## The Twin paradox falsifies SR


First of all, let's stress that absolute motion can be determined
against the CMBR, via the observed dipole:

"If you're looking for an "absolute rest frame", [the CMBR] is an
excellent candidate."
(Eric Prebys, Physics Department, Princeton University)

In the simplest form of the paradox, Terence stays on Earth,
whereas Stella leaves the Earth at .99 c during one of her years
(her outward trip), makes an instantaneous turnaround and goes
back to Terence at the same velocity of .99 c (her return trip).

The LET analysis

Using LET without the ROS (relativity of simultaneity) to analyse
the Twin paradox, it can be demonstrated (see below "A LET analysis of the
Twin paradox") that the outward and return trips of Stella are a function
not only of her velocity v wrt the Earth, but also of the "absolute" Earth
velocity V, measured via an observed dipole in the CMBR.

The obtained LET relation is t(one-way) = t(round-trip) * (1+Vv)/2,
where t(round-trip) = 2t / sqrt(1-v^2), t being the duration of Stella's
one-way trips measured on her watch, i.e. 1 year, whereas
t(one-way) and t(round-trip)  are times observed on Earth by Terence.

Hence, in LET, the duration of the round-trip is independent of the
Earth's velocity wrt the CMBR. This is also the case with SR.

But generally, V<>0, and the duration of Stella's one-way trips is
a function, not only of her velocity v wrt the Earth, but also of the
"absolute" velocity V of the Earth.

Let's analyse 3 different cases. In all cases,  Stella's velocity v
wrt the Earth is - .99 c during her outward trip and + .99 c during
her return trip.

1. The general case:

The Earth moves at V wrt the CMBR, for instance V=.5 c
The duration of the whole journey of Stella is, according to
Terence, 2/sqrt(1-v^2) = 14.1776 years.
The duration of Stella's outward trip is 14.1776 * (1 + .5*(-.99)) /2 =
3.5798 years, and her return trip will take 14.1773 *
(1 + .5*(+.99)) / 2 = 10.5978 years.
And indeed, 3.5798 + 10.5978 = 14.1776.

2. The Earth is at rest wrt the CMBR (V=0)

Let's remember that Stella moves at v = -.99 c wrt the Earth during
her outward trip and at v = +.99 c during har return trip.

The formula giving the one-way times reduces to t(one-way) =
t(round-trip) / 2, or 14.1776 years / 2, hence

Stella's outward trip:                  7.0888 years
Stella's return trip:                      7.0888 years
Duration of the whole journey: 14.1776 years

3. The Earth moves at V=.99 c wrt the CMBR.

As Stella is moving at v = -.99c wrt the Earth during her
outward trip, she is in fact at rest wrt the CMBR.

The duration of Stella's outward trip is (14.1776 / 2) *
(1+ (.99 * - .99)) =.1411 year, and the duration of her return trip
is (14.1776 / 2) *(1+ (.99 * + .99)) = 14.0365 years:

Stella's outward trip:                     .1411 year
Stella's return trip:                     14.0365 years
Duration of the whole journey: 14.1776 years

The SR analysis

(See below "The Twin paradox SR analysis")

According to SR, all inertial frames are equivalent, hence the times
measured on Earth by Terence for Stella's outward and return
trips shoud be the same, whether the reference frame is the
Earth or the frame in which Stella is at rest during her outward trip

This is not the case. Indeed, the durations of Stella's trips (read
on Terence's clock), vary according to the chosen reference frame.

Results of the SR analysis:

1.The reference frame is the Earth frame ( = Terence's frame)

Stella's outward trip:                   7.0888 years
Stella's return trip:                       7.0888 years
Duration of the whole journey: 14.1776 years

The SR results are identical to those of the case 2 of the LET
analysis, where the Earth is at rest wrt the CMBR.

But SR would obtain identical results if the Earth had a
velocity V wrt the CMBR! It couldn't be otherwise, as SR considers
only relative motion. Or LET shows that the duration of
Stella's one-way trips is also a function of V (see case 1 above).

This means that SR's claim according to which only relative
motions have to be taken into consideration is wrong. Indeed,
using the Earth frame, SR analysis is correct only if the Earth
is at rest wrt the CMBR.

2. The reference frame is the one in which Stella is at rest during
her outward trip (called hereafter the outbound frame):

Stella's outward trip:                     .1411 year
Stella's return trip:                     14.0365 years.
Duration of the whole journey: 14.1776 years

The SR results are the same as those of the case 3 of the LET
analysis.

Clearly, the duration of the one-way trips is frame-dependent,

This contradiction is entirely due to the fact that SR considers
only relative motion, hence ignores the physical consequences
of a change of reference frame.

Indeed, the choice of the outbound frame physically implies
that Stella is still moving at - .99 c wrt the the Earth, but the Earth
is moving at V = .99 c wrt the CMBR.
Iow, it implies that Stella is at rest wrt the CMBR! (see above).

The physical situation is thus wholly different between the two frames.
The use of the Earth frame implies an Earth at "absolute" rest,
whereas the the choice of the outbound frame  implies an Earth
moving at V = .99 c wrt the CMBR.

exists only with SR. SR is clearly conceptually false.

The SRists should realize that their so-called change of frame is not
physically neutral.

They should also realize that the ROS doesn't apply in concrete
situations. If it did, the LET analysis would not have obtained
the same results as SR.
Iow, also the validity of the ROS is falsified by the Twin paradox.
This is due to the fact that it is not physically possible to measure
a change of position of the Earth wrt the CMBR. A change of
position can only be measured wrt a material object.

In conclusion, LET is physically correct and SR is a subset of SR, but
SR is also conceptually wrong.

Marcel Luttgens

September 11, 2001

A LET analysis of the Twin paradox

Hereafter, it is assumed that an object at rest wrt the CMBR
is "absolutely" at rest. In other words, a time interval t(V) shown by
a clock moving at a velocity V wrt the CMBR is related to the
time interval t(CMBR) shown by a clock at rest in the CMBR by the
formula t(CMBR) = t(V) / sqrt(1-V^2).

It is also assumed that the turnaround of Stella is instantaneous
(The relativity FAQ description of the Twin paradox is given below).

In the Twin paradox scenario, Terence stays on Earth, whereas
Stella leaves the Earth at a velocity v (measured wrt the Earth)
during one year of her time. This part of her trip will be called
Stella's outward trip. Then Stella makes an instantaneous
U-turn and goes back to Terence at the same velocity v. This
part of her trip will be called Stella's return trip.

Let's suppose that the Earth is moving at V wrt the CMBR.

Analysis of Stella's outward trip

Let's remember that the Earth moves at V wrt the CMBR, and
Stella moves at v wrt the Earth during t years (1 year in the

Applying the LET (SR) addition of velocities, Stella's velocity
v' wrt the CMBR is given by the formula v' = (v+V) / (1+Vv),
hence the time interval t1(CMBR) read on a "CMBR clock"
= t / sqrt(1- v'^2) or t * (1+Vv) / sqrt [(1-V^2) * (1-v^2)].
(The details of the calculation are left to the reader.)
The corresponding time interval t1(Earth) measured on Earth, or by
Terence, as Terence is staying on Earth, is given by
t1(Earth) = t1(CMBR) * sqrt(1-V^2) = t * (1+Vv) / sqrt(1-v^2)

For instance, if V=0 (Earth at rest wrt the CMBR), v = - .99 c
and t (duration of Stella outward trip) = 1 year,
t1(Earth) = t1(Terence) = 1/sqrt(1 - .99^2) = 7.0888 years.
In SR lingo, those 7.0888 years are called Terence's proper
time.

If V=.99 c, v= - .99c and t=1 year, t(Earth) = .1411 years, which
is the time read on Terence's clock if .99 c is the Earth's velocity wrt
the CMBR, or, again in SR lingo, if the "outbound" frame (corresponding
to Stella's outbound trip) is taken as the reference frame.

Analysis of Stella's return trip

The formula giving the time interval read on a "CMBR clock" remains
t2(CMBR) = t * (1+Vv) / sqrt [(1-V^2) * (1-v^2), and t2(Earth) =
t * (1+Vv) / sqrt(1-v^2). Let's remember that t(Earth) is also the
time measured by Terence on his clock.

If V=0, v = .99c and t=1 year, t(Earth) = 1 / sqrt(1 - .99^2) =
7.0888 years as above. Terence will claim that the duration of
Stella's return trip is the same as that of her outward trip, i.e.
7.0888 years, and also that Stella's round-trip took 14.1778 years,
whereas for Stella, it took only 2 years.

Now, wrt the SR "outbound frame" (with V=.99c, v = .99 c and t = 1
year), Terence's "proper time" = 1 year * (1 +.99^2) /
sqrt(1-.99^2) = 14.0366 years, and the whole trip of Stella takes,
according to Terence, .1411+14.0366 = 14.1777 years, which is
the same duration as that obtained hypothetizing an Earth at rest
wrt the CMBR.

Duration of the Stella's round-trip according to Terence:

We have seen above that the duration of Stella's outward trip is
t1(Earth) = t * (1+Vv) / sqrt(1-v^2), with v = -.99 c
and that the duration of her return trip is
t2(Earth) = t * (1+Vv) / sqrt(1-v^2) with v = .99 v
Hence, the round-trip takes t1(Earth) + t2(Earth) = 2t / sqrt(1-v^2)
on Terence's clock , against 2t on Stella's clock.
It is easy to calculate that the duration of the one-way trips is
related to that of the round-trip by the relation
t(one-way) = t(round-trip) * (1+Vv)/2.
Only if the Earth is at rest wrt the CMBR (V=0) are the durations of
the one-way trips identical.

Conclusion:

Using LET, we straightforwardly demonstrate that the duration
of the round trip, as measured by Terence, is independent
of the Earth velocity V wrt the CMBR, but we also see that the
duration of the one-way trips is a function, not only of Stella's velocity,
but also of V.
This means that the claim of SR that all inertial frames are identical
is only valid for round-trips times.

Marcel Luttgens

ANNEX:

Here is an excerpt from the FAQ, describing and discussing a version

"Let's lay out a "standard version" of the paradox in detail, and settle
on some terminology. We'll get rid of Stella's acceleration at the start
and end of the trip. Stella flashes past Terence in her spaceship both
times, coasting along.
Here's the itinerary according to Terence:
Start Event
Stella flashes past. Clocks are synchronized to 0.
Outbound Leg
Stella coasts along at (say) nearly 99 percent of light-speed.
At 99 percent, the time dilation factor is a bit over 7, so let's say the
speed is just a shade under 99 percent and the time dilation factor is 7.
Let's say this part of the trip takes 7 years (according to Terence,
of course).
Turnaround
Stella fires her thrusters for, say, 1 day, until she is coasting back
towards Earth at nearly 99 percent light-speed. (Stella is the hardy sort.)
Some variations on the paradox call for an instantaneous Turnaround;
we'll call that the Turnaround Event.
Inbound Leg
Stella coasts back for 7 years at 99 percent light-speed.
Return Event
Stella flashes past Terence in the other direction, and they compare
clocks, or grey hairs, or any other sign of elapsed time.
According to Terence, 14 years and a day have elapsed between
the Start and Return Events; Stella's clock however reads just a shade
over 2 years.
How much over? Well, Terence says the Turnaround took a day.
Stella's speed was changing throughout the Turnaround, and so her
time dilation factor was changing, varying between 1 and 7. So Stella's
measurement of the Turnaround Time will be something between 1 day
and one-seventh of a day. If you work it out, it turns out to be a bit over
15 hours. "

Here is a summary of the discussion I had with Tim Shuba about
the validity of the assumption of SR that all inertial frames are
equivalent. The Twin paradox was used as a demonstration tool.

I asked Tim Shuba: "Suppose that the time difference between
Terence's clock and Stella's clock is t(cmbr) seconds if the CMBR
is taken as the reference frame, and t(outbound) seconds in the
frame where Stella is at rest during the outbound part of her trip.
If t(outbound)=t(cmbr), SR is right, otherwise it is wrong, because
all inertial frames would not be equivalent."

And Tim Shuba replied:
"You most certain haven't demonstrated that SR is wrong, in fact if you
were to learn about the Lorentz transformation groups you would know
that t(outbound) = t(cmbr) = t(anyFrame)."

We will see hereafter that all inertial frames are only equivalent for
round-trip times, not for one way times. These vary according to the
reference frame used, hence SR should be considered as falsified.

(ML = Marcel Luttgens, TS = Tim Shuba)

ML:
Finally, after so many posts, you present the SR solution, thanks!
TS:
The other things I have mentioned are actually much more important
than this solution.  Again, I recommend you carefully consider them.
ML:
The theory is indeed fundamental, but its validity is best verified by
a concrete application, in this case the thought experiment known as

Here is an excerpt from the FAQ, describing and discussing a version

"Let's lay out a "standard version" of the paradox in detail, and settle
on some terminology. We'll get rid of Stella's acceleration at the start
and end of the trip. Stella flashes past Terence in her spaceship both
times, coasting along.
Here's the itinerary according to Terence:
Start Event
Stella flashes past. Clocks are synchronized to 0.
Outbound Leg
Stella coasts along at (say) nearly 99 percent of light-speed.
At 99 percent, the time dilation factor is a bit over 7, so let's say the
speed is just a shade under 99 percent and the time dilation factor is 7.
Let's say this part of the trip takes 7 years (according to Terence,
of course).
Turnaround
Stella fires her thrusters for, say, 1 day, until she is coasting back
towards Earth at nearly 99 percent light-speed. (Stella is the hardy sort.)
Some variations on the paradox call for an instantaneous Turnaround;
we'll call that the Turnaround Event.
Inbound Leg
Stella coasts back for 7 years at 99 percent light-speed.
Return Event
Stella flashes past Terence in the other direction, and they compare
clocks, or grey hairs, or any other sign of elapsed time.
According to Terence, 14 years and a day have elapsed between
the Start and Return Events; Stella's clock however reads just a shade
over 2 years.
How much over? Well, Terence says the Turnaround took a day.
Stella's speed was changing throughout the Turnaround, and so her
time dilation factor was changing, varying between 1 and 7. So Stella's
measurement of the Turnaround Time will be something between 1 day
and one-seventh of a day. If you work it out, it turns out to be a bit over
15 hours. "

In our discussion, it was assumed that the turnaround is instantaneous,
and also that Terence was at rest wrt the CMBR frame, whereas Stella
moves at v1= .99 c in her outward journey, and at v2 = - .99 c during
her return trip.

Terence's clock will thus read 1/ sqrt(1-v^2) at the turnaround event, and
2/sqrt(1-v^2) when the twins reunite (where v^2 = v1^2 = v2^2).

The corresponding Terence's proper times are:

Stella's outward trip 7.0888 years
Stella's return trip  7.0888 years
Duration of the whole journey: 17.1776 years

Let's stress that those proper times of Terence are obtained using the
CMBR as the frame of reference.

TS:
And my measurements are from clocks synched in the "outbound frame".
How does the picture look from the "outbound" frame?  Well, Terence
moves at a constant velocity of v_T = -.99 the whole time.  Stella moves with a
velocity of zero until the "turnaround marker" meets her.
This turnaround event takes place at approximately one year (as
measured in the outbound frame). At that event, Stella's
clock reads 1 year and Terence's clock as measured *from the
outbound frame* has ticked off only 1/7 or ~.142 year.
It does correspond to Terence's proper time, from when his clock
reads zero until his clock reads .142 year.  Let's say that Terence
fires a flare at that time on his clock.  That flare event is
simultaneous with Stella's turnaround event *in the outbound frame*.
Of course Terence, using clocks synchronized in his frame, does not
agree, because he reckons correctly *in his frame* that Stella has
only made it 1/49 of the way to the turnaround when he lights the
flare.

ML:
Let's notice that, using the "outbound frame", i.e. the frame in which
Stella is at rest during her outward trip, Tim Shuba concludes that,
according to SR, Terence's clock will read 1*sqrt(1-v^2) = .1411 year.

TS:
After the (assumed instantaneous) turnaround, Stella's velocity in the
outbound frame is -tanh(2 * acrtanh(.99) ) = -.99995 and she must keep
going at this velocity until she makes up the spatial difference of .99
light years by which the twins are separated at the turnaround.  Since
the differences in their speeds is .99995 - .99 = .00995, they will meet
after .00995 * t = .99 meaning after 99.5 years.  That happens after the
turnaround, so the time in the outbound frame when the twins reunite is
100.5 years.  We can now compute Terence's wristwatch time, that is
his proper time,  \tau = \integral dt/g = 100.5 / 7 or about 14.4 years.

ML:
If you had used (v+v) / (1+v^2) instead of  -tanh(2 * acrtanh(.99) ) as
you would have obtained 2 / sqrt(1-v^2) for the round-trip time

TS:
Of course I do not disagree that 2 * 7 = 14.  But 2 * \gamma is not a
general result that can be used to solve this type of problem.  It works
in this case only because the FAQ problem was designed to be very simple.

ML:
Using the "Outbound frame" instead of the "CMBR frame", the
calculated proper times of Terence are, according to Tim Shuba
using SR:

Duration of the whole journey: 2 / sqrt(1-v^2) =14.1776 years
(which is exactly the value obtained with the CMBR frame)

But here is the fundamental difference with the CMBR frame:

Stella's outward trip: 1*sqrt(1-v^2) = .1411 year
Stella's return trip: 14.1776 - .1411 = 14.0365 years.

Indeed, using the "CMBR frame", Terence's proper times are:

Stella's outward trip: 1/sqrt(1_v^2) = 7.0888 years
Stella's return trip: 1/sqrt(1-v^2) =  7.0888 years

CONCLUSION:

The time values obtained using the CMBR are certainly correct.
Indeed, as distances and velocities are identical for the 2 parts of
the trip, the times observed by Terence must also be identical.
In fact, they correspond to a situation where the velocities can be
considered as absolute.  Iow, they correspond to the LET solution.

Otoh, the claim of SR that all inertial frames are identical is only valid
for round-trips times, hence, SR, contrary to LET, is not a correct theory.

Marcel Luttgens