| Topic: |
Science > Physics |
| User: |
"Peter Riedt" |
| Date: |
12 Oct 2003 09:41:31 PM |
| Object: |
MMX and Time Dilation |
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. 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
.
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|
| User: "Ken and Vicki" |
|
| Title: Re: MMX and Time Dilation |
13 Oct 2003 09:24:13 AM |
|
|
"Peter Riedt" <riedt1@yahoo.co.uk> wrote in message
news:1d36893d.0310121841.356b73ac@posting.google.com...
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. 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
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
.
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|
| User: "Androcles" |
|
| Title: Re: MMX and Time Dilation |
14 Oct 2003 06:25:41 PM |
|
|
"Peter Riedt" <riedt1@yahoo.co.uk> wrote in message
news:1d36893d.0310121841.356b73ac@posting.google.com...
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. 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
It is not unexplained. Rather, it is unexplained by relativity.
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
.
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| User: "Peter Riedt" |
|
| Title: Re: MMX and Time Dilation |
15 Oct 2003 02:43:18 PM |
|
|
"Androcles" <jp006f9750@blurbblueyonder.co.uk> wrote in message news:<wB%ib.160$WR5.144@news-binary.blueyonder.co.uk>...
"Peter Riedt" <riedt1@yahoo.co.uk> wrote in message
news:1d36893d.0310121841.356b73ac@posting.google.com...
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. 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
It is not unexplained. Rather, it is unexplained by relativity.
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
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
.
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|
| User: "Androcles" |
|
| Title: Re: MMX and Time Dilation |
15 Oct 2003 04:29:52 PM |
|
|
"Peter Riedt" <riedt1@yahoo.co.uk> wrote in message
news:1d36893d.0310151143.602b69ec@posting.google.com...
"Androcles" <jp006f9750@blurbblueyonder.co.uk> wrote in message
news:<wB%ib.160$WR5.144@news-binary.blueyonder.co.uk>...
"Peter Riedt" <riedt1@yahoo.co.uk> wrote in message
news:1d36893d.0310121841.356b73ac@posting.google.com...
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. 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
It is not unexplained. Rather, it is unexplained by relativity.
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
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.
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...
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
.
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| User: "Paul B. Andersen" |
|
| Title: Re: MMX and Time Dilation |
16 Oct 2003 05:11:54 AM |
|
|
"Androcles" <jp006f9750@blurbblueyonder.co.uk> wrote
news:T_ijb.19$4t.17@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
.
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| User: "Peter Riedt" |
|
| Title: Re: MMX and Time Dilation |
16 Oct 2003 10:24:45 PM |
|
|
"Androcles" <jp006f9750@blurbblueyonder.co.uk> wrote in message news:<T_ijb.19$4t.17@news-binary.blueyonder.co.uk>...
"Peter Riedt" <riedt1@yahoo.co.uk> wrote in message
news:1d36893d.0310151143.602b69ec@posting.google.com...
"Androcles" <jp006f9750@blurbblueyonder.co.uk> wrote in message
news:<wB%ib.160$WR5.144@news-binary.blueyonder.co.uk>...
"Peter Riedt" <riedt1@yahoo.co.uk> wrote in message
news:1d36893d.0310121841.356b73ac@posting.google.com...
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. 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
It is not unexplained. Rather, it is unexplained by relativity.
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
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.
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...
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
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
.
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| User: "Androcles" |
|
| Title: Tetrahedral photons, formerly MMX and time dilation |
17 Oct 2003 06:44:25 AM |
|
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"Peter Riedt" <riedt1@yahoo.co.uk> wrote in message
news:1d36893d.0310161924.533bbad9@posting.google.com...
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.
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...
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
Adrocles, perhaps we do see stars more than once. However, my views on
photons and source dependency are these:
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
.
|
|
|
| User: "Peter Riedt" |
|
| Title: Re: Tetrahedral photons, formerly MMX and time dilation |
18 Oct 2003 11:31:27 PM |
|
|
"Androcles" <jp006f9750@blurbblueyonder.co.uk> wrote in message news:<rCQjb.5726$tW2.2248@news-binary.blueyonder.co.uk>...
"Peter Riedt" <riedt1@yahoo.co.uk> wrote in message
news:1d36893d.0310161924.533bbad9@posting.google.com...
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.
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...
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
Adrocles, perhaps we do see stars more than once. However, my views on
photons and source dependency are these:
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
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
.
|
|
|
| User: "Paul B. Andersen" |
|
| Title: Re: Tetrahedral photons, formerly MMX and time dilation |
20 Oct 2003 09:11:21 AM |
|
|
"Peter Riedt" <riedt1@yahoo.co.uk> skrev i melding
news:1d36893d.0310182031.406614de@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
.
|
|
|
|
| User: "Androcles" |
|
| Title: Re: Tetrahedral photons, formerly MMX and time dilation |
19 Oct 2003 08:21:02 AM |
|
|
"Peter Riedt" <riedt1@yahoo.co.uk> wrote in message
news:1d36893d.0310182031.406614de@posting.google.com...
"Androcles" <jp006f9750@blurbblueyonder.co.uk> wrote in message
news:<rCQjb.5726$tW2.2248@news-binary.blueyonder.co.uk>...
"Peter Riedt" <riedt1@yahoo.co.uk> wrote in message
news:1d36893d.0310161924.533bbad9@posting.google.com...
[snip old material]
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
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
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
.
|
|
|
| User: "Peter Riedt" |
|
| Title: Re: Tetrahedral photons, formerly MMX and time dilation |
20 Oct 2003 12:45:03 AM |
|
|
"Androcles" <jp006f9750@blurbblueyonder.co.uk> wrote in message news:<Qcwkb.1129$5b.965@news-binary.blueyonder.co.uk>...
"Peter Riedt" <riedt1@yahoo.co.uk> wrote in message
news:1d36893d.0310182031.406614de@posting.google.com...
"Androcles" <jp006f9750@blurbblueyonder.co.uk> wrote in message
news:<rCQjb.5726$tW2.2248@news-binary.blueyonder.co.uk>...
"Peter Riedt" <riedt1@yahoo.co.uk> wrote in message
news:1d36893d.0310161924.533bbad9@posting.google.com...
[snip old material]
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
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
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
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
.
|
|
|
| User: "Androcles" |
|
| Title: Re: Tetrahedral photons, formerly MMX and time dilation |
20 Oct 2003 03:28:53 AM |
|
|
"Peter Riedt" <riedt1@yahoo.co.uk> wrote in message
news:1d36893d.0310192145.7eb8c932@posting.google.com...
"Androcles" <jp006f9750@blurbblueyonder.co.uk> wrote in message
news:<Qcwkb.1129$5b.965@news-binary.blueyonder.co.uk>...
"Peter Riedt" <riedt1@yahoo.co.uk> wrote in message
news:1d36893d.0310182031.406614de@posting.google.com...
"Androcles" <jp006f9750@blurbblueyonder.co.uk> wrote in message
news:<rCQjb.5726$tW2.2248@news-binary.blueyonder.co.uk>...
"Peter Riedt" <riedt1@yahoo.co.uk> wrote in message
news:1d36893d.0310161924.533bbad9@posting.google.com...
[snip old material]
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
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
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
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.
Good.
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
.
|
|
|
| User: "Peter Riedt" |
|
| Title: Re: Tetrahedral photons, formerly MMX and time dilation |
21 Oct 2003 09:07:33 PM |
|
|
"Androcles" <jp006f9750@blurbblueyonder.co.uk> wrote in message news:<51Nkb.152$HA1.93@news-binary.blueyonder.co.uk>...
"Peter Riedt" <riedt1@yahoo.co.uk> wrote in message
news:1d36893d.0310192145.7eb8c932@posting.google.com...
"Androcles" <jp006f9750@blurbblueyonder.co.uk> wrote in message
news:<Qcwkb.1129$5b.965@news-binary.blueyonder.co.uk>...
"Peter Riedt" <riedt1@yahoo.co.uk> wrote in message
news:1d36893d.0310182031.406614de@posting.google.com...
"Androcles" <jp006f9750@blurbblueyonder.co.uk> wrote in message
news:<rCQjb.5726$tW2.2248@news-binary.blueyonder.co.uk>...
"Peter Riedt" <riedt1@yahoo.co.uk> wrote in message
news:1d36893d.0310161924.533bbad9@posting.google.com...
[snip old material]
[snip old material]
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
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.
Good.
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
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
.
|
|
|
|
|
|
|
| User: "FrediFizzx" |
|
| Title: Re: Tetrahedral photons, formerly MMX and time dilation |
19 Oct 2003 02:56:46 AM |
|
|
"Peter Riedt" <riedt1@yahoo.co.uk> wrote in message
news:1d36893d.0310182031.406614de@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
.
|
|
|
| User: "Peter Riedt" |
|
| Title: Re: Tetrahedral photons, formerly MMX and time dilation |
20 Oct 2003 12:11:49 AM |
|
|
"FrediFizzx" <fredifizzx@hotmail.com> wrote in message news:<bmtg3q$qeggj$1@ID-185976.news.uni-berlin.de>...
"Peter Riedt" <riedt1@yahoo.co.uk> wrote in message
news:1d36893d.0310182031.406614de@posting.google.com...
[snip]
| If photons are true they are emitted from a circular body such as a
| star at an angle from each other. Between the angle is nothing. Two
| neighbouring photons diverge from each other as they travel into
| space. At astronomical distances, the empty space between neighbouring
| photons is vast. It does not contain any photons or light but we can
| see complete images of stars everywhere. Photons as transmitters of
| light do not make sense.
For sure there must be photons traveling parallel to each other from the
star to us. That is no big mystery.
FrediFizzx
Freddi, in my mind it's not a mystery either but an impossibility. How
can photons emitted at an angle from the surface of a spherical body
and continuing in straight, diverging lines bend to realign into
parallel paths?
Peter Riedt
.
|
|
|
| User: "FrediFizzx" |
|
| Title: Re: Tetrahedral photons, formerly MMX and time dilation |
20 Oct 2003 01:10:27 AM |
|
|
"Peter Riedt" <riedt1@yahoo.co.uk> wrote in message
news:1d36893d.0310192111.773c4022@posting.google.com...
| "FrediFizzx" <fredifizzx@hotmail.com> wrote in message
news:<bmtg3q$qeggj$1@ID-185976.news.uni-berlin.de>...
| > "Peter Riedt" <riedt1@yahoo.co.uk> wrote in message
| > news:1d36893d.0310182031.406614de@posting.google.com...
| > [snip]
| > | If photons are true they are emitted from a circular body such as a
| > | star at an angle from each other. Between the angle is nothing. Two
| > | neighbouring photons diverge from each other as they travel into
| > | space. At astronomical distances, the empty space between
neighbouring
| > | photons is vast. It does not contain any photons or light but we can
| > | see complete images of stars everywhere. Photons as transmitters of
| > | light do not make sense.
| >
| > For sure there must be photons traveling parallel to each other from
the
| > star to us. That is no big mystery.
| >
| > FrediFizzx
|
| Freddi, in my mind it's not a mystery either but an impossibility. How
| can photons emitted at an angle from the surface of a spherical body
| and continuing in straight, diverging lines bend to realign into
| parallel paths?
The same way you see the moon or the sun as a flat disk. Photons are
emitted in all directions from each *point* on the sphere. A star is not
a point source. It has many many point sources. We see all the photon
from a star that are nearly parallel to each other emitted from all the
many point sources on the surface of that sphere. Of course we don't see
the ones that go off in a direction not in our direction.
FrediFizzx
.
|
|
|
|
|
|
|
|
|
|
|
| User: "Peter Riedt" |
|
| Title: Re: MMX and Time Dilation |
15 Oct 2003 02:57:44 PM |
|
|
"Androcles" <jp006f9750@blurbblueyonder.co.uk> wrote in message news:<wB%ib.160$WR5.144@news-binary.blueyonder.co.uk>...
"Peter Riedt" <riedt1@yahoo.co.uk> wrote in message
news:1d36893d.0310121841.356b73ac@posting.google.com...
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. 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 | | |
