Twins A and B are stationary at point O of space: (-4)----(-3)----(-2)----(-1)----O----(+1)----(2+)----(+3)----(+ 4)
If a twin moves away at constant relativistic speed, there is no
difference between them, in the sense that twin A can safely say that it
is B that is moving away from him, just as twin B can say (just as
safely) which is A who is moving away.
So, in the reference of the twin A, it is B goes, arrives at point +4 (4 light-years away) and then returns to O.
Instead, in the reference of the twin B, it is he who remains in O while
the twin A goes, arrives at the point -4 and then returns.
If the speed corresponds to gamma=2, the travel the other twin (for BOTH twins) lasts 16 years, 8 for the outward journey and 8 for the return.
Both will be 16 years old!
Where is the conceptual difference that should make the time of a twin different from that of the other, if we are talking about Special
Relativity ONLY?
Twins A and B are stationary at point O of space: (-4)----(-3)----(-2)----(-1)----O----(+1)----(2+)----(+3)----(+ 4)
If a twin moves away at constant relativistic speed, there is no
difference between them, in the sense that twin A can safely say that it
is B that is moving away from him, just as twin B can say (just as
safely) which is A who is moving away.
So, in the reference of the twin A, it is B goes, arrives at point +4 (4 light-years away) and then returns to O.
Instead, in the reference of the twin B, it is he who remains in O while
the twin A goes, arrives at the point -4 and then returns.
If the speed corresponds to gamma=2, the travel the other twin (for BOTH twins) lasts 16 years, 8 for the outward journey and 8 for the return.
Both will be 16 years old!
Where is the conceptual difference that should make the time of a twin different from that of the other, if we are talking about Special
Relativity ONLY?
Twins A and B are stationary at point O of space: (-4)----(-3)----(-2)----(-1)----O----(+1)----(2+)----(+3)----(+ 4)
If a twin moves away at constant relativistic speed, there is no
difference between them, in the sense that twin A can safely say that it
is B that is moving away from him, just as twin B can say (just as
safely) which is A who is moving away.
So, in the reference of the twin A, it is B goes, arrives at point +4 (4 light-years away) and then returns to O.
Instead, in the reference of the twin B, it is he who remains in O while
the twin A goes, arrives at the point -4 and then returns.
If the speed corresponds to gamma=2, the travel the other twin (for BOTH twins) lasts 16 years, 8 for the outward journey and 8 for the return.
Both will be 16 years old!
Where is the conceptual difference that should make the time of a twin different from that of the other, if we are talking about Special
Relativity ONLY?
Twins A and B are stationary at point O of space: (-4)----(-3)----(-2)----(-1)----O----(+1)----(2+)----(+3)----(+ 4)
If a twin moves away at constant relativistic speed, there is no
difference between them, in the sense that twin A can safely say that it
is B that is moving away from him, just as twin B can say (just as
safely) which is A who is moving away.
So, in the reference of the twin A, it is B goes, arrives at point +4 (4 light-years away) and then returns to O.
Instead, in the reference of the twin B, it is he who remains in O while
the twin A goes, arrives at the point -4 and then returns.
If the speed corresponds to gamma=2, the travel the other twin (for BOTH twins) lasts 16 years, 8 for the outward journey and 8 for the return.
Both will be 16 years old!
Where is the conceptual difference that should make the time of a twin different from that of the other, if we are talking about Special
Relativity ONLY?
[...]
Where is the conceptual difference that should make the time of a twin different from that of the other, if we are talking about Special
Relativity ONLY?
On 5/21/20 2:36 AM, Luigi Fortunati wrote:
[...]The difference between the twins comes from the simplicity that applies
Where is the conceptual difference that should make the time of a twin
different from that of the other, if we are talking about Special
Relativity ONLY?
only to inertial frames.
If twin A remains at rest in an inertial frame, it is easy to calculate
the age (elapsed proper time) of each twin: simply integrate
T = \integral sqrt(1-v^2/c^2) dt
where T is the elapsed proper time, the integral is taken over the path
of the twin relative to A's rest frame, v is the speed of the twin
relative to that frame (as a function of t), and t is the time
coordinate of the frame. Note that neither position nor acceleration
appear in this equation; all that matters is the speed of the twin
relative to the inertial frame being used to calculate.
For twin A, v = 0, giving an age that is simply the frame's total
coordinate time of the scenario.
For twin B, 0<v<c so the integrand is strictly less than 1, and it is
clear that for any value of t, B's age will be less than the value of t.
So if twin B follows a path that leaves and then returns to A, when they rejoin B will have aged less than A.
If you want to calculate what happens using twin B as a reference, there
is a problem: B does not remain at rest in any inertial frame, and the
above equation does not apply. B must necessarily accelerate in order to return to A. It is possible to use SR to calculate using the accelerated coordinates of B; this is done in some textbooks, but is beyond the
scope of a newsgroup post. When done correctly (which is non-trivial),
this calculation yields the same answer as the much easier calculation
using A's inertial frame.
Tom Roberts
Twins A and B are stationary at point O of space: (-4)----(-3)----(-2)----(-1)----O----(+1)----(2+)----(+3)----(+ 4)
If a twin moves away at constant relativistic speed, there is no
difference between them, in the sense that twin A can safely say that it
is B that is moving away from him, just as twin B can say (just as
safely) which is A who is moving away.
So, in the reference of the twin A, it is B goes, arrives at point +4 (4 light-years away) and then returns to O.
Instead, in the reference of the twin B, it is he who remains in O while
the twin A goes, arrives at the point -4 and then returns.
Where is the conceptual difference that should make the time of a twin different from that of the other, if we are talking about Special
Relativity ONLY?
(Why there is such a thing as absolute acceleration is, as far as I...
know, an unsolved problem. Some believe that Mach's Principle explains
it, i.e. acceleration is relative to the bulk of the matter in the
universe. That implies that if there were no matter in the universe
other than the two twins, then the resolution of the "paradox" might be different.)
On 22/05/2020 21:15, Tom Roberts wrote:
If twin A remains at rest in an inertial frame, it is easy to calculate
the age (elapsed proper time) of each twin: simply integrate
T = \integral sqrt(1-v^2/c^2) dt
where T is the elapsed proper time, the integral is taken over the path
of the twin relative to A's rest frame, v is the speed of the twin
relative to that frame (as a function of t), and t is the time
coordinate of the frame.
One should note that the proper time is an invariant and as such
completely independent of the coordinates and the parametrization you
choose to calculate it. You can introduce arbitrary generalized
coordinates (most of which then correspond to the description of a non-inertial observer) and an arbitrary parameter to parametrize the worldline. As in Euclidean space the length of a curve also proper
time, which is in a somewhat generalized sense nothing else than the
length of a time-like worldline, is independent of these choices.
Twins A and B are stationary at point O of space: (-4)----(-3)----(-2)----(-1)----O----(+1)----(2+)----(+3)----(+ 4)
If a twin moves away at constant relativistic speed, there is no
difference between them, in the sense that twin A can safely say that it
is B that is moving away from him, just as twin B can say (just as
safely) which is A who is moving away.
So, in the reference of the twin A, it is B goes, arrives at point +4 (4 light-years away) and then return to O.
Instead, in the reference of the twin B, it is he who remains in O while
the twin A goes, arrives at the point -4 and then returns.
If the speed corresponds to gamma=2, the travel the other twin (for BOTH twins) lasts 16 years, 8 for the outward journey and 8 for the return.
Both will be 16 years old!
Where is the conceptual difference that should make the time of a twin different from that of the other, if we are talking about Special
Relativity ONLY?
On 5/21/20 2:36 AM, Luigi Fortunati wrote:
[...]
Where is the conceptual difference that should make the time of a twin different from that of the other if we are talking about Special
Relativity ONLY?
The difference between the twins comes from the simplicity that applies
only to inertial frames.
If twin A remains at rest in an inertial frame, it is easy to calculate
the age (elapsed proper time) of each twin: simply integrate
T = \integral sqrt(1-v^2/c^2) dt
where T is the elapsed proper time, the integral is taken over the path
of the twin relative to A's rest frame, v is the speed of the twin
relative to that frame (as a function of t), and t is the time
coordinate of the frame.
Note that neither position nor acceleration
appear in this equation; all that matters is the speed of the twin
relative to the inertial frame being used to calculate.
If you want to calculate what happens using twin B as a reference, there
is a problem: B does not remain at rest in any inertial frame, and the
above equation does not apply. B must necessarily accelerate in order to return to A. It is possible to use SR to calculate using the accelerated coordinates of B; this is done in some textbooks, but is beyond the
scope of a newsgroup post. When done correctly (which is non-trivial),
this calculation yields the same answer as the much easier calculation
using A's inertial frame.
In article <ra3mu2$1gre$1@gioia.aioe.org>, Luigi Fortunati <fortunati.luigi@gmail.com> writes:
Where is the conceptual difference that should make the time of a twin different from that of the other if we are talking about Special
Relativity ONLY?
This is, of course, the "twin paradox". As you hint, special relativity
is sufficient to explain it.
There is a huge literature on this topic, and it has certainly been
answered better than is possible in a newsgroup post.
The basic explanation is that the twin who accelerates moves through
space while the other doesn't.
While the acceleration is not the CAUSE of the difference,
it does allow one to say which twin really moves and which doesn't.
In other words, the motion is NOT relative in this case.
(Why there is such a thing as absolute acceleration is, as far as I
know, an unsolved problem. Some believe that Mach's Principle explains
it, i.e. acceleration is relative to the bulk of the matter in the
universe. That implies that if there were no matter in the universe
other than the two twins, then the resolution of the "paradox" might be different.)
On Thursday, 21 May 2020 09:36:24 UTC+2, Luigi Fortunati wrote:
Twins A and B are stationary at point O of space:If one twin physical moves away from point O (and returns back to point O)
(-4)----(-3)----(-2)----(-1)----O----(+1)----(2+)----(+3)----(+ 4)
If a twin moves away at constant relativistic speed, there is no
difference between them, in the sense that twin A can safely say that it
is B that is moving away from him, just as twin B can say (just as
safely) which is A who is moving away.
it is only his clock that will show a different clock count compared to
a reference clock which stayed at point O.
(if that is what actually is measured when he returns back to point O.).
So, in the reference of the twin A, it is B goes, arrives at point +4 (4Reference frames are not important. What physical happens is important.
light-years away) and then return to O.
Instead, in the reference of the twin B, it is he who remains in O while
the twin A goes, arrives at the point -4 and then returns.
If the speed corresponds to gamma=2, the travel the other twin (for BOTHOnly in the case that both twins physical move away, it is possible that
twins) lasts 16 years, 8 for the outward journey and 8 for the return.
Both will be 16 years old!
the clock count of both clocks don't coincide with the reference clock (count) at point O.
In that case what the clock at O shows is also important.
Where is the conceptual difference that should make the time of a twinIn order to understand better, the exact details of the physical experiment are most important to known and to predict the outcome and to compare
different from that of the other, if we are talking about Special
Relativity ONLY?
if the prediction is correct.
If the comparison between prediction and actual are wrong the error
can be both in the prediction (calculation) or in the experiment.
Nicolaas Vroom.
On Friday, 22 May 2020 21:15:39 UTC+2, Tom Roberts wrote:
On 5/21/20 2:36 AM, Luigi Fortunati wrote:
[...] Where is the conceptual difference that should make the
time of a twin different from that of the other if we are
talking about Special Relativity ONLY?
The difference between the twins comes from the simplicity that
applies only to inertial frames.
What is the reason that you cannot use one reference frame for the
whole experiment from start to finish i.e. when both clocks again
can be compared at point O?
If twin A remains at rest in an inertial frame, it is easy to
calculate the age (elapsed proper time) of each twin: simply
integrate
T = \integral sqrt(1-v^2/c^2) dt
where T is the elapsed proper time, the integral is taken over the
path of the twin relative to A's rest frame, v is the speed of the
twin relative to that frame (as a function of t), and t is the time
coordinate of the frame.
My understanding is that t represents the time (clock count) of a
clock at rest and T the (time) clock count of a moving clock.
The question is how is v (the speed of the clock) measured?
IMO in order to do that, you need a reference rod in the frame at
rest with two clocks at backend x1 and the frontend x2 of this rod,
also at rest. When you do that you can calculate the speed v by
applying the following formula: v = (x2-x1) / (t2-t1) where t1 is
the time that the moving clock coincides with the backend x1 and t2
the time that the clock coincides with the front end x2 of the
clock.
Note that neither position nor acceleration appear in this
equation; all that matters is the speed of the twin relative to
the inertial frame being used to calculate.
That is correct. But if the speed varies along this path you need
more clocks to calculate the time T correctly.
But there is another issue: How do you know that the equation to
calculate T is correct?
The only way is when the calculated T at the end can be verified by
means of an actual experiment.
IMO experiments are the only way to derive this formula.
Using different experiments you can also test if all clocks with a
different internal physical construction behave the same i.e. show
the same number of clock counts.
On 5/25/20 8:38 AM, Nicolaas Vroom wrote:
What is the reason that you cannot use one reference frame for the
whole experiment from start to finish i.e. when both clocks again
can be compared at point O?
There is no such reason, and my discussion explicitly showed using the inertial frame of A for the entire analysis.
T = \integral sqrt(1-v^2/c^2) dt
My understanding is that t represents the time (clock count) of a
clock at rest and T the (time) clock count of a moving clock.
Not quite. t is the time coordinate of the inertial frame,
So it is clear that the time coordinate
of the inertial frame is the same as the elapsed proper time of a clock
at rest in it (reset the clock to zero at t=0). They are conceptual
different but numerically equal.
A clock can only indicate its proper time, and only
along its worldline; the time coordinate of an inertial
frame can indicate the frame's coordinate time anywhere
the frame is valid. Of course, such coordinates are
models, and to implement them requires physical clocks
located where events of interest occur; all such clocks
must, of course, be synchronized to the coordinate time.
The question is how is v (the speed of the clock) measured?
In the usual way: v=dx/dt, where x(t) is the position of the object at
time t, and t is the time coordinate of the inertial frame.
IMO in order to do that, you need a reference rod in the frame at
rest with two clocks at backend x1 and the frontend x2 of this rod,
also at rest. When you do that you can calculate the speed v by
applying the following formula: v = (x2-x1) / (t2-t1) where t1 is
the time that the moving clock coincides with the backend x1 and t2
the time that the clock coincides with the front end x2 of the
clock.
That works, too.
Note this is a gedanken, and we implicitly presume that when we use a
given inertial frame, we can identify its coordinate values for all
events of interest. This, of course, includes the events occupied by the clock as it moves around.
Note that neither position nor acceleration appears in this
equation; all that matters is the speed of the twin relative to
the inertial frame being used to calculate.
That is correct. But if the speed varies along this path you need
more clocks to calculate the time T correctly.
Hmmm. I repeat: this is a gedanken, and we implicitly presume that when
we use a given inertial frame, we can identify its coordinate values for
all events of interest. This, of course, includes the events occupied by
the clock as it moves around.
If you want to place clocks at rest in the inertial frame, located along
the path of the moving clock, that is fine, but it provides no
additional information than is contained in the coordinates themselves.
But there is another issue: How do you know that the equation to
calculate T is correct?
This is NOT an issue. We are using SR to model this Gedanken, and that
is the correct equation of SR. We know this because it has been derived
many times in many textbooks.
The only way is when the calculated T at the end can be verified by
means of an actual experiment.
Hmmmm. The model itself has no need of experiments because it is merely
a set of postulates and theorems, plus definitions of the symbols that
appear in them.
Testing the model and validating it certainly does require experiments.
For this equation of SR, that has been done many
times, for many different types of clocks and many different physical situations.
Using different experiments etc
Hmmm. If you can apply different types of clocks in the same physical situation, yes. But in general, that is not practical, and what is
required is to show agreement with the theory; after all, it is the
theory we are testing.
On Thursday, 28 May 2020 20:14:03 UTC+2, Tom Roberts wrote:Ad 1) It is wise to formulate a problem in a manifestly covariant
On 5/25/20 8:38 AM, Nicolaas Vroom wrote:In fact, there are two issues:
What is the reason that you cannot use one reference frame for theThere is no such reason, and my discussion explicitly showed using the
whole experiment from start to finish i.e. when both clocks again
can be compared at point O?
inertial frame of A for the entire analysis.
1) Is it wise to select only one reference frame?
2) Which is the best reference frame to select?
Apparently you agree with issue 1.
I would prefer point O, as indicated in the text.
t is the proper time of the resting twin, which is of course identicalThis raises a new question: What is the difference my 't represents theNot quite. t is the time coordinate of the inertial frame,T = \integral sqrt(1-v^2/c^2) dtMy understanding is that t represents the time (clock count) of a
clock at rest and T the (time) clock count of a moving clock.
time of a clock in a reference frame' versus 'the time coordinate of
the inertial frame'
My philosophy is to use as simple as possible concepts, and inSo it is clear that the time coordinateMy philosophy is to use as many concepts as possible
of the inertial frame is the same as the elapsed proper time of a clock
at rest in it (reset the clock to zero at t=0). They are conceptual
different but numerically equal.
Of course the clocks have to be synchronized at one instant, i.e., inA clock can only indicate its proper time, and onlyThat means IMO that all the clocks should be synchronized with
along its worldline; the time coordinate of an inertial
frame can indicate the frame's coordinate time anywhere
the frame is valid. Of course, such coordinates are
models, and to implement them requires physical clocks
located where events of interest occur; all such clocks
must, of course, be synchronized to the coordinate time.
one standard clock. IMO at the origin O
There is not one "nearest physical clock". As this example shows, what aIn practice, this means that t is the time of the nearest physical clock.The question is how is v (the speed of the clock) measured?In the usual way: v=dx/dt, where x(t) is the position of the object at
time t, and t is the time coordinate of the inertial frame.
On 30/05/2020 16:17, Nicolaas Vroom wrote:
Apparently you agree with issue 1.Ad 1) It is wise to formulate a problem in a manifestly covariant
I would prefer point O, as indicated in the text.
way and then use the most convenient reference frame to do the
calculations.
The obviously Lorentz-invariant quantity to be calculated here is
the proper time tau of the moving twin as a function of the
proper time of the twin staying in his inertial frame at rest, which
is the coordinate time, t, of this inertial frame.
Ad 2) In this case, of course, the inertial frame (the restframe of
the twin staying at home all the time).
t is the proper time of the resting twin, which is, of course, identical
with the inertial restframe's coordinate time.
So it is clear that the time coordinateMy philosophy is to use as many concepts as possible
of the inertial frame is the same as the elapsed proper time of a clock
at rest in it (reset the clock to zero at t=0). They are conceptual
different but numerically equal.
My philosophy is to use as simple as possible concepts, and in
relativity the most simple concept is to describe a physical situation
in terms of Lorentz-invariant quantities.
Here we compare two Lorentz invariant proper times:
the proper time of the twin at rest and the proper time of the travelling twin the round-trip of the travelling twin takes.
This resolves all paradoxes with different frames of reference, because
the results do not depend on any frame of reference, and it's allowed
to use any frame of reference to evaluate the quantities.
For that I choose the most convenient reference frame.
That means IMO that all the clocks should be synchronized withOf course, the clocks have to be synchronized at one instant, i.e., in
one standard clock. IMO at the origin O
this case on the beginning of the trip, where both twins are at the
location 0.
That's another important concept: use as long as you can simple local
events to describe a situation in relativity.
In the usual way: v=dx/dt, where x(t) is the position of the object atIn practice, this means that t is the time of the nearest physical clock.
time t, and t is the time coordinate of the inertial frame.
There is not one "nearest physical clock". As this example shows, what
a clock once synchronized with another clock, when both clocks have been
at the same place, shows after some time when you compare these clocks
(again at a common-place) depends on the motion of these clocks.
In SR it's pretty easy to synchronize a set of clocks in one inertial
frame because you can use light signals and the universality of the
speed of light. That's how Einstein's clock synchronization
works for an inertial frame.
In accelerated reference frames or even relativity, clock synchronization
and the definition of "distance"
becomes a complicated issue, but that's not related to the special-relativistic twin paradox at all.
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