Twins, events and transformations

For background information I refer to:
http://www.math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html

We use 3 inertial reference frames.
S: The frame of the "stay at home" twin.
S': The frame of the "outbound part of the trip".
S": The frame of the "inbound part of the trip".

In neither of these frames any form of acceleration is felt.
In order for the travelling twin to make his trip, she must be in frame S'
while going away and in frame S" when coming back. So at the turnaround
event, she must immediately jump from frame S' to frame S" (without getting
hurt!), and she must take over the time her clock is reading to her new
frame. Her clock will then continue ticking until she returns to her twin
brother who remained in his own frame S during the whole process.

Suppose:
- S uses coordinates (t,x,y,z)
- S' uses coordinates (t',x',y',z')
- S" uses coordinates (t",x",y",z")

\t"(worldline of inbound S")
\
\ |t(worldline of S) /t'(worldline of outbound S')
\ | /
\| R: Return event /
[R] (t,x) = (2T,0) /
|\ (t',x')=(2T/gamma,0)
| \ /
| \ /
| \ /
| \ /
| \ /A: Turnaround event
| \ / (t,x)=(T,vT)
| \ / (t',x')=(T/gamma,0) x'_
| [A]._ (t",x")=(T/gamma,0) _.-''
| / `--._ __.-'
| / ``-.__ _.-'
| / _:-.:_
| / _.-' `--._
| / _.-'' ``-.._ x"
| / __.-' `--.
| / _.-'
___[E].=________________________________________
| E: Start event x
| (t,x) = (0,0)
| (t',x') = (0,0)
|

1) The Lorentz transformation between S and S'
==============================================
Suppose S' recedes from S with velocity v along their x- and x'-axes,
and that they synchronize their clocks at event E with coordinates:
E: ( t, x, y, z ) = ( 0, 0, 0, 0 )
E: ( t', x', y', z' ) = ( 0, 0, 0, 0 )

Then the Lorentz Transformation between S and S' is given by:
t' = g(t-vx/c^2)
x' = g(x-vt)
y' = y
z' = z
where:
g = 1/sqrt(1-v^2/c^2) - This is gamma

2) The coordinates of the turnaround event
==========================================
Suppose that the outbound twin (now in frame S') has been travelling
during a time T (according to frame S) and decides to turn back to
her brother. Let's call this event A.
According to frame S this event has the coordinates:
A: ( t, x, y, z ) = ( T, vT, 0, 0 )
To find the coordinates according to the travelling twin who is
in frame S', we apply the Lorentz transformation between S and S'
and find:
A: ( t', x', y', z' ) = ( T/g, 0, 0, 0 )

We see that the travelling twin's clock shows T/g elapsed time
at this event.
She is now ready to (carefully!) jump on frame S" and return to her
brother.

3) The Lorentz transformation between S and S"
==============================================
Frame S" approaches S with velocity v along their x- and x"-axes,
and they synchronize their clocks at event A with coordinates:
A: ( t, x, y, z ) = ( T, vT, 0, 0 )
A: ( t", x", y", z" ) = ( T/gamma, 0, 0, 0 )

Here we have taken over the already elapsed clock time of the travelling
twin since we are interested in the time the twin's clock will show when
they reunite again.

Then the Lorentz Transformation between S and S" is given by:
[t"-T/g] = g( [t-T] + v[x-vT]/c^2 )
x" = g( [x-vT] + v[t-T] )
y" = y
z" = z
where:
g = 1/sqrt(1-v^2/c^2)

This is the same gamma factor as before since
v^ = (-v)^2

4) The coordinates of the return event
======================================
When the twin (now in frame S") has been approaching S during a time T
(again according to frame S), she will reunite with her brother.
Let's call this event R. According to frame S this event has coordinates:
R: ( t, x, y, z ) = ( 2T, 0, 0, 0 )
Applying the Lorentz transformation between frame S and S", we get:
R: ( t"-T/g, x", y", z" ) = ( T/g, 0, 0, 0 )
and thus:
R: ( t", x", y", z" ) = ( 2T/g, 0, 0, 0 )

5) Conclusion
=============
At this event R both clocks of frames S and S" are at the same place
denoted by (x,y,z) = (x",y",z") = (0,0,0), but
- the clock of S shows an elapsed time t = 2T
- the clock of S" shows an elapsed time t" = 2T/g

So if T = 5 years and v = 0.8c, then the stay at home twin will
have aged 10 years while his travelling twin sister will have aged
6 years.

Dirk Vdm

Nicholas Steele

"Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote
in message news:NheA9.17733$Nd....@afrodite.telenet-ops.be...

http://groups.google.com/groups?&as_umsgid=PGdA9.782606$v53.29377199@news3.c

Only in respect to the view of the frame S. You have not shown
reciprocity here. If the travelling twin imagines himself at rest
(which he will). He will see the mirror image of this process and
conclude that T=T'/g. (Something that has been stated many
times generally by using words)

Actually I went through this step by step. The process is quite correct
in every respect, but all you have done is show what the time-dilation is
from the point of view of the stay-at-home, which no one even
disagrees with. You should realize this as you transformed the
frames S' and S'' in relation to S. Of course, with this single viewpoint
there will be no contradiction. But it has nothing to say about
the twin paradox.

H.Ellis Ensle

He was presenting a clear picture of this situation at my request and it
was not meant to be anything more. His explanation was quite straightforward
and easy to follow so why do you say there is a paradox?

Nicholas

[snip]

That is right. I have not show that here.
If you like, I can show reciprocity here.
Let's look at both events E and A as seen by S and S':

We see that both event E takes place on observer S
and on observer S' (since in E x=0 and x'=0). O.t.o.h.
we see that event A takes place on observer S' (since
there x'=0), but *not* on observer S (since x=vT#0).
We also see that S' calculates *her* time of this event
by dividing her brother's time of this event by gamma.

So if we want to look at the reciprocal situation, we
can still use event E (because it happens on S and S'),
but now we must fix our attention to some other event
that takes place on S (x=0) but *not* on S' (x'#0).
Let's say we call this event F and that it has the
coordinates
F: ( t', x', y', z' ) = ( T, -vT, 0, 0 )
Note that I have taken -vT here to make sure that
the event takes place on observer S.
With the Lorentz transformation, we can now calculate
the coordinates of *this* event according to S. We get:
F: ( t, x, y, z ) = ( T/g, 0, 0, 0 )
So we see that in this case the observer S can calculate
his time T/g of event F again by dividing his sister's time
T by gamma.
Beautiful reciprocity.

In the reciprocal situation that I sketched above, nothing
relevant to the travel of the sister twin happened in the
event F, since the event took place on the stay at home
brother. The event F could have been him sneezing or
something.
On the other hand, the event A has some real significance
to the travel: A is the event taking place on observer S'!
It is the event of jumping to another spaceship - dangerous
business!

This situation is about a twin who goes away *and*
comes back. There is no way such a twin can imagine
herself at rest during the whole trip. Either she must
severely decelerate, turn around and accelerate again,
or she must make that dangerous jump from the outbound
spaceship onto the returning one. That is the situation
I am talking about. The "stay at home" twin does not feel
anything during the whole process: he can close his eyes
and wait: he does not have to jump to another spaceship
or something. But the "travelling" twin must do something
to return. There lies the asymmetry of the situation.

[snip]

I have not shown it from the point of view of the stay-at-home,
since I have given all the coordinates of the relevant events as
seen by both twins. The asymmetry that you see in the treatment
of the situation originates in the fact that he feels nothing, where
she must jump to another ship.

I did not transform frames. I gave mathematical relationships
between coordinates, used in different frames.
I have only written down how to calculate (t',x',y',z') and
(t",x",y",z") when (t,x,y,z) are known because that is all we
needed in this problem.
For instance, I have not written down the equations to calculate
(t,x,y,z) when (t',x',y',z') are known because we didn't have
an interesting event where we could use them. However now
with the event F, we *did* have such an event, so I will write
the transformation equations after all:
t = g(t'+vx'/c^2)
x = g(x'+vt')

-----------------------------
As a very interesting exercise, try to deduce the
above equations from these ones:

It takes simple algebra. Try it, it is fun.
-----------------------------

Well, the situation that I have described, is exacly what the twin
paradox is about.
I hope I have made it somewhat clear by looking at the event F
that is not taking place on the travelling twin. I have included this
event in the updated drawing below and called it "F: recip.event"

|\ (t",x")=(2T/g,0) <== had written (t',x') here

F: recip.event \ / (t',x')=(T/g,0) x'_
(t,x) =(T/g,0) [A]._ (t",x")=(T/g,0) _.-''
(t',x')=(T,-vT) / `--._ __.-'
[F] / ``-.__ _.-'

[Note that I have corrected a typo as well.]

Dirk Vdm

Did you actually understand his post? He simply did the problem in respect
to
the stay-at-home, so it does not address the paradox issue. Relativity
requires
that the travelling twin (at least before acceleration, (though techniquely
it really
shouldn't matter)) be able to view himself at rest. He then will see a time
dilation
of the stay-at-home. Van de Moortel simply didn't show it, so what I can't
understand is why you think your question was answered?

H.Ellis Ensle

In the second post I explained that I did not.

I explained that she can't since she must either
- jump to another spaceship (inertial frame), or
- decelerate, turn around and accelerate,
while the stay-at-home twin does not.
That is the very *essence* of the situation: there is no
symmetry.

Like I explained, she can see time dilation of the stay-at-home
but only on events taking place *on* the stay at home. The
crucial event however is the event of turnaround. And that
event takes place not on the stay-at-home, but on her.

I did show it very clearly in the second post.
Perhaps you didn't understand it or perhaps you found
an error in it? If so, feel free to comment.
But keep in mind that I am only explaining how the system
works.
[You know, you can learn something with this. When I see that
you are prepared to try to learn something, I'll be glad to remove
that fumble off my site. No problem.]

Dirk Vdm

Nicholas Steele

news:ar0htu$96m$1...@slb4.atl.mindspring.net...

Do you mean "event according to S' ", here?

H.Ellis Ensle

No, coordinates according to S, namely the numbers t,x,y,z
that are attributed by S to event F, to which S' attributes the
numbers t',x',y',z' with the given values:

or pulled apart:
t' = T
x' = -vT
y' = 0
z' = 0
You can check the result by putting the numbers in the
transformation.

Dirk Vdm

In the first post you admitted that you only showed the view
as calculated from the point of view the stay-at home, which
is fine, but does not address the paradox. In your second post
you _did_ try and address the issue. Nicholas posted here
before your second post.

(I will respond there on the argument itself)

[...........]

H.Ellis Ensle

This is an interesting comment, because if one reads the literature, he will
find two opinions on the subject. Some will actually state that GR is
required to resolve the problem. Some will state that GR is not required
and will go through _various_ methods to resolve it in SR.

This indecision alone should be of concern to those who value
logic and consistency in a physical theory.

All Van de Moorte did (in the first post) was to show the stay-at-home's
view of time for the scenario in a rather belabored fashion. Actually the
time difference can be shown from the stay-at-home's perspective for
any general motion of the travelling twin by using the formula:

Change in time = integral from t1 to t2 of sqrt(1-v^2(t)dt/c^2)

But the issue remains: Why can't the travelling twin apply this formula
to the stay-at-home? Often one gets the response that the acceleration
that is "felt" disallows this view, but if one actually thinks about it, he
will see that it doesn't make sense. Force is not included in the
Lorentz transformations, so while a person might feel it, the equations
do not. They simply map space and time. Plus if each atom of the
travelling twin were accelerated identically, he would feel nothing at all,
but the equations should still perform their mapping function (in some
differential form if need be).
.
Furthermore, even given that the acceleration does something
magical, the twin does not know his view is disallowed until
he accelerates, so up to that point, he still can use the above
formula in reverse, thus when the acceleration occurs, all this
false accumulation of time must be compensated. But there is no
method in acceleration in SR to do this, especially one that would
compensate for all possible trip lengths.

H.Ellis Ensle

Are these Ts supposed to be T' s? Because the way you have
it here, Event F happens for S at the same time as Event A.

H.Ellis Ensle

Almost correct: The way I have it here, is that

the same amount of time (namely T/g seconds) after event E as
Event A happens for S'.
or:
Event A happens for S
the same amount of time (namely T seconds) after event E as
Event F happens for S'.

I will put it more precisely, so look very carefully:

- Events E and A take place on S' (since their x'=0).
- Because A:(t,x) =(T,vT), the other observer
S says that A happens T seconds after E.
- Because A:(t',x') = (T/g,0), the observer S' (on
which event A happens) says that A happens
T/g seconds after E.

- Events E and F take place on S (since their x=0).
- Because F:(t',x') =(T,-vT), the other observer
S' says that F happens T seconds after E.
- Because F:(t,x) = (T/g,0), the observer S (on
which event F happens) says that F happens
T/g seconds after E.

This is full reciprocity in action.
Everything is according to the book (events, transformations
etc...) and we can go from the first block to the second and
vice versa by merely swapping the symbols:
S',t',x' <==> S,t,x
F <==> A
v <==> -v

So you see that I have chosen the event F in a special way.
I could have used another letter like U or T' or T" or whatever,
but I explicitly wanted to show reciprocity.

I notice that you are working on it. Good.
I will not respond yet to your other post. Lets' concentrate
on this subthread.
I will wait for your next question or comment.

Dirk Vdm

This is OK and I have seen it before. (I misread youir notation
and thought you might have been screwing it up even by SR.)

There is as of yet no problem. You have shown here the
reciprocity for the outgoing trip.

(back to main thread)

H.Ellis Ensle

Actually there is. What if each of her atoms is accelerated
equivalently. She would then have no indication that the
acceleration had occurred.

This is the old argument after all. And it just doesn't cut it.

You just agreed above that you had (in the first post). Here you have
shown reciprocity for the outgoing trip and that is all. You still have
not shown it for the entire trip. Of course, what will happen is the same
as always. One has to differentiate by acceleration to dissallow one
view. There is also the issue of accumulated time that is observed. I
suspect you will utilize the time-skip solution, which is certainly goofy
enough, but have at it.

[.....more of the same, which is good enough in a sense, but still
does not address the real issues.....]

BTW you should read my response to Steele, as it might explain
my position better.

PS. I went to the FAQ that you indicated as a source and I couldn't
even believe how lame it was. First he insults anyone who would
seriously complain about the paradox, but then when one actually
reads his solutions, they are a complete bust. What worm hole did
he pull out the Doppler garbage. All one has to do is flip the graphs
to put the travelling twin at rest and all the doppler effects are
reversed. The site does NOT explain the twin paradox. It is just
the usual (actually, a bit more than usual) hot air.

H.Ellis Ensle

[snip]

From the questions you asked earlier on this thread, I got
the impression that you were somehow prepared to look
at things with a slightly open mind.
When I read what you just wrote here, I now realize that
I was wrong.
Sorry for having wasted your time.

Dirk Vdm

Does this mean that you do not understand the Doppler explanation?
One would expect that to be easy to understand for anyone.
The travelling twin will see the change in Doppler shift from
red-shift to blue shift the instant he turns around.
So he will see the blue-shift and the red shift for an equal
long time. Isn't that blatantly obvious?
The home twin will however see the Doppler shift turn from
a red shift to a blue shift at the time he _see_ the travelling
twin turn around, and that will of course be after the time
it takes for the travelling twin to reach the turning point
PLUS the time it takes for the light to reach the home twin.
Thus the home twin will see the red shift for a longer time
than he will see the blue shift.

There is no way this scenario can be "flipped" to give the opposite
result, it is inherently unsymmetrical.

A more detailed description:

A is the home twin, B the travelling twin.
They both have emitters flashing with the frequency once per second.

Observations made by A:
When B goes out, A will receive the pulses with the Doppler shifted
frequency f1 = sqrt((c-v)/(c+v))
He will observe this frequency for the time it takes B to reach
the turning point at some distance L, _plus_ the time it takes for
the light to reach him from the turning point: ta1 = L/v + L/c
Thereafter, A will observe the frequency f2 = sqrt((c+v)/(c-v))
for the time ta2 = L/v - L/c (ta1 + ta2 = 2*L/v, obviously)
A will receive the total number of pulses N = f1*ta1 + f2*ta2
N = (2*L/v)*sqrt(1 - v^2/c^2)
A's total time Ta = 2*L/v
Thus N = Ta*sqrt(1 - v^2/c^2)
Since B's clock has emitted one pulse per second, then it must show:
Tb = Ta*sqrt(1 - v^2/c^2)

Observations made by B:
We assume that B uses the time Tb/2 on his way out, and Tb/2 back.
Since B is the one who does the abrupt turn-around, he will see
the frequency change immediately as he does so.
Thus the number of pulses he will count from A is simply:
N = 0.5*Tb*sqrt((c-v)/(c+v)) + 0.5*Tb*sqrt((c+v)/(c-v)) = Tb/sqrt(1-v^2/c^2)
Since clock A has emitted one pulse per second,
Ta = Tb/sqrt(1-v^2/c^2) when B returns.

Why is this garbage?

Paul

Very neatly put :-)

And again :-)

Dirk Vdm

There is a difference between and open mind and
accepting something uncritically.

H.Ellis Ensle

But the stay-at-home will also see a red-shift to a blue shift.

But isn't it blatantly obvious that the travelling twin can imagine he
is at rest and the shifts will demonstrate exactly the opposite?

Even stranger, the Doppler shift idea seems to imply that the
time is going faster on the return trip. But that doesn't
even agree with the Lorentz transformations which has that little
v^2 in the formula.

I know what you are trying to say here. BUT it cannot be inherently
"unsymmetrical" unless you use the same old argument that one view is
rejected (by acceleration or whatever). When I observe someone
receding from me, they observe me receding from them. That
scenarion is completely and logically symmetric. It does not take
a genius to see it. It is blatently obvious that the scenario is
_inherently_ symmetric. In this sense, the Doppler argument is
also irrelevant, since it is already dependent on the asymmetry of
the choice. The Doppler effect does not create the asymmetry so
it does _not_ resolve the twin paradox....exactly as I stated

[...........]

H.Ellis Ensle

Ensle: I see you guys are playing Monopoly.
Monopoly is a stupid game.
[produces fart]
Those who play it are stupid.
- "Do you know the game?"
Ensle: Yes.
- "What is stupid about it?"
Ensle: It is wrong.
- "What is wrong?"
Ensle: The rules.
- "Do you know the rules?"
Ensle: Yes, I'll tell you about the rules: there are stupid.
- "No, wait, there are not stupid if you really understand them."
Ensle: Sure they are.
- "Wait, I will try to carefully explain the rules.
Let's start with rule 1: throw 2 dice..."
Ensle: Rule 1 is wrong.
[produces fart and looks other way]
- "What can be wrong about throwing 2 dice?"
Ensle: Suppose you throw them away, then you can't play.
So it's a stupid game. And you are stupid too.
- "But you shouldn't throw the dice away.
You should throw them on the table and look at the numbers."
Ensle: That is ridiculous.
- "What is ridiculous about it?"
Ensle: The dice can fall on the floor and the dog can eat them.
- "Then imagine you don't have a dog."
Ensle: I cannot imagine that, since I do have a dog.
- "Then imagine that your wife has taken the dog out."
Ensle: I cannot imagine that, since my wife has left me.
- "Then imagine the dog to be sleeping."
Ensle: That is stupid. He'll wake up when the dice hit him.
Then he will eat them. I told you it's a stupid game.
- "Imagine the dog is sleeping in another room."
Ensle: The dice can roll all the way to the other room.
- "Imagine the room is upstairs."
Ensle: Hmmm, ah yes, okay, I can imagine that. But keep it short.
- "So, where were we... rule 1 was: throw the dice..."
Ensle: That is ridiculous.
- "Why is that ridiculous?"
Ensle: Suppose you throw the dice and they jump up and hit your eyes,
then you can go blind and you have to stop playing.
You see, I told you Monopoly is a stupid game.
[produces fart and looks other way]
Those who play it, are stupid.

http://users.pandora.be/vdmoortel/dirk/Stuff/Monopoly.html

Dirk Vdm

Captures the fruitcake perfectly. Well done!

--
Stephen
s...@speicher.com

Ignorance is just a placeholder for knowledge.

Printed using 100% recycled electrons.
-----------------------------------------------------------

Right. In the case where the relative velocity is .8c and the
travelling twin turns around after 5 years (as measured in the
frame of the stay-at-home twin), then (assuming that each
twin sends out a signal once per second).

1. The stay-at-home twin sees the light from the travelling
twin red-shifted for 9 years, and sees it blue-shifted for
the last 2 years.

2. The travelling twin sees the light from the stay-at-home
twin red-shifted for 3 years and red-shifted for 3 years.

Whether it is "obvious" to you or not, it is false. The twin who
accelerates immediately sees a change from red-shift to blue shift.
The other twin only sees a change after a delay. The situations
are not symmetrical.

Yes, an accelerated coordinate system is not the same as
an unaccelerated coordinate system. That's true for both
Special Relativity and Galilean Relativity.

Yes, but if you suddenly accelerate, then that is observably
different than if the other person suddenly accerates. You
can tell who is doing the acceleration.

Perhaps you should make sure that you understand Galilean
Relativity before you start trying to analyze Special Relativity.
Galilean relativity makes a distinction between unaccelerated and
accelerated observers, as well.

--
Daryl McCullough
Ithaca, NY

xxein: A and B (stay vs. travel). You are quite right in asking
these questions in the context or SR as you understand it. The
problem is that SR (and it's proponents) doesn't understand it
either.

If we block out one femto-second of instant acceleration from
everyone's memory, what is different about the relativity between A
and B? They receed from each other and see each other's clocks as
slower than their own. They approach each other and see each other's
clocks as faster than their own. Symmetrical, yes?

What is it with that femto-second that makes it asymmetrical? If A
and B both blacked out during the femto-second that each would have
observed a frequency shift, under the prescription of SR, each could
expect the other to be the traveling twin (having a slower clock).

In that f-second is an absolute change in velocity that SR cannot
admit has any part other than a difference in relative
velocities---except when it needs to make added interpretation to get
the correct result for the scenario.

Ex: A trackside observer (O), parallel trains, one with the speed 50
(F) and the other (S) at 25. The trains both travel the same length
of time (according to O) to the left, turn around (f-second) and to
the right, turn around and always pass at O at the same time. F will
always have the slower clock wrt O and S. That f-second didn't mean
diddly-squat!

With that blacked out f-second, both F and S see each cycle the same
as S and F. Symmetrical. Either can be either twin. That damned
f-second means something more than a point of view from an inertial
frame. It can be given a relative value depending on an inertial
perspective, but it is more than just that.

According to O it was nothing. But if O's frame were moving wrt P's,
then it is not nothing and the assymetry creeps in for O. What this
essentially means is that there is no symmetry in any frame in a flat
space, but that is SR's claim to fame (symmetry). The key to
understanding symmetry is that SR's symmetry is relative and therein
lies the makings of its unintuitive and paradoxical nature. That
f-second contained an absolute that was immune to relative symmetry.

Velocity is absolute in nature regardless of observational theory.
That makes the speed of light an absolute regardless of its
measurement in a frame. TWLS (every object is a lightclock) is always
measured as c in all inertial frames, but OWLS is NEVER measured
without clocks set as if c was c in that frame (circularity).

Do yourself a favor and just consider an SR space (a really flat space
with no gravity). Make that space homogeneous with light being c
throughout. Temporarily ignore time dilation and set limits on what
you will observe of a sent 1 hertz frequency from other moving
objects. Now give them time dilation for their velocity.

Next, imagine yourself moving wrt a stationary object emitting 1 hz.
First without your time dilation and then with it.

If you make limit functions for each of the four velocity scenarios
above from -c to +c, you will find that you can describe, by way of
observed frequency, all of SR without missing the key ingredients that
SR misses.

The keen part is to graph all of this. Start without time dilation
and you get two sets of curves. But with time dilation you get one
curve consisting of both. That is SR's claim to fame. But now you
know that it is lacking all the absolutes that you have included in
your graph. They are nowhere in the SR formulation. SR missed the
detail in gross relative observation.

(x-axis, velocity from -c to +c. y-axis, limits as to what frequency
you will see by simple doppler without time dilation for any
participant. Then change y to include time dilation.)

This is no surprise to those knowledgable of the subject. A much
lesser number of those will realize that each factor is itself
absolute, and while it can be generalized as SR, acute distinctions
remain that remove all hints of paradox.

Do not even think of gravity until you have completely mastered what
you get out of SR AND the underlying LET version of SR-types. And
please don't confuse length contraction of matter with the
non-existant length contraction of space: you put YOUR contracted
ruler out into space to measure distance.

The context is in reality, not formulations based upon subjective
observation (SR), shortcuts or unintuitive explanations. It took me a
few years to realize this even after voting Einstein out. So long?
Well there is life to live regardless of such worthy endeavors.

You've got to look back at the history of science and say "I don't
want to be as shortsighted as those guys". The doctrines of science
are rebuilt often enough for a healthy doubt their veracity. You're
on a healthy track. Don't screw it up with preconcieved belief.

But most of all, keep it as a fun challenge

[.........story about a monopoly game........]

You really do have a serious problem. You went to
all the work to create a completely irrelevant strawman
argument. In fact, it isn't even analogous.

The truth is, you simply never responded to certain
arguments. The arguments I gave are real and
have purpose (as always). I am not claiming to be
perfect, but these things I post are certainly relevant.
If you are not competent to address them, just
say so. I will not ridicule you because of it.

H.Ellis Ensle

Duh.....obviously, but there is a difference between SR and Galilean (see
below).

This is an interesting point and it may point to a huge misconception
on your part. The galilean transformations are, in fact, always
identically reciprocal regardless of any acceleration in regards
to the mapping of space and time. There _are_ pseudo-forces
that can be generated in an accelerated frame, but the reciprocal
mapping of space and time remains perfectly consistent.

In fact, this may be the root of the problem for SR. With a
constant velocity the Lorentz Transformations do maintain
reciprocity (I have vacilated on this statement in the past,
and I am still not absolutely certain.), but given that, when
acceleration occurs, the reciprocity of the transformations
are broken, not in the sense of simply needing a differential
form (as would work in Galilean transformations), but as
losing a consistent reciprocal mapping of the space. After
all, when the twins reunite, they force what would be mapped
as two different events by mutual Lorentz transformations
into one event...and thus the conflict arises. This would not
occur with galilean transformations mapped from the
traveling and _accelerating_ twin's frame. The travelling
twin who thought he was at rest might feel forces of
mysterious origin, but he would not perceive any conflict
in position or time. He might see other objects move in odd
non-linear ways as well, but their displacement and time
at any instant would still be perfectly reciprocal.

H.Ellis Ensle

It is exactly analogous.

You don't have arguments.
All you have ignorance and arrogance, combined into
a severe attitude problem.

If you think they are relevant, then you could ask questions.
You could call for help.

I cannot possibly be ridiculed by you.
Do not get me wrong: perhaps you *are* not an
arrogant idiot. Perhaps you only *act* like one.
You can do something about it.
But I'm pretty sure that you won't.

Dirk Vdm

Let's go through the same scenario using Galilean relativity. Instead
of light travelling through vacuum, we'll use sound travelling through
the air.

Suppose we have one twin who is sitting at home. We have another twin
who travels away at speed v for T seconds and then travels back at speed
v. At all times, each twin is blowing a horn with
a frequency F. Then let's make the following definitions:

* F1 = Doppler-shifted low frequency of stay-at-home horn as
heard by the travelling twin on his outward trip.
* T1 = time (as measured by the travelling twin)
during which he hears this frequency.
* N1 = total number of cycles heard of this frequency (F1 * T1)
* F2 = Doppler-shifted higher frequency of stay-at-home horn as
heard by the travelling twin during his return trip.
* T2 = time during which he hears this frequency.
* N2 = number of cycles (F2 * T2)
* N_tt = total number of cycles heard by travelling twin (N1 + N2)
* F3 = Doppler shifted low frequency of travelling horn
as heard by the stay-at-home twin, during outward trip.
* T3 = time (as measured by stay-at-home twin) during
which he hears this frequency.
* N3 = cycles heard (F3 * T3)
* F4 = Doppler shifted higher frequency of travelling horn
as heard by stay-at-home twin, during return trip.
* T4 = time (as measured by stay-at-home twin) during
which he hears this frequency.
* N4 = cycles heard (F4 * T4)
* N_st = total number of cycles heard by stay-at-home twin (N3 + N4)

Using Galilean relativity and the Doppler formulas for sound,
we have (where v_c = speed of sound)

F1 = F * (1 - v/v_c)
T1 = T
N1 = F T * (1 - v/v_c)
F2 = F * (1 + v/v_c)
T2 = T
N2 = F T * (1 + v/v_c)
N_tt = 2 F T

F3 = F/(1 + v/v_c)
T3 = T (1 + v/v_c)
N3 = F T
F4 = F/(1 - v/v_c)
T4 = T (1 - v/v_c)
N4 = F T
N_st = 2 F T

Note the differences: (1) The Doppler shift formula is not symmetric
between the two twins; a different formula is used for the twin
travelling relative to the medium and the twin at rest in the
medium. (2) The travelling twin hears the change in frequency
immediately when he turns around, while the stay-at-home twin
doesn't hear the frequency change until later.

In Galilean relativity, these two differences cancel out. The
net effect is that both twins hear the same number of cycles in
total.

If instead, we used light signals and Special Relativity, the
formulas work out like this:

F1 = F * square-root((1 - v/c)/(1 + v/c))
T1 = T * square-root(1 - (v/c)^2)
N1 = F T * (1 - v/c)
F2 = F * square-root((1 + v/c)/(1 - v/c))
T2 = T * square-root(1 - (v/c)^2)
N2 = F T * (1 + v/v_c)
N_tt = 2 F T

F3 = F * square-root((1 - v/c)/(1 + v/c))
T3 = T (1 + v/c)
N3 = F T square-root(1 - v^2/c^2)
F4 = F * square-root((1 + v/c)/(1 - v/c))
T4 = T (1 - v/c)
N4 = F T square-root(1 - v^2/c^2)
N_st = 2 F T square-root(1 - v^2/c^2)

Notice: the relativistic Doppler shift formula (unlike
the Galilean formula) is symmetric between the two twins.
But like the Galilean case, the stay-at-home twin sees
a high Doppler shift for a short time and a low Doppler
shift for a longer time. Like the Galilean case, the
travelling twin sees a high Doppler shift for half the
time and a low Doppler shift for half the time.

There _are_ pseudo-forces

If you think so, you are obviously incompetent.

H.Ellis Ensle

I repeat the conclusion of my previous post:
Perhaps you *are* not an arrogant idiot.

http://users.pandora.be/vdmoortel/dirk/Stuff/Monopoly.html

The important points of the argument ignored as usual.

[........a couple of nice derivations........]

In the SR case, you simply did everything from the point
of view of the stay at home, so obviously it will correlate
with the galilean example where the stay-at-home is at
rest in the medium. I would expect nothing else.

But what does this even have to do with the twin paradox?

This is probably what irritates me the most about the
method. It simply does not address the issue. SRists
_pretend_ that it does, but the choice of preference has
already been made. So instead of illuminating anything,
this is just more smoke and mirrors, to try and prevent
people from questioning the faith.

You can have your religion. I just want to know what is
really happening in the universe.

H.Ellis Ensle

You don't give the impression that you want to know
what is happening. You give the impression that you
are an arrogant idiot.
Perhaps it is not on purpose.
Perhaps you can't help it.
Perhaps you mean well.
Perhaps it is your parent's fault.
All I can tell, is that you give the impression of being
an arrogant idiot.
I don't think that anyone (who knows the subject) on
this forum will contradict that. So this is - de facto -
a property intrinsic to you.

Dirk Vdm

That accounts for his stupidity and obnoxious behavior, but don't
forget his profound ignorance, which accounts for how little he
knows about the universe he pretends he is desirous of
understanding.

Ensle is completely hopeless, and has absolutely no desire to
become educated. Many people tried years ago, but gave up when it
became clear to them, exactly what you identify now. The judgment
of history lies in the archives.

I told you for *each* twin what he measures. I told
you how things look from the standpoint of the travelling
twin, and I told you how things look from the point
of view of the stay-at-home twin.

It is the twin paradox. One twin travels away, and then
travels back. The other twin stays put. It is just one
way of keeping track of the relative ages. Instead of
using the Lorentz transformations, you can instead assume
that each twin sends a light signal once per second (according
to his own clock). Each twin keeps track of the number of
signals received from the other twin.

Using the relativistic Doppler formula, you can show that
the travelling twin receives more signals from the stay-at-home
twin than vice versa. So the stay-at-home twin's clock has
ticked more often for the whole trip than the travelling twin.

SR gives you equations that allow you to predict the results
of experiments. What religion does that for you? What more
do you want to know about the universe?

Show me the equations for Christianity. What predictions does
it make for the energy levels of hydrogen, or for the lifetimes
of muons in an accelerator, or for the relationship among
mass, energy, and momentum for a particle?

What about Buddhism? What about Hinduism?

Your calling relativity a "religion" shows that you know nothing
about science or religion.

I wouldn't have believed it if I didn't see it
happen in front of me. It really is like that
"Monopoly" scenario.

Here's a free clue which you won't use: When somebody
describes what the travelling twin sees, it is a
description of the point of view of the travelling twin.

- Randy

I had planned to tell something about his remark about
points of view. So, anticipating a remark about my having
drawn the axes of stay-at-home as perpendicular, I already
had prepared
http://users.pandora.be/vdmoortel/dirk/Stuff/twins.gif ,
to show other "viewpoints".
But when I noticed his remark about the Doppler explanation,
I decided not to bother.

Dirk Vdm

Yes, but why did only one of them see the delay as to when
the red shift changed to a blue shift? There cannot be
assymmetry in the scenario unless it is introduced at
some point. A symmetrical Doppler will obviously not
introduce the assymmetry, therefore you had to do it
at some point _independent_ of any Doppler argument.
Are you aware that you did so? And do you know how
you did it?

You also never addressed another point. According to
the Lorentz transformations, the returning twin must see
time dilation in the stay-at-home, but with the Doppler
method he sees time going faster for the stay-at-home.
Thus the Doppler method even contradicts the Lorentz
transformations.

Don't you see it?

Only if you bias it at some point in the process. The Doppler
is perfectly symmetrical, so it cannot do the job.

H.Ellis Ensle

Actually no. SR cannot give anything reliable since it is self
contradicting. What has happened is simply that the gamma
factor works in respect to the absolute frame of reference
(that would be the frame where the electro-magnetic tensor
_actually_ has no magnetic component). So all the experiments
work, but not because of SR. SR simply takes the credit for
something it could never do.

A physical theory must have a _consistent_ fundamental
physical model. That rule has been violated not only in SR but also
in QM.

Modern physicists have neglected to find it, and with their
cleverness have convinced themselves that they did find it.
It is all a self-deception. They have somehow lost a
natural filter for rejecting _bad_ ideas. But since they
are clever they can devise arguments (albeit specious)
to support even bad ideas. If problems crop up, they
merely increase the complexity to the point where the
problem is sufficiently confused so they can continue to
accept the originally _bad_ idea.

No. I have called relativity a religion because it is believed by a
large number of people without the theory having proper
scientific backing. For many, it has become an article of faith,
since any argument against it is immediately rejected, not
because the argument is wrong (though it may be in some
cases) but because relativity is _believed_ to be true.
This _belief_ may have come about by the person's previous
experience and training, but that does not guarantee that it
is true.

H.Ellis Ensle

Yes he did, but why did only one of them see the delay as

introduce the assymmetry, therefore he had to do it

Are you aware that he did so? And do you know how
he did it?

You see, this isn't me playing a game of monopoly,
it is you not understanding the argument.

H.Ellis Ensle

No, what really happened is that you used my aside as an excuse
to avoid replying to the real argument. You used my somewhat
charged statement about the FAQ (which did not relate to the
specific argument) to turn the discussion from the topic to the
poster (that would be me).

I see above, however, that you did have something relevant
(hidden away). So I went and looked, they are not too clear
as I am not sure of what your justification is for the different
lengths of the travelling twin's space-time lines (this could just
be due to the lack of a scale). Actually, these may be OK,
for half of the travelling twin's view. The problem is that you
need to show reciprocity for the twin both going and returning.
You must do so, because he is moving at a constant velocity
on return and a previous acceleration cannot dissallow him
from applying the Lorentz transformations to the stay-at-home's
coordinates getting yet another reciprocal view. To show this
I would imagine that you would need some hybrid combination
of your upper right graph and your lower left graph. Of course,
you would end up with a big mess (but what else can one
expect from relativity?).

H.Ellis Ensle

[..................]

Strange....I get that same impression of you.
Who is correct? Are you correct because a thousand
other people would agree with you?

You see, it is not so simple. One of the keys to science
is that it addresses (or tries to address) the issue of how
do we know that we know. How do we determine that
we are right about something? Historically, this is a huge
issue and not one that can be easily solved. Science
went a long way with the problem by using experiment
to verify our concepts, but that does not save us from
misinterpreting them or a large number of other pitfalls.

So...are you really correct? Or do you just think you are
correct.

H.Ellis Ensle

Think about it, Harold. Imagine you have a highway sparsely
filled with cars, each travelling 100 kilometers per hour.
The space between the cars is 1/10 kilometer. Suppose that
you are driving 90 kilometers per hour. Then how often
will one of these cars pass you? Answer: a car will pass
you every 36 seconds.

Now suppose you suddenly turn around, and do a U-turn. Now
you are heading *into* the traffic. How often will cars
pass you *now*? Answer: a car will pass you every 1.9 seconds.

If you turn around, you will *immediately* see a change
in the rate at which cars pass you. In the same way, when
the travelling twin turns around, he *immediately* sees
a change in the frequency of the light signals he is getting
from the stay-at-home twin.

Why doesn't the stay-at-home twin immediately see a
change? Because light takes time to travel. The light
that the stay-at-home twin is seeing now was sent a
long time ago, *before* the travelling twin turned
around.

In summary, there are two reasons that the light
you are receiving from a distant source might change
frequency: (1) Because that distant source changed
its velocity, or (2) because you have changed *your*
velocity. In case of reason number (1), there will
be a delay. In case of reason number (2), there will
*not* be a delay.

Yes, when the travelling twin turns around, his situation
becomes different from that of the stay-at-home twin, who
never changes his velocity.

The Lorentz transformations cannot be applied to a noninertial
coordinate system, which is what the travelling twin has.

If you like, you can describe things from the inertial
coordinate system in which the travelling twin is initially
at rest. From the point of view of this coordinate systems,
things look like this:

1. Initially, the travelling twin is at rest. The
stay-at-home twin is travelling at velocity .8c in the
-x direction. The rate of the travelling twin's clock
is 1. The rate of the stay-at-home twin's clock is
.6.

2. After 3 years, the travelling twin changes velocity,
to have velocity .9756 in the -x direction. At this time,
the stay-at-home clock shows time 1.8 years, while the
travelling twin shows time 3 years.

3. After 13.6667 more years, the travelling twin catches
up to the stay-at-home twin. During this time, the rate
of the travelling twin's clock is .2196. So the travelling
twin's clock advances by 3 years (13.6667 * .2196). So the
total time on the travelling twin's clock is 6 years (3 + 3).
The stay-at-home twin's clock has rate .6, so it advances
by 8.2 years. So the total time shown on the stay-at-home
twin's clock is 10 years (1.8 + 8.2).

So, if you use the outgoing coordinate system of the travelling
twin, you get the same answer: that the stay-at-home clock advances
by 10 years altogether while the travelling twin's clock advances
by only 6 years.

You can also use the incoming coordinate system of the travelling
twin. You'll get the same answer.

So, using the Lorentz transformations in any inertial coordinate
system gives you the same answer as the Doppler shift answer---the
stay-at-home twin's clock advances by more.

May I insert a word of appreciation for your good services. This NG is
so fouled up with cranks, nincompoops and ignoramouses that input of
some one who knows what he is talking about is most welcome.

Just as in the -The Lone Ranger- , who WAS that Masked Man? I wanted to
thank him.

Thank you.

Bob Kolker

Both these theories are mathematically consistent (internally
consisten). Furthermore they predict experimental outcomes correctly. So
what is your objection?

Bob Kolker

Nearly one hundred years of experimental support. What is your problem?
The -only- thing that counts in science is experimental verification
and experimental falsification.

Bob Kolker

Once again, I'd like to review the facts: There are people
(call them "physicists") who use what they believe to be
relativity to calculate predictions for the outcomes of a
huge assortment of experiments. The predictions they calculate
agree with experiment to an extraordinary accuracy.

On the other, there are people who claim that SR is contradictory,
and can't be used to predict anything.

What's a more likely explanation: (A) The physicists don't understand
SR, and are making stupid, incompetent calculations, but by accident,
their results turn out to be perfectly consistent and in perfect
agreement with all experiments. (B) The people who claim that SR
is contradictory don't understand SR, and are making stupid, incompetent
calculations, but by accident, their results turn out to be inconsistent.

What's more likely: that misunderstandings can produce correct predictions,
or that misunderstanding can produce inconsistencies?

Once again, what's more likely: (A) The physicists who use
relativity are stupidly following nonsensical rules out of
blind faith, but somehow they end up with perfectly consistent
results that are in agreement with all known experiments, or
(B) The physicists who use relativity actually understand it.

Your hypothesis (A) looks pretty dog-gone unlikely to me.

No, it doesn't give you equations?

No, people don't calculate the results of experiments
from those equations?

So you say. Yet people take these equations and do these
calculations. What are you saying exactly about the people
who make such calculations and then compare them with
experiment?

- Randy

Just think it through and don't declare the prejudicial
view that "it can't be so".

When the traveling twin turns around, the light that
he sees Doppler shifted is light that is already in
his vicinity. There's no delay.

The earth twin does not see an instantaneous
change in Doppler, because the affected portion of
the light is far away. At the instant (in his frame)
that his twin is turning, he's still seeing light
emitted from the outbound journey. It will be years
before he sees light from the inbound journey.

Think about each twin and the light they are moving
relative to. When the traveling twin turns around,
is the relationship between the earth twin and the
incoming light changed?

There really is an asymmetry.

See the above and rather than tell me it MUST be
symmetric, tell me where my analysis is wrong.

- Randy

Read my lips asshole: I decided not to bother.

Dirk Vdm

Humm - are you saying that SR is wrong becaue it can only predict the
observable values but fails at predicting the unobservable values?

It was also violated by Newton's theory of Gravity. Newton gave no
physical mechanism why every particle should attract every other
particle. Also, the instantaneous action at a distance aspect of
Newton's gravity is odder than anything in SR (or so I think).
Of course QM transcends odd - it's downright weird. But so are
the experimental results. Anyway, QM does a great job of describing
the chemical bond. Let me know when you have a QM replacement that
works at predicting chemical structure or even the bond energy of H2.

That is a great description of my reaction to josX's ASCII diagrams.

> H.Ellis Ensle

Bruce Seiler

news:arlsdk$558$1...@slb1.atl.mindspring.net...

> H.Ellis Ensle

Harold,

Why do you, and so many others, fail to realize that if SR was so full of
"_bad_ ideas" someone would have cropped up in the past 100 years with
_proof_ of this or, better yet, a _viable_ alternative that did not contain
such ideas? Do you not think that the motivation for instantaneous
international recognition (let alone almost certainly a Nobel prize) would
cause some of the brightest in the world to stray from the established
theory to offer a better one? The catch is that someone in the know would
have to have the suspicion that SR is indeed full of "_bad_ ideas".
Unfortunately, the only people who seem to generate these suspicions are
those such as yourself, and needless to say, I doubt I'll be reading about a
Nobel prize being handed your way anytime soon, at least in this universe.

Regards,

Jeff

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Note: It is NOT a question of what one twin _imagine_ the other
twin will see, it is a question of what the travelling twin actually
does see.

And my question was:
Isn't it blatantly obvious that the travelling twin ** will see
the blue shift and the red shift for an equal long time**?
----------------------
So what might "the opposite" be?
That the travelling twin, when he turns around can _imagine_ that
he doesn't see the Doppler shift change from blue to red immediately
as he turns around?

So I will restate my question:
Isn't it blatantly obvious to you that the travelling twin
actually will observe that the Doppler shift change from
a red shift to a blue shift immediately when he turns around?

Is there any way he can _imagine_ to be at rest so that he
will not actually observe this immediate frequency shift?

> Even stranger, the Doppler shift idea seems to imply that the
> time is going faster on the return trip. But that doesn't
> even agree with the Lorentz transformations which has that little
> v^2 in the formula.

Does it? Why?
You know that the light from an approaching object is blue
shifted, don't you?
Why do you think this imply that the clock on the approaching
object is running faster?

> > The home twin will however see the Doppler shift turn from
> > a red shift to a blue shift at the time he _see_ the travelling
> > twin turn around, and that will of course be after the time
> > it takes for the travelling twin to reach the turning point
> > PLUS the time it takes for the light to reach the home twin.
> > Thus the home twin will see the red shift for a longer time
> > than he will see the blue shift.
> >
> > There is no way this scenario can be "flipped" to give the opposite
> > result, it is inherently unsymmetrical.
>
> I know what you are trying to say here. BUT it cannot be inherently
> "unsymmetrical" unless you use the same old argument that one view is
> rejected (by acceleration or whatever).

We are not rejecting any view. Quite the contrary.
Both views are correct, both views agree;
their clocks do not advance equally much.

> When I observe someone
> receding from me, they observe me receding from them. That
> scenarion is completely and logically symmetric. It does not take
> a genius to see it. It is blatently obvious that the scenario is
> _inherently_ symmetric.

Certainly.

It doesn't take a genius to see that the scenario:
"Two objects are receding" is inherently symmetrical.

And it doesn't take a genius to see that the scenario:
"A is inertial all the time while the receding B abruptly
changes his state of motion to approach B"
is inherently asymmetrical.

It doesn't take a genius to know that it is the latter
scenario we are discussing, but it takes a will to
evade the issue to pretend it is the former.

> In this sense, the Doppler argument is
> also irrelevant, since it is already dependent on the asymmetry of
> the choice. The Doppler effect does not create the asymmetry so
> it does _not_ resolve the twin paradox....exactly as I stated

It seems to me like you really don't understand much of
the "Doppler shift scenario".
You seem to have completely missed its basic idea.
It is:
We do NOT ask what each twin _imagine_.
We ask what each twin actually observes!
Each twin continuously visually observe the other twin's clock,
and count every second the other clock ticks. Since the twins
are co-located at the start and at the end, the other clock
MUST have advanced the number of seconds each twin has counted.
This idea shouldn't be hard to grasp.
So why don't you?

So the question is:
How many ticks does each twin observe the other clock to tick
compared to the number of ticks his own clock have ticked?

That was what the part you snipped answered.
I will repost it.

I challenge you to point out the error which must be there
if you are right.

A is the home twin, B the travelling twin.
They both have emitters flashing with the frequency once per second.

Observations made by A:
When B goes out, A will receive the pulses with the Doppler shifted
frequency f1 = sqrt((c-v)/(c+v))
He will observe this frequency for the time it takes B to reach
the turning point at some distance L, _plus_ the time it takes for
the light to reach him from the turning point: ta1 = L/v + L/c
Thereafter, A will observe the frequency f2 = sqrt((c+v)/(c-v))
for the time ta2 = L/v - L/c (ta1 + ta2 = 2*L/v, obviously)
A will receive the total number of pulses N = f1*ta1 + f2*ta2
N = (2*L/v)*sqrt(1 - v^2/c^2)
A's total time Ta = 2*L/v
Thus N = Ta*sqrt(1 - v^2/c^2)
Since B's clock has emitted one pulse per second, then it must show:
Tb = Ta*sqrt(1 - v^2/c^2)

Observations made by B:
We assume that B uses the time Tb/2 on his way out, and Tb/2 back.
Since B is the one who does the abrupt turn-around, he will see
the frequency change immediately as he does so.
Thus the number of pulses he will count from A is simply:
N = 0.5*Tb*sqrt((c-v)/(c+v)) + 0.5*Tb*sqrt((c+v)/(c-v)) = Tb/sqrt(1-v^2/c^2)
Since clock A has emitted one pulse per second,
Ta = Tb/sqrt(1-v^2/c^2) when B returns.

Why is this garbage?
What is wrong?

Paul

>In the first post you admitted that you only showed the view
>as calculated from the point of view the stay-at home, which
>is fine, but does not address the paradox. In your second post
>you _did_ try and address the issue. Nicholas posted here
>before your second post.

The symmetry is obviously broken by the acceleration. Which twin is
accelerated may be determined by asking which twin produces radiation
during the interval between meetings. Accelerated charges radiate.
The origin of the radiation is agreed upon by all observers. If neither
twin produces any radiation (electromagnetic or otherwise), neither
can accelerate. Try to give a counterexample. Note that the thrust of
an ordinary rocket produces electromagnetic radiation by vitue of the
energy dissipated in the chemical reaction. Furthermore, if each twin
is self-contained, then the accelerated twin will have lost mass in
providing the acceleration to make the journey. The inertial twin
need expend no energy whatsoever to remain in an inertial trajectory.

I understand perfectly what you are stating here.
But here is the problem. The assymmetry here is
still based on establishing that one of them accelerated
(exactly as I said).

Imagine that I am the travelling twin. I see that the
stay-at-home accelerated toward me. Now when
I map the Doppler effects I see him as instantly
noticing the change.

Now you would immediately say that the travelling
twin is wrong. Why? because we _know_ that he
was the one who accelerated. Do we? And if we do,
is there something in the Lorentz transformations that
indicate the forces involved that would indicate that
there was an assymmetric acceleration which then
excludes the twin from using the Lorentz
transformations uniformily through the whole trip?

Yes, of course, we must know that he accelerated, but to
properly exclude the Lorentz transformations from use by the
traveling twin through this point, the transformations must
somehow incorporate this knowledge (they currently do not).

Yes we have been through here before (sans Doppler). The problem
here is, in the second part of the trip, this is not what the twin sees
for himself at that time. It is how he would have seen himself when
he was originally at another velocity. But now he is not at that velocity
so this does not show his real self view.

Yes we have been through here before (sans Doppler). The problem
here is, in the first part of the trip, this is not what the twin sees
for himself at that time. It is how he would see himself when
he is at another velocity _in the future_. But now he is not at that
velocity
so this does not show his real self view.

H.Ellis Ensle

[.............]

When I say something must be true, it is not by whim...
or prejudice. It is because there is a reason.
Please see my response to McCullough in answering
this point.

H.Ellis Ensle

[snip]

A beauty :-))
Title: "I see him as instantly noticing the change"
http://users.pandora.be/vdmoortel/dirk/Physics/ImmortalFumbles.html#Instantly
[snip remainder - max one fumble/thread]

Dirk Vdm

Since nobody claimed this, it is a strawman argument. As I have
explained in the past, SR works experimentally by providing an
appropriate gamma. The point of contention is in the reality of
mutual time-dilation. And this has never been experimentally
verified.

Or maybe those who think that SR is _not_ contradictory do not
understand all that SR requires.

Strawman argument noted.

So you are saying that you are making the determination
indirectly by applying a statistics based on popular beliefs.

Slick!

H.Ellis Ensle

I object to mutual time-dilation, acually I object to all intrinsic
spacial/temporal transformations.

I object to particle wave duality.

I object to intrinsic uncertainty.

H.Ellis Ensle

You forgot Monopoly!
http://users.pandora.be/vdmoortel/dirk/Stuff/Monopoly.html

Dirk Vdm

The issue here is mutual time dilation, and there is no experimental
evidence of this SR prediction.

H.Ellis Ensle

SR equations give two contradictory results. People simply
choose the one that works.

H.Ellis Ensle

Already been done. The Twin _contradiction_ has been around
since the beginning. It is just excused away using an incredibly
lame argument.

It turns out that this is contingent on your previous statement.
Since so many are satisfied with SR, there are very few
looking for alternatives. However, I must confess that I
have the working alternative and have already published
it. The book is called "Stationary Field Theory" and can
be obtained through Amazon.com

Only if they believed that a better one existed....and who does?

An expose on your character that you could not remain civil through
an entire post.

Probably not.............( I deserve 8 , but who's counting?)

H.Ellis Ensle