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SR & GR & "twin paradox"
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Chalky
Apr 13, 2013, 4:55:03 PM4/13/13
to
We seem to have established beyond reasonable doubt that a body moving
in interstellar space CAN have its direction reversed via eventual
interaction with one or more massive bodies at the far end of its
trajectory, as in our earlier discussion at
https://groups.google.com/forum/?fromgroups=#!topic/sci.physics.research/H5pvAGNS_aA
The example shown in the second diagram at
http://en.wikipedia.org/wiki/Gravity_assist can, in fact, be simplified
still further by making U = 0, whereupon the velocity of the projectile
is simply reversed (exactly).
This brings us neatly back to my originally intended point. In the well
known twin "paradox" situation, the apparent paradox is resolved by the
relativistic physics that applies relative to the projectile during its
direction reversal. That direction reversal is traditionally assumed to
be caused by the firing of retro rockets on the projectile, thereby
inducing a subjective gravitational field, relative to this projectile,
during the period of acceleration. However, it seems equally possible to
me that such direction reversal can be achieved via gravitational
slingshot at the end of the trajectory, whereupon NO gravitational field
is experienced relative to the projectile during that period.
Given that the outward and return legs of the journey can be made
arbitrarily long, it strikes me that the physics during the period of
direction reversal should remain essentially the same in both cases.
Am I missing something here?
in interstellar space CAN have its direction reversed via eventual
interaction with one or more massive bodies at the far end of its
trajectory, as in our earlier discussion at
https://groups.google.com/forum/?fromgroups=#!topic/sci.physics.research/H5pvAGNS_aA
The example shown in the second diagram at
http://en.wikipedia.org/wiki/Gravity_assist can, in fact, be simplified
still further by making U = 0, whereupon the velocity of the projectile
is simply reversed (exactly).
This brings us neatly back to my originally intended point. In the well
known twin "paradox" situation, the apparent paradox is resolved by the
relativistic physics that applies relative to the projectile during its
direction reversal. That direction reversal is traditionally assumed to
be caused by the firing of retro rockets on the projectile, thereby
inducing a subjective gravitational field, relative to this projectile,
during the period of acceleration. However, it seems equally possible to
me that such direction reversal can be achieved via gravitational
slingshot at the end of the trajectory, whereupon NO gravitational field
is experienced relative to the projectile during that period.
Given that the outward and return legs of the journey can be made
arbitrarily long, it strikes me that the physics during the period of
direction reversal should remain essentially the same in both cases.
Am I missing something here?
Phillip Helbig---undress to reply
Apr 13, 2013, 9:11:52 PM4/13/13
to
In article <fa92e855-6908-4afe...@googlegroups.com>,
However, if I recall correctly, though, as you say, the thought
experiment usually involves the travelling twin accelerating with a
rocket and thus providing, as you say, something that feels like a local
gravitational field, it is wrong to think that this is the reason that
the travelling twin experiences a slower passage of time. Think about
it this way: the longer the journey, the greater the difference, though
the acceleration at the turnaround is independent of the length of the
journey. Also, one can have a universe which is spatially closed and
can be circumnavigated; in this case, there would be nothing which feels
like a local gravitational field. So if the apparent gravitational
field isn't the reason for the difference, the slingshot idea won't
change anything.
Chalky <chalk...@bleachboys.co.uk> writes:
> This brings us neatly back to my originally intended point. In the well
> known twin "paradox" situation, the apparent paradox is resolved by the
> relativistic physics that applies relative to the projectile during its
> direction reversal. That direction reversal is traditionally assumed to
> be caused by the firing of retro rockets on the projectile, thereby
> inducing a subjective gravitational field, relative to this projectile,
> during the period of acceleration. However, it seems equally possible to
> me that such direction reversal can be achieved via gravitational
> slingshot at the end of the trajectory, whereupon NO gravitational field
> is experienced relative to the projectile during that period.
I'm no expert on relativity (though Jonathan is; maybe he'll chime in).
> This brings us neatly back to my originally intended point. In the well
> known twin "paradox" situation, the apparent paradox is resolved by the
> relativistic physics that applies relative to the projectile during its
> direction reversal. That direction reversal is traditionally assumed to
> be caused by the firing of retro rockets on the projectile, thereby
> inducing a subjective gravitational field, relative to this projectile,
> during the period of acceleration. However, it seems equally possible to
> me that such direction reversal can be achieved via gravitational
> slingshot at the end of the trajectory, whereupon NO gravitational field
> is experienced relative to the projectile during that period.
However, if I recall correctly, though, as you say, the thought
experiment usually involves the travelling twin accelerating with a
rocket and thus providing, as you say, something that feels like a local
gravitational field, it is wrong to think that this is the reason that
the travelling twin experiences a slower passage of time. Think about
it this way: the longer the journey, the greater the difference, though
the acceleration at the turnaround is independent of the length of the
journey. Also, one can have a universe which is spatially closed and
can be circumnavigated; in this case, there would be nothing which feels
like a local gravitational field. So if the apparent gravitational
field isn't the reason for the difference, the slingshot idea won't
change anything.
Oliver Jennrich
Apr 14, 2013, 2:06:25 AM4/14/13
to
Chalky <chalk...@bleachboys.co.uk> writes:
> We seem to have established beyond reasonable doubt that a body moving
> in interstellar space CAN have its direction reversed via eventual
> interaction with one or more massive bodies at the far end of its
> trajectory, as in our earlier discussion at
> https://groups.google.com/forum/?fromgroups=#!topic/sci.physics.research/H5pvAGNS_aA
> The example shown in the second diagram at
> http://en.wikipedia.org/wiki/Gravity_assist can, in fact, be simplified
> still further by making U = 0, whereupon the velocity of the projectile
> is simply reversed (exactly).
> This brings us neatly back to my originally intended point. In the well
> known twin "paradox" situation, the apparent paradox is resolved by the
> relativistic physics that applies relative to the projectile during its
> direction reversal.
This is a common and depressingly popular (or vice versa) argument and
> We seem to have established beyond reasonable doubt that a body moving
> in interstellar space CAN have its direction reversed via eventual
> interaction with one or more massive bodies at the far end of its
> trajectory, as in our earlier discussion at
> https://groups.google.com/forum/?fromgroups=#!topic/sci.physics.research/H5pvAGNS_aA
> The example shown in the second diagram at
> http://en.wikipedia.org/wiki/Gravity_assist can, in fact, be simplified
> still further by making U = 0, whereupon the velocity of the projectile
> is simply reversed (exactly).
> This brings us neatly back to my originally intended point. In the well
> known twin "paradox" situation, the apparent paradox is resolved by the
> relativistic physics that applies relative to the projectile during its
> direction reversal.
can be sadly found even in textbooks. But it is, as you just showed
nicely and correctly, dead wrong.
Considering a triangle in Euclidean space (i.e. on a flat sheet of
paper), no one would argue that the fact the path 'around the
other corner' is longer than the straight way between two corners of the
triangle is due to the details on how the test particle navigates its
way around the corner. The distance via the third corner is longer
because it is a different path than the direct one - this is intuitively
evident to anyone who takes a look at a triangle.
The analogy to a triangle in Euclidean geometry holds quite far, if you
think of 'distance traveled' instead of 'time passed' and 'angles'
instead of 'velocity' - much like veolicities in spacetime, angles in
Euclidean geometry are purely relative concepts. A line has no angle as
such, it only has an angle wrt to some other line.
If you start walking from your starting point A to a point C on a
distant wall (where the line AC is orthogonal to the wall), and you keep
walking on the line AC, then this is the shortest possible way. Anyone
(e.g. your twin) that starts in A as well, and with the same walking
speed, but walks at an angle to the line AC, will be at any time further
away from the wall as you. At some point during his walk, your twin
notices his mistaken bearing, turns appropriately and walks from his
present point B dircetly towards C. When you finally meet at C you
compare the walked distances and you find, of course, that he had to
walk a larger distance. You (and anybody else who has seen a triangle
before) will intuitively understand that. But your twin argues that
this cannot be: Yes, you saw him walking away at an angle and those who
walk away at an angle will have to walk further, but angles are relative
and he saw *you* walking away at an angle with repect to his chosen
direction as well. So why did *you* not have to work further than *he*?
[1]
The answer is blindingly obvious: You chose the shortest possible
way. Yes, he saw you walking away at an angle, but that was because *he*
chose a longer way. Angles are relative, but that doesn't mean that all
paths form A to C have the same length.
The solution the the twin paradox is similarly obvious. The
'stay-at-home' twin chose the path through spacetime (from A to C) on
which the most amount of time passes. Any other path through spacetime
will yield in less time passed [2]. And yes, your twin will argue that
he saw *you* flying away with some considerable velocity, so it should be
*your* clock that goes slower. But you chose the path through spacetime
[3] that led to the maximum passage of time. Velocities are relative,
but that doesn't mean that the same amount of time passes between A and
C.
So the solution to your problem is simple: There is no problem. You
identified a flaw in a popular "explanation" to the twin paradox. Cudos
for that - not everybody realises that blaming the turning aound is not
helping. However, just because an apparent explanation of the twin
paradox is wrong, it doesn't mean that the paradox itself is real.
HTH,
Oliver
[1] After some discussion the two of you concluded that he had to walk
further, because when he changed his bearing at point B, he *turned* and
turning is governed by a vastly more complex theory than walking
straight (it involves general rotations which few understand, indeed
Euclid is known to have stated that not more then 3 of his fellow Greeks
was able to understand rotations), so it is probably safe to blame the
additional distance on the turning.
[2] And this is where the analogy with the triangle breaks down. Instead
of the *shortest* possible path (or the *minimum* distance), you get the
*maximum* passage of time. But that is of little to no consequence to
the argument.
[3] Or just through time, if you like. The whole point of the experiment
is that the stay-at-home twin *doesn't* move through space. But it is
still a path through spacetime. Much as a line parallel to the x-axis is
still a line in the x-y-plane.
--
Space - The final frontier
Jonathan Thornburg
Apr 14, 2013, 6:24:06 AM4/14/13
to
From: "Jonathan Thornburg [remove -animal to reply]" <jth...@astro.indiana-zebra.edu>
Subject: Re: SR & GR & "twin paradox"
Newsgroups: sci.physics.research
References: <fa92e855-6908-4afe...@googlegroups.com>
Chalky <chalk...@bleachboys.co.uk> wrote:
> This brings us neatly back to my originally intended point. In the well
> known twin "paradox" situation, the apparent paradox is resolved by the
> relativistic physics that applies relative to the projectile during its
As for the somewhat-longer answer.... I have a number of comments:
*First point*:
The direction reversal you're talking about doesn't actually need a
full-fledged gravitational slingshot, it can be accomplished by
(conceptually) a slightly simpler phenomenon.
That is, the classic gravitational slingshot takes place in a (global)
inertial reference frame, and involves a test particle P (initial
velocity v_P0) having a (Newtonian) gravitational encounter with a
massive body M which is moving at some (in this approximation, constant)
velocity v_M. Thus in M's reference frame the initial velocity of
the test particle is v_P0-v_M. Since Newtonian gravitation is
conservative, the final velocity of the test particle in M's
reference frame must have this same magnitude, but it can have
a different direction. That is, the final velocity of the test
particle in M's reference frame is
Q(v_P0-v_m)
where Q is an orthogonal matrix. Thus in the global inertial frame
the test particle's final velocity is
v_m + Q(v_P0-v_m)
Depending on the choice of Q (which is determined by the encounter
geometry) this final velocity can be very different in both direction
*and* *magnitude* from the initial velocity, i.e., the test particle
can gain or loose kinetic energy as measured in the global inertial
reference frame.
However, the direction reversal that Chalky is considering doesn't
need this full generality: it suffices to take v_m = 0, i.e., to have
M be at rest in our global inertial reference frame. In this case the
test particle can (if M is placed in just the right place) still have
its velocity vector reversed (matrix Q = diag(-1, -1, -1)). In this
case the test particle's final velocity has the same magnitude as the
initial velocity, so there's no change in the test particle's kinetic
energy as measured in the global inertial reference frame.
*Second point*:
You're right that such a gravitational direction reversal can be used
to construct a twin-paradox example where the traveller remains in
(locally-measured) free fall throuhout her travels. (I think that's
what you were trying to get at with the phrase "NO gravitational field
is experienced relative to the projectile", i.e., the projectile is
free-falling in whatever ambient gravitational field may be present.)
*Third point*:
Since we now have gravitational fields around, we need to think about
general relativity, and in this context the twin paradox takes a
particularly simple form: each twin travels along some (timelike)
worldline connecting events A and B, and (in general) each twin
measures a different proper time along her worldline. The general
relativity perspective is that there's no paradox here: you'd no more
expect to measure the same proper time along different worldlines
than you would to (say) measure the same distance along two different
driving routes between Paris and Moscow.
Notice that, in this general relativity perspective, we have NOT thus
far required that either worldline be a geodesic (i.e., that either
twin is in (locally-measured) free-fall throughout her travels). If
we want to require that the worldlines be geodesics, then providing
that A and B are "not too close"
[more precisely, providing that there *are* distinct
(timelike) geodesics from A to B, which there often
[usually? always? I'm not sure of
the precise conditions]
are if the spacetime is non-flat and A and B are far
enough apart]
it's still the case that in general each twin will measure a
different proper time along her worldline. In other words, we still
have the twin "paradox".
The general relativity "explanation" of the twin paradox in the
both-travellers-are-in-free-fall-throughout-their-journeys case is
also still the same, i.e., there's really no paradox here, this is
just how one should expect things to behave.
*Fourth Point*:
One way to think about the twin paradox in special relativity
[I think Taylor & Wheeler's special relativity book
describes this interpretation... but I'm several time
zones away from my copy of T&W right now, so I can't
easily confirm this or provide a precise reference.
The Wikipedia article
http://en.wikipedia.org/wiki/Twin_paradox
also describes this conceptualization.
interprets the time difference between the two twins being due in
large part to the change in the hypersurface-of-simultaneity between
the inertial reference frame the travelling twin is in in her
outbound journey, and the hypersurface-of-simultaneity of the
*different* inertial reference frame she's in on her return journey.
In *this* perspective, the turnaround is crucial for the proper-time
difference. It may be that this is what you (Chalky) meant by the
But I wouldn't say that the turnaround -- or the physics that applies
during it -- *causes* the proper-time difference. The difference is
really a property of the two worldlines as a whole, not of this one
segment of one of them.
*Fifth point*:
In the general-relativity perspective the proper time is integrated
along each worldline, and the "turnaround" is no more or less important
than anywhere else along the worldline.
There's no contradiction between this and the special-relativity
perspective: both predict precisely the same observed results
(each observer's measured proper time at each point along her
worldline). In some sense the special-relativity perspective (with
the changing of inertial reference frames and hypersurfaces-of-simultaneity
during the turnaround) gives an additional insight into why the proper
times are as they are.
And finally.....
*Sixth point*:
I suspect Chalky and most other s.p.r readers already know this,
but for anyone who doesn't: There have been a number of direct
experimental tests of the twin paradox, by flying atomic clocks in
airplanes travelling around various closed paths. The results agree
precisely with the special/general relativity predictions.
ciao,
--
-- "Jonathan Thornburg [remove -animal to reply]" <jth...@astro.indiana-zebra.edu>
Dept of Astronomy & IUCSS, Indiana University, Bloomington, Indiana, USA
on sabbatical in Canada starting August 2012
"Washing one's hands of the conflict between the powerful and the
powerless means to side with the powerful, not to be neutral."
-- quote by Freire / poster by Oxfam
Subject: Re: SR & GR & "twin paradox"
Newsgroups: sci.physics.research
References: <fa92e855-6908-4afe...@googlegroups.com>
Chalky <chalk...@bleachboys.co.uk> wrote:
> This brings us neatly back to my originally intended point. In the well
> known twin "paradox" situation, the apparent paradox is resolved by the
> relativistic physics that applies relative to the projectile during its
> direction reversal. That direction reversal is traditionally assumed to
> be caused by the firing of retro rockets on the projectile, thereby
> inducing a subjective gravitational field, relative to this projectile,
> during the period of acceleration. However, it seems equally possible to
> me that such direction reversal can be achieved via gravitational
> slingshot at the end of the trajectory, whereupon NO gravitational field
> be caused by the firing of retro rockets on the projectile, thereby
> inducing a subjective gravitational field, relative to this projectile,
> during the period of acceleration. However, it seems equally possible to
> me that such direction reversal can be achieved via gravitational
> slingshot at the end of the trajectory, whereupon NO gravitational field
> is experienced relative to the projectile during that period.
>
> Given that the outward and return legs of the journey can be made
> arbitrarily long, it strikes me that the physics during the period of
> direction reversal should remain essentially the same in both cases.
>
> Am I missing something here?
I think the very-short answer is that you're right.
>
> Given that the outward and return legs of the journey can be made
> arbitrarily long, it strikes me that the physics during the period of
> direction reversal should remain essentially the same in both cases.
>
> Am I missing something here?
As for the somewhat-longer answer.... I have a number of comments:
*First point*:
The direction reversal you're talking about doesn't actually need a
full-fledged gravitational slingshot, it can be accomplished by
(conceptually) a slightly simpler phenomenon.
That is, the classic gravitational slingshot takes place in a (global)
inertial reference frame, and involves a test particle P (initial
velocity v_P0) having a (Newtonian) gravitational encounter with a
massive body M which is moving at some (in this approximation, constant)
velocity v_M. Thus in M's reference frame the initial velocity of
the test particle is v_P0-v_M. Since Newtonian gravitation is
conservative, the final velocity of the test particle in M's
reference frame must have this same magnitude, but it can have
a different direction. That is, the final velocity of the test
particle in M's reference frame is
Q(v_P0-v_m)
where Q is an orthogonal matrix. Thus in the global inertial frame
the test particle's final velocity is
v_m + Q(v_P0-v_m)
Depending on the choice of Q (which is determined by the encounter
geometry) this final velocity can be very different in both direction
*and* *magnitude* from the initial velocity, i.e., the test particle
can gain or loose kinetic energy as measured in the global inertial
reference frame.
However, the direction reversal that Chalky is considering doesn't
need this full generality: it suffices to take v_m = 0, i.e., to have
M be at rest in our global inertial reference frame. In this case the
test particle can (if M is placed in just the right place) still have
its velocity vector reversed (matrix Q = diag(-1, -1, -1)). In this
case the test particle's final velocity has the same magnitude as the
initial velocity, so there's no change in the test particle's kinetic
energy as measured in the global inertial reference frame.
*Second point*:
You're right that such a gravitational direction reversal can be used
to construct a twin-paradox example where the traveller remains in
(locally-measured) free fall throuhout her travels. (I think that's
what you were trying to get at with the phrase "NO gravitational field
is experienced relative to the projectile", i.e., the projectile is
free-falling in whatever ambient gravitational field may be present.)
*Third point*:
Since we now have gravitational fields around, we need to think about
general relativity, and in this context the twin paradox takes a
particularly simple form: each twin travels along some (timelike)
worldline connecting events A and B, and (in general) each twin
measures a different proper time along her worldline. The general
relativity perspective is that there's no paradox here: you'd no more
expect to measure the same proper time along different worldlines
than you would to (say) measure the same distance along two different
driving routes between Paris and Moscow.
Notice that, in this general relativity perspective, we have NOT thus
far required that either worldline be a geodesic (i.e., that either
twin is in (locally-measured) free-fall throughout her travels). If
we want to require that the worldlines be geodesics, then providing
that A and B are "not too close"
[more precisely, providing that there *are* distinct
(timelike) geodesics from A to B, which there often
[usually? always? I'm not sure of
the precise conditions]
are if the spacetime is non-flat and A and B are far
enough apart]
it's still the case that in general each twin will measure a
different proper time along her worldline. In other words, we still
have the twin "paradox".
The general relativity "explanation" of the twin paradox in the
both-travellers-are-in-free-fall-throughout-their-journeys case is
also still the same, i.e., there's really no paradox here, this is
just how one should expect things to behave.
*Fourth Point*:
One way to think about the twin paradox in special relativity
[I think Taylor & Wheeler's special relativity book
describes this interpretation... but I'm several time
zones away from my copy of T&W right now, so I can't
easily confirm this or provide a precise reference.
The Wikipedia article
http://en.wikipedia.org/wiki/Twin_paradox
also describes this conceptualization.
interprets the time difference between the two twins being due in
large part to the change in the hypersurface-of-simultaneity between
the inertial reference frame the travelling twin is in in her
outbound journey, and the hypersurface-of-simultaneity of the
*different* inertial reference frame she's in on her return journey.
In *this* perspective, the turnaround is crucial for the proper-time
difference. It may be that this is what you (Chalky) meant by the
"the apparent paradox is resolved by the relativistic physics that
applies relative to the projectile during its direction reversal.".
But I wouldn't say that the turnaround -- or the physics that applies
during it -- *causes* the proper-time difference. The difference is
really a property of the two worldlines as a whole, not of this one
segment of one of them.
*Fifth point*:
In the general-relativity perspective the proper time is integrated
along each worldline, and the "turnaround" is no more or less important
than anywhere else along the worldline.
There's no contradiction between this and the special-relativity
perspective: both predict precisely the same observed results
(each observer's measured proper time at each point along her
worldline). In some sense the special-relativity perspective (with
the changing of inertial reference frames and hypersurfaces-of-simultaneity
during the turnaround) gives an additional insight into why the proper
times are as they are.
And finally.....
*Sixth point*:
I suspect Chalky and most other s.p.r readers already know this,
but for anyone who doesn't: There have been a number of direct
experimental tests of the twin paradox, by flying atomic clocks in
airplanes travelling around various closed paths. The results agree
precisely with the special/general relativity predictions.
ciao,
--
-- "Jonathan Thornburg [remove -animal to reply]" <jth...@astro.indiana-zebra.edu>
Dept of Astronomy & IUCSS, Indiana University, Bloomington, Indiana, USA
on sabbatical in Canada starting August 2012
"Washing one's hands of the conflict between the powerful and the
powerless means to side with the powerful, not to be neutral."
-- quote by Freire / poster by Oxfam
Mike Fontenot
Apr 14, 2013, 1:15:34 PM4/14/13
to
On 04/14/2013 04:24 AM, Jonathan Thornburg wrote:
>
> [...]
> describes this interpretation... [...]
explanation is in their Example 49 on pages 94-95 of their "Spacetime
Physics" book. That is also the explanation given by Brian Greene in
his "The Fabric of the Cosmos" NOVA TV show (and also in his book of the
same title). And it is the explanation that I give on my "CADO
equation" webpage:
https://sites.google.com/site/cadoequation/cado-reference-frame
and in my paper:
"Accelerated Observers in Special Relativity", PHYSICS ESSAYS,
December 1999, p629.
--
Mike Fontenot
>
> [...]
> *Fourth Point*:
>
> One way to think about the twin paradox in special relativity
> [I think Taylor& Wheeler's special relativity book
>
> One way to think about the twin paradox in special relativity
> describes this interpretation... [...]
> The Wikipedia article
> http://en.wikipedia.org/wiki/Twin_paradox
> also describes this conceptualization.
> interprets the time difference between the two twins being due in
> large part to the change in the hypersurface-of-simultaneity between
> the inertial reference frame the travelling twin is in in her
> outbound journey, and the hypersurface-of-simultaneity of the
> *different* inertial reference frame she's in on her return journey.
>
You are correct about Taylor & Wheeler's point of view: that
> http://en.wikipedia.org/wiki/Twin_paradox
> also describes this conceptualization.
> interprets the time difference between the two twins being due in
> large part to the change in the hypersurface-of-simultaneity between
> the inertial reference frame the travelling twin is in in her
> outbound journey, and the hypersurface-of-simultaneity of the
> *different* inertial reference frame she's in on her return journey.
>
explanation is in their Example 49 on pages 94-95 of their "Spacetime
Physics" book. That is also the explanation given by Brian Greene in
his "The Fabric of the Cosmos" NOVA TV show (and also in his book of the
same title). And it is the explanation that I give on my "CADO
equation" webpage:
https://sites.google.com/site/cadoequation/cado-reference-frame
and in my paper:
"Accelerated Observers in Special Relativity", PHYSICS ESSAYS,
December 1999, p629.
--
Mike Fontenot
Jonathan Thornburg [remove -animal to reply]
Apr 14, 2013, 2:42:32 PM4/14/13
to
In article <yg1bo9i...@ID-371.news.uni-berlin.de>,
We don't customarily do "me too" or "thank you" postings in s.p.r,
but in this case I'm going to make an exception:
Thank you, Oliver, for the clearest explanation of the twin paradox
I've read in a very long time. The opportunity to read articles like
this is a key reason why I (and I hope many others) participate in
this newsgroup!
Oliver Jennrich writes:
> Considering a triangle in Euclidean space (i.e. on a flat sheet of
> paper), no one would argue that the fact the path 'around the
> other corner' is longer than the straight way between two corners of the
> triangle is due to the details on how the test particle navigates its
> way around the corner. The distance via the third corner is longer
> because it is a different path than the direct one - this is intuitively
> evident to anyone who takes a look at a triangle.
>
> [[...]]
> Considering a triangle in Euclidean space (i.e. on a flat sheet of
> paper), no one would argue that the fact the path 'around the
> other corner' is longer than the straight way between two corners of the
> triangle is due to the details on how the test particle navigates its
> way around the corner. The distance via the third corner is longer
> because it is a different path than the direct one - this is intuitively
> evident to anyone who takes a look at a triangle.
>
We don't customarily do "me too" or "thank you" postings in s.p.r,
but in this case I'm going to make an exception:
Thank you, Oliver, for the clearest explanation of the twin paradox
I've read in a very long time. The opportunity to read articles like
this is a key reason why I (and I hope many others) participate in
this newsgroup!
Gregor Scholten
Apr 17, 2013, 2:37:34 AM4/17/13
to
Am 13.04.2013 22:55, schrieb Chalky:
> This brings us neatly back to my originally intended point. In the well
> known twin "paradox" situation, the apparent paradox is resolved by the
> relativistic physics that applies relative to the projectile during its
> direction reversal. That direction reversal is traditionally assumed to
> be caused by the firing of retro rockets on the projectile, thereby
> inducing a subjective gravitational field, relative to this projectile,
> during the period of acceleration.
within the framework of *special* relativity. Whereas this:
> This brings us neatly back to my originally intended point. In the well
> known twin "paradox" situation, the apparent paradox is resolved by the
> relativistic physics that applies relative to the projectile during its
> direction reversal. That direction reversal is traditionally assumed to
> be caused by the firing of retro rockets on the projectile, thereby
> inducing a subjective gravitational field, relative to this projectile,
> during the period of acceleration.
> However, it seems equally possible to
> me that such direction reversal can be achieved via gravitational
> slingshot at the end of the trajectory, whereupon NO gravitational field
> is experienced relative to the projectile during that period.
> Given that the outward and return legs of the journey can be made
> arbitrarily long, it strikes me that the physics during the period of
> direction reversal should remain essentially the same in both cases.
remain essentially the same in both cases, just the physics in general
is not essentially the same in both cases. There is an essential
difference in the physics: in the one case, physics is
special-relativistic, in the other case, physics is
general-relativistic. In general relativity, two free-falling observers
can meet twice, with having different proper time intervals elapsed
between the two encounters.
Phillip Helbig---undress to reply
Apr 18, 2013, 2:59:30 AM4/18/13
to
In article <kkkdf8$67v$1...@news.albasani.net>, Gregor Scholten
However, a better definition is that it describes the physics in
Minkowski space-time. Thus, it is often assumed that in special
relativity, while each sees the other twin aging more slowly, there is
no "real" effect since they can never meet up for a static face-to-face
comparison and that to discuss this one needs general relativity. You
seem to agree, though, that it is OK to discuss the rocket-firing
scenario within the context of special relativity, so they could meet up
for a face-to-face comparison. The question is whether the result is
different from the case of using general relativity.
> > However, it seems equally possible to
> > me that such direction reversal can be achieved via gravitational
> > slingshot at the end of the trajectory, whereupon NO gravitational field
> > is experienced relative to the projectile during that period.
>
> is an effect of gravitation, i.e. *general* relativity.
True, however...
> > Given that the outward and return legs of the journey can be made
> > arbitrarily long, it strikes me that the physics during the period of
> > direction reversal should remain essentially the same in both cases.
>
> then it strikes you wrongly. The physics during that period does not
> remain essentially the same in both cases, just the physics in general
> is not essentially the same in both cases. There is an essential
> difference in the physics: in the one case, physics is
> special-relativistic, in the other case, physics is
> general-relativistic. In general relativity, two free-falling observers
> can meet twice, with having different proper time intervals elapsed
> between the two encounters.
...while it is true that twins meeting up after having moved apart goes
beyond uniform unaccelerated motion, it can nevertheless be completely
handled within special relativity; nevertheless, as others in this
thread have pointed out, the resolution of the twin paradox is not that
general relativity is somehow essentially involved. It turns out that
one can discuss it within the framework of special relativity, and
resolve it correctly. I think the easiest way to see this is that the
effect does not depend on the details of the turnaround, be it realized
via gravitational slingshot or firing rockets, which becomes clear when
one considers that the magnitude of the effect depends only on the
distance, speed, time etc during the outward and inward legs of the
journey and not at all on the turnaround (which is essentially the same
for journeys of different lengths, durations or speeds).
In other words, if general relativity were somehow necessary for this
discussion, why does the effect depend only on the details of the
journey (length, duration, speed) during the time of uniform
unaccelerated motion? If the turnaround via a gravitational slingshot
(which one twin experiences and the other doesn't) were somehow
important, then why isn't the effect the same regardless of the length,
duration and speed of the rest of the journey.
Yes, the physics of firing rockets and the gravitational slingshot is
different, but not relevant to the discussion.
<g.sch...@gmx.de> writes:
> > This brings us neatly back to my originally intended point. In the well
> > known twin "paradox" situation, the apparent paradox is resolved by the
> > relativistic physics that applies relative to the projectile during its
> > direction reversal. That direction reversal is traditionally assumed to
> > be caused by the firing of retro rockets on the projectile, thereby
> > inducing a subjective gravitational field, relative to this projectile,
> > during the period of acceleration.
>
> within the framework of *special* relativity.
I have often read that special relativity does not include acceleration.
> > This brings us neatly back to my originally intended point. In the well
> > known twin "paradox" situation, the apparent paradox is resolved by the
> > relativistic physics that applies relative to the projectile during its
> > direction reversal. That direction reversal is traditionally assumed to
> > be caused by the firing of retro rockets on the projectile, thereby
> > inducing a subjective gravitational field, relative to this projectile,
> > during the period of acceleration.
>
> within the framework of *special* relativity.
However, a better definition is that it describes the physics in
Minkowski space-time. Thus, it is often assumed that in special
relativity, while each sees the other twin aging more slowly, there is
no "real" effect since they can never meet up for a static face-to-face
comparison and that to discuss this one needs general relativity. You
seem to agree, though, that it is OK to discuss the rocket-firing
scenario within the context of special relativity, so they could meet up
for a face-to-face comparison. The question is whether the result is
different from the case of using general relativity.
> > However, it seems equally possible to
> > me that such direction reversal can be achieved via gravitational
> > slingshot at the end of the trajectory, whereupon NO gravitational field
> > is experienced relative to the projectile during that period.
>
> is an effect of gravitation, i.e. *general* relativity.
> > Given that the outward and return legs of the journey can be made
> > arbitrarily long, it strikes me that the physics during the period of
> > direction reversal should remain essentially the same in both cases.
>
> then it strikes you wrongly. The physics during that period does not
> remain essentially the same in both cases, just the physics in general
> is not essentially the same in both cases. There is an essential
> difference in the physics: in the one case, physics is
> special-relativistic, in the other case, physics is
> general-relativistic. In general relativity, two free-falling observers
> can meet twice, with having different proper time intervals elapsed
> between the two encounters.
beyond uniform unaccelerated motion, it can nevertheless be completely
handled within special relativity; nevertheless, as others in this
thread have pointed out, the resolution of the twin paradox is not that
general relativity is somehow essentially involved. It turns out that
one can discuss it within the framework of special relativity, and
resolve it correctly. I think the easiest way to see this is that the
effect does not depend on the details of the turnaround, be it realized
via gravitational slingshot or firing rockets, which becomes clear when
one considers that the magnitude of the effect depends only on the
distance, speed, time etc during the outward and inward legs of the
journey and not at all on the turnaround (which is essentially the same
for journeys of different lengths, durations or speeds).
In other words, if general relativity were somehow necessary for this
discussion, why does the effect depend only on the details of the
journey (length, duration, speed) during the time of uniform
unaccelerated motion? If the turnaround via a gravitational slingshot
(which one twin experiences and the other doesn't) were somehow
important, then why isn't the effect the same regardless of the length,
duration and speed of the rest of the journey.
Yes, the physics of firing rockets and the gravitational slingshot is
different, but not relevant to the discussion.
Chalky
Apr 20, 2013, 7:13:37 AM4/20/13
to
On Sunday, 14 April 2013 11:24:06 UTC+1, Jonathan Thornburg wrote:
> From: "Jonathan Thornburg [remove -animal to reply]" <jth...@astro.indiana-zebra.edu>
> Subject: Re: SR & GR & "twin paradox"
> Newsgroups: sci.physics.research
> References: <fa92e855-6908-4afe...@googlegroups.com>
[snip]
> From: "Jonathan Thornburg [remove -animal to reply]" <jth...@astro.indiana-zebra.edu>
> Subject: Re: SR & GR & "twin paradox"
> Newsgroups: sci.physics.research
> References: <fa92e855-6908-4afe...@googlegroups.com>
> I think the very-short answer is that you're right.
respondents have been very reassuring. It has already been mentioned
that this query was based on a particularly early derivation of mine
and, after I had checked through all succeeding steps for logical
consistency, was the one remaining thing that I suspected may not have
been rigorously valid. However, that reservation has now evaporated.
Thanks
> As for the somewhat-longer answer.... I have a number of comments:
time that there is no time paradox to resolve in the twin paradox
situation, having worked that out for myself. There is, however,
something else of interest going on here, and it is the resolution of
that "something else" that I had in mind when I said "the apparent
paradox is resolved by the relativistic physics that applies relative to
the projectile during its direction reversal."
AFAIAC, the important point to come out of this discussion is that that
the projectile during its direction reversal."
relativistic physics remains essentially the same, irrespective of
whether that direction reversal is achieved by proper acceleration, or
is gravitationally induced.
Mike Fontenot
Apr 20, 2013, 7:15:02 AM4/20/13
to
On 04/18/2013 12:59 AM, Phillip Helbig---undress to reply wrote:
> The question is whether the result is
> different from the case of using general relativity.
>
No, the SR and GR results are NOT different.
> The question is whether the result is
> different from the case of using general relativity.
>
The SR result, given in Taylor & Wheeler's Example 49 on pages 94-95 of
their "Spacetime Physics" book, and in the explanation given by Brian
Greene in
his "The Fabric of the Cosmos" NOVA TV show, and on my "CADO
equation" webpage:
https://sites.google.com/site/cadoequation/cado-reference-frame
is the same as the GR result.
--
Mike Fontenot
Tom Roberts
Apr 20, 2013, 7:21:56 AM4/20/13
to
On 4/18/13 4/18/13 - 1:59 AM, Phillip Helbig---undress to reply wrote:
> I have often read that special relativity does not include acceleration.
That is just plain wrong. One merely needs calculus to handle
> I have often read that special relativity does not include acceleration.
accelerating objects and coordinates in SR. In particular, no additional
physics is required.
It can get complicated very fast, however, and one must keep
in mind the difference between Born rigid motion and common
expectations about accelerating objects.
> However, a better definition is that it describes the physics in
> Minkowski space-time.
> Thus, it is often assumed that in special
> relativity, while each sees the other twin aging more slowly, there is
> no "real" effect since they can never meet up for a static face-to-face
> comparison and that to discuss this one needs general relativity.
is: a given twin needs co-moving assistants positioned along the other
twin's trajectory to observe the other twin's ages multiple times
(multiple assistants), and communicate them back to the twin doing the
observing. So the traveling twin cannot be alone, she needs a bunch of
assistants in identical rockets, located far away from her.
And as Bell's paradox shows, it can be challenging to keep
those assistants at rest in the twin's instantaneously
co-moving inertial frame.
> You
> seem to agree, though, that it is OK to discuss the rocket-firing
> scenario within the context of special relativity, so they could meet up
> for a face-to-face comparison. The question is whether the result is
> different from the case of using general relativity.
spacetime) yields the same answer. Indeed, SR is just GR applied to
Minkowski spacetime, to this is a distinction without a difference.
In GR, one can consider the rocket-firing as an equivalent gravitational
field -- the other twin is very high up in this equivalent field, so the
other twin ages more than the accelerating twin during this phase of the
scenario. As in SR, one could consider several scenarios with identical
acceleration profiles but different-length inertial legs between them.
The age difference upon rejoining is related to the length of those
inertial legs, as expected (and proportional to their length if the
duration of the rocket firing can be neglected).
> ...while it is true that twins meeting up after having moved apart goes
> beyond uniform unaccelerated motion, it can nevertheless be completely
> handled within special relativity; nevertheless, as others in this
> thread have pointed out, the resolution of the twin paradox is not that
> general relativity is somehow essentially involved. It turns out that
> one can discuss it within the framework of special relativity, and
> resolve it correctly.
compare their elapsed proper times. In SR the metric is constant,
greatly simplifying the calculation.
In an inertial frame, the integral reduces to
\integral dt / \gamma
where t is the frame's time coordinate and \gamma is the
usual function of the moving object's speed (wrt the frame).
BTW this formula was obtained before GR and the use of a metric.
> I think the easiest way to see this is that the
> effect does not depend on the details of the turnaround, be it realized
> via gravitational slingshot or firing rockets, which becomes clear when
> one considers that the magnitude of the effect depends only on the
> distance, speed, time etc during the outward and inward legs of the
> journey and not at all on the turnaround (which is essentially the same
> for journeys of different lengths, durations or speeds).
of the traveling twin's total elapsed proper time, then the details can
matter. If one can neglect the traveling twin's elapsed proper time
during turnaround, compared to her total elapsed proper time, then the
details don't matter.
> Yes, the physics of firing rockets and the gravitational slingshot is
> different, but not relevant to the discussion.
calculations in GR are difficult and subtle.
Note it is also possible to construct rather counter-intuitive twin
scenarios which show that the result has nothing to do with
acceleration:
A) neither twin experiences any proper acceleration, but they
experience different elapsed proper times between meetings.
B) both twins experience nonzero proper acceleration, but the
one with the larger proper acceleration is the one who
experiences the LARGER elapsed proper time between meetings.
C) both twins experience nonzero proper acceleration, but the
one with the larger proper acceleration is the one who
experiences the SMALLER elapsed proper time between meetings.
A) (GR) put one twin in circular orbit around a massive object, and
put the other into a highly elliptical orbit, arranged so
they periodically meet. The elliptical orbit spends most of
its time higher than the circle, and this twin experiences more
elapsed proper time between meetings.
B) (SR) put the twins onto the rims of two circular rotating disks
whose centers are at rest in some inertial frame, whose rims are
tangential, such that they periodically meet. Vary the radii and
rotational speeds of the disks so the twin with the lower speed
(relative to the inertial frame) has the larger proper acceleration.
A third twin at rest in the inertial frame would experience even
larger elapsed proper time.
C) (SR) as (B) except with different radii and rotational speeds, so
the higher speed has higher proper acceleration.
Thus we see that the result is unrelated to the magnitude(s) of proper
acceleration. The only thing that matters is the integral of the metric
over each twin's path through spacetime. That can vary a lot, depending
on the details of the scenario.
Tom Roberts
Jos Bergervoet
Apr 20, 2013, 3:34:15 PM4/20/13
to
On 4/18/2013 8:59 AM, Phillip Helbig---undress to reply wrote:
..
use GR you get some extra time dilatation by
gravitational time delay. But this Shapiro time
delay presumably is a small fraction. *Almost*
the same would still hold. (And other things
might be slightly different, picking up a small
Lense-Thirring rotation for instance).
> Yes, the physics of firing rockets and the gravitational slingshot is
> different, but not relevant to the discussion.
At least not to the major part of the time
difference in the twin-experiment (do we have
to keep calling it a paradox?)
--
Jos
..
> In other words, if general relativity were somehow necessary for this
> discussion, why does the effect depend only on the details of the
> journey (length, duration, speed) during the time of uniform
> unaccelerated motion? If the turnaround via a gravitational slingshot
> (which one twin experiences and the other doesn't) were somehow
> important, then why isn't the effect the same regardless of the length,
> duration and speed of the rest of the journey.
It is of course not exactly the same.. if you
> discussion, why does the effect depend only on the details of the
> journey (length, duration, speed) during the time of uniform
> unaccelerated motion? If the turnaround via a gravitational slingshot
> (which one twin experiences and the other doesn't) were somehow
> important, then why isn't the effect the same regardless of the length,
> duration and speed of the rest of the journey.
use GR you get some extra time dilatation by
gravitational time delay. But this Shapiro time
delay presumably is a small fraction. *Almost*
the same would still hold. (And other things
might be slightly different, picking up a small
Lense-Thirring rotation for instance).
> Yes, the physics of firing rockets and the gravitational slingshot is
> different, but not relevant to the discussion.
difference in the twin-experiment (do we have
to keep calling it a paradox?)
--
Jos
Gregor Scholten
Apr 20, 2013, 3:35:10 PM4/20/13
to
Phillip Helbig wrote:
>>> This brings us neatly back to my originally intended point. In the well
>>> known twin "paradox" situation, the apparent paradox is resolved by the
>>> relativistic physics that applies relative to the projectile during its
>>> direction reversal. That direction reversal is traditionally assumed to
>>> be caused by the firing of retro rockets on the projectile, thereby
>>> inducing a subjective gravitational field, relative to this projectile,
>>> during the period of acceleration.
>>
>> within the framework of *special* relativity.
>
> I have often read that special relativity does not include acceleration.
> However, a better definition is that it describes the physics in
> Minkowski space-time.
from the historical point of view, SR originally dealed only with
>>> This brings us neatly back to my originally intended point. In the well
>>> known twin "paradox" situation, the apparent paradox is resolved by the
>>> relativistic physics that applies relative to the projectile during its
>>> direction reversal. That direction reversal is traditionally assumed to
>>> be caused by the firing of retro rockets on the projectile, thereby
>>> inducing a subjective gravitational field, relative to this projectile,
>>> during the period of acceleration.
>>
>> within the framework of *special* relativity.
>
> I have often read that special relativity does not include acceleration.
> However, a better definition is that it describes the physics in
> Minkowski space-time.
uniformly moving, i.e. non-accelerated, frames of reference. The
mathematical apparatus to describe accelerated frames of reference was
developed together with the mathematical formalism for GR. However, if
one considers curved spacetime, i.e. the presence of gravity, as the
domain of GR, accelerated frames in flat Minkowski spacetime turn out to
be within SR.
> Thus, it is often assumed that in special
> relativity, while each sees the other twin aging more slowly, there is
> no "real" effect since they can never meet up for a static face-to-face
> comparison and that to discuss this one needs general relativity. You
> seem to agree, though, that it is OK to discuss the rocket-firing
> scenario within the context of special relativity, so they could meet up
> for a face-to-face comparison.
> The question is whether the result is
> different from the case of using general relativity.
considered as paradoxon because due to the relativity principle of SR,
both twins should be equivalent as long as they are moving uniformly
(non-accelerated). So, any solution to the paradoxon needed something
that would break that equivalence.
In that case that one twin must fire his rocketes, it is the firing of
the rockets that breaks the equivalence: while rockets are fired, the
twin is no longer uniformly moving, but accelerated, so SR's relativity
principle no longer applies.
In general relativity, however, SR's relativity principle only holds in
the SR-limit, i.e. with gravitational fields and spacetime curvature
negligible. With the movement of one twin being re-directed by a
gravitational field, it is immediately clear that we are not in
SR-limit. Thus, it is immediately clear that there's no reason to
consider both twins as equivalent. So, in GR, there's no paradoxon at all.
>>> Given that the outward and return legs of the journey can be made
>>> arbitrarily long, it strikes me that the physics during the period of
>>> direction reversal should remain essentially the same in both cases.
>>
>> then it strikes you wrongly. The physics during that period does not
>> remain essentially the same in both cases, just the physics in general
>> is not essentially the same in both cases. There is an essential
>> difference in the physics: in the one case, physics is
>> special-relativistic, in the other case, physics is
>> general-relativistic. In general relativity, two free-falling observers
>> can meet twice, with having different proper time intervals elapsed
>> between the two encounters.
>
> ...while it is true that twins meeting up after having moved apart goes
> beyond uniform unaccelerated motion, it can nevertheless be completely
> handled within special relativity; nevertheless, as others in this
> thread have pointed out, the resolution of the twin paradox is not that
> general relativity is somehow essentially involved. It turns out that
> one can discuss it within the framework of special relativity, and
> resolve it correctly.
construction premises a non-equivalence between the two twins. It
premises that one can ascertain absolutely that the worldline of one
twin is straight-lined whereas the worldline of the other twin has a
buckle somewhere. This non-equivalence results either from the one twin
firing his rockets or from the one twin experiencing a slingshot by a
gravitational field, i.e. from being in GR instead of SR.
In the case of gravitational slingshot, one can assume flat Minkowski
metric everywhere for calculating the lengths of the worldlines of the
two twins, or the passing propor times respectively. However, the fact
that a gravitational slingshot has taken place proves that spacetime is
not really flat everywhere.
> I think the easiest way to see this is that the
> effect does not depend on the details of the turnaround,
details does not yield anything concerning this.
> In other words, if general relativity were somehow necessary for this
> discussion, why does the effect depend only on the details of the
> journey (length, duration, speed) during the time of uniform
> unaccelerated motion?
Additionally, in GR, "uniform unaccelerated motion" is not a proper
denotation for the periods of the journey outside the turnaround period.
In GR, the twin that experiences a gravitational slingshot is always
moving uniformly unaccelerated - in the sense of being free-falling -
even during turnaround phase, or never - in these sense that uniform
unaccelerated motion if not defined in curved spacetime.
> If the turnaround via a gravitational slingshot
> (which one twin experiences and the other doesn't) were somehow
> important, then why isn't the effect the same regardless of the length,
> duration and speed of the rest of the journey.
Phillip Helbig---undress to reply
Apr 20, 2013, 3:35:30 PM4/20/13
to
In article <qLmdnSPQtMF...@giganews.com>, Tom Roberts
confusion on this topic, though, does stem from books which state that
> > I think the easiest way to see this is that the
> > effect does not depend on the details of the turnaround, be it realized
> > via gravitational slingshot or firing rockets, which becomes clear when
> > one considers that the magnitude of the effect depends only on the
> > distance, speed, time etc during the outward and inward legs of the
> > journey and not at all on the turnaround (which is essentially the same
> > for journeys of different lengths, durations or speeds).
>
> This is not quite true. If the turnaround takes a significant fraction
> of the traveling twin's total elapsed proper time, then the details can
> matter. If one can neglect the traveling twin's elapsed proper time
> during turnaround, compared to her total elapsed proper time, then the
> details don't matter.
Right; I should have said "provided the time for rocket-firing is
negligible".
<tjrobe...@sbcglobal.net> writes:
> On 4/18/13 4/18/13 - 1:59 AM, Phillip Helbig---undress to reply wrote:
> > I have often read that special relativity does not include acceleration.
>
> That is just plain wrong. One merely needs calculus to handle
> accelerating objects and coordinates in SR. In particular, no additional
> physics is required.
Just to be clear, I agree with what you say in this post. Much
> On 4/18/13 4/18/13 - 1:59 AM, Phillip Helbig---undress to reply wrote:
> > I have often read that special relativity does not include acceleration.
>
> That is just plain wrong. One merely needs calculus to handle
> accelerating objects and coordinates in SR. In particular, no additional
> physics is required.
confusion on this topic, though, does stem from books which state that
special relativity does not include acceleration.
> The age difference upon rejoining is related to the length of those
> inertial legs, as expected (and proportional to their length if the
> duration of the rocket firing can be neglected).
Right; that is the main point.
> inertial legs, as expected (and proportional to their length if the
> duration of the rocket firing can be neglected).
> > I think the easiest way to see this is that the
> > effect does not depend on the details of the turnaround, be it realized
> > via gravitational slingshot or firing rockets, which becomes clear when
> > one considers that the magnitude of the effect depends only on the
> > distance, speed, time etc during the outward and inward legs of the
> > journey and not at all on the turnaround (which is essentially the same
> > for journeys of different lengths, durations or speeds).
>
> This is not quite true. If the turnaround takes a significant fraction
> of the traveling twin's total elapsed proper time, then the details can
> matter. If one can neglect the traveling twin's elapsed proper time
> during turnaround, compared to her total elapsed proper time, then the
> details don't matter.
negligible".
Tom Roberts
Apr 22, 2013, 1:57:01 PM4/22/13
to
On 4/20/13 4/20/13 2:35 PM, Gregor Scholten wrote:
GR. That includes acceleration. But I believe that the theory was not developed
to fully include acceleration until after the tools of GR were available.
>> It turns out that
>> one can discuss it within the framework of special relativity, and
>> resolve it correctly.
>
> you mean Oliver Jennrich's triangle construction?
As long as the scenario occurs in flat spacetime, SR is sufficient. For the
gravitational slingshot, GR is of course required to obtain a correct answer.
> However, this
> construction premises a non-equivalence between the two twins. It
> premises that one can ascertain absolutely that the worldline of one
> twin is straight-lined whereas the worldline of the other twin has a
> buckle somewhere. This non-equivalence results either from the one twin
> firing his rockets or from the one twin experiencing a slingshot by a
> gravitational field, i.e. from being in GR instead of SR.
Each twin can also determine whether she is in inertial motion, independent of
the other twin or any external observations. In the flat manifold of SR, it is
not possible for the twins to meet twice and both be moving inertially; in the
curved manifolds of GR it is possible.
> In the case of gravitational slingshot, one can assume flat Minkowski
> metric everywhere for calculating the lengths of the worldlines of the
> two twins, or the passing propor times respectively.
That can only give an approximate value, not the correct value.
> Additionally, in GR, "uniform unaccelerated motion" is not a proper
> denotation for the periods of the journey outside the turnaround period.
Yes. "Unaccelerated" applies throughout each twin's journey (each twin has zero
proper acceleration), but I have no idea what "uniform" could mean in GR, for
the general case -- there's no suitable reference ("uniform" is necessarily a
comparison to something -- in Newton's first law it is relative to the absolute
space, which is not part of GR).
Tom Roberts
> Phillip Helbig wrote:
>> I have often read that special relativity does not include acceleration.
>> However, a better definition is that it describes the physics in
>> Minkowski space-time.
>
> from the historical point of view, SR originally dealed only with
> uniformly moving, i.e. non-accelerated, frames of reference. The
> mathematical apparatus to describe accelerated frames of reference was
> developed together with the mathematical formalism for GR.
Max Born wrote about what is now known as Born rigid motion in 1909, well before
>> I have often read that special relativity does not include acceleration.
>> However, a better definition is that it describes the physics in
>> Minkowski space-time.
>
> from the historical point of view, SR originally dealed only with
> uniformly moving, i.e. non-accelerated, frames of reference. The
> mathematical apparatus to describe accelerated frames of reference was
> developed together with the mathematical formalism for GR.
GR. That includes acceleration. But I believe that the theory was not developed
to fully include acceleration until after the tools of GR were available.
>> It turns out that
>> one can discuss it within the framework of special relativity, and
>> resolve it correctly.
>
> you mean Oliver Jennrich's triangle construction?
gravitational slingshot, GR is of course required to obtain a correct answer.
> However, this
> construction premises a non-equivalence between the two twins. It
> premises that one can ascertain absolutely that the worldline of one
> twin is straight-lined whereas the worldline of the other twin has a
> buckle somewhere. This non-equivalence results either from the one twin
> firing his rockets or from the one twin experiencing a slingshot by a
> gravitational field, i.e. from being in GR instead of SR.
the other twin or any external observations. In the flat manifold of SR, it is
not possible for the twins to meet twice and both be moving inertially; in the
curved manifolds of GR it is possible.
> In the case of gravitational slingshot, one can assume flat Minkowski
> metric everywhere for calculating the lengths of the worldlines of the
> two twins, or the passing propor times respectively.
> Additionally, in GR, "uniform unaccelerated motion" is not a proper
> denotation for the periods of the journey outside the turnaround period.
proper acceleration), but I have no idea what "uniform" could mean in GR, for
the general case -- there's no suitable reference ("uniform" is necessarily a
comparison to something -- in Newton's first law it is relative to the absolute
space, which is not part of GR).
Tom Roberts
Gregor Scholten
Apr 22, 2013, 1:57:43 PM4/22/13
to
Tom Roberts wrote:
> Remember how unnatural and artificial this "aging more slowly" actually
> is: a given twin needs co-moving assistants positioned along the other
> twin's trajectory to observe the other twin's ages multiple times
> (multiple assistants), and communicate them back to the twin doing the
> observing. So the traveling twin cannot be alone, she needs a bunch of
> assistants in identical rockets, located far away from her.
the observation of the slow aging of the other twin could also be
> Remember how unnatural and artificial this "aging more slowly" actually
> is: a given twin needs co-moving assistants positioned along the other
> twin's trajectory to observe the other twin's ages multiple times
> (multiple assistants), and communicate them back to the twin doing the
> observing. So the traveling twin cannot be alone, she needs a bunch of
> assistants in identical rockets, located far away from her.
realized in a different way: each twin could observe the other twin
using ligth signals emitted from the other twin. Of course, this does
not directly yield the slower aging caused by time dilatation because it
is additionally affected by effects based on light travelling time, but
each twin can do a correction calculation to dismiss those effects from
his observations if he knows the velocity of the other twin, which he
can detect from redshift.
So, the slower aging of the other twin due to time dilatation is indeed
observable for each twin.
Nicolaas Vroom
Apr 22, 2013, 2:00:38 PM4/22/13
to
Op zaterdag 20 april 2013 21:35:30 UTC+2 schreef Phillip Helbig---undress to reply het volgende:
and energy) the most important parameter mentioned is v/c.
The book Introducing Einstein's Relativity by Ray d'Inverno at page 24
mentions acceleration.
Interested people should study the part after the text:
"Some opponents of SR have argued that the short period of acceleration
should not make such a difference, " etc
The problem is when you study different speeds always acceleration
is involved.
During the acceleration phase the clock starts to tick slower. This rate
is maintained when the speed is constant. During the deceleration
phase the clock rate increases, but the clock time lacks behind.
For length contraction almost the same situation exists. The main
difference is that after deceleration the length returns back to
normal.
> > The age difference upon rejoining is related to the length of those
> > inertial legs, as expected (and proportional to their length if the
> > duration of the rocket firing can be neglected).
>
> Right; that is the main point.
>
Before the above mentioned quote is written:
"But in this case the clock would have to be accelearated when being
transferred to C and so it is no longer inertial."
> > > I think the easiest way to see this is that the
> > > effect does not depend on the details of the turnaround, etc
experiment that the formula T' = T * sqr(1-v^2/c^2) is correct.
You could argue that performing such an experiment with an airplane
(earth based) where gravity is involved and which for allmost
100% consists of turning around (no straight leg is included)
is not very convincing.
Nicolaas Vroom
http://users.pandora.be/nicvroom/
> In article <qLmdnSPQtMF...@giganews.com>, Tom Roberts
> <tjrobe...@sbcglobal.net> writes:
> >
> > That is just plain wrong. One merely needs calculus to handle
> > accelerating objects and coordinates in SR. In particular, no additional
> > physics is required.
>
> Just to be clear, I agree with what you say in this post. Much
> confusion on this topic, though, does stem from books which state that
> <tjrobe...@sbcglobal.net> writes:
> >
> > That is just plain wrong. One merely needs calculus to handle
> > accelerating objects and coordinates in SR. In particular, no additional
> > physics is required.
>
> Just to be clear, I agree with what you say in this post. Much
> special relativity does not include acceleration.
When you study SR (length contraction, time dilation and relativistic mass
and energy) the most important parameter mentioned is v/c.
The book Introducing Einstein's Relativity by Ray d'Inverno at page 24
mentions acceleration.
Interested people should study the part after the text:
"Some opponents of SR have argued that the short period of acceleration
should not make such a difference, " etc
The problem is when you study different speeds always acceleration
is involved.
During the acceleration phase the clock starts to tick slower. This rate
is maintained when the speed is constant. During the deceleration
phase the clock rate increases, but the clock time lacks behind.
For length contraction almost the same situation exists. The main
difference is that after deceleration the length returns back to
normal.
> > The age difference upon rejoining is related to the length of those
> > inertial legs, as expected (and proportional to their length if the
> > duration of the rocket firing can be neglected).
>
> Right; that is the main point.
>
"But in this case the clock would have to be accelearated when being
transferred to C and so it is no longer inertial."
> > > I think the easiest way to see this is that the
> > This is not quite true. If the turnaround takes a significant fraction
> > of the traveling twin's total elapsed proper time, then the details can
> > matter. If one can neglect the traveling twin's elapsed proper time
> > during turnaround, compared to her total elapsed proper time, then the
> > details don't matter.
>
> Right; I should have said "provided the time for rocket-firing is
> negligible".
The major problem with time dilation is to demonstrate with a real
> > of the traveling twin's total elapsed proper time, then the details can
> > matter. If one can neglect the traveling twin's elapsed proper time
> > during turnaround, compared to her total elapsed proper time, then the
> > details don't matter.
>
> Right; I should have said "provided the time for rocket-firing is
> negligible".
experiment that the formula T' = T * sqr(1-v^2/c^2) is correct.
You could argue that performing such an experiment with an airplane
(earth based) where gravity is involved and which for allmost
100% consists of turning around (no straight leg is included)
is not very convincing.
Nicolaas Vroom
http://users.pandora.be/nicvroom/
Tom Roberts
Apr 23, 2013, 1:15:33 AM4/23/13
to
On 4/22/13 4/22/13 1:00 PM, Nicolaas Vroom wrote:
> During the acceleration phase the clock starts to tick slower.
This is not true. The clock ALWAYS ticks at its usual rate, and acceleration
> During the acceleration phase the clock starts to tick slower.
does not change this. Note that your language implies you are talking about the
clock AND NOTHING ELSE, which in relativity means that you must observe it from
its rest frame.
If one observes the clock from the inertial frame from which it began, then one
will OBSERVE it to start to tick more slowly as it accelerates. There is, of
course, nothing affecting the clock itself, it is just your method of
observation that makes this happen.
When you watch a friend walk away, she appears to get smaller,
but does not really do so. This is the same sort of effect,
but based on relative velocity, not distance.
> This rate
> is maintained when the speed is constant. During the deceleration
> phase the clock rate increases, but the clock time lacks behind.
clock AND NOTHING ELSE.
> For length contraction almost the same situation exists. The main
> difference is that after deceleration the length returns back to
> normal.
clock rate. Note that the length of a ruler is a differential measurement, and
the corresponding differential measurement of a clock is the time between ticks,
not its indicated time. Clocks naturally integrate their tick intervals, while
rulers do not; this is merely an historical aspect of these devices.
Consider an odometer, which does integrate....
> The major problem with time dilation is to demonstrate with a real
> experiment that the formula T' = T * sqr(1-v^2/c^2) is correct.
HEP experiments around the world, and has been for > 50 years.
For many references, see
http://math.ucr.edu/home/baez/physics/Relativity/SR/experiments.html
> You could argue that performing such an experiment with an airplane
> (earth based) where gravity is involved and which for allmost
> 100% consists of turning around (no straight leg is included)
> is not very convincing.
With your emphasis on "straight leg", you seem to be hung up on testing SR.
That's silly, as SR is merely the local limit of GR. One should test this aspect
of GR, and airplanes carrying atomic clocks can do that quite well (and have
done so several times). Indeed, even AUTOMOBILES carrying atomic clocks can do
it in a few hours; one could arrange their paths so altitude changes are
negligible.
IIRC Alley et al's measurement included an airplane flying out
and back, with minimal altitude changes and short turn-arounds.
Of course the earth was rotating, but that applies to both clocks.
Tom Roberts
[[Mod. note -- In the airplane experiments described in
@inbook
{
Alley-1983-GR-rods-and-clocks,
author = "Carroll O. Alley",
title = "Proper Time Experiments in Gravitational Fields
With Atomic Clocks, Aircraft, and Laser Light Pulses",
pages = "421--424",
editor = "Pierre Meystre and Marlan O. Scully",
booktitle = "Quantum Optics, Experimental Gravitation,
and Measurement Theory",
publisher = "Plenum",
address = "New York",
year = 1983,
snote = "++good discussion on `rods and clocks' experiments
testing special/general relativity",
}
the airplane flew repeatedly around a "racetrack" shape (generally
oval but with straight sides) over Chesapeake Bay. The apparatus
included multiple atomic clocks both on the ground and on the airplane,
and a laser link for real-time clock comparisons while the aircraft
was in flight. Alas, I don't know of an online version of this
article.
-- jt]]
Nicolaas Vroom
Apr 23, 2013, 5:16:28 PM4/23/13
to
Op dinsdag 23 april 2013 07:15:33 UTC+2 schreef Tom Roberts het volgende:
travelling two straight legs with an allmost "constant" speed) after two
years and the elapsed time on the clock shows a smaller value, then IMO
the clock in this spaceship did tick physical at a slower rate. If this
assumption is correct then the next question is when did this clock
started to tick slower ? IMO immediate during the acceleration phase. If
the captain on board the spaceship wants to return he has to stop the
engines, "change engine direction" and starts the engines for the return
leg. (other return scenarios are possible) During deceleration the clock
rate increases back to normal. During return leg the clock rate changes
in the same pattern as in out going leg. The captain, if he has no
communication with back home, will not realise this during the flight.
> If one observes the clock from the inertial frame from which it began,
> then one
> will OBSERVE it to start to tick more slowly as it accelerates.
IMO an observer back home (assuming he knows at which speed the space
ship travels) will immediate know when the first clock pulse arrives
from the spaceship that the clock on board the space ship ticks slower.
> There is, of
> course, nothing affecting the clock itself, it is just your method of
> observation that makes this happen.
My understanding of this experiment that the elapsed time of the moving
clock is smaller then the elapsed time of clock at "rest"
> > The major problem with time dilation is to demonstrate with a real
> > experiment that the formula T' = T * sqr(1-v^2/c^2) is correct.
above equation then it is correct to raise the question how do you prove
that equation only by using SR. If the answer is NO than "you" should
write as such.
>
> [[Mod. note -- In the airplane experiments described in
>
half-rounds and two straight legs. The half-rounds represents the turn
around points.
Tom Roberts wrote in the past:
significant fraction making the physical interpretation what is caused
by GR (gravitation) and what is caused by SR more complicated.
Nicolaas Vroom
http://users.pandora.be/nicvroom/program4.htm
> On 4/22/13 4/22/13 1:00 PM, Nicolaas Vroom wrote:
>
> > During the acceleration phase the clock starts to tick slower.
>
> This is not true. The clock ALWAYS ticks at its usual rate, and acceleration
> does not change this.
If you put a clock in a spaceship and this spaceship comes back (after
>
> > During the acceleration phase the clock starts to tick slower.
>
> This is not true. The clock ALWAYS ticks at its usual rate, and acceleration
> does not change this.
travelling two straight legs with an allmost "constant" speed) after two
years and the elapsed time on the clock shows a smaller value, then IMO
the clock in this spaceship did tick physical at a slower rate. If this
assumption is correct then the next question is when did this clock
started to tick slower ? IMO immediate during the acceleration phase. If
the captain on board the spaceship wants to return he has to stop the
engines, "change engine direction" and starts the engines for the return
leg. (other return scenarios are possible) During deceleration the clock
rate increases back to normal. During return leg the clock rate changes
in the same pattern as in out going leg. The captain, if he has no
communication with back home, will not realise this during the flight.
> If one observes the clock from the inertial frame from which it began,
> then one
> will OBSERVE it to start to tick more slowly as it accelerates.
ship travels) will immediate know when the first clock pulse arrives
from the spaceship that the clock on board the space ship ticks slower.
> There is, of
> course, nothing affecting the clock itself, it is just your method of
> observation that makes this happen.
clock is smaller then the elapsed time of clock at "rest"
> > The major problem with time dilation is to demonstrate with a real
> > experiment that the formula T' = T * sqr(1-v^2/c^2) is correct.
> With your emphasis on "straight leg", you seem to be hung up on testing SR.
> That's silly, as SR is merely the local limit of GR.
I fully agree with you. However if you read a book SR and you find the
> That's silly, as SR is merely the local limit of GR.
above equation then it is correct to raise the question how do you prove
that equation only by using SR. If the answer is NO than "you" should
write as such.
>
> [[Mod. note -- In the airplane experiments described in
>
> title = "Proper Time Experiments in Gravitational Fields
> With Atomic Clocks, Aircraft, and Laser Light Pulses",
>
> With Atomic Clocks, Aircraft, and Laser Light Pulses",
>
> the airplane flew repeatedly around a "racetrack" shape (generally
> oval but with straight sides) over Chesapeake Bay. The apparatus
> included multiple atomic clocks both on the ground and on the airplane,
> and a laser link for real-time clock comparisons while the aircraft
> was in flight.
When you study a horse racetrack then the track consists of two
> oval but with straight sides) over Chesapeake Bay. The apparatus
> included multiple atomic clocks both on the ground and on the airplane,
> and a laser link for real-time clock comparisons while the aircraft
> was in flight.
half-rounds and two straight legs. The half-rounds represents the turn
around points.
Tom Roberts wrote in the past:
> This is not quite true. If the turnaround takes a significant fraction
> of the traveling twin's total elapsed proper time, then the details can
> matter. If one can neglect the traveling twin's elapsed proper time
> during turnaround, compared to her total elapsed proper time, then the
> details don't matter.
Answer Phillip:
> of the traveling twin's total elapsed proper time, then the details can
> matter. If one can neglect the traveling twin's elapsed proper time
> during turnaround, compared to her total elapsed proper time, then the
> details don't matter.
Right; I should have said "provided the time for rocket-firing is
negligible".
The problem is when you study a racetrack the turnaround can take a
negligible".
significant fraction making the physical interpretation what is caused
by GR (gravitation) and what is caused by SR more complicated.
Nicolaas Vroom
http://users.pandora.be/nicvroom/program4.htm
Nicolaas Vroom
Apr 23, 2013, 5:18:09 PM4/23/13
to
Op zondag 14 april 2013 20:42:32 UTC+2 schreef Jonathan Thornburg [remove -
animal to reply] het volgende:
> Thank you, Oliver, for the clearest explanation of the twin paradox
> I've read in a very long time. The opportunity to read articles like
> this is a key reason why I (and I hope many others) participate in
> this newsgroup!
(I fully agree)
Starting point in the Twin Paradox discussion is a unambiguous
definition of all the concepts used. Concepts like spacetime, paradox,
acceleration and thought experiment should be used with great care.
Science is based on experiments. The more simple the better.
1. In order to explain the problems involved with travelling let us
first study ships in 1492. Our plan is to travel from Lisbon to Cuba
with two ships. We use three pendulum clocks. One stays at the admiralty
in Lisbon and one on each ship. After 20 days (half way) one ship
returns back home. After 40 days the second ship reaches Cuba and
returns back home. When all the ships are back in Lisbon we compare the
time. The elapsed time in days for the three clocks are: 40, 38 and 36.
The explanation is the time at sea and in the quality of the clocks. How
longer the clock is at sea the slower the clocks ticks caused by the
movement of the ships. When we repeat the same experiment using better
clocks we get: 40, 40 and 40. Meaning that the movement of the clocks at
sea does not affect the clock rate, which is constant. See also:
http://en.wikipedia.org/wiki/Marine_chronometer#Today In this particular
section you can read that clocks are synchronized by means of GPS
signals. The fact that this is required means that clocks at sea are
still not 100% perfect.
2. Next we are going to repeat the same experiment using two spaceships
and again three clocks. One clock stays at home. The speed of both
spaceships is identical 0.5*c and they both fly in the same direction
(towards the same star). In order to take care that the spaceships
return, at home two clock pulses are transmitted, one after one year and
a second one after two years . When the first space ship receives the
first pulse (after 2 years) he returns and when the second spaceship
receives the second pulse (after 4 years) he returns. After 4 years the
first space ship returns , however his clock shows 3.5 years. After 8
years the second space ship returns. His clock shows 7 years.
The same phenomena we see as before: moving clocks run slower. Comparing
the three clocks after 8 years we get the following 3 values ( elapsed
times) : 8, 7.5 and 7 (approximate). You can repeat the same experiment
using a different speed. With v=0.9*c we get 40,28 and 16 years and with
v=0.99*c we get: 400,228 and 56 years. This means if you start this
experiment as a baby you will be 56 years when you return but your twin
brother who stayed at home
(if still alive) should be 400 years. When you compare the two problems
there is one major difference: The first problem is a physical problem,
the second problem is primary a thought experiment or a mathematical
problem based on SR. The basic parameter is the speed of light c.
3. Let us study the difficulties using light in more detail. Consider a
person A at a certain position who transmits a light signal. At the same
time a second person B passes A and also transmits a light signal.
Consider a third person C a distance away. The question is will C
observe both signals simultaneous. Assume that the answer is Yes. That
means the speed of light is independent of the speed of the source. You
can also argue is independent of both source and destination. (A
different issue is the frequency measured)
4. However there is one more important issue. When you study both
examples they are both round trips. As such you calculate are average
values i.e. average speeds based on the final elapsed times. That is not
really what we want. What you want are the true speeds. To calculate
that is not so difficult because the distance travelled is based on
fixed points and a fixed distance (within margins). In the space ship
example this is not the case: the starting point, return point and
finish are moving points in space. If the starting point and finish
point move towards the point of return than your return trip will be the
shortest. IMO this same issue could be part of the Pioneer anomaly. See
http://en.wikipedia.org/wiki/Pioneer_anomaly
A more important problem is your speed relative to the speed of light.
How do you know that your speed is 1% of the speed of light? Suppose
that your speed is higher that a spaceship in front of you. Using your
"brakes" you reduce your speed that the speed becomes equal (and the
distance stays equal). Are you in agreement with the other captain what
your speed is (relative to the speed of light)? What experiments are
available to test this? Of course physical identical clocks on both
space ships will tick at the same rate.
All in all the physics of moving clocks is a very interesting issue.
More water should flow through the Rhine until everything is solved.
When you study http://en.wikipedia.org/wiki/Twin_paradox you can read
that there is a difference between twins and identical twins.
Nicolaas Vroom
http://users.pandora.be/nicvroom/
animal to reply] het volgende:
> Thank you, Oliver, for the clearest explanation of the twin paradox
> I've read in a very long time. The opportunity to read articles like
> this is a key reason why I (and I hope many others) participate in
> this newsgroup!
Starting point in the Twin Paradox discussion is a unambiguous
definition of all the concepts used. Concepts like spacetime, paradox,
acceleration and thought experiment should be used with great care.
Science is based on experiments. The more simple the better.
1. In order to explain the problems involved with travelling let us
first study ships in 1492. Our plan is to travel from Lisbon to Cuba
with two ships. We use three pendulum clocks. One stays at the admiralty
in Lisbon and one on each ship. After 20 days (half way) one ship
returns back home. After 40 days the second ship reaches Cuba and
returns back home. When all the ships are back in Lisbon we compare the
time. The elapsed time in days for the three clocks are: 40, 38 and 36.
The explanation is the time at sea and in the quality of the clocks. How
longer the clock is at sea the slower the clocks ticks caused by the
movement of the ships. When we repeat the same experiment using better
clocks we get: 40, 40 and 40. Meaning that the movement of the clocks at
sea does not affect the clock rate, which is constant. See also:
http://en.wikipedia.org/wiki/Marine_chronometer#Today In this particular
section you can read that clocks are synchronized by means of GPS
signals. The fact that this is required means that clocks at sea are
still not 100% perfect.
2. Next we are going to repeat the same experiment using two spaceships
and again three clocks. One clock stays at home. The speed of both
spaceships is identical 0.5*c and they both fly in the same direction
(towards the same star). In order to take care that the spaceships
return, at home two clock pulses are transmitted, one after one year and
a second one after two years . When the first space ship receives the
first pulse (after 2 years) he returns and when the second spaceship
receives the second pulse (after 4 years) he returns. After 4 years the
first space ship returns , however his clock shows 3.5 years. After 8
years the second space ship returns. His clock shows 7 years.
The same phenomena we see as before: moving clocks run slower. Comparing
the three clocks after 8 years we get the following 3 values ( elapsed
times) : 8, 7.5 and 7 (approximate). You can repeat the same experiment
using a different speed. With v=0.9*c we get 40,28 and 16 years and with
v=0.99*c we get: 400,228 and 56 years. This means if you start this
experiment as a baby you will be 56 years when you return but your twin
brother who stayed at home
(if still alive) should be 400 years. When you compare the two problems
there is one major difference: The first problem is a physical problem,
the second problem is primary a thought experiment or a mathematical
problem based on SR. The basic parameter is the speed of light c.
3. Let us study the difficulties using light in more detail. Consider a
person A at a certain position who transmits a light signal. At the same
time a second person B passes A and also transmits a light signal.
Consider a third person C a distance away. The question is will C
observe both signals simultaneous. Assume that the answer is Yes. That
means the speed of light is independent of the speed of the source. You
can also argue is independent of both source and destination. (A
different issue is the frequency measured)
4. However there is one more important issue. When you study both
examples they are both round trips. As such you calculate are average
values i.e. average speeds based on the final elapsed times. That is not
really what we want. What you want are the true speeds. To calculate
that is not so difficult because the distance travelled is based on
fixed points and a fixed distance (within margins). In the space ship
example this is not the case: the starting point, return point and
finish are moving points in space. If the starting point and finish
point move towards the point of return than your return trip will be the
shortest. IMO this same issue could be part of the Pioneer anomaly. See
http://en.wikipedia.org/wiki/Pioneer_anomaly
A more important problem is your speed relative to the speed of light.
How do you know that your speed is 1% of the speed of light? Suppose
that your speed is higher that a spaceship in front of you. Using your
"brakes" you reduce your speed that the speed becomes equal (and the
distance stays equal). Are you in agreement with the other captain what
your speed is (relative to the speed of light)? What experiments are
available to test this? Of course physical identical clocks on both
space ships will tick at the same rate.
All in all the physics of moving clocks is a very interesting issue.
More water should flow through the Rhine until everything is solved.
When you study http://en.wikipedia.org/wiki/Twin_paradox you can read
that there is a difference between twins and identical twins.
Nicolaas Vroom
http://users.pandora.be/nicvroom/
Gregor Scholten
Apr 24, 2013, 1:21:35 AM4/24/13
to
Nicolaas Vroom wrote:
>>> During the acceleration phase the clock starts to tick slower.
>>
>> This is not true. The clock ALWAYS ticks at its usual rate, and acceleration
>> does not change this.
>
> If you put a clock in a spaceship and this spaceship comes back (after
> travelling two straight legs with an allmost "constant" speed) after two
> years and the elapsed time on the clock shows a smaller value, then IMO
> the clock in this spaceship did tick physical at a slower rate. If this
> assumption is correct then the next question is when did this clock
> started to tick slower ?
the answer to this question depends on the frame of reference on that
>>> During the acceleration phase the clock starts to tick slower.
>>
>> This is not true. The clock ALWAYS ticks at its usual rate, and acceleration
>> does not change this.
>
> If you put a clock in a spaceship and this spaceship comes back (after
> travelling two straight legs with an allmost "constant" speed) after two
> years and the elapsed time on the clock shows a smaller value, then IMO
> the clock in this spaceship did tick physical at a slower rate. If this
> assumption is correct then the next question is when did this clock
> started to tick slower ?
you base your consideration. If you take the rest frame of the resting
twin (the twin that is uniformly moving during the whole situation), you
are right when you say:
> IMO immediate during the acceleration phase.
twin accelerates away from the resting twin.
If, however, you take the inertial frame, in which the traveling twin is
resting during the first uniformly moving segment of his travel (after
he stopped accelarating away from the resting twin and before he starts
his deceleration for turnaround), then the traveling twin has his clock
ticking faster that the resting twin's clock during this first segment,
because seen from this frame, the resting's clock is time-delated. The
traveling twin's clock starts to slow down in ticking when this first
segement is over, i.e. when the traveling twin starts his deceleration
phase.
If you take the (accelerated) rest frame of the traveling twin, the
traveling twin's clock always ticks with the same speed. During the two
segments of uniform movement, the resting twin's clock is ticking
slower. During the turnaround phase, the resting twin's clock is ticking
much faster.
It is useful to regard the relativity of simultaneity:
http://en.wikipedia.org/wiki/Relativity_of_simultaneity
The traveling twin's space of simultaneity is rotated against the
resting twin's space of simultaneity during the two segments of the
traveling twin's uniform movement. During the traveling twin's
accelaration phases, his space of simultaneity is rotating.
The matter of fact is, that for answering your question, you need a
procedure to compare spatially seperated clocks, i.e. a mean of
simultaneity (the one twin's clock indicates a clocktime, and *at the
same time* the other twin's clock indicates another clocktime). But in
SR, simultaneity depends on the frame of reference.
In GR, the situation is even worse: the concept of frame of reference
breaks down, except in SR-limit, instead, you can only use the concept
of coordinate system, which is much more arbitraray than SR's frames of
reference. That implies that in GR, even after choosing an observer (a
frame of reference in SR) the mean of simultaneity is still not fixed.
Only in the situation that the two twins meet twice, you can definitely
say whose clock has more proper time elapsed between both encounters.
That is, because the two clocks are not spatially seperated at the
encounters.
>> If one observes the clock from the inertial frame from which it began,
>> then one
>> will OBSERVE it to start to tick more slowly as it accelerates.
>
> IMO an observer back home (assuming he knows at which speed the space
> ship travels) will immediate know when the first clock pulse arrives
> from the spaceship that the clock on board the space ship ticks slower.
either mean that ones measures the times when luminal signals arrive the
observer, or that one additionally does a correction calculation to
eliminate the effects caused by light travelling time.
Imagine a clock that moves away from you. On the one hand, the clock
ticks slower than your own clock due to time dilation. However, if you
watch luminal signals emitted by this moving clock every tick, then the
time difference between the arrivals of two of those luminal signals is
even longer than the increased time interval between two ticks due to
time dilation. Because between two ticks, the distance between the clock
and you increases, causing a longer path for the second luminal signal
to travel, and thus, a longer travelling time.
To observe the correct time dilation for the clock moving aways from
you, you need to apply a correction calculation to your observation, to
compensate this effect.
>> There is, of
>> course, nothing affecting the clock itself, it is just your method of
>> observation that makes this happen.
>
> My understanding of this experiment that the elapsed time of the moving
> clock is smaller then the elapsed time of clock at "rest"
wich clock is moving and which is at rest depends on the frame of
reference, as long as only uniform movement (no acceleration) is considered.
Jonathan Thornburg [remove -animal to reply]
Apr 24, 2013, 1:21:56 AM4/24/13
to
Nicolaas Vroom <nicolaa...@pandora.be> wrote:
> The problem is when you study a racetrack the turnaround can take a
> significant fraction making the physical interpretation what is caused
> by GR (gravitation) and what is caused by SR more complicated.
Quoting Tom Roberts's excellent web page
> The problem is when you study a racetrack the turnaround can take a
> significant fraction making the physical interpretation what is caused
> by GR (gravitation) and what is caused by SR more complicated.
http://math.ucr.edu/home/baez/physics/Relativity/SR/experiments.html#Clock_Hypothesis
| The clock hypothesis states that the tick rate of a clock when
| measured in an inertial frame depends only upon its velocity relative
| to that frame, and is independent of its acceleration or higher
| derivatives.
See that web page for references to experiments which strongly support
this hypothesis.
In the context of atomic-clock-in-airplane experiments, this means that
the shape (racetrack vs circular vs straight-line) of the airplane's
flight path doesn't matter; only the groundspeed and altitude matter.
(If we're interested in the Sagnac effect, then I think the winding
number around the Earth also matters.)
In the experiments described by Alley,
@inbook {
Alley-1983-GR-rods-and-clocks,
author = "Carroll O. Alley",
Alley-1983-GR-rods-and-clocks,
author = "Carroll O. Alley",
title = "Proper Time Experiments in Gravitational Fields
With Atomic Clocks, Aircraft, and Laser Light Pulses",
With Atomic Clocks, Aircraft, and Laser Light Pulses",
pages = "421--424",
editor = "Pierre Meystre and Marlan O. Scully",
booktitle = "Quantum Optics, Experimental Gravitation,
and Measurement Theory",
publisher = "Plenum",
address = "New York",
year = 1983,
snote = "++good discussion on `rods and clocks' experiments
testing special/general relativity",
}
the goal was to measure the GR effects (the gravitational time dialation
editor = "Pierre Meystre and Marlan O. Scully",
booktitle = "Quantum Optics, Experimental Gravitation,
and Measurement Theory",
publisher = "Plenum",
address = "New York",
year = 1983,
snote = "++good discussion on `rods and clocks' experiments
testing special/general relativity",
}
due to the airplane being higher in the Earth's gravitational field than
the ground station); the SR effects were "just" a distraction (indeed,
the SR effects were of the opposite sign to the GR effects). Therefore,
the experimenters arranged for the airplane to fly as slowly as possible
(it wasn't practical to use a helicopter) so as to minimize the SR
effects.
Fortunately, it was easy to separate out the SR and GR effects in
the data -- the GR effects varied with the airplane's altitude,
whereas the SR effects varied with the airplane's groundspeed. Recall
that for these experiments, as well as continuous theodolite tracking
of the aircraft, they had a real-time laser link between the aircraft
and a ground station, so they could monitor the time difference between
the ground and airplane clocks in real time during the flight. It's
been some years since I last read Alley's paper (alas I don't have a
copy accessible right now), but my memory is that the paper has a nice
graph showing the buildup of the measured time difference during a
long (on the order of 10 hour) airplane flight. In that graph you
could clearly see the change in the rate at which the time difference
accumulated when (part way through the flight) the airplane (having
burned off some fuel) was able to climb to a higher altitude.
Nicolaas Vroom
Apr 24, 2013, 2:30:32 AM4/24/13
to
[[Mod. note -- Please limit your text to fit within 80 columns,
preferably around 70, so that readers don't have to scroll horizontally
to read each line. I have manually reformatted this article. -- jt]]
Op zondag 14 april 2013 20:42:32 UTC+2 schreef Jonathan Thornburg
[remove -animal to reply] het volgende:
> Thank you, Oliver, for the clearest explanation of the twin paradox
> I've read in a very long time. The opportunity to read articles like
> this is a key reason why I (and I hope many others) participate in
> this newsgroup!
-- I fully agree with the last sentence --
Starting point in the Twin Paradox discussion is a unambiguous
definition of all the concepts used. Concepts like spacetime,
paradox, acceleration, rest frame, inertial frame and thought
preferably around 70, so that readers don't have to scroll horizontally
to read each line. I have manually reformatted this article. -- jt]]
Op zondag 14 april 2013 20:42:32 UTC+2 schreef Jonathan Thornburg
[remove -animal to reply] het volgende:
> Thank you, Oliver, for the clearest explanation of the twin paradox
> I've read in a very long time. The opportunity to read articles like
> this is a key reason why I (and I hope many others) participate in
> this newsgroup!
Starting point in the Twin Paradox discussion is a unambiguous
definition of all the concepts used. Concepts like spacetime,
Gregor Scholten
Apr 24, 2013, 2:31:17 AM4/24/13
to
[ The following text is in the "UTF-8" character set. ]
[ Your display is set for the "US-ASCII" character set. ]
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Tom Roberts wrote:
>> During the acceleration phase the clock starts to tick slower.
>
> This is not true. The clock ALWAYS ticks at its usual rate, and acceleration
> does not change this. Note that your language implies you are talking about the
> clock AND NOTHING ELSE, which in relativity means that you must observe it from
> its rest frame.
he obviously means the clock of the traveling twin (i.e. the twin that
has some acceleration phases) observed from the inertial frame of the
resting (uniformly moving) twin.
> If one observes the clock from the inertial frame from which it began, then one
> will OBSERVE it to start to tick more slowly as it accelerates. There is, of
> course, nothing affecting the clock itself, it is just your method of
> observation that makes this happen.
that is not that clearly to say. In that frame of reference, from which
the clock is ovserved to tick slower, the clock indeed is affected. Only
in the clock's rest frame the clock is not affected.
> When you watch a friend walk away, she appears to get smaller,
> but does not really do so. This is the same sort of effect,
no! The appearance of the friend is based on the view angle getting
smaller. That is a completely different effect.
[ Your display is set for the "US-ASCII" character set. ]
[ Some characters may be displayed incorrectly. ]
Tom Roberts wrote:
>> During the acceleration phase the clock starts to tick slower.
>
> This is not true. The clock ALWAYS ticks at its usual rate, and acceleration
> does not change this. Note that your language implies you are talking about the
> clock AND NOTHING ELSE, which in relativity means that you must observe it from
> its rest frame.
has some acceleration phases) observed from the inertial frame of the
resting (uniformly moving) twin.
> If one observes the clock from the inertial frame from which it began, then one
> will OBSERVE it to start to tick more slowly as it accelerates. There is, of
> course, nothing affecting the clock itself, it is just your method of
> observation that makes this happen.
the clock is ovserved to tick slower, the clock indeed is affected. Only
in the clock's rest frame the clock is not affected.
> When you watch a friend walk away, she appears to get smaller,
> but does not really do so. This is the same sort of effect,
smaller. That is a completely different effect.
Ken S. Tucker
Apr 24, 2013, 9:31:30 AM4/24/13
to
straightforward explanation of 'time paradox' cause.
The principle, extended to 4 velocity U^u, has an absolute derivative
DU^u=0 , (acceleration vanishes), rendering the geodesics, applied to
the rate of time U^0, such as,
DU^0 = dU^0/ds + {ab,0} U^a U^b, {} are 2nd kind Christoffels.
A change in U^0, is a change in the rate of time and thus the induction
of the cause of the 'time paradox' found by solving the Christoffel, as
far as I know.
Regards
Ken S. Tucker
JohnF
Apr 24, 2013, 3:28:59 PM4/24/13
to
Nicolaas Vroom <nicolaa...@pandora.be> wrote:
> Jonathan Thornburg het volgende:
Why isn't it the standard treatment? I've never seen anything
similar before, but it's so obvious once explained that way.
May I ask if this is OJ's original treatment, or is there
maybe a textbook (or other) reference where it appears?
I'd love to take a closer look at that book. Thanks,
--
John Forkosh ( mailto: j...@f.com where j=john and f=forkosh )
> Jonathan Thornburg het volgende:
>
>> Thank you, Oliver, for the clearest explanation of the twin paradox
>> I've read in a very long time. The opportunity to read articles like
>> this is a key reason why I (and I hope many others) participate in
>> this newsgroup!
>
> (I fully agree)
Ditto. Inspired, slap-yourself-on-the-forehead explanation.
>> Thank you, Oliver, for the clearest explanation of the twin paradox
>> I've read in a very long time. The opportunity to read articles like
>> this is a key reason why I (and I hope many others) participate in
>> this newsgroup!
>
> (I fully agree)
Why isn't it the standard treatment? I've never seen anything
similar before, but it's so obvious once explained that way.
May I ask if this is OJ's original treatment, or is there
maybe a textbook (or other) reference where it appears?
I'd love to take a closer look at that book. Thanks,
--
John Forkosh ( mailto: j...@f.com where j=john and f=forkosh )
Oliver Jennrich
Apr 25, 2013, 12:38:25 AM4/25/13
to
precise: As far as I can remember, it is not in any book. I cannot
exclude the possibility that I read it somewhere or something similar
and got inspired by that. But I wouldn't know *where* I read it or who
planted the idea using that particular analogy.
Certainly the idea to bring the problem down to euclidean geometry is
not new and not mine - Norbert Dragon, who used to write here (and who
was the first to try and bring me closer to SR and GR) used to give
example of the two travelers that drive along a triangular course. The
claim was that yanking the steering wheel in B does not
cause |AB|+|BC| > |AC|. I didn't quite understand that argument then
(about two decades ago) and from the looks I got when I used that
argument myself, few other people (see below) that struggle with the TP
do. In his book/script [1] ND makes comments along this line as well.
The observation that the rapidity in the Lorentz-boost is something like
the angle in a hyperbolic rotation is, I believe, common knowledge among
those, who consider SR/GR common knowledge. This is where I got the idea
of the velocity/angle correspondence from - technically wrong, but
sufficiently correct, I think.
If the 'acceleration free' variant of the TP is not common knowledge, it
should be [2]. And be it only to convince people that the 'turning
around' is a red herring.
When I read the original posting here constructing the force-free TP I
had an idea about where the poster wanted to take that argument, so I
had a bit of time to think about how to present the explanation in a way
that does not involve technical discussions about rocket engines and
uses as few mathematical arguments as possible. The TP is rarely the
topic of any academic discussion but is usually raised by 'educated
laymen' (with varying degrees of education...) I tried to stay away from
anything that requires more than very basic geometrical concepts.
As I'm not teaching, I have little opportunity to spread (and test its
effciency!) this treatment, but anybody wants to give it a go is more
than welcome to use it.
A word of warning though. When I thought about the treatment a few days
later, mentally trying to polish it and to expand it to curved
space-time, I noticed that a potential comeback of a student might
be: 'Well, thank you for that explanation. But now I'm not only confused
about the TP and why it doesn't exist, I'm confused about Euclidean
geometry as well.'
In my view, the strength of the treatment is that it appeals to the
intuition we developed in Euclidean geometry. This might be it's biggest
weakness as well, as the mere whiff of intuition as a base for proof
usually raises the warning flags in the mathematically trained and
inclined mind.
[1] http://www.itp.uni-hannover.de/~dragon/stonehenge/relativ.pdf, in
German.
[2] The easiest version of that is probably given by two test particles
in orbit around a central mass. If the orbits are not identical, but
intersect and have identical semi-major axes, it can always be arranged
that the particles meet at one of the intersection point once per
orbit. More complex cases can be constructed using resonant orbits.
Gregor Scholten
Apr 25, 2013, 4:10:15 AM4/25/13
to
Am 24.04.2013 15:31, schrieb Ken S. Tucker:
> The GR relativity of nonuniform motion apparently enables a fairly
> straightforward explanation of 'time paradox' cause.
there is no relativity of nonuniform motion in GR. GR is generally
> The GR relativity of nonuniform motion apparently enables a fairly
> straightforward explanation of 'time paradox' cause.
covariant, i.e. you can use any coordinate system to express equations,
however, this does not imply a relativity of non-uniform motion that
would be in any way analogous to SR's relativity of uniform motion.
According to SR's relativity of uniform motion, two uniformly moving
observers are equivalent, implying that if the first observers observes
the second observer's clock being time-dilated, the second observer
observes the first observer's clock being time-dilated as well. In GR,
there is no such equivalence for non-uniformly moving observers. There's
is even no such equivalence for free-falling observers. SR's relativity
of uniform motion is not extended in GR, it is contrarily even limited -
to the SR limit, i.e. to local relativity of local free-falling (locally
uniformly moving) observers.
> The principle, extended to 4 velocity U^u, has an absolute derivative
> DU^u=0 , (acceleration vanishes), rendering the geodesics, applied to
> the rate of time U^0, such as,
>
> DU^0 = dU^0/ds + {ab,0} U^a U^b, {} are 2nd kind Christoffels.
>
> A change in U^0, is a change in the rate of time and thus the induction
> of the cause of the 'time paradox' found by solving the Christoffel, as
> far as I know.
coordinate system, in that U^u is indicated, elapses in comparison to
the proper time of the clock whose four-velocity is given by U^u:
U^0 = d x^0 / d(tau) = dt / d(tau)
with t = x^0 being the coordinate time and tau being the proper time.
So, U^0 is highly depending on the chosen coordinate system, or frame of
reference.
Gregor Scholten
Apr 24, 2013, 5:10:48 PM4/24/13
to
Nicolaas Vroom wrote:
> Starting point in the Twin Paradox discussion is a unambiguous
> definition of all the concepts used. Concepts like spacetime, paradox,
> acceleration and thought experiment should be used with great care.
> Science is based on experiments.
however, based on experiments, one develops theories in science. Within
> Starting point in the Twin Paradox discussion is a unambiguous
> definition of all the concepts used. Concepts like spacetime, paradox,
> acceleration and thought experiment should be used with great care.
> Science is based on experiments.
the framework of such a theory, it is of course permitted to make usage
of the concepts of that theory.
Ken S. Tucker
Apr 26, 2013, 1:28:43 AM4/26/13
to
On 4/25/2013 1:10 AM, Gregor Scholten wrote:
> Am 24.04.2013 15:31, schrieb Ken S. Tucker:
>
>> The GR relativity of nonuniform motion apparently enables a fairly
>> straightforward explanation of 'time paradox' cause.
>
> there is no relativity of nonuniform motion in GR.
Are you sure you meant what you wrote?
> Am 24.04.2013 15:31, schrieb Ken S. Tucker:
>
>> The GR relativity of nonuniform motion apparently enables a fairly
>> straightforward explanation of 'time paradox' cause.
>
> there is no relativity of nonuniform motion in GR.
[snip rest pending clarification]
Ken
Nicolaas Vroom
Apr 29, 2013, 3:50:24 PM4/29/13
to
Op woensdag 24 april 2013 07:21:56 UTC+2 schreef Jonathan Thornburg [remove -animal to reply] het volgende:
Of course if in any experiment both SR and GR are involved
you need a clear distinction which is which.
In the past I have made a simulation to discuss the issues involved.
To read more select this link:
http://users.telenet.be/nicvroom/VB%20Train%20operation.htm
The simulation shows a train which runs on either a round
or horse race track. Both time dilation and length contraction
are demonstrated to gether with Terrell rotation.
For the round track two questions should be answered:
A) Is there time dilation involved.
B) Is there length contraction involved.
Similar questions are for the horse racetrack.
Nicolaas Vroom
http://users.telenet.be/nicvroom/
> Nicolaas Vroom <nicolaa...@pandora.be> wrote:
>
> > The problem is when you study a racetrack the turnaround can take a
> > significant fraction making the physical interpretation what is caused
> > by GR (gravitation) and what is caused by SR more complicated.
>
> Quoting Tom Roberts's excellent web page
> http://math.ucr.edu/home/baez/physics/Relativity/SR/experiments.html#Clock_Hypothesis
> | The clock hypothesis states that the tick rate of a clock when
> | measured in an inertial frame depends only upon its velocity relative
> | to that frame, and is independent of its acceleration or higher
> | derivatives.
> See that web page for references to experiments which strongly support
> this hypothesis.
>
> In the context of atomic-clock-in-airplane experiments, this means that
> the shape (racetrack vs circular vs straight-line) of the airplane's
> flight path doesn't matter; only the groundspeed and altitude matter.
> (If we're interested in the Sagnac effect, then I think the winding
> number around the Earth also matters.)
>
My main of point of interest in this discussion is SR.
>
> > The problem is when you study a racetrack the turnaround can take a
> > significant fraction making the physical interpretation what is caused
> > by GR (gravitation) and what is caused by SR more complicated.
>
> Quoting Tom Roberts's excellent web page
> http://math.ucr.edu/home/baez/physics/Relativity/SR/experiments.html#Clock_Hypothesis
> | The clock hypothesis states that the tick rate of a clock when
> | measured in an inertial frame depends only upon its velocity relative
> | to that frame, and is independent of its acceleration or higher
> | derivatives.
> See that web page for references to experiments which strongly support
> this hypothesis.
>
> In the context of atomic-clock-in-airplane experiments, this means that
> the shape (racetrack vs circular vs straight-line) of the airplane's
> flight path doesn't matter; only the groundspeed and altitude matter.
> (If we're interested in the Sagnac effect, then I think the winding
> number around the Earth also matters.)
>
Of course if in any experiment both SR and GR are involved
you need a clear distinction which is which.
In the past I have made a simulation to discuss the issues involved.
To read more select this link:
http://users.telenet.be/nicvroom/VB%20Train%20operation.htm
The simulation shows a train which runs on either a round
or horse race track. Both time dilation and length contraction
are demonstrated to gether with Terrell rotation.
For the round track two questions should be answered:
A) Is there time dilation involved.
B) Is there length contraction involved.
Similar questions are for the horse racetrack.
Nicolaas Vroom
http://users.telenet.be/nicvroom/
Nicolaas Vroom
Apr 29, 2013, 4:24:48 PM4/29/13
to
Am Mittwoch, 24. April 2013 21:28:59 UTC+2 schrieb JohnF:
> Nicolaas Vroom <nicolaa...@pandora.be> wrote:
> > Jonathan Thornburg het volgende:
> >
> >> Thank you, Oliver, for the clearest explanation of the twin paradox
> >> I've read in a vers long time. The opportunity to read articles like
> Nicolaas Vroom <nicolaa...@pandora.be> wrote:
> > Jonathan Thornburg het volgende:
> >
> >> Thank you, Oliver, for the clearest explanation of the twin paradox
> >> this is a key reason why I (and I hope many others) participate in
> >> this newsgroup!
> >
> > (I fully agree)
>
> Ditto. Inspired, slap-yourself-on-the-forehead explanation.
> Why isn't it the standard treatment? I've never seen anything
> similar before, but it's so obvious once explained that way.
The problem is that I fully agree that with the concept
> >> this newsgroup!
> >
> > (I fully agree)
>
> Ditto. Inspired, slap-yourself-on-the-forehead explanation.
> Why isn't it the standard treatment? I've never seen anything
> similar before, but it's so obvious once explained that way.
that many participate in this newsgroup, but that does
agree with the explanation of the twin paradox.
First of all what is the paradox? What is the contradiction ?
I do not see any contradiction when two people I and II
travel from A to B and compare their clocks and observe that
the time is different.
I also see no contradiction if they realise that the clock of
the person who has travelled the longest distance shows the
smallest elapsed time.
The explanation in both cases is physical and depents
on the quality of the clocks used.
There is a problem is the elapsed time of one of the clocks
shows a negative value. But I think this is not the issue.
The explanation of the problem in all cases, even if you use
the most accurate clock, is physical.
The next step is to derive the mathematics to describe the
physical phenomena observed.
For example if atomic clocks are used
(See: http://en.wikipedia.org/wiki/Atomic_clock )
and if atomic clocks use atoms (in some way) then the
accuracy depents on the behaviour of the atoms.
If photons are used then the behavour depents on photons.
The critical part of the problem is the link
between (the physics of) clocks and time.
Have a look at:
http://users.telenet.be/nicvroom/VB%20Train%20operation.htm
The simulation is based on mathematics (SR)
The question is the relation between mathematics and the physical
reality.
Nicolaas Vroom.
Gregor Scholten
May 2, 2013, 12:47:05 AM5/2/13
to
Ken S. Tucker wrote:
>>> The GR relativity of nonuniform motion apparently enables a fairly
>>> straightforward explanation of 'time paradox' cause.
>>
>> there is no relativity of nonuniform motion in GR.
>
> Are you sure you meant what you wrote?
hm, my reply obviously hasn't been published by the moderators for some
>>> The GR relativity of nonuniform motion apparently enables a fairly
>>> straightforward explanation of 'time paradox' cause.
>>
>> there is no relativity of nonuniform motion in GR.
>
> Are you sure you meant what you wrote?
reason. I try again:
Yes, I am absolutely sure that I meant what I wrote.
Hopefully, this will be published now.
Ken S. Tucker
May 3, 2013, 3:32:27 AM5/3/13
to
relativity of accelerated motion?
Ken
Tom Roberts
May 3, 2013, 6:13:57 PM5/3/13
to
On 5/3/13 5/3/13 - 2:32 AM, Ken S. Tucker wrote:
> If I understand you, you disagree with the GR postulate of the
> relativity of accelerated motion?
I have seen no treatment of GR that mentions such a "postulate".
> If I understand you, you disagree with the GR postulate of the
> relativity of accelerated motion?
GR is based on coordinate independence, or better, diffeomorphism invariance
[#]. From that it follows directly that one can use the equations of GR in
coordinates fixed to an accelerating object. But I see nothing that could be
sensibly construed as "relativity of accelerated motion". Indeed, the usual
representation of acceleration in relativity, 4-acceleration, is completely
independent of observer or coordinates.
[#] This is required by the fundamental fact of what science is:
formulating models of physical phenomena. We use a manifold to
model the world, and models using different manifolds must be
equally valid, implying mappings among all possible manifolds
usable for the model; for a theory like GR involving continuous
fields and their derivatives, this is diffeomorphism invariance.
As coordinates are just diffeomorphisms to R^N, coordinate
independence is included.
Tom Roberts
Hendrik van Hees
May 4, 2013, 8:38:46 AM5/4/13
to
On 03/05/13 09:32, Ken S. Tucker wrote:
> If I understand you, you disagree with the GR postulate of the
> relativity of accelerated motion?
> Ken
Far from being an expert in GR, here my 2 cts :-):
> If I understand you, you disagree with the GR postulate of the
> relativity of accelerated motion?
> Ken
I think there is still a remnant of the special principle of relativity
left in GR, because it bases on the postulate of the strong equivalence
principle, which (put in a abstract mathematical way) basically says
that spacetime is described by a pseudo-Riemannian manifold with a
pseudo-metric fundamental form of signature (1,3) (or (3,1) if you
prefer the east-coast convention). This means the tangent space at each
space-time point is the Minkowski space of special relativity.
Physically speaking this means that at each space-time point in the
space-time manifold within an infinitesimal space-time four-volume
element you always find reference frames, where the laws of special
relativity are valid. These are the reference frames of local observers
freely falling. Observers accelerated relative this class of "Galilean
frames" can realize the corresponding inertial forces locally as in SRT.
According to the strong equivalence principle they can not distinguish
these from the local consequences of gravitational forces.
--
Hendrik van Hees
Frankfurt Institute of Advanced Studies
D-60438 Frankfurt am Main
http://fias.uni-frankfurt.de/~hees/
Gregor Scholten
May 6, 2013, 12:14:29 AM5/6/13
to
Ken S. Tucker wrote:
>>>> there is no relativity of nonuniform motion in GR.
>>>
>>> Are you sure you meant what you wrote?
>>
>> hm, my reply obviously hasn't been published by the moderators for some
>> reason. I try again:
>>
>> Yes, I am absolutely sure that I meant what I wrote.
>>
>> Hopefully, this will be published now.
>
> If I understand you, you disagree with the GR postulate of the
> relativity of accelerated motion?
there is no GR postulate of the relativity of accelerated motion.
>>>> there is no relativity of nonuniform motion in GR.
>>>
>>> Are you sure you meant what you wrote?
>>
>> hm, my reply obviously hasn't been published by the moderators for some
>> reason. I try again:
>>
>> Yes, I am absolutely sure that I meant what I wrote.
>>
>> Hopefully, this will be published now.
>
> If I understand you, you disagree with the GR postulate of the
> relativity of accelerated motion?
Therefore there's nothing to agree or disagree with.
Jos Bergervoet
May 6, 2013, 2:35:32 AM5/6/13
to
On 5/4/2013 12:13 AM, Tom Roberts wrote:
> On 5/3/13 5/3/13 - 2:32 AM, Ken S. Tucker wrote:
>> If I understand you, you disagree with the GR postulate of the
>> relativity of accelerated motion?
>
> I have seen no treatment of GR that mentions such a "postulate".
>
> GR is based on coordinate independence, or better, diffeomorphism invariance
> [#]. From that it follows directly that one can use the equations of GR in
> coordinates fixed to an accelerating object. But I see nothing that could be
> sensibly construed as "relativity of accelerated motion". Indeed, the usual
> representation of acceleration in relativity, 4-acceleration, is completely
> independent of observer or coordinates.
Could it be expressed then as the opposite:
> On 5/3/13 5/3/13 - 2:32 AM, Ken S. Tucker wrote:
>> If I understand you, you disagree with the GR postulate of the
>> relativity of accelerated motion?
>
> I have seen no treatment of GR that mentions such a "postulate".
>
> GR is based on coordinate independence, or better, diffeomorphism invariance
> [#]. From that it follows directly that one can use the equations of GR in
> coordinates fixed to an accelerating object. But I see nothing that could be
> sensibly construed as "relativity of accelerated motion". Indeed, the usual
> representation of acceleration in relativity, 4-acceleration, is completely
> independent of observer or coordinates.
the non-relativity of accelerated motion?!
It's often necessary to explain that
"everything is relative" is a completely
wrong summary of the theory of relativity.
In SR, already, we know that things like
charge are non-relative. Does GR give more
of those examples than SR?
--
Jos
Ken S. Tucker
May 6, 2013, 2:35:53 AM5/6/13
to
entitled, "The Need for an Extension of the Postulate of Relativity",
but returning to the topic of Philips thread, I suggested,
a view of the geodesic equation,
DU^0 = dU^0/ds + {ab,0} U^a U^b, {} are 2nd kind Christoffels,
based on the relativity of nonuniform motion (vanishing absolute
acceleration) in space AND time which seems to have caused confusion.
That also relates well to invariance of Planck's h,
"It may be of interest how the Quantum Theory readily reveals the
'time paradox'. Using the invariance of Planck's constant,
h = energy x time units , suppose at rest using a clock of 1 kg,
ticking away. Relative to a moving frame K', suppose the relativistic
mass is 2 kg, then the time rate of the clock is 1/2 ticks, to
maintain the invariance of h."
Regards
Ken
Jonathan Thornburg [remove -animal to reply]
May 7, 2013, 2:58:05 AM5/7/13
to
Jos Bergervoet <jos.ber...@xs4all.nl> wrote:
> In SR, already, we know that things like
> charge are non-relative. Does GR give more
> of those examples than SR?
In GR, each of the following propositions is observer-independent:
(a) some specified (finite or infinite) region of spacetime is flat
(b) some specified (finite or infinite) segment of some specified
worldline is a geodesic, i.e., an observer on (that part of)
the worldline is in free-fall
(c) the total electric charge contained within a certain finite volume
of spacetime is such-and-such
In other words (being very pedantic to avoid possible confusion):
* if you fix the region of spacetime (set of events) for proposition
(a), then regardless of what coordinate system you use to make your
measurements and do your calculations, you'll always get the same
answer for whether or not (a) is true.
* if you fix the segment-of-worldline (again a set of events) for
proposition (b), then regardless of what coordinate system you use
to make your measurements and do your calculations, you'll always
get the same answer for whether or not (b) is true.
* if you fix the volume of spacetime (again a set of events) for
proposition (c), then regardless of what coordinate system you use
to make your measurements and do your calculations, you'll always
get the same value of the total charge in that volume
In this sense, then, the flatness of spacetime, the "free-fallness"
of a worldline, and electric charge are each "absolute"
(coordinate-independent) in GR.
Rich L.
May 7, 2013, 4:43:27 AM5/7/13
to
On Monday, May 6, 2013 1:35:53 AM UTC-5, Ken S. Tucker wrote:
...
s, is not the important frequency. If you consider that a mass has energy =
as m c^2, and per quantum mechanics this energy has an associated frequency=
per Planck's constant, then the wave function of the mass in its own rest =
frame is just a simple frequency with no wavenumber (momentum). If you tra=
nsform this wave function to an observers frame that is moving wrt the rest=
frame of the mass using Special Relativity, the frequency that the observe=
r sees is NOT the "clock frequency", which is a lower freqeuncy, but is gam=
ma times the rest mass frequency, which is the total energy of the mass in =
this frame. This is because the frequency that the observer measures is th=
e rate of change of the wave function phase AT A SINGLE POSITION in the obs=
ervers frame. If you go through the math, you also find that the wave func=
tion in the observers frame also has a spatial variation (wavenumber, propo=
rtional to momentum) that is gamma beta times the rest mass frequency.
This shows that energy and momentum are closely related. Energy is the pro=
jection of the particles rest mass frequency onto the observers time axis, =
and momentum is the projection onto the observers spatial axes.
The math for this is given in detail on my blog page:
http://www.blogger.com/blogger.g?blogID=8801097810586906820#editor/target=
=post;postID=4892629478579302997;onPublishedMenu=pages;onClosedMenu=
=pages;postNum=11;src=postname
or if that doesn't work, just try "makingsenseofphysics.blogspot.com" and l=
ook for the post "A deeper relationship between Energy and Momentum", about=
the third post from the bottom.
Rich L.
[Moderator's note: For best results, please post non-encoded plain
text. -P.H.]
...
> "It may be of interest how the Quantum Theory readily reveals the
> 'time paradox'. Using the invariance of Planck's constant,
> h = energy x time units , suppose at rest using a clock of 1 kg,
> ticking away. Relative to a moving frame K', suppose the relativistic
> mass is 2 kg, then the time rate of the clock is 1/2 ticks, to
> maintain the invariance of h."
> Regards
> Ken
I think the "clock frequency", the frequency of a clock moving with the mas=
> 'time paradox'. Using the invariance of Planck's constant,
> h = energy x time units , suppose at rest using a clock of 1 kg,
> ticking away. Relative to a moving frame K', suppose the relativistic
> mass is 2 kg, then the time rate of the clock is 1/2 ticks, to
> maintain the invariance of h."
> Regards
> Ken
s, is not the important frequency. If you consider that a mass has energy =
as m c^2, and per quantum mechanics this energy has an associated frequency=
per Planck's constant, then the wave function of the mass in its own rest =
frame is just a simple frequency with no wavenumber (momentum). If you tra=
nsform this wave function to an observers frame that is moving wrt the rest=
frame of the mass using Special Relativity, the frequency that the observe=
r sees is NOT the "clock frequency", which is a lower freqeuncy, but is gam=
ma times the rest mass frequency, which is the total energy of the mass in =
this frame. This is because the frequency that the observer measures is th=
e rate of change of the wave function phase AT A SINGLE POSITION in the obs=
ervers frame. If you go through the math, you also find that the wave func=
tion in the observers frame also has a spatial variation (wavenumber, propo=
rtional to momentum) that is gamma beta times the rest mass frequency.
This shows that energy and momentum are closely related. Energy is the pro=
jection of the particles rest mass frequency onto the observers time axis, =
and momentum is the projection onto the observers spatial axes.
The math for this is given in detail on my blog page:
http://www.blogger.com/blogger.g?blogID=8801097810586906820#editor/target=
=post;postID=4892629478579302997;onPublishedMenu=pages;onClosedMenu=
=pages;postNum=11;src=postname
or if that doesn't work, just try "makingsenseofphysics.blogspot.com" and l=
ook for the post "A deeper relationship between Energy and Momentum", about=
the third post from the bottom.
Rich L.
[Moderator's note: For best results, please post non-encoded plain
text. -P.H.]
Tom Roberts
May 7, 2013, 11:43:30 PM5/7/13
to
On 5/6/13 5/6/13 1:35 AM, Jos Bergervoet wrote:
> On 5/4/2013 12:13 AM, Tom Roberts wrote:
>> GR is based on coordinate independence, or better, diffeomorphism invariance
>> [#]. From that it follows directly that one can use the equations of GR in
>> coordinates fixed to an accelerating object. But I see nothing that could be
>> sensibly construed as "relativity of accelerated motion". Indeed, the usual
>> representation of acceleration in relativity, 4-acceleration, is completely
>> independent of observer or coordinates.
>
> Could it be expressed then as the opposite:
> the non-relativity of accelerated motion?!
Not really. As is usual in physics, or any other complex field replete with
> On 5/4/2013 12:13 AM, Tom Roberts wrote:
>> GR is based on coordinate independence, or better, diffeomorphism invariance
>> [#]. From that it follows directly that one can use the equations of GR in
>> coordinates fixed to an accelerating object. But I see nothing that could be
>> sensibly construed as "relativity of accelerated motion". Indeed, the usual
>> representation of acceleration in relativity, 4-acceleration, is completely
>> independent of observer or coordinates.
>
> Could it be expressed then as the opposite:
> the non-relativity of accelerated motion?!
subtleties, sound-byte statements like that are rarely valid.
Your statement depends on what you mean: "acceleration" can have three different
meanings for a given object:
A) dx/dt relative to specified coordinates, called "coordinate acceleration"
B) the object's 4-acceleration
C) the object's proper acceleration
For instance, near earth's surface, neglecting air, a dropped stone has nonzero
coordinate acceleration in coordinates fixed to the surface; it has zero
4-acceleration and zero proper acceleration. But in locally inertial
coordinates, its coordinate acceleration is zero (B and C are invariants).
Coordinate acceleration is usually what neophytes have in mind when they
casually say "acceleration"; it is clearly coordinate dependent (which I suppose
makes it "relative", though I dislike that usage).
> It's often necessary to explain that
> "everything is relative" is a completely
> wrong summary of the theory of relativity.
specifically local Lorentz invariance. Yes, the nomenclature has evolved
considerably.
> In SR, already, we know that things like
> charge are non-relative. Does GR give more
> of those examples than SR?
in every manifold of GR is invariant in SR. Note that "invariant in every
manifold" is the best I can interpret your "non-relative", and it may not be
quite what you actually had in mind.
Tom Roberts
Alfred Einstead
May 7, 2013, 7:33:28 PM5/7/13
to
On Apr 13, 3:55=A0pm, Chalky <chalkys...@bleachboys.co.uk> wrote:
> the velocity of the projectile is simply reversed (exactly).
There is no such thing as "simply reversed". The acceleration is
infinite on moment of reversal. Therefore, you have to include a delta
function in the expression. The integral of a delta function is non-
zero -- and here the time integral is the total change in speed which
(by your description) is non-zero.
Since you have a delta function in the relevant expressions, then the
corresponding shifts in the coordinate frames also involves delta
functions in their expressions. It is in there you will find the usual
results for the twin disparity.
> Given that the outward and return legs of the journey can be made
> arbitrarily long, it strikes me that the physics during the period of
> direction reversal should remain essentially the same in both cases.
>
> Am I missing something here?
Yes, you're missing the acceleration that is STILL there (in the delta
function, now).
If you draw the coordinates natural to the world line (at least for a
1 + 1 dimensional diagram), the times are measured with respect to the
world line's clock. Each event E has a twin light cone that intersects
the world line in 2 places. Call them E- and E+, and call their times
T- and T+. Then the time associated with the event in this -- the
"world line natural" coordinate grid is 1/2 (T- + T+), while the
distance from the world line is c/2 (T+ - T-), where I'm assuming that
T+ is the later time and T- the earlier time.
When you do the plot for the jagged world line, you will see sharp
jagged lines in the coordinate grid that ramify from the corner in the
world line. That's the CORRECT account for what's going on here. And
it is there you can see what effect the delta function in the world
line acceleration actually has.
And the main point is there IS an acceleration, even if it's zero
almost everywhere.
> the velocity of the projectile is simply reversed (exactly).
There is no such thing as "simply reversed". The acceleration is
infinite on moment of reversal. Therefore, you have to include a delta
function in the expression. The integral of a delta function is non-
zero -- and here the time integral is the total change in speed which
(by your description) is non-zero.
Since you have a delta function in the relevant expressions, then the
corresponding shifts in the coordinate frames also involves delta
functions in their expressions. It is in there you will find the usual
results for the twin disparity.
> Given that the outward and return legs of the journey can be made
> arbitrarily long, it strikes me that the physics during the period of
> direction reversal should remain essentially the same in both cases.
>
> Am I missing something here?
Yes, you're missing the acceleration that is STILL there (in the delta
function, now).
If you draw the coordinates natural to the world line (at least for a
1 + 1 dimensional diagram), the times are measured with respect to the
world line's clock. Each event E has a twin light cone that intersects
the world line in 2 places. Call them E- and E+, and call their times
T- and T+. Then the time associated with the event in this -- the
"world line natural" coordinate grid is 1/2 (T- + T+), while the
distance from the world line is c/2 (T+ - T-), where I'm assuming that
T+ is the later time and T- the earlier time.
When you do the plot for the jagged world line, you will see sharp
jagged lines in the coordinate grid that ramify from the corner in the
world line. That's the CORRECT account for what's going on here. And
it is there you can see what effect the delta function in the world
line acceleration actually has.
And the main point is there IS an acceleration, even if it's zero
almost everywhere.
Chalky
May 16, 2013, 5:10:18 PM5/16/13
to
On Wednesday, 8 May 2013 00:33:28 UTC+1, Alfred Einstead wrote:
> > The example shown in the second diagram at
> > http://en.wikipedia.org/wiki/Gravity_assist can, in fact, be simplified
> > still further by making U = 0, whereupon the velocity of the projectile
> The acceleration is infinite on moment of reversal.
There is no such thing as "moment of reversal" here. LOOK at the
diagram. It takes significant time for the projectile to reverse
direction, by going half way round the massive body.
> Therefore, you have to include a delta function in the expression.
If so, where?
> On Apr 13, 3:55 pm, Chalky <chalkys...@bleachboys.co.uk> wrote:
>
> > the velocity of the projectile is simply reversed (exactly).
What I actually wrote, in the first link, was:
>
> > the velocity of the projectile is simply reversed (exactly).
> > The example shown in the second diagram at
> > http://en.wikipedia.org/wiki/Gravity_assist can, in fact, be simplified
> > still further by making U = 0, whereupon the velocity of the projectile
> > is simply reversed (exactly).
> There is no such thing as "simply reversed".
Yes there is. This expresses the difference between velocities of + v and - v.
> There is no such thing as "simply reversed".
> The acceleration is infinite on moment of reversal.
diagram. It takes significant time for the projectile to reverse
direction, by going half way round the massive body.
> Therefore, you have to include a delta function in the expression.
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