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.

The problem you describe is diferent from the original post. Hence,
when you say "this matches your C R ", it doesnt match my CR; we are
describing diferent setups.

As for your setup, well, I will read it in more detail cause I dont
quite follow your description (but I can conclude that it is diferent
from the original poster, and you say so too)

"Pentcho Valev" wrote in message
oups.com...
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


Pentcho:
(1) Given the synchronization of the R and T clocks IN THEIR COMMON FRAME OF
REFERENCE, then, when they meet, the clock at R reads greater than the clock
at C. Or in your notation, RC. If you understand Minkowski diagrams, you
will understand why.
(2) Yes, when the C and T clocks meet a second time, the C clock will read
greater than the T clock. The "standard time dilation analysis" does NOT
say "C must prove slower than T."

Eli Botkin

Androcles wrote:
"Bryan Olson" wrote:
| Androcles wrote:
[...]
| 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.

I'm not talking about, nor interested in, your fabrications. I'm looking
at Einstein's theory of relativity, like all the good universities teach.

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 negative.
[Albert Einstein; Relativity: The Special and General Theory; 1920]


--
--Bryan

Bryan Olson wrote:
Androcles wrote:
"Bryan Olson" wrote:
| Androcles wrote:
[...]
| 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.

I'm not talking about, nor interested in, your fabrications. I'm looking
at Einstein's theory of relativity, like all the good universities teach.

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 negative.
[Albert Einstein; Relativity: The Special and General Theory; 1920]

What university teaches that you can 'show' anything
if you ionize two columms of air, suck them out so the
train can go fast, then move one of the observers to
an invalid position?



Sue...



--
--Bryan

"Sue..." wrote in
oups.com:


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.

But the number of heart beats will NOT be the same, even assuming that
they each spend the same amount of 'proper time' exercising.



--
bz

please pardon my infinite ignorance, the set-of-things-I-do-not-know is an
infinite set.

remove ch100-5 to avoid spam trap

"Bryan Olson" wrote in message
...
| Androcles wrote:
| "Bryan Olson" wrote:
| | Androcles wrote:
| [...]
| | 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.
|
| I'm not talking about, nor interested in, your fabrications.

You whined "I included a URL for exactly that ... Oh look--I did."
and I quoted from your URL, blowhard.
If you are not interested in your own garbage why call others
blowhard, blowhard?


| I'm looking
| at Einstein's theory of relativity, like all the good universities
teach.

Which you do not understand, blowhard, because you are a phuckwit.
I'm looking at it too.

| 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 negative.
| [Albert Einstein; Relativity: The Special and General Theory;
1920]

Well, he's wrong, blowhard, the answer must be in the positive, as
I've shown directly.
Snipping the truth won't make it any less true, blowhard.
You want to argue in favour of relativity, blowhard, do it with math,
snipping doesn't aid your case.


The light leaves the caboose and arrives at the engine, then reflects
back to the caboose.
If you want to **** a relativist off let the light leave the engine,
reflect at the caboose and return 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.

Androcles

bz wrote:
"Sue..." wrote in
oups.com:


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.

But the number of heart beats

You have a URL to the theory's chapter on biology ?

)

Sue...
will NOT be the same, even assuming that
they each spend the same amount of 'proper time' exercising.



--
bz

please pardon my infinite ignorance, the set-of-things-I-do-not-know is an
infinite set.

remove ch100-5 to avoid spam trap

Androcles wrote:
"Bryan Olson" wrote:
[...]
| I'm not talking about, nor interested in, your fabrications.

You whined "I included a URL for exactly that ... Oh look--I did."

There I go again, telling the truth.

and I quoted from your URL, blowhard.

Right up to where you didn't want to face what it says. At that
point you inserted your own "POSITIVE" when the truth of the
matter is that it says "negative".

If you are not interested in your own garbage why call others
blowhard, blowhard?

I'm interested in what correct, and not in your fantasy.

| I'm looking
| at Einstein's theory of relativity, like all the good universities
teach.

Which you do not understand, blowhard, because you are a phuckwit.
I'm looking at it too.

And changing parts you don't want to bother understanding.

| 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 negative.
| [Albert Einstein; Relativity: The Special and General Theory;
1920]

Well, he's wrong, blowhard, the answer must be in the positive, as
I've shown directly.

You've shown you don't understand, so you made up your own
theory.


Snipping the truth won't make it any less true, blowhard.

Sorry, I see no reason to be interested in your silly theory. I'm
just correcting the wrong claims about Einstein's theory.


--
--Bryan

"Sue..." wrote in
ups.com:


bz wrote:
"Sue..." wrote in
oups.com:

.....
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.

But the number of heart beats

You have a URL to the theory's chapter on biology ?

)

sure ;}

http://www.google.com/search?hl=en&q...+space+fl igh
t


Sue...
will NOT be the same, even assuming that
they each spend the same amount of 'proper time' exercising.



--
bz

please pardon my infinite ignorance, the set-of-things-I-do-not-know is
an infinite set.

remove ch100-5 to avoid spam trap







--
bz

please pardon my infinite ignorance, the set-of-things-I-do-not-know is an
infinite set.

remove ch100-5 to avoid spam trap

"Bryan Olson" wrote in message
t...
| Androcles wrote:
| "Bryan Olson" wrote:
| [...]
| | I'm not talking about, nor interested in, your fabrications.
|
| You whined "I included a URL for exactly that ... Oh look--I did."
|
| There I go again, telling the truth.
|
| and I quoted from your URL, blowhard.
|
| Right up to where you didn't want to face what it says.

I know what it says, and what it says is wrong.

| At that
| point you inserted your own "POSITIVE" when the truth of the
| matter is that it says "negative".


Of course I did, I corrected it.


|
| If you are not interested in your own garbage why call others
| blowhard, blowhard?
|
| I'm interested in what correct, and not in your fantasy.

It's your fantasy, phuckwit blowhard.
I'm using logic and math, you are ranting.


|
| | I'm looking
| | at Einstein's theory of relativity, like all the good universities
| teach.
|
| Which you do not understand, blowhard, because you are a phuckwit.
| I'm looking at it too.
|
| And changing parts you don't want to bother understanding.

Define "understanding", blowhard. I'm not indoctrinated, you are.
You want to rant or discuss math, arsehole?
If you can't discuss the math, **** off.


|
| | 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 negative.
| | [Albert Einstein; Relativity: The Special and General Theory;
| 1920]
|
| Well, he's wrong, blowhard, the answer must be in the positive, as
| I've shown directly.
|
| You've shown you don't understand, so you made up your own
| theory.

You want to rant or discuss math, arsehole?
If you can't discuss the math, **** off.


|
| Snipping the truth won't make it any less true, blowhard.
|
| Sorry, I see no reason to be interested in your silly theory. I'm
| just correcting the wrong claims about Einstein's theory.


Start correcting, ****.
The light leaves the caboose and arrives at the engine, then reflects
back to the caboose.
If you want to **** a relativist off let the light leave the engine,
reflect at the caboose and return 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.

Androcles.