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| Guest |
 Posted: Sun Feb 27, 2005 5:59 pm |
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The Symmetry:
Two platforms approach each other at c/3 with a pair of twins on each
one.
One of each twin departs at c/3 relative to his platform and travels
1/2 his invariant time to the other platform and 1/2 his other
invariant time returns with the other platform at c/3.
Before the platforms intersect, from 1000km away they proportionally
reduce their velocity so that the new velocity difference is zero upon
intersection, NOW the platforms are together.
PART_1 of the Impossibility:
The twins do INDEED end-up fully capable of comparing ALL their clocks
simultaneously at the end within the same inertial reference frame. And
it is IMPOSSIBLE TO DETERMINE ANY LOGICAL CLOCK READING.
Just plug-in absolutely any clock rate for both Platform's invariant
times and it will always be impossible...try it.
1. The 2nd half of both voyages each immigrant twin must AGE LESS by
the same amount as the first half of his trip = large percentage = 50%
of aging time.
2. The 2nd half of both voyages each immigrant twin & platform's clock
must tick at the same rate and therefore age at the SAME rate.
Therefore it is IMPOSSIBLE for each immigrant twin & platform to both
age less than the other immigrant twin & platform and as demonstrated
they do all INDEED compare clocks at the END.
(The clocks values ****WILL NOT**** magically change values at the last
1000km stretch when they proportionally reduce their velocities to meet
each other, THE CLOCK RATES AND VALUES ARE MORE AFFECTED DURING THE
ENTIRE VOYAGE).
PART_2 of the Mathematical Impossibililty:
Because during the final 1/2 which is the return voyage each plaform
beats at the same rate as its new immigrant twin therefore:
invariant_t1 = gamma2(invariant_t2)
invariant_t2 = gamma1(invariant_t1)
2. Since gamma1 = gamma2 since both v = c/3 therefore reducing equation
#1:
invartiant_t1/invatiant_t2= invariant_t2/invariant_t1
***Which is MATHEMATICALLY IMPOSSIBLE and therefore the RELATIVISTS are
going crazy since this means the end of RELATIVITY'S TIME DILATION****
Anyone can "try" to plug-in their own numbers to "try" and determine
what the final clock comparison readings will be and IT WILL NOT WORK.
UNTIL THEN THIS MEANS THE END OF RELATIVITY'S TIME DILATION. |
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| Sam Wormley |
| Posted: Sun Feb 27, 2005 6:17 pm |
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Guest
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+------------+ +---------------------------------------------+
| PLEASE | | BEST TO IGNORE ATTENTION SEEKING TROLLS |
| DO NOT | | LIKE CHILDISH guskz@hotmail.com -- THEY DRY |
| FEED | | UP AND BLOW AWAY WITHOUT FEEDBACK |
| DA | | |
| TROLLS | | http://www.angelfire.com/space/usenet/ |
+------------+ +---------------------------------------------+
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`\ '/ / ' / `\ '/ / ' / `\ '/ / ' / |
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| Uncle Al |
| Posted: Sun Feb 27, 2005 6:58 pm |
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Guest
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| Guest |
| Posted: Sun Feb 27, 2005 7:44 pm |
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Uncle Al wrote:
Quote: guskz@hotmail.com wrote:
The Symmetry:
Two platforms approach each other at c/3 with a pair of twins on
each
one.
[snip crap]
UNTIL THEN THIS MEANS THE END OF RELATIVITY'S TIME DILATION.
Hopeless ineducable idiot.
Stop rambling and prove I'm an idiot.
All the variables and factors for the math have been provided, YOU WILL
NOT BE ABLE TO COME TO A LOGICAL NUMERICAL ANSWER to what the four
twins clock readings will be.... IMPOSSIBLE.
FOR THAT REASON Relativity's TIME DILATION THEORY MUST END.
Come to a logical numerical computation for all four twins clock (with
the gvien inital settings) AND YOU WILL NOT READ FROM ME FOR AN ENTIRE
MONTH! |
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| Guest |
| Posted: Sun Feb 27, 2005 7:50 pm |
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g...@hotmail.com wrote:
Quote: Uncle Al wrote:
guskz@hotmail.com wrote:
The Symmetry:
Two platforms approach each other at c/3 with a pair of twins on
each
one.
[snip crap]
UNTIL THEN THIS MEANS THE END OF RELATIVITY'S TIME DILATION.
Hopeless ineducable idiot.
Stop rambling and prove I'm an idiot.
All the variables and factors for the math have been provided, YOU
WILL
NOT BE ABLE TO COME TO A LOGICAL NUMERICAL ANSWER to what the four
twins clock readings will be.... IMPOSSIBLE.
FOR THAT REASON Relativity's TIME DILATION THEORY MUST END.
Come to a logical numerical computation for all four twins clock
(with
the gvien inital settings) AND YOU WILL NOT READ FROM ME FOR AN
ENTIRE
MONTH!
Of course you must also adhere to the same rules and stay away from
posting to this newsgroup for one month if you fail. You will fail.
FOR THAT REASON Relativity's TIME DILATION THEORY MUST END |
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| Guest |
| Posted: Sun Feb 27, 2005 11:33 pm |
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Uncle Al wrote:
Quote: guskz@hotmail.com wrote:
Uncle Al wrote:
guskz@hotmail.com wrote:
The Symmetry:
Two platforms approach each other at c/3 with a pair of twins
on
each
one.
[snip crap]
UNTIL THEN THIS MEANS THE END OF RELATIVITY'S TIME DILATION.
Hopeless ineducable idiot.
Stop rambling and prove I'm an idiot.
All the variables and factors for the math have been provided, YOU
WILL
NOT BE ABLE TO COME TO A LOGICAL NUMERICAL ANSWER to what the four
twins clock readings will be.... IMPOSSIBLE.
FOR THAT REASON Relativity's TIME DILATION THEORY MUST END.
Come to a logical numerical computation for all four twins clock
(with
the gvien inital settings) AND YOU WILL NOT READ FROM ME FOR AN
ENTIRE
MONTH!
SNIP
OK: You know the twin paradox, twin travels at c/3 and returns at -c/3
therefore t_traveler = gamma (t_stay_at_home) where gamma uses v = c/3
Quote:
*****> Acceleration breaks the symmetry of who ages faster. To
accomplish
Quote: that, the acceleration can occur before the clocks (or the twins)
exist. Only reference frames matter.
Do Not be a coward then...you know the arithmetics, Give Us exactly
what will the clocks read if INSTEAD:
as the traveling TWIN returns home he does not DECELARATE but instead
keeps his return velocity contant at -c/3 and the TWIN who
stayed-at-home accelerates to -c/3 just as the traveling TWIN passes
by....WHAT WILL BE BOTH CLOCK READINGS!
You probably don't know and just repeat already known knowledge about
time dilation. There's a difference between reading and doing..... |
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| RP |
| Posted: Mon Feb 28, 2005 2:00 am |
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Guest
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guskz@hotmail.com wrote:
Quote: Uncle Al wrote:
guskz@hotmail.com wrote:
Uncle Al wrote:
guskz@hotmail.com wrote:
The Symmetry:
Two platforms approach each other at c/3 with a pair of twins
on
each
one.
[snip crap]
UNTIL THEN THIS MEANS THE END OF RELATIVITY'S TIME DILATION.
Hopeless ineducable idiot.
Stop rambling and prove I'm an idiot.
All the variables and factors for the math have been provided, YOU
WILL
NOT BE ABLE TO COME TO A LOGICAL NUMERICAL ANSWER to what the four
twins clock readings will be.... IMPOSSIBLE.
FOR THAT REASON Relativity's TIME DILATION THEORY MUST END.
Come to a logical numerical computation for all four twins clock
(with
the gvien inital settings) AND YOU WILL NOT READ FROM ME FOR AN
ENTIRE
MONTH!
SNIP
OK: You know the twin paradox, twin travels at c/3 and returns at -c/3
therefore t_traveler = gamma (t_stay_at_home) where gamma uses v = c/3
*****> Acceleration breaks the symmetry of who ages faster. To
accomplish
that, the acceleration can occur before the clocks (or the twins)
exist. Only reference frames matter.
Do Not be a coward then...you know the arithmetics, Give Us exactly
what will the clocks read if INSTEAD:
as the traveling TWIN returns home he does not DECELARATE but instead
keeps his return velocity contant at -c/3 and the TWIN who
stayed-at-home accelerates to -c/3 just as the traveling TWIN passes
by....WHAT WILL BE BOTH CLOCK READINGS!
You probably don't know and just repeat already known knowledge about
time dilation. There's a difference between reading and doing.....
Assume a single inertial observational frame in which to perform all
measurements. Calculate clock differences from those measurements. All
inertial observers will agree.
_______________________________________
The use of momentary comoving inertial frames, or MCIF's (Newton's
method) is sufficient to account for the behavior of clocks that are
accelerated wrt the observational inertial frame. For uniform
acceleration the average ticking rate offset of the accelerated clock
over the space-like interval is given by:
delta_t' = (r/w) * sqrt(1 - w^2/c^2)
where
w = average (weighted) velocity.
This reduces to
delta_t' = sqrt(r^2/w^2 - r^2/c^2)
Since the weightedness of w hasn't yet been defined, this equation can
be taken as a generalized equation, i.e. there exists some w such that
the equation above gives the correct time offset over a given
space-like interval r, regardless of the nature of the motion of the
clock over r.
When w = v, i.e. when the velocity is constant, the equation reduces to:
delta_t' = sqrt(r^2/v^2 - r^2/c^2) = (r/v)sqrt(1 - v^2/c^2)
= delta_t/gamma
Which you should recognize as the special case solution for a
uniformly moving clock, i.e. as derived directly from the lorentz
transform.
Keeping in mind that w is a weighted average, it is left as a
possibility that even though two clocks may leave x1 at the same
instant t1 wrt K and arrive at x2 at the same instant t2 wrt K, these
clocks won't necessarily have accrued equal elapsed times.
Acceleration involves a tilt of the time axis, and thus differences in
lengths of the respective geodesics. Complicated motions require
integrating over the entire path, which can be extremely difficult
given irregular motion. This is why uniform acceleration is usually
used in the arguments.
Though Dirk supposes that inverting the resultant ticking rate
relationships that were derived via integration provides for events
wrt an accelerated frame, entirely in the context of SR, his
assumption is incorrect. The assumption that some acceleration
dependent component of the relative ticking rates is absolute
(non-reciprocated) is an assumption of GR. Thus though the basics of
GR time offsets can be derived from the logical requirements set forth
by SR, it doesn't follow that SR is sufficient to account for
observations made from accelerated frames, only that SR taken with the
logical requirement of non-reciprocity in the case of relative
acceleration is sufficient to derive GR time offsets. OTOH, this is
exactly how Einstein derived GR offsets as well (Rotating disk
argument), but Einstein had the sense to note that SR wasn't
sufficient, since the derivation required an additional assumption
(non-reciprocity in the gravitational field) in order to resolve the
paradox.
Others are free to make corrections and/or additions to the above
explanation.
Richard Perry |
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| Dirk Van de moortel |
| Posted: Mon Feb 28, 2005 5:44 am |
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"RP" <no_mail_no_spam@yahoo.com> wrote in message news:38ftn5F5n9pjqU1@individual.net...
[snip]
Quote: Though Dirk supposes that inverting the resultant ticking rate
relationships that were derived via integration provides for events
wrt an accelerated frame, entirely in the context of SR, his
assumption is incorrect.
*What* am I supposing?
"that inverting the resultant ticking rate relationships that
were derived via integration provides for events wrt an
accelerated frame, entirely in the context of SR"
?
Perry, you are making progress, but you are still talking
through your back end.
Quote: The assumption that some acceleration
dependent component of the relative ticking rates is absolute
(non-reciprocated) is an assumption of GR.
So you still should have that careful look at
http://www.geocities.com/slithytove5/AccelClocks.htm
[followup set to sci.physics.relativity]
Dirk Vdm |
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| Gregory L. Hansen |
| Posted: Mon Feb 28, 2005 8:03 am |
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Guest
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In article <1109612604.706615.121910@z14g2000cwz.googlegroups.com>,
<guskz@hotmail.com> wrote:
Quote:
Uncle Al wrote:
2) Past acceleration is irrelevant to the running of present clocks,
MUCH MUCH MORE ****IMPORTANTLY*****:
Present acceleration is irrelevant to the running of past clocks.
MEANING:
As the traveling twin returns say at -c/3 to his home brother, HE WILL
STILL REMAIN YOUNGER THAN HIS BROTHER ***IF INSTEAD*** of reducing his
velocity to zero, his brother accelerates to -c/3 to meet him!!!!!
SOLELY FOR THIS REASON THE GIVEN DOUBLE PLATFORM TWIN PARADOX
DEMONSTRATES THE END OF EINSTEIN'S RELATIVITY....the variables,factors
and settings have been given ....do the math where at the end all four
clocks are compared, you will not be able to come to a logical
tabulated answer.
You're aware of the identically accelerated twins paradox, right? The one
in front ages faster.
--
"A nice adaptation of conditions will make almost any hypothesis agree
with the phenomena. This will please the imagination but does not advance
our knowledge." -- J. Black, 1803. |
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| Alex |
| Posted: Mon Feb 28, 2005 12:05 pm |
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Guest
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The twin paradox is not symmetrical.
The symmetrical situation occurs when both twins leave each other and
both turn at the half way point.
Your challenge is, from your knowledge of relativistic phase, to work
out the effects of the asymmetry in the twin paradox compared with the
symmetrical situation. (Clue: look at the tilt of the x-axes of the
twins, their 'lines of simultaneity' in the x direction, and work out
what happens as they change direction of tilt during the turn).
Best Wishes
Alex Green |
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| Guest |
| Posted: Mon Feb 28, 2005 12:19 pm |
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Alex wrote:
Quote: The twin paradox is not symmetrical.
This is not the twin paradox but the double platform twin paradox.
Quote: The symmetrical situation occurs when both twins leave each other and
both turn at the half way point.
Your challenge is,
I already gave a challenge and NO one has computed it because it cannot
be computed therefore it means the End of Time Dilation.
The variables are given, just plug in the numbers if you can...you will
not be able to come to an arithmetic answer....remember according to
the LAWS of Relativity both PLATFORMS have never met each other and by
LAW are both allowed to set their own reference frame and clocks.
Quote: from your knowledge of relativistic phase, to work
out the effects of the asymmetry in the twin paradox compared with
the
symmetrical situation. (Clue: look at the tilt of the x-axes of the
twins, their 'lines of simultaneity' in the x direction, and work out
what happens as they change direction of tilt during the turn).
Best Wishes
Alex Green |
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| Guest |
| Posted: Mon Feb 28, 2005 12:33 pm |
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RP wrote:
Quote:
When w = v, i.e. when the velocity is constant, the equation reduces
to:
delta_t' = sqrt(r^2/v^2 - r^2/c^2) = (r/v)sqrt(1 - v^2/c^2)
= delta_t/gamma
The above equation only works with ONE INERTIAL FRAME, the case given
for the double plaform twin paradox has two reference frames:
1. The LAWS of Relativity allow more than one reference frame as
demonstrated by the equation w = (u+v)/(1+uv/c^2).
2. Likewise not only velocity but also time (clocks) are allowed to
have more than one reference frame.
3. Both platforms of the double platform twin paradox have NEVER met
each other yet and by the LAWS of Relatvitity Both Platforms are
legally permitted to each establish his clock and platform as his
Inertial Reference Frame.
4. Both Platform is legally permitted to expect HIS travelling twin to
age less.
5. Both Platforms come to a relavistic full stop at the End just as
they intersect and compare all their clocks.
The variables have been given and it is impossible to come to an
arithmaticly sound answer for the 4 clock readings at the End, try it,
you can't because it's impossible!
Therefore it means the End of Einsteind Relativity, paving the way for
perhaps absolute space. |
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| Guest |
| Posted: Mon Feb 28, 2005 12:43 pm |
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Uncle Al wrote:
Quote:
2) Past acceleration is irrelevant to the running of present clocks,
MUCH MUCH MORE ****IMPORTANTLY*****:
Present acceleration is irrelevant to the running of past clocks.
MEANING:
As the traveling twin returns say at -c/3 to his home brother, HE WILL
STILL REMAIN YOUNGER THAN HIS BROTHER ***IF INSTEAD*** of reducing his
velocity to zero, his brother accelerates to -c/3 to meet him!!!!!
SOLELY FOR THIS REASON THE GIVEN DOUBLE PLATFORM TWIN PARADOX
DEMONSTRATES THE END OF EINSTEIN'S RELATIVITY....the variables,factors
and settings have been given ....do the math where at the end all four
clocks are compared, you will not be able to come to a logical
tabulated answer. |
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| Guest |
| Posted: Mon Feb 28, 2005 3:38 pm |
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Found out the simple error, I'm shocked no one could point it out.
The time dilation between both platforms clocks occurs during the 1st
1/2 of the travel which is t_dilated = gamma (t_invariant) where
velocity = w = (u+v)/(1+uv/c^2)
u = other's platform velocity = -c/3
v = other's twin traveler velocity = -c/3
Thus w = -2c/3/(1+1/9) = -2c/3(9/10) = -6c/10 = -3c/5 = -0.6c
which is almost twice as fast as the original -c/3 and therefore more
dilation per invariant time unit of each platform's time.
In conclusion both traveling twins may have aged less where as MANY
other's would post no and there must only be one platform as the
Invariant inertial reference frame. |
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