SR question/bending light
Rod Ryker
as observed by the rest system , would the rest system observe
the light beam of the laser bend towards the left ?
Perplexing isn't it .
Rod Ryker
Tom Roberts
> If there is a laser in the primed system traveling towards the right
> as observed by the rest system , would the rest system observe
> the light beam of the laser bend towards the left ?
No. assuming both systems are interial. Relative motion between inertial
frames maps straight lines to straight lines.
> Perplexing isn't it .
Not at all.
Tom Roberts tjro...@lucent.com
Rod Ryker
Tom Roberts wrote:
Rod: Tom I screwed up . :)
No acceleration at all is intended (mathematically speaking) .
If there is a laser in the primed system traveling towards the right
as observed by the rest system , would the rest system observe
the light beam of the laser bend towards the right ?
Sorry Tom ,
Rod Ryker
Bilge
SR question/bending light to usenet:
Since the observer won't see any light that isn't pointed directly
at him, it's not that perplexing. One can always create a conundrum by
hypothesizing something that's physically impossible. One of the keys to
getting the correct description of reality in a model, is not to require
it to explain something that can't physically occur or that it precludes,
a priori, since that assures you it's wrong.
Eric Baird
wrote:
"Bending" ususally refers to the changing of an angle, with position
or with time.
In the case of a coasting object and observer under special
relativity, the lightbeams are supposed to be equally straight for all
observers, =BUT= observers in different frames will usually disagree
about the nominal angle value that should be assigned to these
straight lines.
If the "moving" object aims a laser at what they take to be 90 degrees
to their direction of motion, a background observer will see the same
lightbeam to be moving forward with the object, and to be aimed
slightly forward (SR's angle-aberrration formula is shared with
emission-theories, and appears in Einstein's 1905 paper).
But this effect isn't normally what is meant by "light-bending".
=Erk= (Eric Baird)
: "Standards are important. That's why there are so many of them."
Rod Ryker
Eric Baird wrote:
Rod: All I care about is what the rest system observer records .
Study the below and point out any errors I have , please .
*
*
*
*
(ME)(ME)(ME)(ME) -----> at v
(RO) stationary
1. Above we have a stationary system (RO) and a system always
in non accelerated motion (ME) .
A laser , that comoves with (ME) , emits one pulse vertically that (RO)
observes as propagating to the the right of (RO)'s position .
It is clear that there is a transfer of momentum from the emitter
to the light pulse according to (RO) .
* *
* *
* *
* *
(ME)(ME)----> at v
(RO)
2. In this example , the laser of (ME) system emits a continous
beam . (RO) observes this beam as traveling verically as opposed
to example 1 above . There is a transfer of momentum also , as
I will demonstrate below in example 3 .
*
*
* *
* *
*
*
(ME)---> at v (ME)----> at v
(RO)
3. Here we show the continous beam being emitted and then
not . (RO) observes the beam in vertical formation and still
comoving with system (ME) .
Hence , a transfer of momentum to the light .
Rod Ryker
Rod Ryker
Bilge wrote:
Rod: One must measure EM waves in some fasion or is this
impossible wrt my example ?
I think not .
Rod Ryker
Bilge
Yes. You think not. You measure what lands in your detector. That
happens to be the electromagnetic radiation that propagate toward you.
You can't detect e-m radiation propagating propagating toward something
else.
Rod Ryker
whilst the detector is detecting the detection . :)
Maybe commonsense is far removed from you ,
treat yourself to a game of *hokey-pokey* ,
and turn yourself around . :)
Rod Ryker
Bilge
Re: SR question/bending light to usenet:
>whilst the detector is detecting the detection . :)
to where the detector was located to detect it.
>
>Maybe commonsense is far removed from you ,
Possibly, since my first thought was to be less polite.
>treat yourself to a game of *hokey-pokey* ,
>and turn yourself around . :)
Let's see you show how you detect light which doesn't propagate
to your detector, first.
Rod Ryker
Bilge wrote:
Rod: Let's see where exactly I posted such .
Rod Ryker
Tom Roberts
> If there is a laser in the primed system traveling towards the right
> as observed by the rest system , would the rest system observe
Your phrase "bend towards the right" is ambiguous -- it normally means
"follow a curved trajectory increasingly headed to the right", but it
could mean "follow a straight-line trajectory which is inclined more
to the right".
But for neither interpretation is that correct. Assuming both the
"rest system" and the "primed system" are inertial, then in the rest
system the laser beam will follow a straight-line trajectory inclined
more to the LEFT than it is in the primed system (e.g. a vertical
beam in the primed system will not be vertical when seen from the rest
system, it will lean to the left).
This is so basic -- as you clearly don't understand what is going on,
I really have to wonder why you waste your time around here....
Tom Robverts tjro...@lucnet.com
Rod Ryker
Tom Roberts wrote:
> Rod Ryker wrote:
> > If there is a laser in the primed system traveling towards the right
> > as observed by the rest system , would the rest system observe
> > the light beam of the laser bend towards the right ?
>
> Your phrase "bend towards the right" is ambiguous -- it normally means
> "follow a curved trajectory increasingly headed to the right", but it
> could mean "follow a straight-line trajectory which is inclined more
> to the right".
>
> But for neither interpretation is that correct. Assuming both the
> "rest system" and the "primed system" are inertial, then in the rest
> system the laser beam will follow a straight-line trajectory inclined
> more to the LEFT than it is in the primed system (e.g. a vertical
> beam in the primed system will not be vertical when seen from the rest
> system, it will lean to the left).
>
Rod: Why does it lean to the left ?
There is no acceleration , I meant not to include
any gravity effects nor resistance to the light either .
So , why does this continuous laser beam lean to the left
as observed by the rest system observer ?
>
> This is so basic -- as you clearly don't understand what is going on,
> I really have to wonder why you waste your time around here....
>
> Tom Robverts tjro...@lucnet.com
Rod: There is no waste of time when I learn . :)
Rod Ryker
Rod Ryker
answer the damn question ?
Rod Ryker
Bilge
OK:
----------------------------------------------
From: Rod Ryker <rry...@fuse.net>
Newsgroups: sci.physics.relativity
Subject: SR question/bending light
Date: Sat, 22 Sep 2001 19:40:51 -0400
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If there is a laser in the primed system traveling towards the right
as observed by the rest system , would the rest system observe
the light beam of the laser bend towards the left ?
Perplexing isn't it .
Rod Ryker
--------------------------------------------
Draw a picture for me which indicates the position of the laser,
the path of the light, the position of the observer and since you agree
that you can't see light that doesn't land in your detector, include
the path of the light that does enter the detector. I'll draw what I
think you meant.
You won't be perplexed if you set the problem to obtain what is
physically possible to obtain. Set up many observers along the path of
the light at intervals \Delta x so that a short flash of light
by the laser results in a flash of light by each of these observer as
the light passes by. Now flash the laser light every t = \Delta x/c
seconds so that each of these observers sends a flash of light every
t seconds.
Now decide what constitutes the "path" of the laser light. Your
observer will see these flashes of light. The "path" will be the
flashes that your observer sees at the same time.
. 1 The light from 1,2,3,4,5 arrive at O at _different_
. 2 times.
. 3
. 4
. 5
O
The light that arrives at the _same_ time is:
1 .
2 .
3 .
4 .
5 .
O
--------------------
Set your problem up without trying to take shortcuts.
If you have to use something which doesn't exist, your
result won't br correct.
Rod Ryker
Bilge wrote:
Rod: That does not answer the question of mine . :)
>
> Draw a picture for me which indicates the position of the laser,
> the path of the light, the position of the observer and since you agree
> that you can't see light that doesn't land in your detector, include
> the path of the light that does enter the detector. I'll draw what I
> think you meant.
>
Rod: Then you "draw" incorrectly . :)
Rod: It worked for Einstein in 1905 . :)
Rod Ryker
Bilge
Where is the picture I asked you to draw? Until you provide
that, I will assume not only that I answered your question, but
that you weren't interested enough to know why.
--
Henry Wilson
Tom is completely wrong here. The vertical laser beam remains dead vertical in
all moving frames. Just plot the paths of individual 'wave crests' and the truth
of this is immediately apparent.
Individual infinitesimal elements of the beam DO follow SEPARATE diagonal paths
but they DO NOT constitute light and clearly move at sqrt (c^2+v^2) in the
moving frame. The laser light takes the same time to reach a certain height IN
ALL FRAMES!!!!!
This is a basic misconception that has plagued physics ever since the MMX/aether
fiasco. SR is based on the same elementary error and is therefore also complete
nonsense.
I have been pointing out this mammoth blunder for several years now but SRian
faith prevents its followers from even trying to understand.
They simply don't want to know the truth.
Rod Ryker
moments , that's all . ;)
Why not join in on this similar discussion on the thread :
Re: Why ?
Rod Ryker
Bilge
>nothing
Draw a picture containing the signals actually received
by the observer. Not the ones your shortcuts get wrong.
--
Rod Ryker
Rod Ryker
Henry Wilson
>Rod: Hi Henry , Tom is going through one of his
>moments , that's all . ;)
>Why not join in on this similar discussion on the thread :
>
>Re: Why ?
>
>Rod Ryker
It isn't just Tom. It applies to all of them. They have been deluded into
believing that a vertical beam of light somehow appears diagonally in a moving
frame. That is essentially the origin of gamma, which therefore cannot be
correct.
Nor can any other aspect of SR.
Tom Roberts
> Tom Roberts wrote:
> > Assuming both the
> > "rest system" and the "primed system" are inertial, then in the rest
> > system the laser beam will follow a straight-line trajectory inclined
> > more to the LEFT than it is in the primed system (e.g. a vertical
> > beam in the primed system will not be vertical when seen from the rest
> > system, it will lean to the left).
> Rod: Why does it lean to the left ?
Just think about what it looks like. Imagine a stream of machine-gun
bullets instead. The line formed by the bullets (or short regions of
a light beam) leans to the left, while the individual bullets (short
regions of a light beam) travel vertically. This difference between the
line and the individual bullets confuses Henry Wilson.
Tom Roberts tjro...@lucent.com
Rod Ryker
Tom Roberts wrote:
Rod: I understand this , however , light clock examples
have the lean towards the direction of motion for a pulse
of light .
This contradicts what you state re bullets .
There is no way to tell if a continuous beam is leaning
if your reasoning is based upon trajectory
of individual pulses or photons which can be mapped .
Rod Ryker
Rod Ryker
Henry Wilson wrote:
Rod: I agree that a continuous beam in an inertial system
will not look as if leaning in any inertial frame .
You cannot map the trajectory of a continuous beam .
That is , the x component , but not the y component .
Rod Ryker
Henry Wilson
No Tom, you are talking nonsense. You are saying that the bullets are arranged
like this in the rest frame:
|
|
|
|
and like this in the moving frame:
|
|
|
|
<--v
So tell me, what happens when they are fired so rapidly that the ends touch?
You are forgetting that ALL the bullets START WITH a transverse velocity in the
moving frame. Thus they remain aligned vertically in all frames. In any moving
frame, the whole vertical line itself appears to move sideways but always
remaining vertical, as do the individual bullets.
There only conclusion one can reach from this is that SR is completely wrong
from the very first argument.
Bruce Richmond
to keep moving horizontally when they leave the gun, despite
the barrel pointing straight up. If the gun is moving to the
right, their paths are diagonal to the right in the stationary
frame. Their velocity, as measured in the stationary frame, is
greater then their velocity as measured relative to the gun.
Of course, we don't base our system of measurement on the velocity
of the bullets. If we did things could appear quite different.
The measured path of the photons depends on the synchronization
of the clocks doing the measurements. And the clocks are
synchronized based on light traveling at c in both frames.
Henry Wilson
Oh for christ's sake Bruce, do you think we want to hear about 'c being
constant in both frames' when that is basically the very thing we are
questioning.
Do you think we are stupid enough not to recognize a baltantly circular argument
when we are presented with one?
A vertical laser beam appears vertical in all horizontally moving frames. Just
plot the bloody thing and you will see why!
Why can you SRians not accept simple truths?
Henry Wilson
Yes you can. I did it - and discovered where and why relativity is wrong.
Just let the beam resemble a sine wave and plot the positions of each wave
crest.
You will see that they remain aligned vertically in all frames.
Paul B. Andersen
> A vertical laser beam appears vertical in all horizontally moving frames. Just
> plot the bloody thing and you will see why!
>
> Why can you SRians not accept simple truths?
I think you are suffering from monomania, Henry.
Please answer the following questions:
1. Have anybody ever insisted that a vertical light beam defined
as "the line through the instant position of the consecutive
photons in the bean" is NOT vertical in all horizontally moving frames?
2. Have anybody ever failed to realize this obvious fact?
3. Has this been pointed out to you numerous times before?
Paul
Henry Wilson
wrote:
>Henry Wilson wrote:
>
>> A vertical laser beam appears vertical in all horizontally moving frames. Just
>> plot the bloody thing and you will see why!
>>
>> Why can you SRians not accept simple truths?
>
>
>I think you are suffering from monomania, Henry.
>
>Please answer the following questions:
>1. Have anybody ever insisted that a vertical light beam defined
> as "the line through the instant position of the consecutive
> photons in the bean" is NOT vertical in all horizontally moving frames?
YES!!!! Tom Roberts just did above. Can't you read?
>2. Have anybody ever failed to realize this obvious fact?
>3. Has this been pointed out to you numerous times before?
The only reason you know it is true is because I TOLD YOU SO! Now will you
please pass on this information to Tom Roberts.
Will you also inform him that the SR explanation of the MMX null result is now
known to be completely wrong, as a result of Henry Wilson's revelation that NO
DIAGONAL LIGHT BEAM EXISTS!!!!!!
Every text book MMX diagram has been wrong for 130 years!
The diagonal line conventionally drawn represents the trajectory of a separate
dimensionless element of the light beam, in the moving frame. The next element
DOESN'T FOLLOW IT UP THAT SAME DIAGONAL. It moves up the diagonal which is
infinitesimally adjacent!!!
By what stretch of the imagination would anyone with an IQ greater than that of
an earthworm assign the value c to the velocity of this infinitesimal piece of
nothing as it moves along its own personal diagonal?
Its velocity is obviously sqrt(c^2+v^2).
What is the maximum speed of the cutting point of a very long pair of scissors,
Paul?
>
>Paul
>
>
Rod Ryker
Henry Wilson wrote:
Rod: I agree 100% with your example wrt a continuous beam .
Rod Ryker
Rod Ryker
Bruce Richmond wrote:
> Tom Roberts <tjro...@lucent.com> wrote in message news:<3BBC648E...@lucent.com>...
> > Rod Ryker wrote:
> > > Tom Roberts wrote:
> > > > Assuming both the
> > > > "rest system" and the "primed system" are inertial, then in the rest
> > > > system the laser beam will follow a straight-line trajectory inclined
> > > > more to the LEFT than it is in the primed system (e.g. a vertical
> > > > beam in the primed system will not be vertical when seen from the rest
> > > > system, it will lean to the left).
> > > Rod: Why does it lean to the left ?
> >
> > Just think about what it looks like. Imagine a stream of machine-gun
> > bullets instead. The line formed by the bullets (or short regions of
> > a light beam) leans to the left, while the individual bullets (short
> > regions of a light beam) travel vertically. This difference between the
> > line and the individual bullets confuses Henry Wilson.
> >
>
> If you use bullets, the bullets have momentum that causes them
> to keep moving horizontally when they leave the gun, despite
> the barrel pointing straight up. If the gun is moving to the
> right, their paths are diagonal to the right in the stationary
> frame. Their velocity, as measured in the stationary frame, is
> greater then their velocity as measured relative to the gun.
>
Rod: Light also carries a momentum in the direction of propagation .
Rod Ryker
Tom Roberts
> A vertical laser beam appears vertical in all horizontally moving frames.
Yes. I was wrong earlier. I forgot about aberration -- in a moving frame
the individual regions of the light beam are not moving vertically; they
experience aberration of precisely the right amount to keep them directly
above the source at all times.
Sorry about that.
The same applies to bullets.
For the transformation of velocities in SR, see Schwartz:
_Introduction_to_Special_Relativity_, eq 3-22 and problem 3-3.
I have often admonished others around here to use the _full_ Lorentz
transform when doing analyses. Until now in this thread I did not
follow that advice, and gave wrong answers.
Tom Roberts tjro...@lucent.com
Bruce Richmond
In SR the speed of light is c in all frames. That is a given.
You use that given to construct the coordinate system for each
frame. If you then measure the speed of light to be anything
but c, you must have made a mistake. Call it circular if you
like, but that's the way it works.
> A vertical laser beam appears vertical in all horizontally moving frames. Just
> plot the bloody thing and you will see why!
>
> Why can you SRians not accept simple truths?
You will get no argument from me on that. But I don't consider
myself a SRian either.
I suspect the problem stems from the statement, "Light is always
propagated in empty space with a definite velocity c which is
independent of the state of motion of the emitting body."
Some would interpret this to mean that the path of a photon
emitted by a moving laser must be the same as a photon emitted
by a stationary laser at the same point. If you plot photons
emitted by lasers at rest in the stationary system, that fire
when the x' origin passes to the right, you get photons moving
vertically while arranged in a diagonal line sloping to the left.
I interpret the statement above to mean the speed of light is
always measured to be c regardless of the motion of the emitter.
Consider a gun sitting at the origin of the primed system,
aimed along the y' axis. There is a target some distance from
the origin, also on the y' axis. The principle of relativity
says that any shot fired by the gun will travel along the y'
axis and hit the target. When the shot is fired doesn't matter.
The velocity of the bullet doesn't matter.
Let the bullet be a photon. Have it emitted from the laser gun
when the origins of the two systems coincide. We know that the
photon travels along the y' axis and hits the target. We are
talking about a single photon bullet here, not a circular wave
front. It cannot travel along both the y and y' axis. The
target on the y' axis, which we know must get hit, is not on
the y axis when it is hit. The photon must travel at an angle
to the y axis.
Does this contradict SR? I don't think so. There is a
relativistic correction that must be applied to transverse
velocities measured in a moving system. What it eventually
comes down to is that the moving system measures the photon's
velocity as c along the y' axis, while the stationary system
measures the photon's velocity as c along the diagonal.
Both systems agree that the photon was emitted when the origins
coincided. And both systems agree that it hits the target on
the y' axis. They do not agree on when the hit takes place.
Paul B. Andersen
> On Fri, 05 Oct 2001 15:45:00 +0200, "Paul B. Andersen" <paul.b....@hia.no>
> wrote:
>
>
>>Henry Wilson wrote:
>>
>>
>>>A vertical laser beam appears vertical in all horizontally moving frames. Just
>>>plot the bloody thing and you will see why!
>>>
>>>Why can you SRians not accept simple truths?
>>>
>>
>>I think you are suffering from monomania, Henry.
>>
>>Please answer the following questions:
>>1. Have anybody ever insisted that a vertical light beam defined
>> as "the line through the instant position of the consecutive
>> photons in the bean" is NOT vertical in all horizontally moving frames?
>>
>
> YES!!!! Tom Roberts just did above. Can't you read?
Yes, _I_ can.
Tom said the _trajectory_ of the light was not vertical in the moving
observers frame. And he is right about that, isn't he?
>>2. Have anybody ever failed to realize this obvious fact?
>>3. Has this been pointed out to you numerous times before?
>>
>
> The only reason you know it is true is because I TOLD YOU SO!
You are incredible naive, Henry. :-)
> Now will you
> please pass on this information to Tom Roberts.
>
> Will you also inform him that the SR explanation of the MMX null result is now
> known to be completely wrong, as a result of Henry Wilson's revelation that NO
> DIAGONAL LIGHT BEAM EXISTS!!!!!!
There sure is no "diagonal light beam" in the SR explanation of the MMX.
Why should it be?
> Every text book MMX diagram has been wrong for 130 years!
You are of course referring to the "MMX diagram" explaining Michelson's
prediction of the fringe shift. That diagram has of course been correct
since Michelson first draw it in 1886. That is, it is a correct description
of why Michelson's ether theory predicts a fringe shift.
> The diagonal line conventionally drawn represents the trajectory of a separate
> dimensionless element of the light beam, in the moving frame. The next element
> DOESN'T FOLLOW IT UP THAT SAME DIAGONAL. It moves up the diagonal which is
> infinitesimally adjacent!!!
You are stating obvious trivialities, Henry.
> By what stretch of the imagination would anyone with an IQ greater than that of
> an earthworm assign the value c to the velocity of this infinitesimal piece of
> nothing as it moves along its own personal diagonal?
> Its velocity is obviously sqrt(c^2+v^2).
Indeed?
Nobody with an IQ greater than that of an earthworm can fail
to understand that "the speed you assign to light" depend on
the theory of physics you use.
So which theory did you apply to arrive at the above result?
Is this theory tested?
> What is the maximum speed of the cutting point of a very long pair of scissors,
> Paul?
And why do you think that is relevant?
But the answer is that the instant speed of the cutting point can
be anything, but the average speed (the distance from your fingers
to the cutting point divided by the time since you started moving
your fingers) can never exceed the speed of sound in the scissors.
Now you can chew on that.
Paul
Henry Wilson
>Henry Wilson wrote:
>> A vertical laser beam appears vertical in all horizontally moving frames.
>
>Yes. I was wrong earlier. I forgot about aberration -- in a moving frame
>the individual regions of the light beam are not moving vertically; they
>experience aberration of precisely the right amount to keep them directly
>above the source at all times.
Very noble of you to admit you were wrong, Tom. Not many here would do that.
Now about these 'individual regions'. By what stretch of the imagination should
we conclude that their diagonal velocity is always c?
>
>Sorry about that.
>
>The same applies to bullets.
>
> For the transformation of velocities in SR, see Schwartz:
> _Introduction_to_Special_Relativity_, eq 3-22 and problem 3-3.
>
>
>I have often admonished others around here to use the _full_ Lorentz
>transform when doing analyses. Until now in this thread I did not
>follow that advice, and gave wrong answers.
Where is the justification for the lorentz transform, given the above?
>
>
>Tom Roberts tjro...@lucent.com
Henry Wilson
wrote:
>Henry Wilson wrote:
>
>> On Fri, 05 Oct 2001 15:45:00 +0200, "Paul B. Andersen" <paul.b....@hia.no>
>> wrote:
>>
>>
>>>Henry Wilson wrote:
>>>
>>>
>>>>A vertical laser beam appears vertical in all horizontally moving frames. Just
>>>>plot the bloody thing and you will see why!
>>>>
>>>>Why can you SRians not accept simple truths?
>>>>
>>>
>>>I think you are suffering from monomania, Henry.
>>>
>>>Please answer the following questions:
>>>1. Have anybody ever insisted that a vertical light beam defined
>>> as "the line through the instant position of the consecutive
>>> photons in the bean" is NOT vertical in all horizontally moving frames?
>>>
>>
>> YES!!!! Tom Roberts just did above. Can't you read?
>
>
>Yes, _I_ can.
>Tom said the _trajectory_ of the light was not vertical in the moving
>observers frame. And he is right about that, isn't he?
To has now admitted he was wrong about that.
>
>
>>>2. Have anybody ever failed to realize this obvious fact?
>>>3. Has this been pointed out to you numerous times before?
>>>
>>
>> The only reason you know it is true is because I TOLD YOU SO!
>
>
>You are incredible naive, Henry. :-)
>
>
>
>> Now will you
>> please pass on this information to Tom Roberts.
>>
>> Will you also inform him that the SR explanation of the MMX null result is now
>> known to be completely wrong, as a result of Henry Wilson's revelation that NO
>> DIAGONAL LIGHT BEAM EXISTS!!!!!!
>
>
>There sure is no "diagonal light beam" in the SR explanation of the MMX.
>Why should it be?
SR needs a diagonal light beam TRAVELLING AT C so the its so called prediction
of length and time contractions will cause the path lengths and traveltimes of
both rays being the same, independent of velocity.
Since there is no diagonal light ray traveling at c, Sr's theory must be wrong.
>
>> Every text book MMX diagram has been wrong for 130 years!
>
>
>You are of course referring to the "MMX diagram" explaining Michelson's
>prediction of the fringe shift. That diagram has of course been correct
>since Michelson first draw it in 1886. That is, it is a correct description
>of why Michelson's ether theory predicts a fringe shift.
The diagram has misrepresented the fact, bugger Michelson and what he believed.
>
>
>> The diagonal line conventionally drawn represents the trajectory of a separate
>> dimensionless element of the light beam, in the moving frame. The next element
>> DOESN'T FOLLOW IT UP THAT SAME DIAGONAL. It moves up the diagonal which is
>> infinitesimally adjacent!!!
>
>
>You are stating obvious trivialities, Henry.
Obvious trivialities with far reaching consequences for physics.
The point is, it doesn't move at c!!!!
>
>
>> By what stretch of the imagination would anyone with an IQ greater than that of
>> an earthworm assign the value c to the velocity of this infinitesimal piece of
>> nothing as it moves along its own personal diagonal?
>
>> Its velocity is obviously sqrt(c^2+v^2).
>
>
>Indeed?
>
>Nobody with an IQ greater than that of an earthworm can fail
>to understand that "the speed you assign to light" depend on
>the theory of physics you use.
>
>So which theory did you apply to arrive at the above result?
>Is this theory tested?
Yes. I test it every time I drive along the road. The walls remain vertical.
>
>> What is the maximum speed of the cutting point of a very long pair of scissors,
>> Paul?
>
>
>And why do you think that is relevant?
>But the answer is that the instant speed of the cutting point can
>be anything, but the average speed (the distance from your fingers
>to the cutting point divided by the time since you started moving
>your fingers) can never exceed the speed of sound in the scissors.
>Now you can chew on that.
OK replace the scissors with two laser beams.
What is the maximum speed of their point of intersection as they are brought
into parallel.
>
>Paul
>
>
>
>
>
David Evens
>On Fri, 05 Oct 2001 20:20:25 -0500, Tom Roberts <TomRo...@avenew.com> wrote:
>>Henry Wilson wrote:
>>> A vertical laser beam appears vertical in all horizontally moving frames.
>>
>>Yes. I was wrong earlier. I forgot about aberration -- in a moving frame
>>the individual regions of the light beam are not moving vertically; they
>>experience aberration of precisely the right amount to keep them directly
>>above the source at all times.
>
>Very noble of you to admit you were wrong, Tom. Not many here would do that.
>
>Now about these 'individual regions'. By what stretch of the imagination should
>we conclude that their diagonal velocity is always c?
Since all the light always moves diagonally, and the light always
moves at c, it is obvious that the light must move diagonally at c.
>>Sorry about that.
>>
>>The same applies to bullets.
>>
>> For the transformation of velocities in SR, see Schwartz:
>> _Introduction_to_Special_Relativity_, eq 3-22 and problem 3-3.
>>
>>
>>I have often admonished others around here to use the _full_ Lorentz
>>transform when doing analyses. Until now in this thread I did not
>>follow that advice, and gave wrong answers.
>
>Where is the justification for the lorentz transform, given the above?
The fact that he pointed out that it is not possible to not recognise
it as correct.
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David Evens
>On Sat, 06 Oct 2001 19:40:23 +0200, "Paul B. Andersen" <paul.b....@hia.no>
>wrote:
>>Henry Wilson wrote:
>>> On Fri, 05 Oct 2001 15:45:00 +0200, "Paul B. Andersen" <paul.b....@hia.no>
>>> wrote:
>>>>Henry Wilson wrote:
>>>>
>>>>>A vertical laser beam appears vertical in all horizontally moving frames. Just
>>>>>plot the bloody thing and you will see why!
>>>>>
>>>>>Why can you SRians not accept simple truths?
>>>>>
>>>>
>>>>I think you are suffering from monomania, Henry.
>>>>
>>>>Please answer the following questions:
>>>>1. Have anybody ever insisted that a vertical light beam defined
>>>> as "the line through the instant position of the consecutive
>>>> photons in the bean" is NOT vertical in all horizontally moving frames?
>>>>
>>>
>>> YES!!!! Tom Roberts just did above. Can't you read?
>>
>>
>>Yes, _I_ can.
>>Tom said the _trajectory_ of the light was not vertical in the moving
>>observers frame. And he is right about that, isn't he?
>
>To has now admitted he was wrong about that.
Wherte did you hal;ucinate him saying that the speed of light is
source dependent?
>>>>2. Have anybody ever failed to realize this obvious fact?
>>>>3. Has this been pointed out to you numerous times before?
>>>>
>>>
>>> The only reason you know it is true is because I TOLD YOU SO!
>>
>>
>>You are incredible naive, Henry. :-)
>>
>>
>>
>>> Now will you
>>> please pass on this information to Tom Roberts.
>>>
>>> Will you also inform him that the SR explanation of the MMX null result is now
>>> known to be completely wrong, as a result of Henry Wilson's revelation that NO
>>> DIAGONAL LIGHT BEAM EXISTS!!!!!!
>>
>>
>>There sure is no "diagonal light beam" in the SR explanation of the MMX.
>>Why should it be?
>
>SR needs a diagonal light beam TRAVELLING AT C so the its so called prediction
>of length and time contractions will cause the path lengths and traveltimes of
>both rays being the same, independent of velocity.
>
>Since there is no diagonal light ray traveling at c, Sr's theory must be wrong.
You forgot to support your assumption that the speed of light is
source dependent.
>>> Every text book MMX diagram has been wrong for 130 years!
>>
>>
>>You are of course referring to the "MMX diagram" explaining Michelson's
>>prediction of the fringe shift. That diagram has of course been correct
>>since Michelson first draw it in 1886. That is, it is a correct description
>>of why Michelson's ether theory predicts a fringe shift.
>
>The diagram has misrepresented the fact, bugger Michelson and what he believed.
You forgot to suport your assumption that the light doesn't travel to
the mirror and back in every frame.
>>> The diagonal line conventionally drawn represents the trajectory of a separate
>>> dimensionless element of the light beam, in the moving frame. The next element
>>> DOESN'T FOLLOW IT UP THAT SAME DIAGONAL. It moves up the diagonal which is
>>> infinitesimally adjacent!!!
>>
>>
>>You are stating obvious trivialities, Henry.
>
>Obvious trivialities with far reaching consequences for physics.
>The point is, it doesn't move at c!!!!
Thwn how can it EVER reach the mirror?
>>> By what stretch of the imagination would anyone with an IQ greater than that of
>>> an earthworm assign the value c to the velocity of this infinitesimal piece of
>>> nothing as it moves along its own personal diagonal?
>>
>>> Its velocity is obviously sqrt(c^2+v^2).
>>
>>
>>Indeed?
>>
>>Nobody with an IQ greater than that of an earthworm can fail
>>to understand that "the speed you assign to light" depend on
>>the theory of physics you use.
>>
>>So which theory did you apply to arrive at the above result?
>>Is this theory tested?
>
>Yes. I test it every time I drive along the road. The walls remain vertical.
You forgot to mention where you tested your pretense.
>>> What is the maximum speed of the cutting point of a very long pair of scissors,
>>> Paul?
>>
>>
>>And why do you think that is relevant?
>>But the answer is that the instant speed of the cutting point can
>>be anything, but the average speed (the distance from your fingers
>>to the cutting point divided by the time since you started moving
>>your fingers) can never exceed the speed of sound in the scissors.
>>Now you can chew on that.
>
>OK replace the scissors with two laser beams.
>What is the maximum speed of their point of intersection as they are brought
>into parallel.
This will make an interesting computation, since you have to take into
account the way that the beams curve as the source rotates.
Rod Ryker
David Evens wrote:
> On Sat, 06 Oct 2001 20:26:02 GMT, He...@the.edge(Henry Wilson) wrote:
> >On Fri, 05 Oct 2001 20:20:25 -0500, Tom Roberts <TomRo...@avenew.com> wrote:
> >>Henry Wilson wrote:
> >>> A vertical laser beam appears vertical in all horizontally moving frames.
> >>
> >>Yes. I was wrong earlier. I forgot about aberration -- in a moving frame
> >>the individual regions of the light beam are not moving vertically; they
> >>experience aberration of precisely the right amount to keep them directly
> >>above the source at all times.
> >
> >Very noble of you to admit you were wrong, Tom. Not many here would do that.
> >
> >Now about these 'individual regions'. By what stretch of the imagination should
> >we conclude that their diagonal velocity is always c?
>
> Since all the light always moves diagonally, and the light always
> moves at c, it is obvious that the light must move diagonally at c.
>
Rod: You forgot to provide logic and experimental evidence
that supports as to why a vertical moving laser (to the right) of a rest observer ,
both in inertial systems , where the laser actually leans to the left
in the rest system but because of abberation appears vertical ,
above the comoving emitter in the moving system , where we
would conclude easily from simple experiment that the laser
leaning to the left would bounce off the mirror and travel towards
the left and not to the right .
Your ghost (abberation) contains sUm major ju ju to acomplish
what it does .
Rod Ryker
Henry Wilson
Good Evens. That will give you something useful to think about.
Let us know in a couple of months how you are getting on..
Henry Wilson
>On Sat, 06 Oct 2001 20:26:02 GMT, He...@the.edge(Henry Wilson) wrote:
>>On Fri, 05 Oct 2001 20:20:25 -0500, Tom Roberts <TomRo...@avenew.com> wrote:
>>>Henry Wilson wrote:
>>>> A vertical laser beam appears vertical in all horizontally moving frames.
>>>
>>>Yes. I was wrong earlier. I forgot about aberration -- in a moving frame
>>>the individual regions of the light beam are not moving vertically; they
>>>experience aberration of precisely the right amount to keep them directly
>>>above the source at all times.
>>
>>Very noble of you to admit you were wrong, Tom. Not many here would do that.
>>
>>Now about these 'individual regions'. By what stretch of the imagination should
>>we conclude that their diagonal velocity is always c?
>
>Since all the light always moves diagonally, and the light always
>moves at c, it is obvious that the light must move diagonally at c.
I might be a bit stupid Evens because I cannot see that this follows. All I can
see is that the light takes the same time to traverse the same vertical distance
in all frames.
>
>>>Sorry about that.
>>>
>>>The same applies to bullets.
>>>
>>> For the transformation of velocities in SR, see Schwartz:
>>> _Introduction_to_Special_Relativity_, eq 3-22 and problem 3-3.
>>>
>>>
>>>I have often admonished others around here to use the _full_ Lorentz
>>>transform when doing analyses. Until now in this thread I did not
>>>follow that advice, and gave wrong answers.
>>
>>Where is the justification for the lorentz transform, given the above?
>
>The fact that he pointed out that it is not possible to not recognise
>it as correct.
Why, Evens? Do I have to believe it just because every textbook in the last 100
years states it?
David Evens
wrote:
You forgot to mention anyone (othwer than anti-physics cranks like Ken
Seto and Androcles) who says that the light beam is not aimed
vertically.
David Evens
>On Sun, 07 Oct 2001 09:11:43 GMT, dev...@technologist.com (David Evens) wrote:
>>On Sat, 06 Oct 2001 20:26:02 GMT, He...@the.edge(Henry Wilson) wrote:
>>>On Fri, 05 Oct 2001 20:20:25 -0500, Tom Roberts <TomRo...@avenew.com> wrote:
>>>>Henry Wilson wrote:
>>>>> A vertical laser beam appears vertical in all horizontally moving frames.
>>>>
>>>>Yes. I was wrong earlier. I forgot about aberration -- in a moving frame
>>>>the individual regions of the light beam are not moving vertically; they
>>>>experience aberration of precisely the right amount to keep them directly
>>>>above the source at all times.
>>>
>>>Very noble of you to admit you were wrong, Tom. Not many here would do that.
>>>
>>>Now about these 'individual regions'. By what stretch of the imagination should
>>>we conclude that their diagonal velocity is always c?
>>
>>Since all the light always moves diagonally, and the light always
>>moves at c, it is obvious that the light must move diagonally at c.
>
>I might be a bit stupid Evens because I cannot see that this follows. All I can
>see is that the light takes the same time to traverse the same vertical distance
>in all frames.
Yes, we KNOW that you do not thing that the universe operates in the
manner that it is observed to operate. However, we SEE that the speed
of light is NOT dependent on the motion of the source in the manner
that you pretend.
>>>>Sorry about that.
>>>>
>>>>The same applies to bullets.
>>>>
>>>> For the transformation of velocities in SR, see Schwartz:
>>>> _Introduction_to_Special_Relativity_, eq 3-22 and problem 3-3.
>>>>
>>>>
>>>>I have often admonished others around here to use the _full_ Lorentz
>>>>transform when doing analyses. Until now in this thread I did not
>>>>follow that advice, and gave wrong answers.
>>>
>>>Where is the justification for the lorentz transform, given the above?
>>
>>The fact that he pointed out that it is not possible to not recognise
>>it as correct.
>
>Why, Evens? Do I have to believe it just because every textbook in the last 100
>years states it?
You are stupuid if you do not accept that things that are seen are
things that occur.
David Evens
[I absolutely efuse to answer any of the questions that demonstrate
why I am wrong.]
That would be typical of people like you.
Paul B. Andersen
> On Sat, 06 Oct 2001 19:40:23 +0200, "Paul B. Andersen" <paul.b....@hia.no>
> wrote:
>
>
>>Henry Wilson wrote:
>>
>>
>>>On Fri, 05 Oct 2001 15:45:00 +0200, "Paul B. Andersen" <paul.b....@hia.no>
>>>wrote:
>>>
>>>
>>>
>>>>Henry Wilson wrote:
>>>Will you also inform him that the SR explanation of the MMX null result is now
>>>known to be completely wrong, as a result of Henry Wilson's revelation that NO
>>>DIAGONAL LIGHT BEAM EXISTS!!!!!!
>>>
>>
>>There sure is no "diagonal light beam" in the SR explanation of the MMX.
>>Why should it be?
>>
>
> SR needs a diagonal light beam TRAVELLING AT C so the its so called prediction
> of length and time contractions will cause the path lengths and traveltimes of
> both rays being the same, independent of velocity.
>
> Since there is no diagonal light ray traveling at c, Sr's theory must be wrong.
Sorry.
I am unable to follow your reasoning, so I have no
idea what you are talking about.
What are you trying to say?
That SR really predicts fringe shifts?
>>>Every text book MMX diagram has been wrong for 130 years!
>>>
>>
>>You are of course referring to the "MMX diagram" explaining Michelson's
>>prediction of the fringe shift. That diagram has of course been correct
>>since Michelson first draw it in 1886. That is, it is a correct description
>>of why Michelson's ether theory predicts a fringe shift.
>>
>
> The diagram has misrepresented the fact, bugger Michelson and what he believed.
>
>>
>>>The diagonal line conventionally drawn represents the trajectory of a separate
>>>dimensionless element of the light beam, in the moving frame. The next element
>>>DOESN'T FOLLOW IT UP THAT SAME DIAGONAL. It moves up the diagonal which is
>>>infinitesimally adjacent!!!
>>>
>>
>>You are stating obvious trivialities, Henry.
>>
>
> Obvious trivialities with far reaching consequences for physics.
> The point is, it doesn't move at c!!!!
THAT is the point, of course:
At what speed do the light elements move.
>>>By what stretch of the imagination would anyone with an IQ greater than that of
>>>an earthworm assign the value c to the velocity of this infinitesimal piece of
>>>nothing as it moves along its own personal diagonal?
>>>
>>>Its velocity is obviously sqrt(c^2+v^2).
>>>
>>
>>Indeed?
>>
>>Nobody with an IQ greater than that of an earthworm can fail
>>to understand that "the speed you assign to light" depend on
>>the theory of physics you use.
>>
>>So which theory did you apply to arrive at the above result?
>>Is this theory tested?
> Yes. I test it every time I drive along the road. The walls remain vertical.
Does this mean that you are unable to give a serious answer, Henry?
The questions were:
1. What assumptions do you make to arrive at the conclusion that
the "light elements" always travel at c in the interferometer frame?
2. What assumptions do you make to arrive at the conclusion that
IF the first conclusion is right, then the speed in the moving
observer frame is sqrt(c^2+v^2)?
3. In which theory do these assumptions belong?
4. Is this theory tested?
>>>What is the maximum speed of the cutting point of a very long pair of scissors,
>>>Paul?
>>>
>>
>>And why do you think that is relevant?
>>But the answer is that the instant speed of the cutting point can
>>be anything, but the average speed (the distance from your fingers
>>to the cutting point divided by the time since you started moving
>>your fingers) can never exceed the speed of sound in the scissors.
>>Now you can chew on that.
>>
>
> OK replace the scissors with two laser beams.
> What is the maximum speed of their point of intersection as they are brought
> into parallel.
I still fail to see the relevance.
But the answer is the same. Substitute speed of light for speed of sound.
The instant speed of the intersection can be anything up to infinite,
but you cannot use this to transport information faster than c.
(Like a scissor cannot transport information faster than sound.)
Paul
Tom Roberts
> Now about these 'individual regions'. By what stretch of the imagination should
> we conclude that their diagonal velocity is always c?
No imagination is needed, measurements have been performed, and they
measure c. See the recently-updated experiments page in the FAQ for a
bunch of references to measurements of the speed of light from moving
sources.
> Where is the justification for the lorentz transform, given the above?
In the fact that theories derived from it agree with experiments. This is,
after all, supposed to be _physics_.
Tom Roberts tjro...@lucent.com
Henry Wilson
wrote:
>Henry Wilson wrote:
>>>>
>>>>>Henry Wilson wrote:
>>>>Will you also inform him that the SR explanation of the MMX null result is now
>>>>known to be completely wrong, as a result of Henry Wilson's revelation that NO
>>>>DIAGONAL LIGHT BEAM EXISTS!!!!!!
>>>>
>>>
>>>There sure is no "diagonal light beam" in the SR explanation of the MMX.
>>>Why should it be?
>>>
>>
>> SR needs a diagonal light beam TRAVELLING AT C so the its so called prediction
>> of length and time contractions will cause the path lengths and traveltimes of
>> both rays being the same, independent of velocity.
>>
>> Since there is no diagonal light ray traveling at c, Sr's theory must be wrong.
>
>
>Sorry.
>I am unable to follow your reasoning, so I have no
>idea what you are talking about.
SR provides a seemingly perfect explanation of the null result. The trouble is,
it requires a huge error of interpretation to achieve this. (that of assuming a
light beam travels diagonally at c)
Doesn't that suggest that SR must also be one huge blunder?
>
>
>What are you trying to say?
>
>That SR really predicts fringe shifts?
NO. Sr predicts no fringe shifts because the same error appears in the theory as
in the MMX interpretation.
>
>
>>>>Every text book MMX diagram has been wrong for 130 years!
>>>>
>>>
>>>You are of course referring to the "MMX diagram" explaining Michelson's
>>>prediction of the fringe shift. That diagram has of course been correct
>>>since Michelson first draw it in 1886. That is, it is a correct description
>>>of why Michelson's ether theory predicts a fringe shift.
>>>
>>
>> The diagram has misrepresented the fact, bugger Michelson and what he believed.
>>
>>>
>>>>The diagonal line conventionally drawn represents the trajectory of a separate
>>>>dimensionless element of the light beam, in the moving frame. The next element
>>>>DOESN'T FOLLOW IT UP THAT SAME DIAGONAL. It moves up the diagonal which is
>>>>infinitesimally adjacent!!!
>>>>
>>>
>>>You are stating obvious trivialities, Henry.
>>>
>>
>> Obvious trivialities with far reaching consequences for physics.
>> The point is, it doesn't move at c!!!!
>
>
>THAT is the point, of course:
>At what speed do the light elements move.
They don't actually move. They are dimensionless points, NOTHING!
A diagonal line will show the path of a dimensionless element of a vertical
light beam (vertical in all frames, by the way).
You could say that such an element will have a 'pseudo-velocity' along that
diagonal of sqrt(c^2+v^2), in the rest frame.
The light elements take the same time to reach a vertical height in all
horizontally moving frames.
>
>
>>>>By what stretch of the imagination would anyone with an IQ greater than that of
>>>>an earthworm assign the value c to the velocity of this infinitesimal piece of
>>>>nothing as it moves along its own personal diagonal?
>>>>
>>>>Its velocity is obviously sqrt(c^2+v^2).
>>>>
>>>
>>>Indeed?
>>>
>>>Nobody with an IQ greater than that of an earthworm can fail
>>>to understand that "the speed you assign to light" depend on
>>>the theory of physics you use.
>>>
>>>So which theory did you apply to arrive at the above result?
>>>Is this theory tested?
>
>> Yes. I test it every time I drive along the road. The walls remain vertical.
>
>
>Does this mean that you are unable to give a serious answer, Henry?
It is a serious answer. Even if I drove at .9c, the walls would still be
vertical in my frame.
They wouldn't actually APPEAR vertical. To determine their normality, I would
have to either use an array of detectors to find their positions or correct for
light travel time.
>
>The questions were:
>1. What assumptions do you make to arrive at the conclusion that
> the "light elements" always travel at c in the interferometer frame?
Good question. My answer has been that it travels at c, only in the direction of
its wave axis of symmetry. That axis remains vertical in all frames.
>2. What assumptions do you make to arrive at the conclusion that
> IF the first conclusion is right, then the speed in the moving
> observer frame is sqrt(c^2+v^2)?
No assumptions. This follows directly from the observation that the beam takes
the same time to reach a particular height, IN ALL FRAMES. Inertial movement of
clocks does not affect their 'real' rates.
>3. In which theory do these assumptions belong?
There are no assumption. Certainly not the existence of an aether.
It is all proven logic.
>4. Is this theory tested?
Thoroughly. It is all around us.
>
>>>>What is the maximum speed of the cutting point of a very long pair of scissors,
>>>>Paul?
>>>>
>>>
>>>And why do you think that is relevant?
>>>But the answer is that the instant speed of the cutting point can
>>>be anything, but the average speed (the distance from your fingers
>>>to the cutting point divided by the time since you started moving
>>>your fingers) can never exceed the speed of sound in the scissors.
>>>Now you can chew on that.
>>>
>>
>> OK replace the scissors with two laser beams.
>> What is the maximum speed of their point of intersection as they are brought
>> into parallel.
>
>
>I still fail to see the relevance.
>But the answer is the same. Substitute speed of light for speed of sound.
>The instant speed of the intersection can be anything up to infinite,
>but you cannot use this to transport information faster than c.
>(Like a scissor cannot transport information faster than sound.)
>
>Paul
Nor can you convey information faster than c, using the diagonally moving
infinitesimal elements of light. In fact they cannot convey information, at all.
They are nothing. Information can only be conveyed in the vertical direction.
Henry Wilson
>Henry Wilson wrote:
>> Now about these 'individual regions'. By what stretch of the imagination should
>> we conclude that their diagonal velocity is always c?
>
>No imagination is needed, measurements have been performed, and they
>measure c. See the recently-updated experiments page in the FAQ for a
>bunch of references to measurements of the speed of light from moving
>sources.
>
That's not quite the same problem though Tom.
Consider this.
When they are adjacent, two observers set up their own arrays of evenly spaced
and presynched clocks. These define their respective rest frames.
. . . . . . |
. . . . . . |
. . . . . . |
. . . . . . |
A B <-v
B and his array then move to the left through A's array, shining a vertical
laser beam.
Do you not agree that A will record the same time on his clocks, as B will, when
the beam reaches a certain height.
This relies on the logically proven fact that clocks don't physically change
rate due to inertial movement. (Otherwise they would have to change in an
infinite number of ways depending on the observer velocity).
You will probably argue that the grid distorts so that it depart from right
angle symmetry. Once again that can be disproved on logical grounds. There can
be no physical alteration to the grid due to velocity alone.
>
>> Where is the justification for the lorentz transform, given the above?
>
>In the fact that theories derived from it agree with experiments. This is,
>after all, supposed to be _physics_.
It is not at all surprising that some results DO appear to support many SR
predictions.
>
>
>Tom Roberts tjro...@lucent.com
Rod Ryker
David Evens wrote:
Rod: ?
Put down that intoxicating object that has caught
your eye and focus on the discussion at hand .
You argue with yourself .
Re-read Tom's and my posts on this thread .
Rod Ryker