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TWIN PARADOX AND REDUCTIO AD ABSURDUM
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Pentcho Valev
Dec 8, 2011, 3:14:36 PM12/8/11
to
A twin paradox scenario where acceleration is avoided:
http://www.people.fas.harvard.edu/~djmorin/book.html
Introduction to Classical Mechanics With Problems and Solutions, David
Morin, Cambridge University Press, Chapter 11, p. 44: "11.19. Modified
twin paradox *** Consider the following variation of the twin paradox.
A, B, and C each have a clock. In A's reference frame, B flies past A
with speed v to the right. When B passes A, they both set their clocks
to zero. Also, in A's reference frame, C starts far to the right and
moves to the left with speed v. When B and C pass each other, C sets
his clock to read the same as B's. Finally, when C passes A, they
compare the readings on their clocks. At this moment, let A's clock
read TA, and let C's clock read TC. (a) Working in A's frame, show
that TC = TA/(gamma). (b) Working in B's frame, show again that TC =
TA/(gamma). (c) Working in C's frame, show again that TC = TA/
(gamma)."
Note that in this scenario, up to the moment when C sets his clock to
read the same as B's, A has been a genuine travelling twin who has
completed the outward part of his journey in B's frame. A's clock has
been running slow relative to clocks in B's frame, as judged from B's
frame.
A will remain a genuine travelling twin in C's frame and will perform
the rest of his journey if, as C sets his clock to read the same as
B's, simultaneously (in B's frame), at the location of A, another
clock belonging to C's frame is set to read the same as another clock
belonging to B's frame (it is assumed that clocks in B's frame are
synchronized). Then, as A moves between the two clocks belonging to
C's frame, his clock is running slow as judged from C's frame.
According to Einstein's special relativity, time dilation is
RECIPROCAL. This means that, in a twin paradox scenario where
acceleration is avoided, the sedentary twin CAN be interpreted as a
travelling twin. Needless to say, such an interpretation amounts to
REDUCTIO AD ABSURDUM - at the end of the journey, either twin proves
both younger and older than his brother.
Pentcho Valev
pva...@yahoo.com
http://www.people.fas.harvard.edu/~djmorin/book.html
Introduction to Classical Mechanics With Problems and Solutions, David
Morin, Cambridge University Press, Chapter 11, p. 44: "11.19. Modified
twin paradox *** Consider the following variation of the twin paradox.
A, B, and C each have a clock. In A's reference frame, B flies past A
with speed v to the right. When B passes A, they both set their clocks
to zero. Also, in A's reference frame, C starts far to the right and
moves to the left with speed v. When B and C pass each other, C sets
his clock to read the same as B's. Finally, when C passes A, they
compare the readings on their clocks. At this moment, let A's clock
read TA, and let C's clock read TC. (a) Working in A's frame, show
that TC = TA/(gamma). (b) Working in B's frame, show again that TC =
TA/(gamma). (c) Working in C's frame, show again that TC = TA/
(gamma)."
Note that in this scenario, up to the moment when C sets his clock to
read the same as B's, A has been a genuine travelling twin who has
completed the outward part of his journey in B's frame. A's clock has
been running slow relative to clocks in B's frame, as judged from B's
frame.
A will remain a genuine travelling twin in C's frame and will perform
the rest of his journey if, as C sets his clock to read the same as
B's, simultaneously (in B's frame), at the location of A, another
clock belonging to C's frame is set to read the same as another clock
belonging to B's frame (it is assumed that clocks in B's frame are
synchronized). Then, as A moves between the two clocks belonging to
C's frame, his clock is running slow as judged from C's frame.
According to Einstein's special relativity, time dilation is
RECIPROCAL. This means that, in a twin paradox scenario where
acceleration is avoided, the sedentary twin CAN be interpreted as a
travelling twin. Needless to say, such an interpretation amounts to
REDUCTIO AD ABSURDUM - at the end of the journey, either twin proves
both younger and older than his brother.
Pentcho Valev
pva...@yahoo.com
rotchm
Dec 8, 2011, 3:57:05 PM12/8/11
to
All analysis from any frame yields the same conclusion: The A twin
will be older than the (returning) C twin (A>C). There is no
contradiction. Note that the twins situations are NOT SYMMETRICAL.
Therefore you can not claim that the times on A and C be the same as
they coincide, nor can you claim that A<C.
Can YOU see/say why A and C are not symmetrical?
G. L. Bradford
Dec 8, 2011, 4:32:40 PM12/8/11
to
"rotchm" <rot...@gmail.com> wrote in message
news:e71080c3-f87a-47e7...@k23g2000yqa.googlegroups.com...
And c equals an increasingly infinite velocity because the traveling twin
never leaves the stay at home twin's frame of control.
--------------------
To explain that:
What of the potential *observation* the stay at home twin would have of
the traveling twin going away from that twin (light and time distances
stretching out between the twins, it taking light a longer and ever longer
time to reach back to the stay at home twin, putting the traveling twin --
in the stay at home twin's *observation* -- ever farther behind himself in
*observable time* and, therefore, observable age (altogether an appearance
of the traveling twin's clock slowing down in clocking time: slower, slower,
and slower with every fraction of a light second's increase in distance
between twins))? I answered and you can't answer since you have the
*observation of the traveling twin* from the stay at home twin's frame, or
point of view, as the actual physical reality of the traveling twin in the
traveling twin's own immediate locality (which
*current-universe-at-a-distance* in point of fact is unobservable by the
stay at home twin). You merge observability with reality, thus [your] c
equaling infinite velocity / light equaling an instantaneous transmission of
information.
GLB
====================
micro...@hotmail.com
Dec 8, 2011, 6:39:06 PM12/8/11
to
But there is a case whereby there is no lost time. And
That is when a high speed object is passing something in relatively
still in space.
The high speed object and what it is passing could
see together for who time or clock is going slow. That slow time is
always for the object (like a train) with the high speed
(from weighted acceleration.) The high speed in dimensionality will
see the relatively
stationary with fastest clock tick.
Pentcho Valev
Dec 9, 2011, 5:51:03 AM12/9/11
to
http://www.amazon.com/Relativity-Its-Roots-Banesh-Hoffmann/dp/0486406768
Relativity and Its Roots, Banesh Hoffmann, p. 105: "In one case your
clock is checked against two of mine, while in the other case my clock
is checked against two of yours, and this permits us each to find
without contradiction that the other's clocks go more slowly than his
own."
A scenario in which an "alien" clock moves between clocks at rest in
the measurement system is INDISPENSABLE for a system that is to
measure time dilation (the "alien" clock shows less time elapsed than
clocks at rest in the measurement system, according to special
relativity). Einsteiniana's thought experiments implicitly convert the
sedentary twin's system into a full-blooded measurement system (that
is, capable of measuring time dilation) while the travelling twin's
system is reduced to an "alien" clock moving between clocks at rest in
the sedentary twin's system. So the travelling twin always returns
younger and makes Einsteinians fiercely sing "Divine Einstein" and
"Yes we all believe in relativity, relativity, relativity".
Any scenario converting the sedentary twin's clock into an "alien"
clock moving between clocks at rest in another system, if analysed
correctly, refutes special relativity.
Relativity and Its Roots, Banesh Hoffmann, p. 105: "In one case your
clock is checked against two of mine, while in the other case my clock
is checked against two of yours, and this permits us each to find
without contradiction that the other's clocks go more slowly than his
own."
A scenario in which an "alien" clock moves between clocks at rest in
the measurement system is INDISPENSABLE for a system that is to
measure time dilation (the "alien" clock shows less time elapsed than
clocks at rest in the measurement system, according to special
relativity). Einsteiniana's thought experiments implicitly convert the
sedentary twin's system into a full-blooded measurement system (that
is, capable of measuring time dilation) while the travelling twin's
system is reduced to an "alien" clock moving between clocks at rest in
the sedentary twin's system. So the travelling twin always returns
younger and makes Einsteinians fiercely sing "Divine Einstein" and
"Yes we all believe in relativity, relativity, relativity".
Any scenario converting the sedentary twin's clock into an "alien"
clock moving between clocks at rest in another system, if analysed
correctly, refutes special relativity.
Pentcho Valev
Dec 9, 2011, 9:40:44 AM12/9/11
to
By definition, nothing is absurd in Einsteiniana's schizophrenic world
(REDUCTIO AD ABSURDUM is an obsolete procedure):
http://www.astro.gla.ac.uk/~norman/lectures/A2SR/part3.pdf
University of Glasgow: "A farmer with a 20m ladder holds it
horizontally and runs toward a barn which is 10m deep. The farmer's
wife, standing by the barn door, sees him running at a speed at which
gamma=2. The ladder is therefore length-contracted to have a measured
length of 10m in the barn's frame, so that the ladder will fit
entirely into the barn, and the farmer's wife can slam the door behind
him, with the '20m' ladder entirely (and briefly!) within the 10m
barn..."
http://www.liferesearchuniversal.com/1984-7
George Orwell: "In the end the Party would announce that two and two
made five, and you would have to believe it. It was inevitable that
they should make that claim sooner or later: the logic of their
position demanded it. Not merely the validity of experience, but the
very existence of external reality, was tacitly denied by their
philosophy. The heresy of heresies was common sense. And what was
terrifying was not that they would kill you for thinking otherwise,
but that they might be right. For, after all, how do we know that two
and two make four? Or that the force of gravity works? Or that the
past is unchangeable? If both the past and the external world exist
only in the mind, and if the mind itself is controllable what then?"
Pentcho Valev
pva...@yahoo.com
(REDUCTIO AD ABSURDUM is an obsolete procedure):
http://www.astro.gla.ac.uk/~norman/lectures/A2SR/part3.pdf
University of Glasgow: "A farmer with a 20m ladder holds it
horizontally and runs toward a barn which is 10m deep. The farmer's
wife, standing by the barn door, sees him running at a speed at which
gamma=2. The ladder is therefore length-contracted to have a measured
length of 10m in the barn's frame, so that the ladder will fit
entirely into the barn, and the farmer's wife can slam the door behind
him, with the '20m' ladder entirely (and briefly!) within the 10m
barn..."
http://www.liferesearchuniversal.com/1984-7
George Orwell: "In the end the Party would announce that two and two
made five, and you would have to believe it. It was inevitable that
they should make that claim sooner or later: the logic of their
position demanded it. Not merely the validity of experience, but the
very existence of external reality, was tacitly denied by their
philosophy. The heresy of heresies was common sense. And what was
terrifying was not that they would kill you for thinking otherwise,
but that they might be right. For, after all, how do we know that two
and two make four? Or that the force of gravity works? Or that the
past is unchangeable? If both the past and the external world exist
only in the mind, and if the mind itself is controllable what then?"
Pentcho Valev
pva...@yahoo.com
G. L. Bradford
Dec 9, 2011, 10:24:24 AM12/9/11
to
"Pentcho Valev" <pva...@yahoo.com> wrote in message
news:a140915d-2a89-4200...@f11g2000yql.googlegroups.com...
You have the slowing always going away. You have exactly the opposite
coming back, or coming toward. That movie goes into fast forward (that clock
goes into fast forward), along with the change to blue shifting from red
shifting. The second clock, the *observed clock* of the *observed traveling
twin* by the stay at home twin (which never leaves the stay at home twin's
frame) was always situated between the reals (0) closer to the stay at home
twin in the observed space of the observed *observable universe* but farther
away in time (-) than either of the realtime twins, c essentially,
eventually, becoming an awfully slow transmitter of information with all
increasing distance between the twins. Both of the twins are now historical
figures (-) in time, as *observed*, to the other realtime twin (0). But each
to the other, UPON THE TURN AROUND of the traveling twin, is the unobserved
and unobservable dark matter finish line (0) at the end of road of (-) to
(0) in time.
Where each was increasingly negative (-) to the other in time going away,
each is now increasingly positive (+) in time from that distant negative (-)
in time, in the traveling twin's return to home (0). That second *observed*
traveler (-) (to the stay at home (0)), and second *observed* stay at home
(-) (to the traveler (0)), those second *observed* clocks (-) (-) (one each
belonging to the frames of the traveling twin and the stay at home, residing
BETWEEN (-) the traveling twin and the stay at home in the observed
*OBSERVABLE UNIVERSE (-)*) remain situated between the two in the space-time
of the observable both going (-) (slowing down in time -- into history --
going away) and coming (+) (speeding up in time -- up and out from
history -- oncoming). This is where the idea of the time traveler, and of
time traveling back and forth in time, comes from.
To confuse the issue by now reaching into multi-dimensionality, by now
reaching into the multi-verse, and make severals' head ache from attempted
mind stretching, oddly enough the above VIEW is the same VIEW of *observed*
relative distancing and returning from observed relative distance as
supposedly accelerating up (therefore also an *observation* of slowing down
(going negative) in time) and decelerating down (therefore also an
*observation* of speeding up (going positive) in time) from the speed of
light (c) regarding observed relative velocity....observed from any distant
point of observation. And in some others' view I've read and agree with, the
VIEW from here and now (0) toward the most distant reach (realized by that
some to *be* the universe at the speed of light (c)) (1 light second distant
from here and now (0) toward distant (c), 1 light year distant...toward
distant (c), to 14 billion light years distant from here and now (0)
toward -and in that distant HORIZON, that distant UNIVERSE, the achievement
of- (c)) of the *OBSERVABLE* universe!.....the light coordinate universe,
the negative-time coordinate universe, the universe of nothing but
recessionist histories, observed out and away (-) from every point of an
infinity of heres and nows (0) in the unobserved, the unobservable, *current
Universe* (having a Universal Mean Time, and coordinate space-time point,
that effectively equals 0 (0 = 0 (= 0))).
------------------------
**(In a modern dark age of almost nothing but the shortest of attention
spans and so little retentive ability in so many mental 'pygmies' (evolved
forward in time from Gibbon's Rome's decline and fall), my requirement for a
long hold on a many more dimensioned, and/or a far more complex dimensioned,
thought than a short, much more simplistic, much more one-dimensional
thought (or something broken down into many, many, short, more simplistic,
more one-dimensional thoughts destroying the meat), a requirement for deep
attention to deep detail in a single long thought long held, may be far too
much for far too many (who wouldn't get it, who lose it *shortly*, even if I
did break it down). I don't have to give a damn for such lack of capacity
and capability in so many people anymore. Far too many won't have the
ability to even begin to realize what I *may* be talking about here, much
less what I am talking about here. My hope is for what it has always been, a
long term understanding developing over longer times than the shortest
seconds or minutes. But as I said, too many are incapable now of any long
term itself, much less any 'understanding', ever.)**
-----------------------
GLB
===================
Pentcho Valev
Dec 9, 2011, 4:33:33 PM12/9/11
to
Leonard Susskind is sure that, in the absence of acceleration, the
roles of the twins can be reversed - "then, surely the previously
stationary twin would age less than the previously moving twin":
http://www.lecture-notes.co.uk/susskind/special-relativity/lecture-4/time-dilation/
Leonard Susskind: "The supposed paradox is that surely, since
everything is relative, we can reverse the roles of the twins - the
stationary one is in fact moving at velocity -v relative to the
previously moving twin. Then, surely the previously stationary twin
would age less than the previously moving twin. The problem with this
argument is that the roles cannot be reversed. It must be the case
that the original moving twin has to experience some acceleration
during the motion, whilst the stationary one certainly does not."
Of course, if Susskind knew that no-acceleration versions of the twin
paradox exist, he would not have even thought of a younger stationary
twin and older moving twin:
http://www.liferesearchuniversal.com/1984-17
George Orwell: "Crimestop means the faculty of stopping short, as
though by instinct, at the threshold of any dangerous thought. It
includes the power of not grasping analogies, of failing to perceive
logical errors, of misunderstanding the simplest arguments if they are
inimical to Ingsoc, and of being bored or repelled by any train of
thought which is capable of leading in a heretical direction.
Crimestop, in short, means protective stupidity."
Pentcho Valev
pva...@yahoo.com
roles of the twins can be reversed - "then, surely the previously
stationary twin would age less than the previously moving twin":
http://www.lecture-notes.co.uk/susskind/special-relativity/lecture-4/time-dilation/
Leonard Susskind: "The supposed paradox is that surely, since
everything is relative, we can reverse the roles of the twins - the
stationary one is in fact moving at velocity -v relative to the
previously moving twin. Then, surely the previously stationary twin
would age less than the previously moving twin. The problem with this
argument is that the roles cannot be reversed. It must be the case
that the original moving twin has to experience some acceleration
during the motion, whilst the stationary one certainly does not."
Of course, if Susskind knew that no-acceleration versions of the twin
paradox exist, he would not have even thought of a younger stationary
twin and older moving twin:
http://www.liferesearchuniversal.com/1984-17
George Orwell: "Crimestop means the faculty of stopping short, as
though by instinct, at the threshold of any dangerous thought. It
includes the power of not grasping analogies, of failing to perceive
logical errors, of misunderstanding the simplest arguments if they are
inimical to Ingsoc, and of being bored or repelled by any train of
thought which is capable of leading in a heretical direction.
Crimestop, in short, means protective stupidity."
Pentcho Valev
pva...@yahoo.com
Helmut Wabnig
Dec 9, 2011, 6:41:50 PM12/9/11
to
Even Uncle Al's three-rocket version is not free of acceleration, as
he wants to suggest, because somebody has to get down to the right
corner in order to fly up to the left with his rocket before the
actual experiment starts.
w.
rotchm
Dec 9, 2011, 8:27:52 PM12/9/11
to
> There exists no no-accelerating version of the twin paradox.
> Even Uncle Al's three-rocket version is not free of acceleration,
and Bondi's (and few others) are totally void of accelerations:
B--> A"fixed" <--C
Helmut Wabnig
Dec 10, 2011, 5:01:17 AM12/10/11
to
On Fri, 9 Dec 2011 17:27:52 -0800 (PST), rotchm <rot...@gmail.com>
wrote:
You missed the point.
The acceleration occurs before the actual experiment.
B and C must enter their positions first.
Once they are there, the experiment can begin.
Why that is crucial, is another issue.
w.
wrote:
The acceleration occurs before the actual experiment.
B and C must enter their positions first.
Once they are there, the experiment can begin.
Why that is crucial, is another issue.
w.
rotchm
Dec 10, 2011, 9:33:32 AM12/10/11
to
> The acceleration occurs before the actual experiment.
> B and C must enter their positions first.
> Once they are there, the experiment can begin.
>
> Why that is crucial, is another issue.
>
> w.
Since the exp begins AFTER those accelerations. The exp begins when B
and A coincide; there, A, B and C have been inertial already for a
long time! Moreover, B and A set their clocks to zero as they
coincide (they turn them on); so their past accelerations are of no
consequence/effect on the clocks.
G. L. Bradford
Dec 10, 2011, 9:52:51 AM12/10/11
to
news:371ee3be-34be-4dec...@i6g2000vbe.googlegroups.com...
=======================
You and Suskind start from a radically wrong premise and you flatly refuse
to see it. Why don't you stop quoting Orwell (Blair) since every word you
quote in the above describes you perfectly. Again, neither twin is dealing
in anything real concerning the other twin. No such eventuality as each twin
becoming younger than the other, or older than the other, is in fact
happening to the twin physically within his own physical field of space
local to him.
Each twin is dealing in an observed past universe (the OBSERVABLE
UNIVERSE) in any *observation* of the other twin.
In a mutually shared current universe neither twin (0) is moving as each
is the dead center of the infinite Universe, (0 = 0). The traveling twin in
his vessel, in accelerating, expands his field of space, balloons his field
of space, contracting all the surrounding universe around him into his own
field, then contracts into the field of his destination even if that is his
return home, re-expanding the local universe around him, including his
destination, arriving at his destination. He balloons, inflates, fields of
space and deflates them. If we were dealing in Quantum Mechanics rather than
Relativity the traveling twin would move from being in only one place in the
universe to being in many places in the universe all at once, then back to
being in only one place once more.
But neither of these situations above is the *relative* position of each
unobserved and unobservable *dark* twin, rather than the observed light born
twin (past universe), to the other in time or space-time. When the traveling
twin departs the stay at home twin, he is observed (per the speed of light,
c) to travel into an observable past, the observable past universe.
*Relative* to that observed and observable past universe, and the observable
stay at home twin observed to be increasingly in its depths, the traveling
unobserved / unobservable *dark* twin effectively travels into an
unobservable future, the unobservable future universe. So to does a now
*dark* stay at home twin, now also in an unobservable future universe,
relative to the *dark* traveling twin, relatively speaking that is. The
relative positions of each *dark* twin to the other is (+) to (0) in the
unobservable future universe rather than the (-) to (0) observed of the past
universe or the expansionary / contractionary universe's (0 = 0).
The unobservable *dark* traveler from the direction of the relative future
universe he sailed off into, and the observable light born traveler from the
direction of the relative past universe he sailed off into, re-merge back
into one and the same traveler only when they re-merge back into the frame
of home (strictly regarding the "time traveling" environment and scenario of
the 'twin paradox' that is). The stay at home twin's situation is no
different than this relative to the traveling twin. He also splits up
between an observable past (light born) and an unobservable future (*dark*)
only to re-merge when the traveling twin arrives back home.
That is quite an expansion between *light born* and *dark* (to (-) and (+)
from (0)). And quite a contraction (from (-) and (+) to (0)).
GLB
====================
You and Suskind start from a radically wrong premise and you flatly refuse
to see it. Why don't you stop quoting Orwell (Blair) since every word you
quote in the above describes you perfectly. Again, neither twin is dealing
in anything real concerning the other twin. No such eventuality as each twin
becoming younger than the other, or older than the other, is in fact
happening to the twin physically within his own physical field of space
local to him.
Each twin is dealing in an observed past universe (the OBSERVABLE
UNIVERSE) in any *observation* of the other twin.
In a mutually shared current universe neither twin (0) is moving as each
is the dead center of the infinite Universe, (0 = 0). The traveling twin in
his vessel, in accelerating, expands his field of space, balloons his field
of space, contracting all the surrounding universe around him into his own
field, then contracts into the field of his destination even if that is his
return home, re-expanding the local universe around him, including his
destination, arriving at his destination. He balloons, inflates, fields of
space and deflates them. If we were dealing in Quantum Mechanics rather than
Relativity the traveling twin would move from being in only one place in the
universe to being in many places in the universe all at once, then back to
being in only one place once more.
But neither of these situations above is the *relative* position of each
unobserved and unobservable *dark* twin, rather than the observed light born
twin (past universe), to the other in time or space-time. When the traveling
twin departs the stay at home twin, he is observed (per the speed of light,
c) to travel into an observable past, the observable past universe.
*Relative* to that observed and observable past universe, and the observable
stay at home twin observed to be increasingly in its depths, the traveling
unobserved / unobservable *dark* twin effectively travels into an
unobservable future, the unobservable future universe. So to does a now
*dark* stay at home twin, now also in an unobservable future universe,
relative to the *dark* traveling twin, relatively speaking that is. The
relative positions of each *dark* twin to the other is (+) to (0) in the
unobservable future universe rather than the (-) to (0) observed of the past
universe or the expansionary / contractionary universe's (0 = 0).
The unobservable *dark* traveler from the direction of the relative future
universe he sailed off into, and the observable light born traveler from the
direction of the relative past universe he sailed off into, re-merge back
into one and the same traveler only when they re-merge back into the frame
of home (strictly regarding the "time traveling" environment and scenario of
the 'twin paradox' that is). The stay at home twin's situation is no
different than this relative to the traveling twin. He also splits up
between an observable past (light born) and an unobservable future (*dark*)
only to re-merge when the traveling twin arrives back home.
That is quite an expansion between *light born* and *dark* (to (-) and (+)
from (0)). And quite a contraction (from (-) and (+) to (0)).
GLB
====================
xxein
Dec 11, 2011, 10:44:57 PM12/11/11
to
Drostie
Dec 12, 2011, 6:13:15 AM12/12/11
to
For example, I myself avoided commenting because the paragraph
beginning "A will remain a genuine travelling twin..." contains a
bunch of bizarre discussion. It's not clear what Petcho is claiming
with this paragraph, and it's not clear that he's done the analysis
right.
The paragraph in question starts to invoke an idea that C synchronizes
two clocks separated by a distance L (as measured by B) with two in-
sync clocks in B's own reference frame.
Of course, in special relativity, C does not see these two clocks as
in-sync and does not even think that they are separated by this
distance L. And if B were to "look back" upon C's clocks they would
appear to be ticking slowly.
So it would *appear* to be the case that B would say, "hey, C's clocks
don't represent anything meaningful to me -- they tick too slowly!"
and C would also say, "hey, these clocks don't represent anything
meaningful to me either -- they're out of sync!".
So it's not clear what Pentcho Valev is trying to assert here.
Apparently if nobody uses meaningful clocks, some sort of bizarre
consequence occurs in trying to make a meaningful interpretation of
what's going on. As is very common with scientific theories, there are
two levels: "here is the theory which helps me phrase and investigate
the question" and "here is the model within that theory which I am
testing against experiment." His claimed "reductio" in this case seems
to discard the model being used, and not the theory being used.
He seems to be claiming that special relativity (the idea that we
should admit Lorentz boosts alongside coordinate rotations etc.) is
somehow incoherent. This is actually mathematically very easy to
disprove, via the matrix multiplication:
[g -ug] [g ug] = [1 0]
[-ug g] [ug g] = [0 1]
...when g = 1/sqrt(1 - u^2). This says that if you Lorentz boost with
velocity + c u, and then Lorentz boost with velocity -c u, you get
back to your original coordinate system directly. So Lorentz boosts
have the same mathematical consistency as the classical Galilean
transform:
[1 -u] [1 u] = [1 0]
[0 1] [0 1] = [0 1]
In fact, we can define the Lorentz group as a set of transformations
in a 4-dimensional space preserving a certain metric, much like the
rotations in 3-dimensional space preserve a related metric (actually,
those rotations are part of the Lorentz group). So we've not just got
a theory, but we've got a very nice mathematics to elucidate that
theory.
It also doesn't help that he claims that we're somehow caught up in a
shell of "protective stupidity." We really are normal people, trying
our hardest to understand the world. We choose theories which help us
explain stuff. That's all.
-- Chris Drost
Androcles
Dec 12, 2011, 9:56:58 AM12/12/11
to
"Drostie" <chris....@gmail.com> wrote in message
news:574100a9-fa92-498f...@e2g2000vbb.googlegroups.com...
Bizarre discussion snipped.
The Einstein length expansion is not the Lorentz length contraction,
nor was Lorentz talking about time, so anything you say about Lorentz
is both bizarre and stupid.
Drostie
Dec 12, 2011, 1:16:34 PM12/12/11
to
On Dec 12, 3:56 pm, "Androcles" <Headmas...@Hogwarts.physics.December.
2011> wrote:
> "Drostie" <chris.dros...@gmail.com> wrote in message
Source?
I don't know precisely what you mean by "Einstein length expansion."
Einstein's trains imply Lorentz contractions, of course, and it's a
very simple argument in six steps:
(1) shine a beam of light forward through the train, let it hit a
mirror at the front of the train, let it bounce back to a detector at
the back of the train.
(2) The clock at the back of the train measures a proper time between
events of T = 2 L / c.
(3) Someone on the ground will see this clock running slowly by a
factor of \gamma, but can use it to time events just the same if they
correct for that. So they must measure the time between events as T' =
\gamma T.
(4) Their picture for distance is more complicated, however, as the
distance that the light (moving at c) travels is increasing by v in
the one case and decreasing by v in the other case. It is therefore
the case that they also measure this time as:
T' = L' / (c - v) + L' / (c + v).
(5) Combining denominators gives T' = (c + v + c - v) L' / (c^2 -
v^2) = 2 \gamma^2 L' / c
(6) Comparing the two expressions for T' gives \gamma 2 L / c = 2
\gamma^2 L' / c, or L' = L / \gamma .
So the co-moving observer measures at L, and the stationary observer
measures at L / \gamma, which is exactly what the Lorentz
transformation holds.
I don't know why you don't like my reference to the Lorentz group (the
set of all transformations of (ct, x, y, z) preserving c^2 t^2 - x^2 -
y^2 - z^2) or Lorentz transformations. Those are perfectly standard
terms to describe the things which they describe.
I maintain that it's ridiculously easy to prove that Lorentz boosts
have a coherent mathematics (I already did it for you) and that the
original poster seems to have constructed a clock which is
simultaneously meaningless for two different observers and then has
said (under the assumption that it was a meaningful clock) that it
proves some sort of contradiction in the easily-proven-coherent
mathematics.
With care,
== Drostie ==
2011> wrote:
> "Drostie" <chris.dros...@gmail.com> wrote in message
I don't know precisely what you mean by "Einstein length expansion."
Einstein's trains imply Lorentz contractions, of course, and it's a
very simple argument in six steps:
(1) shine a beam of light forward through the train, let it hit a
mirror at the front of the train, let it bounce back to a detector at
the back of the train.
(2) The clock at the back of the train measures a proper time between
events of T = 2 L / c.
(3) Someone on the ground will see this clock running slowly by a
factor of \gamma, but can use it to time events just the same if they
correct for that. So they must measure the time between events as T' =
\gamma T.
(4) Their picture for distance is more complicated, however, as the
distance that the light (moving at c) travels is increasing by v in
the one case and decreasing by v in the other case. It is therefore
the case that they also measure this time as:
T' = L' / (c - v) + L' / (c + v).
(5) Combining denominators gives T' = (c + v + c - v) L' / (c^2 -
v^2) = 2 \gamma^2 L' / c
(6) Comparing the two expressions for T' gives \gamma 2 L / c = 2
\gamma^2 L' / c, or L' = L / \gamma .
So the co-moving observer measures at L, and the stationary observer
measures at L / \gamma, which is exactly what the Lorentz
transformation holds.
I don't know why you don't like my reference to the Lorentz group (the
set of all transformations of (ct, x, y, z) preserving c^2 t^2 - x^2 -
y^2 - z^2) or Lorentz transformations. Those are perfectly standard
terms to describe the things which they describe.
I maintain that it's ridiculously easy to prove that Lorentz boosts
have a coherent mathematics (I already did it for you) and that the
original poster seems to have constructed a clock which is
simultaneously meaningless for two different observers and then has
said (under the assumption that it was a meaningful clock) that it
proves some sort of contradiction in the easily-proven-coherent
mathematics.
With care,
== Drostie ==
Androcles
Dec 12, 2011, 9:35:58 PM12/12/11
to
"Drostie" <chris....@gmail.com> wrote in message
news:818bdcad-d95e-4030...@f39g2000yqn.googlegroups.com...
Source?
==========================
Dover edition.
I don't know precisely what you mean by "Einstein length expansion."
Then I'll explain.
If we place x'=x-vt, it is clear that a point at rest in the system k must
have a system of values x', y, z, independent of time.
So that you understand what that means here is a little picture.
http://www.androcles01.pwp.blueyonder.co.uk/SR4kids/x'=x-vt.gif
The dotted line is the light travelling from A to B and back again,
a distance x' from coordinate 0' to the coordinate x' as seen from
the stationary system K.
The fool Einstein returns the light to the coordinate 0, so he never gets
to derive his transform, he's already fucked up.
However, he does go on to divide (x-vt) by a denominator that is
less than 1, giving xi = (x-vt)/sqrt(1-v^2/c^2), so xi is greater than
x'.
Now you understand what is meant by "Einstein length expansion",
unless you can't manage simple schoolboy algebra.
Rest of word salad snipped.
Drostie
Dec 13, 2011, 10:46:27 AM12/13/11
to
On Dec 13, 3:35 am, "Androcles" <Headmas...@Hogwarts.physics.December.
2011> wrote:
> "Drostie" <chris.dros...@gmail.com> wrote in message
Aha, I see. You don't actually understand what you've worked out, or
else you don't understand who observes the "contraction." That is
okay. I can rectify both very quickly. Let g = 1/sqrt(1 - (v/c)^2) to
make some of the notation a little simpler.
First off is the easy way to do it. Suppose in K we have two lines,
x(t) = vt, and X(t) = L + vt. In the primed coordinates they become:
x' = 0
X' = g L
This means that the proper length in the co-moving frame is L' = g L
while the observed length in your "K" is L, which means that L = L' /
g, which is the normal formula for Lorentz contraction. This is what's
wrong if you don't understand who observes the "contraction": the
reference length is always measured by the co-moving observer.
But I think that instead your problem is the other one. The reason
that you didn't get L = L' / g, was that you forgot about the
relativity of simultaneity (the time equation in the Lorentz
transform). So you said instead something like,
x'(t') = 0, for all t'
X'(t') = L', for all t'
Therefore, x(t') = g v t' and X(t') = g (L' + v t') and therefore
X(t') - x(t') = g L'.
That is *incorrect* as a result for the length that K believes the
length is, because those two expressions are not evaluated at the same
time t. They're evaluated at the same time t' but at different places,
and so the relativity of simultaneity kicks in.
For x'(t'), we have t = g (t' + v x'/c^2), but x' = 0 so t' = t/g and
x(t) = v t, as it should be.
For X'(t') we have t = g (t' + v X'/c^2), but X' = L' so t' = t/g - v
L'/c^2. Thus:
X(t) = g (L' + v [t/g - v L'/c^2] = v t + L' g [1 - v^2/c^2] = vt +
L' / g.
*NOW* we can perform the subtraction X(t) - x(t) = L' / g, and we get
the same result.
So, please remember that it's only the transform:
x' = g (x - v t)
t' = g (t - v t/c^2)
...which is provably mathematically consistent. When you implicitly
assume that t' = t, then you void that proof and stop talking about
special relativity.
Cheers!
== Drostie ==
2011> wrote:
> "Drostie" <chris.dros...@gmail.com> wrote in message
else you don't understand who observes the "contraction." That is
okay. I can rectify both very quickly. Let g = 1/sqrt(1 - (v/c)^2) to
make some of the notation a little simpler.
First off is the easy way to do it. Suppose in K we have two lines,
x(t) = vt, and X(t) = L + vt. In the primed coordinates they become:
x' = 0
X' = g L
This means that the proper length in the co-moving frame is L' = g L
while the observed length in your "K" is L, which means that L = L' /
g, which is the normal formula for Lorentz contraction. This is what's
wrong if you don't understand who observes the "contraction": the
reference length is always measured by the co-moving observer.
But I think that instead your problem is the other one. The reason
that you didn't get L = L' / g, was that you forgot about the
relativity of simultaneity (the time equation in the Lorentz
transform). So you said instead something like,
x'(t') = 0, for all t'
X'(t') = L', for all t'
Therefore, x(t') = g v t' and X(t') = g (L' + v t') and therefore
X(t') - x(t') = g L'.
That is *incorrect* as a result for the length that K believes the
length is, because those two expressions are not evaluated at the same
time t. They're evaluated at the same time t' but at different places,
and so the relativity of simultaneity kicks in.
For x'(t'), we have t = g (t' + v x'/c^2), but x' = 0 so t' = t/g and
x(t) = v t, as it should be.
For X'(t') we have t = g (t' + v X'/c^2), but X' = L' so t' = t/g - v
L'/c^2. Thus:
X(t) = g (L' + v [t/g - v L'/c^2] = v t + L' g [1 - v^2/c^2] = vt +
L' / g.
*NOW* we can perform the subtraction X(t) - x(t) = L' / g, and we get
the same result.
So, please remember that it's only the transform:
x' = g (x - v t)
t' = g (t - v t/c^2)
...which is provably mathematically consistent. When you implicitly
assume that t' = t, then you void that proof and stop talking about
special relativity.
Cheers!
== Drostie ==
Androcles
Dec 13, 2011, 11:38:52 AM12/13/11
to
news:e07e392d-d060-4741...@p9g2000vbb.googlegroups.com...
============================================
Ok, so you ARE too fucking stupid to understand 2 = 1/0.5 for v = 0.866c
and cannot manage simple schoolboy algebra, as I suspected.
Ok, so you ARE too fucking stupid to understand 2 = 1/0.5 for v = 0.866c
and cannot manage simple schoolboy algebra, as I suspected.
Drostie
Dec 13, 2011, 5:42:53 PM12/13/11
to
On Dec 13, 5:38 pm, "Androcles" <Headmas...@Hogwarts.physics.December.
Look. I have, in fact, done that algebra in complete detail,
identifying two stumbling blocks which you might have fallen upon, to
believe that Lorentz transforms imply L' = gamma L. You might have the
L and L' wrong, or you might have accidentally measured L at constant
t' rather than at constant t.
If you actually have problems with the algebra, as I have written it
out or as you have written it out, you may ask those. My guess is that
your unwillingness to actually do "simple schoolboy algebra" is
connected to your own accusations that other people, who in fact *are*
doing the relevant algebra *in front of your very face*, are incapable
of it.
But if you have more than bluster, feel free to contribute. If your
entire life reduces to being blustery on the internet, rather than
being helpful, then I wouldn't trade lives for the world. ^_^
Cheers!
== Drostie ==
identifying two stumbling blocks which you might have fallen upon, to
believe that Lorentz transforms imply L' = gamma L. You might have the
L and L' wrong, or you might have accidentally measured L at constant
t' rather than at constant t.
If you actually have problems with the algebra, as I have written it
out or as you have written it out, you may ask those. My guess is that
your unwillingness to actually do "simple schoolboy algebra" is
connected to your own accusations that other people, who in fact *are*
doing the relevant algebra *in front of your very face*, are incapable
of it.
But if you have more than bluster, feel free to contribute. If your
entire life reduces to being blustery on the internet, rather than
being helpful, then I wouldn't trade lives for the world. ^_^
Cheers!
== Drostie ==
Drostie
Dec 13, 2011, 6:21:07 PM12/13/11
to
postings. This should say "to believe L = gamma L' " -- the point
being that special relativity in its common sense used in physics does
not derive some sort of "length expansion" of the form you've
postulated, and attributing it to Einstein rather than yourself seems
similarly suspect.)
Androcles
Dec 13, 2011, 9:02:38 PM12/13/11
to
news:58b28952-a79d-4989...@da3g2000vbb.googlegroups.com...
===========================================
No, YOU look, you fucking ignoramus! You have no source for
g = 1/sqrt(anything), you fucking made it up.
Here is the source I'm using:
http://www.fourmilab.ch/etexts/einstein/specrel/www/
I'm not interested in Drostie's relativity and nor is anyone else (except
the crank shuba).
This discussion is EINSTEIN'S relativity, so don't tell me to look.
LOOK, you moron, LOOK, until you DO understand:
http://www.androcles01.pwp.blueyonder.co.uk/SR4kids/x'=x-vt.gif
No, YOU look, you fucking ignoramus! You have no source for
g = 1/sqrt(anything), you fucking made it up.
Here is the source I'm using:
http://www.fourmilab.ch/etexts/einstein/specrel/www/
I'm not interested in Drostie's relativity and nor is anyone else (except
the crank shuba).
This discussion is EINSTEIN'S relativity, so don't tell me to look.
LOOK, you moron, LOOK, until you DO understand:
http://www.androcles01.pwp.blueyonder.co.uk/SR4kids/x'=x-vt.gif
John Gogo
Dec 13, 2011, 9:17:52 PM12/13/11
to
On Dec 13, 8:02 pm, "Androcles" <Headmas...@Hogwarts.physics.December.
Its going to take at least a million more cycles of watching this.
Tony M
Dec 16, 2011, 9:04:09 AM12/16/11
to
On Dec 8, 3:14 pm, Pentcho Valev <pva...@yahoo.com> wrote:
> A twin paradox scenario where acceleration is avoided:
>
> http://www.people.fas.harvard.edu/~djmorin/book.html
> Introduction to Classical Mechanics With Problems and Solutions, David
> Morin, Cambridge University Press, Chapter 11, p. 44: "11.19. Modified
> twin paradox *** Consider the following variation of the twin paradox.
> A, B, and C each have a clock. In A's reference frame, B flies past A
> with speed v to the right. When B passes A, they both set their clocks
> to zero. Also, in A's reference frame, C starts far to the right and
> moves to the left with speed v. When B and C pass each other, C sets
> his clock to read the same as B's. Finally, when C passes A, they
> compare the readings on their clocks.
This is a Twin "Paradox" scenario that does not involve acceleration.
>
> http://www.people.fas.harvard.edu/~djmorin/book.html
> Introduction to Classical Mechanics With Problems and Solutions, David
> Morin, Cambridge University Press, Chapter 11, p. 44: "11.19. Modified
> twin paradox *** Consider the following variation of the twin paradox.
> A, B, and C each have a clock. In A's reference frame, B flies past A
> with speed v to the right. When B passes A, they both set their clocks
> to zero. Also, in A's reference frame, C starts far to the right and
> moves to the left with speed v. When B and C pass each other, C sets
> his clock to read the same as B's. Finally, when C passes A, they
> compare the readings on their clocks.
It is meant to show that acceleration is not the cause of the
dissymmetry and the difference in ageing. What differentiates the
twins is the fact that the stay-at-home twin remains in the same
inertial frame while the travelling twin canges frames.
> At this moment, let A's clock
> read TA, and let C's clock read TC. (a) Working in A's frame, show
> that TC = TA/(gamma). (b) Working in B's frame, show again that TC =
> TA/(gamma). (c) Working in C's frame, show again that TC = TA/
> (gamma)."
>
> Note that in this scenario, up to the moment when C sets his clock to
> read the same as B's, A has been a genuine travelling twin who has
> completed the outward part of his journey in B's frame. A's clock has
> been running slow relative to clocks in B's frame, as judged from B's
> frame.
>
> A will remain a genuine travelling twin in C's frame and will perform
> the rest of his journey if, as C sets his clock to read the same as
> B's, simultaneously (in B's frame),
> at the location of A, another
> clock belonging to C's frame is set to read the same as another clock
> belonging to B's frame (it is assumed that clocks in B's frame are
> synchronized).
meeting point of B and C will read the same time. At any other
location (assuming 1D space) a clock in frame B and a clock in frame C
will NOT read the same time. If the above is done then the clocks in
frame C will no longer be synchronized. The rest is therefore
nonsense.
Pentcho Valev
Dec 23, 2011, 2:23:37 AM12/23/11
to
Twin paradox and doublethink: The turn-around of the travelling twin
is ignored if the sedentary twin evaluates the time-dilation effects
(believers fiercely sing "Divine Einstein" and "Yes we all believe in
relativity, relativity, relativity" and the ecstasy increases) but
then all miracles occur during the turn-around if the travelling twin
evaluates the time-dilation effects (the ecstasy gets uncontrollable -
believers tumble to the floor, start tearing their clothes and go into
convulsions):
http://www.youtube.com/watch?feature=player_embedded&v=3Z9eIEkz-ag#!
http://www.pitt.edu/~jdnorton/teaching/HPS_0410/chapters/spacetime_tachyon/index.html
John Norton: "Now consider the judgments of simultaneity of the
traveling twin, as shown in the spacetime diagram opposite. Since the
traveling twin is moving very rapidly, the traveler's hypersurfaces of
simultaneity are quite tilted. Two hypersurfaces of simultaneity are
shown in the lower part of the diagram for the outward part of the
traveler's journey. These are the hypersurfaces that pass through the
event at which the clock reads 1 day and just before the turn-around
at the traveler's clock time of 2 days. We read from these
hypersurfaces that the traveling twin judges the stay-at-home twin's
clock to be running at half the speed of the travelers. When the
traveler's clock reads 1 day, the stay-at-home twin's reads 1/2 day;
just before the turn around, when the traveler's clock is almost at 2
days, the stay-at-home twin's clock is almost at 1 day. Then, at the
end of the outward leg, the traveler abruptly changes motion,
accelerating sharply to adopt a new inertial motion directed back to
earth. What comes now is the key part of the analysis. The effect of
the change of motion is to alter completely the traveler's judgment of
simultaneity. The traveler's hypersurfaces of simultaneity now flip up
dramatically. Moments after the turn-around, when the travelers clock
reads just after 2 days, the traveler will judge the stay-at-home
twin's clock to read just after 7 days. That is, the traveler will
judge the stay-at-home twin's clock to have jumped suddenly from
reading 1 day to reading 7 days. This huge jump puts the stay-at-home
twin's clock so far ahead of the traveler's that it is now possible
for the stay-at-home twin's clock to be ahead of the travelers when
they reunite."
http://en.wikisource.org/wiki/Dialog_about_objections_against_the_theory_of_relativity
Dialog about Objections against the Theory of Relativity (1918), by
Albert Einstein
"...according to the special theory of relativity the coordinate
systems K and K' are by no means equivalent systems. Indeed this
theory asserts only the equivalence of all Galilean (unaccelerated)
coordinate systems, that is, coordinate systems relative to which
sufficiently isolated, material points move in straight lines and
uniformly. K is such a coordinate system, but not the system K', that
is accelerated from time to time. Therefore, from the result that
after the motion to and fro the clock U2 is running behind U1, no
contradiction can be constructed against the principles of the theory.
(...) During the partial processes 2 and 4 the clock U1, going at a
velocity v, runs indeed at a slower pace than the resting clock U2.
However, this is more than compensated by a faster pace of U1 during
partial process 3. According to the general theory of relativity, a
clock will go faster the higher the gravitational potential of the
location where it is located, and during partial process 3 U2 happens
to be located at a higher gravitational potential than U1. The
calculation shows that this speeding ahead constitutes exactly twice
as much as the lagging behind during the partial processes 2 and 4.
This consideration completely clears up the paradox that you brought
up."
http://www.liferesearchuniversal.com/1984-17
George Orwell: "Doublethink means the power of holding two
contradictory beliefs in one's mind simultaneously, and accepting both
of them. The Party intellectual knows in which direction his memories
must be altered; he therefore knows that he is playing tricks with
reality; but by the exercise of doublethink he also satisfies himself
that reality is not violated. The process has to be conscious, or it
would not be carried out with sufficient precision, but it also has to
be unconscious, or it would bring with it a feeling of falsity and
hence of guilt. (...) It need hardly be said that the subtlest
practitioners of doublethink are those who invented doublethink and
know that it is a vast system of mental cheating. In our society,
those who have the best knowledge of what is happening are also those
who are furthest from seeing the world as it is. In general, the
greater the understanding, the greater the delusion ; the more
intelligent, the less sane."
Pentcho Valev
pva...@yahoo.com
is ignored if the sedentary twin evaluates the time-dilation effects
(believers fiercely sing "Divine Einstein" and "Yes we all believe in
relativity, relativity, relativity" and the ecstasy increases) but
then all miracles occur during the turn-around if the travelling twin
evaluates the time-dilation effects (the ecstasy gets uncontrollable -
believers tumble to the floor, start tearing their clothes and go into
convulsions):
http://www.youtube.com/watch?feature=player_embedded&v=3Z9eIEkz-ag#!
http://www.pitt.edu/~jdnorton/teaching/HPS_0410/chapters/spacetime_tachyon/index.html
John Norton: "Now consider the judgments of simultaneity of the
traveling twin, as shown in the spacetime diagram opposite. Since the
traveling twin is moving very rapidly, the traveler's hypersurfaces of
simultaneity are quite tilted. Two hypersurfaces of simultaneity are
shown in the lower part of the diagram for the outward part of the
traveler's journey. These are the hypersurfaces that pass through the
event at which the clock reads 1 day and just before the turn-around
at the traveler's clock time of 2 days. We read from these
hypersurfaces that the traveling twin judges the stay-at-home twin's
clock to be running at half the speed of the travelers. When the
traveler's clock reads 1 day, the stay-at-home twin's reads 1/2 day;
just before the turn around, when the traveler's clock is almost at 2
days, the stay-at-home twin's clock is almost at 1 day. Then, at the
end of the outward leg, the traveler abruptly changes motion,
accelerating sharply to adopt a new inertial motion directed back to
earth. What comes now is the key part of the analysis. The effect of
the change of motion is to alter completely the traveler's judgment of
simultaneity. The traveler's hypersurfaces of simultaneity now flip up
dramatically. Moments after the turn-around, when the travelers clock
reads just after 2 days, the traveler will judge the stay-at-home
twin's clock to read just after 7 days. That is, the traveler will
judge the stay-at-home twin's clock to have jumped suddenly from
reading 1 day to reading 7 days. This huge jump puts the stay-at-home
twin's clock so far ahead of the traveler's that it is now possible
for the stay-at-home twin's clock to be ahead of the travelers when
they reunite."
http://en.wikisource.org/wiki/Dialog_about_objections_against_the_theory_of_relativity
Dialog about Objections against the Theory of Relativity (1918), by
Albert Einstein
"...according to the special theory of relativity the coordinate
systems K and K' are by no means equivalent systems. Indeed this
theory asserts only the equivalence of all Galilean (unaccelerated)
coordinate systems, that is, coordinate systems relative to which
sufficiently isolated, material points move in straight lines and
uniformly. K is such a coordinate system, but not the system K', that
is accelerated from time to time. Therefore, from the result that
after the motion to and fro the clock U2 is running behind U1, no
contradiction can be constructed against the principles of the theory.
(...) During the partial processes 2 and 4 the clock U1, going at a
velocity v, runs indeed at a slower pace than the resting clock U2.
However, this is more than compensated by a faster pace of U1 during
partial process 3. According to the general theory of relativity, a
clock will go faster the higher the gravitational potential of the
location where it is located, and during partial process 3 U2 happens
to be located at a higher gravitational potential than U1. The
calculation shows that this speeding ahead constitutes exactly twice
as much as the lagging behind during the partial processes 2 and 4.
This consideration completely clears up the paradox that you brought
up."
http://www.liferesearchuniversal.com/1984-17
George Orwell: "Doublethink means the power of holding two
contradictory beliefs in one's mind simultaneously, and accepting both
of them. The Party intellectual knows in which direction his memories
must be altered; he therefore knows that he is playing tricks with
reality; but by the exercise of doublethink he also satisfies himself
that reality is not violated. The process has to be conscious, or it
would not be carried out with sufficient precision, but it also has to
be unconscious, or it would bring with it a feeling of falsity and
hence of guilt. (...) It need hardly be said that the subtlest
practitioners of doublethink are those who invented doublethink and
know that it is a vast system of mental cheating. In our society,
those who have the best knowledge of what is happening are also those
who are furthest from seeing the world as it is. In general, the
greater the understanding, the greater the delusion ; the more
intelligent, the less sane."
Pentcho Valev
pva...@yahoo.com
Pentcho Valev
Dec 23, 2011, 4:36:20 PM12/23/11
to
Spetial attention to the miraculous turnaround (demi-tour) which,
although performed by the travelling twin, somehow "puts the stay-at-
http://www.youtube.com/watch?feature=player_embedded&v=3Z9eIEkz-ag#!
"Vu du vaisseau, tout se passe au moment du demi-tour."
http://www.pitt.edu/~jdnorton/teaching/HPS_0410/chapters/spacetime_tachyon/index.html
John Norton: "...just before the turn around, when the traveler's
Einstein in 1918 which successfully camouflaged the absurdity of
special relativity (special relativity predicts that the travelling
twin returns both younger and older than his sedentary brother).
Einstein's 1918 ad hoc hypothesis is just as idiotic as the Lorentz-
Fitzgerald ad hoc length-contraction hypothesis which successfully
camouflaged the fact that the Michelson-Morley experiment had
unequivocally confirmed the variable speed of light predicted by
Newton's emission theory of light.
Pentcho Valev
pva...@yahoo.com
although performed by the travelling twin, somehow "puts the stay-at-
home twin's clock so far ahead of the traveler's that it is now
possible for the stay-at-home twin's clock to be ahead of the
travelers when they reunite":
possible for the stay-at-home twin's clock to be ahead of the
http://www.youtube.com/watch?feature=player_embedded&v=3Z9eIEkz-ag#!
"Vu du vaisseau, tout se passe au moment du demi-tour."
http://www.pitt.edu/~jdnorton/teaching/HPS_0410/chapters/spacetime_tachyon/index.html
John Norton: "...just before the turn around, when the traveler's
clock is almost at 2 days, the stay-at-home twin's clock is almost at
1 day. Then, at the end of the outward leg, the traveler abruptly
changes motion, accelerating sharply to adopt a new inertial motion
directed back to earth. What comes now is the key part of the
analysis. The effect of the change of motion is to alter completely
the traveler's judgment of simultaneity. The traveler's hypersurfaces
of simultaneity now flip up dramatically. Moments after the turn-
around, when the travelers clock reads just after 2 days, the traveler
will judge the stay-at-home twin's clock to read just after 7 days.
That is, the traveler will judge the stay-at-home twin's clock to have
jumped suddenly from reading 1 day to reading 7 days. This huge jump
puts the stay-at-home twin's clock so far ahead of the traveler's that
it is now possible for the stay-at-home twin's clock to be ahead of
the travelers when they reunite."
The turnaround miracle is an ad hoc auxiliary hypothesis advanced by
1 day. Then, at the end of the outward leg, the traveler abruptly
changes motion, accelerating sharply to adopt a new inertial motion
directed back to earth. What comes now is the key part of the
analysis. The effect of the change of motion is to alter completely
the traveler's judgment of simultaneity. The traveler's hypersurfaces
of simultaneity now flip up dramatically. Moments after the turn-
around, when the travelers clock reads just after 2 days, the traveler
will judge the stay-at-home twin's clock to read just after 7 days.
That is, the traveler will judge the stay-at-home twin's clock to have
jumped suddenly from reading 1 day to reading 7 days. This huge jump
puts the stay-at-home twin's clock so far ahead of the traveler's that
it is now possible for the stay-at-home twin's clock to be ahead of
the travelers when they reunite."
Einstein in 1918 which successfully camouflaged the absurdity of
special relativity (special relativity predicts that the travelling
twin returns both younger and older than his sedentary brother).
Einstein's 1918 ad hoc hypothesis is just as idiotic as the Lorentz-
Fitzgerald ad hoc length-contraction hypothesis which successfully
camouflaged the fact that the Michelson-Morley experiment had
unequivocally confirmed the variable speed of light predicted by
Newton's emission theory of light.
Pentcho Valev
pva...@yahoo.com
Helmut Wabnig
Dec 24, 2011, 4:16:30 AM12/24/11
to
Today is 2011, soon will be 2012.
No progress, Pentcho?
w..
Pentcho Valev
Dec 24, 2011, 6:06:52 AM12/24/11
to
http://www.astrosurf.com/luxorion/relativite-restreinte-ex4.htm
"En fait, tout se déroule normalement tant que le voyageur ne fait pas
demi-tour. Théoriquement, pour tout phénomène se déroulant dans un
référentiel uniforme et non accéléré (galiléen), Einstein nous dit que
le battement des horloges est minimum dans ce référentiel et plus long
(le temps se dilate) dans tous les autres qui voient donc leur durée
s'allonger. Concrètement, tant que la fusée avance par inertie, le
frère jumeau embarqué verra tous les événements se déroulant sur Terre
se dérouler plus lentement (dilatation du temps) qu'à bord de sa
fusée. Le paradoxe n'apparaît en fait que lorsque le pilote fait demi-
tour (décélération et accélération qui ne se produisent plus dans un
référentiel d'inertie, y compris le freinage à destination) et rentre
sur Terre, où il constatera que c'est l'inverse qui s'est produit :
son frère jumeau et tous ses amis ont vieilli... Le temps pour eux
s'est écoulé beaucoup plus rapidement qu'à bord de la fusée. La
théorie de la relativité restreinte ne s'applique qu'à des mouvements
uniformes non accélérés. J'insiste bien : le calcul de gamma et autre
contraction ne fonctionnent pas avec une fusée en accélération car le
référentiel n'est plus inertiel et Einstein nous demande de tenir
compte de la topologie de l'espace-temps où masse et énergie
interviennent également. Dans ce cas on doit appliquer les lois de la
relativité générale."
Mais "Einstein nous demande de tenir compte" des effets de
l'accélération dès 1918 alors qu'en 1905 il calcule l'effet (le
voyageur retrouve son frère plus âgé) sans nous demander de tenir
compte de quoi que ce soit! Comment ça? Est-ce qu'en 1905 Einstein
applique, secrètement, "les lois de la relativité générale"?
Pentcho Valev wrote:
Special attention to the miraculous turnaround (demi-tour) which,
"En fait, tout se déroule normalement tant que le voyageur ne fait pas
demi-tour. Théoriquement, pour tout phénomène se déroulant dans un
référentiel uniforme et non accéléré (galiléen), Einstein nous dit que
le battement des horloges est minimum dans ce référentiel et plus long
(le temps se dilate) dans tous les autres qui voient donc leur durée
s'allonger. Concrètement, tant que la fusée avance par inertie, le
frère jumeau embarqué verra tous les événements se déroulant sur Terre
se dérouler plus lentement (dilatation du temps) qu'à bord de sa
fusée. Le paradoxe n'apparaît en fait que lorsque le pilote fait demi-
tour (décélération et accélération qui ne se produisent plus dans un
référentiel d'inertie, y compris le freinage à destination) et rentre
sur Terre, où il constatera que c'est l'inverse qui s'est produit :
son frère jumeau et tous ses amis ont vieilli... Le temps pour eux
s'est écoulé beaucoup plus rapidement qu'à bord de la fusée. La
théorie de la relativité restreinte ne s'applique qu'à des mouvements
uniformes non accélérés. J'insiste bien : le calcul de gamma et autre
contraction ne fonctionnent pas avec une fusée en accélération car le
référentiel n'est plus inertiel et Einstein nous demande de tenir
compte de la topologie de l'espace-temps où masse et énergie
interviennent également. Dans ce cas on doit appliquer les lois de la
relativité générale."
Mais "Einstein nous demande de tenir compte" des effets de
l'accélération dès 1918 alors qu'en 1905 il calcule l'effet (le
voyageur retrouve son frère plus âgé) sans nous demander de tenir
compte de quoi que ce soit! Comment ça? Est-ce qu'en 1905 Einstein
applique, secrètement, "les lois de la relativité générale"?
Pentcho Valev wrote:
Special attention to the miraculous turnaround (demi-tour) which,
Pentcho Valev
Dec 24, 2011, 10:29:45 AM12/24/11
to
A large number of clocks are scattered on and rotate with the
periphery of a rotating disc. A single inertial clock stands outside
the disc but so close against the rotating periphery that its reading
can be compared with the readings of rotating clocks passing by. Will
the single inertial clock go slower or faster than rotating clocks?
By increasing the perimeter of the disc while keeping both the linear
speed of the periphery and the distance between adjacent rotating
clocks constant, one converts the clocks rotating with the periphery
into VIRTUALLY INERTIAL clocks (the "gravitational field" they
experience is reduced to zero). Then special relativity predicts that
the single inertial clock will be seen running SLOWER than the
virtually inertial clocks passing it. That is, the difference between
the reading of the single inertial clock and the reading of any clock
on the periphery will DECREASE with time.
Another prediction based on special relativity is that the single
inertial clock will be seen running FASTER than the clocks on the
periphery ( http://www2.bartleby.com/173/23.html ). That is, the
difference between the reading of the single inertial clock and the
reading of any clock on the periphery will INCREASE with time.
Clearly we have REDUCTIO AD ABSURDUM showing that some postulate of
special relativity is false.
Pentcho Valev
pva...@yahoo.com
periphery of a rotating disc. A single inertial clock stands outside
the disc but so close against the rotating periphery that its reading
can be compared with the readings of rotating clocks passing by. Will
the single inertial clock go slower or faster than rotating clocks?
By increasing the perimeter of the disc while keeping both the linear
speed of the periphery and the distance between adjacent rotating
clocks constant, one converts the clocks rotating with the periphery
into VIRTUALLY INERTIAL clocks (the "gravitational field" they
experience is reduced to zero). Then special relativity predicts that
the single inertial clock will be seen running SLOWER than the
virtually inertial clocks passing it. That is, the difference between
the reading of the single inertial clock and the reading of any clock
on the periphery will DECREASE with time.
Another prediction based on special relativity is that the single
inertial clock will be seen running FASTER than the clocks on the
periphery ( http://www2.bartleby.com/173/23.html ). That is, the
difference between the reading of the single inertial clock and the
reading of any clock on the periphery will INCREASE with time.
Clearly we have REDUCTIO AD ABSURDUM showing that some postulate of
special relativity is false.
Pentcho Valev
pva...@yahoo.com
Tom Roberts
Dec 24, 2011, 11:50:01 AM12/24/11
to
this case it is clearly the assumption that merely increasing the radius of a
circle turns it into a straight line. That is MANIFESTLY false.
Tom Roberts
Pentcho Valev
Dec 24, 2011, 12:09:28 PM12/24/11
to
MANIFESTLY true):
"By increasing the perimeter of the disc while keeping the linear
speed of the periphery constant, one converts the clocks rotating with
the periphery into VIRTUALLY INERTIAL clocks (the "gravitational
field" they experience is reduced to zero)."
Pentcho Valev
field" they experience is reduced to zero)."
pva...@yahoo.com
Pentcho Valev
Dec 25, 2011, 12:14:02 PM12/25/11
to
Confusing believers never stops in Einsteiniana's schizophrenic world:
http://www.physorg.com/news163738003.html
"Physicist Marek Abramowicz of Goteborg University in Sweden and
astronomer Stanislaw Bajtlik of the Nicolaus Copernicus Astronomical
Center in Warszawa, Poland, have proposed the surprising new version
of the twin paradox, which at first seems to run contrary to the
traditional version. However, the scientists show that the traditional
version is actually a specific case of a more general concept. "In the
best known version of the twin paradox, the twin who is accelerated is
younger," Abramowicz and Bajtlik told PhysOrg.com. "In the version
discussed by us the accelerated twin is older. It is quite surprising.
It is almost as to say that 'the older twin is younger'."
http://www.damtp.cam.ac.uk/research/gr/members/gibbons/gwgPartI_SpecialRelativity2010.pdf
Gary W. Gibbons FRS: "In other words, by simply staying at home Jack
has aged relative to Jill. There is no paradox because the lives of
the twins are not strictly symmetrical. This might lead one to suspect
that the accelerations suffered by Jill might be responsible for the
effect. However this is simply not plausible because using identical
accelerating phases of her trip, she could have travelled twice as
far. This would give twice the amount of time gained."
Pentcho Valev
pva...@yahoo.com
http://www.physorg.com/news163738003.html
"Physicist Marek Abramowicz of Goteborg University in Sweden and
astronomer Stanislaw Bajtlik of the Nicolaus Copernicus Astronomical
Center in Warszawa, Poland, have proposed the surprising new version
of the twin paradox, which at first seems to run contrary to the
traditional version. However, the scientists show that the traditional
version is actually a specific case of a more general concept. "In the
best known version of the twin paradox, the twin who is accelerated is
younger," Abramowicz and Bajtlik told PhysOrg.com. "In the version
discussed by us the accelerated twin is older. It is quite surprising.
It is almost as to say that 'the older twin is younger'."
http://www.damtp.cam.ac.uk/research/gr/members/gibbons/gwgPartI_SpecialRelativity2010.pdf
Gary W. Gibbons FRS: "In other words, by simply staying at home Jack
has aged relative to Jill. There is no paradox because the lives of
the twins are not strictly symmetrical. This might lead one to suspect
that the accelerations suffered by Jill might be responsible for the
effect. However this is simply not plausible because using identical
accelerating phases of her trip, she could have travelled twice as
far. This would give twice the amount of time gained."
Pentcho Valev
pva...@yahoo.com
Pentcho Valev
Dec 27, 2011, 7:44:08 AM12/27/11
to
Einsteinians don't like the pole-barn scenario because it is
destructive (but not absurd):
http://www-personal.umich.edu/~jcv/pdfs/chapter6.pdf
John C. Vander Velde, Physics Department, University of Michigan: "An
athlete, running with a 15 ft long pole, gets momentarily captured in
a barn that is only 9 ft long. How is that possible? It's easy. The
athlete simply runs at 80% of the speed of light so that his Lorentz
contraction factor is 0.6. The door-keeper at the back of the barn
agrees to close the door at the instant the front of the pole hits the
front of the barn. Since the door-keeper has, effectively, marked the
front and back of the pole simultaneously in the barn system the pole
will be only 15ft X 0.6 = 9 ft long. It will just exactly fit in the
barn at that instant. This problem is entirely equivalent to Flo and
Joe simultaneously painting spots on the passing rocket ship. The
spots end up 50 ft apart on the ship even thought Flo and Joe were
only 30 ft apart. (I like the painting spots version better than the
pole-vaulter problem because it is less destructive. Think of what
happens to the poor pole-vaulter in the next instant when he slams
into the front of the barn. Ugh.)"
Einsteinians,
Have you ever considered all "destructive" implications of this
particular prediction of special relativity? For instance, a suitable
geometry of the pole and the barn would allow you to reduce the volume
of the pole as much as you wish. Is this absurd? No? Just awfully
destructive? Ugh ugh ugh? Bravo, Einsteinians!
Pentcho Valev
pva...@yahoo.com
destructive (but not absurd):
http://www-personal.umich.edu/~jcv/pdfs/chapter6.pdf
John C. Vander Velde, Physics Department, University of Michigan: "An
athlete, running with a 15 ft long pole, gets momentarily captured in
a barn that is only 9 ft long. How is that possible? It's easy. The
athlete simply runs at 80% of the speed of light so that his Lorentz
contraction factor is 0.6. The door-keeper at the back of the barn
agrees to close the door at the instant the front of the pole hits the
front of the barn. Since the door-keeper has, effectively, marked the
front and back of the pole simultaneously in the barn system the pole
will be only 15ft X 0.6 = 9 ft long. It will just exactly fit in the
barn at that instant. This problem is entirely equivalent to Flo and
Joe simultaneously painting spots on the passing rocket ship. The
spots end up 50 ft apart on the ship even thought Flo and Joe were
only 30 ft apart. (I like the painting spots version better than the
pole-vaulter problem because it is less destructive. Think of what
happens to the poor pole-vaulter in the next instant when he slams
into the front of the barn. Ugh.)"
Einsteinians,
Have you ever considered all "destructive" implications of this
particular prediction of special relativity? For instance, a suitable
geometry of the pole and the barn would allow you to reduce the volume
of the pole as much as you wish. Is this absurd? No? Just awfully
destructive? Ugh ugh ugh? Bravo, Einsteinians!
Pentcho Valev
pva...@yahoo.com
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