moving with because...?
> Look, pick up a pencil, hold it vertically, then move it sideways.
> Does it tip over diagonally?
It certainly TRIES to, and you have to apply a torque to keep it upright
while you accelerate it. Light beams, being composed of disconnected
photons, are not subject to torques any more than a stream of water.
> >> I suggest you plot the paths individual wave crests and you might be
> >> able to understand the significance of my discovery.
> >
> >You REALLY should do this, since it could not avoid demonstrating that
> >you are stupidly wrong.
> >
> >> >> Anyway, why don't you concentrate on the parallel beam. It is much
> >> >> more interesting!
> >> >
> >
> >> I will try to draw it using fixed pitch fonts. I did this
> >> before but you obviously didn't follow.
> >
> >Just because your ignorance of how to use your newsreader matches your
> >ignorance of physics does not mean that you will be able to support your
> >delusions.
> Do you actually know how to switch to fixed pitch fonts to dsplay
> diagrams correctly on all newsreaders. I suppose you are using crappy
> old MS software.
Why would you assume that I am using the software you favour? After
all, I use somewhat better stuff.
I see you didn't look at the diagrams you mindless attack which include
the light being reflected from a MIRROR.
> What I have drawn here is the way a
> short section of upward moving beam look like. The mirror is always
> situated directly above each infinitesimally small vertical element.
You forgot to support your assumption that photons are divisible.
> >> This is the crucial point of my revelation. The wave axis of each
> >> element moves diagonally but remains upright, like an infinitesimal
> >> rigid vertical rod. This is why the concept of the light beam moving
> >> diagonally at c has no credibility.
> >
> >And the OBSERVATION that the beam moves diagonally at c is explained
> >by...?
> Ignorance of basic geometry and lack of spatial ability on the part of
> those who make such incorrect claims.
That is to say, those who do not ignore their own observations because
they do not adhere to your ignorant prejudices.
> >> >L is the length of both arms in the REST FRAME OF THE APPARATUS. In any
> >> >OTHER frame, there is a length contraction of the arm in the direction
> >> >of motion, but the arm normal to the direction of motion does NOT
> >> >contract. This eliminates your fantasy that the transverse arm
> >> >contracts.
> >> I didn't say it did.
> >
> >So, you explain your claim that there is a contraction involved by...?
> Will you please read up on the standard analysis of the MMX before you
> post any more embarassing nonsense.
Indeed you should, since you have humiliated yourself with your
publication of your delusions about the MMX.
> >> I said the standard treatment claims that the number of wavelengths in
> >> the cross beam varies with horizontal velocity - in proportion to the
> >> length of the diagonal, to be exact. This is not true!
> >
> >Indeed your claim is not true, since it is the CONSTANCY of the number
> >of wavelength in the transverse beam that inevitably leads to the
> >conclusion that clocks in other frames must appear to be running slow.
> Oh really???
Yes. Were you familiar with the MMX, you could not avoid knowing that
the constancy of the number of wavelengths means that the NUMBER of
waves along the beam path in BOTH frames is the same. Since the photon
path is the frame in which the apparatus is NOT at rest is LONGER than
it is in the rest frame in which the apparatus IS at rest, and we SEE
that the speed of light is identical in both frames, the ONLY way that
the number of waves can be the same is if the observed rate of time flow
drops as relative speed rises, since this allows the frequency to change
in the required manner.
> >> I hope you can follow the diagram and now understand the point I have
> >> been trying to get across.
> >
> >I have followed your diagrams, and understand perfectly well what your
> >mistakes are. You have IGNORED the fact that you ARE making mistakes.
> I have corrected one basic mistake in the MMX, that being the fact
> that no diagonal beam of light ever moves at velocity c.
Pretending that there was a beam that did something that no beam in the
MMX diagrams does doesn't help you.
> Now that I have cleaned up the interpretive errors of the cross beam,
> lets talk about the parallel one because that is much more
> controversial.
Your inability to understand the physics of the cross beam does not
constitute an interpretative error on the part of anyone but yourself.
David Evens
Henry Wilson wrote:
> On Wed, 07 Jun 2000 23:28:51 -0400, David Evens
So you are NOW claiming that your previous claim of a superluminal light
motion in the MMX is wrong.
> They are infinitesimally small, which is why
> I used the chainsaw to demonstrate the principle involved. This slows
> everything down and greatly magnifies and assimilates the movement of
> each minute photon length element.
YOu MUST stop contradicting yourself. Are you talking about photons, or
are you talking about your magical 'infinitesimally small light beam
elements'?
> PS. How's the paper going? Please give me an acknowledgement, will
> you!
Acknowledgments are not normally given in comedy writing.
Bilge
>
>Acknowledgments are not normally given in comedy writing.
He could always be acknowledged for the incite gained from posting
all of the errata and arguing about it. There's lttle doubt that
in trying to find an explanation that he'll absorb without chiseling
it into his forehead, I've probably come up with another half-dozen
points that will help explain the same sorts of things to people
that don't have the explanation hard-coded.
Bilge
>I have said all along, a wave crest moves diagonally at >c. This is a
>'phase velocity', as Bilgey so rightly agreed. I used "wavecrest"
>because it is easier to type than "points of equal phase belonging to
>the intrinsic photon waveform which always has its axis pointing
>vertically".
The phase velocity in vacuum is equal to the group velocity and both
are equal to c. The only velocity which I claimed was faster than
c, was the phase between fourier components of two different pulses
seen by different observers which I also said shouldn't be considered
to be a velocity and that in general phase velocity was a poor name
for what it is, since nothing moves. Henry, If you could correlate,
e.g., see an image, with the part of the disturbance travelling at
the phase velocity, it would violate sr. So, if that is your starting
point, then you've assumed this from the outset. Therefore, you differ
with sr before ever getting off the ground. You haven't shown anything,
you've asserted it as a premise. Since your premise is incompatible
with relativity, you cant expect it to be correct. N.B. I think
relativity predicts that the rotating "beacon" will appear to
an observer to shrink to a point along the diameter in the direction
that ends up perpendicular to the observer as wr -> c.
Henry Wilson
McGregor) wrote:
>On Mon, 12 Jun 2000 03:31:54 GMT, HWilson@..(Henry Wilson) wrote:
>
>>On Sun, 11 Jun 2000 07:28:39 GMT, ro...@radioactivex.lebesque-al.net
>>(Bilge) wrote:
>>
>>>Henry Wilson said some stuff about
>>>
>>> >Of course this whole discussion is hypothetical and only applies to
>>> >infinitesimally thin laser beams or very long distances. Still the
>>> >spot, or part of the spot, will eventually appear to move at greater
>>> >than c.
>>>
>>> That isn't the case if you unless you ignore arrival times, in
>>> which case you didn't start with a relativistic situation anyway.
>>> The detector large enough to see the spot, will instead see a
>>> shorter version of it. By using a continuous beam, the only thing
>>> that you could claim moves in excess of c is the phase velocity,
>>> which can't be a velocity in anything but nomenclature. Sheesh.
>>> Some unfortunate terminology, held over from someone that couldn't
>>> read the future and see that some people people are so unable
>>> to separate physics from the terminology and viola, the era of
>>> grammar based phyical laws. Where yesterdays mistaken semantics
>>> are put forth as physical laws.
>>I indeed used this example to show that phase velocity can exceed c.
>>Just like the diagonal speed of a point on a photon wave element in
>>the MMX cross beam.
>>>
>O.K. let me try to explain.
>Henry, the 'diagonal speed' you talk of has two components.
>One is the speed along the beam, and the other is the component due to
>the othrogonal motion of the observer.
>
>Now think about it.
>
>Say you have two orthogonally moving observers, travelling at greatly
>different speed.
>
>By your reckoning they will see different 'diagonal speeds'.
>
>You can have any number of observers who can measure whatever
>'diagonal speed' you might desire.
>
>Does this make sense?
Yes and no.
It depends how it is observed.
As I have pointed out many times in this thread, no light beam in the
MMX ever moves diagonally, which is the reason why SR has all the
wrong equations. the time taken for the cross beam to reach the top
mirror is independent of apparatus speed.
>
>Does this mean that the speed actually changes each time some new
>observer measures it? Of course not, they could all
>measure the same photon. How can it be travelling at different speeds
>at the same time? It is different relative to each observer, but to
>ascribe any meaningful reality to observer differences in terms of the
>observed interaction is irrational.
>
>What IS constant, i.e. what is potentially calculable by all observers
>to be the same, is the relative speed between the two ends of the
>beam, which is c. The SR transorms allow us to make this calculation
>based on measurements we can make.
>
>It is not just constant, but is also physically meaningful since it
>concerns the observed interaction.
>
>I'd started a thread where I hoped to use energy and momentum to help
>illustrate the confusion which can arise between calculations of
>potential only energy transformations between observer and observed
>interaction entities as opposed to that which actually does occur
>during the observed interaction. This type of confusion can occur even
>at non-relativistic speeds
>
>I'd found it useful at one stage, to consider this and get my head
>straight on it at at non relativistic speeds. However the response to
>the thread was disappointing.
>
>If I may give the following non relativistic speed example, it may
>illuminate you:-
>
>An engine in a siding. It is fitted at the front with long buffers
>which are unusual in that they have regular and numbered gradations
>and they are designed to stay in where-ever they get depressed to so
>that a reading may be taken (e.g. let's say they are extentions to
>pistons which a small aperture and which are pushed through an oil
>filled cylinder).
>
>The siding is at right angles to two main lines close by.
>
>The siding train runs at constant speed until it hits a rigid buffer
>at the end of the track.
>
>While this is happening, there are two trains travelling at different
>speeds on the main-line tracks, they of course contain observers.
>
>The observers each have at their disposal a pair of binoculars (with
>which to read the buffer gradation), a radar gun with which to measure
>the speed of the siding train relative to themselves, a railway manual
>giving the mass of the siding train and the energy dissipation per
>gradation of the buffer. As usual, track resistence etc. is
>negligible.
>
>
>Now the first thing that we must agree on is that each observer sees
>the same final gradation on the buffer, i.e. they agree that the
>energy dissipated is the same, (assuming their manuals have the same
>conversion factors and reported train mass). Right?
I once suggested doing something similar with clocks to measure oneway
light speed.
Instead of reading the clock directly, one observes the reading the
clock recorded about itself. The clock records a time event in its own
frame then sends that information to an observer in another frame, who
would get a different picture by direct communication.
>
>Armed with this information they can both calculate what the siding
>train speed relative to the siding track was, and both will agree.
>
>However, if both use their radar guns on the siding train they will
>get different answers for the siding train speed.
>
>In other words, in terms of the observed interaction, the apparent
>component of velocity orthogonal to the siding track is in fact
>entirely due to the observer's own velocity, is therefore arbitrary
>and is totally irrelevant to the interaction. As far as the two
>interacting entities go, orthogonal components of observer velocity do
>not exist.
>In terms of MMX the observer speed is irrelevant to the interacting
>entities, i.e. the orthgonally pathed light beams.
>
>Note, this observation does not mean that C is everywhere constant,
>c+v would produce the same results.
>
>MMX does not contravene c+v, what it does do is require extra
>conditions for an ether theory e.g. ether dragging or length
>contraction which knocks the ether theory a notch down the Occam
>league.
The MMX definitely supports c+v, at least locally. It is the most
obvious explanation.
>
>To extend the above analogy to the ether case. This puts an additional
>boundary condition in play, namely that the siding train must always
>move at the same speed relative to the ether. If the ether had say a
>velocity at right angles to the siding track, then the component of
>train speed along the track must slow down and the train would take
>longer to reach the end of the track. This would be discernible in MMX
>so to explain a lack of said discernibility the track must either be
>decreed to shrink appropriately or a theory that the ether is always
>dragged along locally with the track must come into play.
Well, I don't accept that. I believe that the time taken for anything
to move from A to B is not affected my any orthogonal movement.
that's what happens in the case of raindrops passing your moving car
window and that's what my chainsaw demo shows.
>
>If an ether was present with a velocity with asymetric components to
>the two beam paths, then the light would travel in a 'diagonal' or
>more properly at an angle anomolous to that expected from mirror
>orientation.
>
>regards
>chic
I'm not really into ether. I don't see any need for it and it doesn't
support my theory. But I don't entirely rule it out either. I have
suggested that there are two more mass dimensions that might explain
many current mysteries.
Henry Wilson
(Bilge) wrote:
>Henry Wilson said some stuff about
>
> >I have said all along, a wave crest moves diagonally at >c. This is a
> >'phase velocity', as Bilgey so rightly agreed. I used "wavecrest"
> >because it is easier to type than "points of equal phase belonging to
> >the intrinsic photon waveform which always has its axis pointing
> >vertically".
>
> The phase velocity in vacuum is equal to the group velocity and both
> are equal to c.
By phase velocity, I was refering to the movement of a point of equal
phase in whatever periodicity a photon has, be it temporal spatial or
both. To clarify that, Imagine a photon to have a physical form
involving a sine wave like the teeth of a saw blade. When the photon
moves along something, a time frequency is generated as each tooth
goes past a reference point.. In the case of an operating chainsaw
moving sideways, the diagonal phase velocity would be that of a
particular point on each tooth.
If you follow my demo, you will see that this velocity, in the MMX =
sqrt(c^2+v^2)
>The only velocity which I claimed was faster than
> c, was the phase between fourier components of two different pulses
> seen by different observers which I also said shouldn't be considered
> to be a velocity and that in general phase velocity was a poor name
> for what it is, since nothing moves.
I'm not quite with you here. Light can only have a fundamental
frequency. Otherwise the photoelectric effect would show thresholds
corresponding to each harmonic. That is something I deduced for myself
and has probably got nothing to do with what you are talking about but
it is interesting anyway..
>Henry, If you could correlate,
> e.g., see an image, with the part of the disturbance travelling at
> the phase velocity, it would violate sr. So, if that is your starting
> point, then you've assumed this from the outset. Therefore, you differ
> with sr before ever getting off the ground.
Of course I differ with SR. I am trying to tell everyone why it is
completely wrong, right from its first assumption!
I don't, however, claim to be able to see an image resulting from a
phase velocity. Quite the opposite. I have stressed that the entity
which moves >c is not a light beam. It is an infinitesimal element and
contains no information.
>You haven't shown anything,
> you've asserted it as a premise. Since your premise is incompatible
> with relativity, you cant expect it to be correct.
On the contrary. I cannot expect SR to be correct. for every bit of
so-called proof of SR, I can substitute an equation involving either
1-c/v or 1-(c/v)^2, that will give just as good an answer within
experimental error.
My main grouch with SR is the sqrt of gamma. THERE IS NO SQUARE ROOT
TERM!!!!!!!!! This is what I have shown here. The triangulation
normally assumed with the cross beam of the MMX is a fallacy.
I haven't made up my mind about the parallel beam however. It is more
interesting.
>N.B. I think
> relativity predicts that the rotating "beacon" will appear to
> an observer to shrink to a point along the diameter in the direction
> that ends up perpendicular to the observer as wr -> c.
hhhmmm! I wouldn't be surprised. How would the spot appear to a grid
of photodetectors on a distant wall, I wonder?
>
>
Charles McGregor
Your chainsaw and the raindrops are both examples where observer only
dependant sideways velocity components 'exist' or seem to exist. These
observer only components are irrelevant to that which is observed
therefore in those cases the 'time from A to B' would be unaffected.
They are extraneous, illusory and arbitrary and therefore not
elliminating them from the calculations totally invalidates analysis
that purports to any kind of consistency across observers.
However MMX was designed to measure the lab speed w.r.t. the ether.
The ether was the hypothetical medium for EM, the EM should have a
CONSTANT speed relative to, AND DEFINED BY, it's propagating medium.
Therefore for any given direction, provided of course, it is not
aligned with ether velocity, the component along that direction MUST
be less than the medium defined relative speed .
In other words, observer velocity and ether velocity are not
equivalent. The ether is part of the interaction.
Let's try the more traditional river example.
A swimmer will take the same time to cross a river regardless of
flow (ignoring choppiness, panic, constant water speed etc. all the
usual stuff), provided he remains oriented orthogonally to the river
banks. He appears to move diagonally, w.r.t the observer on the bank,
but consider this. If the observer on the bank were to run to match
the river speed, then he would observe the swimmer swimming straight
across, and not diagonally. In other words, this is yet just another
cunningly disguised eample where observer only velocity components
w.r.t the interaction have been included. except this time, for added
confusion we also moving riverbanks which play no part in the
interaction observed (i.e. how long it takes to get across).
If, however, as in the case of MMX, we place posts on the banks
opposite each other, at A and B, and now we tell the swimmer to make
sure he is always heading directly towards B, as in the case of MMX,
then because his water speed is the same as before (A torpedo might
have been a more plausible analogy) he will take longer to cross the
river. And this time, if the observer runs to match the river flow,
he will see that the swimmer does move diagonally w.r.t. river
velocity. This time the riverbank has been made an integral part of
the interaction.
The two cases are not equivalent.
What COULD produce a result similar to what you expect, i.e. that
could contrive that MMX was really analogous to river case 1 but with
added posts. i.e. the swimmer is still told to swim orthogonally to
the banks at all times BUTwhere the swimmer is very wide AND the timer
doesn't care what part of the swimmer touches post B. So regardless
of river/ether flow, the timing/phase change would be the same,
provided the swimmer/light beam was wide enough..
This would presume, for MMX, that
1) The beam width was wide enough.
2) That mirror angle was insufficiently determinate to provide an
independent assessment of beam angle.
3) That the 'wave front' was sufficiently planar
In the MMX case 'the timer' wouldn't care, because it is
interference/phase shift.
regarding 1)
If we assume that beam length was say 3 m
and the minimum findable ether speed was the speed of the Earth around
the sun, quick mental arithmetic 500, 000,000 km/y, ....15 km/s so we
have what... 3m = 10^-8 seconds trip time... approx 1.5e-5 m, about
15 microns...shit that's not a lot. Almost certainly the beam would
be wider than that. Anyone know if M&M looked for missing fringe at
beam edges?:-)
regarding 2)
1.5e-5/3 =5e-6 is approx 7.5e-4 degrees, how tough is that to
measure?, probably not very.
regarding 3)
Possibly
You COULD be kinda right Henry, :-) at least in terms of possibly
explaining MMX in terms of still having an ether, but not for the
analogous reasons you cite. In this case any ether effect would be
additive to light speed. But I guess you wouldn't be unhappy with
that either.:-)
OTOH I expect M&M were aware of this possibility and allowed for it
somehow in their set-up and calculations ( multiple angle measurements
with same mirror settings etc.)
It should be noted ether flow is not 100% analagous to that of a
river, despite the frequency with which such analogies are drawn.
Whereas the swimmer has the same speed w.r.t. the river analogous to
the ether case it is the swimmer that determines what that speed he
travels w.r.t the water, NOT the river.
>>
>>If an ether was present with a velocity with asymetric components to
>>the two beam paths, then the light would travel in a 'diagonal' or
>>more properly at an angle anomolous to that expected from mirror
>>orientation.
>>
>>regards
>>chic
>I'm not really into ether. I don't see any need for it and it doesn't
>support my theory. But I don't entirely rule it out either. I have
>suggested that there are two more mass dimensions that might explain
>many current mysteries.
regards
chic
Bilge
>By phase velocity, I was refering to the movement of a point of equal
>phase in whatever periodicity a photon has, be it temporal spatial or
>both. To clarify that, Imagine a photon to have a physical form
>involving a sine wave like the teeth of a saw blade. When the photon
A photon has no such form. A phase velocity, by definition
requires more than a single frequency. The definition is
v_phase = c/n(k). In vacuum, n(k) == n == 1.
>I'm not quite with you here. Light can only have a fundamental
>frequency. Otherwise the photoelectric effect would show thresholds
>corresponding to each harmonic. That is something I deduced for myself
>and has probably got nothing to do with what you are talking about but
>it is interesting anyway..
>
Phases refer to more than one frequency and can't refer to a
single photon. The phase velocity is what characterizes
dispersion. 1 photon cant disperse.
>Of course I differ with SR. I am trying to tell everyone why it is
>completely wrong, right from its first assumption!
In that case, it's an issue for an experiment. Since the fundamental
problem you have with relativity is one of the postulates, you can't
really argue with what follows from it as a consequence without
actually having some data to point at that illustrates clearly what
failed and why.
>I don't, however, claim to be able to see an image resulting from a
>phase velocity. Quite the opposite. I have stressed that the entity
I didn't.
>> you've asserted it as a premise. Since your premise is incompatible
>> with relativity, you cant expect it to be correct.
>On the contrary. I cannot expect SR to be correct. for every bit of
That's really what I meant. If you discard the postulates, then I
can't expect your description to match the one from relativity.
I was under the impression that you were also claiming this is
what relativity should predict. Regardless of the fact that I still
think you're wrong, that determination can't be made from an
argument or thought experiment. It's not very likely anyone will
build such a device soon to see which one of us is correct and
regardlee of the fact that I don't believe an experiment would
bear out your assertion, I'll always believe what competently
acquired and analyzed data comes up with. I'm not going to hold
my breath, however....
>so-called proof of SR, I can substitute an equation involving either
>1-c/v or 1-(c/v)^2, that will give just as good an answer within
>experimental error.
Well, there are several things to think about before getting
carried away. Like, why is that a surprise? If you rewrite
v as a fraction c, v = nc and ask by what ratio those differ
for 0 < n < 1:
r^2 = (1-n)^2/(1-n^2) = (1-n)/(1+n)
for n = .2 [velocities of 6x10^7 m/s], r^2 =~ .83 or the
percent difference ( |r^2 - 1|/(r^2 + 1), is the simplest
expression I see right offhand to express %diff compactly),
is about 10%. For a velocity of .001c (a mere 3x10^5 m/s),
r^2 =~ .05, for a %diff of 3%. What experiment would you
perform that you wouldn't say requires "unwarranted inferences"
so that you can determine which one of those (or from your
viewpoint, both) are incorrect? If you are basing the incor-
rectness on the fact that both are consistent with experiment,
describe an experiment to discern the two (that can be done
and has measured quantities you'd be satisfied don't "hide"
anything.
>My main grouch with SR is the sqrt of gamma. THERE IS NO SQUARE ROOT
>TERM!!!!!!!!! This is what I have shown here. The triangulation
>normally assumed with the cross beam of the MMX is a fallacy.
>
Even with no square root, the apparatus doesn't stay vertical. For
it to remain vertical, c = oo.
>I haven't made up my mind about the parallel beam however. It is more
>interesting.
>hhhmmm! I wouldn't be surprised. How would the spot appear to a grid
>of photodetectors on a distant wall, I wonder?
Actually, as I started to describe the result, there turned out to be
some interesting things about this that I need to calculate to actually
get the correct result. If the spot is to move as a continuous spot,
i.e., each detector in a line records the spot passing, then the first
thing is to define a spot which is continuous. I would define it such
that at two points the observers at each describe the same thing (color,
intensity). To be considered superluminal, the observers would have
to describe the same thing at times faster than the light between
them travels. I believe that I can show no such spot even can realized
Even if you subtract the time required for each to communicate the
result to the other, or to an independent observer:
A B Arrange the distance AC to be the same as BC. The
spot moving from A->B. Light travels from A-B in t_AB.
If anything travels from A->B in time T, then for
T <= t_AB C will receive communications from A&B such
C that t_ac + t_ab <= t_bc + T. This should be sufficient
to determine if the spot is superluminal. The detectors must see
the same thing. Exactly what that is, I have to calculate, but I
don't think it will be anything that can be continuous between
points you place detectors. At some point, I think you have to
violate liouvilles theorem or the uncertainty principle to get
that result. The uncertainty principle sort of being the bottom
line.
Henry Wilson
(Bilge) wrote:
>Henry Wilson said some stuff about
>
> >By phase velocity, I was refering to the movement of a point of equal
> >phase in whatever periodicity a photon has, be it temporal spatial or
> >both. To clarify that, Imagine a photon to have a physical form
> >involving a sine wave like the teeth of a saw blade. When the photon
>
>
> A photon has no such form. A phase velocity, by definition
> requires more than a single frequency. The definition is
> v_phase = c/n(k). In vacuum, n(k) == n == 1.
>
> >I'm not quite with you here. Light can only have a fundamental
> >frequency. Otherwise the photoelectric effect would show thresholds
> >corresponding to each harmonic. That is something I deduced for myself
> >and has probably got nothing to do with what you are talking about but
> >it is interesting anyway..
> >
> Phases refer to more than one frequency and can't refer to a
> single photon. The phase velocity is what characterizes
> dispersion. 1 photon cant disperse.
>
Bilgey, how come you know so much about photons when nodody else seems
to know anything conclusive? Have you actually seen a photon?
> >Of course I differ with SR. I am trying to tell everyone why it is
> >completely wrong, right from its first assumption!
>
> In that case, it's an issue for an experiment. Since the fundamental
> problem you have with relativity is one of the postulates, you can't
> really argue with what follows from it as a consequence without
> actually having some data to point at that illustrates clearly what
> failed and why.
>
The MMX is a good start, then. If that is wrong, then the whole of SR
is also wrong!
> >I don't, however, claim to be able to see an image resulting from a
> >phase velocity. Quite the opposite. I have stressed that the entity
>
> I didn't.
>
>
> >> you've asserted it as a premise. Since your premise is incompatible
> >> with relativity, you cant expect it to be correct.
> >On the contrary. I cannot expect SR to be correct. for every bit of
>
> That's really what I meant. If you discard the postulates, then I
> can't expect your description to match the one from relativity.
> I was under the impression that you were also claiming this is
> what relativity should predict. Regardless of the fact that I still
> think you're wrong, that determination can't be made from an
> argument or thought experiment. It's not very likely anyone will
> build such a device soon to see which one of us is correct and
> regardlee of the fact that I don't believe an experiment would
> bear out your assertion, I'll always believe what competently
> acquired and analyzed data comes up with. I'm not going to hold
> my breath, however....
I am basically saying that relativity, in its present form, is
incorrect. I don't claim have an answer. I don't go in for aether
models and I can see the difficulties with absolutism and source
dependency. I think we need a new angle on the whole subject, frankly.
>
> >so-called proof of SR, I can substitute an equation involving either
> >1-c/v or 1-(c/v)^2, that will give just as good an answer within
> >experimental error.
>
> Well, there are several things to think about before getting
> carried away. Like, why is that a surprise? If you rewrite
> v as a fraction c, v = nc and ask by what ratio those differ
> for 0 < n < 1:
>
> r^2 = (1-n)^2/(1-n^2) = (1-n)/(1+n)
>
> for n = .2 [velocities of 6x10^7 m/s], r^2 =~ .83 or the
> percent difference ( |r^2 - 1|/(r^2 + 1), is the simplest
> expression I see right offhand to express %diff compactly),
> is about 10%. For a velocity of .001c (a mere 3x10^5 m/s),
> r^2 =~ .05, for a %diff of 3%. What experiment would you
> perform that you wouldn't say requires "unwarranted inferences"
> so that you can determine which one of those (or from your
> viewpoint, both) are incorrect? If you are basing the incor-
> rectness on the fact that both are consistent with experiment,
> describe an experiment to discern the two (that can be done
> and has measured quantities you'd be satisfied don't "hide"
> anything.
GPS clocks. The SR speed correction is tiny anyway. The difference in
the correction, itself, would be around .01% and negligible compared
with other corrections in GPS system.
In any instance where v is <<c, the differences are negligible. eg the
precession of Mercury.
>
> >My main grouch with SR is the sqrt of gamma. THERE IS NO SQUARE ROOT
> >TERM!!!!!!!!! This is what I have shown here. The triangulation
> >normally assumed with the cross beam of the MMX is a fallacy.
> >
> Even with no square root, the apparatus doesn't stay vertical. For
> it to remain vertical, c = oo.
>
> >I haven't made up my mind about the parallel beam however. It is more
> >interesting.
>
>
> >hhhmmm! I wouldn't be surprised. How would the spot appear to a grid
> >of photodetectors on a distant wall, I wonder?
>
> Actually, as I started to describe the result, there turned out to be
> some interesting things about this that I need to calculate to actually
> get the correct result. If the spot is to move as a continuous spot,
> i.e., each detector in a line records the spot passing, then the first
> thing is to define a spot which is continuous. I would define it such
> that at two points the observers at each describe the same thing (color,
> intensity). To be considered superluminal, the observers would have
> to describe the same thing at times faster than the light between
> them travels. I believe that I can show no such spot even can realized
I reckon there would be line with a series of light and dark areas
corresponding to wave crests and troughs.
> Even if you subtract the time required for each to communicate the
> result to the other, or to an independent observer:
>
>
> A B Arrange the distance AC to be the same as BC. The
> spot moving from A->B. Light travels from A-B in t_AB.
> If anything travels from A->B in time T, then for
> T <= t_AB C will receive communications from A&B such
> C that t_ac + t_ab <= t_bc + T. This should be sufficient
>
> to determine if the spot is superluminal. The detectors must see
> the same thing. Exactly what that is, I have to calculate, but I
> don't think it will be anything that can be continuous between
> points you place detectors. At some point, I think you have to
> violate liouvilles theorem or the uncertainty principle to get
> that result. The uncertainty principle sort of being the bottom
> line.
>
But each detector could have its own synchronized clock which would
record the time the light strikes it. That is easy enough. You are
thinking of a twoway system. All you need is a clock on each
photodetector. They can be a kilometre apart if you wish. Time
difference about 33 microsecs at light speed. Rotating the beam in the
opposite direction will eliminate any out-of-synch.
Now there's an easy experiment for you Bilgey. If I still had lab, I
would set it up now.
Henry Wilson
McGregor) wrote:
>On Tue, 13 Jun 2000 01:12:46 GMT, HWilson@..(Henry Wilson) wrote:
>
>>On Mon, 12 Jun 2000 18:07:40 GMT, chi...@zetnet.co.uk (Charles
>>McGregor) wrote:
snip
These models are completely wrong. In the MMX equivalent, there is no
water in the river and the whole system moves as one piece, as seen by
someone on Mars. There are posts positioned directly opposite on
either bank and these are connected by a wire flying fox, over which
the frustrated swimmer is transported, always with his body aligned
with the wire and the two posts.
>This would presume, for MMX, that
>1) The beam width was wide enough.
>2) That mirror angle was insufficiently determinate to provide an
>independent assessment of beam angle.
>3) That the 'wave front' was sufficiently planar
>
This is the misapprehension that I am trying to clear up. The beam
always points directly at the centre of the mirror, no matter who or
what speed the observer.
The standard MMX analysis incorporates this: (fixed pitch fonts)
__
/\
/ \
/ \
/ \
This is misleading.
It should be like this:
_ _ _ _ _ _ _
|
| |
| |
| |
showing how the top mirror is always aligned with the axis of the
light beam. The mirror moves with the beam.
>
>regarding 1)
>If we assume that beam length was say 3 m
>and the minimum findable ether speed was the speed of the Earth around
>the sun, quick mental arithmetic 500, 000,000 km/y, ....15 km/s so we
>have what... 3m = 10^-8 seconds trip time... approx 1.5e-5 m, about
>15 microns...shit that's not a lot. Almost certainly the beam would
>be wider than that. Anyone know if M&M looked for missing fringe at
>beam edges?:-)
>
>regarding 2)
>1.5e-5/3 =5e-6 is approx 7.5e-4 degrees, how tough is that to
>measure?, probably not very.
>
>regarding 3)
>Possibly
>
>You COULD be kinda right Henry, :-) at least in terms of possibly
>explaining MMX in terms of still having an ether, but not for the
>analogous reasons you cite. In this case any ether effect would be
>additive to light speed. But I guess you wouldn't be unhappy with
>that either.:-)
There is no need for an aether! You have missed the point of the
chainsaw example.
Here's another way to look at it. Let's say that a number of clocks
are aligned at points along the crossbeam. These clocks are capable
of registering and indicating the time taken for a pulse to travel
from the 45 mirror.
Firstly, to any moving observer, the clocks will always appear to
remain in line with the two mirrors. Secondly, no matter what the
observer speed, the times registered by the clocks will be the same.
If the moving observers use their own clocks, they will have to
correct for the time taken for information to reach them. This is
where a good deal of confusion originates because the line joining the
two mirrors can appear curved or leaning, depending on the position of
the observer. In a moving perpendicular plane, however, the path of
each infinitesimal element of the beam is diagonal but the beam itself
is always aligned exactly vertically.
>
>OTOH I expect M&M were aware of this possibility and allowed for it
>somehow in their set-up and calculations ( multiple angle measurements
>with same mirror settings etc.)
>
>It should be noted ether flow is not 100% analagous to that of a
>river, despite the frequency with which such analogies are drawn.
>Whereas the swimmer has the same speed w.r.t. the river analogous to
>the ether case it is the swimmer that determines what that speed he
>travels w.r.t the water, NOT the river.
>
>>>
>>>If an ether was present with a velocity with asymetric components to
>>>the two beam paths, then the light would travel in a 'diagonal' or
>>>more properly at an angle anomolous to that expected from mirror
>>>orientation.
>>>
>>>regards
>>>chic
>>I'm not really into ether. I don't see any need for it and it doesn't
>>support my theory. But I don't entirely rule it out either. I have
>>suggested that there are two more mass dimensions that might explain
>>many current mysteries.
>regards
>chic
Einstein must have accepted that an Aether exists because he assumed
that the light would move diagonally at c, just as sound would. He
then devised his crazy theory to match this incorrect assumption.
Subsequently, he and his followers abolished the aether but kept the
assumption.
Paul Stowe
writes:
> I am basically saying that relativity, in its present form, is
>incorrect. I don't claim have an answer. I don't go in for aether
>models and I can see the difficulties with absolutism and source
>dependency. I think we need a new angle on the whole subject, frankly.
Can you descibe what you see as the problem with source dependency?
Paul Stowe
David Evens
Do READ THINGS THAT YOU POST TO, HENRY!
Had you READ THE POSTING, you could not have avoided noticing that he
EXPLAINED HOW THE MMX IS ACTUALLY SUPPOSED TO DO ITS THING!
> There are posts positioned directly opposite on
> either bank and these are connected by a wire flying fox, over which
> the frustrated swimmer is transported, always with his body aligned
> with the wire and the two posts.
This, of course, is totally unrelated to the actual effects that are
possible in the MMX, and thus is not a valid analogy. Your assumption
is that there was a WAVE GUIDE in the MMX, which there simply is not.
> >This would presume, for MMX, that
> >1) The beam width was wide enough.
> >2) That mirror angle was insufficiently determinate to provide an
> >independent assessment of beam angle.
> >3) That the 'wave front' was sufficiently planar
> >
> This is the misapprehension that I am trying to clear up. The beam
> always points directly at the centre of the mirror, no matter who or
> what speed the observer.
This, of course, is not relevant to your assumption that the light does
not hit the mirror. You DO realize that you ARE claiming this, since if
the light DOESN'T travel diagonally from the place where the source is
when it is emitted ot the place the mirror WILL BE when it reaches it,
the light cannot avoid MISSING the mirror.
> The standard MMX analysis incorporates this: (fixed pitch fonts)
> __
> /\
> / \
> / \
> / \
>
> This is misleading.
> It should be like this:
>
>
>
> _ _ _ _ _ _ _
> |
> | |
> | |
> | |
>
> showing how the top mirror is always aligned with the axis of the
> light beam. The mirror moves with the beam.
You forgot to draw a difference in the diagrams. All you did was make
the mirror wider, which cannot make a difference, since any finitely
large mirror will ALWAYS eventually move out of the beam aimed other
than at it (as you assume the light is aimed) if the speed is high
enough.
> >regarding 1)
> >If we assume that beam length was say 3 m
> >and the minimum findable ether speed was the speed of the Earth around
> >the sun, quick mental arithmetic 500, 000,000 km/y, ....15 km/s so we
> >have what... 3m = 10^-8 seconds trip time... approx 1.5e-5 m, about
> >15 microns...shit that's not a lot. Almost certainly the beam would
> >be wider than that. Anyone know if M&M looked for missing fringe at
> >beam edges?:-)
> >
> >regarding 2)
> >1.5e-5/3 =5e-6 is approx 7.5e-4 degrees, how tough is that to
> >measure?, probably not very.
Not in the lab, with modern instruments. The hard part is isolating the
apparatus from vibration.
> >regarding 3)
> >Possibly
> >
> >You COULD be kinda right Henry, :-) at least in terms of possibly
> >explaining MMX in terms of still having an ether, but not for the
> >analogous reasons you cite. In this case any ether effect would be
> >additive to light speed. But I guess you wouldn't be unhappy with
> >that either.:-)
> There is no need for an aether! You have missed the point of the
> chainsaw example.
No, YOU did. He pointed out your mistake.
> Here's another way to look at it. Let's say that a number of clocks
> are aligned at points along the crossbeam. These clocks are capable
> of registering and indicating the time taken for a pulse to travel
> from the 45 mirror.
> Firstly, to any moving observer, the clocks will always appear to
> remain in line with the two mirrors. Secondly, no matter what the
> observer speed, the times registered by the clocks will be the same.
>
> If the moving observers use their own clocks, they will have to
> correct for the time taken for information to reach them. This is
> where a good deal of confusion originates because the line joining the
> two mirrors can appear curved or leaning, depending on the position of
> the observer. In a moving perpendicular plane, however, the path of
> each infinitesimal element of the beam is diagonal but the beam itself
> is always aligned exactly vertically.
In reality, of course, the only possible way that the line between
source and mirror can appear curved is because of classical
perspective. If you look from a long way away, this can be made to
vanish below the threshold of delectability.
As for your assumption that the line can appear to be a diagonal, this
is simply false in the MMX. The only way for this to occur is for the
relative motion to not be orthogonal to the line, which is not an
interesting case.
> >OTOH I expect M&M were aware of this possibility and allowed for it
> >somehow in their set-up and calculations ( multiple angle measurements
> >with same mirror settings etc.)
> >
> >It should be noted ether flow is not 100% analagous to that of a
> >river, despite the frequency with which such analogies are drawn.
> >Whereas the swimmer has the same speed w.r.t. the river analogous to
> >the ether case it is the swimmer that determines what that speed he
> >travels w.r.t the water, NOT the river.
> >
> >>>
> >>>If an ether was present with a velocity with asymetric components to
> >>>the two beam paths, then the light would travel in a 'diagonal' or
> >>>more properly at an angle anomolous to that expected from mirror
> >>>orientation.
> >>>
> >>>regards
> >>>chic
> >>I'm not really into ether. I don't see any need for it and it doesn't
> >>support my theory. But I don't entirely rule it out either. I have
> >>suggested that there are two more mass dimensions that might explain
> >>many current mysteries.
> >regards
> >chic
> Einstein must have accepted that an Aether exists because he assumed
> that the light would move diagonally at c, just as sound would.
I see that you are STILL totally ignorant of SR. Were you NOT so
ignorant, you could not avoid knowing that the light is assumed to move
at c along the diagonal because light ALWAYS moves at c.
> He
> then devised his crazy theory to match this incorrect assumption.
What incorrect assumption are you pretending to have found? All you
have mentioned in an incorrect assumption YOU made ABOUT SR.
> Subsequently, he and his followers abolished the aether but kept the
> assumption.
What assumption are you assuming was made, despite the fact that there
isn't one?
David Evens
Henry Wilson wrote:
> On Tue, 13 Jun 2000 21:49:34 GMT, ro...@radioactivex.lebesque-al.net
> (Bilge) wrote:
> >Henry Wilson said some stuff about
> >
> > >By phase velocity, I was refering to the movement of a point of equal
> > >phase in whatever periodicity a photon has, be it temporal spatial or
> > >both. To clarify that, Imagine a photon to have a physical form
> > >involving a sine wave like the teeth of a saw blade. When the photon
> >
> >
> > A photon has no such form. A phase velocity, by definition
> > requires more than a single frequency. The definition is
> > v_phase = c/n(k). In vacuum, n(k) == n == 1.
> >
> > >I'm not quite with you here. Light can only have a fundamental
> > >frequency. Otherwise the photoelectric effect would show thresholds
> > >corresponding to each harmonic. That is something I deduced for myself
> > >and has probably got nothing to do with what you are talking about but
> > >it is interesting anyway..
> > >
> > Phases refer to more than one frequency and can't refer to a
> > single photon. The phase velocity is what characterizes
> > dispersion. 1 photon cant disperse.
> >
> Bilgey, how come you know so much about photons when nodody else seems
> to know anything conclusive? Have you actually seen a photon?
Henry, YOU are the one who doesn't seem to know anything about photons!
EVERYONE sees photons, ALL THE TIME. They are the energy quanta of
LIGHT.
> > >Of course I differ with SR. I am trying to tell everyone why it is
> > >completely wrong, right from its first assumption!
> >
> > In that case, it's an issue for an experiment. Since the fundamental
> > problem you have with relativity is one of the postulates, you can't
> > really argue with what follows from it as a consequence without
> > actually having some data to point at that illustrates clearly what
> > failed and why.
> >
> The MMX is a good start, then. If that is wrong, then the whole of SR
> is also wrong!
How unfortunate (for you) that you have been incapable of finding
anything wrong with it.
> > >I don't, however, claim to be able to see an image resulting from a
> > >phase velocity. Quite the opposite. I have stressed that the entity
> >
> > I didn't.
> >
> >
> > >> you've asserted it as a premise. Since your premise is incompatible
> > >> with relativity, you cant expect it to be correct.
> > >On the contrary. I cannot expect SR to be correct. for every bit of
> >
> > That's really what I meant. If you discard the postulates, then I
> > can't expect your description to match the one from relativity.
> > I was under the impression that you were also claiming this is
> > what relativity should predict. Regardless of the fact that I still
> > think you're wrong, that determination can't be made from an
> > argument or thought experiment. It's not very likely anyone will
> > build such a device soon to see which one of us is correct and
> > regardlee of the fact that I don't believe an experiment would
> > bear out your assertion, I'll always believe what competently
> > acquired and analyzed data comes up with. I'm not going to hold
> > my breath, however....
> I am basically saying that relativity, in its present form, is
> incorrect.
A bold claim for someone without any understanding of it.
> I don't claim have an answer.
Now, to make you see that you don't have a QUESTION, either...
> I don't go in for aether
> models and I can see the difficulties with absolutism and source
> dependency. I think we need a new angle on the whole subject, frankly.
And what do you ASSUME that angle to be?
> > >so-called proof of SR, I can substitute an equation involving either
> > >1-c/v or 1-(c/v)^2, that will give just as good an answer within
> > >experimental error.
> >
> > Well, there are several things to think about before getting
> > carried away. Like, why is that a surprise? If you rewrite
> > v as a fraction c, v = nc and ask by what ratio those differ
> > for 0 < n < 1:
> >
> > r^2 = (1-n)^2/(1-n^2) = (1-n)/(1+n)
> >
> > for n = .2 [velocities of 6x10^7 m/s], r^2 =~ .83 or the
> > percent difference ( |r^2 - 1|/(r^2 + 1), is the simplest
> > expression I see right offhand to express %diff compactly),
> > is about 10%. For a velocity of .001c (a mere 3x10^5 m/s),
> > r^2 =~ .05, for a %diff of 3%. What experiment would you
> > perform that you wouldn't say requires "unwarranted inferences"
> > so that you can determine which one of those (or from your
> > viewpoint, both) are incorrect? If you are basing the incor-
> > rectness on the fact that both are consistent with experiment,
> > describe an experiment to discern the two (that can be done
> > and has measured quantities you'd be satisfied don't "hide"
> > anything.
> GPS clocks. The SR speed correction is tiny anyway. The difference in
> the correction, itself, would be around .01% and negligible compared
> with other corrections in GPS system.
And highly detectable, too.
> In any instance where v is <<c, the differences are negligible. eg the
> precession of Mercury.
Also highly detectable. In fact, detected in the 19th century.
> > >My main grouch with SR is the sqrt of gamma. THERE IS NO SQUARE ROOT
> > >TERM!!!!!!!!! This is what I have shown here. The triangulation
> > >normally assumed with the cross beam of the MMX is a fallacy.
> > >
> > Even with no square root, the apparatus doesn't stay vertical. For
> > it to remain vertical, c = oo.
> >
> > >I haven't made up my mind about the parallel beam however. It is more
> > >interesting.
> >
> >
> > >hhhmmm! I wouldn't be surprised. How would the spot appear to a grid
> > >of photodetectors on a distant wall, I wonder?
> >
> > Actually, as I started to describe the result, there turned out to be
> > some interesting things about this that I need to calculate to actually
> > get the correct result. If the spot is to move as a continuous spot,
> > i.e., each detector in a line records the spot passing, then the first
> > thing is to define a spot which is continuous. I would define it such
> > that at two points the observers at each describe the same thing (color,
> > intensity). To be considered superluminal, the observers would have
> > to describe the same thing at times faster than the light between
> > them travels. I believe that I can show no such spot even can realized
> I reckon there would be line with a series of light and dark areas
> corresponding to wave crests and troughs.
Only if you can get the beam to interfere with itself, which you might
be able to do, since it will ACT as if it is coming from a slit. Of
course, what you ACTUALLY get in that kind of interference would be a
series of concentric RINGS (since the beam is radially symmetric), NOT
light and dark spots.
> > Even if you subtract the time required for each to communicate the
> > result to the other, or to an independent observer:
> >
> >
> > A B Arrange the distance AC to be the same as BC. The
> > spot moving from A->B. Light travels from A-B in t_AB.
> > If anything travels from A->B in time T, then for
> > T <= t_AB C will receive communications from A&B such
> > C that t_ac + t_ab <= t_bc + T. This should be sufficient
> >
> > to determine if the spot is superluminal. The detectors must see
> > the same thing. Exactly what that is, I have to calculate, but I
> > don't think it will be anything that can be continuous between
> > points you place detectors. At some point, I think you have to
> > violate liouvilles theorem or the uncertainty principle to get
> > that result. The uncertainty principle sort of being the bottom
> > line.
> >
> But each detector could have its own synchronized clock which would
> record the time the light strikes it. That is easy enough. You are
> thinking of a twoway system. All you need is a clock on each
> photodetector. They can be a kilometre apart if you wish. Time
> difference about 33 microsecs at light speed. Rotating the beam in the
> opposite direction will eliminate any out-of-synch.
> Now there's an easy experiment for you Bilgey. If I still had lab, I
> would set it up now.
When did YOU have a lab?
Charles McGregor
Henry! Behave. The 'hypothetical' I used in describing the ether at
the start of this article should have indicated that I am less than
enthusiastic about that theory. The stance taken was purely in the
context of illustrating what MMX was trying to prove at the time,(i.e.
the existence and motion of an ether), and how the frig factors
required to retain the concept of an ether came about. You seemed to
be attacking MMX, i.e. questioning that there might be an ether after
all? I must have picked you up wrong. I was attempting to show that
even with an ether your analogies were still wrong, i.e. that you
weren't consistent. In the end I conceded that even with an ether, you
could still be correct in the basic premise I thought you had,( that a
'light beam' travelling from A to B would have an A to B velocity
component relative to A, determined only at source and that any ether
speed would not effect this.) provided the 'light beam' was wide
enough to 'bridge' the orthogonal displacement caused by the alleged
ether.
If you concede no ether, then there would be no relative velocity
component orthogonal to the AB line.MMX produces it's zippo result,
everybody's happy, what's all the fuss?
O.K. that is clearer about what you intend. But if there is no ether
then there will be no meaningful diagonal movement. Looking at these
diagrams alone one could again easily conclude that you believed in an
ether of some kind (or an absolute velocity).
It should look something like
_____________________
||
||
||
||
||
||
If we talk about diagonals at all we're back to erroneously including
observer only (i.e. not taking part in observed interaction) velocity
components again.
The orthogonal component in your diagrams is entirely due to relative
observer velocity and has NOTHING to do with the observed interaction
between A and B.
The orientation you illustrate therefore has no meaning.
It would ONLY have meaning if the diagonal were caused by ether
velocity rather than observer velocity and would result in a
difference in A to B time. In the first diagram the time would be
longer and dependent on ether flow (expected by M&M) in the second it
would equal c regardless of ether flow, provided the beam were wide
enough to bridge the displacement due to ether flow.
With no ether, there would be no diagonal and both your diagrams would
be the same, i.e. like the one I produced.
You have introduced a relative observer velocity to the MMX experiment
in order to produce an 'apparent' diagonal which actually bears no
more than a cosmetic similarity to the hypothesised diagonal that
would exist if there really were an ether and which was tested(at
least by intention) in MMX.
Even IF one assumed there such a thing as absolute velocity, since one
cannot detect it, the proposition can ALWAYS be made without fear of
disproof, that the apparatus sits, by coincidence, at zero absolute
velocity.
IOW until the day dawns that someone establishes the existence of
absolute motion and a means of measuring it, it can always be
contended that two relatively stationary points in any inertial frame
are AT the said absolute zero point and that THEREFORE the relative
speed of anything going between them, would then also be absolutely
correct. i.e. that there was no 'hidden extra' component that the
absolute speed and relative speedwere the same. IOW2 if zero
orthogonal velocity component can ALWAYS be contended for light moving
between two relatively stationary points, then it should never be
considered to exist (unless and until a detectable ether or
detectable Absolute motion arises) and certainly not as a consequence
of observer movement.
Now I have made the above semi-philosophical point to try to give you
a clue as to how the orientation of photons difference you percieve,
is illusory.
However, it is every bit as valid to contend that the apparatus IS
moving, in fact it may have any velocity, v, we might wish to assign
it. This is the same thing as introducing a relative observer
velocity -v.
"Hold it!" you say, "I thought observer velocity was supposed to be
irrelevant?."
Yes, it IS irrelevant in terms of what physically happens between the
apparatus parts, but it is very relevant to the observers perception
of what is happening which is of course an interaction which involves
observer relative velocity. and so observer perspective based
transformations which ideally cope with all possible v are a valid and
sensible thing to derive since it enables very useful calculation.
All that can be derived from the semi-philosophical stuff above (aside
from, I hope some insight into your orientation thing) is that only
relative motion is knowable.
It is sensible to talk about light moving diagonally between
relatively moving objects where it's direction can be quantified and
benchmarked to the measurable relative velocity, but it is not usually
useful to consider this between stationary objects in the same
inertial frame, even if it appears to exist as a consequence of
observer motion.
The exception is in the derivation of general. i.e. non specific
observer perspective transformations as remarked on already
and of course the application of them at relativistic speeds where
the observer's relative velocity effectively becomes a no longer
ignorable part of the interaction being observed.
regards
chic
Henry Wilson
<dev...@technologist.com> wrote:
>
>Henry Wilson wrote:
>> On Tue, 13 Jun 2000 18:34:12 GMT, chi...@zetnet.co.uk (Charles
>> McGregor) wrote:
>> These models are completely wrong. In the MMX equivalent, there is no
>> water in the river and the whole system moves as one piece, as seen by
>> someone on Mars.
>
>Do READ THINGS THAT YOU POST TO, HENRY!
>
>Had you READ THE POSTING, you could not have avoided noticing that he
>EXPLAINED HOW THE MMX IS ACTUALLY SUPPOSED TO DO ITS THING!
I am rapidly coming to the conclusion that you are a complete idiot.
>
>> There are posts positioned directly opposite on
>> either bank and these are connected by a wire flying fox, over which
>> the frustrated swimmer is transported, always with his body aligned
>> with the wire and the two posts.
>
>This, of course, is totally unrelated to the actual effects that are
>possible in the MMX, and thus is not a valid analogy. Your assumption
>is that there was a WAVE GUIDE in the MMX, which there simply is not.
If you had any kind of brain you would be able to understand a few
facts. As it is, your arguments carry about as much weight as a bible
basher trying to flog the bible.
Why don't you stick to something like alt.tiddlywinks.forkids.
You really are a complete fuckwit ,excuse the expression, You have
zero intelligence. The mirror moves with the beam like I said.
The dashes represent the positions of the mirror, always directly over
the beam. That should be obvious.
Go out and buy a chainsaw and araldite the bloody thing on top of you
car, then start it up, drive off and ask a bystander to watch what has
happens.
>
>> >regarding 1)
>> >If we assume that beam length was say 3 m
>> >and the minimum findable ether speed was the speed of the Earth around
>> >the sun, quick mental arithmetic 500, 000,000 km/y, ....15 km/s so we
>> >have what... 3m = 10^-8 seconds trip time... approx 1.5e-5 m, about
>> >15 microns...shit that's not a lot. Almost certainly the beam would
>> >be wider than that. Anyone know if M&M looked for missing fringe at
>> >beam edges?:-)
>> >
>> >regarding 2)
>> >1.5e-5/3 =5e-6 is approx 7.5e-4 degrees, how tough is that to
>> >measure?, probably not very.
>
>Not in the lab, with modern instruments. The hard part is isolating the
>apparatus from vibration.
>
>> >regarding 3)
>> >Possibly
>> >
>> >You COULD be kinda right Henry, :-) at least in terms of possibly
>> >explaining MMX in terms of still having an ether, but not for the
>> >analogous reasons you cite. In this case any ether effect would be
>> >additive to light speed. But I guess you wouldn't be unhappy with
>> >that either.:-)
>> There is no need for an aether! You have missed the point of the
>> chainsaw example.
>
>No, YOU did. He pointed out your mistake.
Nonsense, He is trying to fathom the (il)logic of the standard
treatment, too.
>
>> Here's another way to look at it. Let's say that a number of clocks
>> are aligned at points along the crossbeam. These clocks are capable
>> of registering and indicating the time taken for a pulse to travel
>> from the 45 mirror.
>> Firstly, to any moving observer, the clocks will always appear to
>> remain in line with the two mirrors. Secondly, no matter what the
>> observer speed, the times registered by the clocks will be the same.
>>
>> If the moving observers use their own clocks, they will have to
>> correct for the time taken for information to reach them. This is
>> where a good deal of confusion originates because the line joining the
>> two mirrors can appear curved or leaning, depending on the position of
>> the observer. In a moving perpendicular plane, however, the path of
>> each infinitesimal element of the beam is diagonal but the beam itself
>> is always aligned exactly vertically.
>
>In reality, of course, the only possible way that the line between
>source and mirror can appear curved is because of classical
>perspective. If you look from a long way away, this can be made to
>vanish below the threshold of delectability.
That's what I was indicating, dummy!
>
>As for your assumption that the line can appear to be a diagonal, this
>is simply false in the MMX. The only way for this to occur is for the
>relative motion to not be orthogonal to the line, which is not an
>interesting case.
Now what the hell? You have reversed your fucking argument. you have
been trying to convince me for weeks that the light beam moves
diagonally, now you are saying it doesn't. Are you drunk?
In the MMX, it is not LIGHT that moves diagonally.
>
>> He
>> then devised his crazy theory to match this incorrect assumption.
>
>What incorrect assumption are you pretending to have found? All you
>have mentioned in an incorrect assumption YOU made ABOUT SR.
>
>> Subsequently, he and his followers abolished the aether but kept the
>> assumption.
>
>What assumption are you assuming was made, despite the fact that there
>isn't one?
The fact that light behaves like sound when it's convenient but not
otherwise.
Henry Wilson
>In <3946bd6d...@nsw.nnrp.telstra.net> HWilson@..(Henry Wilson)
>writes:
>>
>> I am basically saying that relativity, in its present form, is
>>incorrect. I don't claim have an answer. I don't go in for aether
>>models and I can see the difficulties with absolutism and source
>>dependency. I think we need a new angle on the whole subject, frankly.
>
>Can you descibe what you see as the problem with source dependency?
>
>Paul Stowe
>
Not very much, can you? Pions, maybe?
I'm working on the binary star business now.
I think there might be 'local' source dependency - but what is local?
Henry Wilson
McGregor) wrote:
>On Wed, 14 Jun 2000 02:32:03 GMT, HWilson@..(Henry Wilson) wrote:
>
snip
>>>>>However MMX was designed to measure the lab speed w.r.t. the ether.
>>>The ether was the hypothetical medium for EM, the EM should have a
>>>Let's try the more traditional river example.
Sorry, You are correct as far as the original intentions of the MMX
were. The river idea works for that. I was a bit hasty in condemning
it. My motives might not be very clear if you came in late on this
thread.
The fact is, even without an ether, the MMX takes a lot of explaining.
It would appear that the number of wavelengths in the two paths would
still vary with velocity, unless one accepts the notion of source
dependency of light velocity. This is rejected outright because of the
binary star argument (something which I would question).
SR came up with the Lorentz contraction idea and ended up with an
equation that involved a sqrt of gamma.
I am not really arguing against the notion of an apparent contraction.
I merely claim that the standard analysis is wrong because it assumes
that the cross beam follows a diagonal path, as seen in a rest frame,
when the apparatus moves. It also assumes that the velocity of this
fictitious light beam is c. I have used my chainsaw demo to show that
no light beam ever moves diagonally at c. Infinitesimal elements
follow diagonal paths but their velocity is sqrt(c^2+v^2).
>>This is the misapprehension that I am trying to clear up. The beam
>>always points directly at the centre of the mirror, no matter who or
>>what speed the observer.
>>The standard MMX analysis incorporates this: (fixed pitch fonts)
>> __
>> /\
>> / \
>> / \
>> / \
>>
>>This is misleading.
>>It should be like this:
>>
>>
>>
>> _ _ _ _ _ _ _
>> |
>> | |
>> | |
>> | |
>>
>>showing how the top mirror is always aligned with the axis of the
>>light beam. The mirror moves with the beam.
>>
>
>O.K. that is clearer about what you intend. But if there is no ether
>then there will be no meaningful diagonal movement. Looking at these
>diagrams alone one could again easily conclude that you believed in an
>ether of some kind (or an absolute velocity).
>
>It should look something like
>
> _____________________
> ||
> ||
> ||
> ||
> ||
> ||
>
>
>If we talk about diagonals at all we're back to erroneously including
>observer only (i.e. not taking part in observed interaction) velocity
>components again.
>
Yes but that has been the standard explanation. According to the
establishment, when the apparatus moves, the light beam is supposed to
tilt over diagonally wrt the rest frame. That does not happen, the
beam always remains upright in all observer frames.
>The orthogonal component in your diagrams is entirely due to relative
>observer velocity and has NOTHING to do with the observed interaction
>between A and B.
Don't you mean 'diagonal component?
>The orientation you illustrate therefore has no meaning.
>It would ONLY have meaning if the diagonal were caused by ether
>velocity rather than observer velocity and would result in a
>difference in A to B time. In the first diagram the time would be
>longer and dependent on ether flow (expected by M&M) in the second it
>would equal c regardless of ether flow, provided the beam were wide
>enough to bridge the displacement due to ether flow.
>
>With no ether, there would be no diagonal and both your diagrams would
>be the same, i.e. like the one I produced.
There are diagonals involved. Each infinitesimal element of the cross
beam moves diagonally wrt the rest frame. Each element has a different
diagonal. The light axis remains vertical and it is light that moves
at c, not the elements. That is my point and that is why the term
involving the sqrt gamma is wrong!
>
>You have introduced a relative observer velocity to the MMX experiment
>in order to produce an 'apparent' diagonal which actually bears no
>more than a cosmetic similarity to the hypothesised diagonal that
>would exist if there really were an ether and which was tested(at
>least by intention) in MMX.
That's correct - and that is what the whole of SR is effectively based
on.
>
>Even IF one assumed there such a thing as absolute velocity, since one
>cannot detect it, the proposition can ALWAYS be made without fear of
>disproof, that the apparatus sits, by coincidence, at zero absolute
>velocity.
>
>IOW until the day dawns that someone establishes the existence of
>absolute motion and a means of measuring it, it can always be
>contended that two relatively stationary points in any inertial frame
>are AT the said absolute zero point and that THEREFORE the relative
>speed of anything going between them, would then also be absolutely
>correct. i.e. that there was no 'hidden extra' component that the
>absolute speed and relative speedwere the same. IOW2 if zero
>orthogonal velocity component can ALWAYS be contended for light moving
>between two relatively stationary points, then it should never be
>considered to exist (unless and until a detectable ether or
>detectable Absolute motion arises) and certainly not as a consequence
>of observer movement.
That would be correct IF there is absolute space. I think we have to
look for better theories than absolutism. I ouldn't rule it out but it
involves that ever present question of infinity. (So does any other
theory, I suppose).
>Now I have made the above semi-philosophical point to try to give you
>a clue as to how the orientation of photons difference you percieve,
>is illusory.
>
Well, It is not illusory because there is no reorietation. The photon
'axes' always point vertically.
The diagonal business is due to observer movement but that does not
involve a light BEAM.
Like you say, it is nonsensical to think that the number of
wavelengths in the arms of the apparatus would change just because an
observer happened to move past it.
>However, it is every bit as valid to contend that the apparatus IS
>moving, in fact it may have any velocity, v, we might wish to assign
>it. This is the same thing as introducing a relative observer
>velocity -v.
>"Hold it!" you say, "I thought observer velocity was supposed to be
>irrelevant?."
>Yes, it IS irrelevant in terms of what physically happens between the
>apparatus parts, but it is very relevant to the observers perception
>of what is happening which is of course an interaction which involves
>observer relative velocity. and so observer perspective based
>transformations which ideally cope with all possible v are a valid and
>sensible thing to derive since it enables very useful calculation.
I agree - but I say the standard equations are wrong. (1-v/c) and
(1-(v/c)^2) should appear but not the sqrt.
>
>All that can be derived from the semi-philosophical stuff above (aside
>from, I hope some insight into your orientation thing) is that only
>relative motion is knowable.
>
>>Einstein must have accepted that an Aether exists because he assumed
>>that the light would move diagonally at c, just as sound would. He
>>then devised his crazy theory to match this incorrect assumption.
>>Subsequently, he and his followers abolished the aether but kept the
>>assumption.
>It is sensible to talk about light moving diagonally between
>relatively moving objects where it's direction can be quantified and
>benchmarked to the measurable relative velocity, but it is not usually
>useful to consider this between stationary objects in the same
>inertial frame, even if it appears to exist as a consequence of
>observer motion.
Agreed but what the standard MMX treatment does.
>The exception is in the derivation of general. i.e. non specific
>observer perspective transformations as remarked on already
>and of course the application of them at relativistic speeds where
>the observer's relative velocity effectively becomes a no longer
>ignorable part of the interaction being observed.
>
As long as the correct equations are used. I don't believe SR provides
them.
>regards
>chic
Bilge
>>
>Bilgey, how come you know so much about photons when nodody else seems
>to know anything conclusive? Have you actually seen a photon?
>The MMX is a good start, then. If that is wrong, then the whole of SR
>is also wrong!
Set up your own interferometer. You can get a laser for < $10.00.
Go buy a laser pointer and pry open the case. There's a little
tiny circuit board inside. You can solder some pins onto it and
mount it however you like and run it off three volts. I don't know
what the exact wavelength is, but I would guess whatever it is,
it's visible, relatively monochromatic, coherent and good enough
to provide a reasonable degree of sensitivity. How much can a few
mirrors cost? We aren't talking about high-dollar optics for this
experiment. Precision ground optics aren't a prerequisite until
you get to a point that nothing will convince you anyway. Since the
fringe shift you are looking for is depends upon path length,
you can always make the interferometer longer. The ability to
align the mirrors is pretty much the quality level issue in a
nutshell. No matter what you have, you shouldn't see a change from
rotating 90 deg. If, you do, you should write it up immediately,
(immediately after checking the mechanical integrity of your
equipment, that is).
> I am basically saying that relativity, in its present form, is
>incorrect. I don't claim have an answer. I don't go in for aether
>models and I can see the difficulties with absolutism and source
>dependency. I think we need a new angle on the whole subject, frankly.
I'm the opposite. The more things I see posted as discrepancies,
the more obvious I think it is that those discrepancies make the
picture clearer. Primarily, because after stripping away the
obfuscatory cladding of the most rube-goldberg like mechanisms,
there are the same, fundamentally wrong assumptions about what
to expect. Most of the confusion is the confused's own successfully
outwitting of himself instead of relativity.
>GPS clocks. The SR speed correction is tiny anyway. The difference in
>the correction, itself, would be around .01% and negligible compared
>with other corrections in GPS system.
>In any instance where v is <<c, the differences are negligible. eg the
>precession of Mercury.
I haven't looked into gps clocks. The fact that almost no aspect
of day-to-day life exists that didn't involve results, predictions
etc, from relativity. (I consider not relying on ad hoc-ery for
equipment like magnetic resonance imaging which use the fact that
fermions have intrinsic spins to be a plus). Any theory that would
supercede it will have to incorporate it as some lind of limit.
Unlike the people that consider the fact that the sagnac effect
can be explained classically some fault with relativity, I consider
the fact that relativity has to give the same result that classical
physics does for things which are moving classically, to be a plus
as well. Relativity wont quit explaining the electron magnetic moment
the second something would replace relativity.
>>
>>
>> A B Arrange the distance AC to be the same as BC. The
>> spot moving from A->B. Light travels from A-B in t_AB.
>> If anything travels from A->B in time T, then for
>> T <= t_AB C will receive communications from A&B such
>> C that t_ac + t_ab <= t_bc + T. This should be sufficient
>>
>> to determine if the spot is superluminal. The detectors must see
>> the same thing. Exactly what that is, I have to calculate, but I
>> don't think it will be anything that can be continuous between
>But each detector could have its own synchronized clock which would
>record the time the light strikes it. That is easy enough. You are
So what? I am assuming that you can measure this any way you want.
The only thing you have to not do is make up numbers.
>thinking of a twoway system. All you need is a clock on each
>photodetector. They can be a kilometre apart if you wish. Time
>difference about 33 microsecs at light speed. Rotating the beam in the
>opposite direction will eliminate any out-of-synch.
The confusion is yours.
>Now there's an easy experiment for you Bilgey. If I still had lab, I
>would set it up now.
>
How would a lot of controlled chemicals and glassware, hidden in
remote, accessibility-challenged terrain help?
David Evens
Henry Wilson wrote:
> On Wed, 14 Jun 2000 01:53:29 -0400, David Evens
> <dev...@technologist.com> wrote:
> >Henry Wilson wrote:
> >> On Tue, 13 Jun 2000 18:34:12 GMT, chi...@zetnet.co.uk (Charles
> >> McGregor) wrote:
>
> >> These models are completely wrong. In the MMX equivalent, there is no
> >> water in the river and the whole system moves as one piece, as seen by
> >> someone on Mars.
> >
> >Do READ THINGS THAT YOU POST TO, HENRY!
> >
> >Had you READ THE POSTING, you could not have avoided noticing that he
> >EXPLAINED HOW THE MMX IS ACTUALLY SUPPOSED TO DO ITS THING!
> I am rapidly coming to the conclusion that you are a complete idiot.
Being humiliated is starting to get you, isn't it, Henry.
> >> There are posts positioned directly opposite on
> >> either bank and these are connected by a wire flying fox, over which
> >> the frustrated swimmer is transported, always with his body aligned
> >> with the wire and the two posts.
> >
> >This, of course, is totally unrelated to the actual effects that are
> >possible in the MMX, and thus is not a valid analogy. Your assumption
> >is that there was a WAVE GUIDE in the MMX, which there simply is not.
> If you had any kind of brain you would be able to understand a few
> facts. As it is, your arguments carry about as much weight as a bible
> basher trying to flog the bible.
> Why don't you stick to something like alt.tiddlywinks.forkids.
You are REALLY starting to feel stupid, aren't you, Henry.
This is not related to your diagrams.
> The dashes represent the positions of the mirror, always directly over
> the beam. That should be obvious.
The beam is not the path of the light. Had you any understanding of the
MMX, you could not avoid knowing this.
> Go out and buy a chainsaw and araldite the bloody thing on top of you
> car, then start it up, drive off and ask a bystander to watch what has
> happens.
And this is relevant because...? (Other than the fact that you don't
understand that it proves you wrong, that is.)
What illogic are you assuming is present in the valid analysis? All
your previous pretenses have been utterly destroyed, although you don't
seem to care about being humiliated.
> >> Here's another way to look at it. Let's say that a number of clocks
> >> are aligned at points along the crossbeam. These clocks are capable
> >> of registering and indicating the time taken for a pulse to travel
> >> from the 45 mirror.
> >> Firstly, to any moving observer, the clocks will always appear to
> >> remain in line with the two mirrors. Secondly, no matter what the
> >> observer speed, the times registered by the clocks will be the same.
> >>
> >> If the moving observers use their own clocks, they will have to
> >> correct for the time taken for information to reach them. This is
> >> where a good deal of confusion originates because the line joining the
> >> two mirrors can appear curved or leaning, depending on the position of
> >> the observer. In a moving perpendicular plane, however, the path of
> >> each infinitesimal element of the beam is diagonal but the beam itself
> >> is always aligned exactly vertically.
> >
> >In reality, of course, the only possible way that the line between
> >source and mirror can appear curved is because of classical
> >perspective. If you look from a long way away, this can be made to
> >vanish below the threshold of delectability.
> That's what I was indicating, dummy!
Then why didn't you talk about it?
> >As for your assumption that the line can appear to be a diagonal, this
> >is simply false in the MMX. The only way for this to occur is for the
> >relative motion to not be orthogonal to the line, which is not an
> >interesting case.
> Now what the hell? You have reversed your fucking argument. you have
> been trying to convince me for weeks that the light beam moves
> diagonally, now you are saying it doesn't. Are you drunk?
Why do you assume that I am making mistakes for the reasons that you
are?
In any event, the BEAM (which is NOT the path of the light in the beam)
cannot appear to be diagonal unless motion is neither along it nor
perpendicular to it, which isn't part of the situation under
consideration. The LIGHT, of course, ALWAYS moves diagonally.
Then what DOES move diagonally? After all, the only thing that DOES
move in the apparatus IS the light, so what ELSE could be bouncing from
the mirror?
> >> He
> >> then devised his crazy theory to match this incorrect assumption.
> >
> >What incorrect assumption are you pretending to have found? All you
> >have mentioned in an incorrect assumption YOU made ABOUT SR.
>
> >
> >> Subsequently, he and his followers abolished the aether but kept the
> >> assumption.
> >
> >What assumption are you assuming was made, despite the fact that there
> >isn't one?
> The fact that light behaves like sound when it's convenient but not
> otherwise.
That is to say, that light behaves like like is OBSERVED to behave.
Having declared observation to not be a valid basis for formulating a
theory, what do you propose to replace it with?
David Evens
You would PROBABLY do better if you paid attention when people explain
it to you, then.
> It would appear that the number of wavelengths in the two paths would
> still vary with velocity, unless one accepts the notion of source
> dependency of light velocity. This is rejected outright because of the
> binary star argument (something which I would question).
> SR came up with the Lorentz contraction idea and ended up with an
> equation that involved a sqrt of gamma.
You would also probably do better if you showed ANY familiarity with SR
AT ALL. At least you wouldn't make the kind of bonehead mistakes that
show you know NOTHING about it.
> I am not really arguing against the notion of an apparent contraction.
> I merely claim that the standard analysis is wrong because it assumes
> that the cross beam follows a diagonal path, as seen in a rest frame,
> when the apparatus moves.
That is to say, ti assumes that the light actually REACHES the mirror.
> It also assumes that the velocity of this
> fictitious light beam is c.
That is to say, it assumes that the light beam behaves like a light
beam.
> I have used my chainsaw demo to show that
> no light beam ever moves diagonally at c. Infinitesimal elements
> follow diagonal paths but their velocity is sqrt(c^2+v^2).
That is to say, ALL the light ALWAYS travels diagonally. Assuming that
it travels at a speed light cannot travel at doesn't help your claims,
however.
And the light manages to move diagonally so that it doesn't miss the
mirror because...? (Answer: The light is not moving in the beam at
all, since all the LIGHT moves diagonally!)
> >The orthogonal component in your diagrams is entirely due to relative
> >observer velocity and has NOTHING to do with the observed interaction
> >between A and B.
> Don't you mean 'diagonal component?
> >The orientation you illustrate therefore has no meaning.
> >It would ONLY have meaning if the diagonal were caused by ether
> >velocity rather than observer velocity and would result in a
> >difference in A to B time. In the first diagram the time would be
> >longer and dependent on ether flow (expected by M&M) in the second it
> >would equal c regardless of ether flow, provided the beam were wide
> >enough to bridge the displacement due to ether flow.
> >
> >With no ether, there would be no diagonal and both your diagrams would
> >be the same, i.e. like the one I produced.
> There are diagonals involved. Each infinitesimal element of the cross
> beam moves diagonally wrt the rest frame.
That is to say, ALL the light ALWAYS moves diagonally.
> Each element has a different
> diagonal. The light axis remains vertical and it is light that moves
> at c, not the elements. That is my point and that is why the term
> involving the sqrt gamma is wrong!
MAke up your alleged mind: Does the light all move along the diagonals,
or does it move along the beam? Does the light move at c or at the
never-seen superluminal speed you claim it moves at? Are you going to
CONTINUE to demonstrate that you have NEVER seen ANY equation of SR?
> >You have introduced a relative observer velocity to the MMX experiment
> >in order to produce an 'apparent' diagonal which actually bears no
> >more than a cosmetic similarity to the hypothesised diagonal that
> >would exist if there really were an ether and which was tested(at
> >least by intention) in MMX.
> That's correct - and that is what the whole of SR is effectively based
> on.
You mean, the whole of sR is based upon YOUR (apparently intention)
misunderstanding of the analysis of the MMX, an experiment about which
Einstein was scarcely aware at the time he developed SR?
So you are claiming that the light ALWAYS misses the mirror.
> The diagonal business is due to observer movement but that does not
> involve a light BEAM.
Well, the light need not travel along the beam, so this makes no
difference.
> Like you say, it is nonsensical to think that the number of
> wavelengths in the arms of the apparatus would change just because an
> observer happened to move past it.
Then why do you CLAIM it does? After all, the constancy of the number
of wavelengths is RATHER important to the SR analysis of the MMX.
You REALLY must read things. Had you READ it, you would have noticed
the 'usually' in the posting you replied to. This 'usually' refers to
the fact that you can (and, indeed, MUST) consider the diagonal motion
of the light for the general case of the relatively moving observer in
order to explain what is seen.
> >The exception is in the derivation of general. i.e. non specific
> >observer perspective transformations as remarked on already
> >and of course the application of them at relativistic speeds where
> >the observer's relative velocity effectively becomes a no longer
> >ignorable part of the interaction being observed.
> >
> As long as the correct equations are used. I don't believe SR provides
> them.
And you support this assumption (about equations you have demonstrated a
complete lack of knowledge of) by...?
> >regards
> >chic
Henry Wilson
<dev...@technologist.com> wrote:
>
>Henry Wilson wrote:
>
>You are REALLY starting to feel stupid, aren't you, Henry.
Yes, From wasting time conversing with a closed minded person
completely devoid of imagination or creative and spatial ability.
>
>> >
>> >> The standard MMX analysis incorporates this: (fixed pitch fonts)
>> >> __
>> >> /\
>> >> / \
>> >> / \
>> >> / \
>> >>
>> >> This is misleading.
>> >> It should be like this:
>> >>
>> >>
>> >>
>> >> _ _ _ _ _ _ _
>> >> |
>> >> | |
>> >> | |
>> >> | |
>> >>
>> >> showing how the top mirror is always aligned with the axis of the
>> >> light beam. The mirror moves with the beam.
>> >
>> The dashes represent the positions of the mirror, always directly over
>> the beam. That should be obvious.
>
>The beam is not the path of the light. Had you any understanding of the
>MMX, you could not avoid knowing this.
>
>> Go out and buy a chainsaw and araldite the bloody thing on top of you
>> car, then start it up, drive off and ask a bystander to watch what has
>> happens.
>
>And this is relevant because...? (Other than the fact that you don't
>understand that it proves you wrong, that is.)
Can't you make any kind of intelligent comment?
>
>> >
>> >No, YOU did. He pointed out your mistake.
>> Nonsense, He is trying to fathom the (il)logic of the standard
>> treatment, too.
>
>What illogic are you assuming is present in the valid analysis? All
>your previous pretenses have been utterly destroyed, although you don't
>seem to care about being humiliated.
I can't recall anyone actually discussing what I have actually said.
You are all too busy trying any desperate measure to defend your blind
faith in what is, at best, a very suspect theory.
>
>> Now what the hell? You have reversed your fucking argument. you have
>> been trying to convince me for weeks that the light beam moves
>> diagonally, now you are saying it doesn't. Are you drunk?
>
>Why do you assume that I am making mistakes for the reasons that you
>are?
>
>In any event, the BEAM (which is NOT the path of the light in the beam)
>cannot appear to be diagonal unless motion is neither along it nor
>perpendicular to it, which isn't part of the situation under
>consideration. The LIGHT, of course, ALWAYS moves diagonally.
What do you mean by 'light'? You seem to be implying that 'light' is
some kind of point particle with zero dimension. That is the only
condition which might make my theory questionable. Light is obviously
more than that.
>
>Then what DOES move diagonally? After all, the only thing that DOES
>move in the apparatus IS the light, so what ELSE could be bouncing from
>the mirror?
A particular diagonal line of zero width represents the path taken by
an infinitesimal element of a photon. That should be simple enough for
you to understand, assuming you have some knowledge of calculus.
>> >> Subsequently, he and his followers abolished the aether but kept the
>> >> assumption.
>> >
>> >What assumption are you assuming was made, despite the fact that there
>> >isn't one?
>> The fact that light behaves like sound when it's convenient but not
>> otherwise.
>
>That is to say, that light behaves like like is OBSERVED to behave.
>
>Having declared observation to not be a valid basis for formulating a
>theory, what do you propose to replace it with?
You sound drunk again.