The rocket of the travelling twin from the famous twin paradox has two
clocks: one at the front end and another at the back end. As the front
end clock (T) passes the single clock (C) of the twin at rest, both
readings are set to zero (T=C=0):

RockeT-..........
......C..........

Then the back end clock (R) is synchronized with T and, as R passes C,
their readings are compared. Which reading is greater? RC? CR?

After R has passed C, the rocket returns as quickly as possible, R at
the front end this time, so that the time interval between the two
consecutive meetings of R and C can be minimal:

........-RockeT......
......C.............

In a normal science, further analysis of this setup would amount to
reductio ad absurdum. As C and T meet for the second time, C must prove
slower than T, in accordance with the standard time dilation analysis,
but at the same time C must run faster than T precisely by a factor of
gamma because Einstein said so in 1905. In relativity absurdity is
called paradox: instead of being rejected, the theory becomes
sensational and brings prestige and money to both initiated and
zombies.

Pentcho Valev

On 20 Oct 2005 01:56:39 -0700, "Pentcho Valev"
wrote:

The rocket of the travelling twin from the famous twin paradox has two
clocks: one at the front end and another at the back end. As the front
end clock (T) passes the single clock (C) of the twin at rest, both
readings are set to zero (T=C=0):

RockeT-..........
.....C..........

Then the back end clock (R) is synchronized with T and, as R passes C,
their readings are compared. Which reading is greater? RC? CR?

After R has passed C, the rocket returns as quickly as possible, R at
the front end this time, so that the time interval between the two
consecutive meetings of R and C can be minimal:

.......-RockeT......
.....C.............

In a normal science, further analysis of this setup would amount to
reductio ad absurdum. As C and T meet for the second time, C must prove
slower than T, in accordance with the standard time dilation analysis,
but at the same time C must run faster than T precisely by a factor of
gamma because Einstein said so in 1905. In relativity absurdity is
called paradox: instead of being rejected, the theory becomes
sensational and brings prestige and money to both initiated and
zombies.

Pentcho Valev


Loon.

PLONK!

Einstein's Twin Idiocy Again

*******************

I thought his was a single birth?????????

Androcles wrote:
"Bryan Olson" [to put it simply] wrote:

| Sigh. Synchronized in what frame?
|
| http://www.bartleby.com/173/9.html
|
| It might be a good idea to learn the basics of the theory before
| trying to criticise it.

| Science accepts theories that predict what actually happens.
| We're not going to reject a theory just because some Usenet
| blowhard cannot or will not comprehend it.
|
|
| --
| --Bryan

Science accepts theories that predict what actually happens.
We're not going to accept a theory just because some Usenet
blowhard cannot or will not comprehend it.
So, blowhard, what are the basics of the theory?
Androcles.

I thought I included a URL for exactly that ... Oh look--I did.

Hey, check it out: "Synchronized in what frame?" turns out to
be a really good question.

Any clues yet?


--
--Bryan

wrote:
First part: CR. Solving from C frame or from rocket frame, we get the
same conclusion. As Rear and C coincide, CR.

Second part: Here, why did you introduce the first part, where there
was a synch? Your second part is independent of that. You have a
traveler (T) who passes C (at time synch 0) goes away and comes back,
No matter what happens on R. So why you need to synck for this?


And, here is a reminder. Front and Rear are in synch in the rocket
frame. They are not in synch in C frame. As Front=C=0, Rear is NOT 0 in
C frame (it is in T or R frame).

For the first part, do you mean CR? Excuse me for my ignorance on
Relativity, but according to the time dilation, the moving clock runs
slower. That means clock R runs slower than clock C, so CR. Does my
interpretation wrong or do I miss anything?

I think u r missing something. As I say, CR.

This is because the synch has been done between The front and rear
clock in *their* (rocket) frame. The clocks are not in synch in the C
frame.
This, combined with LorentzContraction of the lenght of the rocket
(wrt C) and the speed of the rocket produces and time contraction
(slowdown) of R (wrt C) makes a combined effect of CR ... if I am not
missing anything

That is, if the initial synch was done on the rockets frame as I assume
since the initial poster did not specify it.

snapdragon31 wrote:
wrote:
First part: CR. Solving from C frame or from rocket frame, we get the
same conclusion. As Rear and C coincide, CR.

Second part: Here, why did you introduce the first part, where there
was a synch? Your second part is independent of that. You have a
traveler (T) who passes C (at time synch 0) goes away and comes back,
No matter what happens on R. So why you need to synck for this?


And, here is a reminder. Front and Rear are in synch in the rocket
frame. They are not in synch in C frame. As Front=C=0, Rear is NOT 0 in
C frame (it is in T or R frame).

For the first part, do you mean CR? Excuse me for my ignorance on
Relativity, but according to the time dilation, the moving clock runs
slower. That means clock R runs slower than clock C, so CR. Does my
interpretation wrong or do I miss anything?

Compare the mechanism of the hypothetical light-clock,
used to describe Minkowski space with a clock viewed
over a path of changing length.

The only thing the 'twins paradox' illustrates is that
the clocks *by definition* are not the same.

... But the discovery, of Minkowski, which was of
importance for the formal development of the theory
of relativity, does not lie here. It is to be found
rather in the fact of his recognition that the four-dimensional
space-time continuum of the theory of
relativity, in its most essential formal properties,
shows a pronounced relationship to the three-dimensional continuum of
Euclidean geometrical space. 1 In order
to give due prominence to this relationship, however,
we must replace the usual time co-ordinate t by an
imaginary magnitude
http://www.bartleby.com/173/M11.GIF (sqrt -1)

ct proportional to it. Under these conditions, the
natural laws satisfying the demands of the (special)
theory of relativity assume mathematical forms, in
which the time co-ordinate plays exactly the same
rôle as the three space co-ordinates.
http://www.bartleby.com/173/17.html
http://mathworld.wolfram.com/ImaginaryAxis.html

Sue...

"Bryan Olson" wrote in message
t...
| Androcles wrote:
| "Bryan Olson" [to put it simply] wrote:
|
| | Sigh. Synchronized in what frame?
| |
| | http://www.bartleby.com/173/9.html
| |
| | It might be a good idea to learn the basics of the theory before
| | trying to criticise it.
|
| | Science accepts theories that predict what actually happens.
| | We're not going to reject a theory just because some Usenet
| | blowhard cannot or will not comprehend it.
| |
| |
| | --
| | --Bryan
|
| Science accepts theories that predict what actually happens.
| We're not going to accept a theory just because some Usenet
| blowhard cannot or will not comprehend it.
| So, blowhard, what are the basics of the theory?
| Androcles.
|
| I thought I included a URL for exactly that ... Oh look--I did.

Sigh... It might be a good idea to learn the basics of the theory before
trying to promote it, blowhard.

"Are two events (e.g. the two strokes of lightning A and B) which are
simultaneous with reference to the railway embankment also simultaneous
relatively to the train? We shall show directly that the answer must be
in the" POSITIVE.
|
| Hey, check it out: "Synchronized in what frame?" turns out to
| be a really good question.
|
| Any clues yet?

More than you will ever realise, blowhard.

One lightning flash leaves the caboose and arrives at the engine, then
reflects
back to the caboose.

The other lightning flash leaves the engine and arrives at the caboose,
then reflects
back to the engine.

The diagram is like this (fixed font needed).

|
|
| C'
| /
| B /
| ____________Mirror
| /\ /
| / \ /
C / \ /
|\ / \ /
| \ / \A'
| \ / | /
| \ / /
| \ / /
| \ / | /
| \ / /
| \/ |
| /\ /
| / \ / |
| / \ /
| / \ / |
| / ____\/__________Mirror
| /
| / D |
|/
/ ____________|____|________________
A D B A' C'


[quote]
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'].
[end quote]
Ref: http://www.fourmilab.ch/etexts/einstein/specrel/www/

We establish by definition that the "time" required by light to travel
from C to D equals the "time" it requires to travel from D to C'.

Distance between mirrors is x'

Einstein's equation:

½[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))] = tau(x',0,0,t+x'/(c-v))

What it means in the diagram:

½[tau(A,t)+tau(A',t+x'/(c-v)+x'/(c+v))] = tau(B,t+x'/(c-v))
½[tau(C,t)+tau(C',t+x'/(c-v)+x'/(c+v))] = tau(D,t+x'/(c PLUS v))

So the time at B, the engine, equals the time at D, the caboose,
but it doesn't. Ergo Einstein was a phuckwit.

Got a clue now, blowhard?
Androcles.

wrote:
I think u r missing something. As I say, CR.

This is because the synch has been done between The front and rear
clock in *their* (rocket) frame. The clocks are not in synch in the C
frame.
This, combined with LorentzContraction of the lenght of the rocket
(wrt C) and the speed of the rocket produces and time contraction
(slowdown) of R (wrt C) makes a combined effect of CR ... if I am not
missing anything

That is, if the initial synch was done on the rockets frame as I assume
since the initial poster did not specify it.

Thank you. Yes, I should have to include the effect of Lorentz
Contraction into consideration. But when I include the Lorentz
Contraction I get a strange result.

If the speed is 0.866 c then the length contraction is 0.5. In the
example below, the speed used is 100 m/s but the factor of 0.5 is being
kept.

1. Assume the original length of the moving rocket is 100 m.
Assume the Lorentz contraction factor is 0.5
Legnth of the rocket is x = 50 m (View from the stationary frame C).
Legnth of the rocket is x' = 100 m (View from the moving frame T).
2. The rocket is travelling at a speed of 100 m/s
Let v be the speed of (moving) clock T wrt (stationary) clock C
Let v' be the speed of (stationary) clock C wrt (moving) clock T
v = v' = 100 m/s
3. In the moving frame T,
Length of rocket x' = 100 m
Vel. of the rocket v' = 100 m/s
Time t' = 1 sec (Takes 1 moving sec. for the rocket to pass clock C)
4. In the stationary frame C,
Length of rocket x = 50 m (Rocket length view from frame C)
Vel. of the rocket v = 100 m/s
Time t = 0.5 sec (Takes 0.5 stationary sec. the rocket to pass clock
C)
5. t = 0.5 sec; t' = 1 sec. i.e. t t', this matches your C R
But according to time dilation, if t=0.5 sec then t' should be equal
to 0.25 second. i.e t t' (Moving clock runs slower). Because of this
effect, the moving twin is younger. Every body including Einstein say
so.

In this example, there is no clock synchronization. I might miss
something bigger than before. This reply does not go to the
sci.physics, sci.skeptic and sci.philosophy.tech.

snapdragon31 wrote:
wrote:
I think u r missing something. As I say, CR.

This is because the synch has been done between The front and rear
clock in *their* (rocket) frame. The clocks are not in synch in the C
frame.
This, combined with LorentzContraction of the lenght of the rocket
(wrt C) and the speed of the rocket produces and time contraction
(slowdown) of R (wrt C) makes a combined effect of CR ... if I am not
missing anything

That is, if the initial synch was done on the rockets frame as I assume
since the initial poster did not specify it.

Thank you. Yes, I should have to include the effect of Lorentz
Contraction into consideration. But when I include the Lorentz
Contraction I get a strange result.

If the speed is 0.866 c then the length contraction is 0.5. In the
example below, the speed used is 100 m/s but the factor of 0.5 is being
kept.

1. Assume the original length of the moving rocket is 100 m.
Assume the Lorentz contraction factor is 0.5
Legnth of the rocket is x = 50 m (View from the stationary frame C).
Legnth of the rocket is x' = 100 m (View from the moving frame T).
2. The rocket is travelling at a speed of 100 m/s
Let v be the speed of (moving) clock T wrt (stationary) clock C
Let v' be the speed of (stationary) clock C wrt (moving) clock T
v = v' = 100 m/s
3. In the moving frame T,
Length of rocket x' = 100 m
Vel. of the rocket v' = 100 m/s
Time t' = 1 sec (Takes 1 moving sec. for the rocket to pass clock C)
4. In the stationary frame C,
Length of rocket x = 50 m (Rocket length view from frame C)
Vel. of the rocket v = 100 m/s
Time t = 0.5 sec (Takes 0.5 stationary sec. the rocket to pass clock
C)
5. t = 0.5 sec; t' = 1 sec. i.e. t t', this matches your C R
But according to time dilation, if t=0.5 sec then t' should be equal
to 0.25 second. i.e t t' (Moving clock runs slower). Because of this
effect, the moving twin is younger. Every body including Einstein say

No...*Everybody* agress that both twins will see
each and every frame of a commonly viewed motion
picture, and will see each and every orbit of Jupiters
moons, regardless of their travels.

That doesn't make much of a case for any difference
in age.

Sue...

so.

In this example, there is no clock synchronization. I might miss
something bigger than before. This reply does not go to the
sci.physics, sci.skeptic and sci.philosophy.tech.