a relativity paradox
RichD
field, he is an inertial frame.
Adam syncs his clock with Bob, who is orbiting the earth.
Bob is in free fall, he feels no gravitational field, he is an
inertial frame.
From Adam's viewpoint, Bob accelerates during every
revolution, in speed and direction. Adam appears the
same, to Bob.
What do their clocks say, after each encirclement by Bob?
--
Rich
Darwin123
> Adam sits at the center of the earth. He feels no gravitational
> field, he is an inertial frame.
>
> Adam syncs his clock with Bob, who is orbiting the earth.
> Bob is in free fall, he feels no gravitational field, he is an
> inertial frame.
>
> From Adam's viewpoint, Bob accelerates during every
> revolution, in speed and direction. Adam appears the
> same, to Bob.
spherical surrounding Adam. Bob is outside the spherical mass.
>
> What do their clocks say, after each encirclement by Bob?
>
> --
> Rich
Adams clock will have run slower than Bobs because he is at a
smaller gravitational potential than Bob. The gravitational potential
slows down Bobs clock.
The situation is not symmetrical. You have to take into account
the mass of the earth. Adam and Bob are not in equivalent positions
with respect to the earth's mass.
BURT
Einstein said contractile curvature is even strength gravity all the
way to the center of the Earth. There is no inner drop off of gravity
strength inside the Earth.
Mitch Raemsch
dlzc
On Aug 26, 12:38 pm, RichD <r_delaney2...@yahoo.com> wrote:
> Adam sits at the center of the earth. He feels no
> gravitational field, he is an inertial frame.
... he is at some gravitational potential...
> Adam syncs his clock with Bob, who is orbiting the
> earth. Bob is in free fall, he feels no
> gravitational field, he is an inertial frame.
... Bob is also in motion, and he is at some gravitaitonal potential.
> From Adam's viewpoint, Bob accelerates during
> every revolution, in speed and direction.
No change in speed is necessary, if his orbit is circular.
> Adam appears the same, to Bob.
Except for motion, and difference in gravitational potential...
> What do their clocks say, after each
> encirclement by Bob?
It depends.
David A. Smith
Androcles
"RichD" <r_dela...@yahoo.com> wrote in message
news:61f1d039-a8a7-4beb...@p11g2000prf.googlegroups.com...
Androcles
"Darwin123" <drose...@yahoo.com> wrote in message
news:fe9da5d1-ba3e-4725...@q1g2000yqg.googlegroups.com...
On Aug 26, 3:38 pm, RichD <r_delaney2...@yahoo.com> wrote:
> Adam sits at the center of the earth. He feels no gravitational
> field, he is an inertial frame.
>
> Adam syncs his clock with Bob, who is orbiting the earth.
> Bob is in free fall, he feels no gravitational field, he is an
> inertial frame.
Neither is in a true inertial frame.
>
> From Adam's viewpoint, Bob accelerates during every
> revolution, in speed and direction. Adam appears the
> same, to Bob.
There is a big difference between the two. There is a large mass
spherical surrounding Adam. Bob is outside the spherical mass.
>
> What do their clocks say, after each encirclement by Bob?
>
> --
> Rich
Adams clock will have run slower than Bobs because he is at a
smaller gravitational potential than Bob. The gravitational potential
slows down Bobs clock.
==============================================
Do you have any empirical evidence for a clock at the centre of the
Earth (other than the drool oozing from your lower lip)?
Henry Wilson DSc
wrote:
Good clocks are not affected in any way by movement or gravity.
Henry Wilson...
.......Einstein's Relativity...The religion that worships negative space.
waitedavidmsphysics
Since you probably are not considering general relativistic effects
and are only looking at special relativity look at "example for the
use of 5.4.4" at
http://www.modernrelativitysite.com/chap5.htm#BM5_4
If you are wanting the general relativistic answer, neither can be
considered local inertial frames in the sence that you are comparing
clocks between them that are in no way local to each other.
rotchm
> |
> | What do their clocks say, after each encirclement by Bob?
> |
> "tick tock"
:)
artful
So they have talking clocks? :):)
If we were ignoring gravity (and just had Bob moving around Adam in a
circle), then Adam would see Bob's clock always an earlier time than
his after each orbit (overall it had ticked slower), and Bob would see
Adams clock ticking as a later time than his after each orbit (overall
it had ticked faster). This is effectively a variation of the twins
paradox.
If you include the effect of gravitational potential, then that has
the opposite effect.
RichD
> > Adam sits at the center of the earth. He feels no gravitational
> > field, he is an inertial frame.
>
> > Adam syncs his clock with Bob, who is orbiting the earth.
> > Bob is in free fall, he feels no gravitational field, he is an
> > inertial frame.
>
> Neither is in a true inertial frame.
Why?
> > From Adam's viewpoint, Bob accelerates during every
> > revolution, in speed and direction. Adam appears the
> > same, to Bob.
>
> There is a big difference between the two. There is a large mass
> spherical surrounding Adam. Bob is outside the spherical mass.
So?
> > What do their clocks say, after each encirclement by Bob?
>
> Adams clock will have run slower than Bobs because he is at a
> smaller gravitational potential than Bob. The gravitational potential
> slows down Bobs clock.
I find the logical relationship between these
2 assertions disturbing, for some reason...
> The situation is not symmetrical. You have to take into account
> the mass of the earth. Adam and Bob are not in equivalent positions
> with respect to the earth's mass.
--
Rich
RichD
>
> > Adam sits at the center of the earth. He feels no gravitational
> > field, he is an inertial frame.
>
> > Adam syncs his clock with Bob, who is orbiting the earth.
> > Bob is in free fall, he feels no gravitational field, he is an
> > inertial frame.
>
> > From Adam's viewpoint, Bob accelerates during every
> > revolution, in speed and direction. Adam appears the
> > same, to Bob.
>
> > What do their clocks say, after each encirclement by Bob?
>
> way to the center of the Earth. There is no inner drop off of gravity
> strength inside the Earth.
Mimsy were the borogroves,
and the mome graths outgrabe.
--
Rich
RichD
On Aug 26, dlzc <dl...@cox.net> wrote:
> > Adam sits at the center of the earth. He feels no
> > gravitational field, he is an inertial frame.
>
> ... he is at some gravitational potential...
But he doesn't feel it.
He is at rest in a gravity free environment.
(the earth is opaque)
(except he can see Bob somehow)
> > Adam syncs his clock with Bob, who is orbiting the
> > earth. Bob is in free fall, he feels no
> > gravitational field, he is an inertial frame.
>
> ... Bob is also in motion, and he is at some gravitaitonal potential.
But he doesn't feel it.
He is at rest in a gravity free environment.
(his space shuttle has no windows)
(except he can see Adam somehow)
> > From Adam's viewpoint, Bob accelerates during
> > every revolution, in speed and direction.
>
> No change in speed is necessary, if his orbit is circular.
Elliptical, most likely.
> > Adam appears the same, to Bob.
>
> Except for motion, and difference in gravitational potential...
Adam watches Bob, Bob watches Adam.
By symmetry, they observe the same motion.
(neither is rotating)
> > What do their clocks say, after each
> > encirclement by Bob?
>
> It depends.
On what?
--
Rich
Androcles
--
Rich
Einstein said "It is essential to have time defined by means of stationary
clocks in the stationary system, and the time now defined being appropriate
to the stationary system we call it ``the time of the stationary system''. "
I say it is essential to have borogroves defined by means of stationary mome
raths in the stationary outgrabe, and the borogrove now defined being
appropriate to the stationary outgrabe we call it ``the borogrove of the
stationary outgrabe.''
Androcles
"RichD" <r_dela...@yahoo.com> wrote in message
news:de82b1e4-89a7-4747...@x20g2000pro.googlegroups.com...
--
Rich
Einstein said "It is essential to have time defined by means of stationary
RichD
>
> > Adam sits at the center of the earth. He feels no gravitational
> > field, he is an inertial frame.
>
> > Adam syncs his clock with Bob, who is orbiting the earth.
> > Bob is in free fall, he feels no gravitational field, he is an
> > inertial frame.
>
> > From Adam's viewpoint, Bob accelerates during every
> > revolution, in speed and direction. Adam appears the
> > same, to Bob.
>
> > What do their clocks say, after each encirclement by Bob?
>
> circle), then Adam would see Bob's clock always an earlier time than
> his after each orbit (overall it had ticked slower), and Bob would see
> Adams clock ticking as a later time than his after each orbit (overall
> it had ticked faster). This is effectively a variation of the twins
> paradox.
No, that's the point.
In the twins paradox, the traveler experiences a
shorter time duration, because he turns around,
thus feels acceleration (grav.) effects at that time.
In the above circumstance, neither feels any such
effects. So the twins paradox explanation doesn't
apply.
> If you include the effect of gravitational potential, then that has
> the opposite effect.
?
--
Rich
artful
> On Aug 26, artful <artful...@hotmail.com> wrote:
>
>
>
>
>
>
>
> > > Adam sits at the center of the earth. He feels no gravitational
> > > field, he is an inertial frame.
>
> > > Adam syncs his clock with Bob, who is orbiting the earth.
> > > Bob is in free fall, he feels no gravitational field, he is an
> > > inertial frame.
>
> > > From Adam's viewpoint, Bob accelerates during every
> > > revolution, in speed and direction. Adam appears the
> > > same, to Bob.
>
> > > What do their clocks say, after each encirclement by Bob?
>
> > If we were ignoring gravity (and just had Bob moving around Adam in a
> > circle), then Adam would see Bob's clock always an earlier time than
> > his after each orbit (overall it had ticked slower), and Bob would see
> > Adams clock ticking as a later time than his after each orbit (overall
> > it had ticked faster). This is effectively a variation of the twins
> > paradox.
>
> No, that's the point.
What point?
> In the twins paradox, the traveler experiences a
> shorter time duration, because he turns around,
> thus feels acceleration (grav.) effects at that time.
Not in the SR analysis of it .. there's no need to talk about
gravitational-equivalent effects (indeed, SR doesn't handle them). In
a GR analysis you can take than into account and it results in the
same effect.
> In the above circumstance, neither feels any such
> effects.
Not as I described .. read what I wrote again.
> So the twins paradox explanation doesn't
> apply.
Yes .. it DOES apply to what *I* just described.
See what I wrote immediately here for *your* scenario .....
> > If you include the effect of gravitational potential, then that has
> > the opposite effect.
>
> ?
You haven't heard of the effects of different gravitational potential
on clock rates in GR?
The lower the gravitational potential (closer to the center of a
massive object), the more slowly time passes.
With the scenario you are devising, that is critical, as Adam is AT
the center of a massive object, so time goes slower than for Bob who
is some distance from it. This is the opposite effect to what SR
predicts (when no massive earth present, and just one object moving
around another)
Please do some reading on the subject so you understand what you're
asking, and why its a bit more complicated than you think.
Try here for a start:
http://en.wikipedia.org/wiki/Gravitational_time_dilation
BURT
>
> - Show quoted text -
In the twin paradox both the fast moving twin and stay at home see the
fast moving clock tick slow. Both see the same thing. The clock that
ticks fast is still and the clock that ticks slow is moving fast
through space.
Mitch Raemsch
Daryl McCullough
Before I get to the answer, let me describe some analogous
"paradoxes" in geometry.
Suppose you have two cities, A and B, with B to the west of A.
There are two different routes connecting them. One route, call
it Route 1, goes straight west from A to B. The other route,
Route 2, goes Northwest (at a 45 degree angle to Route 1)
for half its length, then makes a right-angle turn, and
goes Southwest for the second half of its length. The
lengths of these two roads are different, even though they
both connect the same two points. But you can see that there
is a difference between them: Route 1 is straight, while
Route 2 is bent. The bent route is longer.
Now, let's make things trickier. Suppose that A and B
are on the equator, with B to the west of A. Again,
there are two routes from A to B. Route 1 goes west
from A to B. Route 2 goes *east* from A to B (around
the equator). Now, the lengths of these two routes
are different. But what broke the symmetry? They *both*
go straight.
This geometry lesson teaches two facts about geodesics
(a geodesic is a generalization of a "straight line"
that applies to curved spaces such as the surface of
the Earth):
(1) For short routes, we have a simple rule:
the shortest route between two points is a geodesic
(or "straight line").
(2) For longer routes, there can be two *different*
geodesics connecting the same two point, and they can
have different lengths. In the latter case, you have
to actually know the details of the geometry in order
to figure out which of two geodesics is longer.
Why did I take that digression through geometry?
Because there is an analogous lesson for relativity.
In relativity, an "unaccelerated path" is analogous
to a geodesic. An "accelerated path" is analogous
to a bent path. The proper time is analogous to
the length of a path. For relativity, we have
analogous facts about spacetime paths:
(1) For short paths, we have a simple rule:
the *longest* proper time between two
spacetime points is a geodesic (unaccelerated
path).
(2) For longer paths, there can be two *different*
geodesics connecting the same two points, and they
can have different proper times. You have to look
at the details of the geometry top figure out
which is longer.
--
Daryl McCullough
Ithaca, NY
artful
>
> > - Show quoted text -
>
> In the twin paradox both the fast moving twin and stay at home see the
> fast moving clock tick slow.
Depends on what you mean by 'see'. They will both 'see' each others
clocks ticking slower for outward journey (due to doppler) .. and they
will both 'see' each others clocks ticking faster for the return
journey. But overall both will agree the there were more ticks for
the stay at home twin during the trip than for the travelling twin.
ie. They certainly must agree on what the clocks read when they return
together.
> Both see the same thing. The clock that
> ticks fast is still
'Still' makes no sense in physics without a frame of reference. Its
certainly not still according to the travelling twin during the
journey.
> and the clock that ticks slow is moving fast
> through space.
Probably is for at least half its journey, or all of it, depending on
which inertial frame is measuring it.
You really are not very good at physics, are you? Yet you still feel
the need to stalk me and pretend that you know what you're talking
about. Odd obsession you have there. Maybe its some sort of sexual
fantasy of yours?
artful
dlzc
On Aug 26, 6:00 pm, RichD <r_delaney2...@yahoo.com> wrote:
> Dear David A. Smith,
>
> On Aug 26, dlzc <dl...@cox.net> wrote:
>
> > > Adam sits at the center of the earth. He feels no
> > > gravitational field, he is an inertial frame.
>
> > ... he is at some gravitational potential...
>
> But he doesn't feel it.
They both did, getting there.
> He is at rest in a gravity free environment.
It is not gravity free, that is the point. That the "acceleration of
gravity" is zero, does not mean that a price has not been paid.
> (the earth is opaque)
> (except he can see Bob somehow)
X-ray vision.
> > > Adam syncs his clock with Bob, who is orbiting the
> > > earth. Bob is in free fall, he feels no
> > > gravitational field, he is an inertial frame.
>
> > ... Bob is also in motion, and he is at some
> > gravitaitonal potential.
>
> But he doesn't feel it.
He did getting into orbit.
> He is at rest in a gravity free environment.
> (his space shuttle has no windows)
> (except he can see Adam somehow)
>
> > > From Adam's viewpoint, Bob accelerates during
> > > every revolution, in speed and direction.
>
> > No change in speed is necessary, if his orbit
> > is circular.
>
> Elliptical, most likely.
Doesn't have to be. You don't need to make your thought experiment
more difficult.
> > > Adam appears the same, to Bob.
>
> > Except for motion, and difference in
> > gravitational potential...
>
> Adam watches Bob, Bob watches Adam.
> By symmetry, they observe the same motion.
> (neither is rotating)
Then they are not looking at each other. Bob and Adam must rotate
360° for each of Bob's orbits, as compared to "the fixed stars".
> > > What do their clocks say, after each
> > > encirclement by Bob?
>
> > It depends.
>
> On what?
The altitude of Bob's orbit. Why not put him in geosync?
David A. Smith
BURT
>
> > > - Show quoted text -
>
> > In the twin paradox both the fast moving twin and stay at home see the
> > fast moving clock tick slow.
>
> Depends on what you mean by 'see'. They will both 'see' each others
> clocks ticking slower for outward journey (due to doppler) .. and they
> will both 'see' each others clocks ticking faster for the return
> journey. But overall both will agree the there were more ticks for
> the stay at home twin during the trip than for the travelling twin.
> ie. They certainly must agree on what the clocks read when they return
> together.
>
> > Both see the same thing. The clock that
> > ticks fast is still
>
> 'Still' makes no sense in physics without a frame of reference. Its
> certainly not still according to the travelling twin during the
> journey.
>
> > and the clock that ticks slow is moving fast
> > through space.
>
> Probably is for at least half its journey, or all of it, depending on
> which inertial frame is measuring it.
>
> You really are not very good at physics, are you? Yet you still feel
> the need to stalk me and pretend that you know what you're talking
> about. Odd obsession you have there. Maybe its some sort of sexual
>
> - Show quoted text -
If one clock goes faster than the other there is nothing else to be
seen. Observers will agree that the fast clock is fast and the slow
clock is slow. There is no lost time to justify two clocks both seen
to go slower than one another. The slow clock is always the one that
accelerated to speed.
Mitch Raemsch
Daryl McCullough
Oops! I never did give the answer. Here is an *approximate* answer,
that works for mild gravitational fields and rocket ships that
travel slowly compared with the speed of light:
Elapsed proper time =
Integral of [1 - 1/2(v^2/c^2 - U_g/c^2)] dt
where v is the speed of the rocket, and U_g is the
gravitational potential (-GM/r above the Earth,
and decreases from -GM/R at the surface to -3/2 GM/R
at the center of the Earth).
Tom Roberts
> Adam sits at the center of the earth. He feels no gravitational
> field, he is an inertial frame.
Hmmm. He most definitely is not at rest in an inertial frame in the sense of SR
(the earth orbits the sun). But he is at rest in a LOCALLY inertial frame in the
sense of GR. This difference is important in this situation.
> Adam syncs his clock with Bob, who is orbiting the earth.
> Bob is in free fall, he feels no gravitational field, he is an
> inertial frame.
Bob most definitely is not at rest in an inertial frame in the sense of SR (he
is orbiting the earth, which orbits the sun). But he is at rest in a LOCALLY
inertial frame in the sense of GR. This difference is important in this situation.
Moreover, gravitation is important, because it is what permits Bob to orbit the
earth.
So this gedanken must be analyzed using GR; SR is inadequate.
> From Adam's viewpoint, Bob accelerates during every
> revolution, in speed and direction. Adam appears the
> same, to Bob.
Sure. But it turns out that acceleration does not affect their clocks. Nor does
it directly affect how their readings compare [#].
[#] The "twin paradox" in SR shows that acceleration can IN-directly
affect how clocks compare.
Moreover, this is GR and one must be careful about the meaning of
"acceleration". For both Adam and Bob, their 4-acceleration is zero, as is their
proper acceleration. The non-zero acceleration you mention is merely their
COORDINATE acceleration when using the other person's coordinates. Such
coordinate-dependent quantities are risky to use, because it is all too easy to
confuse aspects of the coordinates with physical aspects of the situation. That
is the case here: both follow geodesic paths, which in GR are analogous to the
unaccelerated straight lines of SR.
Even in SR, such straight-line inertial trajectories have nonzero
coordinate acceleration if one uses accelerated coordinates.
> What do their clocks say, after each encirclement by Bob?
To answer questions like this, in either SR or GR, one must integrate the metric
along the path of each clock, because that is the only way to determine the
elapsed proper time of a clock. In SR the integral is usually easier than in GR,
because the math in SR's flat manifold is simpler than that in the curved
manifolds of GR.
In this case, it is difficult to determine what you mean, because the clocks are
never collocated -- only when they are collocated can they be compared
definitively and unambiguously. So let me construct ECI (earth centered
inertial) coordinates as in the GPS; Adam is at rest at the origin, and each
turn of Bob's orbit can be unambiguously determined from his coordinate
position. Let me assume the clocks are compared via light signals, with a
correction for the propagation of the light. There are two competing effects: a)
due to Bob's motion relative to the ECI, and b) due to his altitude -- (a) makes
his clock compare with less elapsed proper time than Adam's, while (b) makes
Bob's clock compare with more elapsed proper time than Adam's. So we cannot
determine the result of the comparison easily, a real computation must be made.
The result is that for low orbits the effect from speed outweighs the effect
from altitude and Bob's clock experiences less elapsed proper time in each
encirclement, but for high enough orbits Bob's clock experiences more elapsed
proper time in each encirclement. There is a unique circular orbit for which
they are equal; there are infinitely many elliptical orbits for which they are
equal.
A better gedanken is to consider two orbiting clocks, A in a circular orbit and
B in a highly eccentric elliptical orbit, arranged so they repeatedly intersect
and the clocks can be compared when they are collocated (the periods of the two
orbits must have a ratio that is a rational number). This is a "twin paradox"
with both twins following geodesic (locally inertial) paths. The orbits can be
arranged so B experiences more elapsed proper time between meetings than A, by
making B's orbit to be mostly outside A's. The orbits can be arranged so B
experiences less elapsed proper time between meetings than A, by making B's
orbit to be mostly inside A's. And they can experience equal elapsed proper
times by choosing B's orbit to be an appropriate mixture of inside and outside
A's orbit.
Bottom line: the elapsed proper time of a clock between two designated points
depends on its path between those points. This is true in both SR and GR. It is
directly analogous to the obvious geometrical fact that the length of a path
between two designated points depends on the path (e.g. two sides of a triangle
always have a longer total length than the third side; the distance from New
York to Chicago is longer if one detours through New Orleans).
Note that this is related to "time dilation" in SR, but is not the same. There
is no way to apply "time dilation" and obtain the correct answer; one must
integrate the clocks' elapsed proper times over their trajectories.
Tom Roberts
Tom Roberts
> In the twins paradox, the traveler experiences a
> shorter time duration, because he turns around,
> thus feels acceleration (grav.) effects at that time.
Hmmm. While it's true that the traveler must accelerate in order to return to
the homebound twin, the acceleration does not "cause" the difference in their ages.
> In the above circumstance, neither feels any such
> effects. So the twins paradox explanation doesn't
> apply.
Yes, the SR explanation does not apply. See my other post. But this is not about
acceleration, it is about path lengths.
Here's a related gedanken that illustrates this, using SR:
Put two clocks on the edges of two rotating turntables such that they
periodically come directly adjacent to each other; both centers of rotation are
at rest in the same inertial frame. This implies that the tables' centers are
separated by the sum of their radii, and the ratio of their periods is a
rational number. The clocks' readings can be directly compared at each meeting.
With appropriate values of the radii and rotation rates, one can arrange for the
clock with lower acceleration to experience either less or more elapsed proper
time between meetings. If acceleration somehow "caused" the result of the "twin
paradox" that would not be possible. In this case, the elapsed proper time of
each clock is related to its speed relative to the inertial frame, not its
acceleration.
Both "time dilation" and the "twin paradox" are subtle. No simplistic notion
like "moving clocks run slow", or "accelerated clocks run fast", can capture the
essence of either.
Tom Roberts
Daryl McCullough
>
>Daryl McCullough says...
>
>Oops! I never did give the answer. Here is an *approximate* answer,
>that works for mild gravitational fields and rocket ships that
>travel slowly compared with the speed of light:
>
>Elapsed proper time =
>Integral of [1 - 1/2(v^2/c^2 - U_g/c^2)] dt
That should be
Elapsed proper time =
Integral of [1 - (1/2(v^2/c^2) - U_g/c^2)] dt
PD
> Adam sits at the center of the earth. He feels no gravitational
> field, he is an inertial frame.
>
> Bob is in free fall, he feels no gravitational field, he is an
> inertial frame.
This is inconsistent with what you say below about a completed orbit.
Whether a frame is inertial or not doesn't have to do with whether the
observer "feels" a gravitational field. It has to do with whether, in
the scope of the laboratory, gravitational effects can be detected. On
the scale of an orbit, of course, the anisotropy of the field CAN be
detected.
>
> From Adam's viewpoint, Bob accelerates during every
> revolution, in speed and direction. Adam appears the
> same, to Bob.
>
> What do their clocks say, after each encirclement by Bob?
>
> --
> Rich
BURT
>
> - Show quoted text -
If time itself ends by definition all forms of time end including the
appearence of proper time at an event horizon of a black hole.
The idea that time ends at a point in space defeats black hole
science. Proper time from times end is stupid and rediculous.
Mitch Raemsch
RichD
> >Adam sits at the center of the earth. He feels
> >no gravitational field, he is an inertial frame.
>
> >Adam syncs his clock with Bob, who is orbiting
> >the earth. Bob is in free fall, he feels no
> >gravitational field, he is an
> >inertial frame.
>
> This is inconsistent with what you say below about a
> completed orbit.
> Whether a frame is inertial or not doesn't have to do
> with whether the
> observer "feels" a gravitational field. It has to
> do with whether, in the scope of the laboratory,
> gravitational effects can be detected. On the
> scale of an orbit, of course, the anisotropy of
> the field CAN be detected.
Bob, in free fall, can detect a G field, without
looking out the window? How?
> > From Adam's viewpoint, Bob accelerates during every
> > revolution, in speed and direction. Adam appears the
> > same, to Bob.
Rich
RichD
> > In the twins paradox, the traveler experiences a
> > shorter time duration, because he turns around,
> > thus feels acceleration (grav.) effects at that time.
>
> Hmmm. While it's true that the traveler must accelerate
> in order to return to the homebound twin, the
> acceleration does not "cause" the difference in their
> ages.
That is the stock explanation, to explain
away the paradox.
> > In the above circumstance, neither feels any such
> > effects. So the twins paradox explanation doesn't
> > apply.
>
> Yes, the SR explanation does not apply. See
> my other post. But this is not about
> acceleration, it is about path lengths.
You claim a discrepancy between SR and GR?
hmmm... what would the white haired old hippie say?
> Here's a related gedanken that illustrates thi
> Both "time dilation" and the "twin paradox" are subtle.
> No simplistic notion like "moving clocks run slow", or
> "accelerated clocks run fast", can capture the
> essence of either.
--
Rich
Inertial
news:87a5d8b2-3e15-4a16...@s17g2000prh.googlegroups.com...
>
>On Aug 26, Tom Roberts <tjroberts...@sbcglobal.net> wrote:
>> > In the twins paradox, the traveler experiences a
>> > shorter time duration, because he turns around,
>> > thus feels acceleration (grav.) effects at that time.
>>
>> Hmmm. While it's true that the traveler must accelerate
>> in order to return to the homebound twin, the
>> acceleration does not "cause" the difference in their
>> ages.
>
>That is the stock explanation, to explain
>away the paradox.
You mean the correct solution of the apparent paradox using relativity
theory?
And no .. what he said is not a 'stock explanation' of the twins paradox at
all. Clearly you don't know your physics
Also it is very clear that you are one of those crackpots who picks and
chooses part of relativity, and disregard the rest and so end up with an
incomplete and inconsistent subset which DOES have paradoxes, and then
irrationally claim that that prove relativity is wrong. Sorry .. that
doesn't work .. either you use the whole theory or you go play some other
game. You can't pick and choose what parts of the theory to use. As it
stands not, you lose.
Bye
PD
> On Aug 27, PD <thedraperfam...@gmail.com> wrote:
>
>
>
> > >Adam sits at the center of the earth. He feels
> > >no gravitational field, he is an inertial frame.
>
> > >Adam syncs his clock with Bob, who is orbiting
> > >the earth. Bob is in free fall, he feels no
> > >gravitational field, he is an
> > >inertial frame.
>
> > This is inconsistent with what you say below about a
> > completed orbit.
> > Whether a frame is inertial or not doesn't have to do
> > with whether the
> > observer "feels" a gravitational field. It has to
> > do with whether, in the scope of the laboratory,
> > gravitational effects can be detected. On the
> > scale of an orbit, of course, the anisotropy of
> > the field CAN be detected.
>
> Bob, in free fall, can detect a G field, without
> looking out the window? How?
You're not paying attention. The fact that Bob completes an orbit is
information enough.
Your confusion is that you have two different scales of laboratory for
Bob. One has the scale of his spaceship alone. The other is the scale
of his orbit around the Earth. The equivalence principle, carefully
stated, relies on the scale of the laboratory.
The simplest notion of this is as follows. If you have a laboratory
that is sufficiently big enough and that is in free-fall in the
Earth's field, then two objects released at opposite ends of the
laboratory will not remain equidistant. The reason is that both
objects fall toward the center of the Earth, and those are not
parallel lines. As the laboratory falls, the two objects fall along
converging lines, which means that they will get closer to each other.
This is where the equivalence principle starts to break down and you
begin to observe the *tidal* effects of gravity. You have to have a
laboratory that is sufficiently small (local) so that those tidal
effects are beneath your measurement resolution and THEN the
equivalence principle holds. But for a laboratory the size of a
geocentric orbit, this is certainly NOT the case.
This is the difference between LOCAL equivalence and GLOBAL
equivalence that Tom was referring to.
PD
PD
> On Aug 26, Tom Roberts <tjroberts...@sbcglobal.net> wrote:
>
> > > In the twins paradox, the traveler experiences a
> > > shorter time duration, because he turns around,
> > > thus feels acceleration (grav.) effects at that time.
>
> > Hmmm. While it's true that the traveler must accelerate
> > in order to return to the homebound twin, the
> > acceleration does not "cause" the difference in their
> > ages.
>
> That is the stock explanation, to explain
> away the paradox.
Then you are reading poor stock explanations.
Tom Roberts
> On Aug 26, Tom Roberts <tjroberts...@sbcglobal.net> wrote:
>> While it's true that the traveler must accelerate
>> in order to return to the homebound twin, the
>> acceleration does not "cause" the difference in their
>> ages.
>
> That is the stock explanation, to explain
> away the paradox.
Only in excessively naive discussions. If acceleration somehow did "cause" the
difference in elapsed proper times, then the gedanken with clocks on rotating
turntables could not possibly have the clock with lower acceleration have the
lower elapsed proper time between meetings. Nor could the twin scenario in which
the traveling twin has long inertial paths and a short turnaround have a
difference proportional to the duration of the inertial paths and independent of
the acceleration at turnaround. In this last case, the essential difference is
that the traveling twin changed inertial frames while the homebound twin did not.
Moreover, we know EXPERIMENTALLY that acceleration does not affect
the ticking of clocks to very high accuracy: the experiment of Bailey
et al shows that muons' internal decay "clocks" tick unperturbed
with accelerations of ~10^18 gee (1 gee = 9.8 m/s^2) [see the FAQ
for reference]. This experiment can be interpreted as a direct
implementation of the twin scenario (the traveling twin is a muon on
an approximately circular path with speed ~0.9999 c).
>> Yes, the SR explanation does not apply. See
>> my other post. But this is not about
>> acceleration, it is about path lengths.
>
> You claim a discrepancy between SR and GR?
NO! You are supposed to read what I write: SEE MY OTHER POST. The SR explanation
of this gedanken does not apply, because SR itself does not apply -- one must
use GR to describe this gedanken because gravitation is essential. In the usual
twin scenario of SR, and in this scenario using GR, the result depends on the
different path lengths, not the acceleration (in this scenario, both clocks have
zero 4-acceleration, so acceleration cannot possibly explain the difference).
Tom Roberts
Androcles
"Inertial" <relat...@rest.com> wrote in message
news:4c791db9$0$11110$c3e...@news.astraweb.com...
Fuck off, moron.
Inertial
> Fuck off, moron.
Hayek
> Adam sits at the center of the earth. He feels no gravitational
> field, he is an inertial frame.
Mach's Principle, which Einstein included in General
Realtivity (GR) says that surrounding mass creates
inertia. Thus, the stars that surround us, or the mass
of the Earth, when you are sitting at the center of the
Earth, give rise to inertia.
A clock is an inertiameter, and will tick slower at the
center of the Earth. Easy to see, as the escapement, or
the quartz undergoes higher inertia, and so goes slower
back and forth.
A persons molecules also move slower, so his metabolic
rate also is slower, and it seems that "time" itself
moves slower, but as some intelligent people know, there
is no such thing as time.
> Adam syncs his clock with Bob, who is orbiting the earth.
> Bob is in free fall, he feels no gravitational field, he is an
> inertial frame.
Now, moving through this inertial field, causes inertial
increase. Clocks just measure this. Bob has two
components slowing down his clock :
1: The inertial field caused by the Earth,
which is lower than Adam's, because he is further away
from the Earth's mass. Bob's clock is ticking faster
than an Earth's surface clock.
2: the inertial field increase because he moves through
the field (of the universe). The higher his speed, the
more his clock will slow down, but his speed is limited,
because he has to stay in orbit around the Earth.
> From Adam's viewpoint, Bob accelerates during every
> revolution, in speed and direction. Adam appears the
> same, to Bob.
>
> What do their clocks say, after each encirclement by Bob?
Find the formulas and put in the numbers.
Uwe Hayek.
--
We are fast approaching the stage of the ultimate
inversion : the stage where the government is free to do
anything it pleases, while the citizens may act only by
permission; which is the stage of the darkest periods of
human history. -- Ayn Rand
I predict future happiness for Americans if they can
prevent the government from wasting the labors of the
people under the pretense of taking care of them. --
Thomas Jefferson.
Socialism is a philosophy of failure, the creed of
ignorance, and the gospel of envy, its inherent virtue
is the equal sharing of misery. -- Winston Churchill.
Tom Roberts
> Mach's Principle, which Einstein included in General Realtivity (GR)
> says that surrounding mass creates inertia.
This is just plain not true. For two rather different reasons:
A) that is not really Mach's Principle, though it is a loose statement
of something related to it (actually, Mach's Principle itself is not
well defined).
B) Mach's principle is most definitely NOT included in GR. Nor is what you
say. In particular, even in manifolds devoid of mass, test particles
exhibit all of the properties commonly included in the notion "inertia".
A "shadow" of Mach's Principle is included in GR: the set of locally-inertial
frames at any given point depends on all the mass-energy throughout the universe
(at all places and times, plus boundary conditions). But in GR, mass is
intrinsic to an object, as are the properties commonly associated with
"inertia"; it is the locally inertial frames that are related to global aspects
of the manifold.
> Thus, the stars that
> surround us, or the mass of the Earth, when you are sitting at the
> center of the Earth, give rise to inertia.
Not in any mainstream theory of physics. Certainly not in GR.
> A clock is an inertiameter,
Might as well say it is a blzftqrbx. There is no commonly accepted meaning of
"inertia" that appears in any mainstream theory of physics (mass is well
defined; "inertia" is not]. Ditto for "inertiameter".
> and will tick slower at the center of the
> Earth.
This may or may not be true, depending on the altitude of the satellite clock to
which it is being compared. See my earlier post in this thread.
> [... more such nonsense]
Tom Roberts
Jonathan Doolin
> field, he is an inertial frame.
>
> Bob is in free fall, he feels no gravitational field, he is an
> inertial frame.
>
> revolution, in speed and direction. Adam appears the
> same, to Bob.
>
> What do their clocks say, after each encirclement by Bob?
>
> Rich
I think I'll have something definitive to say on this topic after I
get the integral of sqrt(c^2-a^2 t^2) dt. I know, right? I'm that
lazy. I don't feel like looking up my integral tables. Maybe
tomorrow.
Jonathan Doolin
Tom Roberts
> I think I'll have something definitive to say on this topic after I
> get the integral of sqrt(c^2-a^2 t^2) dt.
Tom Roberts
Jonathan Doolin
Thanks, Tom.
I'm not sure whether I have anything truly "definitive" to say, but I
have just done a calculation that leads me to doubt some of the common
knowledge about the Principle of Equivalence.
It occurs to me that if acceleration is equivalent to acceleration in
some way, then one should say that a person standing on the ground on
earth is, at least locally, equivalent to a person standing on a
platform accelerating at 9.8 m/s^2.
If we are to make an analysis of this situation, it would be
appropriate to imagine an accelerating platform, just at the moment
that it is at rest relative to an inertial frame.
The appropriate equations of motion are:
v= a*t.
y=1/2 a t^2
At time t=0, then, the platform has position y=0, and velocity v=0.
So at that instant, it is at relative rest in the inertial reference
frame.
We have a bit of a conundrum here, if we want to compare the speed of
a clock on board a platform which is only *instantaneously* at rest
with our inertial frame, because there is no way to measure time in
any *instantaneous* fashion.
My own method then, (and this might be where my analysis is open to
further consideration) was to consider (1) the time passed for a clock
on the platform, (2) the time passed for a clock in the inertial
frame, (3) the ratio of the two, in the limit as that time is reduced
toward 0.
The time on the platform is given by (the path integral of ds)/c
where
ds^2=c^2 dt^2 - dy^2
Where y=1/2 a t^2, so dy= a t dt
So s = integral(ds) = integral(sqrt(c^2 - a^2 t^2)dt)
In particular we want is the definite integral from -t0 to t0. While
in the inertial frame, we have we will have a time equal to 2*t0.
In the end, I found that the limit as t0->0 of (s(t0))/(2 c t0) was
equal to exactly 1. This shows that there should not be any
difference between the speed of time for the instantaneously at rest
platform, and the inertial reference frame.
Therefore, if the principle of equivalence holds, there also should
not be any difference in the speed of time for a person standing on
the ground in a gravitational field, vs. the speed of time for a
person out in space, far from any gravitational field, but stationary
(at relative rest) with the person on the ground.
This is contrary to what I have heard (and parroted) in the past, that
time should slow down for bodies deep in gravitational wells.
Jonathan Doolin
Koobee Wublee
the acceptable resolution to the twins' paradox is.
The so-called resolution put into the mouth of Einstein the nitwit,
the plagiarist, and the liar relying on acceleration to break the
symmetry has been criticized by Professor Roberts in the past few
years, and yours truly is totally in agreement poo-poo-ing that
nonsense sprouting from a nitwit, a plagiarist, and a liar. What do
you expect? <shrug>
However, the self-styled physicists rebelled against their idol
Einstein the nitwit, the plagiarist, and the liar seem to have no
other choice but to embrace an equally stupid assumption that the
measurement of aging is not time but this some crap called proper
time. Hey! Mysticism, anyone?
After all, both Larmor's (supporting the Aether) and the Lorentz
(satisfying the principle of relativity) transforms share the same
overall mathematics below but with very different details.
c^2 dTau^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2 = c^2 dt'^2 - dx'^2 - dy'^2
- dz'^2
Where
** dTau = So-called proper time
The above spacetime equation only describes the magnitude. The other
information in vector form is lost. The vector form where both primed
and unprimed frames are not necessarily moving in parallel relative to
the absolute frame is lost in the Lorentz transform where the primed
and the unprimed must be moving in parallel. This mistake initiated
by Poincare that Larmor's transform and the Lorentz transform are the
same was plagiarized by Einstein the nitwit, the plagiarist, and the
liar because the nitwit did not know any better but plagiarism excused
through lies. <shrug>
I suppose most of the self-styled physicists are just so dumb. They
have been indeed trained to be zombies (such as PD, the janitor from
Cornell, etc.) They could no longer execute any rationalizations to
save their lives. Thus, I don't expect them to understand this basic
and simple point. However, there must exist some self-styled
physicists that can see the mistakes in the past 100 years but choose
to brush all issues under the rug to continue their reign of this
occult physics because they cannot admit any mistakes. If so, their
lifetimes of devotion in the studies of physics would represent
wasteful endeavors. After all, they have hordes of zombies to support
their lies, and they are succeeding in their spread of mysticism in
the past 100 years. Yours truly is thus hoping some young, bright,
and innocent scholar of physics in his or her early career still yet
uncorrupted by the simple mistakes made by the founding fathers of SR
and GR would reverse the trend of ever deepening mysticism. Without
investing a lifetime of nonsense, they should have no problems in
demystifications. <shrug>
Back to this so-called proper time. What is it? It is exactly what
the equation above says. Claiming everyone ages in proper time is
totally absurd and stupid. Does the primed frame age in proper time?
Does the unprimed frame also age in this proper time? Any grade
schools kids would rightfully tell the self-styled physicists that
aging is measured by time not the mythical entity called proper time.
<shrug>
Traditionally, the smarter kids would go into engineering. Only the
ones who cannot do any math (such as PD), schizophrenic in mathemagics
(such as that janitor from Cornell), or babbling loud mouths capable
of conjuring up bush-loads of word salad would carry on to be the
usurpers of science. I strongly hope the trend would reverse or
balance out for the sake of science. <shrug>
Koobee Wublee
something. Finally understood the Newtonian law of gravity
(generously giving the nitwit the benefit of the doubt), the same
nitwit, the same plagiarist, and the same liar managed to rediscover
the principle of equivalence through reverse engineering. You see.
Only armed with the principle of equivalence discovered by Galileo,
Newton was able to launch the law of gravity. Self-styled physicists
would be totally bedazzled by that feat of reverse engineering the
principle of equivalence, but true scholars of physics would not
subscribe to that mysticism. <shrug>
Of course, you can argue that the principle equivalence described by
Galileo is different from the principle of equivalence reverse-
engineered after finally understood the Newtonian law of gravity. The
difference lies in some local bullshit blah, blah, blah, and the
nitwit's version would not exhibit time dilation through more curved
spacetime, but somehow, Galileo's would. Do you want to go there?
Jonathan Doolin
I gather that there are two competing effects. One being the
integration of the path integral ct= int (ds), which can be applied
from the Earth Centered Inertial (ECI)coordinates, and another that
comes somehow from the gravitational field.
Then, I gather, in fact, this path integral is not quite correct; "in
either SR or GR, one must integrate the metric along the path of each
clock, because that is the only way to determine the elapsed proper
time of a clock."
You "construct ECI (earth centered inertial) coordinates as in the
GPS)" and then you get into the meat of the matter:
"each turn of Bob's orbit can be unambiguously determined from his
coordinate position. Let me assume the clocks are compared via light
signals, with a correction for the propagation of the light."
This suggests that you have some confidence in the metric in
gravitational fields--that a meter is a meter, and a second is a
second; or at least you have confidence that the ratio of m/s is
constant for the speed of light.
"There are two competing effects: a) due to Bob's motion relative to
the ECI, and b) due to his altitude -- (a) makes his clock compare
with less elapsed proper time than Adam's, while (b) makes Bob's clock
compare with more elapsed proper time than Adam's."
Here is where I am a little bit unclear. Adam is at the center of the
earth. If I'm not mistaken, the gravitational potential function
looks like this (where C is the center, and S is the surface).
____
\ -
\ /
\ /
\/
C S
Adam Bob
Allow me to beg for correction here, because I think the graph is
right, but I must be mixing more than one idea, because the tendency
should be to fall to decreasing gravitational potential, yet I am
thinking the potential must be zero at the center, since there is no
pull in any direction.
Tom continued: "So we cannot determine the result of the comparison
easily, a real computation must be made. The result is that for low
orbits the effect from speed outweighs the effect from altitude"
From that, I gather that
(1) higher speed implies slower clocks, (implicit in special
relativity)
(2) Higher altitude implies faster clocks. (or another way to say it
would be deeper gravity well implies slower clocks.)
Okay, one final question; since geodesics are supposed to lead to the
highest possible amount of time, is it assumed in general relativity
that these two effects *exactly* cancel out for objects in free-fall?
Jonathan Doolin
Daryl McCullough
[Cut to the chase]
>In the end, I found that the limit as t0->0 of (s(t0))/(2 c t0) was
>equal to exactly 1. This shows that there should not be any
>difference between the speed of time for the instantaneously at rest
>platform, and the inertial reference frame.
That's right. An accelerating clock ticks at the same rate as
a nearby unaccelerated clock that is instantaneously traveling at the
same velocity.
>Therefore, if the principle of equivalence holds, there also should
>not be any difference in the speed of time for a person standing on
>the ground in a gravitational field, vs. the speed of time for a
>person out in space, far from any gravitational field, but stationary
>(at relative rest) with the person on the ground.
I think you've completely missed the point of gravitational time
dilation. It's not for comparing accelerated clocks to unaccelerated
clocks, it's for comparing two different clocks at different heights
in the gravitational field. You didn't calculate the corresponding
effect for accelerating rockets.
Let's assume that we have a rocket of length L, with a clock at
the front and the rear. If the rocket is undergoing constant
acceleration, then the two clocks will go out of sync; the
clock in the front will get ahead of the clock in the rear.
Why would that happen? Well, from the point of view of
the initial rest frame of the rocket, an accelerated rocket
undergoes length contraction. So the distance between the
front clock and the rear clock is shrinking with time.
That means that the front clock is traveling slower than
the rear clock. That means that it is experiencing less
time dilation. That means that the front clock ticks
faster than the rear clock.
We can calculate the magnitude of this effect by
using light signals to compare the rates of the
front and rear clocks. Assume that the rear clock
sends out a light signal once every T seconds (as
measured by the rear clock). Let T' be the time
between signals received by the front clock
(as measured by the front clock). Then if
T' is greater than T, that means that the front
clock will see the rear clock fall farther and
farther behind.
We can compute T' in terms of T, approximately,
to see the magnitude of the acceleration effect.
Suppose the rocket starts accelerating at time
t=0, and the rear clock sends a signal at that
time. This first signal will arrive at time
t_1 = L/c, approximately (because the rocket
will not have had time to accelerate much
during the time it takes light to travel from
the bottom of the rocket to the top).
Now, at a time T later, the rear clock
releases a second signal. Let t_2 be the
time at which the signal reaches the
front clock. We can compute this approximately
by using the approximate equations of motion
for the front clock and the light signal.
Letting y_rear be the location of the rear
clock, y_front be the location of the front
clock, and y_light be the location of the
second light signal, we have, for t > T:
y_light(t) = y_light(T) + c(t-T)
Since the light signal starts at the rear
clock at time T, we have:
y_light(T) = y_rear(T) = 1/2 a T^2 (approximately)
So
y_light(t) = 1/2 a T^2 + c (t-T)
The position of the front clock will be given
approximately by
y_front(t) = y_front(0) + 1/2 a t^2
But y_front(0) is just L, the height of the rocket. So
y_front(t) = L + 1/2 a t^2
To find t_2, the time at which the light signal
reaches the front clock, we set y_front(t) equal
to y_light(t):
L + 1/2 a t^2 = 1/2 a T^2 + c (t-T)
We could solve this exactly using the quadratic
equation, but we've already made some approximations,
so let's just solve it approximately, as a power
series in a, and keep just the leading term. The
answer is:
t_2 = T + L/c (1 + aT/c)
So the first signal arrives at the front clock at time L/c,
and the second signal arrives at time T + L/c(1+aT/c). So
the time between signals is
T' = T (1 + aL/c^2)
So now, let's switch views to an accelerated coordinate
system in which the two clocks are at "rest".
The rear clock is *supposed* to be sending
signals at a rate of 1 signal per T seconds, but
instead appears to be sending signals at a slower rate,
1 signal per T' seconds. The front clock concludes that
the rear clock is running slow, by a factor of (1+aL/c^2).
Now, we apply the equivalence principle to get the analogous
case of a rocket sitting at rest in a gravitational field
of strength g. We conclude that clocks at an altitude L
will measure clocks on the ground to be running slower
by a factor of (1+gL/c^2).
Daryl McCullough
>I gather that there are two competing effects. One being the
>integration of the path integral ct= int (ds), which can be applied
>from the Earth Centered Inertial (ECI)coordinates, and another that
>comes somehow from the gravitational field.
No, ds already takes into account the gravitational field. For
slow moving clocks in a spherically symmetric, weak gravitational
field, ds is given approximately by:
ds = square-root(1 - (2GM/r + v^2)/c^2) dt
So the time on a clock moving in a gravitational field
is approximately given by:
T = Integral of (1 - (GM/r + 1/2 v^2)/c^2) dt
(where I've used the approximation: square-root(1+x) == 1 + 1/2x + higher
order terms)
So the proper time is decreased by making v larger (moving faster)
and by making GM/r larger (by decreasing your distance from the
planet).
spudnik
of Lorenz and Larmor?... as for Newton,
he stole that from Hooke, then burnt all
of his portraits -- "ahahaha,
on the shoulders of that clever little dwarf!" (viz,
Sir I., the plagiarist, Freemason, alchemist-
who-burnt-his-"Principles"-in-an-accident-and-
had-it-"reconstructed"-by-the-RS-with-the-dydx-rectangle
etc. ad vomitorium .-)
thus:
wright-on; Euclid's proof is so simple, that
it takes a truly linguistically challenged ibdividual
to dyss it; after all,
all math problems are really wordproblemmas!
thus:
Melder-Mead peters-out around 20 dimensions?
as for ordinary spatial finite elements,
you really need tetrahedronometry; eh?
thus: if you can configure polysignosis,
you will ipso facto have found the cure --
dood can type!
> enough to jot down here.
--les ducs d'Enron!
http://tarpley.net
--Light, A History!
http://wlym.com/~animations/fermat/index.html
Hayek
> Hayek wrote:
>> Mach's Principle, which Einstein included in General Realtivity (GR)
>> says that surrounding mass creates inertia.
>
> This is just plain not true.
I beg to differ.
And so does Gravitation, from Misner, Thorne Wheeler :
http://www.xs4all.nl/~notime/inert/gravp543.html
For two rather different reasons:
> A) that is not really Mach's Principle, though it is a loose statement
> of something related to it (actually, Mach's Principle itself is not
> well defined).
> B) Mach's principle is most definitely NOT included in GR. Nor is what you
> say. In particular, even in manifolds devoid of mass, test particles
> exhibit all of the properties commonly included in the notion
> "inertia".
>
> A "shadow" of Mach's Principle is included in GR: the set of
> locally-inertial frames at any given point depends on all the
> mass-energy throughout the universe (at all places and times, plus
> boundary conditions). But in GR, mass is intrinsic to an object, as are
> the properties commonly associated with "inertia"; it is the locally
> inertial frames that are related to global aspects of the manifold.
>
>
>> Thus, the stars that surround us, or the mass of the Earth, when you
>> are sitting at the center of the Earth, give rise to inertia.
>
> Not in any mainstream theory of physics. Certainly not in GR.
GR defines clock rates. and,
>> A clock is an inertiameter,
So GR defines inertia. Einstein was disappointed in GR,
bescause it did not describe inertia. It did. But you
cannot keep time and inertia twice in a formula when
they mean the same thing.
> Might as well say it is a blzftqrbx.
Not really.
> There is no commonly accepted
> meaning of "inertia" that appears in any mainstream theory of physics
You said it : "commonly accepted" or "mainstream", these
are not scientific arguments.
> (mass is well defined; "inertia" is not]. Ditto for "inertiameter".
Clock rate is well defined, ergo is inertia, a clock
being an inertiameter.
Uwe Hayek.
>
>> and will tick slower at the center of the Earth.
>
> This may or may not be true, depending on the altitude of the satellite
> clock to which it is being compared. See my earlier post in this thread.
>
>
>> [... more such nonsense]
>
>
> Tom Roberts
Tom Roberts
> Tom Roberts wrote:
>> Hayek wrote:
>>> Mach's Principle, which Einstein included in General Realtivity (GR)
>>> says that surrounding mass creates inertia.
>>
>> This is just plain not true.
>
> I beg to differ.
> And so does Gravitation, from Misner, Thorne Wheeler :
> http://www.xs4all.nl/~notime/inert/gravp543.html
They merely state what I said below, in different words: GR includes only a
"shadow" of Mach's principle. In particular, the mass of an object is completely
unrelated to other objects in the universe, contrary to what most of us think as
the essence of Mach's principle.
Yes, there is no definitive statement of Mach's principle,
because he made about a dozen statements of similar "principles"
that differ in minor but important ways. That's a major part of
the problem in identifying whether this or that theory includes
"Mach's Principle".
>>> Thus, the stars that surround us, or the mass of the Earth, when you
>>> are sitting at the center of the Earth, give rise to inertia.
>>
>> Not in any mainstream theory of physics. Certainly not in GR.
>
> GR defines clock rates. and,
>
>>> A clock is an inertiameter,
>
> So GR defines inertia.
Hmmmm. The only way I can make sense of this is if you are re-defining the word
"inertia" here. That's why I said "Might as well say it is a blzftqrbx." -- your
meaning of "inertia" is COMPLETELY DIFFERENT from what anyone else means by that
word. Specifically: clocks measure time, and NOBODY ELSE would claim that time
is "inertia" -- the concepts are incommensurate.
Yes, inertia is not well defined in any mainstream theory of physics (but
locally inertial frames are well defined). Inertia is a rather nebulous concept
related to mass, but distinct from it. Mass appears in lots of equations,
inertia appears in none, which is why it is not well defined.
Tom Roberts
Tom Roberts
> I'm not sure whether I have anything truly "definitive" to say, but I
> have just done a calculation that leads me to doubt some of the common
> knowledge about the Principle of Equivalence.
>
> It occurs to me that if acceleration is equivalent to acceleration in
> some way, then one should say that a person standing on the ground on
> earth is, at least locally, equivalent to a person standing on a
> platform accelerating at 9.8 m/s^2.
I assume your 7th word should be "gravity". Then yes, this is just the principle
of equivalence.
> [...]
You made multiple mistakes, the first of which is that the analysis must be
relativistic.
This is discussed in many textbooks, including those by Rindler, by Mould, and
by Misner, Thorne, and Wheeler. Personally I find this last one to be the most
comprehensive (the textbook is _Gravitation_, and this comes in the section on
SR, before gravity is introduced).
A correct, relativistic analysis of an isolated, uniformly accelerating system
does indeed show that clocks "lower down" experience less elapsed proper time
than clocks "higher up", when compared via EM signals, slow clock transport, or
simultaneity in any inertial frame. The analysis uses SR only, and obtains the
same formula as does GR for a uniform gravitational field (the equality is NOT
happenstance).
A simple demonstration is contained in Einstein's famous gedanken, in which the
energy of a photon/light-ray is proportional to its frequency -- if
gravitational redshift did not occur, one could construct a perpetual motion
machine.
Tom Roberts
G. L. Bradford
"Jonathan Doolin" <good4...@gmail.com> wrote in message
news:0aaa2ea5-17c5-45e9...@f6g2000yqa.googlegroups.com...
========================
(1) Higher speed implies a larger space occupied, a different, higher, plane
of universe being occupied or transitioned to. Relatively speaking
concerning certain observers versus other observers, an ever increasing
ability to position versus an ever increasing inability to position. An
implication of depth...or verticality...existing to space. There being a
dimension of depth (0-point never being reachable) as well as breadth
existing to space. And there being velocity less than zero in the universe
as well as velocity greater than zero.
If I occupy an enormously larger space in the universe than you do, doing
an enormously greater velocity than you, dealing in an enormously greater
uncertainty of position than you (relative to you as an observer of the
universe, not to me as an observer of the universe), I might be very close
to occupying both position A and position B at the same time while you are
only occupying position A at that time....at precisely that same time (of
course unobservable by you -- being at position A only you being so
one-sidedly, so single-sidedly, positioned dimensionally, that same second
and length of second precisely).
GLB
========================
Tom Roberts
> [...]
> From that, I gather that
> (1) higher speed implies slower clocks, (implicit in special
> relativity)
> (2) Higher altitude implies faster clocks. (or another way to say it
> would be deeper gravity well implies slower clocks.)
Both of these "gatherings" are wrong. In all cases the clocks tick at their
usual rates. It is the COMPARISONS of two clocks that are affected.
Tom Roberts
Inertial
Further, a difference is that which object has 'higher speed' depends on the
(inertial) observer, so different observers may disagree on which clock is
measured as ticking slower (so you can have mutual time dilation being
measured).
But which one has higher gravitational potential (or altitude) does not
depend on the observer, so all observers will agree on which clock is
measured as ticking slower.
Tom Roberts
> I gather that there are two competing effects. One being the
> integration of the path integral ct= int (ds), which can be applied
> from the Earth Centered Inertial (ECI)coordinates, and another that
> comes somehow from the gravitational field.
No. In GR one simply integrates the path integral. Gravitation is included,
because the path integral is
\integral sqrt(g_ij dx^i dx^j)
In GR gravitation is described by the metric and its components {g_ij}.
In SR the same applies, except the {g_ij} are the Minkowski metric components
relative to an inertial frame, diag(-1,1,1,1) (i.e. there is no gravitation). In
SR, the path moving relative to an inertial frame is different from the path
that is at rest in that frame, and this completely describes the "twin paradox".
In addition in GR, the variation of the {g_ij} due to gravitation gives rise to
gravitational redshift.
> Then, I gather, in fact, this path integral is not quite correct; "in
> either SR or GR, one must integrate the metric along the path of each
> clock, because that is the only way to determine the elapsed proper
> time of a clock."
Your "gather" is wrong. The path integral is correct, and is the way to
determine the elapsed proper time of a clock following that path.
> This suggests that you have some confidence in the metric in
> gravitational fields--that a meter is a meter, and a second is a
> second; or at least you have confidence that the ratio of m/s is
> constant for the speed of light.
All that is true in GR. So this is not merely "confidence", it is part and
parcel of the model being used.
> If I'm not mistaken, the gravitational potential function
> looks like this (where C is the center, and S is the surface).
>
>
> ____
> \ -
> \ /
> \ /
> \/
> C S
> Adam Bob
You are mistaken. Remember that F=-grad phi, so between C and S your graph has
the gravitational force pointing OUTWARD, which is wrong. The correct plot has
phi DECREASING even more as one goes from S to the center C. Indeed for a
uniform-density sphere the potential at the center is 1.5 times the potential at
its surface, and it is a parabola from there to the surface.
The "gravitational force" is zero at the center; the potential
is not. But grad phi = 0 at the center.
> From that, I gather that
> (1) higher speed implies slower clocks, (implicit in special
> relativity)
> (2) Higher altitude implies faster clocks. (or another way to say it
> would be deeper gravity well implies slower clocks.)
This is not correct. In all cases the clocks tick at their usual rates. It is
the COMPARISON of clocks that is affected by both their relative motion and by
gravitation.
> Okay, one final question; since geodesics are supposed to lead to the
> highest possible amount of time, is it assumed in general relativity
> that these two effects *exactly* cancel out for objects in free-fall?
Not at all. Given two timelike-separated events, there will be a timelike
geodesic between them that has the largest elapsed proper time. But there can be
multiple geodesics between a given pair of events, and they can have different
elapsed proper times (because a timelike geodesic is merely a LOCAL maximum of
elapsed proper time, not necessarily a global maximum [here local and global
apply to the set of paths, not regions of the manifold]).
In GR, one cannot really "separate" those two effects; one merely integrates
proper time along the path, and both effects are inherently included. For some
(simple) cases involving weak gravitation and approximately-Minkowski
coordinates, the result is the sum of two terms which can be identified as those
two cases, but that most definitely is not general.
Consider two satellites in orbit around an isolated spherical mass. Let one
orbit be circular and one be highly eccentric, and let them be such that they
periodically intersect (among other conditions for this, the ratio of their
periods must be a rational number). Both are geodesics, but they can have
different elapsed proper times between intersections. This is a "twin paradox"
in GR that uses only geodesic paths.
Tom Roberts
BURT
If time dilation is mutual (and equal) one clock cannot age more.
Mitch Raemsch
Inertial
news:af9c5893-e789-4b04...@f20g2000pro.googlegroups.com...
>If time dilation is mutual (and equal) one clock cannot age more.
That's right .. that's what you've been told over and over. One clock does
NOT age more in cases of mutual time dilation
BURT
The twins have different ages.
Mitch Raemsch
Inertial
news:ae7db0c4-6407-4173...@p22g2000pre.googlegroups.com...
As you've been told before .. they don't .. because you cannot absolutely
compare them (note to readers .. he is referring not to the twins paradox,
but to a twin on a train passing a twin on a station)
Just because you use the word 'twin' in a relativistic gedanken, that does
not mean you automatically must have them with different ages.
Mitch Raemsch
G. L. Bradford
"Tom Roberts" <tjrobe...@sbcglobal.net> wrote in message
news:2_idnSI9dIW...@giganews.com...
=======================
I didn't write this that you mistakenly connected me directly with. I
addressed it in my own response concerning space in [SPACE]-time......in
which its total equal in SPACE will offset TIME in the merger of the two
ABSOLUTE EQUALS into a third. As to those who can't perceive and deal in the
simultaneous existence of vastly (so to speak) harder 'local' universes and
a vastly (again so to speak) softer 'non-local' universe....well, "You can
lead a horse to water but you can't make it think!"
GLB
=======================
Jonathan Doolin
wrote:
I think your set-up below is pretty good, but there is no length
contraction in your analysis. You have assumed the equations of
motion for the front and rear of the rocket are co-moving.
(It took me some extra time because I kept looking for how you were
accounting for de-synchronization of the accelerations.)
Positions of the (comoving) front and rear of the rocket:
y_rear= .5 a t^2
y_front = L + .5 a t^2
Positions of the light signals released at t=0 and t=T
y_sig1 = ct ; t > 0
y_sig2 = c(t-T); t > T
We want to find t_1 when the first signal reaches the front of the
rocket, and t_2 when the second signal reaches the front of the
rocket.
Then the time between ticks for the rear clock, as seen from the front
clock is:
T' = t_2 - t_1
Plugging in
y_sig1(t_1) = y_front(t_1);
y_sig2(t_2) = y_front(t_2);
I got
c t_1 = .5 a t_1^2 + L
c (t_2 - T) = .5 a t_2^2 + L
So we must find T'=t_2-t_1 and compare it to T; yielding a function of
a, L, and c.
I'm getting there, but I need to review how estimate a solution to a
quadratic equation from a power-series.
Jonathan Doolin
Tom Roberts
> If time dilation is mutual (and equal) one clock cannot age more.
This is not true, basically because it is not precise enough. To make it correct
one must say: If time dilation is mutual (and equal) one clock cannot age more
than another clock BETWEEN A SPECIFIED PAIR OF EVENTS.
In SR is it not possible for two different inertial trajectories to intersect
more than once. And in SR mutual time dilation occurs ONLY for inertial
trajectories. So the corrected statement cannot be applied to any physical
situation modeled by SR (except the trivial case where both clocks move together
along the same trajectory).
Note that GR does not have mutual time dilation, except for situations in which
SR is applicable and has it. So the corrected statement cannot be applied to any
physical situation modeled by GR (except the trivial case).
As I have described several times recently, in GR it is possible
for two inertial (freefall) trajectories to intersect multiple
times, and their elapsed proper times can differ between meetings.
There is no mutual time dilation, however.
Tom Roberts
Jonathan Doolin
y_front should account for the delay somehow.
Suggested correction: y_front= L + .5 a (t-L/c)^2
>
> Positions of the light signals released at t=0 and t=T
> y_sig1 = ct ; t > 0
> y_sig2 = c(t-T); t > T
y_sig2 must start higher than y_sig1 because the rocket moved.
Suggested correction: y_sig2 = c(t-T)+.5 aT^2
>
> We want to find t_1 when the first signal reaches the front of the
> rocket, and t_2 when the second signal reaches the front of the
> rocket.
>
> Then the time between ticks for the rear clock, as seen from the front
> clock is:
> T' = t_2 - t_1
>
> Plugging in
> y_sig1(t_1) = y_front(t_1);
> y_sig2(t_2) = y_front(t_2);
>
> I got
> c t_1 = .5 a t_1^2 + L
> c (t_2 - T) = .5 a t_2^2 + L
>
Correction with above changes:
c t_1 = L + .5 a(t_1 -L/c)^2
c (t_2-T) + .5 a T^2 = L + .5 a (t_2-L/c)^2
> So we must find T'=t_2-t_1 and compare it to T; yielding a function of
> a, L, and c.
>
> I'm getting there, but I need to review how estimate a solution to a
> quadratic equation from a power-series.
>
> Jonathan Doolin- Hide quoted text -
>
> - Show quoted text -
Sorry, it is taking me a few tries to get everything right.
Jonathan Doolin
>
> - Show quoted text -- Hide quoted text -
>
> - Show quoted text -
LOL.
And with these corrections, the answer is:
(drumroll please?)
t2 - t1 = T
Exactly the same clock rate as we have below.
So... what did I do wrong?
spudnik
in which the front accelerates differently
from tthe back for significant times, that is,
beyond the elasticity of the ship, or what ever.
about redshift, I just want to note that
the "light ray" is the same thing as the "photon,"
being the "geometrical optics" treatment
of a normal to the wavefront (of Huyghens).
> A simple demonstration is contained in Einstein's famous gedanken, in which the
> energy of a photon/light-ray is proportional to its frequency -- if
> gravitational redshift did not occur, one could construct a perpetual motion
> machine.
thus:
well, not overnight, since you can't light fusion
without fission (excepting possibly "cold" fusion, but
even that requires highly refined metallic lattices) e.g,.
besides, it ain't fossils; that is just a tradename!
> Even better, team up with China and do it together, for all our sakes.
thus: I'd like to review the status quo ante
on Fermat's "miracle" theorem, but, of course,
not being able to do *any* of his elementary stuff,
is a sure sign of ne'er being able to prove ****,
in numbertheorie. and, here it is, again; but
it might go better to begin with the "geometrical
fragments," specifically the "reconstruction
of Euclid's porisms," because they *are* constructive,
synthetic & not specifically dimensionalized (that is,
the measurements of the figures and assignation
of values don't have to be made, just scribed,
using a "pair" of compasses):
Jonathan Doolin
Here is where Daryl's equations differs with mine:
"To find t_2, the time at which the light signal reaches the front
clock, we set y_front(t) equal to y_light(t):
Daryl's: L + 1/2 a t^2 = 1/2 a T^2 + c (t-T)"
Mine: c (t_2-T) + .5 a T^2 = L + .5 a (t_2-L/c)^2
Okay; In Daryl's analysis he DOES have the front and back comoving.
There's no delay in the impulse. I can't exactly come up with a
convincing argument either way, but perhaps this one will do: IF you
really had a rocket that just started to accelerate, there would be at
least a speed of light delay before the top of the rocket began to
accelerate. But under constant acceleration, eventually, the two will
match accelerations. So I think the better equation is Daryl's.
> > So we must find T'=t_2-t_1 and compare it to T; yielding a function of
> > a, L, and c.
>
> > I'm getting there, but I need to review how estimate a solution to a
> > quadratic equation from a power-series.
>
> > Jonathan Doolin- Hide quoted text -
>
> > - Show quoted text -
>
> Sorry, it is taking me a few tries to get everything right.- Hide quoted text -
>
> - Show quoted text -- Hide quoted text -
BURT
> "BURT" wrote in message
>
> news:ae7db0c4-6407-4173...@p22g2000pre.googlegroups.com...
>
>
>
> >On Sep 1, 7:25 pm, "Inertial" <relativ...@rest.com> wrote:
> >> "BURT" wrote in message
>
> >>news:af9c5893-e789-4b04...@f20g2000pro.googlegroups.com...
>
> >> >If time dilation is mutual (and equal) one clock cannot age more.
>
> >> That's right .. that's what you've been told over and over. One clock
> >> does
> >> NOT age more in cases of mutual time dilation
>
> >The twins have different ages.
>
> As you've been told before .. they don't .. because you cannot absolutely
> compare them (note to readers .. he is referring not to the twins paradox,
> but to a twin on a train passing a twin on a station)
>
> Just because you use the word 'twin' in a relativistic gedanken, that does
> not mean you automatically must have them with different ages.
>
> Mitch Raemsch
Have you forgot why Einstein called it the twin paradox?
Have you forgot that Einstein pointed out their different aging?
Mitch Raemsch
Hayek
> Hayek wrote:
>> Tom Roberts wrote:
>>> Hayek wrote:
>>>> Mach's Principle, which Einstein included in General Realtivity (GR)
>>>> says that surrounding mass creates inertia.
>>>
>>> This is just plain not true.
>>
>> I beg to differ.
>> And so does Gravitation, from Misner, Thorne Wheeler :
>> http://www.xs4all.nl/~notime/inert/gravp543.html
>
> They merely state what I said below, in different words: GR includes
> only a "shadow" of Mach's principle.
May be you should actually read those pages.
> In particular, the mass of an
> object is completely unrelated to other objects in the universe,
If inertia has external causes, then we should make a
difference between mass, as in count the atoms, and
inertial mass, as in difficulty to de- or accelerate.
The number of atoms of a test mass does not vary, but
the ease or difficulty of de/accelerating it, varies.
And the latter IS related to the other objects in the
universe.
The fact that we see inertia different form gravitation,
is that the inertia of the planets caused by the
universe is very much larger than the inertia caused by
the sun. But for mercury it is already necessary to
explain its perihelion advance.
> contrary to what most of us think as the essence of Mach's principle.
>
> Yes, there is no definitive statement of Mach's principle,
> because he made about a dozen statements of similar "principles"
> that differ in minor but important ways. That's a major part of
> the problem in identifying whether this or that theory includes
> "Mach's Principle".
GR works, one only needs to see "proper time" as
inertial field strength. GR is the key to the correct
Mach's principle. Note that clocks measure "proper
time", thus they measure inertial field strength.
>
>>>> Thus, the stars that surround us, or the mass of the Earth, when you
>>>> are sitting at the center of the Earth, give rise to inertia.
>>>
>>> Not in any mainstream theory of physics. Certainly not in GR.
>>
>> GR defines clock rates. and,
>>
>>>> A clock is an inertiameter,
>>
>> So GR defines inertia.
>
> Hmmmm. The only way I can make sense of this is if you are re-defining
> the word "inertia" here.
Not at all. It is plain old inertia as Newton saw it.
What is new is that inertia has an external cause, as
Mach said it : "mass over there causes inertia over
here". And what is also new is that it is variable, as
in "more mass over there causes more inertia over here".
> That's why I said "Might as well say it is a
> blzftqrbx."
I do not think that was defined by Newton.
> -- your meaning of "inertia" is COMPLETELY DIFFERENT from
> what anyone else means by that word.
I do not see where it is different. Please explain.
> Specifically: clocks measure time,
I see. And time is "blzftqrbx"?
> and NOBODY ELSE would claim that time is "inertia" --
I am quite glad to hear that. It is better than
"everybody knows that". I had that as an answer too.
> the concepts are
> incommensurate.
Accept for a moment that inertia could be variable.
Then you can try to build an inertiameter. Let some mass
go back and forth and measure the time it takes for the
cycle to complete. Then realize you just build a clock.
GR makes mass cause gravitation, and gravitation causes
clocks to slow down. How magnificent.
Thus mass causes clocks to slow down. But mass also
causes inertia.
Where do clocks slow down ? exactly where Mach says
where we should have high(er) inertia.
>
> Yes, inertia is not well defined in any mainstream theory of physics
It is there alright, but hidden by ignorance.
> (but locally inertial frames are well defined).
> Inertia is a rather
> nebulous concept related to mass, but distinct from it.
Empty statement.
> Mass appears in
> lots of equations, inertia appears in none, which is why it is not well
> defined.
Inertia influences ALL of physics. It is THE external
variable. That is why you can never measure it in a
local frame, you have to look at another frame to
measure its "relative" strength. The inertial field
strength is expressed by the gamma factor. A lab with
gamma=2 will have half the inertia than a lab frame with
gamma=1. clocks will run twice as slow in the lab frame
with gamma 2. The clocks will just react to this
inertia, that is why I call them inertiameters.
HTH.
Uwe Hayek.
Inertial
news:2cb1366f-4c02-45ea...@d8g2000yqf.googlegroups.com...
>Have you forgot why Einstein called it the twin paradox?
>Have you forgot that Einstein pointed out their different aging?
You are NOT talking about the twins paradox, because you are talking about
mutual time dilation.
If you ARE talking about the twins paradox, fine, but that is not caused by
mutual time dilation.
However, you can't even understand a train passing a station, so there is no
point in going into the more complicated twins paradox.
In fact, no point in discussing physics with you are you refuse to
understand.
BURT
> "BURT" wrote in message
>
> news:2cb1366f-4c02-45ea...@d8g2000yqf.googlegroups.com...
>
> >Have you forgot why Einstein called it the twin paradox?
> >Have you forgot that Einstein pointed out their different aging?
>
> You are NOT talking about the twins paradox, because you are talking about
> mutual time dilation.
If there is time dilation according to Einstein it is always with
the twins and every other experiment with one fast observer and one
not.
Einstein's lost time was for the twins but because it is not true for
passing twins this proves it wrong.
Mitch Raemsch
Inertial
news:e87e4600-c8e6-4837...@n3g2000yqb.googlegroups.com...
>
>On Sep 2, 4:28 pm, "Inertial" <relativ...@rest.com> wrote:
>> "BURT" wrote in message
>>
>> news:2cb1366f-4c02-45ea...@d8g2000yqf.googlegroups.com...
>>
>> >Have you forgot why Einstein called it the twin paradox?
>> >Have you forgot that Einstein pointed out their different aging?
>>
>> You are NOT talking about the twins paradox, because you are talking
>> about
>> mutual time dilation.
>> by
>> mutual time dilation.
>> However, you can't even understand a train passing a station, so there is
>> no
>> point in going into the more complicated twins paradox.
>> In fact, no point in discussing physics with you are you refuse to
>> understand.
>
>the twins
Nope .. there is nothing about Einstein and relativity that requires twins,
or even siblings or close relatives. I'm not sure why you have this
obsession with twins.
> and every other experiment with one fast observer and one
>not.
Nope .. The twins paradox occurs for a single observer, who notes the
difference in ages when the travelling twin returns.
> Einstein's lost time
Nope .. No time is lost in the 'twins paradox'. Just different amounts of
elapsed time in two paths between two events .. just like there can be
different distances for two paths between two events.
>was for the twins
Nope .. Einstein's first example of it in his 1905 paper was for a
travelling clock .. no twins required
But as you seem to have a twin obsession, we'll talk twins again
In the twins paradox, the travelling twin does not remain at rest in any
inertial frame, the stay at home one does. This lack of symmetry (NON
mutual) is what results in the difference in aging.
> but because it is not true for
> passing twins
Which are both at rest in their own inertial frames .. their motion is
symmetric and so time dilation is mutual. You cannot say one ages more or
less than the other.
> this proves it wrong.
No .. it proves you're a moron who comments with pretend authority on that
which he does not understand
BURT
>
> - Show quoted text -
You will find out that I have won the Nobel Prize
for this year IN PHYSICS with the theory of Maximum Gravity.
The second is for location of wave function collapse in the
two slit experiment.
The official anouncements are in the second week of next month.
Look forward to them!
Mitch RAEMSCH
Jonathan Doolin
I still don't like this explanation. I think, in reality, if you have
a rocket accelerating, then the back end has always been accelerating
for a slightly longer time than the front end. There will always be
impulses from the back that have not yet reached the front. So,
essentially the two ends can *never* be co-moving.
So we should use:
y_back(t)= .5 a t^2
y_front(t)= L + .5 a (t-L/c)^2
And after checking the math, we should conclude that with
acceleration, we wouldn't see any difference in the clocks.
On the other hand, if we are talking about gravity, we have to assume
the front and the back *are* comoving, and we can use Daryl's equation
for the front.
y_back(t)= .5 a t^2
y_front(t)= L + .5 a t^2
So, assuming Daryl's approach is correct, the derivation actually
relies on a fundamental difference between acceleration and gravity.
PD
This is where those who respond to Mitch at any length should pause
and consider who they are engaging with.
It's like trying to have a rational discussion with someone on a park
bench with drool coming from his mouth and a pigeon on each shoulder.
PD
Androcles
"PD" <thedrap...@gmail.com> wrote in message
news:f990774f-3b73-4f0b...@t20g2000yqa.googlegroups.com...
PD
============================================
That describes you to a tee, Duck.
> > > > > > > Let's say you have a machine gun on a carousel, Phuckwit Duck,
> > > > > > > and the tangential velocity of the gun is 200 m/s. The bullets
> > > > > > > emerge
> > > > > > > from the gun and have a speed of 800 m/s with respect to the
> > > > > > > muzzle.
> > > > > > > The target has a negligible velocity of 0.0001 m/s toward the
> > > > > > > carousel.
> > > > > > > What do you "guess" the speed of the bullets with respect to
> > > > > > > the
> > > > > > > target are, and why, Mr Phuckwit "Closing Speed" Duck ?
>
> > > > > > It's a number close to, but not exactly, 1000.0001 m/s,
> > > > > > Androcles,
> > > > > > in
> > > > > > the target frame.
Daryl McCullough
>I still don't like this explanation. I think, in reality, if you have
>a rocket accelerating, then the back end has always been accelerating
>for a slightly longer time than the front end. There will always be
>impulses from the back that have not yet reached the front. So,
>essentially the two ends can *never* be co-moving.
That's correct. What we can do is to define an *accelerated* coordinate
system for use inside the rocket, and demand that it has the following
features: (for simplicity, let's just worry about one spatial coordinate,
X, whose axis is in the direction of the rocket's motion, and one time
coordinate, T)
1. For an object "at rest" on board the rocket, the X coordinate of
that object does not change with time.
2. If a light signal is sent from location X_1 to location X_2,
then the travel time (as measured by the coordinate T) for the
light signal does not change in time. In particular, when the
rocket is at rest, the travel time from the bottom of the rocket
to the top of the rocket is L/c. We want this to be the travel
time even after the rocket has been accelerating for a while.
3. The X coordinate agrees with the inertial coordinate x of
the rocket's initial rest frame when the rocket first launches.
4. The T coordinate for an object at "rest" in the accelerated
coordinate system is proportional to the time shown on a clock
at the same location.
5. An object at "rest" in the accelerated coordinate system
feels constant acceleration. That is, its acceleration as
measured in a comoving inertial frame is always the same.
The coordinate system that has all these properties is this
one: (I think they're called
X = square-root(x^2 - (ct)^2)
T = c/g arctanh(ct/x)
with the inverse transforms:
x = X cosh(gT/c)
t = X/c sinh(gT/c)
where g is the acceleration measured by the rear of the rocket.
The origin of this coordinate system is actually not at the rear
of the rocket, but is at a distance c^2/g behind the rear of the
rocket. The significance of this origin is that a light signal
that is sent from this origin at the moment the rocket launches
will *never* reach the rocket (as long as the rocket continues
to accelerate). The distance between the light signal and the
rocket will decrease with time, but will never vanish. The
location x of the rear of the rocket as a function of time t,
as measured in the initial inertial frame is given by:
x = square-root((ct)^2 + (c^4/g^2))
while the location of the light signal sent at the same time
is given by
x = ct
The rocket will always be ahead of the light signal, even
though the rocket travels slower, because the rocket has
a head start of c^2/g.
Anyway, if you have an object at "rest" in the accelerated
coordinate system, then the acceleration it "feels" as
measured by an accelerometer, will be given by:
g(X) = c^2/X
So the rear of the rocket, which is at location X = c^2/g,
will feel acceleration g, but the front of the rocket will
feel slightly less acceleration, since it has a larger value
of X.
The front and the rear are *not* comoving, as measured in
an inertial reference frame; the front travels slightly slower
than the rear, with slightly less acceleration. The fact that
they are not comoving is hinted at by the fact that the clocks
at the top and bottom do not run at the same rate.
>On the other hand, if we are talking about gravity, we have to assume
>the front and the back *are* comoving, and we can use Daryl's equation
>for the front.
Even in a gravitational field, the front and back (or top and bottom)
are not comoving, as measured in a free-falling coordinate system. But
they are at rest in the noninertial coordinate system of the Earth,
just like the front and rear are at rest in the noninertial coordinate
system of an accelerating rocket. Just like on an accelerated rocket,
clocks at the top and bottom of a mountain run at slightly different
rates.
Dono.
>
> So, assuming Daryl's approach is correct, the derivation actually
> relies on a fundamental difference between acceleration and gravity.
"fundamental difference between acceleration and gravity" falls in the
category "not even wrong".
You don't know what you are talking about.
Androcles
"Dono." <sa...@comcast.net> wrote in message
news:ecf649f3-8cab-4990...@p22g2000pre.googlegroups.com...
Forces can be balanced, I have a force on my arse which is
compressing my flab but I am not accelerating vertically.
What has Doolin befuddled is the idiot Einstein's "Principle of
Equivalence", wherein he confuses acceleration with an
unbalanced force rather than the result of an unbalanced
force. The reverse is also true, in the "elevator in space"
example, acceleration produces a force which is balanced
by inertia of the arse and compresses flab. This does not
make gravity equivalent to acceleration, it makes gravity
equivalent to what it is, a force.
spudnik
and don't say, the curvature of space, which is a)
probably a result of gravity, and b)
very well measured (as opposed to the "curvature
of spacetime," which is a bit of an absurdity).
> it makes gravity equivalent to what it is, a force.
--les ducs d'Enron!
http://tarpley.net
--Light, A History!
http://wlym.com/~animations/fermat/index.html
BURT
One twin will age more than the other and this is how it will appear
all along
to both taking Doppler into account Transverse Doppler will do the
job.
Mitch Raemsch
Jonathan Doolin
wrote:
First, I want to make the point that I would like to make sure we
don't overlook the question of my previous analysis. Namely, when we
take into account the impulse delay from the back to the front of the
rocket, the difference in clock-speeds disappeared altogether.
However, when we assumed the front and back are co-moving, there was a
difference in clockspeed. You came up with an estimate, which was a
power-series expansion in a (which I have not yet reproduced.)
"We conclude that clocks at an altitude L will measure clocks on the
ground to be running slower by a factor of (1+gL/c^2)." -Daryl
McCullough August 31, 2010
I am guessing, this would be a valid approximation for a low earth-
like acceleration. You gave a formula for Elapsed proper time:
"Here is an *approximate* answer, that works for mild gravitational
fields and rocket ships that travel slowly compared with the speed of
light: Elapsed proper time = Integral of [1 - 1/2(v^2/c^2 - U_g/c^2)]
dt where v is the speed of the rocket, and U_g is the
gravitational potential (-GM/r above the Earth,
and decreases from -GM/R at the surface to -3/2 GM/R
at the center of the Earth)." -Daryl McCullough August 26, 2010
In the case we have been considering, we have the top of the building
sitting still, so v = 0, and the elapsed proper time would simplify to
integral [1 + (G M)/(R*c^2)] dt. (verify-I'm prone to math error)
So, these are steps that I don't want to overlook. (1) in getting the
power series expansion; especially why you chose to get series
expansion in terms of a, rather than in T. (2) Going from T'/T =
(1+gL/c^2) to T'/T = integral [1 + (G M)/(R*c^2)] dt
----------
Now you are entering analysis using special relativity, which is also
interesting. Let's highlight what I think is the major difference
Nonrelativistic estimation:
top and bottom of an accelerating rocket are NOT comoving due to a
delay in impulse.
top and bottom of a building in gravity ARE comoving.
Relativistic (exact analysis)
top and bottom of an accelerationg rocket are NOT comoving due to
a delay in impulse (+additional relativistic effects)
top and bottom of building in intense gravity ARE NOT comoving due
to relativistic effects.
Your analysis of the relativistic effects then lies around producing a
stationary origin in an accelerated frame... or not so much producing
it, but using it.
I made a demonstration some time ago where this event becomes somewhat
apparent.
http://www.wiu.edu/users/jdd109/stuff/relativity/LT.html
If you press "constant acceleration" and then "pass time" you will see
the blue line move slowly into an equilibrium position. The origin
you spoke of is where this blue line eventually comes to intersect the
horizontal (x) axis.
If you try to place an event directly at this point, it will stay
there for some time before moving off. If you are careful, you can
place a light-like interval; when the slope is just right, the line
should turn red. If you make the photon going to the right, it will
hover around that point before eventually moving off to the right.
Okay, enough playing with the demo; Let's talk about the position of
that event. It is a precise distance from the back of the rocket, and
stays at that distance from the back of the rocket, as long as the
rocket stays at the same acceleration. The distance between the back
of the rocket and the stationary event is a function of the
acceleration.
However, if the front of the rocket is going with the same
acceleration, then the associated stationary event for the front of
the rocket will be offset by exactly the length of the rocket.
So what you're saying--(or rather, what will produce the same
mathematical result to what you're saying)--is that the acceleration
of the front of the rocket is somewhat slower than the back--so that
the front and back of the rocket share the very same stationary event
for the origin of their accelerated frame.
Some questions then to work on;
1) What is the distance to the origin (stationary event) from the back
of the rocket as a function of a, c, etc?
2) precisely what is the difference in accelerations, according to the
back rocket and the front rocket?
3) Can this difference be neglected for earth gravity; so we can say
the front and back are comoving?
4) Does the nonrelativistic approximation done earlier handle the job,
to go from T'/T = (1+gL/c^2) to T'/T = integral [1 + (G M)/(R*c^2)]
dt?
5) You lined up 5 criteria, "demands" or "features" of the accelerated
frame. Let's make some list of why these demands must be met. Are
they experimentally determined, or do they come from some
"principle."
6) You gave the equations:
> I think they're called
> X = square-root(x^2 - (ct)^2)
> T = c/g arctanh(ct/x)
> with the inverse transforms:
>
> x = X cosh(gT/c)
> t = X/c sinh(gT/c)
I still need to become more comfortable with exactly what is going on
with these equations; I was able to visualize the inverse transforms
(x(X,T),t(X,T)) I think; imagining X1 and X2 being constants, as the
front and back of the rocket, and T increasing over time, making two
hyperbolic arcs; with events of equal T being co-linear with the
"stationary event" but I'm still not quite comfortable with how the
equations meet the five criteria. My mind still balks at the non-
inertial frame. X and T are okay to put on physical objects; labeling
the meter sticks, and ticking off the clock, but extending those
sticks and out to infinite length behind the rocket to where the
clocks stand still or go backwards still bugs me (*despite the fact
that I have produced a demonstration that actually shows it happens.*)
I think the trouble is that I think of that origin that you're using
as an anomoly (when talking about actual acceleration); it's like
where half of the mass of the universe going in one direction meets a
single photon going the other direction in a single eternal instant.
It's just such a wierd thing to say is the origin of your coordinate
system. But we're not talking about actual acceleration; we're
talking about gravity. So, if it's the only way to meet all of your
criteria, and the criteria are experimentally determined, then it
still makes sense to use that virtual origin to make predictions about
gravitation.
Jonathan Doolin
Jonathan Doolin
> On Sep 3, 2:57 am, Jonathan Doolin <good4us...@gmail.com> wrote:
>
>
>
> > So, assuming Daryl's approach is correct, the derivation actually
> > relies on a fundamental difference between acceleration and gravity.
>
> "fundamental difference between acceleration and gravity" falls in the
> category "not even wrong".
"Not even wrong" is roughly synonymous with "ambiguous." Stated
without context, a "fundamental difference between acceleration and
gravity" is ambiguous.
However, I had precisely defined what I meant by the fundamental
difference between acceleration and gravity. The difference appeared
in the equation of motion for the front of the rocket.
Acceleration--> y_front= L + .5 a (t-L/c)^2
Gravity--> y_front= L + .5 a (t)^2
If you have another idea for the equation of motion for the front of
the rocket under acceleration, then you should be able to argue that
my equation is "wrong." I would like to know if there is something
quantitatively significant that I have overlooked.
Jonathan Doolin
Androcles
"Jonathan Doolin" <good4...@gmail.com> wrote in message
news:fbb99337-dcc0-4596...@y11g2000yqm.googlegroups.com...
On Sep 3, 11:20 am, "Dono." <sa...@comcast.net> wrote:
> On Sep 3, 2:57 am, Jonathan Doolin <good4us...@gmail.com> wrote:
>
>
>
> > So, assuming Daryl's approach is correct, the derivation actually
> > relies on a fundamental difference between acceleration and gravity.
>
> "fundamental difference between acceleration and gravity" falls in the
> category "not even wrong".
"Not even wrong" is roughly synonymous with "ambiguous." Stated
without context, a "fundamental difference between acceleration and
gravity" is ambiguous.
However, I had precisely defined what I meant by the fundamental
difference between acceleration and gravity. The difference appeared
in the equation of motion for the front of the rocket.
Acceleration--> y_front= L + .5 a (t-L/c)^2
http://www1.somerset.gov.uk/stcsymbols/STCsymbolswm/Y-Fronts.jpg
Daryl McCullough
>First, I want to make the point that I would like to make sure we
>don't overlook the question of my previous analysis. Namely, when we
>take into account the impulse delay from the back to the front of the
>rocket, the difference in clock-speeds disappeared altogether.
That's not correct.
>However, when we assumed the front and back are co-moving, there was a
>difference in clockspeed. You came up with an estimate, which was a
>power-series expansion in a (which I have not yet reproduced.)
>
>"We conclude that clocks at an altitude L will measure clocks on the
>ground to be running slower by a factor of (1+gL/c^2)." -Daryl
>McCullough August 31, 2010
>
>I am guessing, this would be a valid approximation for a low earth-
>like acceleration. You gave a formula for Elapsed proper time:
>
>"Here is an *approximate* answer, that works for mild gravitational
>fields and rocket ships that travel slowly compared with the speed of
>light: Elapsed proper time = Integral of [1 - 1/2(v^2/c^2 - U_g/c^2)]
>dt where v is the speed of the rocket, and U_g is the
>gravitational potential (-GM/r above the Earth,
>and decreases from -GM/R at the surface to -3/2 GM/R
>at the center of the Earth)." -Daryl McCullough August 26, 2010
>
>In the case we have been considering, we have the top of the building
>sitting still, so v = 0,
Well, if we ignore the rotation of the Earth, that's true.
>and the elapsed proper time would simplify to
>integral [1 + (G M)/(R*c^2)] dt. (verify-I'm prone to math error)
I think you got the sign wrong. It should be
integral [1 - (G M)/(R*c^2)] dt
>So, these are steps that I don't want to overlook. (1) in getting the
>power series expansion; especially why you chose to get series
>expansion in terms of a, rather than in T. (2) Going from T'/T =
>(1+gL/c^2) to T'/T = integral [1 + (G M)/(R*c^2)] dt
I see that there is a confusion here. In T'/T = (1+gL/c^2),
T is the time between signals sent by the rear clock, as
measured by the rear clock, and T' is the time between signals,
received by the front clock, as measured by the front clock.
In the formula T = integral [1 + (G M)/(R*c^2)] dt,
T is the proper time for a clock, and t is the coordinate
time.
So the meaning of T is not the same in the two formulas.
We can use the latter formula to compute the former ratio,
though. On the Earth, you have a clock that is producing
signals once every T seconds, as measured by that clock.
At a height of L above the surface of the Earth, you have
a clock that is receiving those signals.
Let's consider the following 4 events:
e1 : the first signal is sent from lower clock.
e2 : the first signal arrives at the higher clock.
e3 : the second signal is sent from the lower clock.
e4 : the second signal arrives at the higher clock.
Let (r1,t1), (r2, t2), (r3, t3) and (r4, t4) be the
coordinates of these 4 events. Then using the formula
for proper time:
Proper time between e1 and e3 = T =
integral from t1 to t3 of [1 - (G M)/(R*c^2)] dt
Since R is constant, the integral is simple:
T = (1-GM/(R c^2)) (t3 - t1).
Proper time between e2 and e4 = T' =
integral from t2 to t4 of [1 - (G M)/((R+L)*c^2)] dt
So
T' = (1-GM/((R+L) c^2)) (t4 - t2).
Now, if we let delta-t be the coordinate time it takes
for light to travel upwards from r=R to r=(R+L)
(which will be approximately L/c), then
t2 = t1 + delta-t
t4 = t3 + delta-t
So
(t4 - t2) = (t3 - t1)
So
T'/T = (1 - GM/((R+L) c^2))/(1 - GM/(R c^2))
If L is small compared with R, we can write
1/(R+L) = 1/R - L/R^2 + negligible terms
So
T'/T = (1-GM/(Rc^2) + GM/(R^2 c^2) L)/(1-GM/Rc^2)
= 1 + GM/(R^2c^2)/(1-GM/Rc^2) L
GM/R^2 is just the surface gravity, g. So we can write this as
T'/T = (1 + gL/(c^2(1-GM/Rc^2)))
If GM/Rc^2 is much less than 1, then we get:
T'/T = (1+gL/c^2)
So for small distances L above the surface of the Earth,
the General Relativity prediction is the same as for a
rocket undergoing constant proper acceleration g.
>5) You lined up 5 criteria, "demands" or "features" of the accelerated
>frame. Let's make some list of why these demands must be met. Are
>they experimentally determined, or do they come from some
>"principle."
A choice of coordinates is a convention. You choose a coordinate
system that is convenient for calculations. If you want a clock
to be able measure coordinate time, then you want coordinate time
to be proportional to the elapsed time on the clock. If you want
to be able to measure distances inside a rocket ship, then you
want the distance between the top of the rocket and the bottom
of the rocket to be constant, etc.
Now, there is one assumption that is experimental, and not just
a convention. That is, that when an object undergoes constant
acceleration, it will tend to reach an equilibrium length, as
measured by the time (as measured by onboard clocks) required
for a light signal to go from one end to the other, and back.
BURT
>
> - Show quoted text -
If you move toward a wall how does it set up space movement
coordinates toward you?
Mitch Raemsch
BURT
>
> - Show quoted text -
An acccelerated frame not by gravity experiences an opposite direction
weight and an appearence around it of opposite direction acceleration.
Mitch Raemsch
GogoJF
> "PD" <thedraperfam...@gmail.com> wrote in message
A paradox means knowledge in opposition to convention.
Androcles
"GogoJF" <jfgo...@yahoo.com> wrote in message
news:de8ed96a-73c5-48dc...@g17g2000yqe.googlegroups.com...
=================================================
Ever heard of a dictionary?
Definition of PARADOX
1
: a tenet contrary to received opinion
2
a : a statement that is seemingly contradictory or opposed to common sense
and yet is perhaps true b : a self-contradictory statement that at first
seems true c : an argument that apparently derives self-contradictory
conclusions by valid deduction from acceptable premises
3
: one (as a person, situation, or action) having seemingly contradictory
qualities or phases
Examples of PARADOX
1.. It is a paradox that computers need maintenance so often, since they
are meant to save people time.
2.. As an actor, he's a paradox-he loves being in the public eye but also
deeply values and protects his privacy.
Nothing there about knowledge, son.
"an argument that apparently derives self-contradictory conclusions by valid
deduction from acceptable premises"
In mathematics the hypothetically "acceptable" premises are unacceptable.
This is the unacceptable premise:
Ref:
http://www.fourmilab.ch/etexts/einstein/specrel/www/figures/img22.gif
What kind of lunacy prompted Einstein to say
the speed of light from A to B is c-v,
the speed of light from B to A is c+v,
the "time" each way is the same?
Answer that and you'll understand what a paradox is.
You can always claim he didn't say that as most do, but then he has no
argument either.
That's a paradox.
You can always claim the idiot was a genius, that's another paradox.
1) But it is not possible without further ASSUMPTION to compare, in respect
of time, an event at A with an event at B.
2) We ASSUME that this definition of synchronism is free from
contradictions, and possible for any number of points;
3) In agreement with experience we further ASSUME the quantity 2AB/(t'A-tA)
= c
4) Current kinematics tacitly ASSUMES that the lengths determined by these
two operations are precisely equal
5) and where for brevity it is ASSUMED that at the origin of k, tau = 0,
when t=0.
6) If no ASSUMPTION whatever be made as to the initial position of the
moving system and as to the zero point of tau
7) We now have to prove that any ray of light, measured in the moving
system, is propagated with the velocity c, if, as we have ASSUMED, this is
the case in the stationary system
8) If we ASSUME that the result proved for a polygonal line is also valid
for a continuously curved line,
9) and our equations ASSUME the form
10) When phi = 0 the equation ASSUMES the perspicuous form
11) the equation for phi' ASSUMES the form
12) for the law of motion of which we ASSUME as follows
13) we may and will ASSUME that the electron, at the moment when we give it
our attention
14) From the above ASSUMPTION, in combination with the principle of
relativity
All the above taken from Einstein's SR.
In agreement with experience we further ASSUME that SR is a collection of
assumptions.
In agreement with experience we further ASSUME that ASSUMPTIONS can be wrong
assumptions.
That's a paradox.
Jonathan Doolin
I appreciate the calculations. I'll come back later to ask about what
is the sign convention on R? What happens at R = G M / c^2? (The
integral becomes zero.) Is this the Rindler Horizon? Should this be
a lower-case g=9.8 m/s^2 or the G=universal gravitational constant.
i.e. I need to work through it, I guess.
But I don't want to lose focus on the acceleration vs. gravity issue.
I'm not convinced the "change in equilibrium length" phenomenon would
show up in any experimentally achievable acceleration.
Your analysis treats the equations of motion for the front and back of
the rocket as two hyperbolic arcs which are centered on the t=0 axis.
In a real acceleration, though, the front of the rocket would be
represented by a hyperbolic arc centered on t=L/c; i.e. the front of
the rocket would stop at a *later time* than the front of the rocket.
I'm pretty sure (I haven't done a detailed analysis), if this were the
case, the rocket would eventually be crushed in its own reference
frame, but you have said: "Now, there is one assumption that is
experimental, and not just a convention. when an object undergoes
constant acceleration, it will tend to reach an equilibrium length"
However, I doubt that the tiny amount of "constant" acceleration we
have been able to achieve by rocket fuel would create an effect large
enough to be measured. I'd like to see the magnitude of this effect
(the shrinking of the rocket due to offset hyperbolic arcs)
calculated, and compared to the results of the actual experimental
results you cite.
My hypothesis is that if you take into account the impulse delay, the
times on the clocks will remain synchronous, but (possibly) the rocket
will slowly crush. But in gravity, you'll find the building does not
crush, but the clocks become asynchronous.
Jonathan
Daryl McCullough
>I appreciate the calculations. I'll come back later to ask about what
>is the sign convention on R? What happens at R = G M / c^2?
That's the "event horizon" of a black hole. In the case of the Earth,
you never get to that radius, because on the surface of the Earth,
R >> GM/c^2. The significance of the event horizon is that a light
signal sent from within the event horizon will never reach outside
the horizon.
>(The integral becomes zero.) Is this the Rindler Horizon?
The Rindler horizon is at c^2/g behind an accelerating rocket
ship. Like the event horizon of a black hole, if a light signal
is sent from "below" the event horizon, it will never reach
the rocket that is always in front of the horizon.
>Should this be
>a lower-case g= 9.8 m/s^2 or the G=universal gravitational constant.
GM/R^2 is g. The event horizon is at R = GM/c^2
>I'm not convinced the "change in equilibrium length" phenomenon would
>show up in any experimentally achievable acceleration.
Almost certainly not. You can't accelerate a physical object to speeds
close to the speed of light (except for atomic particles). Length
contraction is not measurable in normal circumstances.
>Your analysis treats the equations of motion for the front and back of
>the rocket as two hyperbolic arcs which are centered on the t=0 axis.
>In a real acceleration, though, the front of the rocket would be
>represented by a hyperbolic arc centered on t=L/c;
No, that's not correct. The equations of motion for front
and rear are:
x_rear = square-root((ct)^2 + c^4/g^2)
x_front = square-root((ct)^2 + (c^4/g'^2))
where c^2/g'^2 = c^2/g^2 + L. (For convenience, I chose
the origin at the point c^2/g behind the rear of the rocket).
If instead we used your suggestion:
x_rear = square-root((ct)^2 + c^4/g^2)
x_front = square-root(c^2(t-L/c)^2 + (c^4/g'^2))
then we would find that eventually x_rear would overtake
x_front; asymptotically,
x_rear --> ct
x_front --> c(t-L/c)
That's not possible; it's not possible for x_rear to be greater
than x_front.
You are right, that the front will not start moving until a time
t = L/c. However, during that time, the rocket will become compressed,
and that compression will cause the front to accelerate a bit more,
and the rear to accelerate a bit less. The upshot, when the length
stabilizes, is that the front and rear will be following hyperbolic
paths with the same origin, but with different values of g.
>I'm pretty sure (I haven't done a detailed analysis), if this were the
>case, the rocket would eventually be crushed in its own reference
>frame, but you have said: "Now, there is one assumption that is
>experimental, and not just a convention. when an object undergoes
>constant acceleration, it will tend to reach an equilibrium length"
>
>However, I doubt that the tiny amount of "constant" acceleration we
>have been able to achieve by rocket fuel would create an effect large
>enough to be measured.
Definitely not. We can't measure the time dilation between front and
rear of a rocket. But we can measure the time dilation between bottom
and top of a tall mountain, and that has approximately the same magnitude.
>My hypothesis is that if you take into account the impulse delay, the
>times on the clocks will remain synchronous,
Definitely not.
>but (possibly) the rocket will slowly crush.
No, it won't. If there are stresses in the rocket, then those
stresses will cause the front and rear to accelerate slightly
more or less to compensate. It's not going to crush, any more
than having a rocket sitting in a gravitational field will
crush it. Now it is true that an accelerating rocket will be
slightly smaller than an unaccelerating rocket, in the same
way that a spring gets smaller when you push on it. But it's
not a big effect.
>But in gravity, you'll find the building does not
>crush, but the clocks become asynchronous.
The accelerating rocket behaves in the same way: the rocket
doesn't crush, and the clocks get out of sync.
Look, Einstein predicted gravitational time dilation from
considering the equivalence between acceleration and gravity
*BEFORE* he developed the full theory of General Relativity.
He had done the calculation for accelerating rockets, and he
assumed that the same effect would show up for a rocket sitting
on the Earth.
"Gravitational time dilation" was *FIRST* computed for an accelerating
rocket, and *THEN* was applied to real gravity.
harald
> RichD wrote:
> > Adam sits at the center of the earth. He feels no gravitational
> > field, he is an inertial frame.
>
> Hmmm. He most definitely is not at rest in an inertial frame in the sense of SR
> (the earth orbits the sun). But he is at rest in a LOCALLY inertial frame in the
> sense of GR. This difference is important in this situation.
>
> > Adam syncs his clock with Bob, who is orbiting the earth.
> > Bob is in free fall, he feels no gravitational field, he is an
> > inertial frame.
>
> Bob most definitely is not at rest in an inertial frame in the sense of SR (he
> is orbiting the earth, which orbits the sun). But he is at rest in a LOCALLY
> inertial frame in the sense of GR. This difference is important in this situation.
>
> Moreover, gravitation is important, because it is what permits Bob to orbit the
> earth.
>
> So this gedanken must be analyzed using GR; SR is inadequate.
>
> > From Adam's viewpoint, Bob accelerates during every
> > revolution, in speed and direction. Adam appears the
> > same, to Bob.
>
> Sure. But it turns out that acceleration does not affect their clocks. Nor does
> it directly affect how their readings compare [#].
>
> [#] The "twin paradox" in SR shows that acceleration can IN-directly
> affect how clocks compare.
>
> Moreover, this is GR and one must be careful about the meaning of
> "acceleration". For both Adam and Bob, their 4-acceleration is zero, as is their
> proper acceleration. The non-zero acceleration you mention is merely their
> COORDINATE acceleration when using the other person's coordinates. Such
> coordinate-dependent quantities are risky to use, because it is all too easy to
> confuse aspects of the coordinates with physical aspects of the situation. That
> is the case here: both follow geodesic paths, which in GR are analogous to the
> unaccelerated straight lines of SR.
>
> Even in SR, such straight-line inertial trajectories have nonzero
> coordinate acceleration if one uses accelerated coordinates.
>
> > What do their clocks say, after each encirclement by Bob?
>
> To answer questions like this, in either SR or GR, one must integrate the metric
> along the path of each clock, because that is the only way to determine the
> elapsed proper time of a clock. In SR the integral is usually easier than in GR,
> because the math in SR's flat manifold is simpler than that in the curved
> manifolds of GR.
>
> In this case, it is difficult to determine what you mean, because the clocks are
> never collocated -- only when they are collocated can they be compared
> definitively and unambiguously. So let me construct ECI (earth centered
> inertial) coordinates as in the GPS; Adam is at rest at the origin, and each
> turn of Bob's orbit can be unambiguously determined from his coordinate
> position. Let me assume the clocks are compared via light signals, with a
> correction for the propagation of the light. There are two competing effects: a)
> due to Bob's motion relative to the ECI, and b) due to his altitude -- (a) makes
> his clock compare with less elapsed proper time than Adam's, while (b) makes
> Bob's clock compare with more elapsed proper time than Adam's. So we cannot
> determine the result of the comparison easily, a real computation must be made.
> The result is that for low orbits the effect from speed outweighs the effect
> from altitude and Bob's clock experiences less elapsed proper time in each
> encirclement, but for high enough orbits Bob's clock experiences more elapsed
> proper time in each encirclement. There is a unique circular orbit for which
> they are equal; there are infinitely many elliptical orbits for which they are
> equal.
>
> A better gedanken is to consider two orbiting clocks, A in a circular orbit and
> B in a highly eccentric elliptical orbit, arranged so they repeatedly intersect
> and the clocks can be compared when they are collocated (the periods of the two
> orbits must have a ratio that is a rational number). This is a "twin paradox"
> with both twins following geodesic (locally inertial) paths. The orbits can be
> arranged so B experiences more elapsed proper time between meetings than A, by
> making B's orbit to be mostly outside A's. The orbits can be arranged so B
> experiences less elapsed proper time between meetings than A, by making B's
> orbit to be mostly inside A's. And they can experience equal elapsed proper
> times by choosing B's orbit to be an appropriate mixture of inside and outside
> A's orbit.
>
> Bottom line: the elapsed proper time of a clock between two designated points
> depends on its path between those points. This is true in both SR and GR. It is
> directly analogous to the obvious geometrical fact that the length of a path
> between two designated points depends on the path (e.g. two sides of a triangle
> always have a longer total length than the third side; the distance from New
> York to Chicago is longer if one detours through New Orleans).
>
> Note that this is related to "time dilation" in SR, but is not the same. There
> is no way to apply "time dilation" and obtain the correct answer; one must
> integrate the clocks' elapsed proper times over their trajectories.
>
> Tom Roberts
Note: Hafele and Keating applied kinematic and gravitational time
dilation, and the same has been done with GPS satellites.
See also:
http://en.wikipedia.org/wiki/Hafele-Keating_experiment
Harald
harald
(if I didn't overlook it) the last ones:
On Aug 31, 1:46 pm, Jonathan Doolin <good4us...@gmail.com> wrote:
> On Aug 26, 11:51 pm, Tom Roberts <tjroberts...@sbcglobal.net> wrote:
[..]
> From that, I gather that
>
> (1) higher speed implies slower clocks, (implicit in special
> relativity)
>
> (2) Higher altitude implies faster clocks. (or another way to say it
> would be deeper gravity well implies slower clocks.)
Although those conclusions are "wrong" according to Roberts, they are
simply correct according to Einstein as well as Hafele-Keating - take
your pick!
> Okay, one final question; since geodesics are supposed to lead to the
> highest possible amount of time, is it assumed in general relativity
> that these two effects *exactly* cancel out for objects in free-fall?
Those effects do NOT cancel out for objects in free-fall!
See "gravity-probe A"*: for the modeling they worked with the
assumption that, after correcting for true Doppler, the measured rate
of an atomic clock in a falling rocket slows down due to increasing
speed AND due to increasing gravitational potential.
The same assumption is applied to clocks in GPS satellites in strongly
elliptic trajectories: the highest speed is attained at the lowest
altitude, so that the two effects add up.
*http://en.wikipedia.org/wiki/Gravity_Probe_A
Of course, clocks in free-fall appear to run exactly at the usual rate
for any co-falling observer.
A very special and interesting case is on the surface of the earth,
where those effects DO cancel out: the higher kinetic energy at the
equator goes along with a higher potential energy thanks to the
flexibility of the earth's surface, so that --contrary to what
Einstein expected in 1905-- clocks all around the globe stay in tune
(neglecting other factors).
Harald
maxwell
A 'paradox' means a logical contradiction. However, paradox is used
in a metaphorical sense, hence the other 'definitions'.
Mathematicians use 'reductio ad absurdum' as a standard method of
proof - the impossibility of two conflicting conclusions. When
physicists try this game & end up with a similar result, they grace it
with the term 'paradox' - and move on.
Inertial
>in a metaphorical sense, hence the other 'definitions'.
>Mathematicians use 'reductio ad absurdum' as a standard method of
>proof - the impossibility of two conflicting conclusions. When
>physicists try this game & end up with a similar result, they grace it
>with the term 'paradox' - and move on.
And then claim whatever theory predicted that is refuted. Just as
mathematics does.
There are no paradoxes (ie contradictions) in SR. There are scenarios which
are informally and popularly labelled as 'paradoxes' because they APPEAR to
be paradoxical to the casual reader, or if the theory is incorrectly used.
BURT
You're the man that claims there is no paradox when Einstein proved
there is.
Mitch Raemsch
Inertial
news:c8f03922-5eda-4f92...@v35g2000prn.googlegroups.com...
Amazing .. you worked that out from me saying "There are no paradoxes in
SR".
> when Einstein proved there is.
He never did .. No one ever has. You're a liar or a moron. My bet is that
you're both.
BURT
Why would he use the term then?
Was it his imagination?
Mitch Raemsch
Inertial
news:12395f12-e3a1-4c56...@a4g2000prm.googlegroups.com...
Where did you see him use the term? Have you read anything he wrote?
Did you miss what I wrote above .. couldn't you get past the first sentence?
Here it is again for you.
==
There are scenarios which are informally and popularly labelled as
'paradoxes' because they APPEAR to be paradoxical to the casual reader, or
if the theory is incorrectly used.
==
>Was it his imagination?
No .. just your ignorance
BURT
>
> - Show quoted text -
Einstein did talk about his twin paradox. You cannot say he did not.
Mitch Raemsch
PD
He didn't.
You've heard the term "twin paradox". You've heard it associated with
relativity. You've heard that Einstein formulated relativity. And so
you assume that Einstein used the term "twin paradox". It was in fact
invented by OTHER people working on relativity -- Langevin,
originally.
Get yer facts straight.