Martin Hogbin wrote:
wrote in message oups.com...
Hi,
I am not a physics, just an amateur. However, I treat myself as a
logician, so the paradoxes are in my interest. After I read the
discussions about the special relativity(SR) and twin's paradox,
believe that it can not be resolved within SR. The defenders of SR
claims that there is an asymmetry in the conditions of the twins and so
there is no paradox. However, one can easily think of an experiment in
which the twins are in symmetric conditions.

Let A B are the twins and C is a third observer. The twins move away
from the earth in opposite directions and C remains rest on the earth.
Assume their clock is synchronized just before the movement. After
accelaration twins continue to their journey in a constant speed. After
some time they return back and land on the earth. Let from the C's
point of view, the twins all accelarate at the same time and with the
same amount. The only difference between the twins is that they travel
in opposite directions. After returning back they compare their
clocks. If SR is true, according to C, the twins clock will slow down
with the same amount and their clocks must be the same.

Correct.

On the other
hand, according to the twin A, twin B's clock must be slow down and not
the same with his/her clock.

Wrong! SR predicts that both twins' clocks will show the same time in
the symmetrical case you describe.

As others have pointed out, you must understand a theory before
you criticise it.

I don't suppose Maxwell's equations has anything to do with SR does it?

http://farside.ph.utexas.edu/teaching.html
http://web.mit.edu/8.02t/www/802TEAL...ight/index.htm
http://en.wikipedia.org/wiki/Multiple_integral

Sue...




Martin Hogbin

Let A B are the twins and C is a third observer. The twins move away
from the earth in opposite directions and C remains rest on the earth.
Assume their clock is synchronized just before the movement. After
accelaration twins continue to their journey in a constant speed. After
some time they return back and land on the earth. Let from the C's
point of view, the twins all accelarate at the same time and with the
same amount. The only difference between the twins is that they travel
in opposite directions. After returning back they compare their
clocks. If SR is true, according to C, the twins clock will slow down
with the same amount and their clocks must be the same.

Correct. As A,B and C coincide, C concludes, as you noted by the
symmetry principle that A and B bust be the same.

On the other
hand, according to the twin A, twin B's clock must be slow down and not
the same with his/her clock.

Careful. B's clock slows down wrt A reference frame. (as measured/as
observed, but these are misleading words)) But the distance that B
traveled wrt A's frame is also lessened. Therefore, although B's clock
runs slower, B also travels a shorter distance, negating the effect and
thus A and B's clock have the same value as they coincide.

You can completely remove the accelerations: A and B travel in
opposite direction (face-to-face) and coincide at point C (a clock on
the ground). They both travel a distance D (wrt the ground) where they
coincide with two other 'twins' A' and B' in opposite direction with
speed v wrt ground. These two A' and B' continue travel till they
coincide with C.

From C point of view, all is symmetric and A' and b' clocks indicate
the same value on coincidence with C. Now you must take the point of
view of A (and/or A') to analyze the situation.
If you do the math correctly, you will conclude that A' will conclude
that B' clock has the same value as A'.

Dear Sue:

Sue... wrote:
N:dlzc D:aol T:com (dlzc) wrote:
Dear lkoluk2003:

wrote in message
oups.com...
...
Let A B are the twins and C is a third observer. The twins
move away from the earth in opposite directions and C
remains rest on the earth. Assume their clock is
synchronized just before the movement. After
accelaration twins continue to their journey in a constant
speed. After some time they return back and land on the
earth. Let from the C's point of view, the twins all
accelarate at the same time and with the same amount.
The only difference between the twins is that they travel
in opposite directions. After returning back they
compare their clocks. If SR is true,

"true" is not part of science. "what is observed, within our
ability to measure" is part of science.

according to C, the twins clock will slow down with the
same amount and their clocks must be the same. On
the other hand, according to the twin A, twin B's clock
must be slow down and not the same with his/her
clock. The same thing is true for twin B. Of course
there can be only one result. However, according to
SR, three observer comes out with three different
results.

If you did the math, observer A sees observer B's clock
go very slowly on A's outbound, continue slowly for some
time with A inbound (since B's start-of-return doesn't reach
A until later) then sees B's clock rate increase until it
meets A at Earth, with net 0 difference at Earth.

So three observers in a symmetric situation yields two
results, because only two unique paths are described.

It is the words that trick your common sense, not the facts.
Use the math.

In response to your complaints about the elastic properties
of today's garden hose material and the possibility of
"bead bunching", we are proud to announce the all new and
improved

Three clocks, two different elapsed times. Argument? Beads, hoses?
I'm not looking to buy Manhattan...

David A. Smith

Lokman Kolukisa ) wrote:
[...]
If SR is true, according to C, the twins clock will slow down
with the same amount and their clocks must be the same.

That's correct. Since C is unaccelerated during the entire
voyages by A and B, C can make use of the time-dilation
result for the entire trips: According to C, A and B will each
age at the same constant rate, which is less than the rate of
ageing of C.

On the other
hand, according to the twin A, twin B's clock must be slow down and not
the same with his/her clock.

That's only true during the two periods where A is not accelerating.
The time-dilation result only holds during periods of time where
the observer is unaccelerated. For A, there are two such periods:
during his outbound leg, and during his inbound leg.

When A accelerates at his turnaround point, he will conclude that
B suddenly ages. When he adds the three components of B's total
ageing (B's ageing during A's outbound leg + B's ageing during A's
turnaround + B's ageing during A's inbound leg), he will get the
same total that C got. (And B will come to the same conclusions
about A's ageing, as A concluded about B's ageing.)

The amount of the ageing of the object that occurs during the
acceleration of the observer is proportional to their separation.
That's why the initial and final accelerations by the observer at
the beginning and end of the trips don't cause any ageing of the
object, but the acceleration of the observer at the turnaround
does cause a sudden ageing of the object (according to the
accelerating observer).

The abrupt ageing during the turnaround occurs both for
instantaneous speed changes, and for finite accelerations
(whenever the separation at the turnaround is sufficiently
great). For a detailed example with 1g accelerations, see my
web page:

http://home.comcast.net/~mlfasf

For more information about the abrupt change in the age of the
object during accelerations by the observer, see my paper:

"Accelerated Observers in Special Relativity",
PHYSICS ESSAYS, December 1999, p629.

Mike Fontenot

dlzc wrote:
Dear Sue:

Sue... wrote:
N:dlzc D:aol T:com (dlzc) wrote:
Dear lkoluk2003:

wrote in message
oups.com...
...
Let A B are the twins and C is a third observer. The twins
move away from the earth in opposite directions and C
remains rest on the earth. Assume their clock is
synchronized just before the movement. After
accelaration twins continue to their journey in a constant
speed. After some time they return back and land on the
earth. Let from the C's point of view, the twins all
accelarate at the same time and with the same amount.
The only difference between the twins is that they travel
in opposite directions. After returning back they
compare their clocks. If SR is true,

"true" is not part of science. "what is observed, within our
ability to measure" is part of science.

according to C, the twins clock will slow down with the
same amount and their clocks must be the same. On
the other hand, according to the twin A, twin B's clock
must be slow down and not the same with his/her
clock. The same thing is true for twin B. Of course
there can be only one result. However, according to
SR, three observer comes out with three different
results.

If you did the math, observer A sees observer B's clock
go very slowly on A's outbound, continue slowly for some
time with A inbound (since B's start-of-return doesn't reach
A until later) then sees B's clock rate increase until it
meets A at Earth, with net 0 difference at Earth.

So three observers in a symmetric situation yields two
results, because only two unique paths are described.

It is the words that trick your common sense, not the facts.
Use the math.

In response to your complaints about the elastic properties
of today's garden hose material and the possibility of
"bead bunching", we are proud to announce the all new and
improved

Three clocks, two different elapsed times. Argument? Beads, hoses?
I'm not looking to buy Manhattan...

)

Uhmm Kimosabe, Manhattan cost you heapum wampum.

I'll let the web publishing of Unnikrishnan, Fitzpatrick and MIT
serve as my argument for *this* thread because the OP doesn't
mention resolution of the postulates and that is where an SR
discussion needs to go to be productive for the OP.

We can discuss equal length path techniques in a thread
where electromagnetism is not the focus. I was just injecting
some humor, not intending to do anything to interfere with
the remote possibility that the discussion might wander off
into the physics arena. ;-)


Regards,

Sue...


David A. Smith

wrote:
Hi,
I am not a physics, just an amateur. However, I treat myself as a
logician, so the paradoxes are in my interest. After I read the
discussions about the special relativity(SR) and twin's paradox,
believe that it can not be resolved within SR. The defenders of SR
claims that there is an asymmetry in the conditions of the twins and so
there is no paradox. However, one can easily think of an experiment in
which the twins are in symmetric conditions.

Let A B are the twins and C is a third observer. The twins move away
from the earth in opposite directions and C remains rest on the earth.
Assume their clock is synchronized just before the movement. After
accelaration twins continue to their journey in a constant speed. After
some time they return back and land on the earth. Let from the C's
point of view, the twins all accelarate at the same time and with the
same amount. The only difference between the twins is that they travel
in opposite directions. After returning back they compare their
clocks. If SR is true, according to C, the twins clock will slow down
with the same amount and their clocks must be the same. On the other
hand, according to the twin A, twin B's clock must be slow down and not
the same with his/her clock. The same thing is true for twin B. Of
course there can be only one result. However, according to SR, three
observer comes out with three different results.

How did you arrive at this conclusion?
Show your calculations, please.
The LT should be simple for one who prides himself of being
a logician.

Paul


For just objections relating to accelarations: the accelaration times
of the twins can be made arbitrararily smaller according to the
duration in which the twins travel at a constant speed, so their
affects can be vanished without a significant error.

Lokman Kolukisa

Sue... wrote:
dlzc wrote:
Dear Sue:

Sue... wrote:
N:dlzc D:aol T:com (dlzc) wrote:
Dear lkoluk2003:

wrote in message
oups.com...
If you did the math, observer A sees observer B's clock
go very slowly on A's outbound, continue slowly for some
time with A inbound (since B's start-of-return doesn't reach
A until later) then sees B's clock rate increase until it
meets A at Earth, with net 0 difference at Earth.

So three observers in a symmetric situation yields two
results, because only two unique paths are described.

It is the words that trick your common sense, not the facts.
Use the math.
In response to your complaints about the elastic properties
of today's garden hose material and the possibility of
"bead bunching", we are proud to announce the all new and
improved
Three clocks, two different elapsed times. Argument? Beads, hoses?
I'm not looking to buy Manhattan...

)

Uhmm Kimosabe, Manhattan cost you heapum wampum.

I'll let the web publishing of Unnikrishnan, Fitzpatrick and MIT
serve as my argument for *this* thread because the OP doesn't
mention resolution of the postulates and that is where an SR
discussion needs to go to be productive for the OP.

We can discuss equal length path techniques in a thread
where electromagnetism is not the focus. I was just injecting
some humor, not intending to do anything to interfere with
the remote possibility that the discussion might wander off
into the physics arena. ;-)


Regards,

Sue...

There is no risk that you ever will wander into the physics arena.
You are babbling a lot about hoses and marbles, but you are
utterly unable to explain how you would use them to prove
your claims.

You flee like hell every time someone asks you to do so.
Like you did now.

Paul

Paul B. Andersen wrote:
wrote:
Hi,
I am not a physics, just an amateur. However, I treat myself as a
logician, so the paradoxes are in my interest. After I read the
discussions about the special relativity(SR) and twin's paradox,
believe that it can not be resolved within SR. The defenders of SR
claims that there is an asymmetry in the conditions of the twins and so
there is no paradox. However, one can easily think of an experiment in
which the twins are in symmetric conditions.

Let A B are the twins and C is a third observer. The twins move away
from the earth in opposite directions and C remains rest on the earth.
Assume their clock is synchronized just before the movement. After
accelaration twins continue to their journey in a constant speed. After
some time they return back and land on the earth. Let from the C's
point of view, the twins all accelarate at the same time and with the
same amount. The only difference between the twins is that they travel
in opposite directions. After returning back they compare their
clocks. If SR is true, according to C, the twins clock will slow down
with the same amount and their clocks must be the same. On the other
hand, according to the twin A, twin B's clock must be slow down and not
the same with his/her clock. The same thing is true for twin B. Of
course there can be only one result. However, according to SR, three
observer comes out with three different results.

How did you arrive at this conclusion?
Show your calculations, please.
The LT should be simple for one who prides himself of being
a logician.

Paul


For just objections relating to accelarations: the accelaration times
of the twins can be made arbitrararily smaller according to the
duration in which the twins travel at a constant speed, so their
affects can be vanished without a significant error.

Lokman Kolukisa


Dear Sir Poul B

One can assert numbers and calculations to any wird and
absurd thing

What makes you think that you don't waste your calculations
on absurd wird things?

Paul B. Andersen wrote:
Sue... wrote:
dlzc wrote:
Dear Sue:

Sue... wrote:
N:dlzc D:aol T:com (dlzc) wrote:
Dear lkoluk2003:

wrote in message
oups.com...
If you did the math, observer A sees observer B's clock
go very slowly on A's outbound, continue slowly for some
time with A inbound (since B's start-of-return doesn't reach
A until later) then sees B's clock rate increase until it
meets A at Earth, with net 0 difference at Earth.

So three observers in a symmetric situation yields two
results, because only two unique paths are described.

It is the words that trick your common sense, not the facts.
Use the math.
In response to your complaints about the elastic properties
of today's garden hose material and the possibility of
"bead bunching", we are proud to announce the all new and
improved
Three clocks, two different elapsed times. Argument? Beads, hoses?
I'm not looking to buy Manhattan...

)

Uhmm Kimosabe, Manhattan cost you heapum wampum.

I'll let the web publishing of Unnikrishnan, Fitzpatrick and MIT
serve as my argument for *this* thread because the OP doesn't
mention resolution of the postulates and that is where an SR
discussion needs to go to be productive for the OP.

We can discuss equal length path techniques in a thread
where electromagnetism is not the focus. I was just injecting
some humor, not intending to do anything to interfere with
the remote possibility that the discussion might wander off
into the physics arena. ;-)


Regards,

Sue...

There is no risk that you ever will wander into the physics arena.
You are babbling a lot about hoses and marbles, but you are
utterly unable to explain how you would use them to prove
your claims.

I've made no claims in this tread other than beads on a
string count as well as marbles in a hose. I can understand
why you might disagree... your telekinesis teacher told
you too.


You flee like hell every time someone asks you to do so.
Like you did now.

Perhaps I find this more interesting than telekinesis and snake
worship.

http://farside.ph.utexas.edu/teaching.html
http://web.mit.edu/8.02t/www/802TEAL...ight/index.htm
http://en.wikipedia.org/wiki/Multiple_integral

Learn some physics so you're not such a bore.


Sue...


Paul

Paul B. Andersen wrote:
wrote:
Hi,
I am not a physics, just an amateur. However, I treat myself as a
logician, so the paradoxes are in my interest. After I read the
discussions about the special relativity(SR) and twin's paradox,
believe that it can not be resolved within SR. The defenders of SR
claims that there is an asymmetry in the conditions of the twins and so
there is no paradox. However, one can easily think of an experiment in
which the twins are in symmetric conditions.

Let A B are the twins and C is a third observer. The twins move away
from the earth in opposite directions and C remains rest on the earth.
Assume their clock is synchronized just before the movement. After
accelaration twins continue to their journey in a constant speed. After
some time they return back and land on the earth. Let from the C's
point of view, the twins all accelarate at the same time and with the
same amount. The only difference between the twins is that they travel
in opposite directions. After returning back they compare their
clocks. If SR is true, according to C, the twins clock will slow down
with the same amount and their clocks must be the same. On the other
hand, according to the twin A, twin B's clock must be slow down and not
the same with his/her clock. The same thing is true for twin B. Of
course there can be only one result. However, according to SR, three
observer comes out with three different results.

How did you arrive at this conclusion?
Show your calculations, please.
The LT should be simple for one who prides himself of being
a logician.

Clocks don't measure time and don't respond in any predictable
manner to motion so only fools would consicer making calculations.

Sue...


Paul


For just objections relating to accelarations: the accelaration times
of the twins can be made arbitrararily smaller according to the
duration in which the twins travel at a constant speed, so their
affects can be vanished without a significant error.

Lokman Kolukisa