Contraction has been abolished by Special Relativity
Peter Riedt
Lorentz published his contraction hypothesis in 1899 in an attempt to
explain the null result of the Michelson-Morley (MMX). He defined the
amount of contraction of moving objects by the formula L'=L*sqrt(1-v^2/
c^2).
Since then there has been a fundamental change in the meaning of v in
the formula of Lorentz. This has effectively invalidated his
explanation of the null result of MMX.
Lorentz: L = absolute length of the object
SR: L = proper length
Lorentz: L' = local coordinate length of the object
SR: L' = length measured in the observer's frame
Lorentz: v = the speed of the object relative to the (absolute) rest
frame of the ether
SR: v = the speed of the object relative to the inertial frame of the
observer
Lorentz: c = speed of light = 300000km/sec
SR: c = speed of light = 300000km/sec
Michelson expected a fringe shift in MMX. When it did not occur,
Lorentz assumed that the parallel axis of the equipment contracted
physically according to the value of v used by Michelson (30km/sec),
producing the null result. SR now says v in the frame of the
experiment (observer co-moving with equipment) is 0, giving a
contraction of 0 (zero, nada, zilch, nix) i.e. L'=L*sqrt(1-0^0/c^2).
So if there is no contraction, what is the reason for the null result?
Peter Riedt
Dirk Van de moortel
96f2d1ea-b145-47c4...@a18g2000pra.googlegroups.com
GOOD GRIEF!
In the frame of the experiment:
proper length of the rod: L
coordinate length of the rod: L
speed of the rod: 0
speed of light: c
so:
<coordinate length> = <proper length> * sqrt(1-v^2/c^2)
i.o.w.
L = L sqrt(1-0)
i.o.w.
L = L
What a surprise!
> So if there is no contraction, what is the reason for the null result?
Perhaps it would be interesting to first try to find out the
meanings of the variables. The way it is now, I estimate
your knowledge about this is 0 (zero, nada, zilch, niks).
Dirk Vdm
Spaceman
Lightspeed is constant to all "at rest" frames only.
The entire experiment was "at rest" to each and every internal frame.
The experiment was doomed to not find anything except a constant
speed.
Use the same experiment with sound and tada... sound is also
constant to all "at rest" frames.
Next.. do it with water waves..
TADA!
same results.
The TimeLord
<rie...@yahoo.co.uk> in
96f2d1ea-b145-47c4...@a18g2000pra.googlegroups.com:
> Contraction has been abolished by Special Relativity
>
> Lorentz published his contraction hypothesis in 1899 in an attempt to
> explain the null result of the Michelson-Morley (MMX). He defined the
> amount of contraction of moving objects by the formula L'=L*sqrt(1-v^2/
> c^2).
>
> Since then there has been a fundamental change in the meaning of v in
> the formula of Lorentz. This has effectively invalidated his explanation
> of the null result of MMX.
>
> Lorentz: L = absolute length of the object SR: L = proper length
>
> Lorentz: L' = local coordinate length of the object SR: L' = length
> measured in the observer's frame
>
> Lorentz: v = the speed of the object relative to the (absolute) rest
> frame of the ether
> SR: v = the speed of the object relative to the inertial frame of the
> observer
>
> Lorentz: c = speed of light = 300000km/sec SR: c = speed of light =
> 300000km/sec
c = 299792.458 km/s exactly
cf
http://physics.nist.gov/cgi-bin/cuu/Value?c
>
> Michelson expected a fringe shift in MMX. When it did not occur, Lorentz
> assumed that the parallel axis of the equipment contracted physically
> according to the value of v used by Michelson (30km/sec), producing the
> null result. SR now says v in the frame of the experiment (observer
> co-moving with equipment) is 0, giving a contraction of 0 (zero, nada,
> zilch, nix) i.e. L'=L*sqrt(1-0^0/c^2).
The last part is not true.
>
> So if there is no contraction, what is the reason for the null result?
SR says that the apparent contraction comes from the fact that what is
simultaneous in one frame can not be simultaneous in another frame that
is moving with respect to the first one. The formula you cited above is
actually
L'=L*sqrt(1-v^2/c^2).
And if v=0 then L'=L. However this is true only if the two ends of L or
L' are observed at the same time, ie dt=dt'=0. When that happens then
the length contraction formula comes out of the Lorentz Transformation.
The null result in the MMX is due to the fact that the interferometer
was not moving with respect to the lab (v=0). Since this was an
intentional feature of the MMX to show the existence of the aether, the
null result proved that the aether does not exist.
--
// The TimeLord says:
// Pogo 2.0 = We have met the aliens, and they are us!
Darwin123
> Contraction has been abolished by Special Relativity
>
> Michelson expected a fringe shift in MMX.
That's right. A fringe SHIFT. The fringes don't disappear either
in Galilean or Lorentzian relativity. What is being measured is a
shift, or lack of same, from some center.
When it did not occur,
> Lorentz assumed that the parallel axis of the equipment contracted
> physically according to the value of v used by Michelson (30km/sec),
> producing the null result.
Lorentz did not "assume" such a shift to explain the MMX result.
He derived the shift based on the electromagnetic force laws known at
the time. He then showed that it would explain certain experiments,
including the MMX experiment. There was another experiment concerning
the torque on an electrical coil that bothered him even more though.
In any case, Lorentz actually did a force analysis. The results
weren't actually a fit the way you suggest.
Fitzgerald hypothesized a change in length to explain the MMX
experiment. Fitzgerald came up with the length contraction that
explains the MMX experiment. However, Fitzgerald did not do a force
analysis like Lorentz did. Fitzgerald did not consider issues of
simultaneity like Einstein did. However, he did come up with a
phenomenological, ad hoc explanation for the null result of the MMX
experiment. It was a formula used to explain the MMX experiment, and
was not generalizable. He had the right formula for the length
contraction though, just no justification. He did it earlier than
either Einstein or Lorentz. Hence it is sometimes called the "Lorentz-
Fitzgerald" length contraction.
It occurs to me that just like you confuse Lorentz with
Einstein, you may be confusing Lorentz with Fitzgerald. Lorentz did a
force analysis. It turns out the that the force that does the
contraction is observer dependent. It is impossible for an
electromagnetic force to not be observer dependent.
>
> So if there is no contraction, what is the reason for the null result?
L'=L in the equipment frame. The distance traveled in each arm is
the same in the equipment frame. The wavelength of light in each arm
is the same. Therefore, there is no shift.
>
Lorentz did not use the MMX experiment to derive the length
contraction. Although the null results of experiment motivated him, he
derived the existence of both time dilation, length contraction, and
simultaneity shift from first principles. He did not make an ad hoc
hypothesis to fit an experiment. He started with Maxwell's equations
and the force laws for electromagnetism.
Lorentz made one ad hoc assumption. He assumed that the only
nonelectrical forces were those that held the electron in a spheroidal
shape. However, the length contraction was already evident in the
known laws of electromagnetism. There was no assumption of "the
constancy of the speed of light." The invariance emerged from the
force laws he used, not the other way around. Why are you ignoring
this?
Peter Riedt
>
> - Show quoted text -
Dirk,
you confirmed the point I was making: L'=L - no contraction. However
without contraction, there should be a phase shift or another reason
for its absence.
Peter Riedt
Peter Riedt
> Am Thu, 25 Sep 2008 06:17:27 -0700 schrieb Peter Riedt
> <rie...@yahoo.co.uk> in
>
>
>
>
>
> > Contraction has been abolished by Special Relativity
>
> > Lorentz published his contraction hypothesis in 1899 in an attempt to
> > explain the null result of the Michelson-Morley (MMX). He defined the
> > amount of contraction of moving objects by the formula L'=L*sqrt(1-v^2/
> > c^2).
>
> > Since then there has been a fundamental change in the meaning of v in
> > the formula of Lorentz. This has effectively invalidated his explanation
> > of the null result of MMX.
>
> > Lorentz: L = absolute length of the object SR: L = proper length
>
> > Lorentz: L' = local coordinate length of the object SR: L' = length
> > measured in the observer's frame
>
> > Lorentz: v = the speed of the object relative to the (absolute) rest
> > frame of the ether
> > SR: v = the speed of the object relative to the inertial frame of the
> > observer
>
> > Lorentz: c = speed of light = 300000km/sec SR: c = speed of light =
> > 300000km/sec
>
> c = 299792.458 km/s exactly
>
>
>
> > Michelson expected a fringe shift in MMX. When it did not occur, Lorentz
> > assumed that the parallel axis of the equipment contracted physically
> > according to the value of v used by Michelson (30km/sec), producing the
> > null result. SR now says v in the frame of the experiment (observer
> > co-moving with equipment) is 0, giving a contraction of 0 (zero, nada,
> > zilch, nix) i.e. L'=L*sqrt(1-0^0/c^2).
>
> The last part is not true.
>
>
>
> > So if there is no contraction, what is the reason for the null result?
>
> SR says that the apparent contraction comes from the fact that what is
> simultaneous in one frame can not be simultaneous in another frame that
> is moving with respect to the first one. The formula you cited above is
> actually
>
> L'=L*sqrt(1-v^2/c^2).
>
> And if v=0 then L'=L. However this is true only if the two ends of L or
> L' are observed at the same time, ie dt=dt'=0. When that happens then
> the length contraction formula comes out of the Lorentz Transformation.
>
> The null result in the MMX is due to the fact that the interferometer
> was not moving with respect to the lab (v=0). Since this was an
> intentional feature of the MMX to show the existence of the aether, the
> null result proved that the aether does not exist.
>
> --
> // The TimeLord says:
>
> - Show quoted text -
Time Lord,
yes, L'=L*sqrt(1-0km/sec^2/299792.458km/s^2). However, the issue is
not if the ether exists or not. The issue is if v=0 and contraction is
0, what is the reason for the absence of the predicted phase shift?
The problem is Lorentz; not MMX.
Peter Riedt
Peter Riedt
Darwin,
I am ignoring most of what you say because it is irrelevant or
splitting hairs or both. The facts are:
Voigt, Larmor, Heaviside, Fitzgerald and Lorentz contributed to the
theory of length contraction and time dilation. Irrelevant. What is
relevant is that the Lorentz transformations if applied to MMX and
using v = 0 do not cause contraction notwithstanding that contraction
is the historic SR explanation for the null result of MMX. If you use
v = 0 in the formula you obtain L' = L, confirmed by Dirk. It is not
L' = L - x. Simple logic: if contraction is absent but necessary to
produce the null result, something is wrong.
Peter Riedt
Dirk Van de moortel
Imbecile.
Dirk Vdm
Peter Riedt
Dirk,
no arguments, only abuse. If I compared the level of your mind to that
of an chimpanzee, I would have to apologize to that animal.
Peter Riedt
Dirk Van de moortel
That's what you live for, so that's what you get.
Don't thank me.
Dirk Vdm
Russell
...
>
> Darwin,
> I am ignoring most of what you say because it is irrelevant or
> splitting hairs or both. The facts are:
> Voigt, Larmor, Heaviside, Fitzgerald and Lorentz contributed to the
> theory of length contraction and time dilation. Irrelevant. What is
> relevant is that the Lorentz transformations if applied to MMX and
> using v = 0 do not cause contraction notwithstanding that contraction
> is the historic SR explanation for the null result of MMX.
Exactly wrong. SR is the theory that *did away* with
the need for contraction to explain MMX; indeed SR became
famous ("historic") for exactly that reason. No ether in
SR, hence no need for any gimmick to explain why the ether
is not observed.
If you disagree, you are welcome to cite here any text,
historic or otherwise, that might support your claim.
Good luck with that.
rbwinn
>
> - Show quoted text -
The Galilean transformation equations show a phase shift.
x'=x-vt
y'=y
z'=z
t'=t
If the speed of light measures c in both frames of reference, then the
velocity of light directed opposite to the motion of S' is -c relative
to both frames of reference.
w=velocity of light
x=wt
x'=wn'
x'=x-vt
wn'=wt-vt
n'=t(1-v/w)
n' is the time it takes a photon to travel a distance of x'. If a
photon has a velocity of c relative to both frames of reference, then
n'=t(1-v/c)
If a photon has a velocity of -c relative to both frames of reference,
then
n'=t(1-v/(-c)= t(1+v/c)
What this means is that if light is emitted simultaneously at two
points in S, then photons from each point will meet halfway between
the two points in S.
What happens in S'?
Exactly the same thing. The light is also emitted simultaneously in
S', and the light meets in S' halfway between the two points where the
light was emitted in S'. That means there is a phase shift. If you
try to do anything else, then you have relativity of simultaneity,
which means there has to be a length contraction.
n' is t' as it relates to a photon traveling a distance of x'. There
is a relativity of time as it relates to photons in each frame of
reference, but, other than that, I do not see any difference in times,
meaning that any difference in clock rates would depend on
gravitation, and without gravitation, which was how Einstein defined
Special Relativity, a clock in S' would read the same as a clock in S.
Robert B. Winn
rbwinn
SR has a length contraction just the same as Lorentz's ether theory.
The only difference is the amount of contraction. Both theories are
convoluted efforts to make experimental results match pre-conceived
ideas of scientists.
Robert B. Winn
PD
No, that's not right, Peter. There would only be a phase shift if c
were not constant. The fact that c is constant completely accounts for
the lack of the phase shift.
PD
Dono
<idiocies snipped>
> So if there is no contraction, what is the reason for the null result?
>
> Peter Riedt
Peter,
Read here: http://www.savefile.com/files/1774004
I wrote it for all you imbeciles deniers. I would not be surprised if
you continued arguing after you read the explanation. This is not
because the explanation is not correct, it is because you and your ilk
are too stupid to understand it. Surprise me !
harry
"Peter Riedt" <rie...@yahoo.co.uk> wrote in message
news:96f2d1ea-b145-47c4...@a18g2000pra.googlegroups.com...
> Contraction has been abolished by Special Relativity
>
> Lorentz published his contraction hypothesis in 1899 in an attempt to
> explain the null result of the Michelson-Morley (MMX). He defined the
> amount of contraction of moving objects by the formula L'=L*sqrt(1-v^2/
> c^2).
>
> Since then there has been a fundamental change in the meaning of v in
> the formula of Lorentz. This has effectively invalidated his
> explanation of the null result of MMX.
Certainly not: after 1905, he and quite some others stuck to that
interpretation. See for example
http://en.wikipedia.org/wiki/Kennedy-Thorndike_experiment
> Lorentz: L = absolute length of the object
> SR: L = proper length
Proper length in the Newtonian coordinate system that is chosen to be "in
rest" - exactly as done in Newtonian mechanics.
Note: the same shift in meaning existed from the very start in Newtonian
mechanics. Please study that first!
> Lorentz: L' = local coordinate length of the object
> SR: L' = length measured in the observer's frame
L' = moving length as measured relative to either the ether or the "rest
system".
> Lorentz: v = the speed of the object relative to the (absolute) rest
> frame of the ether
> SR: v = the speed of the object relative to the inertial frame of the
> observer
Of the chosen Newtonian "rest system"...
> Lorentz: c = speed of light = 300000km/sec
> SR: c = speed of light = 300000km/sec
Lorentz: c = relative to the ether
SR: c = return speed, relative to the chosen coordinate system; and one-way
speed by synchronization convention.
> Michelson expected a fringe shift in MMX. When it did not occur,
> Lorentz assumed that the parallel axis of the equipment contracted
> physically according to the value of v used by Michelson (30km/sec),
> producing the null result.
30 km/s was the *difference* in speed delta_v; v was guessed at several 100
km/s.
> SR now says v in the frame of the
> experiment (observer co-moving with equipment) is 0, giving a
> contraction of 0 (zero, nada, zilch, nix) i.e. L'=L*sqrt(1-0^0/c^2).
Not SR but some *people* say such things. SR instead says that the observer
may choose *any* inertial coordinate system that is convenient (again, just
as in classical mechanics), so that (apparent) contraction is relative in
the same way as (apparent) momentum is relative. However, SR (and classical
mechanics) *always* said that v of an object that is co-moving with a chosen
coordinate system relative to that same system is zero - it can't be
otherwise! Similarly, "proper" length already implies that the length is
compared to a "local" standard, i.e. in the same state of motion.
> So if there is no contraction, what is the reason for the null result?
I would not know a plausible alternative, but you can count on a mixed bag
of interesting attempts by anti-contraction people. ;-)
Cheers,
Harald
Russell
> On Sep 26, 5:10 am, Russell <rblacka...@yahoo.com> wrote:
> > On Sep 25, 8:44 pm, Peter Riedt <rie...@yahoo.co.uk> wrote:
> > ...
>
> > > Darwin,
> > > I am ignoring most of what you say because it is irrelevant or
> > > splitting hairs or both. The facts are:
> > > Voigt, Larmor, Heaviside, Fitzgerald and Lorentz contributed to the
> > > theory of length contraction and time dilation. Irrelevant. What is
> > > relevant is that the Lorentz transformations if applied to MMX and
> > > using v = 0 do not cause contraction notwithstanding that contraction
> > > is the historic SR explanation for the null result of MMX.
>
> > ^^^^^^^^^^^^^^^^^^^^^^^^^^^
>
> > Exactly wrong. SR is the theory that *did away* with
> > the need for contraction to explain MMX; indeed SR became
> > famous ("historic") for exactly that reason. No ether in
> > SR, hence no need for any gimmick to explain why the ether
> > is not observed.
>
> > If you disagree, you are welcome to cite here any text,
> > historic or otherwise, that might support your claim.
> > Good luck with that.
>
> SR has a length contraction just the same as Lorentz's ether theory.
True enough; the two theories give identical predictions
for measurements taken on moving objects.
But you seem to have missed the point of this thread, such
as it is. Peter Riedt is exercised over the (shocking!)
news that in SR, an interferometer traveling at v=0 in the
lab frame is predicted to have no contraction in that frame.
(Lorentz theory, of course, gives the same prediction, for
what is *measured*). Somehow Reidt thinks this is a problem
for SR's explanation of the MMX, but he is grossly confused.
> The only difference is the amount of contraction.
No, there's no difference in the predictions, only in
the reasoning behind those predictions. It is Lorentz
theory, *not* SR, that posits a physical but unmeasurable contraction
(in the ether frame) as the reason for the
MMX result.
Both theories are
> convoluted efforts to make experimental results match pre-conceived
> ideas of scientists.
This seems to be the day for getting things backwards.
Experimental results are what they are -- no change to
any theory can change them. The trick is to come up with
a theory that is not in conflict with experiments, either
those of the past or those suggested to be performed in
the future. Ideally, you want the *least* convoluted
theory with this property. And currently, SR (in its
applicable domain) fills the bill. It's admirable that
you seek to do better, but frankly, sooner or later you
really ought to admit that you're not up to the task.
Eric Gisse
Show us.
[snip]
Darwin123
> On Sep 26, 1:26 am, The TimeLord <math-n-physics-...@att.com> wrote:
>
>
>
> > Am Thu, 25 Sep 2008 06:17:27 -0700 schrieb Peter Riedt
> > <rie...@yahoo.co.uk> in
> > 96f2d1ea-b145-47c4-823d-682f4bbd3...@a18g2000pra.googlegroups.com:
>
> The problem is Lorentz; not MMX.
>
> Peter Riedt
I will provide a long rambling essay to explain, yet again, what
Lorentz had done. Lorentz is relevant only because he gives a specific
example, all details included, of a system where the dichotomy that
bothers you is unavoidable. The Lorentz model is a counterexample to
your statements about relativity being nebulous. Here is a specific
system with very specific answers, but the dichotomy is there and
unavoidable scientifically.
Lorentz assumed that the "aether" had a certain velocity
relative to the equipment. While the speed of light is isotropic to an
observer with respect to this frame, the relative velocity between a
moving object and light is not isotropic. The relative speed of light,
as measured from this hypothetical observer, is not the same in all
directions. Note that the rulers in the aether frame of reference are
not compressed, and the clocks are not slowed down for this aether-
stationary observer. His analysis was tedious but completely
classical. Very little unknown physics was used in his analysis.
There was a nonelectrical force in the electron that he somehow
attributed to aether wind, but there wasn't even any direct connection
between this force and the aether wind. This force was ad hoc, but not
really strange. There was no a priori assumption about the nature of
space and time. H.A. Lorentz was doing what could be called an
engineering problem for systems moving at high speed. This force is
not a complete description of the electron, because real electrons
obey quantum mechanics as well as relativity. However, this force
could very well have explained most of the electrical experiments done
to the date of publication. I suspect that this spheroidal electron
may still may be a good approximation of the 1s state in hydrogen.
Whether it is good all the time is irrelevant. Lorentz's hypothetical
system is merely an example of the type of system that obeys
relativity as described by Einstein.
An interesting consequence of his analysis is that the rulers
moving at any velocity relative to the aether experience a
foreshortening, and the clocks moving at any velocity relative to the
aether experience a slowing down. Constancy of speed of light was not
a priori assumed in his analysis.
An interesting but unavoidable dichotomy dropped from this very
specific analysis. The fundamental problem was in working the problem
backward. Assuming all the clocks experienced both time dilation and
calibration using signals of finite speed, and assuming all the rulers
experienced this compression, what would the objects in the aether
frame of reference look like to the object in motion? The answer is
one you haven't been able to accept.
The objects in motion will observe the same type of length
contraction and time dilation from objects that are stationary with
respect to the aether. As long as their motion is uniform, they have
to observe the exact same dilation. If they try to measure the speed
of light from their position using their "foreshortened" and "slowed"
instruments, they will measure an isotropic and constant speed of
light. What the "aether moving" objects will observe from the "aether
frozen" objects is exactly that which the "aether" frozen" objects
will observe from the "aether moving objects." Assuming that the
fastest moving signals in their world move at the speeed of light,
there is no possible way the "aether moving objects" can tell they are
moving. However, by the same logic, there is no way the "aether
frozen" objects can tell that they really are "aether frozen."
The reason an "aether frozen" observer gives for the null result
is that the arms have a different length. To the aether frozen
observer, light travels isotropically in his coordinates. However, the
relative speed between equipment and light is not c as far as the
aether frozen observer is concerned. The speed of light as measured
relative to the aether frozen observer is c, and is always c. However,
the relative velocity between equipment and light does not have to be
c.
This should explain why Androcles tiresome quotation is taken
out of context. The "c+v" and "c-v" are relative velocities as
measured from another frame. Since v is different in each frame, the "c
+v" and "c-v" are frame dependent. They are not the speed of light as
measured from the inertial frame doeing the measurements. But I
digress.
The problem is that if the "aether moving" observer tries to do
the same measurements, he is going to be biased by the speed of signal
propagation. Well, maybe not biased. You see, his point of view is
just as valid as the aether stationary point of view. There is no way
to do an "unbiased" measurement the way you want it. Not in this
specific system described by Lorentz.
No where was there any a priori assumption made about the shape
of space. The treatment is consistent to as large an extent possible
with both Newton and Maxwell. Of course, there is a hidden
contradiction between Newton and Maxwell (look carefully, you'll find
it).
The only thing that makes Lorentz's theory "not general" is that
peculiar force that holds the electron together. That has turned out
to be wrong. However, the Lorentz transform he derived using that
force is always right as far as has been determined. Lorentz didn't
generalize his theory right. Einstein got most of the credit. In the
subject of theory, the fellow who generalizes best often gets the
credit. But the guy who comes up with specific examples shouldn't be
ignored (in my opinion).
To sum up: The twin paradox is not a real paradox. It should be
called the twin dichotomy. If you want to have a concrete idea of how
this dichotomy works, you should study H. A. Lorentz. The rest of us
will study special relativity and general relativity in the wider
universe.
Dono
Very nice !
Spaceman
> To sum up: The twin paradox is not a real paradox.
There goes the bullshit dump.
It is not a paradox even though two different times
of "away" time have been determined.
You truly are a wonderous joke to science when you
state such crap like a paradox is not really a paradox.
Sheesh.
---
The stupid ass MMX can be simply explained by stating...
All the components were at rest to each other so they
were doomed to not find any differences in anything.
The twins paradox can simple be debunked by stating...
Both twins are the same revolutions of The Earth wrt the Sun old.
The traveling clock malfunctioned according to the Entire Universe.
---
So it takes six lines to laugh at your long winded twisted joke to science
:)
--
James M Driscoll Jr
Creator of the Clock Malfunction Theory
Spaceman
Peter Riedt
wrote:
> A few glitches to point out:
>
> "Peter Riedt" <rie...@yahoo.co.uk> wrote in message
>
> news:96f2d1ea-b145-47c4...@a18g2000pra.googlegroups.com...
>
> > Contraction has been abolished by Special Relativity
>
> > Lorentz published his contraction hypothesis in 1899 in an attempt to
> > explain the null result of the Michelson-Morley (MMX). He defined the
> > amount of contraction of moving objects by the formula L'=L*sqrt(1-v^2/
> > c^2).
>
> > Since then there has been a fundamental change in the meaning of v in
> > the formula of Lorentz. This has effectively invalidated his
> > explanation of the null result of MMX.
>
> Certainly not: after 1905, he and quite some others stuck to that
Harry,
you are contradicting youself:
1. "However, SR (and classical mechanics) *always* said that v of an
object that is co-moving with a chosen coordinate system relative to
that same system is zero". That means L' = L.
2. "I would not know a plausible alternative" (to contraction). That
means L' NE L.
The same inconsistency is symptomatic of SR - 'it does and it does not
contract!'
Peter Riedt
Peter Riedt
>
> - Show quoted text -
PD,
your assertion that constant c proves lack of the phase shift is
equivalent to saying the existence of the universe proves the
existence of god. Both are beliefs, not physics.
Peter Riedt
rbwinn
>
> - Show quoted text -
No one has given me a reason to make such an admission. It seems to
me that the Galilean transformation equations give just as good an
explanation as the Lorentz equations. If light is measured to be
going at c=300,000 km/sec. in any frame of reference, then if a photon
goes a distance of x', the time it takes for a photon to travel that
distance is
n'=x'/w = t(1-v/w)
where w is the velocity of the photon. If the photon is going in the
positive direction relative to the x and x' axis, then the photon has
a velocity of c relative to S and S', but if it is going in the
negative direction, then its velocity is -c relative to the two frames
of reference.
This does away with the length contraction altogether. My
understanding of Lorentz's interpretation of the length contraction
was that he said that the arm of the interferometer was shortened in
the direction of motion of the interferometer, which was said to be
moving through ether. Consequently, that would mean that the moving
frame of reference was contracted, not the ether frame of reference.
Einstein's interpretation was that from S, S' was contracted, while
from S', S was contracted.
I do not believe in a length contraction, but I believe the
Galilean transformation equations show that light can be measured to
be traveling at c in S or S' without a length contraction.
Robert B. Winn
harry
Instead I stressed there (but not loud enough?) that L' = L' and L = L. As
this is as obvious as one = one, I added that it can't be otherwise. I even
repeated it in other words as follows: " "proper" length already implies
that the length is compared to a "local" standard, i.e. in the same state of
motion."
I also emphasized that the handy use of co-moving frames is not any
different in Newtonian mechanics, so that you indirectly claim that
classical mechanics is self-contradictory.
: 2. "I would not know a plausible alternative" (to contraction). That
: means L' NE L.
You mean that the LT are valid? Indeed, that is very plausible.
: The same inconsistency is symptomatic of SR - 'it does and it does not
contract!'
It depends on relative to what, as you yourself indicated. As no absolute
motion can be measured, we can only talk about apparent (or "relative")
motion. Newtonian mechanics is a big help in understanding these issues, as
he discussed them in detail: they were even at the basis for his work.
For example, does the earth move in an ellips or not? According to Newton,
it does and it doesn't!
Harald
PD
I didn't say constant c proves the lack of phase shift. There is no
proof necessary for lack of phase shift, since it is an observational
fact -- one doesn't have to prove observational facts that are plainly
evident. Perhaps you meant that lack of phase shift doesn't prove
constancy of c, and that is correct. No single experimental result
proves any theory, and in fact theories are never *proven* even by the
collective body of experimental evidence from many, many experiments
(as is the case with relativity). But *proving* a theory is not the
aim of science. In science, the (unproven) model that is most
successful in accurately predicting the largest number of different
experimental observations wins.
PD
Russell
> On Sep 26, 11:15 am, Russell <rblacka...@yahoo.com> wrote:
> > This seems to be the day for getting things backwards.
> > Experimental results are what they are -- no change to
> > any theory can change them. The trick is to come up with
> > a theory that is not in conflict with experiments, either
> > those of the past or those suggested to be performed in
> > the future. Ideally, you want the *least* convoluted
> > theory with this property. And currently, SR (in its
> > applicable domain) fills the bill. It's admirable that
> > you seek to do better, but frankly, sooner or later you
> > really ought to admit that you're not up to the task.- Hide quoted text -
>
> > - Show quoted text -
>
> No one has given me a reason to make such an admission.
It's not something someone else can do for you. You have
to come to that realization yourself. It will take a bit
of self-awareness and courage.
It seems to
> me that the Galilean transformation equations give just as good an
> explanation as the Lorentz equations.
But the two are incompatible! So they can't be "just
as good" as each other. (Except in a trivial sense, if
*both* turned out to be wrong.)
If light is measured to be
> going at c=300,000 km/sec. in any frame of reference,
Impossible, if the Galilean transformation holds.
Velocities are *additive* under the Galilean transformation.
E.g. if a light beam is moving at c in some frame S, and
frame S is moving at speed c/2 in the same direction relative
to frame S', then the Galilean transformation gives (3/2)c for
the speed of that same beam in S'. Not c.
then if a photon
> goes a distance of x', the time it takes for a photon to travel that
> distance is
>
> n'=x'/w = t(1-v/w)
>
> where w is the velocity of the photon. If the photon is going in the
> positive direction relative to the x and x' axis, then the photon has
> a velocity of c relative to S and S', but if it is going in the
> negative direction, then its velocity is -c relative to the two frames
> of reference.
Again, that last sentence is incompatible with the
Galilean transformation, unless S and S' happen to be
the same frame. I really can't figure out what you're
trying to do in the rest of all that.
...
rbwinn
>
Well, S' is not the same frame as S. S' is moving with a velocity of
v relative to S. Light is emitted on the x axis of S at two points
equal distances from the origin of S when the origin of S' is at the
origin of S.
So where is the light relative to the origin of S'?
Well, since the origin of S' is at the origin of S when the light is
emitted, the two points where light is emitted are equal distances
from the origin of S'.
Where will the light emitted on one side of the origin on the x' axis
of S' meet light emitted on the other side of the origin of S' on the
x' axis as seen in S'?
It will meet at the origin of S'.
Now scientists do not believe that, so you might want to go talk to
the other scientists about what scientists believe.
Robert B. Winn
Russell
Ok, that would be the case if Galilean relativity applied.
(Just for the record, SR says that in S' the emissions are
not simultaneous and are not equidistant from the origin of
S' in that frame.)
> Where will the light emitted on one side of the origin on the x' axis
> of S' meet light emitted on the other side of the origin of S' on the
> x' axis as seen in S'?
> It will meet at the origin of S'.
That would be the case if light traveled at speed c in
frame S'. But according to the Galilean transformation,
if we take as given that light speed is c in frame S, then
the beam must travel at c+v in one direction and c-v in the
other. So, your reasoning is wrong here no matter which
transformation you use.
> Now scientists do not believe that, so you might want to go talk to
> the other scientists about what scientists believe.
I find it odd that you think the same two beams will come
together at two different places. (The origin of S will not
be at the same place as the origin of S' when the beams
meet.)
I have to admit I am not up to the task of educating you.
From now on, I'll leave that to others.
rbwinn
>
> - Show quoted text -
Well, there you have it. Some people believe that light is energy.
Scientists believe that light is particles or some other kind of
material. I would compare scientists of today with the scientists
that Michael Faraday was working for as a lab assistant. The had an
electrical current going through a wire, and a magnetic compass would
point to the wire. Faraday told them that there must be an
electromagnetic field around the wire. They told him that all
scientists were agreed that the electricity was contained in the wire
and did not extend past the surface of the wire.
As I said before, you will probably want to talk to other
scientists about what scientists believe.
Robert B. Winn
Eric Gisse
Yes robert, we get it. You hate scientists. Now would shut the fuck up
about it please?
Darwin123
You do realize that if the speed of light relative to the equipment is
isotropic, L=L' automatically produces a null result. In the frame of
the center of mass of equipment, the speed of light relative to the
equipment is isotropic. The reason that a contraction was necessary is
that in the "aether frame," or in the center of earth frame, the speed
of light relative to the equipment is not isotropic. That is when we
need
L'-L(1-v^2/c^2)^0.5
to produce a null result.
Read "The Theory of Electrons" by H. A. Lorentz. Go through the
math. Tell me how he defines a frame. I expect to hear from you having
fully understood the book in about ten years.
rbwinn
>
> - Show quoted text -
Profanity is the attempt of a weak mind to make a strong statement.
Robert B. Winn
Peter Riedt
wrote:
>
> - Show quoted text -
Harry,
very good but my point is the inconsistency of SR. SR spports the
contraction formula which in the thinking of Lorentz was the
mathematical explanation of the null result of MMX. However, the
meaning of v in the formula has changed. Lorentz used the value of
Michelson (v = 30km/sec against the ether) but SR assumes a value of
zero which gives a contraction of zero. So there is no contraction of
the parallel arm of the interferometer in MMX according to Lorentz's
formula and its interpretation by today's SR. However, SR still
continues to apply the Lorentz transforms in the original sense. It
makes no sense. Contraction and SR are dead.
Peter Riedt
Eric Gisse
No response other than the canned phrase, robert? Thanks for shutting
the fuck up like I requested.
rbwinn
Russell
...
> Harry,
> very good but my point is the inconsistency of SR. SR spports the
> contraction formula which in the thinking of Lorentz was the
> mathematical explanation of the null result of MMX.
...because he *also* assumed the existence of an ether
that made light propagation anisotropic in the Earth
frame. The two assumptions go *together* in Lorentz
theory. Einstein didn't throw out just one of them, he
threw out both.
However, the
> meaning of v in the formula has changed. Lorentz used the value of
> Michelson (v = 30km/sec against the ether) but SR assumes a value of
> zero which gives a contraction of zero. So there is no contraction of
> the parallel arm of the interferometer in MMX according to Lorentz's
> formula and its interpretation by today's SR. However, SR still
> continues to apply the Lorentz transforms in the original sense. It
> makes no sense. Contraction and SR are dead.
Your argument stands or falls on your phrase "in the original
sense". How, in your opinion, is the absence of ether in SR
not a change of "the original sense"?
rbwinn
The difficulty encountered is shown by Einstein's train and lightning
problem. If the front of the train is x'=1, and the rear of the train
if x'=(-1) in S', the frame of reference of the train, then suppose
that lightning strikes the front and rear of the train instead of some
distance ahead of and behind the train as described by Einstein.
If the bolts of lightning are simultaneous from the frame of
reference of the track, then supposedly you have a little short train
as observed by an observer at x=0, the origin of S. So if the
lightning leaves marks on the front and rear of the train and marks on
the railroad track, how far apart are the marks on the railroad track?
The problem is compounded further by the fact that relativity of
simultaneity says that in the frame of reference of the train, the
lightning strikes first at the front of the train, meaning that the
train travels some distance after the front mark is made on the track
before the second mark is made as seen by an observer on the train.
So according to SR, an observer by the track sees the lightning strike
both ends of a short train simultaneously, leaving marks on the track
a distance of
L sqrt(1-v^2/c^2) apart. An observer on the train sees the bolt of
lightning at the front of a normal length train strike first, then the
train travels some distance before the other bolt of lightning
strikes, leaving two marks on the track a distance of L sqrt(1-v^2/
c^2) apart.
If the bolts of lightning are simultaneous in the frame of
reference of the train, then to an observer on the ground, the bolt of
lightning at the back of the train strikes first, etc., and the marks
on the track are further apart than the length of the train.
I decided long ago that this was all nonsense, as Peter Reidt is
trying to do, and that the correct equations were the Galilean
transformation equations.
Robert B. Winn
bsr...@my-deja.com
What he obviously meant was SR must be able to explain the observed
result or the theory is wrong. Some here have said there is no real
contraction taking place in SR.
Take the MMX and put it on a train. Have the train accelerate up to
the velocity v. Measured in the track frame the closing speed between
light and the train is no longer isotropic. If the length of both
arms are still L then there should be a fringe shift due to the speed
of light no longer being isotropic relative to the apperatus. Perhaps
someone who claims there is no real contraction taking place would
like to explain how a null result is arrived at without a
contraction. Explaining it from the train frame is not allowed, do it
from the track frame :)
To Peter, SR did not do away with contraction. A moving opject will
be measured to be shorter in its direction of motion than it was when
at rest.
As for the meaning of v changing, Lorentz showed in his 1904 paper
that if you translate from one moving frame to the aether frame, and
then from the aether frame to a second moving frame, the result is the
same as if you had just considered the first frame to be the aether
frame. From this he concluded there was no way to identify the aether
frame. So the v in his transformation is the same as the v in SR.
Bruce
rbwinn
>
> - Show quoted text -
What Lorentz did is what scientists still do today. They flip frames
of reference any time they cannot explain something.
Robert B. Winn
bsr...@my-deja.com
Put the MMX on a train moving at v while we observe from the track
frame. We observe speed of light is not isotropic relative to the
equipment now. Please explain the null result from the track frame
using SR and no contraction.
Bruce
Russell
promise to respond further in this thread -- nevertheless I
did read your posting (quoted below) and have a few comments
to make.
On Sep 28, 5:18 am, rbwinn <rbwi...@juno.com> wrote:
...
> The difficulty encountered is shown by Einstein's train and lightning
> problem. If the front of the train is x'=1, and the rear of the train
> if x'=(-1) in S', the frame of reference of the train, then suppose
> that lightning strikes the front and rear of the train instead of some
> distance ahead of and behind the train as described by Einstein.
For the record, in Einstein's gedanken the strikes were
*at* the front and back of the train, same as in yours.
> If the bolts of lightning are simultaneous from the frame of
> reference of the track, then supposedly you have a little short train
> as observed by an observer at x=0, the origin of S. So if the
> lightning leaves marks on the front and rear of the train and marks on
> the railroad track, how far apart are the marks on the railroad track?
The length of the "little short train", of course. I don't
see what the "problem" is, that you refer to in the following:
> The problem is compounded further by the fact that relativity of
> simultaneity says that in the frame of reference of the train, the
> lightning strikes first at the front of the train, meaning that the
> train travels some distance after the front mark is made on the track
> before the second mark is made as seen by an observer on the train.
Correct, well done. Exactly what "problem" does this "compound"?
> So according to SR, an observer by the track sees the lightning strike
> both ends of a short train simultaneously, leaving marks on the track
> a distance of
> L sqrt(1-v^2/c^2) apart.
Yes, where L is the length of the train in S', namely 2
using the example you gave above.
An observer on the train sees the bolt of
> lightning at the front of a normal length train strike first, then the
> train travels some distance
Strictly speaking, it is of course the *track* that travels
in this frame.
before the other bolt of lightning
> strikes, leaving two marks on the track a distance of L sqrt(1-v^2/
> c^2) apart.
That's their distance apart in the track frame; but of course
in the train observer's frame the distance is contracted by
an additional factor 1/gamma, i.e. it is L(1-v^2/c^2).
> If the bolts of lightning are simultaneous in the frame of
> reference of the train, then to an observer on the ground, the bolt of
> lightning at the back of the train strikes first, etc., and the marks
> on the track are further apart than the length of the train.
Yes, that's what SR says.
> I decided long ago that this was all nonsense, as Peter Reidt is
> trying to do, and that the correct equations were the Galilean
> transformation equations.
Take care, if you "decide" something like that, that you
don't close your mind completely to (at least consideration
of) other points of view, or to well-reasoned critiques of
your own position. I note that you never addressed the
substantive point of my earlier post, that contrary to what
you claimed, the same light beam *cannot* have the same
velocity in two different frames under the Galilean
transformation. Do you *not* have an answer?
Might I add that IMO you seem to have learned one or two
things in the years that you have been posting here --
bravo for that -- and so, perhaps a decision taken long
ago might be profitably reexamined now. Just a thought.
Russell
an excellent post, but some comments...
> What he obviously meant was SR must be able to explain the observed
> result or the theory is wrong. Some here have said there is no real
> contraction taking place in SR.
In fairness, the discussion here was entirely in the context
of the frame of the MMX interferometer. In SR there is no
contraction (real or otherwise) of the interferometer in
*that* frame.
That said, I admit that that point is easily missed given
the title of the thread, and apparently *is* being missed by
Reidt.
>
> Take the MMX and put it on a train. Have the train accelerate up to
> the velocity v. Measured in the track frame the closing speed between
> light and the train is no longer isotropic. If the length of both
> arms are still L then there should be a fringe shift due to the speed
> of light no longer being isotropic relative to the apperatus. Perhaps
> someone who claims there is no real contraction taking place would
> like to explain how a null result is arrived at without a
> contraction. Explaining it from the train frame is not allowed, do it
> from the track frame :)
Sure, there is "real" contraction in the sense that it has
consequences on, say, the number of meter sticks you have
to lay down along the track to measure the parallel arm.
But that number depends as well on how one synchronizes the
two clocks (one at the front and one at the back) needed
for marking the arm's position on the track at a given time.
In other words, it's coordinate dependent. IMO it's a matter
of taste whether one's definition of "real" includes such
quantities or does not. To argue over it is pointless; and
to avoid provoking such arguments it's probably best to avoid
the term if one can.
>
> To Peter, SR did not do away with contraction. A moving opject will
> be measured to be shorter in its direction of motion than it was when
> at rest.
>
> As for the meaning of v changing, Lorentz showed in his 1904 paper
> that if you translate from one moving frame to the aether frame, and
> then from the aether frame to a second moving frame, the result is the
> same as if you had just considered the first frame to be the aether
> frame. From this he concluded there was no way to identify the aether
> frame. So the v in his transformation is the same as the v in SR.
Exactly. I thought of making that point myself but could not
find such a felicitous way of putting it. Thanks.
Spaceman
> To Peter, SR did not do away with contraction. A moving opject will
> be measured to be shorter in its direction of motion than it was when
> at rest.
SR incorrectly uses a constant lightspeed ot all to come
to this conclusion.
They have ignored the relative motion across the waves and
used the percieved wavelength instead of a physical wavelength
to determine the "relative" speed.
Using the constant speed of light gives them only one
conclusion to the "shorter" waves, and of course that studid
conclusion is that the "waves" shrunk physically even though
they did not and the fact is you simplly percieved them
to shrink when moving across them in a non "at rest" condition
of the observer.
A moving object is not actually measured shorter and of course
the distance it is traveling does not "shrink standards" either.
Darwin123
wrote:
> Put the MMX on a train moving at v while we observe from the track
> frame. We observe speed of light is not isotropic relative to the
> equipment now. Please explain the null result from the track frame
> using SR and no contraction.
>
> Bruce
From the track frame, there is no explanation without contraction. As
measured from the track frame, one arm of the MMX equipment has
shrunk. The explanation for the null result in the track frame has to
include both length contraction and time dilation.
Special relativity (SR) does not specify the type of forces that
cause this contraction of equipment. SR does place a constraint on the
type of forces holding the equipment together. To get more detail
about the forces, one needs a specific model. Now a days, we would use
a model that includes quantum mechanics. However, quantum mechanics
was not available in the days of H. A. Lorentz. Lorentz only had
Newtons Laws and Maxwell's equations.
If the individual atoms that make up the equipment satisfy all the
conditions in the Lorentz model, the physical reason for the
contraction is obvious from the reference frame centered on the track.
The velocity dependent forces that hold the atoms of the equipment
together caused a stress that shrunk the arm.
If the individual atoms that make up the equipment satisfy all
the conditions in the Lorentz model, the physical reason for the
contraction is obvious from the reference frame centered on the MMX
equipment. The equipment is standing still in this frame. There were
no velocity dependent forces that could cause a stress that shrunk the
arm.
There is no need to provide the "same" explanation in both the
track and the equipment frames. Part of this is there is a dichotomy
built right into Maxwell's equations between magnetic force and
electrical force. For a moving charge, the electromagnetic force on a
moving particle is caused by an electrical field in one frame and a
magnetic force in another frame. The physical force is unambiguous,
but the cause of the force is slightly ambiguous. This is an
unavoidable consequence of the Maxwell's equations.
There may really be an aether involved in all this. However, the
aether provides only "hidden variables" to the Lorentz model. Sure,
Lorentz did a priori assume the existence of an aether. However, his
analysis showed that the velocity of the aether is unmeasurable in his
model. The velocity of the aether can be anything, and the null result
will still be there. The track frame scientists can blame on a aether
wind for the length contraction. The equipment scientists can deny
that there is a length contraction, and deny there is an aether wind.
The aether is therefore redundant, unnecessary, undetectable,
unphysical or any other word you may wish to use on something
impossible to characterize. The idea of an aether may still be useful
in certain contexts, but it does not provide a useful alternative to
special relativity.
Thus, Lorentz had a model that explained in detail how the arms of
the MMX experiment behaved. It was fully consistent with SR, but
included many assumptions that we would consider slightly wrong today.
rbwinn
Well, that is why I said the problem is compounded. Scientists do not
even seem to agree. The distance being contracted from the frame of
reference of the train is the length of the track. What you just said
is the opposite of what Peter Riedt is trying to do. Peter is trying
to get back to no distance contraction at all. You are trying to
compound it to agree with what the observer on the ground is said to
see, L sqrt(1-v^2/c^2). But I see what Peter is saying. If the
length of the track is contracted from the frame of reference of the
train, then the train is occupying a length of track longer than the
length of the train. So the lightning at the front strikes first and
the lightning at the rear of the train second, putting the marks on
the track the length of the train apart as seen from the frame of
reference of the contracted track.
But scientists already said that the train was shorter, so they agree
with you.
>
> > � � If the bolts of lightning are simultaneous in the frame of
> > reference of the train, then to an observer on the ground, the bolt of
> > lightning at the back of the train strikes first, etc., and the marks
> > on the track are further apart than the length of the train.
>
> Yes, that's what SR says.
>
> > � � I decided long ago that this was all nonsense, as Peter Reidt is
> > trying to do, and that the correct equations were the Galilean
> > transformation equations.
>
> Take care, if you "decide" something like that, that you
> don't close your mind completely to (at least consideration
> of) other points of view, or to well-reasoned critiques of
> your own position. �I note that you never addressed the
> substantive point of my earlier post, that contrary to what
> you claimed, the same light beam *cannot* have the same
> velocity in two different frames under the Galilean
> transformation. �Do you *not* have an answer?
Well, yes I did have an answer. I had equations for it. If light is
traveling at c in both frames of reference, then
w=velocity of light
x=wt
x'=wn'
Light traveling on the x axis can have a velocity of c or
-c depending on which way it is directed. x' cannot be ct' as
Einstein defined it if using the Galilean transformation equations
because t'=t. So we define n' as the time it takes light to travel a
distance of x'.
wn'=wt-vt
n'=t(1-v/w)
So what this shows is that light has the same wavelength and frequency
in S' as it does in S. Evidently, this makes light a little more
complex than scientists want it to be, but it is definitely traveling
at a speed of c in both frames of reference.
> Might I add that IMO you seem to have learned one or two
> things in the years that you have been posting here --
> bravo for that -- and so, perhaps a decision taken long
> ago might be profitably reexamined now. �Just a thought.
Oh, I re-examine it all the time, as I just did with Peter Riedt's
idea, but I think the mathematics will show the Lorentz equations
still have marks on the track at
L sqrt(1-v^2/c^2) rather than at the length of the train as Peter is
trying to say. In any event, I do not get as emotional about this as
scientists do. To me it is just something that I pass time doing, my
version of a video game.
I always return to the Galilean transformation equations because
they seem to show light as energy instead of as an observer controlled
entity. The light meets at the origins of both frames of reference,
which seems to me like a very energetic thing to do. It shows light
to be more than just an electromagnetic wave. My idea is founded on
lack of relativity of simultaneity, which puts the marks on the
railroad track the length of the train apart in both frames of
reference. In the frame of reference of the track, the light from the
bolts of lightning meets halfway between the two marks on the track.
In the frame of reference of the train, the light meets at the middle
of the train. That is the way I have to call it.
Robert B. Winn
harry
Do you mean that you *really* claim that SR *and* classical mechanics are
inconsistent?!
- If yes, I'm afraid you're beyond help and I give up.
- If no, you can learn - as I stressed - from classical mechanics what your
mistake is with SR, since they both have "relative measures" and in exactly
the same way; only SR has expanded the class of "relative" variables.
Thus, IF (and only if!) you follow my advice and comment on such relatives
in Newtonian mechanics, I'm willing to advice you more on this.
: SR spports the
: contraction formula which in the thinking of Lorentz was the
: mathematical explanation of the null result of MMX.
So, in line with my above advice, please put here your corresponding
statements of the thinking of Newton about velocity. As I asked, does the
earth move in an ellips or not? Which two meanings did Newton attach to "v"?
: However, the
: meaning of v in the formula has changed. Lorentz used the value of
: Michelson (v = 30km/sec against the ether)
Unbelievable! I corrected that here above, how is it possible that after my
correction you put the same error back in??
: but SR assumes a value of
: zero which gives a contraction of zero. So there is no contraction of
: the parallel arm of the interferometer in MMX according to Lorentz's
: formula and its interpretation by today's SR.
NO, I *also* corrected that erroneous statement of you here above. Thus, why
do you put the same error back in??!
: However, SR still
: continues to apply the Lorentz transforms in the original sense. It
: makes no sense. Contraction and SR are dead.
I'm afraid that it makes no sense to talk to someone who is deaf, sorry.
Harald
Russell
What's to disagree? That result is an easy calculation.
You yourself (a couple paragraphs above) correctly gave
the distance between the marks as L*sqrt(1-v^2/c^2) in
the track frame. That bit of track is moving at velocity
-v in the train frame, so in the train frame that bit of
track is contracted by 1/gamma. Hence my correct result.
It may look odd to you that the 1/gamma factor is in there
twice, but that's the correct result for your gedanken.
The distance being contracted from the frame of
> reference of the train is the length of the track. What you just said
> is the opposite of what Peter Riedt is trying to do.
Gosh, I certainly hope so! Oh, I get it, you think Peter
Reidt is a *scientist*! Heh heh. That idea never occurred
to me.
Peter is trying
> to get back to no distance contraction at all. You are trying to
> compound it to agree with what the observer on the ground is said to
> see, L sqrt(1-v^2/c^2). But I see what Peter is saying. If the
> length of the track is contracted from the frame of reference of the
> train, then the train is occupying a length of track longer than the
> length of the train.
I don't think that's what Peter is saying, but no matter,
let's look at it. You use the term "length" twice in that
final clause without stating what reference frame those
lengths are measured in. Very sloppy! Not something a
*scientist* would say. Your statement would be correct if
you qualified each usage as, respectively, track frame for
the first and train frame for the second. But then the
seeming contradiction vanishes, doesn't it? You're
comparing different things.
By the way, what you just did in your sloppy language is
called a "frame jump". It's something that amateurs often
do, but not physicists. How ironic that you did exactly what
you (falsely) accused physicists of doing.
So the lightning at the front strikes first and
> the lightning at the rear of the train second, putting the marks on
> the track the length of the train apart as seen from the frame of
> reference of the contracted track.
^^^^^^^^^^?
To avoid confusing yourself, say simply "in the frame of
reference of the track". After all, in that frame, it is the
*train* and not the track that is contracted. And yes, in the
track frame the distance between the marks is the length of
the train in the track frame, namely, L*sqrt(1-v^2/c^2). As
you yourself wrote above.
> But scientists already said that the train was shorter, so they agree
> with you.
Right. So where is this disagreement among scientists that
you allege?
> > > If the bolts of lightning are simultaneous in the frame of
> > > reference of the train, then to an observer on the ground, the bolt of
> > > lightning at the back of the train strikes first, etc., and the marks
> > > on the track are further apart than the length of the train.
>
> > Yes, that's what SR says.
>
> > > I decided long ago that this was all nonsense, as Peter Reidt is
> > > trying to do, and that the correct equations were the Galilean
> > > transformation equations.
>
> > Take care, if you "decide" something like that, that you
> > don't close your mind completely to (at least consideration
> > of) other points of view, or to well-reasoned critiques of
> > your own position. I note that you never addressed the
> > substantive point of my earlier post, that contrary to what
> > you claimed, the same light beam *cannot* have the same
> > velocity in two different frames under the Galilean
> > transformation. Do you *not* have an answer?
>
> Well, yes I did have an answer. I had equations for it. If light is
> traveling at c in both frames of reference,
STOP RIGHT THERE and explain how this is possible using
a Galilean transformation. I don't want to consider
anything else until you do this.
then
>
> w=velocity of light
> x=wt
> x'=wn'
> Light traveling on the x axis can have a velocity of c or
> -c depending on which way it is directed. x' cannot be ct' as
> Einstein defined it if using the Galilean transformation equations
> because t'=t. So we define n' as the time it takes light to travel a
> distance of x'.
>
> wn'=wt-vt
> n'=t(1-v/w)
>
> So what this shows is that light has the same wavelength and frequency
> in S' as it does in S. Evidently, this makes light a little more
> complex than scientists want it to be, but it is definitely traveling
> at a speed of c in both frames of reference.
>
> > Might I add that IMO you seem to have learned one or two
> > things in the years that you have been posting here --
> > bravo for that -- and so, perhaps a decision taken long
> > ago might be profitably reexamined now. Just a thought.
>
> Oh, I re-examine it all the time, as I just did with Peter Riedt's
> idea, but I think the mathematics will show the Lorentz equations
> still have marks on the track at
> L sqrt(1-v^2/c^2) rather than at the length of the train as Peter is
> trying to say.
Again, IMO he's not trying to say that, but anyhow I'm glad
to see you have more sense than he does.
In any event, I do not get as emotional about this as
> scientists do. To me it is just something that I pass time doing, my
> version of a video game.
That's ok by me, go for it. For my part, I don't find lack
of emotion necessarily so desirable. Passion is what drives
science, moreover I don't think it's your place as an amateur
to pass judgement on how scientists behave. In any case, what
matters in the end is whether the science stands up after all
the emotional dust clears.
> I always return to the Galilean transformation equations because
> they seem to show light as energy instead of as an observer controlled
> entity.
Does light have a speed or does it not? If it does, then
you must apply the Galilean transform to *it* if you apply
it to anything else.
The light meets at the origins of both frames of reference,
> which seems to me like a very energetic thing to do.
No, the light from two flashes meets at a single point in
space at a single time, i.e. a single event. If what *you*
said were true, the meeting point would depend on who was
looking at it -- exactly the "observer controlled entity"
you wish to avoid. I.e. it is you who are making the
mistake.
It shows light
> to be more than just an electromagnetic wave. My idea is founded on
> lack of relativity of simultaneity, which puts the marks on the
> railroad track the length of the train apart in both frames of
> reference. In the frame of reference of the track, the light from the
> bolts of lightning meets halfway between the two marks on the track.
> In the frame of reference of the train, the light meets at the middle
> of the train. That is the way I have to call it.
You have a fundamental misunderstanding here. The light
beams meet at a fixed event, *one* event in spacetime.
No matter who is watching this happen, all observers must
agree it is the same event, otherwise you have an observer
dependent reality (which you tell me you don't want--good!).
The problem with what you say is that the S and S' origins
cannot both be at that event, since everyone agrees they are
at *different* places when the beams meet. It makes no
difference what you think the nature of light is, the
contradiction remains.
PD
wrote:
You said it more accurately the first time when you said the *closing*
velocity of light and the train-bound observer is not isotropic. But
closing velocity is not a measurable velocity. It is the result of a
linear subtraction of two velocities, and it's therefore dubious
whether it corresponds to any physical velocity at all.
Spaceman
> You said it more accurately the first time when you said the *closing*
> velocity of light and the train-bound observer is not isotropic. But
> closing velocity is not a measurable velocity. It is the result of a
> linear subtraction of two velocities, and it's therefore dubious
> whether it corresponds to any physical velocity at all.
So now we know that PD does not think basic math actually
comes up with "physical" differences.
In short PD thinks that taking away 4 physical apples from 10
physical apples will not prove there should be 6 physical apples
left until you can physically measure the 6 apples that are left.
Poor PD.
He does not trust basic math anymore.
He would rather trust basic math that proves basic math is wrong.
LOL
:)
PD
wrote:
> PD wrote:
> > You said it more accurately the first time when you said the *closing*
> > velocity of light and the train-bound observer is not isotropic. But
> > closing velocity is not a measurable velocity. It is the result of a
> > linear subtraction of two velocities, and it's therefore dubious
> > whether it corresponds to any physical velocity at all.
>
> So now we know that PD does not think basic math actually
> comes up with "physical" differences.
Please define "physical difference". Is closing velocity a *measurable
velocity*?
> In short PD thinks that taking away 4 physical apples from 10
> physical apples will not prove there should be 6 physical apples
> left until you can physically measure the 6 apples that are left.
No, I didn't say that. You just have the impression that a subtraction
will always correspond with a measurable quantity, and this impression
is mistaken.
Spaceman
> On Sep 28, 4:39 pm, "Spaceman" <space...@yourclockmalfunctioned.duh>
> wrote:
>> PD wrote:
>>> You said it more accurately the first time when you said the
>>> *closing* velocity of light and the train-bound observer is not
>>> isotropic. But closing velocity is not a measurable velocity. It is
>>> the result of a linear subtraction of two velocities, and it's
>>> therefore dubious whether it corresponds to any physical velocity
>>> at all.
>>
>> So now we know that PD does not think basic math actually
>> comes up with "physical" differences.
>
> Please define "physical difference". Is closing velocity a *measurable
> velocity*?
>
>> In short PD thinks that taking away 4 physical apples from 10
>> physical apples will not prove there should be 6 physical apples
>> left until you can physically measure the 6 apples that are left.
>
> No, I didn't say that. You just have the impression that a subtraction
> will always correspond with a measurable quantity, and this impression
> is mistaken.
So you are saying I am mistaken about how to subtract physical apples
or physical miles?
LOL
Poor PD.
He will never get it.
LOL
Darwin123
What PD actually said was that there is no way for the observer
to directly measure the closing velocity. One has to determine the
closing velocity by an arithmentic operation, such as you described.
The observer can measure difference in velocity between the
origin of his frame and an object using the Doppler shift. That is a
direct measurement. There is no requirement of first determining the
velocities and then subtracting them. However, there is no way an
observer traveling relative to the equipment to directly measure the
closing velocity.
An arithmetic calculation is not a direct measurement. There are
several methods by which an observer can determine a velocity of an
object relative to the origin. Relative to the origin of the frame, or
relative to any clock and ruler in the frame, the speed of light is
constant. However, the arithmetic difference between the speed of
light and the speed of any object does not have to be constant.
Spaceman
> Yes.
Then you are a moron that has no clue about single standards for
measurement and even more stupid you think basic math is wrong
and yet use it for such stupid ass proof of yours.
LOL
Dono
>
> Put the MMX on a train moving at v while we observe from the track
> frame. We observe speed of light is not isotropic relative to the
> equipment now.
No, you don't. Light speed is isotropic in all frames. You are making
the same error as Marcel Lattkes (and other cranks) , you are
confusing light speed with "closing speed".
Please explain the null result from the track frame
> using SR and no contraction.
>
> Bruce
You can't, you need to use the contraction in the longitudinal
direction.
See here the explanation from both the track frame PoV and the train
frame PoV:
Spaceman
> On Sep 28, 8:22 am, "bsr3...@my-deja.com" <bsr3...@my-deja.com> wrote:
>
>>
>> Put the MMX on a train moving at v while we observe from the track
>> frame. We observe speed of light is not isotropic relative to the
>> equipment now.
>
>
> No, you don't. Light speed is isotropic in all frames.
No it is not.
It is a constant speed from it's source only to "at rest" wrt
the source, frames.
Sheesh.
Get a freakin clue someyear Dono.
Dono
wrote:
> Dono wrote:
> > On Sep 28, 8:22 am, "bsr3...@my-deja.com" <bsr3...@my-deja.com> wrote:
>
> >> Put the MMX on a train moving at v while we observe from the track
> >> frame. We observe speed of light is not isotropic relative to the
> >> equipment now.
>
> > No, you don't. Light speed is isotropic in all frames.
>
> No it is not.
Spaceman
> On Sep 28, 6:15 pm, "Spaceman" <space...@yourclockmalfunctioned.duh>
> wrote:
>> Dono wrote:
>>> On Sep 28, 8:22 am, "bsr3...@my-deja.com" <bsr3...@my-deja.com>
>>> wrote:
>>
>>>> Put the MMX on a train moving at v while we observe from the track
>>>> frame. We observe speed of light is not isotropic relative to the
>>>> equipment now.
>>
>>> No, you don't. Light speed is isotropic in all frames.
>>
>> No it is not.
> Eat shit, SpaceShit
Poor Dono,
He just can not grasp that a speed, no matter what speed it is,
can not be constant to all frames.
It seems he loves to ignore the actual facts that relativity has shown
to support the ignorance he can not understand is wrong about
a part of it.
LOL
bsr...@my-deja.com
> On Sep 28, 8:03 am, "bsr3...@my-deja.com" <bsr3...@my-deja.com> wrote:
> ...
> an excellent post, but some comments...
>
> > What he obviously meant was SR must be able to explain the observed
> > result or the theory is wrong. Some here have said there is no real
> > contraction taking place in SR.
>
> In fairness, the discussion here was entirely in the context
> of the frame of the MMX interferometer. In SR there is no
> contraction (real or otherwise) of the interferometer in
> *that* frame.
I don't think so. If it was only being considered in that frame there
would be no v. Looking back at his first post he defined L, L' and v
in the context of both Lorentz and SR.
> That said, I admit that that point is easily missed given
> the title of the thread, and apparently *is* being missed by
> Reidt.
>
If he is missing it this has got to be one of the best examples I have
ever seen of not seeing the forest for the trees ;)
>
> > Take the MMX and put it on a train. Have the train accelerate up to
> > the velocity v. Measured in the track frame the closing speed between
> > light and the train is no longer isotropic. If the length of both
> > arms are still L then there should be a fringe shift due to the speed
> > of light no longer being isotropic relative to the apperatus. Perhaps
> > someone who claims there is no real contraction taking place would
> > like to explain how a null result is arrived at without a
> > contraction. Explaining it from the train frame is not allowed, do it
> > from the track frame :)
>
> Sure, there is "real" contraction in the sense that it has
> consequences on, say, the number of meter sticks you have
> to lay down along the track to measure the parallel arm.
> But that number depends as well on how one synchronizes the
> two clocks (one at the front and one at the back) needed
> for marking the arm's position on the track at a given time.
>
> In other words, it's coordinate dependent. IMO it's a matter
> of taste whether one's definition of "real" includes such
> quantities or does not. To argue over it is pointless; and
> to avoid provoking such arguments it's probably best to avoid
> the term if one can.
I didn't like using the word "real" but I wasn't sure how else to get
the idea across that it isn't just an artifact of using a different
coordinate system. The synchronization of the clocks on the train has
no effect on the measurements made by the track observers. The track
observers are using clocks synchronized in the track frame.
Dono
wrote:
> He just can not grasp that a speed, no matter what speed it is,
> can not be constant to all frames.
Light speed IS, SpaceSHIT. :-)
Spaceman
So it can violate relative speed rules huh?
Boy you relativists will make up anything to support the
scifi parts of your theory, including violating your own theory!
LOL
bsr...@my-deja.com
> On Sep 28, 11:22 am, "bsr3...@my-deja.com" <bsr3...@my-deja.com>
> wrote:
>
> > Put the MMX on a train moving at v while we observe from the track
> > frame. We observe speed of light is not isotropic relative to the
> > equipment now. Please explain the null result from the track frame
> > using SR and no contraction.
>
> > Bruce
>
> From the track frame, there is no explanation without contraction. As
> measured from the track frame, one arm of the MMX equipment has
> shrunk. The explanation for the null result in the track frame has to
> include both length contraction and time dilation.
We are in agreement. That challenge was for those that claim there is
no contraction involved in the SR explaination. That is somehow
supposed to make SR better than aether theories that use a
contraction ;)
Whether there is an aether or not SR obviously still works. The two
theories are not mutually exclusive.
> Thus, Lorentz had a model that explained in detail how the arms of
> the MMX experiment behaved. It was fully consistent with SR, but
> included many assumptions that we would consider slightly wrong today.
So do we have a current model that explains in detail how the arms of
the MMX experiment behave?
Bruce
Spaceman
Of course we do.
0 length contraction and 0 time dilation and a simple all "at rest"
problem where nothing is in relative motion to find anything at all.
The same type experiment using soundwaves or waterwaves
will also prove soundwaves and waterwaves are supposedly
constant to all also, and would not find any "aether" since the entire
experiment is moving either with the aether is is trying to detect
or sealed away from the aether by the entire atomosphere of Earth.
The experiment is set up like an experiment on a closed
train car that is trying to verify the wind outside the sealed traincar.
It was doomed to fail and also did not prove anything about
lightspeed being constant to all frames.
:)
--
James M Driscoll Jr
Creator of the Clock Malfunction Theory
Spaceman
Dono
wrote:
> Dono wrote:
> > On Sep 28, 6:26 pm, "SpaceSHIT" <space...@yourclockmalfunctioned.duh>
> > wrote:
>
> >> He just can not grasp that a speed, no matter what speed it is,
> >> can not be constant to all frames.
>
> > Light speed IS, SpaceSHIT. :-)
>
> So it can violate relative speed rules huh?
No, it doesn't:
u'=(u+V)/(1+uV/c^2)
If u=c then:
c'=c
Species of shit.
Spaceman
> On Sep 28, 6:41 pm, "SpaceSHIT" <space...@yourclockmalfunctioned.duh>
> wrote:
>> Dono wrote:
>>> On Sep 28, 6:26 pm, "SpaceSHIT"
>>> <space...@yourclockmalfunctioned.duh> wrote:
>>
>>>> He just can not grasp that a speed, no matter what speed it is,
>>>> can not be constant to all frames.
>>
>>> Light speed IS, SpaceSHIT. :-)
>>
>> So it can violate relative speed rules huh?
>
> No, it doesn't:
>
> u'=(u+V)/(1+uV/c^2)
>
> If u=c then:
>
> c'=c
>
> Species of shit.
Poor Dono,
He goes way off what was said and posts some completely
unrelated bullshit.
apparently he can not understand that the relative speed of
light would be using a simple thing such as
c+v = relative speed of waves passing the observer that is heading
towards the lightsource.
or
c-v = relative speed of waves passing the observer that is heading
away from the light source..
It is far too simple for the brainwashed type relativist to understand
so Dono will never get it.
And since he still can not get that percieved wavelength times frequency
ignores the actual relative motion of the observer completely he will
never get it.
:)
bsr...@my-deja.com
Just as the second postulate does not eliminate closing speeds the
fact that every observer finds the speed of light to be isotropic in
his own rest frame does not mean that it is isotropic when measured
using a different coordinate system. I know my statement differs from
how most SR experts here would state it, but it is correct. I used
that wording on purpose to get the idea across that what we measure
depends on the coordinate system used to make the measurements.
SR uses closing speeds in many calculations. If you doubt its
validity here you are throwing all those other calculations into doubt
as well. While a closing speed is not a direct measurement it is
calculated using measurements all made in the same frame, so no
conversions should be needed.
Bruce
Dono
wrote:
> Dono wrote:
> > On Sep 28, 6:41 pm, "SpaceSHIT" <space...@yourclockmalfunctioned.duh>
> > wrote:
> >> Dono wrote:
> >>> On Sep 28, 6:26 pm, "SpaceSHIT"
> >>> <space...@yourclockmalfunctioned.duh> wrote:
>
> >>>> He just can not grasp that a speed, no matter what speed it is,
> >>>> can not be constant to all frames.
>
> >>> Light speed IS, SpaceSHIT. :-)
>
> >> So it can violate relative speed rules huh?
>
> > No, it doesn't:
>
> > u'=(u+V)/(1+uV/c^2)
>
> > If u=c then:
>
> > c'=c
>
> > Species of shit.
>
> Poor Dono,
> He goes way off what was said and posts some completely
> unrelated bullshit.
I agree, now you can go ahead and eat what I posted. It is steaming
fresh. Time for your dinner, imnate!
Dono
>
> Just as the second postulate does not eliminate closing speeds
Why would the second postulate eliminate closing speeds? What gave you
the bright idea to even connect the two?
> the
> fact that every observer finds the speed of light to be isotropic in
> his own rest frame does not mean that it is isotropic when measured
> using a different coordinate system.
Yes, light speed is coordinate dependent. This is at the foundation of
parametrized theories like the SME 9and the older Mansouri-Sexl) that
take advantage of this property in order to devise experiments that
could dtect such light speed anisotropy. So, the experiments in cause
are not judged from the SR perspective since they are devised to TEST
SR. To date, no such isotropy has been detected.
> I know my statement differs from
> how most SR experts here would state it, but it is correct.
No, it isn't correct, it is just somewhat ignorant.
> SR uses closing speeds in many calculations.
Yes, it does. From your taking up on PD it appears that you don't know
the difference between light speed anisotropy and closing speed. Just
like the Marcel Lattkes crank.
> If you doubt its
> validity here you are throwing all those other calculations into doubt
> as well.
PD isn't but you don't know very well what you are talking about.
Spaceman
> I agree, now you can go ahead and eat what I posted. It is steaming
> fresh. Time for your dinner, imnate!
Poor Dono,
He can not grasp that
c + v is what an observer moving towards the source would measure
for relative wave speed.
and
c - v is what an observer moving away from the source would measure
for the relative wavespeed.
And of course he is too stupid to figure such out because he is still
stuck using "percieved wavelength times frequency" to try and measure
relative speed when it basically ignores the relative speed completely.
Any fool can check such using percieved wavelength times frequency
with water waves or sound waves and find out it ignores the observers
speed competely.
Dono likes proving he is a brainwashed fool and has no clue about
percieved wavelength vs physical wavelengths.
:)
bsr...@my-deja.com
> On Sep 28, 8:22 am, "bsr3...@my-deja.com" <bsr3...@my-deja.com> wrote:
>
>
>
> > Put the MMX on a train moving at v while we observe from the track
> > frame. We observe speed of light is not isotropic relative to the
> > equipment now.
>
> No, you don't. Light speed is isotropic in all frames. You are making
> the same error as Marcel Lattkes (and other cranks) , you are
> confusing light speed with "closing speed".
I am not! Light speed is isotropic in all frames for the observers at
rest in the coordinate system used to make the measurements. An
observer on the train is not at rest with respect to the tracks so
light speed as measured in track coordinates is not isotropic with
respect to him.
> Please explain the null result from the track frame
>
> > using SR and no contraction.
>
> > Bruce
>
> You can't, you need to use the contraction in the longitudinal
> direction.
> See here the explanation from both the track frame PoV and the train
> frame PoV:
>
> http://www.savefile.com/files/1774004
Thank you for the link but I already knew the explaination. My
challenge was to those here that claim there is no contraction in SR.
Bruce
Spaceman
> Yes, light speed is coordinate dependent. This is at the foundation of
> parametrized theories like the SME 9and the older Mansouri-Sexl) that
> take advantage of this property in order to devise experiments that
> could dtect such light speed anisotropy. So, the experiments in cause
> are not judged from the SR perspective since they are devised to TEST
> SR. To date, no such isotropy has been detected.
Actually it has, but you are too stupid to read the data correctly.
All the data that has used percieved wavelength times frequency to
determine a "constant speed of light" basically ignored the relative
speed of the observer.
If anyone with a brain uses the "physical wavelength" instead of the
percieved wavelength in any of the supposed "proof experiments"
that used percieved wavelenth times frequency to determine speed,
they will actually find the true "relative" speed of light for the
experiment instead of the constant "wrong" speed.
But Dono is one of the idiots that won't be doing that anytime soon.
LOL
bsr...@my-deja.com
> On Sep 28, 7:28 pm, "bsr3...@my-deja.com" <bsr3...@my-deja.com> wrote:
>
>
>
> > Just as the second postulate does not eliminate closing speeds
>
> Why would the second postulate eliminate closing speeds? What gave you
> the bright idea to even connect the two?
>
> > the
> > fact that every observer finds the speed of light to be isotropic in
> > his own rest frame does not mean that it is isotropic when measured
> > using a different coordinate system.
>
> Yes, light speed is coordinate dependent. This is at the foundation of
> parametrized theories like the SME 9and the older Mansouri-Sexl) that
> take advantage of this property in order to devise experiments that
> could dtect such light speed anisotropy. So, the experiments in cause
> are not judged from the SR perspective since they are devised to TEST
> SR. To date, no such isotropy has been detected.
So you are saying that if I use c-v in calculating how long it will
take a signal I sent to reach a space ship that I am wrong, that I
should just use c. Give that a try and get back to me. Any sane
person can see that v makes a difference. You are trying to use
statements that apply under specific conditions and apply them across
the board. There is no reason to expect you get the same result when
using tow different coordinate systems.
> > I know my statement differs from
> > how most SR experts here would state it, but it is correct.
>
> No, it isn't correct, it is just somewhat ignorant.
What's next, name calling?
> > SR uses closing speeds in many calculations.
>
> Yes, it does. From your taking up on PD it appears that you don't know
> the difference between light speed anisotropy and closing speed. Just
> like the Marcel Lattkes crank.
>
> > If you doubt its
> > validity here you are throwing all those other calculations into doubt
> > as well.
>
> PD isn't but you don't know very well what you are talking about.
Oh, I see, closing speeds are only invalid when inconvenient to your
view of things.
Bruce
rbwinn
to be a scientist, and was posting here the first time I ever did, and
he definitely disagrees with you.
>
>
>
> > > > If the bolts of lightning are simultaneous in the frame of
> > > > reference of the train, then to an observer on the ground, the bolt of
> > > > lightning at the back of the train strikes first, etc., and the marks
> > > > on the track are further apart than the length of the train.
>
> > > Yes, that's what SR says.
>
> > > > I decided long ago that this was all nonsense, as Peter Reidt is
> > > > trying to do, and that the correct equations were the Galilean
> > > > transformation equations.
>
> > > Take care, if you "decide" something like that, that you
> > > don't close your mind completely to (at least consideration
> > > of) other points of view, or to well-reasoned critiques of
> > > your own position. I note that you never addressed the
> > > substantive point of my earlier post, that contrary to what
> > > you claimed, the same light beam *cannot* have the same
> > > velocity in two different frames under the Galilean
> > > transformation. Do you *not* have an answer?
>
> > Well, yes I did have an answer. �I had equations for it. �If light is
> > traveling at c in both frames of reference,
>
> STOP RIGHT THERE and explain how this is possible using
> a Galilean transformation. �I don't want to consider
> anything else until you do this.
>
> �then
>
So we call what the clock in S' is showing as measured by a photon
traveling in a given direction on the x, x' axis by the variable n'.
x'=cn'
for a photon traveling in the +x direction,
x'=(-c)n'
for a photon traveling in the -x direction.
So you might think of it this way. t'=t denotes a common measurement
of time in S and S', such as degrees of rotation of the earth, which
would be the same from either frame of reference. n' is measured by
transitions of a
cesium isotope molecule, which scientists say would be different
measured in S' than in S. So with regard to the train and two bolts
of lightning, when light from the rear of the train reaches the
observer by the track, a time of t'=t has elapsed in both frames of
reference as measured by the rotation of the earth, but a shorter time
of n' has elapsed in S' as measured by transitions of a cesium isotope
molecule in S'. When light from the front flash of lightning reaches
the origin of S', light from the rear flash of lightning is also
reaching that point in S'. When light from the front flash of
lightning reaches the origin of S, according to the rotation of the
earth, t'=t, but according to transitions of a cesium isotope molecule
in S', a greater time than t'=t has elapsed because n'=x'/c, and x' is
a greater distance than x from the points towards the front of the
train where light was emitted by the lightning in each frame of
reference. In the frame of reference of the train, light was emitted
at the front of the train. In the frame of reference of the track, it
was emitted where the lightning left a mark on the track.
>
>
>
>
> > � � � � � � � � w=velocity of light
Well, light does have a speed, but speed is the magnitude of velocity,
and even the Lorentz equations are using velocity of light, not
speed. This is disguised by the fact that everywhere speed of light
appears in the Lorentz equations, it appears as c^2. There is a
reason for that. Wherever c appears in the Lorentz equations, it is
really w, or velocity of light, meaning that if it is talking about
light going in a -x direction, then the velocity of the light relative
to S and S' is -c. In other words, if a photon is emitted at 10 and
travels to 5 on the x axis, then its velocity is -c in both frames of
reference, just as a baseball thrown from 10 to 5 on the x axis has a
negative velocity relative to the frame of reference.
This can be shown in the Lorentz equations by trying to substitute
Einstein's two little equations in
x=ct
x'=ct'
They will not work if x and x' are negative. In that case, either c
or t has to be negative as also with c and t'. A negative time
indicates an event that occurred before t=t'=0. A negative velocity
of light indicates light traveling in the -x direction.
> ��The light meets at the origins of both frames of reference,
>
> > which seems to me like a very energetic thing to do.
>
> No, the light from two flashes meets at a single point in
> space at a single time, i.e. a single event. �If what *you*
> said were true, the meeting point would depend on who was
> looking at it -- exactly the "observer controlled entity"
> you wish to avoid. �I.e. it is you who are making the
> mistake.
No, observers would have nothing to do with events. If an event was
to take place when light met at the origin of S, then the event would
occur when light met at the origin of S and could only be determined
from that frame of reference, but could be calculated from any frame
of reference. In other words, if a bell rings when activated by
light from both directions, it would happen when an observer in S sees
light from both directions meet at the origin of S. The bell rings in
all frames of reference, but in frame of reference S', light from the
back of the train reaches the origin of S before the bell rings, and
light from the front of the train reaches the bell after the bell
rings. The bell rings at a time of t=.5L/c as measured by transitions
of a cesium isotope molecule in S. In S' it will ring when the origin
of S' is a distance of vt from the origin of S, regardless of what an
observer in S' thinks the light is doing.
>
> ��It shows light
>
> > to be more than just an electromagnetic wave. �My idea is founded on
> > lack of relativity of simultaneity, which puts the marks on the
> > railroad track the length of the train apart in both frames of
> > reference. �In the frame of reference of the track, the light from the
> > bolts of lightning meets halfway between the two marks on the track.
> > In the frame of reference of the train, the light meets at the middle
> > of the train. That is the way I have to call it.
>
> You have a fundamental misunderstanding here. �The light
> beams meet at a fixed event, *one* event in spacetime.
> No matter who is watching this happen, all observers must
> agree it is the same event, otherwise you have an observer
> dependent reality (which you tell me you don't want--good!).
The Galilean transformation equations show an event dependent
reality. The bell will ring when the distance between the origin of S
and the origin of S' is vt regardless of how the observer in S' is
seeing the light. When the observer in S sees the light meet at the
origin of S, the bell rings in both frames of reference. What the
observer in S' observes depends on what can be determined concerning
his cesium clock, which we have already seen is not doing the same
thing as a cesium clock in S. It appears to me that the cesium clock
in S' does not have the same phase as a cesium clock in S.
Robert B. Winn
Eric Gisse
> On Sep 27, 9:56 pm, Eric Gisse <jowr...@gmail.com> wrote:
>
>
>
> > On Sep 27, 6:54 pm, rbwinn <rbwi...@juno.com> wrote:
>
> > > On Sep 27, 6:04 pm, Eric Gisse <jowr...@gmail.com> wrote:
>
> > > > On Sep 27, 4:06 pm, rbwinn <rbwi...@juno.com> wrote:
>
> > > > > On Sep 27, 4:49 pm, Russell <rblacka...@yahoo.com> wrote:
>
> > > > > > On Sep 27, 4:24 pm, rbwinn <rbwi...@juno.com> wrote:
>
> > > > > > > On Sep 27, 3:08 pm, Russell <rblacka...@yahoo.com> wrote:
>
> > > > > > > > On Sep 26, 6:30 pm, rbwinn <rbwi...@juno.com> wrote:
>
> > > > > > > > > On Sep 26, 11:15 am, Russell <rblacka...@yahoo.com> wrote:
> > > > > > > > ...
> > > > > > > > > > This seems to be the day for getting things backwards.
> > > > > > > > > > Experimental results are what they are -- no change to
> > > > > > > > > > any theory can change them. The trick is to come up with
> > > > > > > > > > a theory that is not in conflict with experiments, either
> > > > > > > > > > those of the past or those suggested to be performed in
> > > > > > > > > > the future. Ideally, you want the *least* convoluted
> > > > > > > > > > theory with this property. And currently, SR (in its
> > > > > > > > > > applicable domain) fills the bill. It's admirable that
> > > > > > > > > > you seek to do better, but frankly, sooner or later you
> > > > > > > > > > really ought to admit that you're not up to the task.- Hide quoted text -
>
> > > > > > > > > > - Show quoted text -
>
> > > > > > > > > No one has given me a reason to make such an admission.
>
> > > > > > > > It's not something someone else can do for you. You have
> > > > > > > > to come to that realization yourself. It will take a bit
> > > > > > > > of self-awareness and courage.
>
> > > > > > > > It seems to
>
> > > > > > > > > me that the Galilean transformation equations give just as good an
> > > > > > > > > explanation as the Lorentz equations.
>
> > > > > > > > But the two are incompatible! So they can't be "just
> > > > > > > > as good" as each other. (Except in a trivial sense, if
> > > > > > > > *both* turned out to be wrong.)
>
> > > > > > > > If light is measured to be
>
> > > > > > > > > going at c=300,000 km/sec. in any frame of reference,
>
> > > > > > > > Impossible, if the Galilean transformation holds.
> > > > > > > > Velocities are *additive* under the Galilean transformation.
> > > > > > > > E.g. if a light beam is moving at c in some frame S, and
> > > > > > > > frame S is moving at speed c/2 in the same direction relative
> > > > > > > > to frame S', then the Galilean transformation gives (3/2)c for
> > > > > > > > the speed of that same beam in S'. Not c.
>
> > > > > > > > then if a photon
>
> > > > > > > > > goes a distance of x', the time it takes for a photon to travel that
> > > > > > > > > distance is
>
> > > > > > > > > n'=x'/w = t(1-v/w)
>
> > > > > > > > > where w is the velocity of the photon. If the photon is going in the
> > > > > > > > > positive direction relative to the x and x' axis, then the photon has
> > > > > > > > > a velocity of c relative to S and S', but if it is going in the
> > > > > > > > > negative direction, then its velocity is -c relative to the two frames
> > > > > > > > > of reference.
>
> > > > > > > > Again, that last sentence is incompatible with the
> > > > > > > > Galilean transformation, unless S and S' happen to be
> > > > > > > > the same frame. I really can't figure out what you're
> > > > > > > > trying to do in the rest of all that.
> > > > > > > > ...- Hide quoted text -
>
> > > > > > > Well, S' is not the same frame as S. S' is moving with a velocity of
> > > > > > > v relative to S. Light is emitted on the x axis of S at two points
> > > > > > > equal distances from the origin of S when the origin of S' is at the
> > > > > > > origin of S.
> > > > > > > So where is the light relative to the origin of S'?
> > > > > > > Well, since the origin of S' is at the origin of S when the light is
> > > > > > > emitted, the two points where light is emitted are equal distances
> > > > > > > from the origin of S'.
>
> > > > > > Ok, that would be the case if Galilean relativity applied.
> > > > > > (Just for the record, SR says that in S' the emissions are
> > > > > > not simultaneous and are not equidistant from the origin of
> > > > > > S' in that frame.)
>
> > > > > > > Where will the light emitted on one side of the origin on the x' axis
> > > > > > > of S' meet light emitted on the other side of the origin of S' on the
> > > > > > > x' axis as seen in S'?
> > > > > > > It will meet at the origin of S'.
>
> > > > > > That would be the case if light traveled at speed c in
> > > > > > frame S'. But according to the Galilean transformation,
> > > > > > if we take as given that light speed is c in frame S, then
> > > > > > the beam must travel at c+v in one direction and c-v in the
> > > > > > other. So, your reasoning is wrong here no matter which
> > > > > > transformation you use.
>
> > > > > > > Now scientists do not believe that, so you might want to go talk to
> > > > > > > the other scientists about what scientists believe.
>
> > > > > > I find it odd that you think the same two beams will come
> > > > > > together at two different places. (The origin of S will not
> > > > > > be at the same place as the origin of S' when the beams
> > > > > > meet.)
>
> > > > > > I have to admit I am not up to the task of educating you.
> > > > > > From now on, I'll leave that to others.- Hide quoted text -
>
> > > > > > - Show quoted text -
>
> > > > > Well, there you have it. Some people believe that light is energy.
> > > > > Scientists believe that light is particles or some other kind of
> > > > > material. I would compare scientists of today with the scientists
> > > > > that Michael Faraday was working for as a lab assistant. The had an
> > > > > electrical current going through a wire, and a magnetic compass would
> > > > > point to the wire. Faraday told them that there must be an
> > > > > electromagnetic field around the wire. They told him that all
> > > > > scientists were agreed that the electricity was contained in the wire
> > > > > and did not extend past the surface of the wire.
> > > > > As I said before, you will probably want to talk to other
> > > > > scientists about what scientists believe.
> > > > > Robert B. Winn
>
> > > > Yes robert, we get it. You hate scientists. Now would shut the fuck up
> > > > about it please?- Hide quoted text -
>
> > > > - Show quoted text -
>
> > > Profanity is the attempt of a weak mind to make a strong statement.
> > > Robert B. Winn
>
> > No response other than the canned phrase, robert? Thanks for shutting
> > the fuck up like I requested.- Hide quoted text -
>
> > - Show quoted text -
>
> Profanity is the attempt of a weak mind to make a strong statement.
> Robert B. Winn
Since you have nothing to say, why not shut the fuck up?
rbwinn
> The problem with what you say is that the S and S' origins
> cannot both be at that event, since everyone agrees they are
> at *different* places when the beams meet. �It makes no
> difference what you think the nature of light is, the
> contradiction remains.
Well, scientists see a contradiction here. I do not see one. If a
bell at the origin of S is to ring when light from both ends of the
train reaches it, then the bell rings when the origin of S' is a
distance of vt from the origin of S.
Now at lower velocities, what this interpretation of the Galilean
transformation equations is saying is almost mathematically identical
to what the Lorentz equations say. For instance, at the velocity of
the planet Mercury, 30 miles per second, times and distances agree to
several decimal places.
So regardless of what scientists may think, the equations are good
enough for a welder. What is lacking so far are irrefutable reasons
why the equations are wrong. For instance, your objection that light
cannot be meeting at two different places appears to me to be overcome
by the fact that the event triggered by meeting of light at the origin
of S is shown in S' by the fact that the event occurs when the origin
of S' is a distance of vt from the origin of S'. This is the only
objection to the equations that has been given. It does not appear to
hold up to me.
Robert B. Winn
rbwinn
PD
No. You can certainly perform the mathematical operation of
subtraction. The question is whether the result of that operation
corresponds to something that can be independently measured. In the
case of apples, yes. In the case of velocities, no.
PD
Depends on what you mean by "find to be isotropic". I mean "by
measurement of the speed". I discount "by subtraction of speeds,"
because the result of that operation is not something that can be
measured in any frame.
And in fact, one could reasonably argue that the problem is that the
calculation for closing speed is simply wrong. Rather than calculating
closing speed to be v1-v2, and instead choosing (v1-v2)/(1+v1*v2/c^2),
one finds that this expression for closing speed is both measurable in
some frame and in fact restores the experimental observation of the
isotropy of light speed.
> I know my statement differs from
> how most SR experts here would state it, but it is correct. I used
> that wording on purpose to get the idea across that what we measure
> depends on the coordinate system used to make the measurements.
>
> SR uses closing speeds in many calculations. If you doubt its
> validity here you are throwing all those other calculations into doubt
> as well.
No, I don't think so. Using a *calculated* quantity is fine, as long
it is clear what you're doing, and how that is distinct from any
statement about *measurable* quantities. Isotropy of light speed is
generally taken to be a statement about a *measurable* quantity.
"Closing speed" uses that adjective precisely to set it apart from
"relative speed", where the latter is measurable and the former is
not.
PD
wrote:
> Dono wrote:
> > On Sep 28, 6:26 pm, "Spaceman" <space...@yourclockmalfunctioned.duh>
> > wrote:
>
> >> He just can not grasp that a speed, no matter what speed it is,
> >> can not be constant to all frames.
>
> > Light speed IS, SpaceSHIT. :-)
>
> So it can violate relative speed rules huh?
You mean, YOUR relative speed rules -- the one that says that changing
reference frame *always* changes the value of an object's velocity.
That's your rule and no one else's. Just so we're clear on that.
PD
>
> Well, for instance, I have discusswed this with Androcles, who claims
> to be a scientist, and was posting here the first time I ever did, and
> he definitely disagrees with you.
>
Well, this is part of the problem here, isn't it, Bobby? No wonder
you've been scammed by business partners.
Androcles has never been a scientist.
Please reboot your thinking process and try again.
PD
YBM
> This can be shown in the Lorentz equations by trying to substitute
> Einstein's two little equations in
> x=ct
> x'=ct'
>
> They will not work if x and x' are negative.
Reminder : it has been proven to you numerous times here (with full
mathematical details) that it works in all cases.
rbwinn
Well, no, YBM, it hasn't. Here are the equations:
x'=(x-vt)/sqrt(1-v^2/c^2)
y'=y
z'=z
t'=(t-vx/c^2)/sqrt(1-v^2/c^2)
Substituting in Einstein's two little equations:
ct'=(ct-vt)/sqrt(1-v^2/c^2)
y'=y
z'=z
t'=(t-vct)/sqrt(1-v^2/c^2)
So we would have
c(t-vct)/sqrt(1-v^2/c^2)=(ct-vt)/sqrt(1-v^2/c^2)
c(t-vct)=ct-vt
ct-vc^2t =ct-vt
vc^2t = -vt
c^2 = -1
c= sqrt(-1)
Perhaps you could find some application for your idea in electrical
engineering.
Robert B. Winn
Dono
>
> I am not! Light speed is isotropic in all frames for the observers at
> rest in the coordinate system used to make the measurements. An
> observer on the train is not at rest with respect to the tracks so
> light speed as measured in track coordinates is not isotropic with
> respect to him.
>
This is a mistake (and you keep repeating it) . There are plenty of
experiments that falsify your claim.
Dono
> On Sep 28, 10:41 pm, Dono <sa...@comcast.net> wrote:
>
>
>
> > On Sep 28, 7:28 pm, "bsr3...@my-deja.com" <bsr3...@my-deja.com> wrote:
>
> > > Just as the second postulate does not eliminate closing speeds
>
> > Why would the second postulate eliminate closing speeds? What gave you
> > the bright idea to even connect the two?
>
> > > the
> > > fact that every observer finds the speed of light to be isotropic in
> > > his own rest frame does not mean that it is isotropic when measured
> > > using a different coordinate system.
>
> > Yes, light speed is coordinate dependent. This is at the foundation of
> > parametrized theories like the SME 9and the older Mansouri-Sexl) that
> > take advantage of this property in order to devise experiments that
> > could dtect such light speed anisotropy. So, the experiments in cause
> > are not judged from the SR perspective since they are devised to TEST
> > SR. To date, no such isotropy has been detected.
>
> So you are saying that if I use c-v in calculating how long it will
> take a signal I sent to reach a space ship that I am wrong, that I
> should just use c. Give that a try and get back to me.
I am saying that you don't know the difference between closing speed
(c-v,c+v) and anisotropic light speed. Based on your repeatedly
posting the same error, I am quite confident that you still don't know
the difference.
YBM
> On Sep 29, 5:51�am, YBM <ybm...@nooos.fr> wrote:
>> rbwinn wrote:
>>> � � This can be shown in the Lorentz equations by trying to substitute
>>> Einstein's two little equations in
>>> � � � � � � � � � x=ct
>>> � � � � � � � � � x'=ct'
>>> They will not work if x and x' are negative. �
>> Reminder : it has been proven to you numerous times here (with full
>> mathematical details) that it works in all cases.
>
> Well, no, YBM, it hasn't. Here are the equations:
>
> x'=(x-vt)/sqrt(1-v^2/c^2)
> y'=y
> z'=z
> t'=(t-vx/c^2)/sqrt(1-v^2/c^2)
>
> Substituting in Einstein's two little equations:
>
> ct'=(ct-vt)/sqrt(1-v^2/c^2)
> y'=y
> z'=z
ok.
> t'=(t-vct)/sqrt(1-v^2/c^2)
not ok : t'=(t-vct/c^2)/sqrt(1-v^2/c^2)
vct/c^2 is vt/c and NOT vct !
so : t'=(t-vt/c)/sqrt(1-v^2/c^2)
> So we would have
>
> c(t-vct)/sqrt(1-v^2/c^2)=(ct-vt)/sqrt(1-v^2/c^2)
> c(t-vct)=ct-vt
> ct-vc^2t =ct-vt
> vc^2t = -vt
> c^2 = -1
> c= sqrt(-1)
you should have noticed your error when you obtain unhomogeneous
terms such as cv and vc^2t summed together.
c(t-vt/c)/sqrt(...) = (ct-vt)/sqrt(...)
ct-vt=ct-vt
0=0
> Perhaps you could find some application for your idea in electrical
> engineering.
Perhaps you should lost some delusion about your mathematical skills.
rbwinn
> rbwinn a �crit :
>
> - Show quoted text -
I have no illusions about my mathematical abilities. Just talking to
you people introduces errors into my mathematics. I noticed that at a
very young age.
I figure something out and show it to you. You ignore it until I
forget what it was. Then you get me to try to show it again after I
have forgotten, resulting in a mistake. I went the wrong direction
with the substitution.
x'=(x=vt)/srt(1-v^2/c^2)
y'=y
z'=z
t'=(t-vx/c^2)/sqrt(1-v^2/c^2)
x=ct
t'=(t-vct/c^2)/sqrt(1-v^2/c^2)
t'=(t-vt/c)/sqrt(1-v^2/c^2)
This equation for t' gives the wrong answer for t' if x is
negative. That is why scientists leave the equation in its unreduced
form so that velocity of light is introduced into the equation
implicitly if x is a negative number.
If x= (-10), then x does not equal ct. It equals (-c)t.
Robert B. Winn
Eric Gisse
Yes, there is something about french people that ruins my mathematical
ability.
But at least there is ability to ruin, and the ruined ability might
have something to do with the alcohol consumed with the french people.
[...]
YBM
Amazing ! You're not only suffering a deep cognitive dissonance
but you are deeply paranoid too.
> I went the wrong direction
> with the substitution.
>
> x'=(x=vt)/srt(1-v^2/c^2)
> y'=y
> z'=z
> t'=(t-vx/c^2)/sqrt(1-v^2/c^2)
> x=ct
> t'=(t-vct/c^2)/sqrt(1-v^2/c^2)
> t'=(t-vt/c)/sqrt(1-v^2/c^2)
>
> This equation for t' gives the wrong answer for t' if x is
> negative.
Of course not. Let's have a look :
*You* wrote x=ct, so you are considering a light ray which is at the
origin of S at time t=0. Right ?
t'=t(1-v/c)/sqrt(1-v^2/c^2)
What do we get ? If x is negative, given that x=ct, so is t, and so
it t'. If the light ray is at the origin of S at time t=0, which is
also when the origin of S and S' coincides, given that the motion
is in the "+x" direction, it is quite obvious that the light
ray is, in S, at x<0 positions when t<0 and that it is, in S',
at x'<0 positions when t'<0.
> That is why scientists leave the equation in its unreduced
> form so that velocity of light is introduced into the equation
> implicitly if x is a negative number.
There is nothing implicit in using a velocity in a equation of
motion. This is even part of its definition.
> If x= (-10), then x does not equal ct. It equals (-c)t.
A light ray is moving in the "+x" direction, being at x=0 at t=0.
At time t=-10/c its position is -10 (check : x=ct)
At time t=-5/c its position is -5 (check : x=ct)
At time t=5/c its position is 5 (check : x=ct)
and so on...
Russell
> On Sep 28, 11:27 am, Russell <rblacka...@yahoo.com> wrote:
[big snip]
> > Right. So where is this disagreement among scientists that
> > you allege?
>
> Well, for instance, I have discusswed this with Androcles, who claims
> to be a scientist, and was posting here the first time I ever did, and
> he definitely disagrees with you.
Ah, so now we see, when you said "scientist" what you really
meant was "Usenet poster *claiming* to be a scientist."
Touche again -- it never occurred to me that you would
think Androcles is a scientist.
Look, there is *no* disagreement among physics professionals
regarding what SR predicts. Though not one myself, I am
confident that what I wrote above is in agreement with the
views of any physics professionals who might be reading.
(And if it is not, I would welcome any corrections.)
>
>
>
>
>
> > > > > If the bolts of lightning are simultaneous in the frame of
> > > > > reference of the train, then to an observer on the ground, the bolt of
> > > > > lightning at the back of the train strikes first, etc., and the marks
> > > > > on the track are further apart than the length of the train.
>
> > > > Yes, that's what SR says.
>
> > > > > I decided long ago that this was all nonsense, as Peter Reidt is
> > > > > trying to do, and that the correct equations were the Galilean
> > > > > transformation equations.
>
> > > > Take care, if you "decide" something like that, that you
> > > > don't close your mind completely to (at least consideration
> > > > of) other points of view, or to well-reasoned critiques of
> > > > your own position. I note that you never addressed the
> > > > substantive point of my earlier post, that contrary to what
> > > > you claimed, the same light beam *cannot* have the same
> > > > velocity in two different frames under the Galilean
> > > > transformation. Do you *not* have an answer?
>
> > > Well, yes I did have an answer. I had equations for it. If light is
> > > traveling at c in both frames of reference,
>
> > STOP RIGHT THERE and explain how this is possible using
> > a Galilean transformation. I don't want to consider
> > anything else until you do this.
>
> > then
>
> It is possible if a clock in S' is showing something other than t'=t.
I see, so you use the Galilean transformation *except when you
don't*? Why didn't you say that at the beginning? (Maybe you
were afraid we would laugh?)
> So we call what the clock in S' is showing as measured by a photon
> traveling in a given direction on the x, x' axis by the variable n'.
>
> x'=cn'
> for a photon traveling in the +x direction,
>
> x'=(-c)n'
>
> for a photon traveling in the -x direction.
>
> So you might think of it this way. t'=t denotes a common measurement
> of time in S and S', such as degrees of rotation of the earth, which
> would be the same from either frame of reference. n' is measured by
> transitions of a
> cesium isotope molecule, which scientists say would be different
> measured in S' than in S.
No, a good physicist would not use such sloppy language
at all. The only thing that makes sense for a *single*
clock is to count how many times it ticks between some
event A (at which the clock is present) and some event B
(at which the same clock is present). That count is
*invariant*, i.e. not dependent on frame -- provided of
course that you keep the events A and B and don't consider
different events. (Should go without saying.) Note, it
doesn't matter what clock you use; it's invariant for a
beating heart, a rotating planet, anything.
The difference between frames comes when you have to use
*multiple* clocks for some measurement -- i.e. one clock
at some event A and a *different* clock at some event B in
a different location. In such cases you have to worry
about how to synchronize those clocks, and in our universe
this turns out to be not as trivial a matter as physicists
before 1905 believed.
So with regard to the train and two bolts
> of lightning, when light from the rear of the train reaches the
> observer by the track, a time of t'=t has elapsed in both frames of
> reference as measured by the rotation of the earth, but a shorter time
> of n' has elapsed in S' as measured by transitions of a cesium isotope
> molecule in S'. When light from the front flash of lightning reaches
> the origin of S', light from the rear flash of lightning is also
> reaching that point in S'. When light from the front flash of
> lightning reaches the origin of S, according to the rotation of the
> earth, t'=t, but according to transitions of a cesium isotope molecule
> in S', a greater time than t'=t has elapsed because n'=x'/c, and x' is
> a greater distance than x from the points towards the front of the
> train where light was emitted by the lightning in each frame of
> reference. In the frame of reference of the train, light was emitted
> at the front of the train. In the frame of reference of the track, it
> was emitted where the lightning left a mark on the track.
But those emission points are the *same* point. (Same *event*,
we should really say, since we talking about a specified time
as well, the time of the lightning strike.) We set up the whole
scenario so that the front of the train would be exactly at that
point on the track when the lightning struck. Now you want to
change it? But only for *one* of the frames, it seems? This
analysis won't fly. Look, it's as simple as this: either the
light beams meet at S or they meet at S', it can't be both.
Anyhow, I don't see how your system would be an improvement on SR
even in the very unlikely case you can show that it agrees as well
with existing experiments as SR does. Having two different times
(one for 1 kind of device and 1 for others) is going to be
cumbersome, and at every turn in the road you'll need to specify
which time is the right one to be using (and why). SR is much
simpler than that.
[snip]
rbwinn
Well, actually a light ray can be emitted at any point and can go any
direction, not just emitted at the origin of S going in a positive
direction.. So suppose that the light is emitted at x=10 and goes to
x=5. Then x2-x1= x =(-5). Or light can be reflected by a mirror,
which changes its velocity from +c to -c relative to a set of
Cartesian coordinates.
>
> > That is why scientists leave the equation in its unreduced
> > form so that velocity of light is introduced into the equation
> > implicitly if x is a negative number.
>
> There is nothing implicit in using a velocity in a equation of
> motion. This is even part of its definition.
If a negative value for x is introduced into the Lorentz equation for
time, it is understood that the velocity of light is negative because
that is the only way the equation will work unless time is negative.
If time is negative and light was emitted at t=0, then the equation is
talking about events that took place before the light was emitted.
If light is going in a negative direction on the x axis, then its
velocity is -c relative to the two sets of coordinates, as the
equations show.
> > If x= (-10), then x does not equal ct. �It equals (-c)t.
>
> A light ray is moving in the "+x" direction, being at x=0 at t=0.
> At time t=-10/c its position is -10 (check : x=ct)
> At time t=-5/c �its position is -5 �(check : x=ct)
> At time t=5/c � its position is �5 �(check : x=ct)
> and so on...- Hide quoted text -
>
That is wonderful, but light can also travel the opposite direction
relative to a set of coordinates. If light is emitted at the origin
of S, then at t=0, the light is at the origin. At a time of -10/-c,
which is a positive time, the light is at -10.
In the example I gave, light was emitted at x=10 and went to x=5.
x2-x1=x = 5-10 = -5
If x is negative and t is positive, then the velocity of light has to
be negative.
Robert B. Winn
stric...@gmail.com
<dirkvandemoor...@nospAm.hotmail.com> wrote:
> Peter Riedt <rie...@yahoo.co.uk> wrote in message
>
> 75989097-8e8b-44bb-bf29-02fc62579...@v16g2000prc.googlegroups.com
>
>
>
>
>
> > On Sep 25, 9:26 pm, "Dirk Van de moortel"
> > <dirkvandemoor...@nospAm.hotmail.com> wrote:
> >> Peter Riedt <rie...@yahoo.co.uk> wrote in message
>
> >> 96f2d1ea-b145-47c4-823d-682f4bbd3...@a18g2000pra.googlegroups.com
>
> >>> Contraction has been abolished by Special Relativity
>
> >>> explain the null result of the Michelson-Morley (MMX). He defined the
> >>> amount of contraction of moving objects by the formula L'=L*sqrt(1-v^2/
> >>> c^2).
>
> >>> Since then there has been a fundamental change in the meaning of v in
> >>> the formula of Lorentz. This has effectively invalidated his
> >>> explanation of the null result of MMX.
>
> >>> Lorentz: L = absolute length of the object
> >>> SR: L = proper length
>
> >>> Lorentz: L' = local coordinate length of the object
> >>> SR: L' = length measured in the observer's frame
>
> >>> Lorentz: v = the speed of the object relative to the (absolute) rest
> >>> frame of the ether
> >>> SR: v = the speed of the object relative to the inertial frame of the
> >>> observer
>
> >>> Lorentz: c = speed of light = 300000km/sec
> >>> SR: c = speed of light = 300000km/sec
>
> >>> Michelson expected a fringe shift in MMX. When it did not occur,
> >>> Lorentz assumed that the parallel axis of the equipment contracted
> >>> physically according to the value of v used by Michelson (30km/sec),
> >>> experiment (observer co-moving with equipment) is 0, giving a
> >>> contraction of 0 (zero, nada, zilch, nix) i.e. L'=L*sqrt(1-0^0/c^2).
>
> >> GOOD GRIEF!
>
> >> In the frame of the experiment:
> >> proper length of the rod: L
> >> coordinate length of the rod: L
> >> speed of the rod: 0
> >> speed of light: c
> >> so:
> >> <coordinate length> = <proper length> * sqrt(1-v^2/c^2)
> >> i.o.w.
> >> L = L sqrt(1-0)
> >> i.o.w.
> >> L = L
>
> >> What a surprise!
>
> >>> So if there is no contraction, what is the reason for the null result?
>
> >> meanings of the variables. The way it is now, I estimate
> >> your knowledge about this is 0 (zero, nada, zilch, niks).
>
> >> Dirk Vdm- Hide quoted text -
>
> >> - Show quoted text -
>
> > you confirmed the point I was making: L'=L - no contraction. However
> > without contraction, there should be a phase shift or another reason
> > for its absence.
>
> Imbecile.
>
> Dirk Vdm- Hide quoted text -
>
> - Show quoted text -
The squeak reply of somebody who has been stumped.
stric...@gmail.com
> On Sep 26, 3:27 pm, "Dirk Van de moortel"
> no arguments, only abuse. If I compared the level of your mind to that
> of an chimpanzee, I would have to apologize to that animal.
>
> Peter Riedt- Hide quoted text -
>
> - Show quoted text -
Well, like humans, there are retarded chimps too.
YBM
This is irrelevant : if the light ray with equation of motion
x=ct is reflected at some point, then its equation of motion is no
longer x=ct for later times...
>>> If x= (-10), then x does not equal ct. �It equals (-c)t.
>> A light ray is moving in the "+x" direction, being at x=0 at t=0.
>> At time t=-10/c its position is -10 (check : x=ct)
>> At time t=-5/c �its position is -5 �(check : x=ct)
>> At time t=5/c � its position is �5 �(check : x=ct)
>> and so on...- Hide quoted text -
>>
> That is wonderful, but light can also travel the opposite direction
> relative to a set of coordinates.
I don't care : *you* have chosen a light ray with equation of
motion x=ct.
> If light is emitted at the origin
> of S, then at t=0, the light is at the origin. At a time of -10/-c,
> which is a positive time, the light is at -10.
Yep, but this is *again* a different case than the one *you* first
chose. Here x=-ct, not ct.
> In the example I gave, light was emitted at x=10 and went to x=5.
> x2-x1=x = 5-10 = -5
> If x is negative and t is positive, then the velocity of light has to
> be negative.
How could you be sooooo confused on such basic issues ?
stric...@gmail.com
> On Sep 26, 4:50 am, rbwinn <rbwi...@juno.com> wrote:
>
> > > > - Show quoted text -
>
> > > you confirmed the point I was making: L'=L - no contraction. However
> > > without contraction, there should be a phase shift or another reason
> > > for its absence.
>
>
> > > - Show quoted text -
>
>
> Show us.
>
Brazen request from Eric the crook, who cannot quote his data
reference.
stric...@gmail.com
>
> - Show quoted text -
Perhaps that explains why Eric the crook cannot find his data source.
stric...@gmail.com
>
> Since you have nothing to say, why not shut the fuck up?-
Since you can't quote your references, just shut up, Eric the crook.
Spaceman
> Russell <rblac...@yahoo.com> wrote:
>>
>> Look, there is *no* disagreement among physics professionals
>> regarding what SR predicts. Though not one myself, I am
>> confident that what I wrote above is in agreement with the
>> views of any physics professionals who might be reading.
>> (And if it is not, I would welcome any corrections.)
>
> well-established science with an abundance of experimental proof to
> back it up. Scientists and engineers depend on a knowledge of SR to
> construct devices and experiments that work.
Bullshit!
Scientists that have brains, and engineers that actually create
real life objects used by everyone in the world, do not use SR bullshit
at all to design anything.
And in fact, not one freakin' engineer that created Particle accelerator
desing came close to using the "multiple standards" of SR for time or
distance to design nor built particle accelerators.
Go learn how a clock works Mr physicist,
You are only proving you are clueless about reality and the most
basics of using single standards in measurment systems.
--
James M Driscoll Jr
Creator of the Clock Malfunction Theory
Spaceman
Russell
>
> > On Sep 28, 9:43 am, rbwinn <rbwi...@juno.com> wrote:
>
> > You have a fundamental misunderstanding here. The light
> > beams meet at a fixed event, *one* event in spacetime.
> > No matter who is watching this happen, all observers must
> > agree it is the same event, otherwise you have an observer
> > dependent reality (which you tell me you don't want--good!).
> > The problem with what you say is that the S and S' origins
> > cannot both be at that event, since everyone agrees they are
> > at *different* places when the beams meet. It makes no
> > difference what you think the nature of light is, the
> > contradiction remains.
>
> Well, scientists see a contradiction here. I do not see one. If a
> bell at the origin of S is to ring when light from both ends of the
> train reaches it, then the bell rings when the origin of S' is a
> distance of vt from the origin of S.
Fine, so I gather now that you are *not* claiming that a
similar bell on the train (at origin S') would ring, since it
is as you say, distance vt from where the meeting occurs.
Pardon my misunderstanding of what you wrote.
The problem with that is, in the train frame, you are claiming
that the lightning strikes front and back of the train at the
same time, and moreover that the light of these flashes travels
at velocity -c and c, respectively, in this frame; and yet from
the above you say these beams do not meet at origin of S'. Sorry,
that doesn't work out.
> Now at lower velocities, what this interpretation of the Galilean
> transformation equations is saying is almost mathematically identical
> to what the Lorentz equations say. For instance, at the velocity of
> the planet Mercury, 30 miles per second, times and distances agree to
> several decimal places.
> So regardless of what scientists may think, the equations are good
> enough for a welder. What is lacking so far are irrefutable reasons
> why the equations are wrong.
They are wrong for high velocities. If all you care about
is low velocities, then it's fine to keep using the easier
transform of course. That's what physicists themselves do
when making their calculations. But if you are looking for
a general law that will cover all cases, Galilean transform
won't do.
For instance, your objection that light
> cannot be meeting at two different places appears to me to be overcome
> by the fact that the event triggered by meeting of light at the origin
> of S is shown in S' by the fact that the event occurs when the origin
> of S' is a distance of vt from the origin of S'. This is the only
> objection to the equations that has been given. It does not appear to
> hold up to me.
My point is you are making incompatible claims. If you agree
with me on one of them (meeting point at x=0 in S and not at
x'=0 in S') and disagree with me on the other (simultaneity of
the flashes in S') then there is a problem, and one of us is
wrong. (And I know who. ;-) You can't have it both ways.