twin-paradoxon
Leif Kuschel
I have a little problem with the the twinparadoxon. Maybe someone can give
me a solution.
The known problem is...
We have 2 systems A and B
One person stay at the system A and the other person remain at B.
If the system B leave the system A with a very high speed and comes back to
system A,
the person at A will be older than the person at system B.
Now my question...
thinking in relative systems it's the same, if B leaves A or A leaves B and
it's also
the same if B comes back to A or A comes back to B. If I think about the
"reverse problem"
the person at A leaves B with a very high speed and comes back. So the
person
in the system B will be older than the person in the system A. !!!
this is a contradiction.
Where is my mistake ?
thanks in advance,
Leif
Jon Bell
Perhaps the Usenet Relativity FAQ can help you.
<http://www2.corepower.com:8080/~relfaq/twin_paradox.html>
--
Jon Bell <jtb...@presby.edu> Presbyterian College
Dept. of Physics and Computer Science Clinton, South Carolina USA
Spaceman
"Leif Kuschel" <Leif.K...@stak.de> wrote in message
news:9kbhsd$3nioo$2...@ID-21059.news.dfncis.de...
> this is a contradiction.
yes,
that is why it is a paradox
It does not work both ways like it should.
The true time is the same for both A and B no matter thier speed or motion
or placement in
the Universe.
You don't have it wrong.
you have it right.
you found the problem with the problem.
There is no time dilation and it's all in incorrect thoughts and clocks
only.
Arky
One undegoes acceleration (they can feel it), the other doesn't.
--
Arky, truth seeker
http://www.jarkazz.co.uk/musings/
Leif Kuschel
>
> <http://www2.corepower.com:8080/~relfaq/twin_paradox.html>
>
thanks for the URL
I read something of this, but I think it explains only the one position...
Terence is at home on Earth and Stella flies off in a spaceship.
but if I think in a reverse problem... Stella is " at home " on the
spaceship and
Terence flies off " with the Earth ". Ok, it's not practical, but the
problem is a theroretical one.
In this case, Stella will be older than Terence.
Leif
Leif Kuschel
> One undegoes acceleration (they can feel it), the other doesn't.
>
> --
> Arky, truth seeker
Hi Arky,
sounds good, but now the case..
system A and system B are spaceships, and both have a very high speed.
Now system B stops. Is this an acceleration (B can feel it) ? After that B
will be so fast,
that he can fly back to A.
Is now A older than B?
Perhaps, it's only a question of acceleration, what happens if one twin (B)
lives on a swing and
swings the whole time. The other twin (A) sit on a chair and looks at B. If
it's a question of
acceleration, the time goes on faster for A than for B :-)
Leif
Matthew Lybanon
it will answer your question. Bohm devotes almost an entire chapter to
this problem, I seem to recall.
Jon Bell
Leif Kuschel <Leif.K...@stak.de> wrote:
>Jon Bell wrote:
>> Perhaps the Usenet Relativity FAQ can help you.
>>
>> <http://www2.corepower.com:8080/~relfaq/twin_paradox.html>
>>
>
>thanks for the URL
>
>I read something of this, but I think it explains only the one position...
>Terence is at home on Earth and Stella flies off in a spaceship.
>
>but if I think in a reverse problem... Stella is " at home " on the
>spaceship and Terence flies off " with the Earth ". Ok, it's not
>practical, but the problem is a theroretical one.
If you could strap a _huge_ rocket engine to the Earth, and use it to
send the Earth and Terence off on a round trip while Stella and her
spaceship remain "stationary" (coast along in free fall) without using her
rocket engines at all, then Terence _would_ be the younger one when he
returns.
But that's physically different from the situation in which the Earth
remains "stationary" and Stella fires her rocket engines to go on a round
trip, and watches the Earth recede behind her and then "return". The
difference is in who uses his/her rocket engines.
Ken H. Seto
<Leif.K...@stak.de> wrote:
The twin paradox is derived from the SR concept of mutal time
dilation. It is a bogus concept. The proper interpretation is that one
clock will run fast and the other clock runs slow. Why? The rate of
the clock is dependent on the state of absolute motion of the clock.
SR rejects the idea of an absolute frame so it came up with the mutual
time dilation concept. To make their math come out OK they invented
yet another bogus concept called relativity of simutlaneity (RS). RS
explains why the twin's clocks have different elapsed time when they
rejoin. I have a new theory of motion called Doppler Relativity
Theory (DRT). DRT can resolve whose clock is running fast (or slow) as
follows:
A and B are in relative motion:
Tab=Taa/DFa
Tab = clock reading of clock B as predicted by A
Taa = clock reading of clock A as read by A
DFa = Doppler Factor as determined by A
DFa= Faa/Fab
Faa= frequency of an emiiter in A's frame as measured by A
Fab =Frequency of an identical emitter in B's frame as measured by A.
If Fab is not constant then measure Fab at regular time intervals.
Plot the curve and integrate to obtain a mean value for Fab for use in
the equation.
Notice that DFa can be <or> than one. That means that clock B can be
running faster or slower than clock A.
For more information on DRT please visit my website and click on to
the section entitled Doppler Relativity Theory.
<http://www.erinet.com/kenseto/book.html>
Alan McIntire
Say A and B are flying apart at 12/13 the speed of light. As long as
they are receeding from each other, they will see each other as moving
at ((1 - 12/13)/(1+ 12/13))^(1/2) = 1/5 normal speed. When they
approach each other, they will
see each other as moving at 5 times normal speed.
If they are separated by 10 light years when A changes direction, A
will see B approaching immediately, and see B aging at 5 times speed
immediately. B, who hasn't done anything to change relative motion,
will see A continue to recede at 1/5 speed, and won't see A approach
at 5 times normal speed until 10 years have passed, the 10 years it
takes light from A to reach B. Whoevher changes direction, A or B,
will IMMEDIATELY see the other change direction, and start aging more
rapidly than normal. The other party will not see the change until
the light reaches him or her.
- A. McInire
and...@attglobal.net
If one of them accelerates and the other doesn't, then
the world lines are not equivalent modulo a Lorentz
transformation, so there's no reason for SR to have
to predict that the elapsed clock times are the same.
You calculate the elapsed time on a clock between
two events on its world line by integrating
ds = sqrt((dt)^2 - (dx/c)^2)
along the world line between the events.
This will be different for the inertial and non-inertial
twins.
That doesn't mean that acceleration causes this, but that's
another story.
John Anderson
and...@attglobal.net
>
> "Leif Kuschel" <Leif.K...@stak.de> wrote in message
> news:9kbhsd$3nioo$2...@ID-21059.news.dfncis.de...
> > this is a contradiction.
>
> yes,
> that is why it is a paradox
>
No, that's not correct. See my reply to Leif Kuschel.
John Anderson
Vertner Vergon
<and...@attglobal.net> wrote in message news:3B6A17...@attglobal.net...
> Leif Kuschel wrote:
> >
> > Hi physics,
> >
> > I have a little problem with the the twinparadoxon. Maybe someone can
give
> > me a solution.
> >
> > The known problem is...
> >
> > We have 2 systems A and B
> > One person stay at the system A and the other person remain at B.
> > If the system B leave the system A with a very high speed and comes back
to
> > system A,
> > the person at A will be older than the person at system B.
> >
> > Now my question...
> >
> > thinking in relative systems it's the same, if B leaves A or A leaves B
and
> > it's also
> > the same if B comes back to A or A comes back to B. If I think about the
> > "reverse problem"
> > the person at A leaves B with a very high speed and comes back. So the
> > person
> > in the system B will be older than the person in the system A. !!!
> >
> > this is a contradiction.
> >
> > Where is my mistake >
> two events on its world line by integrating
>
> ds = sqrt((dt)^2 - (dx/c)^2)
>
> along the world line between the events.
>
> This will be different for the inertial and non-inertial
> twins.
>
> That doesn't mean that acceleration causes this, but that's
> another story.
> John Anderson
***************************************************
VERGON:
You say
If one of them accelerates and the other doesn't, then
> the world lines are not equivalent modulo a Lorentz
> transformation, so there's no reason for SR to have
> to predict that the elapsed clock times are the same.
And then you say
> That doesn't mean that acceleration causes this, but that's
> another story.
So the reason you give for the cure is not the reason the cure is effected
!!!!!
Somebody's way out in wooopsie land.
Stand back and while I give this poor fellow the real answer.
Time rate and Doppler are the same thing.
An atomic clock never varies in frequency or time rate.
So when you observe a shift in Doppler you are observing a shift in
time rate.
That means time rate observation is vector altered.
Now just for fun run an example where the velocity is sqrt(.75)c.
There the Lorentz factor is .5 -- so the total elapsed time for
the Twin journey is (say) one year for the voyager and two years for
the stay at home.
The outward time rate is .268 and the inward is the reciprocal 3.732.
Now assume the length of the trip to be sqrt(.75) Light Second (LS).
So A will observe B for 1.86603 seconds at .268 = .5 seconds of B time
(It takes A one second to the turn around point -- and then .86603 second
for the signal of that event to return to A. All that represents A's view of
B.)
Also, A is on his way back while the signal of his arrival at turn around is
on its way to A.
B arrives two seconds (A time) after he left. So A will observe his approach
for 2 - 1.86603 = .134 second. And the time rate observed on B is 3.732.
.134 x 1.86603 = .5.
So the trip has been 2 seconds for A and 1 second for B.
THAT"S the so called Einstein time rate. It is actually the total
transit time differential -- and in this case is .5 (B = .5 A)
Now if you use the same technique to calculate B's observations
during the trip YOU WILL GET THE SAME RESULT.
(That's your home work assignment.)
But remember, because of length contraction the distance traversed
from B's point of view is only 1/2 .86603 LS.
This is where the significance of acceleration comes in. It is to the
accelerated observer that the distance traversed is shortened and
his observations appropriately altered.
So that's the story, kiddies. The true time rates are Doppler but the
net transit times are as Al said.
Regards
Vert
Harold Ensle
You have made no mistake.
Anyone who believes that SR is true is either stupid or a coward.
Because he cannot understand your simple argument or if
he does, he is too afraid to speak out lest his peers should
ridicule him.
H.Ellis Ensle
and...@attglobal.net
Well, he ignored that inertial and non-inertial world lines
aren't Lorentz equivalent, so that no reciprocity
argument needs to follow. And he ignored relativity of
simultaneity. Say, you do that too, don't you?
It's easy not to believe SR when you don't understand it.
> Anyone who believes that SR is true is either stupid or a coward.
> Because he cannot understand your simple argument or if
> he does, he is too afraid to speak out lest his peers should
> ridicule him.
>
I understand why the simple argument is untrue about SR, so I'm
not stupid. You're pretending that SR explains this scenario
without using relativity of simultaneity. And that's wrong.
So you're either ignorant or dishonest.
And I don't see how stating what I believe is dishonest.
John Anderson
Vertner Vergon
_______________________________________________________________
EVERYBODY LOVES A SWAP MEET
Profitable, Fun & Informal. Buy - Sell - Swap - Trade - Barter *** ANYTHING.
And
<and...@attglobal.net> wrote in message news:3B7201...@attglobal.net...
VERGON:
Of course he has. He's made two.
One, he failed to recognize that time rate is equal to Doppler rate
and is vector responsive.
And the other is that the distance traveled is Lorentz responsive,
and is therefore less for the accelerated observer than for the
inertial observer.
Run these two factors through the round trip and you will find
that the inertial observer has experienced more time just as
Einstein's time rate calls for. And you will find that both observers
come up with the identical time disparities.
****************************************************
> Well, he ignored that inertial and non-inertial world lines
> aren't Lorentz equivalent, so that no reciprocity
> argument needs to follow. And he ignored relativity of
> simultaneity. Say, you do that too, don't you?
>
> It's easy not to believe SR when you don't understand it.
>
> > Anyone who believes that SR is true is either stupid or a coward.
> > Because he cannot understand your simple argument or if
> > he does, he is too afraid to speak out lest his peers should
> > ridicule him.
> >
>
> I understand why the simple argument is untrue about SR, so I'm
> not stupid. You're pretending that SR explains this scenario
> without using relativity of simultaneity. And that's wrong.
> So you're either ignorant or dishonest.
>
> And I don't see how stating what I believe is dishonest.
**************************************************
VERGON:
It's not dishonest ------- just inadequate.
You keep running around telling everyone they don't understand
relativity -- as though you do, when if fact you are a relativity toady,
clinging to everything you learned in school and clinging slavishly
to what you had difficulty learning.
The fact is, a new thought would kill you.
**************************************************
*************************************************
> John Anderson
Harold Ensle
How can one tell which one is non-inertial? What if the
travelling twin has each atom in his body accelerated
at an equivalent rate, in which case, he wouldn't feel
a thing. Besides we are talking kinematics and force
is not a dependency of the Lorentz transformations.
In fact, when doing a time dilation with a rocket problem,
one merely considers the instantaneous velocity over the
whole trip and sums up everything. If you used that method
with the travelling twin, you would still have a contradiction.
>And he ignored relativity of
> simultaneity. Say, you do that too, don't you?
No I don't, besides, on reunion, simultaneity is not an issue.
That is exactly why the paradox is such a problem. In a one
way measurement, the SRists can always use simultaneity as
an out (as ridiculous as it is), but with the twins, there is no out.
> It's easy not to believe SR when you don't understand it.
You can *only* believe in it if you don't understand it.
> > Anyone who believes that SR is true is either stupid or a coward.
> > Because he cannot understand your simple argument or if
> > he does, he is too afraid to speak out lest his peers should
> > ridicule him.
> >
>
> I understand why the simple argument is untrue about SR, so I'm
> not stupid. You're pretending that SR explains this scenario
> without using relativity of simultaneity. And that's wrong.
> So you're either ignorant or dishonest.
>
> And I don't see how stating what I believe is dishonest.
Where did I write that you were dishonest???? Hmmmm
perhaps your subconscious is telling you something.
H.Ellis Ensle
Harold Ensle
Vertner Vergon <VVe...@prodigy.net> wrote in message
news:Uumc7.5263$Bx2.58...@newssvr17.news.prodigy.com...
What the #$%$#@ is Doppler "rate"??
> and is vector responsive.
What the #$%@# is this? Some psychological term related to Cupid's arrow??
> And the other is that the distance traveled is Lorentz responsive,
What the #$%#@ do you mean here......length contraction?
> and is therefore less for the accelerated observer than for the
> inertial observer.
>
> Run these two factors through the round trip and you will find
> that the inertial observer has experienced more time just as
> Einstein's time rate calls for. And you will find that both observers
> come up with the identical time disparities.
In your dreams.
H.Ellis Ensle
and...@attglobal.net
All terminology that has nothing to do with physics.
> And the other is that the distance traveled is Lorentz responsive,
> and is therefore less for the accelerated observer than for the
> inertial observer.
>
> Run these two factors through the round trip and you will find
> that the inertial observer has experienced more time just as
> Einstein's time rate calls for. And you will find that both observers
> come up with the identical time disparities.
You're welcome to your own pet theory, but unless you can post
a better explanation of it's predictions than this, you might
as well be pulling rabbits out of hats.
John Anderson
and...@attglobal.net
The usual scenario for the twin paradox has one twin remaining
inertial and the other moving inertially for a while, turning
around (ie. accelerating) and coming back inertially.
> What if the
> travelling twin has each atom in his body accelerated
> at an equivalent rate, in which case, he wouldn't feel
> a thing.
That doesn't have anything to do with which one actual accelerates.
Anyway, acceleration really isn't the cause of the effect.
The cause of time dilation is that the elapsed proper times
along the two world lines aren't the same. SR predicts the
elpased proper times to be the elapsed clock times and proper
time is Lorentz invariant, so all observers agree on which
clock has which elpased time.
> Besides we are talking kinematics and force
> is not a dependency of the Lorentz transformations.
> In fact, when doing a time dilation with a rocket problem,
> one merely considers the instantaneous velocity over the
> whole trip and sums up everything. If you used that method
> with the travelling twin, you would still have a contradiction.
>
No, you don't. You don't consider local clock rates since they
involve relativity of simultaneity. They involve comparing
clocks at different events. They are well understood to give
mutual time dilation and that is explained by comparing
elapsed clock times over different parts of the clocks'
world lines.
> >And he ignored relativity of
> > simultaneity. Say, you do that too, don't you?
>
> No I don't, besides, on reunion, simultaneity is not an issue.
> That is exactly why the paradox is such a problem. In a one
> way measurement, the SRists can always use simultaneity as
> an out (as ridiculous as it is), but with the twins, there is no out.
>
It is when you you try to explain which clock is running slower
over the inertial parts of the world lines.
> > It's easy not to believe SR when you don't understand it.
>
> You can *only* believe in it if you don't understand it.
>
Since you have repeatedly failed to post an SR analysis, as
opposed to word arguments, that shows what you're claiming,
I would suggest that you're full of shit.
> > > Anyone who believes that SR is true is either stupid or a coward.
> > > Because he cannot understand your simple argument or if
> > > he does, he is too afraid to speak out lest his peers should
> > > ridicule him.
> > >
> >
> > I understand why the simple argument is untrue about SR, so I'm
> > not stupid. You're pretending that SR explains this scenario
> > without using relativity of simultaneity. And that's wrong.
> > So you're either ignorant or dishonest.
> >
> > And I don't see how stating what I believe is dishonest.
>
> Where did I write that you were dishonest???? Hmmmm
> perhaps your subconscious is telling you something.
>
I have posted replies to you over several years in which I asked
you to post the results of simple SR calculations that demonstrate
how SR explains these effects. I even showed you how to
do the calculations. I didn't get any response.
If there is a paradox in SR, you ought to be able to demonstrate
it using SR reasoning and showing contradictory results. Word
arguments based on parodies of SR don't cut it. I have
shown you how SR resolves this and you have never worked through
what I posted and shown what's wrong with it. You just
crawl back under your rock and re-emerge later with the
same claim.
You never responded to my challenge, you just avoided it.
People can judge that for themselves.
John Anderson
Harold Ensle
<and...@attglobal.net> wrote in message news:3B7C9F...@attglobal.net...
You have??????????????????????
Tell you what. You show me right here and now how to do the
twin paradox problem. I want the time as seen by each twin, and
I want it derived directly from the Lorentz transformations.
If you do, it will be the *first* time.
So.....put up or shut up.
H.Ellis Ensle
Alan McIntire
*****
Say A and B are flying apart at 12/13 the speed of light. As long as
they are receeding from each other, they will see each other as moving
at ((1 - 12/13)/(1+ 12/13))^(1/2) = 1/5 normal speed. When they
approach each other, they will
12/13)^(1/2).
If they are separated by 10 light years when A changes direction, A
will see B approaching immediately, and see B aging at 5 times speed
immediately. B, who hasn't done anything to change relative motion,
will see A continue to recede at 1/5 speed, and won't see A approach
at 5 times normal speed until 10 years have passed, the 10 years it
takes light from A to reach B. Whoevher changes direction, A or B,
will IMMEDIATELY see the other change direction, and start aging more
rapidly than normal. The other party will not see the change until
rocket or feel acceleration, but if they're ever going to meet again,
one of them must do so.
With special relativity, two observers traveling apart will see each
other appear to age at a rate slower than normal, each will see the
other age at the same slower rate. Two observers approaching will see
each other appear to age at a rate faster than normal, the same fast
rate. The DIFFERENCE is, if they do nothing, they will continue to
travel on different paths, never meeting again, so no one could tell
who is "REALLY" aging faster. One of the two will have to fire his
rocket and change direction. This is the observer who will age less.
Incidentally, if you presume light travels at c+ or -v , where v is
the velocity of the source either + towards, or - away, you will
discover that it is possible for some observers to see effects before
causes. This is never possible with special relativity. Also, with
the above assumption about additive velocities, it will sometimes
possible to see multiple images of objects, a "before" and "after"
image at the same time. This doesn't happen with special relativity.
Finally using the above addition of velocity assumption, it's
impossible to
construct a consistent travel diagram, with twins traveling apart and
togeter, without first knowing their "absolute" motion through space.
Under the additive velocity assumption, the twins will see each other
age at different rates while they see each other receding, and/or they
will see each other age at different rates while they see each other
approaching. Try working this out for yourself.
- A. McInire
Harold Ensle
Alan McIntire <mcin...@earthlink.net> wrote in message
news:48658b64.01081...@posting.google.com...
You call this an answer????
I want to see the Lorentz transformations!
I don't want to see any numbers (they only confuse me).
I want to see the problem solved with some generality.
The equations are sufficient, if correct.
H.Ellis Ensle
and...@attglobal.net
Huunnhhhh???
I have asked you several times to back up your erroneous
assertions about what SR predicts by posting an analysis
that involves Lorentz transformations. You never did
this. In fact, you claimed that you didn't need to demonstrate
your understanding of SR by actually posting an analysis
that involves equations instead of words.
You're dipstick challenged. It's still coming out all brown.
You're full of shit.
John Anderson
Harold Ensle
<and...@attglobal.net> wrote in message news:3B80B3...@attglobal.net...
> Harold Ensle wrote:
> > You call this an answer????
> > I want to see the Lorentz transformations!
> > I don't want to see any numbers (they only confuse me).
> > I want to see the problem solved with some generality.
> > The equations are sufficient, if correct.
> >
>
> Huunnhhhh???
>
> I have asked you several times to back up your erroneous
> assertions about what SR predicts by posting an analysis
> that involves Lorentz transformations. You never did
> this. In fact, you claimed that you didn't need to demonstrate
> your understanding of SR by actually posting an analysis
> that involves equations instead of words.
Hey all you have to do is show me how to do the problem
correctly, but you always avoid it.
Maybe you don't know.
> You're dipstick challenged. It's still coming out all brown.
> You're full of shit.
What sophisticated language you use.
H.Ellis Ensle
Alan McIntire
> <and...@attglobal.net> wrote in message news:3B80B3...@attglobal.net...
> > Harold Ensle wrote:
> > > You call this an answer????
> > > I want to see the Lorentz transformations!
> > > I don't want to see any numbers (they only confuse me).
> > > I want to see the problem solved with some generality.
> > > The equations are sufficient, if correct.
> > >
>
What each observer SEES: T' = ((c+v)/(c-v))^(1/2) when approaching
each other,
T' = ((c-v)/(c+v))^(1/2) when receeding
from each other.
What each observer computes, allowing for the finite speed of light:
x' = ((x-vt)/(sqrt (1- (v^2/c^2)),
t' = ( t -(vx/(c^2)))/(sqrt(1-(v^2)/c^2))
The twin who changes directions sees the other receeding and
approaching for equal amounts of personal time, or {D/v) receeding and
(D/v) approaching.
The twin who remains in an inertial frame sees the other receeding
for
(D/v + D) of the time, and sees the other approaching for (D/v -D) of
the
time, where D is the distance in light units or time in the same
units, i.e.
light years and years, light minutes and minutes, etc.
Remember, the twin who changes direction sees the other approaching
IMMEDIATELY, the twin who remains in an inertial frame doesn't see the
chage until time D has passed, the time it takes the light signal to
get to earth.-A.McIntire
Alan McIntire
> "Harold Ensle" <hen...@ix.netcom.com> wrote in message news:<9mkkqg$f73$1...@slb1.atl.mindspring.net>...
> > <and...@attglobal.net> wrote in message news:3B80B3...@attglobal.net...
> > > Harold Ensle wrote:
> > > > You call this an answer????
> > > > I want to see the Lorentz transformations!
> > > > I don't want to see any numbers (they only confuse me).
> > > > I want to see the problem solved with some generality.
> > > > The equations are sufficient, if correct.
> > > >
> >
> > H.Ellis Ensle
Pardon me, I left out a term- A. McIntire
> What each observer SEES: T' = T*((c+v)/(c-v))^(1/2) when approaching
> each other,
> T' = T*((c-v)/(c+v))^(1/2) when receeding