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Science Forum Index » Physics - Relativity Forum » Was Einstein a fake?
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| Author |
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| Pentcho Valev |
Posted: Mon Apr 28, 2008 5:54 am |
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Guest
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On Apr 28, 5:21 pm, Mike <elea...@yahoo.gr> wrote:
Quote: On Apr 28, 10:53 am, Pentcho Valev <pva...@yahoo.com> wrote:
On Apr 28, 3:28 pm, Mike <elea...@yahoo.gr> wrote:
On Apr 28, 7:15 am, Pentcho Valev <pva...@yahoo.com> wrote:
On Apr 26, 1:48 pm, "Dirk Van de moortel" <dirkvandemoor...@ThankS-NO-
SperM.hotmail.com> wrote:
Nice read:
http://www.cosmosmagazine.com/node/1162
"There's nothing quite like Einstein and his theories of
relativity to bring out the doubters, the cranks and the
outright crackpots. Do they have a point? Was Einstein
a fake?"
Moortel Moortel are you getting masochistic? Of course Divine Albert
was a fake - the following text clearly shows that - just analyze it
carefully:
http://www.bartleby.com/173/22.html
Divine Albert: "In the second place our result shows that, according
to the general theory of relativity, the law of the constancy of the
velocity of light in vacuo, which constitutes one of the two
fundamental assumptions in the special theory of relativity and to
which we have already frequently referred, cannot claim any unlimited
validity. A curvature of rays of light can only take place when the
velocity of propagation of light varies with position. Now we might
think that as a consequence of this, the special theory of relativity
and with it the whole theory of relativity would be laid in the dust..
But in reality this is not the case."
Pentcho Valev
pva...@yahoo.com
So why do you think the above statement by Dr. Al shows he was a
fake?
Be explicit.
Because, since the speed of light varies with the gravitational
potential, V, in accordance with the equation c'=c(1+V/c^2), it also
varies with the relative speed of the light source and the observer,
v, in accordance with the EQUIVALENT equation c'=c+v.
Can you please explain what your equations mean and how they are
related? Also, what is an EQUIVALENT equation?
Einstein's 1911 equation c'=c(1+V/c^2) and the eqution c'=c+v given by
Newton's emission theory of light (have you ever noticed these two
equations?) are EQUIVALENT in the sense that, if you start from the
one and apply Einstein's EQUIVALENCE principle, you obtain the other.
You have never noticed the texbook problem involving a tower (presence
of a gravitational field) and an accelerating rocket (absence of a
gravitational field), have you?
Pentcho Valev
pvalev@yahoo.com |
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| Androcles |
Posted: Mon Apr 28, 2008 9:15 am |
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Guest
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--
This message is brought to you by Androcles
http://www.androcles01.pwp.blueyonder.co.uk/
"Alen" <alen1@westserv.net.au> wrote in message
news:b1fd38f4-b8f3-4372-b5d2-399aa74cc2d8@s33g2000pri.googlegroups.com...
On Apr 28, 4:15 pm, Bryan Olson <fakeaddr...@nowhere.org> wrote:
Quote: Alen wrote:
[...]
I dispute Minkowski spacetime. Therefore I have a
self-inflicted blindness, won't learn, am unteachable,
a kook, a crackpot, etc., etc.
Actually, Alen, you thought that the theory predicts, in one
case you described, two observers *in the same place* will each
see the other to be younger. Minkowski spacetime and SR say
otherwise; 'twas your own error. No big deal getting something
wrong, but to this day there has been no correction on your part.
Thus we see that in your case, Alen, "won't learn" is not
something "orthodox supporters" merely assume. It is what you
yourself demonstrate.
| Not at all! In your view, yes, that may well be what I
| demonstrate. But, in my view, I demonstrate no such
| thing. Minkowski spacetime (not SR) may well say
| otherwise than what I say, but that is no imperative to
| me that I should 'correct' myself. My alternative to
| Minkowski spacetime works perfectly well with the
| SR equations, and that is all that is required, as far
| as I am concerned. I am under no obligation to satisfy
| the requirements of a theory I dispute, and to which
| I suggest an alternative. But this stance of mine
| appears to be impossible for the orthodox supporters
| to grasp. They appear to be incapable of saying 'we
| don't agree with your version of SR', but only things
| like 'your error', 'your failure to learn', etc., etc.
| Alen
Take the stupid kook up on it.
The theory predicts two observers *in the same place* will each
see the other to be younger. Minkowski spacetime and SR say
exactly that. |
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| zzbunker@netscape.net |
Posted: Mon Apr 28, 2008 9:23 am |
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Guest
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On Apr 28, 7:15 am, Pentcho Valev <pva...@yahoo.com> wrote:
Quote: On Apr 26, 1:48 pm, "Dirk Van de moortel" <dirkvandemoor...@ThankS-NO-
SperM.hotmail.com> wrote:
Nice read:
http://www.cosmosmagazine.com/node/1162
"There's nothing quite like Einstein and his theories of
relativity to bring out the doubters, the cranks and the
outright crackpots. Do they have a point? Was Einstein
a fake?"
Moortel Moortel are you getting masochistic? Of course Divine Albert
was a fake - the following text clearly shows that - just analyze it
carefully:
http://www.bartleby.com/173/22.html
Divine Albert: "In the second place our result shows that, according
to the general theory of relativity, the law of the constancy of the
velocity of light in vacuo, which constitutes one of the two
fundamental assumptions in the special theory of relativity and to
which we have already frequently referred, cannot claim any unlimited
validity. A curvature of rays of light can only take place when the
velocity of propagation of light varies with position. Now we might
think that as a consequence of this, the special theory of relativity
and with it the whole theory of relativity would be laid in the dust.
But in reality this is not the case."
Well, it is very much the case.
Since we also we invented robots in honor of Einstein.
Since he apparently the first phycist since Maxwell
who knew what "in vaccuo" meant.
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| BradGuth |
Posted: Mon Apr 28, 2008 2:08 pm |
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Guest
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On Apr 28, 4:15 am, Pentcho Valev <pva...@yahoo.com> wrote:
Quote: On Apr 26, 1:48 pm, "Dirk Van de moortel" <dirkvandemoor...@ThankS-NO-
SperM.hotmail.com> wrote:
Nice read:
http://www.cosmosmagazine.com/node/1162
"There's nothing quite like Einstein and his theories of
relativity to bring out the doubters, the cranks and the
outright crackpots. Do they have a point? Was Einstein
a fake?"
Moortel Moortel are you getting masochistic? Of course Divine Albert
was a fake - the following text clearly shows that - just analyze it
carefully:
http://www.bartleby.com/173/22.html
Divine Albert: "In the second place our result shows that, according
to the general theory of relativity, the law of the constancy of the
velocity of light in vacuo, which constitutes one of the two
fundamental assumptions in the special theory of relativity and to
which we have already frequently referred, cannot claim any unlimited
validity. A curvature of rays of light can only take place when the
velocity of propagation of light varies with position. Now we might
think that as a consequence of this, the special theory of relativity
and with it the whole theory of relativity would be laid in the dust.
But in reality this is not the case."
Pentcho Valev
pva...@yahoo.com
Lots of power groups that are not hardly smart enough to blow their
own nose, but otherwise collectively rich and in control of far more
than their fair share, have been known to select upon a likely
candidate that'll make for a good enough puppet that'll work nicely on
their behalf. Why not Einstein?
.. - Brad Guth |
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| Alen |
Posted: Mon Apr 28, 2008 6:54 pm |
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Guest
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On Apr 29, 12:15 am, "Androcles" <Headmas...@Hogwarts.physics> wrote:
Quote: --
This message is brought to you by Androcles
http://www.androcles01.pwp.blueyonder.co.uk/
"Alen" <al...@westserv.net.au> wrote in message
[...]
| Not at all! In your view, yes, that may well be what I
| demonstrate. But, in my view, I demonstrate no such
| thing. Minkowski spacetime (not SR) may well say
| otherwise than what I say, but that is no imperative to
| me that I should 'correct' myself. My alternative to
| Minkowski spacetime works perfectly well with the
| SR equations, and that is all that is required, as far
| as I am concerned. I am under no obligation to satisfy
| the requirements of a theory I dispute, and to which
| I suggest an alternative. But this stance of mine
| appears to be impossible for the orthodox supporters
| to grasp. They appear to be incapable of saying 'we
| don't agree with your version of SR', but only things
| like 'your error', 'your failure to learn', etc., etc.
| Alen
Take the stupid kook up on it.
The theory predicts two observers *in the same place* will each
see the other to be younger. Minkowski spacetime and SR say
exactly that.
You might try to argue that, but I think they will come
back with the supposed permanent nonsimultaneity
relationship between inertial frames, or use a third reference
frame, or both, and you will then have to contest that, etc.
Since I dispute Minkowski spacetime for fundamental reasons,
I have little interest in engaging in arguments about what
it says about anything. Obviously, if a theory is false, then
the solutions it gives to particular problems automatically
become irrelevant, and a waste of time.
Alen |
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| Bob |
Posted: Mon Apr 28, 2008 10:12 pm |
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Guest
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On Apr 29, 1:08 am, BradGuth <bradg...@gmail.com> wrote:
Quote: On Apr 28, 4:15 am, Pentcho Valev <pva...@yahoo.com> wrote:
On Apr 26, 1:48 pm, "Dirk Van de moortel" <dirkvandemoor...@ThankS-NO-
SperM.hotmail.com> wrote:
Nice read:
http://www.cosmosmagazine.com/node/1162
"There's nothing quite like Einstein and his theories of
relativity to bring out the doubters, the cranks and the
outright crackpots. Do they have a point? Was Einstein
a fake?"
Moortel Moortel are you getting masochistic? Of course Divine Albert
was a fake - the following text clearly shows that - just analyze it
carefully:
http://www.bartleby.com/173/22.html
Divine Albert: "In the second place our result shows that, according
to the general theory of relativity, the law of the constancy of the
velocity of light in vacuo, which constitutes one of the two
fundamental assumptions in the special theory of relativity and to
which we have already frequently referred, cannot claim any unlimited
validity. A curvature of rays of light can only take place when the
velocity of propagation of light varies with position. Now we might
think that as a consequence of this, the special theory of relativity
and with it the whole theory of relativity would be laid in the dust.
But in reality this is not the case."
Pentcho Valev
pva...@yahoo.com
Lots of power groups that are not hardly smart enough to blow their
own nose, but otherwise collectively rich and in control of far more
than their fair share, have been known to select upon a likely
candidate that'll make for a good enough puppet that'll work nicely on
their behalf. Why not Einstein?
. - Brad Guth
And so I unsubscribe to this ridiculous group. |
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| Back to top |
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| Bryan Olson |
Posted: Mon Apr 28, 2008 10:45 pm |
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Guest
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Alen wrote:
Quote: Bryan Olson wrote:
Alen wrote:
[...]
I dispute Minkowski spacetime. Therefore I have a
self-inflicted blindness, won't learn, am unteachable,
a kook, a crackpot, etc., etc.
Actually, Alen, you thought that the theory predicts, in one
case you described, two observers *in the same place* will each
see the other to be younger. Minkowski spacetime and SR say
otherwise; 'twas your own error. No big deal getting something
wrong, but to this day there has been no correction on your part.
Thus we see that in your case, Alen, "won't learn" is not
something "orthodox supporters" merely assume. It is what you
yourself demonstrate.
Not at all! In your view, yes, that may well be what I
demonstrate. But, in my view, I demonstrate no such
thing. Minkowski spacetime (not SR) may well say
otherwise than what I say, but that is no imperative to
me that I should 'correct' myself.
You are under no imperative to correct your error, that's true.
You attributed to the the orthodox theory a prediction that it
does not make. I'm not saying you have to correct it; I'm saying
not correcting it demonstrates the "won't learn" trait.
Quote: My alternative to
Minkowski spacetime works perfectly well with the
SR equations, and that is all that is required, as far
as I am concerned.
"As far as I am concerned"... ah, I see.
Quote: I am under no obligation to satisfy
the requirements of a theory I dispute, and to which
I suggest an alternative. But this stance of mine
appears to be impossible for the orthodox supporters
to grasp. They appear to be incapable of saying 'we
don't agree with your version of SR', but only things
like 'your error', 'your failure to learn', etc., etc.
And you will not grasp that you got a physics problem wrong. The
theory does not predict what you thought it does.
--
--Bryan |
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| Don Stockbauer |
Posted: Tue Apr 29, 2008 12:15 am |
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Guest
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Was Einstein a fake?
************************************************
It's much, much worse than anyone could possibly imagine.
He was just a cardboard cutout. |
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| rbwinn |
Posted: Tue Apr 29, 2008 1:22 am |
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Guest
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On Apr 29, 12:47Â am, Bryan Olson <fakeaddr...@nowhere.org> wrote:
Quote: rbwinn wrote:
On Apr 28, 12:10 am, Bryan Olson <fakeaddr...@nowhere.org> wrote:
rbwinn wrote:
On Apr 27, 6:14 pm, Bryan Olson <fakeaddr...@nowhere.org> wrote:
rbwinn wrote:
Bryan Olson wrote:
rbwinn wrote:
The Galilean transformation equations can be applied in any context.
And they describe the universe well in contexts where the motion
is uniform and the velocity is small.
All it means is that a clock in the stationary frame of reference
represents the equation t'=t. �Any clocks running at some other rate,
including cesium clocks in the moving frame of reference, have to be
represented by some other variable than t' because t' is already
defined to be t in the Galilean transformation equations.
If clocks at rest in a coordinate system do not measure the time
in that coordinate system, what does?
Time is relative, Bryan. Â A cesium clock in S' defines one measurement
of time. Â t'=t defines another measurement of time. Â The coordinates
of the Galilean transformation equation will not work unless t'=t. Â n'
is the rate of a clock running slower than t'=t. Â A clock can be made
to run at any speed. Â Two clocks running at any speed can be used to
represent t'=t as long as they are running at the same rate.
So no clue then?
The definition of t' is the time dimension in coordinate system S'.
t' = t is an assertion of the theory, not a definition. The Winn
transform simply contradicts the Galilean transform, and unlike both
the Galilean transform and Lorentz transform, the Winn transform
contradicts the principle of relativity.
Well, it is easy to make a statement, more difficult to prove one.
Good that you understand that much. You've posted a lot of novel
science, while any proof or evidence remains to be seen.
Here are the Galilean transformation equations:
         x'=x-vt
         y'=y
         z'=z
         t'=t
Agreed.
What the equations say is that if S is the frame of reference at rest,
then time on a cesium clock in S has to be used as t'=t.
What a mess.
The equations pertain to events identifiable in space and time, and
to two coordinate systems in which me can specify positions in space
and time. Call the coordinate systems S and S'; for any event, if
(x, y, z, t) are the coordinate of that event in S, then
(x', y', z', t') are, by definition, the coordinates of that same
event in S'.
The equations further assume that the origins of the coordinate
systems coincide. An event at (x=0, y=0, z=0, t=0) in S has
coordinates (x'=0, y'=0, z'=0, t'=0) in S'. Furthermore, the axis
of the two systems must be parallel; and any movement of one
coordinate system with respect to the other must be entirely in
the x direction (same as the x' direction); and such movement must
be uniform, meaning constant velocity.
There is no particular reference to the element cesium in the
Galilean transformation. The element was discovered in 1860, some
218 yeas after Galileo's death. The coordinate system has a time
dimension; a cesium clock is relevant insofar as it measures that
dimension, and no farther.
So there is another cesium clock in S' running at a slower rate, say
scientists.
No, scientists insist on saying things unambiguously, accurately,
and precisely as possible.
They want to call that cesium clock t'.
Mr. Winn has that wrong. Scientists understand t' is, by definition,
the time axis in S'.
No, sorry, if
scientists want to use these correct equations, they will have to call
time on the cesium clock in S' by something other than t'. Â t' is
already defined to be t'=t. Â We refuse to do this, say scientists.
The world in Winn's head has diverged so far from the real one -- and
his idea of scientific understanding has gone so wrong, that we can do
nothing better than put a big red X through Winn's ramblings. To learn
what scientists say, listen to scientists.
  Well, then, play with your imaginary equations.  They actually do
give you a fairly accurate answer. Â However, to find out what time on
the cesium clock in S' actually says, we can use the Galilean
transformation equations and call the time on the clock in S' by the
variable n'.
         n'=t(1-v/c)
The Galilean transformation has no "n'", and such is greatly to the
credit of the Galilean transformation.
Look, after ten or twelve times of trying to get a scientist to admit
that two clocks running at different rates cannot be represented by
the equation t'=t, I think that this basic principle has been
emphasized enough.
There's no question that the same variable cannot represent two
non-equal quantities. The problem is that you claim to use the
Galilean transform, then you contradict the Galilean transform.
I understand what you are saying. Â You are a
respected scientist and Party member,
You must have me confused with someone else.
Spoken like a scientist. Only a scientist could claim that I am
contradicting the Galilean transformation equations by using them.
Exactly how am I contradicting the Galilean transformation equations?
Robert B. Winn |
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| Bryan Olson |
Posted: Tue Apr 29, 2008 2:47 am |
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Guest
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rbwinn wrote:
Quote: On Apr 28, 12:10 am, Bryan Olson <fakeaddr...@nowhere.org> wrote:
rbwinn wrote:
On Apr 27, 6:14 pm, Bryan Olson <fakeaddr...@nowhere.org> wrote:
rbwinn wrote:
Bryan Olson wrote:
rbwinn wrote:
The Galilean transformation equations can be applied in any context.
And they describe the universe well in contexts where the motion
is uniform and the velocity is small.
All it means is that a clock in the stationary frame of reference
represents the equation t'=t. �Any clocks running at some other rate,
including cesium clocks in the moving frame of reference, have to be
represented by some other variable than t' because t' is already
defined to be t in the Galilean transformation equations.
If clocks at rest in a coordinate system do not measure the time
in that coordinate system, what does?
Time is relative, Bryan. A cesium clock in S' defines one measurement
of time. t'=t defines another measurement of time. The coordinates
of the Galilean transformation equation will not work unless t'=t. n'
is the rate of a clock running slower than t'=t. A clock can be made
to run at any speed. Two clocks running at any speed can be used to
represent t'=t as long as they are running at the same rate.
So no clue then?
The definition of t' is the time dimension in coordinate system S'.
t' = t is an assertion of the theory, not a definition. The Winn
transform simply contradicts the Galilean transform, and unlike both
the Galilean transform and Lorentz transform, the Winn transform
contradicts the principle of relativity.
Well, it is easy to make a statement, more difficult to prove one.
Good that you understand that much. You've posted a lot of novel
science, while any proof or evidence remains to be seen.
Here are the Galilean transformation equations:
x'=x-vt
y'=y
z'=z
t'=t
Agreed.
What the equations say is that if S is the frame of reference at rest,
then time on a cesium clock in S has to be used as t'=t.
What a mess.
The equations pertain to events identifiable in space and time, and
to two coordinate systems in which me can specify positions in space
and time. Call the coordinate systems S and S'; for any event, if
(x, y, z, t) are the coordinate of that event in S, then
(x', y', z', t') are, by definition, the coordinates of that same
event in S'.
The equations further assume that the origins of the coordinate
systems coincide. An event at (x=0, y=0, z=0, t=0) in S has
coordinates (x'=0, y'=0, z'=0, t'=0) in S'. Furthermore, the axis
of the two systems must be parallel; and any movement of one
coordinate system with respect to the other must be entirely in
the x direction (same as the x' direction); and such movement must
be uniform, meaning constant velocity.
There is no particular reference to the element cesium in the
Galilean transformation. The element was discovered in 1860, some
218 yeas after Galileo's death. The coordinate system has a time
dimension; a cesium clock is relevant insofar as it measures that
dimension, and no farther.
So there is another cesium clock in S' running at a slower rate, say
scientists.
No, scientists insist on saying things unambiguously, accurately,
and precisely as possible.
They want to call that cesium clock t'.
Mr. Winn has that wrong. Scientists understand t' is, by definition,
the time axis in S'.
No, sorry, if
scientists want to use these correct equations, they will have to call
time on the cesium clock in S' by something other than t'. t' is
already defined to be t'=t. We refuse to do this, say scientists.
The world in Winn's head has diverged so far from the real one -- and
his idea of scientific understanding has gone so wrong, that we can do
nothing better than put a big red X through Winn's ramblings. To learn
what scientists say, listen to scientists.
Well, then, play with your imaginary equations. They actually do
give you a fairly accurate answer. However, to find out what time on
the cesium clock in S' actually says, we can use the Galilean
transformation equations and call the time on the clock in S' by the
variable n'.
n'=t(1-v/c)
The Galilean transformation has no "n'", and such is greatly to the
credit of the Galilean transformation.
Look, after ten or twelve times of trying to get a scientist to admit
that two clocks running at different rates cannot be represented by
the equation t'=t, I think that this basic principle has been
emphasized enough.
There's no question that the same variable cannot represent two
non-equal quantities. The problem is that you claim to use the
Galilean transform, then you contradict the Galilean transform.
Quote: I understand what you are saying. You are a
respected scientist and Party member,
You must have me confused with someone else.
--
--Bryan |
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| Androcles |
Posted: Tue Apr 29, 2008 4:22 am |
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Guest
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This message is brought to you by Androcles
http://www.androcles01.pwp.blueyonder.co.uk/
"Alen" <alen1@westserv.net.au> wrote in message
news:1b453aa0-45f7-416b-a052-693719aebcef@w8g2000prd.googlegroups.com...
On Apr 29, 12:15 am, "Androcles" <Headmas...@Hogwarts.physics> wrote:
Quote: --
This message is brought to you by Androcles
http://www.androcles01.pwp.blueyonder.co.uk/
"Alen" <al...@westserv.net.au> wrote in message
[...]
| Not at all! In your view, yes, that may well be what I
| demonstrate. But, in my view, I demonstrate no such
| thing. Minkowski spacetime (not SR) may well say
| otherwise than what I say, but that is no imperative to
| me that I should 'correct' myself. My alternative to
| Minkowski spacetime works perfectly well with the
| SR equations, and that is all that is required, as far
| as I am concerned. I am under no obligation to satisfy
| the requirements of a theory I dispute, and to which
| I suggest an alternative. But this stance of mine
| appears to be impossible for the orthodox supporters
| to grasp. They appear to be incapable of saying 'we
| don't agree with your version of SR', but only things
| like 'your error', 'your failure to learn', etc., etc.
| Alen
Take the stupid kook up on it.
The theory predicts two observers *in the same place* will each
see the other to be younger. Minkowski spacetime and SR say
exactly that.
| You might try to argue that, but I think they will come
| back with the supposed permanent nonsimultaneity
| relationship between inertial frames, or use a third reference
| frame, or both, and you will then have to contest that, etc.
| Since I dispute Minkowski spacetime for fundamental reasons,
| I have little interest in engaging in arguments about what
| it says about anything. Obviously, if a theory is false, then
| the solutions it gives to particular problems automatically
| become irrelevant, and a waste of time.
Ok, well, nevertheless the theory SPECIFICALLY predicts two
observers *in the same place* will each see the other to be younger.
I don't need to argue it, I can point it out and say "here it is, look".
Your argument is to simply deny it, leading to the pantomime:
Oh yes it does!
Oh no it doesn't!
ad infinitum.
Here it is, look:
"If we assume that the result proved for a polygonal line is also valid for
a continuously curved line, we arrive at this result: If one of two
synchronous clocks at A is moved in a closed curve with constant velocity
until it returns to A, the journey lasting t seconds, then by the clock
which has remained at rest the travelled clock on its arrival at A will be
1/2t v^2/c^2 second slow"
It'll be 1/2t v^2/c^2 years slow by now too.
Fundamentally,
The speed of light from A to B is c-v and the time taken is tau.
The speed of light from B to A is c+v and the time taken is tau.
Kook Einstein said so and that's what his crank equations are for.
Minkowski is just the tail wagging the dog.
Here it is, look:
" we establish by definition that the ``time'' required by light to travel
from A to B equals the ``time'' it requires to travel from B to A. "
http://www.fourmilab.ch/etexts/einstein/specrel/www/
If A is Earth and B is Saturn, the time is 1 hour and 10 minutes.
The Cassini spacecraft is currently there, JPL have not been shouting
about how the proved Einstein right. They've got more sense.
But ok, Alen, you argue with the kook that he's wrong, you'll both enjoy it
more than proving it and getting on to real physics. |
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| Koobee Wublee |
Posted: Tue Apr 29, 2008 7:51 am |
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Guest
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On Apr 28, 12:23 pm, "zzbun...@netscape.net" wrote:
Quote: On Apr 28, 7:15 am, Pentcho Valev wrote:
Moortel Moortel are you getting masochistic? Of course Divine Albert
was a fake - the following text clearly shows that - just analyze it
carefully:
http://www.bartleby.com/173/22.html
Divine Albert: "In the second place our result shows that, according
to the general theory of relativity, the law of the constancy of the
velocity of light in vacuo, which constitutes one of the two
fundamental assumptions in the special theory of relativity and to
which we have already frequently referred, cannot claim any unlimited
validity. A curvature of rays of light can only take place when the
velocity of propagation of light varies with position. Now we might
think that as a consequence of this, the special theory of relativity
and with it the whole theory of relativity would be laid in the dust.
But in reality this is not the case."
Well, it is very much the case.
Since we also we invented robots in honor of Einstein.
Since he apparently the first phycist since Maxwell
who knew what "in vaccuo" meant.
Is this a joke? Einstein through his only contribution in science
gave us the Cosmological constant. The concept subscribes to negative
mass density in vacuum. It is utterly absurd. Yet, it does not
prevent physicists from swallowing this bullsh*t after faithfullyh
equating it to negative energy density in vacuum as if they do no
longer understand (E = m c^2).
Any robot built in honor of Einstein also represent the nitwit, the
plagiarist, and the liar in Einstein and of the following Orwelling
qualities.
** MYSTICISM IS WISDOM
** FAITH IS THEORY
** CONJECTURE IS TRUTH
** PLAGIARISM IS CREATIVITY
** LYING IS TEACHING |
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| BradGuth |
Posted: Tue Apr 29, 2008 9:34 am |
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On Apr 29, 1:12 am, Bob <lizard.lo...@gmail.com> wrote:
Quote:
And so I unsubscribe to this ridiculous group.
Are we having another bad Usenet/Group kind of day?
As puppets of the past, present and future tend to go, why not
Einstein?
This isn't saying that Einstein wasn't a very smart puppet.
.. - Brad Guth |
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| Guest |
Posted: Tue Apr 29, 2008 10:35 am |
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On Apr 28, 3:15 am, Pentcho Valev <pva...@yahoo.com> wrote:
Quote: On Apr 26, 1:48 pm, "Dirk Van de moortel" <dirkvandemoor...@ThankS-NO-
SperM.hotmail.com> wrote:
Nice read:
http://www.cosmosmagazine.com/node/1162
"There's nothing quite like Einstein and his theories of
relativity to bring out the doubters, the cranks and the
outright crackpots. Do they have a point? Was Einstein
a fake?"
Moortel Moortel are you getting masochistic? Of course Divine Albert
was a fake - the following text clearly shows that - just analyze it
carefully:
http://www.bartleby.com/173/22.html
Divine Albert: "In the second place our result shows that, according
to the general theory of relativity, the law of the constancy of the
velocity of light in vacuo, which constitutes one of the two
fundamental assumptions in the special theory of relativity and to
which we have already frequently referred, cannot claim any unlimited
validity. A curvature of rays of light can only take place when the
velocity of propagation of light varies with position. Now we might
think that as a consequence of this, the special theory of relativity
and with it the whole theory of relativity would be laid in the dust.
But in reality this is not the case."
Pentcho Valev
pva...@yahoo.com
Light moves at c in slower time.
Mitch Raemsch |
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| Eric Gisse |
Posted: Tue Apr 29, 2008 6:37 pm |
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On Apr 29, 3:22Â am, rbwinn <rbwi...@juno.com> wrote:
Quote: On Apr 29, 12:47Â am, Bryan Olson <fakeaddr...@nowhere.org> wrote:
rbwinn wrote:
On Apr 28, 12:10 am, Bryan Olson <fakeaddr...@nowhere.org> wrote:
rbwinn wrote:
On Apr 27, 6:14 pm, Bryan Olson <fakeaddr...@nowhere.org> wrote:
rbwinn wrote:
Bryan Olson wrote:
rbwinn wrote:
The Galilean transformation equations can be applied in any context.
And they describe the universe well in contexts where the motion
is uniform and the velocity is small.
All it means is that a clock in the stationary frame of reference
represents the equation t'=t. �Any clocks running at some other rate,
including cesium clocks in the moving frame of reference, have to be
represented by some other variable than t' because t' is already
defined to be t in the Galilean transformation equations.
If clocks at rest in a coordinate system do not measure the time
in that coordinate system, what does?
Time is relative, Bryan. Â A cesium clock in S' defines one measurement
of time. Â t'=t defines another measurement of time. Â The coordinates
of the Galilean transformation equation will not work unless t'=t. Â n'
is the rate of a clock running slower than t'=t. Â A clock can be made
to run at any speed. Â Two clocks running at any speed can be used to
represent t'=t as long as they are running at the same rate.
So no clue then?
The definition of t' is the time dimension in coordinate system S'.
t' = t is an assertion of the theory, not a definition. The Winn
transform simply contradicts the Galilean transform, and unlike both
the Galilean transform and Lorentz transform, the Winn transform
contradicts the principle of relativity.
Well, it is easy to make a statement, more difficult to prove one.
Good that you understand that much. You've posted a lot of novel
science, while any proof or evidence remains to be seen.
Here are the Galilean transformation equations:
         x'=x-vt
         y'=y
         z'=z
         t'=t
Agreed.
What the equations say is that if S is the frame of reference at rest,
then time on a cesium clock in S has to be used as t'=t.
What a mess.
The equations pertain to events identifiable in space and time, and
to two coordinate systems in which me can specify positions in space
and time. Call the coordinate systems S and S'; for any event, if
(x, y, z, t) are the coordinate of that event in S, then
(x', y', z', t') are, by definition, the coordinates of that same
event in S'.
The equations further assume that the origins of the coordinate
systems coincide. An event at (x=0, y=0, z=0, t=0) in S has
coordinates (x'=0, y'=0, z'=0, t'=0) in S'. Furthermore, the axis
of the two systems must be parallel; and any movement of one
coordinate system with respect to the other must be entirely in
the x direction (same as the x' direction); and such movement must
be uniform, meaning constant velocity.
There is no particular reference to the element cesium in the
Galilean transformation. The element was discovered in 1860, some
218 yeas after Galileo's death. The coordinate system has a time
dimension; a cesium clock is relevant insofar as it measures that
dimension, and no farther.
So there is another cesium clock in S' running at a slower rate, say
scientists.
No, scientists insist on saying things unambiguously, accurately,
and precisely as possible.
They want to call that cesium clock t'.
Mr. Winn has that wrong. Scientists understand t' is, by definition,
the time axis in S'.
No, sorry, if
scientists want to use these correct equations, they will have to call
time on the cesium clock in S' by something other than t'. Â t' is
already defined to be t'=t. Â We refuse to do this, say scientists.
The world in Winn's head has diverged so far from the real one -- and
his idea of scientific understanding has gone so wrong, that we can do
nothing better than put a big red X through Winn's ramblings. To learn
what scientists say, listen to scientists.
  Well, then, play with your imaginary equations.  They actually do
give you a fairly accurate answer. Â However, to find out what time on
the cesium clock in S' actually says, we can use the Galilean
transformation equations and call the time on the clock in S' by the
variable n'.
         n'=t(1-v/c)
The Galilean transformation has no "n'", and such is greatly to the
credit of the Galilean transformation.
Look, after ten or twelve times of trying to get a scientist to admit
that two clocks running at different rates cannot be represented by
the equation t'=t, I think that this basic principle has been
emphasized enough.
There's no question that the same variable cannot represent two
non-equal quantities. The problem is that you claim to use the
Galilean transform, then you contradict the Galilean transform.
I understand what you are saying. Â You are a
respected scientist and Party member,
You must have me confused with someone else.
Spoken like a scientist. Â Only a scientist could claim that I am
contradicting the Galilean transformation equations by using them.
Except you aren't using them - you are using them in addition to some
crap you have added.
Quote: Exactly how am I contradicting the Galilean transformation equations?
Why ask the question when you won't understand the answer?
> Robert B. Winn |
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