MMX and Time Dilation
Peter Riedt
The interferometer experiment by Michelson in 1881 did not find the
predicted fringe shifts which would have confirmed the existence of an
aether permeating space. Fitzgerald in 1882 attempted to explain the
null result. He proposed a contraction of the parallel arm of the
interferometer equipment in the direction of its motion through space.
Michelson and Morley repeated the experiment in 1887 with the same
null result. It became known as MMX. Voigt in 1887 and Lorentz and
Poincare in 1904 elaborated on the Fitzgerald contraction hypothesis.
Lorentz established the mathematical relationships for the Fitzgerald
contractions. They became known as the Lorentz transformations. Larmor
found in 1900 that a consequence of contraction was another effect -
time dilation, a shortening of time experienced by bodies in motion.
The Lorentz transformations are used by Special Relativity (SR) to
support its contention that the speed of light is constant. Every body
has the coordinates x, y, z and t. In the Lorentz transformations,
coordinates x and t of moving bodies are affected by a factor, Gamma,
while y and z remain the same. SR assumes that in moving bodies, the x
axis contracts by sqrt(1-(c^2/v^2) and t dilates by sqrt(1-(c^2/v^2).
In the case of MMX, contraction applies to the x axis of the
interferometer but time dilation affects the whole of the apparatus.
However, the Fitzgerald explanation of the null result of MMX takes
into account only parallel arm contraction. Time dilation is ignored.
There are also time components in the perpendicular arm motion through
space (vt and ct). If time dilation is applied to these components,
the distances in the perpendicular arm shrink proportionally to the
contraction of the parallel arm and the original dimensional
differences are restored. Whatever was gained through contraction of
the parallel arm is lost through time dilation affecting the
perpendicular arm.
Let me restate the argument differently. There are two different times
compared in MMX; t1 (parallel arm) and t2 (perpendicular arm). If
contraction equals time dilation so time dilation equals contraction.
Time dilation applies to the whole of a moving body. It affects
equally all dimensions of the body. Coordinates x, y and z of the
interferometer shrink by the same amount as a consequence of time
dilation. The t1, t2 differences existing at rest are proportionally
preserved throughout any velocity v of the equipment.
The null result of MMX remains unexplained; the contraction hypothesis
of Fitzgerald has been falsified by Larmor.
Peter Riedt
The Ghost In The Machine
<rie...@yahoo.co.uk>
wrote
on 12 Oct 2003 19:41:31 -0700
<1d36893d.03101...@posting.google.com>:
Erm...could you give us a hint?
A Google search on "Larmor disproves Fitzgerald" coughs up a character
named Mr. Joseph C. "No Task Too Difficult" Keller at
http://kellerphysics.com/index.php?p=resume
and that's about it. His claims apparently include
Pioneer 10 somehow generating a space warp. (Is NASA
holding out on us? :-) )
Googling for "Larmor falsifies Fitzgerald" coughs up two nearly
identical pages; the first is probably the more interesting of
the two as it includes diagrams. :-)
http://www.wcug.wwu.edu/~erikba/ziegler/1-2.html
http://www.olywa.net/unifieduniverse/I2.html
Larmor is mentioned as one of many theories, but that's all.
--
#191, ewi...@earthlink.net
It's still legal to go .sigless.
Mathew Orman
"Peter Riedt" <rie...@yahoo.co.uk> wrote in message
news:1d36893d.03101...@posting.google.com...
Must be a very small world that you live in!
You can repeat the MMX and get the phase shift proportional to angular
velocity.
The rotation speed of the Earth is 0.000277 Hz.
If you spin the MM interferometer at 100 Hz you will see the phase shift.
Please do not post 100 year old fallacy!
Sincerely,
Mathew Orman
www.ultra-faster-than-light.com
www.radio-faster-than-light.com
Dirk Van de moortel
"Peter Riedt" <rie...@yahoo.co.uk> wrote in message news:1d36893d.03101...@posting.google.com...
[snip]
> The Lorentz transformations are used by Special Relativity (SR) to
> support its contention that the speed of light is constant. Every body
> has the coordinates x, y, z and t. In the Lorentz transformations,
> coordinates x and t of moving bodies are affected by a factor, Gamma,
> while y and z remain the same. SR assumes that in moving bodies, the x
> axis contracts by sqrt(1-(c^2/v^2) and t dilates by sqrt(1-(c^2/v^2).
Just curious: what do you think happens with the
product x*t ?
Dirk Vdm
Tom Roberts
> You can repeat the MMX and get the phase shift proportional to angular
> velocity.
> The rotation speed of the Earth is 0.000277 Hz.
> If you spin the MM interferometer at 100 Hz you will see the phase shift.
Not true. Or rather, true only if you construct your "MMX"
interferometer very poorly and give it a net area. But the non-null
result is due to the similarity to SAGNAC'S experiment, not the MMX.
Topologically, the light paths in Michelson's interferometer
are the same as in Sagnac's interferometer. But Michelson's
encloses 0 area while Sagnac's has a large area. Sagnac's
signal is proportional to the area of the interferometer
(independent of shape).
BTW even for 100 Hz and a modest area the phase shift would be too small
for the naked eye to detect; a modern laser interferometer could easily
do it for much slower rotations, of course -- that's how laser
gyroscopes work. And, of course, the fringe shift is constant for a
constant angular velocity, and thus impossible for the eyeball to
discern unless you can vary the angular velocity; the laser phase
detector does not have this problem (and it is billions of times more
sensitive than the human eye).
Tom Roberts tjro...@lucent.com
Ken and Vicki
I think that your conclusion is flawed because it wrongly assumes that time
dilation and length contraction are the only two 'effects' evidenced under
the Lorentz Transform. There is a crucial third effect which is usually
termed "the relativity of simultaneity", which when correctly applied,
nullifies your assertion that MMX proved nothing substantive. See
www.ezrelativity.com for explanation and example of all three relativistic
effects taken together. It is a common error to forget about time
dissynchronicity (ie. "the relativity of simultaneity") and when so
forgotton makes the twin paradox quite impossible to sort out!
-KJS
Paul Cardinale
> MMX and Time Dilation
>
[crap snipped]
If you could do the math, you would see that you are moron.
Paul Cardinale
Mathew Orman
"Tom Roberts" <tjro...@lucent.com> wrote in message
news:bmeaa6$4...@netnews.proxy.lucent.com...
The non null result is due to the fact that light travels in straight lines
only.
Uncle Al
>
> MMX and Time Dilation
>
> The interferometer experiment by Michelson in 1881 did not find the
> predicted fringe shifts which would have confirmed the existence of an
> aether permeating space.
Phys. Rev. Lett. 90 060403 (2003)
Phys. Rev. Lett. 42(9) 549 (1979)
Phys. Bull. 21 255 (1970)
Europhysics Lett. 56(2) 170 (2001)
Gen. Rel. Grav. 34(9) 1371 (2002)
> The null result of MMX remains unexplained; the contraction hypothesis
> of Fitzgerald has been falsified by Larmor.
Bullshit.
http://arXiv.org/abs/hep-th/0307140
GR structure, especially Part 4/p. 7
<http://rattler.cameron.edu/EMIS/journals/LRG/Articles/Volume4/2001-4will/index.html>
Experimental constraints on General Relativity.
http://arxiv.org/abs/gr-qc/0308010
Nature 425 374 (2003)
<http://rattler.cameron.edu/EMIS/journals/LRG/Articles/Volume6/2003-1ashby/index.html>
http://www.eftaylor.com/pub/projecta.pdf
Relativity in the GPS system
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
"Quis custodiet ipsos custodes?" The Net!
John C. Polasek
>MMX and Time Dilation
>
snip
>
>The null result of MMX remains unexplained; the contraction hypothesis
>of Fitzgerald has been falsified by Larmor.
>
>Peter Riedt
Have you ever seen the expression x + ict? How about its derivative,
xdot + ic?
This indicates that MMX velocities always add to the speed of light
orthogonally. Therefore you get the same result xdot + ic or -xdot +
ic. The sum is not only 2nd order but it's invariant. That's my
opinion for the "failure" of the MMX. How say you?
Mr. Dual Space
(If you have something to say, write an equation.
If you have nothing to say, write an essay).
Peter Riedt
Paul, can you give me a non-moronic explanation of time dilation
a. in general
b. for MMX?
Peter Riedt
Dirk Van de moortel
Are you thinking about answering my question to you?
Knowing what happens with x and with t, what do you
Paul Cardinale
> pcard...@volcanomail.com (Paul Cardinale) wrote in message news:<64050551.03101...@posting.google.com>...
> > rie...@yahoo.co.uk (Peter Riedt) wrote in message news:<1d36893d.03101...@posting.google.com>...
> > > MMX and Time Dilation
> > >
> >
> > [crap snipped]
> >
> > If you could do the math, you would see that you are moron.
> >
> > Paul Cardinale
>
> Paul, can you give me a non-moronic explanation of time dilation
>
> a. in general
>
> b. for MMX?
>
Sure. But you have to earn it first. Do the math.
Paul Cardinale
Androcles
To explain it, all one needs to do is to allow the velocity of light is
added to the velocity of the source.
Apply Occam's razor.
The simplest solution is invariably the correct one.
Androcles
John C. Polasek
I just told you, the velocities are always added at right angles, so
there is no difference either direction.
Peter Riedt
Dirk,
same as x'*t'
Peter Riedt
Peter Riedt
John,
the 1st order is more appropiate: ic/c+xdot/xot+ic/c+xdot/xd. However,
MMX did not fail; it is merely misunderstood.
Peter Riedt
Androcles
"John C. Polasek" <jpol...@cfl.rr.com> wrote in message
news:3f8cb779.46112819@news-server...
I hadn't even read your post, and my reply was to Peter.
However, now that I have, I fail to see what evidence you
have for your strange method of adding velocities.
(If you have something to say, provide the evidence.
If you have nothing to say, make an assertion)
Androcles
Dirk Van de moortel
"Peter Riedt" <rie...@yahoo.co.uk> wrote in message news:1d36893d.03101...@posting.google.com...
> "Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote in message news:<3f8bb624$1...@usenet01.boi.hp.com>...
> > "Peter Riedt" <rie...@yahoo.co.uk> wrote in message news:1d36893d.03101...@posting.google.com...
> > > pcard...@volcanomail.com (Paul Cardinale) wrote in message news:<64050551.03101...@posting.google.com>...
> > > > rie...@yahoo.co.uk (Peter Riedt) wrote in message news:<1d36893d.03101...@posting.google.com>...
> > > > > MMX and Time Dilation
> > > > >
> > > >
> > > > [crap snipped]
> > > >
> > > > If you could do the math, you would see that you are moron.
> > > >
> > > > Paul Cardinale
> > >
> > > Paul, can you give me a non-moronic explanation of time dilation
> > >
> > > a. in general
> > >
> > > b. for MMX?
> > >
> > > Peter Riedt
> >
> > Are you thinking about answering my question to you?
> > Knowing what happens with x and with t, what do you
> > think happens with the product x*t?
> >
> >
> > Dirk Vdm
>
> Dirk,
>
> same as x'*t'
So x t = x' t'
[Sometimes I omit the * for multiplication
and replace it with a space]
Let's see... the Lorentz transformation is
{ x' = g( x - v t )
{ t' = g( t - v x / c^2 )
with g = 1/sqrt(1-v^2/c^2)
So your equation
x t = x' t'
is the same as
x t = g (x - v t) g (t - v x/c^2)
or, using the expression for g:
x t = (x - v t) * (t - v x/c^2) / (1 - v^2/c^2)
which is - with some elementary algebra - 100% equivalent with
x = t * ( v +/- Sqrt( v^2 - c^2 ) )
and also with
t = x /c^2 * ( v +/- Sqrt( v^2 - c^2 ) ).
But these equations have no solution for any speed v that is
smaller than c.
So apart from the trivial event x=t=x'=t' = 0,
there is no event in the whole of spacetime that
satisfies
x t = x' t'
So, what do you think went wrong?
Dirk Vdm
John C. Polasek
(I meant this for Androcles, but deleted his message)
To repeat: xdot + ic = -xdot + ic.
The Lorentz transform is based on a coordinate system expressible as x
+ ict (leaving out y, z for brevity). That's why we have expressions
like (if you'll excuse me)
m = m0/V(1 - v2/c2) where V is the sqrt sign
Draw the triangle, c on hypotenuse, v opposite , reduced c on the x
axis. Events on the hypotenuse represent the frame moving at v. It is
ROTATED by theta via the Lorentz transform.
That frame's physics resolves by cos theta to the (our) x axis, for
example clock rate (GPS, twins).
Likewise a force on x (our frame) reduces to f cos theta onto the
hypotenuse, making m0 look bigger.
There is NO Lorentz contraction.
Inescapable: xdot and c are always at right angles, hence
+-xdot + ic
would be the governing principle in the MMX experiment. The magnitude
of the sum xdot + ic is the same as -xdot + ic and the experiment is
doomed to a null conclusion.
Peter Riedt
Dirk, I didn't say x t = x' t'. I said the same happens to x'*t' as to
x*t. I cannot see what you were getting at with your question. In any
event your response to my post has nothing to do with my argument that
according to Larmor's time dilation, the whole of the MMX apparatus is
subject to gamma t and therefore t2 (perpendicular arm) shrinks by
gamma just as t1 (parallel arm).
Peter Riedt
Dirk Van de moortel
You said:
| "SR assumes that in moving bodies, the x axis contracts
| by sqrt(1-(c^2/v^2) and t dilates by sqrt(1-(c^2/v^2).
So, I wonder what you think happens with the product x*t.
I don't understand your answer: " the same as with x'*t' " then.
If I would have asked what happened with x, would you have
replied " the same as with x' "?
Dirk Vdm
Peter Riedt
Androcles, Occam got it the wrong way round. Riedt's razor: The
correct solution is the simplest one. You are correct in saying that
SR does not explain MMX. In fact it does not explain anything.
However, you cannot add c to v. C is independent of v but not because
of Einstein. As soon as light departs from the source it is governed
by the qualities of the aether which acts as the media of
transmission. Inability to identify the aether experimentally does not
mean it does not exist.
Peter Riedt
Peter Riedt
Androcles,
you can safely ignore John. He and some other correspondents use
pseudo scientific gibberish to trivialise and confuse understanding of
posts that do not kotow to SR.
Peter Riedt
Peter Riedt
Androcles, Occam got it the wrong way round. Riedt's razor: The
Androcles
people that appear to be pseudo-intelligent to have a shot at me. If one of
my posts has been responded to, I answer. He can have a try at explaining
how velocities add at right-angles. I'm still puzzling over how Ken Seto
thinks the MMX apparatus jumps up in the air when it hears a photon coming
down the turnpike.
Androcles
Androcles
a star emitted light that was added to the source, that star would appear in
multiple positions when the light arrived here. It has also long been held
that this is never seen.
The reason it was never seen was that the telescopes weren't good enough,
and turbulence in Earth's atmosphere confused the image anyway.
Now that we have Hubble, well...
http://www.astro.uiuc.edu/stardial/variables/mira.html
Sorry, but you'll need to explain what is going on with Mira. My explanation
is that the velocity of light is added to the velocity of the source, and we
DO see the star in multiple positions. In time, more and more stars will be
discovered that have this effect.
I'd give up on aether, if I were you. Light traverses space as photons, not
waves.
I will agree with you, though, that media such as air, water, crystal, glass
and any other transparent substance will control the speed of light within
them, refraction is ample evidence. Interaction of the photons with the
molecules makes it appear like a wave.
Androcles
Paul B. Andersen
"Androcles" <jp006...@blurbblueyonder.co.uk> wrote
news:T_ijb.19$4t...@news-binary.blueyonder.co.uk...
>
> Sorry Peter, but I have to disagree. You see, it has long been held that if
> a star emitted light that was added to the source, that star would appear in
> multiple positions when the light arrived here. It has also long been held
> that this is never seen.
> The reason it was never seen was that the telescopes weren't good enough,
> and turbulence in Earth's atmosphere confused the image anyway.
> Now that we have Hubble, well...
.. we can get fine pictures of Mira:
http://cfa-www.harvard.edu/cfa/hotimage/mira.html
But since these pictures doesn't show Mira in multiple positions,
Androcles prefers pictures taken from the ground
with a very simple CCD camera with a 50mm(!) lense.
> http://www.astro.uiuc.edu/stardial/variables/mira.html
Which take pictures each 15. minute with a fixed exposure time,
and thus will badly overexpose bright stars, like in these arbitrary
selected pictures taken with the same camera:
http://www.astro.uiuc.edu/stardial/archive/jpg/2003/05/05/05051615.jpg
http://www.astro.uiuc.edu/stardial/archive/jpg/2002/02/04/02040530.jpg
http://www.astro.uiuc.edu/stardial/archive/jpg/2002/02/12/02121230.jpg
And then Androcles can imagine that he sees all the bright stars
in multiple positions. :-)
> Sorry, but you'll need to explain what is going on with Mira. My explanation
> is that the velocity of light is added to the velocity of the source, and we
> DO see the star in multiple positions. In time, more and more stars will be
> discovered that have this effect.
It is VERY obvious that we see the variable Mira in multiple
positions when it is bright, isn't it? :-)
Paul
Paul B. Andersen
"Peter Riedt" <rie...@yahoo.co.uk> skrev i melding news:1d36893d.03101...@posting.google.com...
> In any
> event your response to my post has nothing to do with my argument that
> according to Larmor's time dilation, the whole of the MMX apparatus is
> subject to gamma t and therefore t2 (perpendicular arm) shrinks by
> gamma just as t1 (parallel arm).
So why don't you prove your point?
So far you have not presented any argument at all,
only an assertion.
If I understand you correctly, you are claiming that according
to the Lorentz transform, a moving observer will measure
both arms of the Michelson interferometer to contract equally.
Show it then.
If your claim is true, it should be an easy task.
Paul
Peter Riedt
> "Peter Riedt" <rie...@yahoo.co.uk> skrev i melding news:1d36893d.03101...@posting.google.com...
> > In any
> > event your response to my post has nothing to do with my argument that
> > according to Larmor's time dilation, the whole of the MMX apparatus is
> > subject to gamma t and therefore t2 (perpendicular arm) shrinks by
> > gamma just as t1 (parallel arm).
>
> So why don't you prove your point?
> So far you have not presented any argument at all,
> only an assertion.
>
> If I understand you correctly, you are claiming that according
> to the Lorentz transform, a moving observer will measure
> both arms of the Michelson interferometer to contract equally.
>
inertial systems are subject to time dilation, then t1 and t2 are both
subject to gamma, contract equally and therefore any benefit from only
parallel arm contraction is nullified and MMX remains unexplained,
unresolved and cannot be used to make any claims or be used to derive
gamma. SR has to be consistent with time dilation, it cannot apply it
selectively.
Peter Riedt
Peter Riedt
> http://www.astro.uiuc.edu/stardial/variables/mira.html
Adrocles, perhaps we do see stars more than once. However, my views on
photons and source dependency are these:
PHOTONS
A star shines across millions of light years to be seen by us. If its
light is transmitted by photons, two questions should be answered by
photon experts:
1. Each quartet of conjoining photons emitted from the spherical
surface of a star forms the vertical edges of a pyramid. The apex of
the pyramid rests on the surface, the vertical edges diverge at angles
determined by the radius of the star and the base reaches out evermore
distant into space. The space enclosed by the pyramid does not contain
any photons. The base dimensions of the photon pyramid will surely
exceed the diameter of the earth yet we can see the star from any
point of observation? How is that possible?
2. Where does the energy come from to drive each photon to
destinations millions of light years away?
SOURCE DEPENDENCY
The speed of light c is constant because it is transmitted by the
medium of the aether which may be assumed to have the same properties
throughout the universe. It is constant while travelling in the aether
between source and target. The source of the light does not affect the
speed of light but the speed of the target has an indirect influence
on the speed of light. It is correct to use formulas that include
expressions such as c+v (speed of light plus speed of approaching
target) and c-v (speed of light minus speed of target going away). In
these cases, the speed of light remains the same but the distances it
has to travel between source and target are shortened or lengthened.
To allow for the change in the distances, c and v are added to or
subtracted from each other; v being the speed of the target.
Peter Riedt
Starblade Darksquall
that stupid High School level SR crap. That's extremely basic, and it
doesn't explain things very well.
That must be what you're having trouble with.
>
>
> > Show it then.
> > If your claim is true, it should be an easy task.
> >
> > Paul
(...Starblade Riven Darksquall...)
Dirk Van de moortel
Peter, maybe you haven't really understood what is
meant with time dilation and length contraction.
That is why I asked you what you think happens
with the product x*t, taking into account what
you said earlier:
| "SR assumes that in moving bodies, the x axis contracts
| by sqrt(1-(c^2/v^2) and t dilates by sqrt(1-(c^2/v^2).
It is clear that your first attempt " same as with x'*t' "
isn't really satisfactory.
Can you give this question an honest try?
Dirk Vdm
Androcles
Hmm... someone once told me that
dtau/dt = 1/g,
[g =sqrt(1-v^2/c^2)],
by taking the derivative with respect to time of
= t/g - (xv/c^2)/g
because x was a constant.
Now that you've woken up again, Paul, would you care to correct it, it
should be an easy task?
(For the benefit of Peter, x' = x-vt, hence the so-called 'constant' x =
x'+vt, and x' is taken to be infinitessimally small by Einstein)
I just thought I'd mention it, since the thread title includes 'time
dilation'.
Seems to me that if we make x' vanish, we are left with x = vt, hence
tau = (t-tv^2/c^2)/g
= t(1-v^2/c^2)/g
=t.g
from which dtau/dt = 1.g, <> 1/g....
Which is right?
How say you, Peter?
How about a similar calculation for length?
I'll leave that one for the experts, my math isn't very good, I've been
told.
Androcles
Dirk Van de moortel
"Androcles" <jp006...@blurbblueyonder.co.uk> wrote in message news:0tPjb.4479$tW2....@news-binary.blueyonder.co.uk...
[snip]
> I'll leave that one for the experts, my math isn't very good, I've been
> told.
If I'm not mistaken, it was you who did the telling:
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/PartialDiff.html
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/SqrtAnswers.html
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/AndersenLogic.html
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Gibberish.html
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Equation.html
Dirk Vdm
Androcles
I'm withdrawing my suggestion that you examine Mira, Paul Andersen has
correctly pointed out that the data is suspect. It's a pity he hasn't
withdrawn his suggestion that
dtau/dt = 1/sqrt(1-v^2/c^2) and admit he is wrong.
> PHOTONS
> A star shines across millions of light years to be seen by us. If its
> light is transmitted by photons, two questions should be answered by
> photon experts:
>
> 1. Each quartet of conjoining photons emitted from the spherical
> surface of a star forms the vertical edges of a pyramid.
Ahhhh... I like it!
The tetrahedral crystalline photon structure!
Good insight, I was struggling with that. with FOUR points, we can have
N,S, +ve and -ve at each corner! Of course! Well done!
> The apex of
> the pyramid rests on the surface, the vertical edges diverge at angles
> determined by the radius of the star and the base reaches out evermore
> distant into space. The space enclosed by the pyramid does not contain
> any photons.
Why not? It should be loaded with them, surely?
> The base dimensions of the photon pyramid will surely
> exceed the diameter of the earth yet we can see the star from any
> point of observation? How is that possible?
Because the interior is loaded with photons, of course.
>
> 2. Where does the energy come from to drive each photon to
> destinations millions of light years away?
Bullets from a gun, once fired, don't require further energy to continue on
their path.
They stop as a result of air resistance, but in the absence of anything in
their path, inertia carries them on indefinitely. So do photons. The energy
that 'fired' them in the first place was provided by the mass of the star,
E=mc^2. Our sun, like all stars, is gradually losing its mass.
>
> SOURCE DEPENDENCY
> The speed of light c is constant because it is transmitted by the
> medium of the aether which may be assumed to have the same properties
> throughout the universe. It is constant while travelling in the aether
> between source and target. The source of the light does not affect the
> speed of light but the speed of the target has an indirect influence
> on the speed of light. It is correct to use formulas that include
> expressions such as c+v (speed of light plus speed of approaching
> target) and c-v (speed of light minus speed of target going away). In
> these cases, the speed of light remains the same but the distances it
> has to travel between source and target are shortened or lengthened.
> To allow for the change in the distances, c and v are added to or
> subtracted from each other; v being the speed of the target.
>
> Peter Riedt
No, no, no... much too complicated. I like the crystalline photons, though.
The E-field of each photon varies as sine, the B field varies as cosine, and
when an E-field penetrates the B field of a neighbouring photon, they drive
each other. We have a macroscopic wave, analogous to ripples on water! The
water molecules only bob up and down, yet they carry the wave. Likewise, the
electric field, which you are likening to aether, only bobs up and down. And
like ripples on a moving river, the disturbance travels downstream, yet
still leaves perfect circles of troughs and crests.
Individual photons can peel off from the others when they are reflected or
absorbed by matter. The ones that don't get through leave shadows. No aether
is needed. A static electric field is, but that isn't fixed in universal
space. There are two waves in action here, one is longitudinal and the other
is transverse. The B-field is one analogous to compressive sound waves, the
E-field is the other, analogous to the water bobbing up and down, and they
can interchange their orientation too. No magical undetectable aether, yet
waves existing, photons existing, we have a model. Both the photon and the
wave can exist, and we don't need any aether! Each field creates the other.
The crystalline photon structure IS the aether. The aether isn't a
substance, it is a pair of fields, and it only exists where the photon is.
NOW consider the base of the tetrahedron. The photons are spread over an
ever increasing area, yet they are finite in number. A gap opens up, and a
photon from the layer 'beneath' (nearer to the star) fills it, there are
plenty more coming up from behind, and new ones being created. The net
effect is that the base has its speed reduced, which we perceive as
red-shift. The emitted frequency hasn't changed, the wavelength hasn't
changed, the velocity has.The further away the star is, the greater the
apparent red-shift.
We also have some dispersion, too. A laser beam will spread, albeit
slightly, as the tip of the ray fans out, just as you would expect if you
had a substantial aether. And we can explain polarization with this model as
well. I like it. Crystalline photons...
Androcles
Paul B. Andersen
"Peter Riedt" <rie...@yahoo.co.uk> skrev i melding news:1d36893d.03101...@posting.google.com...
> "Paul B. Andersen" <paul.b....@hia.no> wrote in message news:<bmm70u$l05$1...@dolly.uninett.no>...
> > "Peter Riedt" <rie...@yahoo.co.uk> skrev i melding news:1d36893d.03101...@posting.google.com...
> > > In any
> > > event your response to my post has nothing to do with my argument that
> > > according to Larmor's time dilation, the whole of the MMX apparatus is
> > > subject to gamma t and therefore t2 (perpendicular arm) shrinks by
> > > gamma just as t1 (parallel arm).
> >
> > So why don't you prove your point?
> > So far you have not presented any argument at all,
> > only an assertion.
> >
> > If I understand you correctly, you are claiming that according
> > to the Lorentz transform, a moving observer will measure
> > both arms of the Michelson interferometer to contract equally.
> >
> Paul, not at all. What I am claiming is that if SR believes that
> inertial systems are subject to time dilation, then t1 and t2 are both
> subject to gamma, contract equally and therefore any benefit from only
> parallel arm contraction is nullified and MMX remains unexplained,
> unresolved and cannot be used to make any claims or be used to derive
> gamma. SR has to be consistent with time dilation, it cannot apply it
> selectively.
You assert.
Use the LT and prove you are right.
> > Show it then.
> > If your claim is true, it should be an easy task.
So why don't you?
Paul
Paul B. Andersen
"Androcles" <jp006...@blurbblueyonder.co.uk> skrev i melding
news:0tPjb.4479$tW2....@news-binary.blueyonder.co.uk...
>
> Hmm... someone once told me that
> dtau/dt = 1/g,
> [g =sqrt(1-v^2/c^2)],
> by taking the derivative with respect to time of
> = t/g - (xv/c^2)/g
> because x was a constant.
dtau/dt = 1/g when x = constant
Correct.
> Now that you've woken up again, Paul, would you care to correct it, it
> should be an easy task?
> (For the benefit of Peter, x' = x-vt, hence the so-called 'constant' x =
> x'+vt, and x' is taken to be infinitessimally small by Einstein)
> I just thought I'd mention it, since the thread title includes 'time
> dilation'.
>
> Seems to me that if we make x' vanish, we are left with x = vt, hence
" x' vanish" can only mean x' = 0.
> tau = (t-tv^2/c^2)/g
> = t(1-v^2/c^2)/g
> =t.g
> from which dtau/dt = 1.g, <> 1/g....
dtau/dt = g when x' = constant
Correct.
> Which is right?
Both derivations are right.
Well done, Androcles.
An example of a WRONG derivation would be:
| We consider a stationary clock in the "Latin frame", x = constant.
| dtau/dt = d/dt ((t-vx/c^2)/sqrt(1-v^2/c^2)) = -1/sqrt(1-v^2/c^2))
| thus:
| dt/dtau = -sqrt(1-v^2/c^2)
|
| This is the infamously incomprehensible and not understood "time
| contraction".
| Need I go on?
But nobody in the real world would actually make such
a blunder and draw such a nonsensical conclusion.
Or what do you think, Androcles?
Paul
Dirk Van de moortel
"Paul B. Andersen" <paul.b....@hia.no> wrote in message news:bmou63$ebl$1...@dolly.uninett.no...
>
> "Androcles" <jp006...@blurbblueyonder.co.uk> skrev i melding
> news:0tPjb.4479$tW2....@news-binary.blueyonder.co.uk...
> >
[snip]
> Well done, Androcles.
>
> An example of a WRONG derivation would be:
> | We consider a stationary clock in the "Latin frame", x = constant.
> | dtau/dt = d/dt ((t-vx/c^2)/sqrt(1-v^2/c^2)) = -1/sqrt(1-v^2/c^2))
> | thus:
> | dt/dtau = -sqrt(1-v^2/c^2)
> |
> | This is the infamously incomprehensible and not understood "time
> | contraction".
> | Need I go on?
>
> But nobody in the real world would actually make such
> a blunder and draw such a nonsensical conclusion.
Marcel Luttgens would, remember?
Dirk Vdm
Androcles
So dtau/dt = g when x is constant,
and
dtau/dt = 1/g when x' is constant.
Got it.
Androcles
Paul B. Andersen
"Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> skrev i melding
news:3f8ffe1d$1...@usenet01.boi.hp.com...
Actually, I don't think so.
Marcel did a lot of weird things, but I don't think he
would make the stupid blunders with the signs as Androcles
frequently does.
Paul
Paul B. Andersen
"Androcles" <jp006...@blurbblueyonder.co.uk> skrev i melding news:8MTjb.7685$tW2....@news-binary.blueyonder.co.uk...
>
> "Paul B. Andersen" <paul.b....@hia.no> wrote in message
> news:bmou63$ebl$1...@dolly.uninett.no...
> >
> > "Androcles" <jp006...@blurbblueyonder.co.uk> skrev i melding
> > news:0tPjb.4479$tW2....@news-binary.blueyonder.co.uk...
> > >
> > > Hmm... someone once told me that
> > > dtau/dt = 1/g,
> > > [g =sqrt(1-v^2/c^2)],
> > > by taking the derivative with respect to time of
> > > = t/g - (xv/c^2)/g
> > > because x was a constant.
> >
> > dtau/dt = 1/g when x = constant
> > Correct.
> >
> > > Now that you've woken up again, Paul, would you care to correct it, it
> > > should be an easy task?
> > > (For the benefit of Peter, x' = x-vt, hence the so-called 'constant' x =
> > > x'+vt, and x' is taken to be infinitessimally small by Einstein)
> > > I just thought I'd mention it, since the thread title includes 'time
> > > dilation'.
> > >
> > > Seems to me that if we make x' vanish, we are left with x = vt, hence
> >
> > " x' vanish" can only mean x' = 0.
Well.
Androcles is bringing in new variables without explaining what he
mean by them. He is however referring to the x' in Einstein's derivation,
where the x' is a stationary point in the "Greek frame".
If x' = 0, this point is the origo of the Greek frame.
That is xi = 0
The coordinates of this point in the Latin frame
can be written x = vt
The transform equation:
tau = (t - vx/c^2)/g
then becomes:
> > > tau = (t-tv^2/c^2)/g
> > > = t(1-v^2/c^2)/g
> > > =t.g
> > > from which dtau/dt = 1.g, <> 1/g....
> >
> > dtau/dt = g when x' = constant
> > Correct.
This will however be true not only for xi = 0,
but for any constant xi.
Thus:
dtau/dt = g when xi = constant.
> > > Which is right?
> >
> > Both derivations are right.
> >
> > Well done, Androcles.
> Thanks.
> So dtau/dt = g when x is constant
> and
> dtau/dt = 1/g when x' is constant.
> Got it.
> Androcles
Screwing up your own calculations deliberately, Androcles?
What YOU succeeded in calculating for us, was:
dtau/dt = 1/sqrt(1-v^2/c^2) when x = constant
or:
dt/dtau = sqrt(1-v^2/c^2) when x = constant (1)
dtau/dt = sqrt(1-v^2/c^2) when xi = constant (2)
So if we ask the question:
What is the rate of a moving clock?
The answers are:
x = constant means that the clock is stationary in
the Latin frame, and moving in the Greek frame.
According to (1) the rate of a moving clock in the Greek frame is:
dt/dtau = sqrt(1-v^2/c^2)
The clock is running slow.
xi = constant means that the clock is stationary in
the Greek frame, and moving in the Latin frame.
According to (2) the rate of a moving clock in the Latin frame is:
dtau/dt = sqrt(1-v^2/c^2)
The clock is running slow.
Generally: The moving clock is running slow.
Time dilation.
Got it now when you have proven it yourself, Androcles?
Paul
Dirk Van de moortel
"Paul B. Andersen" <paul.b....@hia.no> wrote in message news:bmrbaq$3ja$1...@dolly.uninett.no...
>
[snip]
> x = constant means that the clock is stationary in
> the Latin frame, and moving in the Greek frame.
> According to (1) the rate of a moving clock in the Greek frame is:
> dt/dtau = sqrt(1-v^2/c^2)
> The clock is running slow.
>
> xi = constant means that the clock is stationary in
> the Greek frame, and moving in the Latin frame.
> According to (2) the rate of a moving clock in the Latin frame is:
> dtau/dt = sqrt(1-v^2/c^2)
> The clock is running slow.
>
> Generally: The moving clock is running slow.
> Time dilation.
>
> Got it now when you have proven it yourself, Androcles?
>
> Paul
He does not understand shit of it, so the only option he's
got left, is to pretend *he* is playing a game with *you*.
Besides, suppose he suddenly *does* understand. That
would imply that all the energy he has been spending up
to now to safely remain ignorant would have been wasted.
I'm pessimistic. But one never knows ;-)
Dirk Vdm
Androcles
>
> "Androcles" <jp006...@blurbblueyonder.co.uk> skrev i melding
news:8MTjb.7685$tW2....@news-binary.blueyonder.co.uk...
> >
> > "Paul B. Andersen" <paul.b....@hia.no> wrote in message
> > news:bmou63$ebl$1...@dolly.uninett.no...
> > >
> > > "Androcles" <jp006...@blurbblueyonder.co.uk> skrev i melding
> > > news:0tPjb.4479$tW2....@news-binary.blueyonder.co.uk...
> > > >
> > > > Hmm... someone once told me that
> > > > dtau/dt = 1/g,
> > > > [g =sqrt(1-v^2/c^2)],
> > > > by taking the derivative with respect to time of
> > > > = t/g - (xv/c^2)/g
> > > > because x was a constant.
> > >
PAUL #1)
quote
dtau/dt = 1/g when x = constant
Correct.
end quote
PAUL #2)
quote
dtau/dt = g when x' = constant
Correct.
end quote
>
> This will however be true not only for xi = 0,
> but for any constant xi.
> Thus:
> dtau/dt = g when xi = constant.
>
> > > > Which is right?
> > >
> > > Both derivations are right.
> > >
> > > Well done, Androcles.
>
> > Thanks.
My summation of PAUL #1)
quote
So dtau/dt = g when x is constant
end quote
> > and
My summation of PAUL #2)
quote
dtau/dt = 1/g when x' is constant.
end quote
> > Got it.
> > Androcles
>
> Screwing up your own calculations deliberately, Androcles?
> What YOU succeeded in calculating for us, was:
>
PAUL #3) dtau/dt = 1/sqrt(1-v^2/c^2) when x = constant
> or:
> dt/dtau = sqrt(1-v^2/c^2) when x = constant (1)
>
PAUL #4) dtau/dt = sqrt(1-v^2/c^2) when xi = constant (2)
>
> So if we ask the question:
> What is the rate of a moving clock?
> The answers are:
>
> x = constant means that the clock is stationary in
> the Latin frame, and moving in the Greek frame.
> According to (1) the rate of a moving clock in the Greek frame is:
> dt/dtau = sqrt(1-v^2/c^2)
> The clock is running slow.
>
> xi = constant means that the clock is stationary in
> the Greek frame, and moving in the Latin frame.
> According to (2) the rate of a moving clock in the Latin frame is:
> dtau/dt = sqrt(1-v^2/c^2)
> The clock is running slow.
>
> Generally: The moving clock is running slow.
> Time dilation.
>
> Got it now when you have proven it yourself, Androcles?
>
> Paul
Well done, Paul.
You see, you've proved (once again, and who knows how many more times) what
I've already accepted by Gardner when he says
Part 1)
"Of course, the situation is perfectly symmetrical. If you send a beam up
and down inside your ship, he will see its path as a V-shape. He will deduce
that your clock is slower. "
If you really want to be helpful, you can explain the results I find at
http://www.androc1es.pwp.blueyonder.co.uk/Gardner.htm
Part 2)
"If you send a beam on a V-shaped path inside your ship, he will see its
path as a vertical line. He will deduce that your clock is faster."
You see, I've got it (Part 1), since I proved it to myself, but I still have
n't quite got all of it (Part 2).
Androcles
Peter Riedt
>
> The null result of MMX remains unexplained; the contraction hypothesis
> of Fitzgerald has been falsified by Larmor.
>
> Peter Riedt
TO ALL RESPONDENTS HERTOBEFORE,
THE ISSUE IS TIME DILATION AFFECTING THE PERPENDICULAR ARM OF MMX. IF
YOU HAVE ANY ARGUMENTS THAT TIME DILATION DOES NOT ALTER THE TIME IT
TAKES TO TRAVERSE THE PERPENDICLUAR ARM AND WHY NOT LET ME KNOW. ANY
OTHER NUMBER ACROBATICS ARE NOT NECESSARY.
Peter Riedt
Peter Riedt
Androcles,
I disagree with everything you said except perhaps that what I said is
too complicated for you. I will not simplify my views on source
dependency but the photons perhaps can be made clearer. Let me restate
it thus:
If photons are true they are emitted from a circular body such as a
star at an angle from each other. Between the angle is nothing. Two
neighbouring photons diverge from each other as they travel into
space. At astronomical distances, the empty space between neighbouring
photons is vast. It does not contain any photons or light but we can
see complete images of stars everywhere. Photons as transmitters of
light do not make sense.
Peter Riedt
FrediFizzx
| If photons are true they are emitted from a circular body such as a
| star at an angle from each other. Between the angle is nothing. Two
| neighbouring photons diverge from each other as they travel into
| space. At astronomical distances, the empty space between neighbouring
| photons is vast. It does not contain any photons or light but we can
| see complete images of stars everywhere. Photons as transmitters of
| light do not make sense.
For sure there must be photons traveling parallel to each other from the
star to us. That is no big mystery.
FrediFizzx
Dirk Van de moortel
[There is no reason to use capitals - that is the Usenet
equivalent of shouting. People tend not to listen to shouters.]
Peter, maybe you haven't really understood what is
meant with time dilation and length contraction.
That is why I asked you what you think happens
with the product x*t, taking into account what
you said earlier:
| "SR assumes that in moving bodies, the x axis contracts
| by sqrt(1-(c^2/v^2) and t dilates by sqrt(1-(c^2/v^2).
It is clear that your first attempt " same as with x'*t' "
isn't really satisfactory.
Can you give this question an honest try?
If you can't, or if you refuse to even try, we can only
decide that you do not know what you are talking about.
Dirk Vdm
Androcles
relativists will attempt to squirm and wriggle through in order promote
their religion, Peter, and no amount of logic will sway them from their
path. They are not going to give you a straight answer, are they? Why not?
Because they don't have one, of course. They'll just sit around, waiting for
the next newbie to come along to demonstrate their wonderful intelligence to
impress him.
Those of us that can see through their charade try to pump some sense into
the newbie as well, but you have to realize we are at a disadvantage. The
newbie is excited by the exotic idea of changing time, so he is already ripe
for picking.
The arrogant ones will tell you that you don't understand relativity, and to
read a book on the subject.
There is no time dilation, that's fairly simple to prove mathematically. The
relativist will
claim that a light beam that is directed vertically toward a mirror and back
to the source will be seen by an observer that is passing by horizontally to
follow a V-shaped path. The angle enclosed by the V is 2.arctan(v/c), and
the path length is therefore longer in the frame of the observer than its
vertical counterpart. From his stupid insistence that the velocity of the
light is c in both frames of reference, he'll conclude that less time has
passed in the frame of the shorter path. It doesn't matter that it is the
observer that is moving, but never mind to that. All you need do is set up a
second light source, this one angled at arctan(v/c), so that the observer
sees it as a vertical line. Now the entire concept breaks down, the same
observer sees two light beams, one vertical and the other following a
V-Shaped path. At this point the relativist starts to babble about the light
not returning to the source. That is, of course, entirely outside the scope
of his original argument and irrelevant. The original beam didn't return to
the same place in the observer's frame either. But the babble goes on. And
on. And on.
In MMX, whether you have light that's speed is source dependent or medium
(air)dependent doesn't matter. Source, medium and detector are all
relatively at rest.
To see any effect (and it IS theoretically possible, although not practical)
you would have to move the medium past the apparatus at 66,600 mph, the
speed of the Earth through the supposed aether. Such a wind would of course
destroy the blades of the impellor driving it long before it attained
anything close to that speed. Nobody would put delicate optical instruments
in the exhaust stream of a jet engine and expect sensible results, and no
engineer can construct a wind tunnel that could stand such forces resulting
from the speed. Perform MMX on the moon and you are left with
no time dilation (hence no length contraction)
no medium (the aether has not been found)
light is source dependent.
Now look for evidence of source dependency.
Plenty of that, but it isn't believed.
If you want to see it,
http://www.androc1es.pwp.blueyonder.co.uk/
Androcles
Androcles
Androcles
"Peter Riedt" <rie...@yahoo.co.uk> wrote in message
news:1d36893d.03101...@posting.google.com...
> "Androcles" <jp006...@blurbblueyonder.co.uk> wrote in message
news:<rCQjb.5726$tW2....@news-binary.blueyonder.co.uk>...
> > "Peter Riedt" <rie...@yahoo.co.uk> wrote in message
> > news:1d36893d.03101...@posting.google.com...
You are saying that I won't get wet from the shower in my bathroom if I
stand far enough below the nozzle, the water droplets diverge to miss me.
If there were only a few paths for the droplets to follow, that would be
true.
Let's choose your angle. 180 degrees, say? So off go the photons, and they
miss us. At least one of them always will, it's going the wrong way.
Probably the other one will too.
The next photon pair is emitted is at 90 degrees to the first. The third
pair are at 90 degrees also, on the third axis (x,y and z to consider), the
next one is 45 degrees to the first, the one after that at 22.5, the one
after that at 11.25 degrees and so on. All these photons are emitted within
femto-seconds of each other, a huge number at the same instant, and there
are an uncountable number of them. I can always divide the angle and fit a
photon in there. What you've said is also true. Astronomers use long time
exposures, even on ccd's, not jsut photographic plates, to collect photons
from stars that only arrive sporadically. Eventually enough arrive to form
an image.
You seem to be saying that because you missed the bus, you have to walk.
There'll be another one along in a minute, trust me. Photons are
transmitters of light, and behave as you say, but they follow all possible
paths.
Androcles
kenseto
WRT the horizontal light rays the apparatus is moving vertically (up or
down). This is the only direction of motion of the apparatus that will give
the null result for all the orientations of the horizontal arms. This means
what I said is confirmed by experiments. BTW, this explanation also explains
the KTX null results; the Compton shift result and why the speed of light is
isotropic.
Ken Seto
kenseto
news:<3f8b19bc.20413584@news-server>...
> > On 12 Oct 2003 19:41:31 -0700, rie...@yahoo.co.uk (Peter Riedt) wrote:
> >
> > >MMX and Time Dilation
> > >
> > snip
> > >
> > >The null result of MMX remains unexplained; the contraction hypothesis
> > >of Fitzgerald has been falsified by Larmor.
> > >
> > >Peter Riedt
> > xdot + ic?
> > This indicates that MMX velocities always add to the speed of light
> > orthogonally. Therefore you get the same result xdot + ic or -xdot +
> > ic. The sum is not only 2nd order but it's invariant. That's my
> > opinion for the "failure" of the MMX. How say you?
> >
> > Mr. Dual Space
> > (If you have something to say, write an equation.
> > If you have nothing to say, write an essay).
>
>
> the 1st order is more appropiate: ic/c+xdot/xot+ic/c+xdot/xd. However,
> MMX did not fail; it is merely misunderstood.
The MMX null result can be explained as follows:
The MMX apparatus is moving in the vertical direction (Up or Down) wrt the
horizontal light rays. This is the reason why there is no fringe shift for
the different orientations of the horizontal arms. Notice that different
locations on earth have different meaning for the horizontal direction of
the light rays. That means that vertical wrt to the horizontal light rays
would also have different meaning in different locations.
Ken Seto
John C. Polasek
wrote:
>
>"Androcles" <jp006...@blurbblueyonder.co.uk> wrote in message
>news:S_ijb.18$4t...@news-binary.blueyonder.co.uk...
>>
>> "Peter Riedt" <rie...@yahoo.co.uk> wrote in message
>> news:1d36893d.03101...@posting.google.com...
>> > "Androcles" <jp006...@blurbblueyonder.co.uk> wrote in message
>> news:<UE7jb.6740$WR5....@news-binary.blueyonder.co.uk>...
>> > > "John C. Polasek" <jpol...@cfl.rr.com> wrote in message
>> > > news:3f8cb779.46112819@news-server...
>> > > > On Wed, 15 Oct 2003 00:25:41 +0100, "Androcles"
>> > > > <jp006...@blurbblueyonder.co.uk> wrote:
>> > > >
>> > > > >
>> > > > >"Peter Riedt" <rie...@yahoo.co.uk> wrote in message
>> > > > >news:1d36893d.03101...@posting.google.com...
>> > > > >> MMX and Time Dilation
>> > > > JP
>> > > > I just told you, the velocities are always added at right angles, so
>> > > > there is no difference either direction.
>> > > >
>> > > > Mr. Dual Space
>> > > > (If you have something to say, write an equation.
>> > > > If you have nothing to say, write an essay).
>> > > No, you didn't just tell me anything.
>> > > I hadn't even read your post, and my reply was to Peter.
>> > > However, now that I have, I fail to see what evidence you
>> > > have for your strange method of adding velocities.
JP:
The velocity of light is in the 4th dimension which is always at right
angles to any velocity we can generate.
>> > > (If you have something to say, provide the evidence.
>> > > If you have nothing to say, make an assertion)
>> > > Androcles
>> >
>> > Androcles,
>> > you can safely ignore John. He and some other correspondents use
>> > pseudo scientific gibberish to trivialise and confuse understanding of
>> > posts that do not kotow to SR.
>> >
>> > Peter Riedt
>> Thanks, but I try to be fair, if somewhat blunt. I usually allow one or
>two
>> people that appear to be pseudo-intelligent to have a shot at me. If one
>of
>> my posts has been responded to, I answer. He can have a try at explaining
>> how velocities add at right-angles.
JP:
As explained above, c is in the 4th dimension. Why do you think we
have x, y, z, ict? ct is normal to any of x, y or z. (But it's really
xdot and ic, the velocities).
snip
Paul B. Andersen
"Androcles" <jp006...@blurbblueyonder.co.uk>
skrev i melding news:V0ekb.143$W16...@news-binary.blueyonder.co.uk...
>
> PAUL #1)
> quote
> dtau/dt = 1/g when x = constant
> Correct.
> end quote
> PAUL #2)
> quote
> dtau/dt = g when x' = constant
> Correct.
> end quote
> My summation of PAUL #1)
> quote
> So dtau/dt = g when x is constant
> end quote
> My summation of PAUL #2)
> quote
> dtau/dt = 1/g when x' is constant.
> end quote
Thanks for the demonstration.
Paul
Paul B. Andersen
>
> "Paul B. Andersen" <paul.b....@hia.no> wrote in message news:bmrbaq$3ja$1...@dolly.uninett.no...
> >
>
> [snip]
>
> > x = constant means that the clock is stationary in
> > the Latin frame, and moving in the Greek frame.
> > According to (1) the rate of a moving clock in the Greek frame is:
> > dt/dtau = sqrt(1-v^2/c^2)
> > The clock is running slow.
> >
> > xi = constant means that the clock is stationary in
> > the Greek frame, and moving in the Latin frame.
> > According to (2) the rate of a moving clock in the Latin frame is:
> > dtau/dt = sqrt(1-v^2/c^2)
> > The clock is running slow.
> >
> > Generally: The moving clock is running slow.
> > Time dilation.
> >
> > Got it now when you have proven it yourself, Androcles?
> >
> > Paul
>
> He does not understand shit of it, so the only option he's
> got left, is to pretend *he* is playing a game with *you*.
Of course he understands nothing.
You didn't think I seriously believed I had got it, did you? :-)
>
> Besides, suppose he suddenly *does* understand. That
> would imply that all the energy he has been spending up
> to now to safely remain ignorant would have been wasted.
> I'm pessimistic. But one never knows ;-)
Yes. I DO know.
Androcles will succeed in staying ignorant.
Paul
Dirk Van de moortel
"Paul B. Andersen" <paul.b....@hia.no> wrote in message news:bmus80$1ft$1...@dolly.uninett.no...
>
> "Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> skrev i melding
> news:unakb.85920$mW1.4...@phobos.telenet-ops.be...
> >
> > "Paul B. Andersen" <paul.b....@hia.no> wrote in message news:bmrbaq$3ja$1...@dolly.uninett.no...
> > >
> >
> > [snip]
> >
> > > x = constant means that the clock is stationary in
> > > the Latin frame, and moving in the Greek frame.
> > > According to (1) the rate of a moving clock in the Greek frame is:
> > > dt/dtau = sqrt(1-v^2/c^2)
> > > The clock is running slow.
> > >
> > > xi = constant means that the clock is stationary in
> > > the Greek frame, and moving in the Latin frame.
> > > According to (2) the rate of a moving clock in the Latin frame is:
> > > dtau/dt = sqrt(1-v^2/c^2)
> > > The clock is running slow.
> > >
> > > Generally: The moving clock is running slow.
> > > Time dilation.
> > >
> > > Got it now when you have proven it yourself, Androcles?
> > >
> > > Paul
> >
> > He does not understand shit of it, so the only option he's
> > got left, is to pretend *he* is playing a game with *you*.
>
> Of course he understands nothing.
> You didn't think I seriously believed I had got it, did you? :-)
Nah, not really.
>
> >
> > Besides, suppose he suddenly *does* understand. That
> > would imply that all the energy he has been spending up
> > to now to safely remain ignorant would have been wasted.
> > I'm pessimistic. But one never knows ;-)
>
> Yes. I DO know.
> Androcles will succeed in staying ignorant.
Indeed, of course. Saying that one never knows was too
optimistic. Silly me. Got carried away I'm afraid.
But my point is that he *is* devious enough to try (!) to twist
his ignorance (and his sad inability to cure it) into a game of
merely *pretending* that he is ignorant and stupid.
I'm sure -and I've seen- that this tactics work with submorons
like Wilson and Seto, since they obviously are even more
stupid than he is. Of course he mistakenly thinks that it works
with everyone, since he is clearly not aware of his place in the
stupidity hierarchy.
In fact, this is quite an interesting and amuzing psychological
phenomenon.
I think I'm going to show this to my son's friend who is in his
second year of psychology - he might be interested.
So, as long as you feel up to it, do carry on :-)
Dirk Vdm
Androcles
Androcles
Peter Riedt
> "Peter Riedt" <rie...@yahoo.co.uk> wrote in message news:1d36893d.03101...@posting.google.com...
> > rie...@yahoo.co.uk (Peter Riedt) wrote in message news:<1d36893d.03101...@posting.google.com>...
> > > MMX and Time Dilation
>
> > >
> > > The null result of MMX remains unexplained; the contraction hypothesis
> > > of Fitzgerald has been falsified by Larmor.
> > >
> > > Peter Riedt
> >
>
> [There is no reason to use capitals - that is the Usenet
> equivalent of shouting. People tend not to listen to shouters.]
>
> Peter, maybe you haven't really understood what is
> meant with time dilation and length contraction.
> That is why I asked you what you think happens
> with the product x*t, taking into account what
> you said earlier:
> | "SR assumes that in moving bodies, the x axis contracts
> | by sqrt(1-(c^2/v^2) and t dilates by sqrt(1-(c^2/v^2).
>
> It is clear that your first attempt " same as with x'*t' "
> isn't really satisfactory.
> Can you give this question an honest try?
>
> If you can't, or if you refuse to even try, we can only
> decide that you do not know what you are talking about.
>
> Dirk Vdm
Dirk, very sorry for shouting. Let me repeat it more civilly:
The issue is time dilation affecting the perpendicular arm of MMX.
Please do not sidetrack. You are free to decide anything.
Peter Riedt
Peter Riedt
Freddi, in my mind it's not a mystery either but an impossibility. How
can photons emitted at an angle from the surface of a spherical body
and continuing in straight, diverging lines bend to realign into
parallel paths?
Peter Riedt
Peter Riedt
Androcles, I understand that you are saying that a succession of
photons are emitted at different angles and therefore fill the gaps
between the first pair.
In practical terms I might question why a non simultaneous image could
be perceived by us as a complete picture. Furthermore, I think no
matter how many photons are ejected under your scheme, they are not of
a quantity sufficient to fill the huge sphere extending for millions
of light years around a star from any point of which sphere the star
is visible. The size of the sphere is astronomical and there must be
still huge gaps. Even if the photons are emitted at angles of a
trillionth of a second or more, any angle will diverge. Your
comparison with a shower is deficient, if you use a larger scale
garden sprinkler system, there are gaps galore between the points
where the droplets impact.
As to your view that photons can travel on forever whithout the need
of a driving force let me ask you why a candle looses its light
intensity by the square of the distance from the observer but star
light does not seem to be subject to this law notwithstanding the
differences in light output. (A larger body has more energy but looses
it in greater amounts too). What do you think now?
Peter Riedt
FrediFizzx
news:1d36893d.03101...@posting.google.com...
| "FrediFizzx" <fredi...@hotmail.com> wrote in message
news:<bmtg3q$qeggj$1...@ID-185976.news.uni-berlin.de>...
| > "Peter Riedt" <rie...@yahoo.co.uk> wrote in message
| > news:1d36893d.03101...@posting.google.com...
| > [snip]
| > | If photons are true they are emitted from a circular body such as a
| > | star at an angle from each other. Between the angle is nothing. Two
| > | neighbouring photons diverge from each other as they travel into
| > | space. At astronomical distances, the empty space between
neighbouring
| > | photons is vast. It does not contain any photons or light but we can
| > | see complete images of stars everywhere. Photons as transmitters of
| > | light do not make sense.
| >
| > For sure there must be photons traveling parallel to each other from
the
| > star to us. That is no big mystery.
| >
| > FrediFizzx
|
| Freddi, in my mind it's not a mystery either but an impossibility. How
| can photons emitted at an angle from the surface of a spherical body
| and continuing in straight, diverging lines bend to realign into
| parallel paths?
The same way you see the moon or the sun as a flat disk. Photons are
emitted in all directions from each *point* on the sphere. A star is not
a point source. It has many many point sources. We see all the photon
from a star that are nearly parallel to each other emitted from all the
many point sources on the surface of that sphere. Of course we don't see
the ones that go off in a direction not in our direction.
FrediFizzx
Androcles
> In practical terms I might question why a non simultaneous image could
> be perceived by us as a complete picture.
That merely requires leaving the shutter of your camera open for more than a
microsecond...and no practical camera can open and close instantaneously.
That applies to your eyes as well. Look at a bright light and look away,
you'll
have an after image 'burned' into your retina.
Did you ever wonder how television works? Does you monitor have a tube?
Three electron beams scan horizontal lines, gradually moving down the
screen,
striking and the image is created sequentially. Note 'sequentially'. If you
have cable, then the information must come along the wire, sequentially. To
illuminate a million pixels simultaneously requires a million parallel
conductors. That is just not practical.
In the USA and Canada, 525/2 lines are scanned horizontally, 30 frames a
second, interlaced.
In Europe, 625/2 lines, 25 frames a second. The reason for interlacing is
that the image at the top would fade before the bottom of the image could be
painted, causing flicker.
Development has since taken place that permits longer persistence phosphors,
and a computer monitor can operate without interlacing; a typical monitor
displays 1024 pixels, 768 lines, although this is just one option. I hope
that answers your practicality question.
> Furthermore, I think no
> matter how many photons are ejected under your scheme, they are not of
> a quantity sufficient to fill the huge sphere extending for millions
> of light years around a star from any point of which sphere the star
> is visible.
I agreed. I said
"Astronomers use long time exposures, even on ccd's, not just photographic
plates, to collect photons from stars that only arrive sporadically.
Eventually enough arrive to form an image."
You *cannot* see a star that is along way off, except that you wait and
collect enough photons to do so. Surely you've seen time-exposed pictures of
city scenes where the cars head and tail lights are streaks of white and
red? You cannot see the car, it is moving and doesn't leave enough photons,
at night, to be seen by the camera. The buildings can only be seen because
enough photons are collected over time to affect the film in the camera.
Have you never heard of Olber's Paradox?
http://www.schoolsobservatory.org.uk/study/sci/cosmo/internal/olbers.htm
http://newton.ex.ac.uk/aip/glimpse.txt/physnews.23.4.html
> The size of the sphere is astronomical and there must be
> still huge gaps. Even if the photons are emitted at angles of a
> trillionth of a second or more, any angle will diverge. Your
> comparison with a shower is deficient, if you use a larger scale
> garden sprinkler system, there are gaps galore between the points
> where the droplets impact.
Yes, I agreed. See above.
>
> As to your view that photons can travel on forever whithout the need
> of a driving force
Interrupting in mid sentence...
Why don't the planets, in their orbit around the sun, slow down and stop?
Why doesn't the Moon spiral into the Earth? Do these massive objects have
a driving force? This is very basic physics...
> let me ask you why a candle looses its light
> intensity by the square of the distance from the observer
The number of photons at 1 unit from the candle, spread over 1 square unit,
is the the same number of photons at 2 units from the candle, spread over 4
square units. Simple projection of a square based pyramid, height 2 units,
sides 2 units. It would apply to a cone as well. That is elementary
geometry.
> but star
> light does not seem to be subject to this law
Of course it is subject to the same law, why would you think it wasn't?
Stars are clearly dimmer than our own sun, which is a star.
> notwithstanding the differences in light output. (A larger body has more
energy but looses
> it in greater amounts too). What do you think now?
The same as I always thought.
Your reasoning isn't faulty, but your predilection for waves is.
Ever considered how far a ripple on a pond can travel? Make the pond as big
as you like, the amplitude of the wave always decreases. Eventually the
height of the ripple
must be less than the size of a molecule of water. From that point on,
brownian motion takes over, the wave as we know it has ceased. The energy is
still there, the water molecules move faster, hence the water is heated.
This will eventually be lost as radiation, the molecules will collide, one
will absorb enough energy from a neighbour to bump an electron into a higher
energy state, then lose it, emitting a photon.
If light were waves, as you presume, then there is a limit to how far we
could obtain the wave. It would be the edge of the *observable* universe,
under a wave model, and closer to us than the present limit, found by
collecting enough individual photons, over time, to obtain an image of a
distant galaxy. And no, we cannot see individual stars at that distance, the
galaxy is a fuzzy blur.
Androcles
Dirk Van de moortel
"Peter Riedt" <rie...@yahoo.co.uk> wrote in message news:1d36893d.0310...@posting.google.com...
The issue is clearly your not understanding what time
dilation is to begin with.
Feel free to remain ignorant about it, but don't be
surprised at the reactions.
Dirk Vdm
Paul B. Andersen
"Peter Riedt" <rie...@yahoo.co.uk> skrev i melding
news:1d36893d.03101...@posting.google.com...
> If photons are true they are emitted from a circular body such as a
> star at an angle from each other. Between the angle is nothing. Two
> neighbouring photons diverge from each other as they travel into
> space. At astronomical distances, the empty space between neighbouring
> photons is vast. It does not contain any photons or light but we can
> see complete images of stars everywhere. Photons as transmitters of
> light do not make sense.
Your thoughts are qualitative correct.
What you miss, however, is how vast the number of photons
emitted from a star is.
Lets make a few calculations.
Let's assume that all the photons have the same energy, in the red
end of the visible spectrum. (Not true, but close enough for
our purpose.) The energy of such a photon is ca. 2*10^-19 Joule.
The intensity of light from the Sun at the Earth is ca. 1.6 kW/m^2,
which means that ca.10^22 photons from the Sun will pass through
a square metre every second. A Sun like star 1000 LY away,
will be ca. 10^8 AU away, and will thus be 10^16 times fainter.
That means that ca. 10^6 - a million - photons from such a star will
pass through a square metre every second.
And note that a Sun like star at 1000 LY will be much too faint
to be visible with the naked eye. (Not surprising, as only
a few photons per second would enter your pupil.)
But the effect you mention is notable when very faint objects
are observed. That's why telescopes must be so big, many
square metres aperture. And that's why the exposure time must be
so long for the real faint objects - it may be hours.
(The Hubble telescope have occasionally used exposure times of
several days.)
Modern CCDs detects more than half of the photons.
The telescope is quite literally collecting the few photons that
arrives every now and then, and builds up the picture gradually.
So yes, photons make sense.
Do a few calculations, and you will see that exactly as many
photons as expected will hit us.
Paul
Peter Riedt
> "Peter Riedt" <rie...@yahoo.co.uk> wrote in message
> news:1d36893d.03101...@posting.google.com...
> > "Androcles" <jp006...@blurbblueyonder.co.uk> wrote in message
> news:<Qcwkb.1129$5b....@news-binary.blueyonder.co.uk>...
> > > "Peter Riedt" <rie...@yahoo.co.uk> wrote in message
> > > news:1d36893d.03101...@posting.google.com...
> > > > "Androcles" <jp006...@blurbblueyonder.co.uk> wrote in message
> news:<rCQjb.5726$tW2....@news-binary.blueyonder.co.uk>...
> > > > > "Peter Riedt" <rie...@yahoo.co.uk> wrote in message
> > > > > news:1d36893d.03101...@posting.google.com...
> > > [snip old material]
[snip old material]
Andy, a physics teacher once said: We teach Mondays, Wednesdays and
Fridays that light is transmitted by waves but on Tuesdays and
Thursdays by photons.
I think only one or the other applies and waves make more sense. How
and when does light decide to travel as a photon and how and when by
waves? You have clever arguments but you use illustrations that cannot
be applied to photons. For instance, your shower example. If you stand
10cm under the shower head and with gravity, you sure will get wet.
But if the shower is 1km above your head and no gravity applies, I
doubt that you will be hit by any droplets. Perhaps you will now reply
that yes gravity is a good cause for photons to congregate towards
physical objects in the universe, these being collection points for
bundles of photons. I am not clear if you are saying that unimpeded
photons continue indefinitely without loss of energy or that
eventually they cease to exist. If the former, could we observe the
universe to its physical limit?
Peter Riedt
Peter Riedt
Dirk, I am attacking the basic premises of SR. There is no need for me
to take up your invitation to use the incorrect conclusions and maths
of SR to show that SR is a fallacy other than to point out that SR's
time dilation negates length contraction of objects in inertial
systems and that therefore SR defeats itself. I am doubting that you
are able to give reasonable arguments that this is not so.
Peter Riedt
Dirk Van de moortel
"Peter Riedt" <rie...@yahoo.co.uk> wrote in message news:1d36893d.03102...@posting.google.com...
I have asked you an extremely simple question.
They don't come any simpler than that.
You cannot answer it. That shows that not only
you don't understand the basic premises of SR.
It shows that you don't even understand what
comes *before*. So for you, there is only one
thing to attack. That is your own ignorance.
If you really had been interested and had looked
a bit more closely you could even have repeated
a part of what I had told you.
You have an opportunity to learn and you throw
it away. So again, feel free to remain ignorant.
With Androcles, our Number One Champion
Of Self-Inflicted Ignorance you are in excellent
company :-)
Dirk Vdm
J?rgen Clade
> Dirk, I am attacking the basic premises of SR. There is no need for me
> to take up your invitation to use the incorrect conclusions and maths
> of SR to show that SR is a fallacy other than to point out that SR's
> time dilation negates length contraction of objects in inertial
> systems and that therefore SR defeats itself. I am doubting that you
> are able to give reasonable arguments that this is not so.
So far, you didn´t show that SR contains any incorrect conclusions,
incorrect math or fallacies. The reason is that you didn´t look into
your physics textbook and thus don´t know how the MMX is explained
correctly. Please make this up.
regards,
Jürgen