SR can be used to calculate the time of the spaceship twin during constant v
and the addition of the clock postulate during ac/deceleration from the
point of view of the earth twin. What does the space-ship twin use during
it's ac/deceleration to calculate the time of the earth twin?
Thanks,
jmc

Dear jmc:

"jmc" wrote in message
...
SR can be used to calculate the time of the spaceship twin during
constant v
and the addition of the clock postulate during ac/deceleration from the
point of view of the earth twin. What does the space-ship twin use during
it's ac/deceleration to calculate the time of the earth twin?

How about periodic signals from the earth twin?

David A. Smith

jmc wrote:
SR can be used to calculate the time of the spaceship twin during constant v
and the addition of the clock postulate during ac/deceleration from the
point of view of the earth twin.

Yes. Assuming the spaceship twin's clock is unaffected by acceleration,
one can integrate sqrt(1-v^2/c^2) over the spaceship twin's path to get
the elapsed proper time of the spaceship twin's clock (all as measured
in the earth twin's inertial frame[#]).

[#] For discussions like this we traditionally neglect the minor
non-inertial nature of the earth's orbit and rotation.

In geometrical terms: one integrates the metric over the spaceship
twin's path to obtain the elapsed proper time of the spaceship twin's clock.


What does the space-ship twin use during
it's ac/deceleration to calculate the time of the earth twin?

This is a difficult problem, unless one states it in geometrical terms:
one integrates the metric over the earth twin's path to obtain the
elapsed proper time of the earth twin's clock.

In the first case above, one expresses the spaceship's trajectory in
terms of the inertial earthbound coordinates, and integrates the metric
components wrt those coordinates over that path. To do the computation
as seen by the spaceship twin is complex, but in principle no more
complicated -- simply express the trajectory of the earth twin in terms
of the accelerated coordiantes (quite messy!), and then integrate the
metric components wrt those coordinates (also messy!).

But if the spaceship twin is smart, she will relaize the earth twin is
inertial, and use earthbound coordinates for the computation, not her
own accelerated coordinates -- as elapsed proper time is a scalar any
observer can use any coordinates for the computation and obtain the same
answer.


Tom ROberts

Tom Roberts wrote in message ...
jmc wrote:
SR can be used to calculate the time of the spaceship twin during constant v
and the addition of the clock postulate during ac/deceleration from the
point of view of the earth twin.

Yes. Assuming the spaceship twin's clock is unaffected by acceleration,
one can integrate sqrt(1-v^2/c^2) over the spaceship twin's path to get
the elapsed proper time of the spaceship twin's clock (all as measured
in the earth twin's inertial frame[#]).

[#] For discussions like this we traditionally neglect the minor
non-inertial nature of the earth's orbit and rotation.

In geometrical terms: one integrates the metric over the spaceship
twin's path to obtain the elapsed proper time of the spaceship twin's clock.


What does the space-ship twin use during
it's ac/deceleration to calculate the time of the earth twin?

This is a difficult problem, unless one states it in geometrical terms:
one integrates the metric over the earth twin's path to obtain the
elapsed proper time of the earth twin's clock.

In the first case above, one expresses the spaceship's trajectory in
terms of the inertial earthbound coordinates, and integrates the metric
components wrt those coordinates over that path. To do the computation
as seen by the spaceship twin is complex, but in principle no more
complicated -- simply express the trajectory of the earth twin in terms
of the accelerated coordiantes (quite messy!), and then integrate the
metric components wrt those coordinates (also messy!).

But if the spaceship twin is smart, she will relaize the earth twin is
inertial, and use earthbound coordinates for the computation, not her
own accelerated coordinates -- as elapsed proper time is a scalar any
observer can use any coordinates for the computation and obtain the same
answer.


Tom ROberts

Tom, SR states that no frame has preference over another. Whatever
frame A observes of or does to frame B, frame B observes of or does to
frame A according to the principle of reciprocity. Therefore, both
twins age at the same rate and SR contradicts itself.

Peter Riedt

"Tom Roberts" wrote in message
...
jmc wrote:
SR can be used to calculate the time of the spaceship twin during
constant v
and the addition of the clock postulate during ac/deceleration from the
point of view of the earth twin.

Yes. Assuming the spaceship twin's clock is unaffected by acceleration,

This is an erroneous assumption. The twin's clock will run slower compared
to before acceleration (the earth clock).

Ken Seto


one can integrate sqrt(1-v^2/c^2) over the spaceship twin's path to get
the elapsed proper time of the spaceship twin's clock (all as measured
in the earth twin's inertial frame[#]).

[#] For discussions like this we traditionally neglect the minor
non-inertial nature of the earth's orbit and rotation.

In geometrical terms: one integrates the metric over the spaceship
twin's path to obtain the elapsed proper time of the spaceship twin's
clock.


What does the space-ship twin use during
it's ac/deceleration to calculate the time of the earth twin?

This is a difficult problem, unless one states it in geometrical terms:
one integrates the metric over the earth twin's path to obtain the
elapsed proper time of the earth twin's clock.

In the first case above, one expresses the spaceship's trajectory in
terms of the inertial earthbound coordinates, and integrates the metric
components wrt those coordinates over that path. To do the computation
as seen by the spaceship twin is complex, but in principle no more
complicated -- simply express the trajectory of the earth twin in terms
of the accelerated coordiantes (quite messy!), and then integrate the
metric components wrt those coordinates (also messy!).

But if the spaceship twin is smart, she will relaize the earth twin is
inertial, and use earthbound coordinates for the computation, not her
own accelerated coordinates -- as elapsed proper time is a scalar any
observer can use any coordinates for the computation and obtain the same
answer.


Tom ROberts

Peter Riedt wrote:
SR states that no frame has preference over another.

True. That is a loose statement of the Principle of Relativity. Note in
particular that in SR this principle holds ONLY for inertial frames.


Whatever
frame A observes of or does to frame B, frame B observes of or does to
frame A according to the principle of reciprocity.

But there is no "principle of reciprocity", in SR, or in physics in general.

Yes, one can apply the Principle of Relativity to the case of two
observers each in an INERTIAL frame, and conclude that what observer A
measures about observer B's clocks and rulers, observer B will
necessarily make similar measurements of observer A's clocks and rulers.
So this is "reciprocity" for this specific case. But it does not hold
for the twin scenario, because in that scenario one twin is necessarily
not in an inertial frame.


Therefore, both
twins age at the same rate and SR contradicts itself.

No. Your invalid GUESS of what SR says contradicts itself. Not SR.


Tom Roberts

jmc wrote:

[...] What does the space-ship twin use during
it's ac/deceleration to calculate the time of the earth twin?

Years ago, I derived a simple equation that
relates the current ages of the twins, according
to each twin. To save writing, I write "the
current age of a distant object", where the
"distant object" is the stay-at-home twin, as
the "CADO". The CADO has a value for each age t
of the traveling twin, CADO(t). The traveler
and the stay-at-home twin get different values
of CADO(t), at any given age t of the traveler.
Call the two values CADO_T(t) for the traveler's
view, and CADO_H(t) for the stay-at-home twin's
view. (And in both cases, CADO is the age of the
home twin, and t is the age of the traveler).

My simple equation says that

CADO_T(t) = CADO_H(t) - L*v/(c*c),

where

L is their current distance apart, in lightyears,
according to the home twin,

and

v is their current relative speed, in lightyears/year,
according to the home twin. v is positive
when the twins are moving apart.

The factor (c*c) has value 1 for these units, and
is needed only to make the dimensionality correct.

The equation immediately shows how the home twin
suddenly ages during the traveler's turnaround
(according to the traveler, not the home twin).

For example, suppose the home twin believes that she
is 40 when the traveler is 20, immediately before
he turns around. So CADO_H(20-) = 40. (Denote his
age immediately before the turnaround as t = 20-,
and immediately after the turnaround as t = 20+.)

Suppose they are 30 ly apart (according to the home
twin), and that their relative speed is +0.9 ly/y (i.e.,
0.9c), when the traveler's age is 20-. Then the traveler
will conclude that the home twin is

CADO_T(20-) = 40 - 0.9*30 = 27

years old immediately before his turnaround.
Immediately after his turnaround (assumed here
to occur in zero time), their relative speed
is -0.9 ly/y. The home twin concludes that
neither of them ages during the turnaround,
so that CADO_H(20+) is still 40. But according
to the traveler,

CADO_T(20+) = 40 - (-0.9)*30 = 67,

so he concludes that his twin ages 40 years
during his instantaneous turnaround.

The equation works for arbitrary accelerations,
not just the idealized instantaneous speed change
assumed above. I've got an example with
1-g accelerations on my web page:

http://home.comcast.net/~mlfasf

The derivation of the equation is given in
"Accelerated Observers in Special Relativity",
PHYSICS ESSAYS, Sept 1999, p629.

Mike Fontenot

jmc wrote:

[...] What does the space-ship twin use during
it's ac/deceleration to calculate the time of the earth twin?

In my earlier posting, I made a careless
error in my arithmetic. Here is the
corrected version:
________________________________________________

Years ago, I derived a simple equation that
relates the current ages of the twins, according
to each twin. To save writing, I write "the
current age of a distant object", where the
"distant object" is the stay-at-home twin, as
the "CADO". The CADO has a value for each age t
of the traveling twin, CADO(t). The traveler
and the stay-at-home twin get different values
of CADO(t), at any given age t of the traveler.
Call the two values CADO_T(t) for the traveler's
view, and CADO_H(t) for the stay-at-home twin's
view. (And in both cases, CADO is the age of the
home twin, and t is the age of the traveler).

My simple equation says that

CADO_T(t) = CADO_H(t) - L*v/(c*c),

where

L is their current distance apart, in lightyears,
according to the home twin,

and

v is their current relative speed, in lightyears/year,
according to the home twin. v is positive
when the twins are moving apart.

The factor (c*c) has value 1 for these units, and
is needed only to make the dimensionality correct.

The equation explicitly shows how the home twin's
age will change abruptly (according to the traveler,
not the home twin), whenever the relative
speed changes abruptly.

For example, suppose the home twin believes that she
is 40 when the traveler is 20, immediately before
he turns around. So CADO_H(20-) = 40. (Denote his
age immediately before the turnaround as t = 20-,
and immediately after the turnaround as t = 20+.)

Suppose they are 30 ly apart (according to the home
twin), and that their relative speed is +0.9 ly/y (i.e.,
0.9c), when the traveler's age is 20-. Then the traveler
will conclude that the home twin is

CADO_T(20-) = 40 - 0.9*30 = 13

years old immediately before his turnaround.

Immediately after his turnaround (assumed here
to occur in zero time), their relative speed
is -0.9 ly/y.

The home twin concludes that their distance apart
doesn't change during the turnaround: it's
still 30 ly.

And the home twin concludes that
neither of them ages during the turnaround,
so that CADO_H(20+) is still 40.

But according to the traveler,

CADO_T(20+) = 40 - (-0.9)*30 = 67,

so he concludes that his twin ages 54 years
during his instantaneous turnaround.

The equation works for arbitrary accelerations,
not just the idealized instantaneous speed change
assumed above. I've got an example with
1-g accelerations on my web page:

http://home.comcast.net/~mlfasf

The derivation of the equation is given in
"Accelerated Observers in Special Relativity",
PHYSICS ESSAYS, Sept 1999, p629.

Mike Fontenot

Mike Fontenot wrote in message ...
jmc wrote:

[...] What does the space-ship twin use during
it's ac/deceleration to calculate the time of the earth twin?

In my earlier posting, I made a careless
error in my arithmetic. Here is the
corrected version:
________________________________________________

Years ago, I derived a simple equation that
relates the current ages of the twins, according
to each twin. To save writing, I write "the
current age of a distant object", where the
"distant object" is the stay-at-home twin, as
the "CADO". The CADO has a value for each age t
of the traveling twin, CADO(t). The traveler
and the stay-at-home twin get different values
of CADO(t), at any given age t of the traveler.
Call the two values CADO_T(t) for the traveler's
view, and CADO_H(t) for the stay-at-home twin's
view. (And in both cases, CADO is the age of the
home twin, and t is the age of the traveler).

My simple equation says that

CADO_T(t) = CADO_H(t) - L*v/(c*c)

SNIP for compactness

The equation explicitly shows how the home twin's
age will change abruptly (according to the traveler,
not the home twin), whenever the relative
speed changes abruptly.
SNIP

But according to the traveler,

CADO_T(20+) = 40 - (-0.9)*30 = 67,

so he concludes that his twin ages 54 years
during his instantaneous turnaround.

Nice. Just one remark:
This traveller believes what his measurement instruments tell him, if
it makes sense or not. Surely nobody of sound mind will believe that
he can change the progress of physical processes in all other bodies
in the universe by simply changing his own speed, and with increasing
effect at greater distance. If he has a rocket pilot licence, it
should be withdrawn.

Harald

Harry wrote:

[...] Surely nobody of sound mind will believe that
he can change the progress of physical processes in all other bodies
in the universe by simply changing his own speed, and with increasing
effect at greater distance.

It IS widely believed that the accelerating observer should
realize that his own bizarre conclusions about the
current age of a distant object should be ignored,
in favor of the "true" conclusions of the inertial
distant object. This is the conventional
recommendation, even though his own conclusions are
consistent with his own measurements (and the
conclusions of the distant object are NOT consistent
with his (the traveler's) own measurements).

But this view cannot be consistently maintained.
In my paper, I spend a considerable amount of time
showing that the above recommendation is untenable.

The easiest and quickest way to indicate the
problem with the above view is the following:

At any instant in his life, the accelerating
observer COULD choose to stop his acceleration,
and to thereafter coast at a constant speed
relative to the distant object (i.e., to coast
at whatever speed he happened to have when
he quit accelerating). If he did that, should
he continue to forever maintain that the
distant object's view of the correspondence
between their two ages is the "true" one?

In other words, if he believed the distant object's
view was the "true" view immediately before
he stopped accelerating, should he continue
to have that same belief, immediately after he
stops accelerating? If so, he will discover that
inertial observers with whom he is now
stationary do NOT agree with the distant
object's view. If fact, those inertial
observers will have the same view of simultaneity
that the accelerating observer had immediately
before he quit accelerating.

If he remains
unaccelerated forever thereafter, should he
continue forever to disagree with those
inertial observers, in favor of the "true"
view of the distant object? If so, he
will forever have to ignore what his own
measurements tell him. And why is the
distant object's view more "true" than
the view of the inertial observers with whom
the traveler is now stationary? It clearly
is not.

On the other hand, if the traveler should
EVENTUALLY decide to adopt the view of
the inertial observers with whom he is now
stationary, and finally reject the view of the
distant object, exactly WHEN should he
make this change? After a year? After a
month? After a day? The only rational
possibility is that, at each instant of
his life (whether accelerating or not),
he must adopt the conclusions of the
inertial reference frame in which he
is momentarily stationary at that instant.

For a much more complete discussion of these
issues, see my paper

"Accelerated Observers in Special Relativity",
PHYSICS ESSAYS, Sept 1999, p629.

An example of the bizarre conclusions that
must be adopted by an observer accelerating
at 1 g is given on my web page:

http://home.comcast.net/~mlfasf

Mike Fontenot