Time dilation scenarios

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Guest

Posted: Thu Apr 17, 2008 8:03 pm

I probably have some flawed understanding, but I can't work out what
it is. I have put this up on two different science forums, but so far
this has stumped both. Below are two scenarios, which combined seem
to defy the resolution of the twins paradox.

In both scenarios, the level of acceleration and time under
acceleration are the same.

Scenario 1a)

A and B accelerate and coast several times in sync. Then A accelerates
then coasts away at constant velocity while B waits at the previous
location. After a while, B accelerates towards A until they are
matching velocities, although some distance away. Is there a
difference in their clocks?

Solution: Similar to the twins paradox, I would expect that, since A
accelerates before B, this breaks the symmetry first. Therefore, time
for A slows down wrt B while B remains unaccelerated. So, when they
match velocities, TA < TB.

Scenario 1b)
A and B accelerate away from C. A accelerates back to C while B ceases
to accelerate and coasts. B later accelerates back toward A and C,
matching velocity with them. Is there a difference in their clocks?

Solution: This is the twins paradox twice over - therefore A and B
break the symmetry, leading to a time dilation. B is in that state
for a longer period than A, therefore, when B matches velocity with A
and C again, TC > TA > TB

Problem: Since we have A and B accelerating in sync in scenario 1a),
all the velocities and accelerations in these two scenarios can be
made to be exactly the same. But they result in opposite
relationships in clocktimes for TA and TB. TA > TB is incompadible
with TB > TA.

Androcles

Posted: Fri Apr 18, 2008 2:12 am

Guest

<matscienceman@yahoo.com> wrote in message
news:27945819-8850-409f-a114-6b1c3e8c062b@e39g2000hsf.googlegroups.com...
|I probably have some flawed understanding, but I can't work out what
| it is. I have put this up on two different science forums, but so far
| this has stumped both. Below are two scenarios, which combined seem
| to defy the resolution of the twins paradox.
|
| In both scenarios, the level of acceleration and time under
| acceleration are the same.
|
| Scenario 1a)
|
| A and B accelerate and coast several times in sync. Then A accelerates
| then coasts away at constant velocity while B waits at the previous
| location. After a while, B accelerates towards A until they are
| matching velocities, although some distance away. Is there a
| difference in their clocks?
|
| Solution: Similar to the twins paradox, I would expect that, since A
| accelerates before B, this breaks the symmetry first. Therefore, time
| for A slows down wrt B while B remains unaccelerated. So, when they
| match velocities, TA < TB.
|
|
| Scenario 1b)
| A and B accelerate away from C. A accelerates back to C while B ceases
| to accelerate and coasts. B later accelerates back toward A and C,
| matching velocity with them. Is there a difference in their clocks?
|
| Solution: This is the twins paradox twice over - therefore A and B
| break the symmetry, leading to a time dilation. B is in that state
| for a longer period than A, therefore, when B matches velocity with A
| and C again, TC > TA > TB
|
|
| Problem: Since we have A and B accelerating in sync in scenario 1a),
| all the velocities and accelerations in these two scenarios can be
| made to be exactly the same. But they result in opposite
| relationships in clocktimes for TA and TB. TA > TB is incompadible
| with TB > TA.

Answer:
'we establish by definition that the "time" required by
light to travel from A to B equals the "time" it requires
to travel from B to A' because I FUCKIN' SAY SO. -- Rabbi Albert Einstein

http://www.androcles01.pwp.blueyonder.co.uk/Smart/tAB=tBA.gif

Problem solved.

--
This message is brought to you by Androcles
http://www.androcles01.pwp.blueyonder.co.uk/

|

rotchm@gmail.com

Posted: Fri Apr 18, 2008 3:05 am

Guest

Similar/equivalent scenarios have already been treated:

Boughn S P 1989 "The case of the identically accelerated twins"
Am. J. Phys. 57 791–3

Quote:

Scenario 1a)

A and B accelerate and coast several times in sync. Then A accelerates
then coasts away at constant velocity while B waits at the previous
location. After a while, B accelerates towards A until they are
matching velocities, although some distance away. Is there a
difference in their clocks?

Solution: Similar to the twins paradox, I would expect that, since A
accelerates before B, this breaks the symmetry first. Therefore, time
for A slows down wrt B while B remains unaccelerated. So, when they
match velocities, TA < TB.

Scenario 1b)
A and B accelerate away from C. A accelerates back to C while B ceases
to accelerate and coasts. B later accelerates back toward A and C,
matching velocity with them. Is there a difference in their clocks?

Solution: This is the twins paradox twice over - therefore A and B
break the symmetry, leading to a time dilation. B is in that state
for a longer period than A, therefore, when B matches velocity with A
and C again, TC > TA > TB

Problem: Since we have A and B accelerating in sync in scenario 1a),
all the velocities and accelerations in these two scenarios can be
made to be exactly the same. But they result in opposite
relationships in clocktimes for TA and TB. TA > TB is incompadible
with TB > TA.

Posted: Fri Apr 18, 2008 4:35 am

Guest

On Apr 18, 1:03 am, matscience...@yahoo.com wrote:

Quote:

The purpose of the twins paradox is to teach that speeding up is not
the origin of time dilation. The simplest explanation is that proper
time between two events depends on the path through spacetime. Put in
very coarse (deceptively simple) terms, in Euclidean space the
shortest distance between two points is a straight line, but in
Minkowski spacetime the *longest* interval between to events is a
straight line. The reason is that minus sign that is in the Minkowski
equivalent of the Pythagorean theorem. A good spacetime diagram on an
FAQ will go along way to helping you understand this.
http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox..html

PD

Tom Roberts

Posted: Fri Apr 18, 2008 9:53 am

Guest

matscienceman@yahoo.com wrote:

Quote:

Your scenario is not completely specified, in particular you don't state
how the final clocks are to be compared. Let me re-phrase your scenario:

Initially, spaceships A and B are at rest in inertial frame S1 at the
same location, and their on-board clocks are synchronized with each
other and with standard coordinate clocks in S1. At time tA=0 indicated
on clock A, spaceship A accelerates with a profile a(tA), ultimately
coming to rest in inertial frame S2. At time tB=T indicated on clock B
(T>0), spaceship B accelerates with the same profile a(tB), ultimately
coming to rest in inertial frame S2 -- this is clearly so because they
used the same profile a(.) as a function of their respective proper
times. Due to the difference in their trajectories, A and B are not
located together in S2. But if one compares clock A and clock B by
comparison with coordinate clocks of S2, they compare equal.
Equivalently, one could use any of Einstein's synchronization methods to
show that in S2 the clocks A and B are synchronized with each other
(without adjustment). This is independent of the delay T.

This conclusion is based on the obvious symmetry between A and B --
their trajectories are time translations of each other, and nothing
depends on such a translation in time (because this is SR and both S1
and S2 are inertial frames).

This example shows that the simplistic phrase "moving clocks run slow"
does not fully capture the phenomenon. Your "solution" is wrong, because
it attempts to apply that phrase directly, without understanding its
limitations.

Quote:

[...]

Similarly for your other scenarios.

Tom Roberts

harry

Posted: Sat Apr 19, 2008 12:48 am

Guest

On Apr 18, 8:03 am, matscience...@yahoo.com wrote:

Quote:

The Twin paradox scenario concerns fact that according to "relativity"
theory, if two clocks can be separated and brought back to each other,
they can be out of sync in a way that is "absolute" in every sense of
the word. Your scenario 1a) is NOT a twin paradox scenario; it relates
to standard relativity of simultaneity. For relative time at a
distance you have to specify the reference frame you use (you didn't).
Let's call the frame in which they were in rest before separating S,
and the one in which they are both at rest in the end S'. Time
dilation is a function of speed, and there is only symmetry of
description between inertial (constant speed) reference frames. For
simplicity I'll omit the clock accelerations from the descriptions.

- Description in S: A and B are synchronized and in rest; next A
leaves which speed slows down its clock rate. Later B departs at the
same speed. Thus as observed in S, clock A is now behind on clock B
(TA < TB).

- Description in S': A and B are synchronized and moving together;
next A stops which brings its clock back to the normal (higher) rate.
Later B stops as well. Thus as observed in S', clock B is now behind
on clock A (TB > TA).

Your above solution is correct for the description in S.

Quote:

Scenario 1b)
A and B accelerate away from C. A accelerates back to C while B ceases
to accelerate and coasts. B later accelerates back toward A and C,
matching velocity with them. Is there a difference in their clocks?

Solution: This is the twins paradox twice over - therefore A and B
break the symmetry, leading to a time dilation. B is in that state
for a longer period than A, therefore, when B matches velocity with A
and C again, TC > TA > TB

Problem: Since we have A and B accelerating in sync in scenario 1a),
all the velocities and accelerations in these two scenarios can be
made to be exactly the same. But they result in opposite
relationships in clocktimes for TA and TB. TA > TB is incompadible
with TB > TA.

Scenario 1b) is a Twin paradox scenario (NOT "twice over" but once!);
there is no need to specify a frame - the easiest to use is the one in
which all are in rest at the start and in the end. Your solution to
1b) is correct.

About your problem: Different problem = different solution.
In scenario 1a), in S: clock A has been longer at a high speed than
clock B . Thus TA < TB.
In scenario 1b), in S: clock B has been longer at a high speed than
clock A . Thus TA > TB.

Regards,
Harald

jem

Posted: Sat Apr 19, 2008 9:29 am

Guest

Quote:

According to SR, there's no absolute sense in which time accumulates
differently for relatively moving (ideal) clocks - any accumulated
time difference will depend on the aggregate speed of each clock
relative to the reference frame in which the clocks are ultimately
compared.

Quote:

So, when they match velocities, TA < TB.

Think so? Which of the two travelers had a greater aggregate speed
relative to the reference frame in which TA and TB were compared?

Quote:

Except that TA > TB in scenario 1a.

tussock

Posted: Sun Apr 20, 2008 11:00 am

Guest

matscienceman wrote:

Quote:

Irrelivant.

Quote:

Then A accelerates then coasts away at constant velocity while B waits
at the previous location. After a while, B accelerates towards A until
they are matching velocities, although some distance away. Is there a
difference in their clocks?

Perhaps not in a particular reference frame that's halfway between
the two above. Otherwise, yes.

Quote:

Solution: Similar to the twins paradox, I would expect that, since A
accelerates before B, this breaks the symmetry first. Therefore, time
for A slows down wrt B while B remains unaccelerated. So, when they
match velocities, TA < TB.

You can't compare their clocks directly, as they're not in the same
place; observers at various speeds will compare them differently.

Quote:

A accelerates and returns from a distance, so shows less time
passing, B does the same at a greater distance, so shows less still.

Quote:

Solution: This is the twins paradox twice over - therefore A and B
break the symmetry, leading to a time dilation.

You don't seem to know what some of those words mean.

Quote:

B is in that state for a longer period than A, therefore, when B
matches velocity with A and C again, TC > TA > TB

Yes, once they're all back with C it makes measurement more
meaningful.

Quote:

Problem: Since we have A and B accelerating in sync in scenario 1a), all
the velocities and accelerations in these two scenarios can be made to
be exactly the same.

It's trivial to add a non-accelerating C to the first problem to show
your total accelerations are not measured from the same reference frame.
If you'd like to add a C to the first problem, that they return to and
stop at, it'd be the same.
Measured from the frame of C, it always is the same, A and B just
happen to agree when they accelerate into C's reference frame.

Quote:

But they result in opposite relationships in clocktimes for TA and TB.
TA > TB is incompadible with TB > TA.

For different reference frames with A and B at a distance it's the
expected behaviour.

--
tussock

I'm like a box of chocolates; you never know what you're gunna get.

Guest

Posted: Mon Apr 28, 2008 6:37 pm

Thanks to all those that answered. There are a variety of responses,
including
• A reference to a journal article (rotchm)
• A reference to a faq (PD)
• TA = TB in both scenarios (Tom Roberts)
• TA < TB and TA > TB in scenario 1a) and 1b) respectively, but the
scenarios are different problems and therefore can have different
solutions (harry)
• “any accumulated time difference will depend on … the frame in which
the clocks are ultimately compared.”(jem)
• “You can't compare their clocks directly, as they're not in the same
place” (tussock)

To quote the faq that PD linked :
"Your Honor, I will show first, that my client never borrowed the Ming
vase from the plaintiff; second, that he returned the vase in perfect
condition; and third, that the crack was already present when he
borrowed it."

(By the way, PD, I very much appreciate your reference. It has helped
me to identify what my scenarios seem to be grappling with – they are
similar to the “distance dependence objection”. Rotchm, thanks for
your reference – I will look it up some time soon.)

Rotchm and PD aside, all the above answers can’t all be correct. I’ll
address them in the order I believe was easiest to most difficult.

“the scenarios are different problems and therefore can have different
solutions”. I believe that you missed the essential point about the
scenarios. There exists a set of scenarios in which 1)a) is exactly
the same as 1)b). In this set, the velocity profiles are the same in
both scenarios, including the penultimate acceleration and coast in
1a). This is why the initial sentence was included in 1a) – to ensure
that the scenarios could be the same.

“You can't compare their clocks directly, as they're not in the same
place” Really? Imagine this: two rockets, containing clocks that
were previously synchronised in the same reference frame, are once
again moving at zero velocity relative to each other. They are
separated by a noticible distance. We should be able to compare
clocks by the following method: Firstly, establish the distance
between them by determining the delay required to carry a message
between them. Secondly, have one rocket, A, send its current time to
the other, B. B notes down the time on his clock at the exact point
that he receives the time message from A. The difference between the
proper times, deltaT, can be determined by deltaT = TB – TA –
delay.

“any accumulated time difference will depend on … the frame in which
the clocks are ultimately compared.” This, and your follow up
questions made me think. However, this statement imposes a
restriction on the universe that I am not yet sure is physically
justified – it suggests that clocks cannot be compared at intermediate
stages. Consider the following scenarios for example.

1c) A and B start apart some distance, but are at rest with respect
to each other and communicate as describe earlier in this post. On a
signal from one, they both accelerate in the same magnitude and
direction for the same period of time. When they finish accelerating,
they are at rest with respect to each other, and can therefore compare
proper times. Has their proper time (the time accumulated on their
individual clocks) differed? I would hazard a guess that you would
agree that TA = TB in this scenario.

1d) A, B and C start together, at the same velocity and location, and
with synchronised clocks. A and B accelerate in sync and then coast
for a period of time at a relativistic velocity which matches D. D
can synchronise its clock with A and B. A then accelerates back
toward C, matching velocity with it. After some time, B follows
suit. A, B and C can now compare clocks because they are matching
velocities (see technique above). This will be referred to as the
midpoint of the scenario. At a given signal, both A and B accelerate
back to matching velocities with D. Once again, A and B can now
compare clocks, and this time D can join in.

Now, if we assume that, as you stated, “any accumulated time
difference will depend on the aggregate speed of each clock relative
to the reference frame in which the clocks are ultimately compared”,
then the first comparison should be the reference frame of C and the
second comparison should be wrt the reference frame of D. According
to C, TC > TA > TB at the midpoint comparison. Similarly, the final
comparison according to D gives TD > TB > TA. To resolve the time
change between TA and TB after the midpoint, even though the flight
paths were the same except for a translation, one has to resort to the
time/distance arguments similar to that given in Micheal Weiss’s faq.
This essentially says that, because A spent more time than B in a
different reference frame to D, when both accelerate, A experience a
greater proper time dilation (slower time) than B, that more than
compensates for the original difference between the two at the
midpoint comparison. This result contradicts the result of scenario
1c)

The essence of all scenarios so far is that 1a) and 1c) can be sub-
scenarios of 1b) and 1d) respectively. Furthermore, the scenarios 1a)
and 1c) both seem complete (if properly defined – see below), both
creating a set of boundary conditions and establishing a sequence of
events. All scenarios are physically possible, and the proper times
are potentially measurable and comparable at specific points. And yet
scenarios 1a) and 1c) yield different comparable results to 1b) and
1d), even though nothing is physically different to A and B in the
respective scenarios. If scenarios require historic information
referenced outside of the basic structure of the scenario, then it
seems to thwart the principle of invariance, upon which relativity is
based.

I left Tom Roberts until last, for a couple of reasons. Firstly,
apart from ignoring the first line, which is an essential ingredient
for comparing the first to the scenario to the next, this was the best
rewrite of the scenario that I have read. Secondly, he pinpointed a
couple of issues that needed to be tightened, specifically, how the
final clocks are compared and the potential on my part to
misunderstand the phrase “moving clocks run slow”. In this scenario,
the clocks are only compared if they are in an inertial reference
frame whereby the relative velocities (wrt each other) is zero. If
this means that they are separated by a distance, then the technique
described above is used. These restrictions mean that the proper time
can be compared – with no attempt made to compare times when in
different reference frames (though this is not necessarily
prohibited).

My understanding of “moving clocks run slow” is this – when two clocks
are in different reference frames, moving at relativistic speeds away
from each other, then the changing distance between them makes an
apparent slowing. However, if the two clocks were to return to the
same reference frame, then there may be a permanent difference
introduced to their proper times, depending on who accelerated and
(possibly) when. At this point in time, I am not sure whether the
time difference is accumulated throughout the journey, or only during
the acceleration periods, but this distinction shouldn't affect the
scenarios, because each comparison is done when they are at rest
relative to each other.

Tom’s solution, that TA = TB in scenario 1a), irrespective of the
delay, is an interesting one. If, as suggested, we apply this
interpretation to scenario 1b), then TA = TB will resolve both
scenarios. I did consider this option in a previous forum, and the
general conclusion there was that there is sufficient similarity
between 1b) and the twins paradox that we needed to consider the same
approaches for this scenario. But the problem is, tradition doesn’t
resolve the twins paradox by suggesting that a delay in the turnaround
time has no effect on the resultant age difference between the
twins.

So, in conclusion, I am still struggling to understand how to resolve
these scenarios without resorting to some non-canonical physics
(which, at this point in time, I would prefer to find out whether my
logic is faulty and why).

Guest

Posted: Tue Apr 29, 2008 12:11 pm

On Apr 29, 2:37 pm, matscience...@yahoo.com wrote:

Quote:

The essence of all scenarios so far is that 1a) and 1c) can be sub-
scenarios of 1b) and 1d) respectively. Furthermore, the scenarios 1a)
and 1c) both seem complete (if properly defined – see below), both
creating a set of boundary conditions and establishing a sequence of
events. All scenarios are physically possible, and the proper times
are potentially measurable and comparable at specific points. And yet
scenarios 1a) and 1c) yield different comparable results to 1b) and
1d), even though nothing is physically different to A and B in the
respective scenarios. If scenarios require historic information
referenced outside of the basic structure of the scenario, then it
seems to thwart the principle of invariance, upon which relativity is
based.

This leads to the question of how do you know when a scenario has
sufficient information to be considered complete? Is there a list to
tick off? If some more information changes the simple scenarios, how
do we know if some further information wouldn't change the more
complex scenarios?

Guest

Posted: Tue Apr 29, 2008 4:40 pm

On Apr 30, 10:41 am, Tom Roberts <tjroberts...@sbcglobal.net> wrote:

Quote:

matscience...@yahoo.com wrote:
[...] But the problem is, tradition doesn’t
resolve the twins paradox by suggesting that a delay in the turnaround
time has no effect on the resultant age difference between the
twins.

But this ISN'T the twin scenario!

The traditional twin scenario has one twin remain at rest in an inertial
frame, and the other twin blasts off with rapid acceleration to travel
inertially to a distant location, turns around rapidly and returns
inertially to the first twin, de-accelerating rapidly to come to rest
nearby [#]. In this scenario, their age difference DOES depend upon how
long the trip takes (assuming the same speed of travel wrt the first
inertial frame).

[#] Periods of acceleration are short enough to be ignored,
to keep the math simple. It can be shown this does not affect
the basic conclusion, though it does affect the numerical
values.

That has little in common with the scenario you and I discussed in this
thread. In particular, one twin remains inertial throughout, and the
other turns around; neither applies to your scenario.

While in the original post, I did state that scenario 1b) was the
twin, I moderated that in the followup. This is what I said:

Quote:

the general conclusion there was that there is sufficient similarity
between 1b) and the twins paradox that we needed to consider the same
approaches for this scenario.

So let's look at the commonality of 1b) with the twins paradox:

1. C and A start in a common reference frame.
2. C remains inertial throughout
3. A, accelerates away and then coasts for some time.
4. A then accelerates back toward C

So far, alot of commonality. In fact, when the relative velocity of A
matches that of C, we are exactly at the halfway mark of the twins
paradox. Now, according to this faq[1], the general relativity
explanation suggests "by uniform "gravitational" time dilation, he
(the stay at home twin) ages years during Stella's (the travelling
twin) Turnaround." Furthermore, it suggests that "Terence's
Turnaround ageing should depend on how far he is from Stella when it
(the Turnaround) happens". So far, we are at the halfway mark of the
turnaround, and therefore should have accumulated half of the time
difference, if this explanation were true.

[1]http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/
twin_gr.html

We can create a similar comparison between C and B, only B coasts for
longer than A. If this coasting delay is factored in as an increased
'"gravitational" potential difference' in the full twins paradox, then
it should similarly be factored in at the halfway mark. It would
seem, according to the arguments presented in the faq, that TC > TA >
TB, not TA = TB.

If your argument is with the way the faq is written, it would be
better to come up with an improved faq for the use of this forum.

Quote:

I suggest you learn about SR, rather than "struggle" to make up some
other "solution", or try to meld a bunch of wrong answers into a
"correct" one. A good textbook is:

Taylor and Wheeler, _Spacetime_Physics_.

This is rather strawman of you Tom. But the reference may be useful -
I'll see if the university library has it.

Tom Roberts

Posted: Tue Apr 29, 2008 7:41 pm

Guest

matscienceman@yahoo.com wrote:

Quote:

Thanks to all those that answered. [...]

This is not politics, and no "vote" is needed. "Knowledge is not gained
by pooling ignorance." -- Ayn Rand.

Quote:

Tom’s solution, that TA = TB in scenario 1a), irrespective of the
delay, is an interesting one.

Of course. Correct solutions are always more interesting than incorrect
ones.

Quote:

[...] But the problem is, tradition doesn’t
resolve the twins paradox by suggesting that a delay in the turnaround
time has no effect on the resultant age difference between the
twins.

But this ISN'T the twin scenario!

The traditional twin scenario has one twin remain at rest in an inertial
frame, and the other twin blasts off with rapid acceleration to travel
inertially to a distant location, turns around rapidly and returns
inertially to the first twin, de-accelerating rapidly to come to rest
nearby [#]. In this scenario, their age difference DOES depend upon how
long the trip takes (assuming the same speed of travel wrt the first
inertial frame).

[#] Periods of acceleration are short enough to be ignored,
to keep the math simple. It can be shown this does not affect
the basic conclusion, though it does affect the numerical
values.

That has little in common with the scenario you and I discussed in this
thread. In particular, one twin remains inertial throughout, and the
other turns around; neither applies to your scenario.

Quote:

So, in conclusion, I am still struggling to understand how to resolve
these scenarios without resorting to some non-canonical physics
(which, at this point in time, I would prefer to find out whether my
logic is faulty and why).

Quote:

There is no general "list". What one must do is analyze the scenario
using a given model (aka theory). That will unambiguously show whether
or not there is sufficient information about the scenario USING THAT
MODEL. But different models might well require different information.
And different scenarios will in general require different information
for a given model (e.g. your original scenario vs the twin scenario).

Tom Roberts

Guest

Posted: Wed Apr 30, 2008 1:36 am

On Apr 30, 8:14 pm, "Dirk Van de moortel" <dirkvandemoor...@ThankS-NO-
SperM.hotmail.com> wrote:

Quote:

matscience...@yahoo.com <matscience...@yahoo.com> wrote in message

2f9366c4-2c50-40ac-bd3e-0d6b18bb7...@y18g2000pre.googlegroups.com

On Apr 29, 2:37 pm, matscience...@yahoo.com wrote:

The essence of all scenarios so far is that 1a) and 1c) can be sub-
scenarios of 1b) and 1d) respectively. Furthermore, the scenarios 1a)
and 1c) both seem complete (if properly defined – see below), both
creating a set of boundary conditions and establishing a sequence of
events. All scenarios are physically possible, and the proper times
are potentially measurable and comparable at specific points. And yet
scenarios 1a) and 1c) yield different comparable results to 1b) and
1d), even though nothing is physically different to A and B in the
respective scenarios. If scenarios require historic information
referenced outside of the basic structure of the scenario, then it
seems to thwart the principle of invariance, upon which relativity is
based.

This leads to the question of how do you know when a scenario has
sufficient information to be considered complete? Is there a list to
tick off? If some more information changes the simple scenarios, how
do we know if some further information wouldn't change the more
complex scenarios?

You can use the following list:

1) was the scenario posted in sci.physics.relativity?
2) was it posted by a newcomer?
3) did the poster ignore the replies from knowlegable non-crackpots?

If you have 3 checks, you know that the scenario has insufficient
information to be considered complete.
Moreover, you know you have discovered a new crackpot.

Dirk Vdm- Hide quoted text -

- Show quoted text -

Lol. Does the fact that I ignored Androcles unusual reply give me the
third check? Smile

I wouldn't want to miss out on crackpot status,
now, would I?

Dirk Van de moortel

Posted: Wed Apr 30, 2008 5:14 am

Guest

matscienceman@yahoo.com <matscienceman@yahoo.com> wrote in message
2f9366c4-2c50-40ac-bd3e-0d6b18bb72b7@y18g2000pre.googlegroups.com

Quote:

On Apr 29, 2:37 pm, matscience...@yahoo.com wrote:

The essence of all scenarios so far is that 1a) and 1c) can be sub-
scenarios of 1b) and 1d) respectively. Furthermore, the scenarios 1a)
and 1c) both seem complete (if properly defined – see below), both
creating a set of boundary conditions and establishing a sequence of
events. All scenarios are physically possible, and the proper times
are potentially measurable and comparable at specific points. And yet
scenarios 1a) and 1c) yield different comparable results to 1b) and
1d), even though nothing is physically different to A and B in the
respective scenarios. If scenarios require historic information
referenced outside of the basic structure of the scenario, then it
seems to thwart the principle of invariance, upon which relativity is
based.

This leads to the question of how do you know when a scenario has
sufficient information to be considered complete? Is there a list to
tick off? If some more information changes the simple scenarios, how
do we know if some further information wouldn't change the more
complex scenarios?

You can use the following list:

1) was the scenario posted in sci.physics.relativity?
2) was it posted by a newcomer?
3) did the poster ignore the replies from knowlegable non-crackpots?

If you have 3 checks, you know that the scenario has insufficient
information to be considered complete.
Moreover, you know you have discovered a new crackpot.

Dirk Vdm

Guest

Posted: Wed Apr 30, 2008 2:25 pm

On Apr 19, 12:35 am, PD <TheDraperFam...@gmail.com> wrote:

Quote:

On Apr 18, 1:03 am, matscience...@yahoo.com wrote:

I probably have some flawed understanding, but I can't work out what
it is. I have put this up on two different science forums, but so far
this has stumped both. Below are two scenarios, which combined seem
to defy the resolution of the twins paradox.

In both scenarios, the level of acceleration and time under
acceleration are the same.

Scenario 1a)

A and B accelerate and coast several times in sync. Then A accelerates
then coasts away at constant velocity while B waits at the previous
location. After a while, B accelerates towards A until they are
matching velocities, although some distance away. Is there a
difference in their clocks?

Solution: Similar to the twins paradox, I would expect that, since A
accelerates before B, this breaks the symmetry first. Therefore, time
for A slows down wrt B while B remains unaccelerated. So, when they
match velocities, TA < TB.

Scenario 1b)
A and B accelerate away from C. A accelerates back to C while B ceases
to accelerate and coasts. B later accelerates back toward A and C,
matching velocity with them. Is there a difference in their clocks?

Solution: This is the twins paradox twice over - therefore A and B
break the symmetry, leading to a time dilation. B is in that state
for a longer period than A, therefore, when B matches velocity with A
and C again, TC > TA > TB

Problem: Since we have A and B accelerating in sync in scenario 1a),
all the velocities and accelerations in these two scenarios can be
made to be exactly the same. But they result in opposite
relationships in clocktimes for TA and TB. TA > TB is incompadible
with TB > TA.

The purpose of the twins paradox is to teach that speeding up is not
the origin of time dilation. The simplest explanation is that proper
time between two events depends on the path through spacetime. Put in
very coarse (deceptively simple) terms, in Euclidean space the
shortest distance between two points is a straight line, but in
Minkowski spacetime the *longest* interval between to events is a
straight line. The reason is that minus sign that is in the Minkowski
equivalent of the Pythagorean theorem. A good spacetime diagram on an
FAQ will go along way to helping you understand this.http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_...

PD- Hide quoted text -

- Show quoted text -

PD suggested this FAQ. Is this an official FAQ for this forum, or is
there another?

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