Requirement for Clock Synchronization

[Justintruth]

I had always assumed that the synchronization of clocks at widely
separated locations to arbitrary accuracy would require a signal,
something that carries information, to travel between the two
locations arbitrarily fast.

If I assume that many entangled particles are being sent
simultaneously to two separated locations and measurements made at
both locations then my understanding of Bell's theorem is that no
information is conveyed on the conditions of measurement at one to the
other. B could not see an action of A in the data from his site. My
understanding is that this is because the effect underlying Bells
theorem is only visible in the correlation between events at the
distant sites and is invisible in the data from just one site. It
therefore requires a sub luminous message to be sent in order to see
the change in correlation.

However it now seems to me that by sending a simple record of the
actions taken at location A to location B and the record of times on
A's local clock that the actions were taken, and by pre-arranging the
actions taken, that B could just lay the data next to his own when he
gets it and see where they changed correlation. B would be able to set
his clock to arbitrary accuracy then and that would be independent of
the time it took the sub luminous message to arrive because he is
basing it solely on the correlations.

Therefore it seems to me that the times at which widely separated
events occurred could be established absolutely even though there is
no signal with which to do the correlating as the correlating
procedure is independent of the speed at which the information was
transmitted but only dependent on its content.

As an experiment I took a spread sheet and filled a set of integers
from 0 to 100 down a column. In the next column I set random bits. In
the next column I set a set of bits that were completely uncorrelated
with the second column for the first half of the rows and completely
correlated with the second column for the last half of the rows.

I then assumed that I did not know what bit went with what bit but I
did know that somewhere the correlation should change and I assumed
that I was also told that it should change at the "50" count of the
local clock at A.. By sliding one set of numbers (B's) past the other
(A's) I got one point at which there was this property of correlation
emerging. I then had the bits matched up. I could therefore set B's
clock to the 50 at the point where the bits became correlated..

No super luminous message was required in this procedure but still I
could set the clocks to arbitrary accuracy?

So if two frames of reference moving relative to one another disagree
on the present can't I use this procedure to establish a time on the
clock that is independent of frame of reference to disambiguate?

What am I doing wrong?

[Jonathan Thornburg [remove -animal to reply]]

Justintruth <justi...@hotmail.com> wrote:
> I had always assumed that the synchronization of clocks at widely
> separated locations to arbitrary accuracy would require a signal,
> something that carries information, to travel between the two
> locations arbitrarily fast.

There are (at least) two different clock-synchronization methods
that don't require "arbitrarily fast" signalling:
* slow transport of a clock from one location to the other
(here "slow" means "slow enough that special-relativistic time
dialation effects are negligible)
* Einstein clock synchronisation: send a time-stamped signal (often,
but not necessarily, light) round-trip from A to B and back to A
again; this allows the round-trip travel time T to be measured, after
which B can set her clock to the time-tagged signal received from A,
offset by -T/2 to correct for time-of-flight
http://en.wikipedia.org/wiki/Einstein_synchronisation

According to special relativity, these give the same result.
Various experiments have confirmed the self-consistency of this aspect
of special relativity to very high precision, e.g.

Wolf and Petit
"Satellite test of special relativity using the global positioning system"
Physical Review A, volume 56, pages 4405-4409
http://link.aps.org/doi/10.1103/PhysRevA.56.4405

--
-- "Jonathan Thornburg [remove -animal to reply]" <jth...@astro.indiana-zebra.edu>
Dept of Astronomy & IUCSS, Indiana University, Bloomington, Indiana, USA
"Washing one's hands of the conflict between the powerful and the
powerless means to side with the powerful, not to be neutral."
-- quote by Freire / poster by Oxfam

[Tom Roberts]

On 2/4/12 2/4/12 - 5:16 AM, Justintruth wrote:
> I had always assumed that the synchronization of clocks at widely
> separated locations to arbitrary accuracy would require a signal,
> something that carries information, to travel between the two
> locations arbitrarily fast.

Hmmmm. The world we inhabit does not seem to admit "arbitrarily fast" signals.
Certainly no human has ever observed such signals, much less used them for anything.

The world we inhabit also imposes an important restriction on one's ability to
synchronize a given pair of clocks and have them remain synchronized: they MUST
both be at rest in a single inertial frame. Note that inertial frames are of
limited extent and accuracy in our world, so this is a strong limitation.

One can use signals of any speed, as long as either the speed of the signal is
known, or it is known that the speed of the signals is isotropic. Since "speed"
is ALWAYS measured relative to some coordinate system, and since clock
synchronization is impossible unless the clocks are at rest in the same inertial
frame, it's clear that this is only needed relative to the coordinates of that
inertial frame.

Einstein, of course, used light in vacuum because we know its speed, and we know
it is isotropic relative to any inertial frame.

The knowledge comes from experiments; depending on one's
outlook and application, one might consider this speed to
have an errorbar from pre-1983 measurements, or one might
accept it as exact.


> [... many entangled particles used to synchronize a pair of clocks...]

> What am I doing wrong?

Your description was a bit vague so I assume that A and B start out measuring
perpendicular spin polarizations, so there is no correlation, and at some
instant A switches to measuring polarizations parallel to B so there is
correlation, and this can be determined after-the-fact from the records. Note
also that the particles must be sent to A and B from their midpoint, and at the
same speed relative to their common inertial frame.

Yes, A and B can use that determine the correct match-up of the sequence of
particles, and can use this to synchronize their clocks. But it would be MUCH
easier to modulate the particles being sent to both places and use that to match
them up.

NOTE: there is nothing "absolute" about this, because two clocks cannot remain
in synch unless they are at rest in the same inertial frame [#] -- in that case
it is clear that the clocks are synchronized in that inertial frame AND NO
OTHER. Using a quantum correlation to initially synchronize the clocks does not
change this basic fact about the world we inhabit.

[#] If one is willing to accept non-standard meanings of
"in synch", then I suppose that other physical situations
could be said to be "in synch". The standard meaning of
this phrase necessarily includes an inertial frame.


Tom Roberts

[harald]

"Justintruth" <justi...@hotmail.com> wrote in message
news:ec9bd8d4-820d-4944...@t30g2000vbx.googlegroups.com...

>I had always assumed that the synchronization of clocks at widely
> separated locations to arbitrary accuracy would require a signal,
> something that carries information, to travel between the two
> locations arbitrarily fast.

[..]

> However it now seems to me that by sending a simple record of the
> actions taken at location A to location B and the record of times on
> A's local clock that the actions were taken, and by pre-arranging the
> actions taken, that B could just lay the data next to his own when he
> gets it and see where they changed correlation. B would be able to set
> his clock to arbitrary accuracy then and that would be independent of
> the time it took the sub luminous message to arrive because he is
> basing it solely on the correlations.

[..]

> As an experiment I took a spread sheet and filled a set of integers
> from 0 to 100 down a column. In the next column I set random bits. In
> the next column I set a set of bits that were completely uncorrelated
> with the second column for the first half of the rows and completely
> correlated with the second column for the last half of the rows.
>
> I then assumed that I did not know what bit went with what bit but I
> did know that somewhere the correlation should change and I assumed
> that I was also told that it should change at the "50" count of the
> local clock at A.. By sliding one set of numbers (B's) past the other
> (A's) I got one point at which there was this property of correlation
> emerging. I then had the bits matched up. I could therefore set B's
> clock to the 50 at the point where the bits became correlated..
>
> No super luminous message was required in this procedure but still I
> could set the clocks to arbitrary accuracy?

Not sure if this is what you have in mind, but there is no issue whatsoever
about clocks that slide next to each other; they can always be compared.
However, different systems disagree on what is the "correct"
synchronization.

It is unclear to me how you simulate an alternative procedure with your
sliding number sets. Perhaps you can make a sketch of what you have in mind?
In particular how do you quantify (in an "absolute" way) your statement:
"B could just lay the data next to his own WHEN he gets it and see WHERE
they changed correlation."

> So if two frames of reference moving relative to one another disagree
> on the present can't I use this procedure to establish a time on the
> clock that is independent of frame of reference to disambiguate?

They don't necessarily disagree on the present but on distant simultaneity
along one direction.

> What am I doing wrong?

Perhaps you misunderstand the problem?

[Thomas Smid]

On Feb 7, 5:16am, Tom Roberts <tjrob...@sbcglobal.net> wrote:

> two clocks cannot remain
> in synch unless they are at rest in the same inertial frame

That it is obviously not correct: if one has two clocks going into
opposite directions with the same (arbitrary) speed, then the clocks
must stay synchronized if they were synchronized at the origin
(provided the two clocks are physically identical). This is a logical
consequence of the symmetry of the situation.

Thomas

[Tom Roberts]

Which is why I added a footnote, which you ignored:

[#] If one is willing to accept non-standard meanings of
"in synch", then I suppose that other physical situations
could be said to be "in synch". The standard meaning of
this phrase necessarily includes an inertial frame.

The case you cite is one of those mentioned in the footnote.

But in EVERY case, including the exceptions of the footnote and your example,
the synchronization is valid only in a single inertial frame. If the clocks
aren't at rest in that frame, then they must be moving at the same speed
relative to that frame, or they cannot remain in synch in that frame.

As I said, the phrase "in synch" normally applies only to clocks at rest in a
specific inertial frame, and the synchronization applies only in that frame.


Tom Roberts

[Thomas Smid]

On Feb 14, 9:53 am, Tom Roberts <tjrob...@sbcglobal.net> wrote:
> On 2/13/12 2/13/12 - 7:58 AM, Thomas Smid wrote:
>
> > On Feb 7, 5:16am, Tom Roberts<tjrob...@sbcglobal.net>  wrote:
> >>   two clocks cannot remain
> >> in synch unless they are at rest in the same inertial frame
>
> > That it is obviously not correct: if one has two clocks going into
> > opposite directions with the same (arbitrary) speed, then the clocks
> > must stay synchronized if they were synchronized at the origin
> > (provided the two clocks are physically identical). This is a logical
> > consequence of the symmetry of the situation.
>
> Which is why I added a footnote, which you ignored:
>          [#] If one is willing to accept non-standard meanings of
>          "in synch", then I suppose that other physical situations
>          could be said to be "in synch". The standard meaning of
>          this phrase necessarily includes an inertial frame.

I ignored the footnote because you didn't give any proof there for
your assertion that 'two clocks cannot remain in synch unless they are
at rest in the same inertial frame'. Saying that 'The standard meaning
of this phrase necessarily includes an inertial frame' is circular
reasoning if you want to support your initial statement that 'there is
nothing "absolute" about this' (because if you define 'in synch' as a
frame dependent concept, then it can obviously not be absolute).

But let's spin my example of the two clocks moving with the same
velocity into opposite directions a bit further. Assume that the
clocks hit a brick wall after some distance x (and -x) and that this
stops both clocks at the corresponding times. The question whether the
clocks were 'in synch' or not can then simply be answered by picking
the clocks up from the foot of the brick wall and comparing the times
they show. And this will be a unique and absolute result, independent
of any external observers and their state of motion.

Thomas

[harald]

"Tom Roberts" <tjro...@sbcglobal.net> wrote in message
news:ROednWhKqd1...@giganews.com...

> On 2/13/12 2/13/12 - 7:58 AM, Thomas Smid wrote:
>> On Feb 7, 5:16am, Tom Roberts<tjrob...@sbcglobal.net> wrote:
>>> two clocks cannot remain
>>> in synch unless they are at rest in the same inertial frame
>>
>> That it is obviously not correct: if one has two clocks going into
>> opposite directions with the same (arbitrary) speed, then the clocks
>> must stay synchronized if they were synchronized at the origin
>> (provided the two clocks are physically identical). This is a logical
>> consequence of the symmetry of the situation.
>
> Which is why I added a footnote, which you ignored:
> [#] If one is willing to accept non-standard meanings of
> "in synch", then I suppose that other physical situations
> could be said to be "in synch". The standard meaning of
> this phrase necessarily includes an inertial frame.
>
> The case you cite is one of those mentioned in the footnote.
>
> But in EVERY case, including the exceptions of the footnote and your
> example, the synchronization is valid only in a single inertial frame. If
> the clocks aren't at rest in that frame, then they must be moving at the
> same speed relative to that frame, or they cannot remain in synch in that
> frame.

That is still imprecise: clocks that are moving at a speed relative to
an inertial frame only remain in sync with clocks that are at rest in
that frame if their rate is adjusted. And apart of imperfections, clocks
in circular orbit can remain continuously in synch with clocks on earth
as well as with other clocks in a different circular orbit if their
rates have been properly adjusted, such that they all remain in synch in
the ECI frame. There is nothing impossible about that.

[holog]

> But let's spin my example of the two clocks moving with the same
> velocity into opposite directions a bit further. Assume that the
> clocks hit a brick wall after some distance x (and -x) and that this
> stops both clocks at the corresponding times. The question whether the
> clocks were 'in synch' or not can then simply be answered by picking
> the clocks up from the foot of the brick wall and comparing the times
> they show. And this will be a unique and absolute result, independent
> of any external observers and their state of motion.
>
> Thomas

to say , motion slows matter , where did you bring them together to
observe this, back at the starting point or not?

holog

[holog]

> That is still imprecise: clocks that are moving at a speed relative to
> an inertial frame only remain in sync with clocks that are at rest in
> that frame if their rate is adjusted. And apart of imperfections, clocks
> in circular orbit can remain continuously in synch with clocks on earth
> as well as with other clocks in a different circular orbit if their
> rates have been properly adjusted, such that they all remain in synch in
> the ECI frame. There is nothing impossible about that.

since nothing in our universe is truly at rest , and there is no
reference point to adjust to , time is dependent on where the the
observation is made. unless one can get beyond the electromagnet
spectrum ,

holog

[Jos Bergervoet]

On Feb 15, 12:56=A0am, Thomas Smid <thomas.s...@gmail.com> wrote:

> On Feb 14, 9:53=A0am, Tom Roberts <tjrob...@sbcglobal.net> wrote:
> > On 2/13/12 2/13/12 - 7:58 AM, Thomas Smid wrote:

...

> > > That it is obviously not correct: if one has two clocks going into
> > > opposite directions with the same (arbitrary) speed, then the clocks
> > > must stay synchronized if they were synchronized at the origin
> > > (provided the two clocks are physically identical). This is a logical
> > > consequence of the symmetry of the situation.
>
> > Which is why I added a footnote, which you ignored:

> > =A0 =A0 =A0 =A0 =A0[#] If one is willing to accept non-standard meaning=
s of
> > =A0 =A0 =A0 =A0 =A0"in synch", then I suppose that other physical situa=
tions
> > =A0 =A0 =A0 =A0 =A0could be said to be "in synch". The standard meaning=
of
> > =A0 =A0 =A0 =A0 =A0this phrase necessarily includes an inertial frame.

>
> I ignored the footnote because you didn't give any proof there for
> your assertion that 'two clocks cannot remain in synch unless they are
> at rest in the same inertial frame'.

No proof is needed if one refers to a "standard meaning"
of a concept. Otherwise every word in the posts in this
newsgroup would require an added proof.

> Saying that 'The standard meaning
> of this phrase necessarily includes an inertial frame' is circular
> reasoning if you want to support your initial statement that 'there is
> nothing "absolute" about this' (because if you define 'in synch' as a
> frame dependent concept, then it can obviously not be absolute).

Do not accuse Thomas Smid. *Others* have defined it in that way
since it is the standard meaning..

> But let's spin my example of the two clocks moving with the same
> velocity into opposite directions a bit further.

Apparently you want to extend the standard meaning and also
allow opposite movement (instead of only equal movement).

> Assume that the
> clocks hit a brick wall after some distance x (and -x)

The distance is dependent on the movement of the observer.
Suppose the observer moves along with one of the clocks,
then he will say that clock has travelled no distance at all
when hitting your wall! (While the other certainly has.)

> and that this
> stops both clocks at the corresponding times. The question whether the
> clocks were 'in synch' or not can then simply be answered by picking
> the clocks up from the foot of the brick wall and comparing the times
> they show. And this will be a unique and absolute result, independent
> of any external observers and their state of motion.

No. Only for the *special* observer that saw them moving
distances -x and +x. Other observers see unequal distances.
They would say that you positioned the brick walls incorrectly
and that with correct placement the clocks would *not* be in
sync according to your own definition.

--
Jos

[Tom Roberts]

On 2/14/12 2/14/12 5:56 PM, Thomas Smid wrote:
> But let's spin my example of the two clocks moving with the same

> velocity into opposite directions a bit further. Assume that the
> clocks hit a brick wall after some distance x (and -x) and that this

> stops both clocks at the corresponding times. The question whether the
> clocks were 'in synch' or not can then simply be answered by picking
> the clocks up from the foot of the brick wall and comparing the times
> they show. And this will be a unique and absolute result, independent
> of any external observers and their state of motion.

Perhaps, depending on the meanings of your words. If both clocks stopped
ticking but could still be read, then yes (but the two objects you
called "clocks" ceased to be clocks when they stopped ticking). If they
keep ticking, then no.

Does your "this stops both clocks" mean they stopped moving or
stopped ticking?

And still, if an observer were traveling past as this scenario unfolds,
that observer would not agree that the clocks hit their walls
simultaneously to her.

This synchronization among clocks is valid ONLY in a single inertial
frame. As best we know, this seems to be a property of the world we
inhabit, so our best theories of physics also have this property.

Tom Roberts

[Thomas Smid]

I doesn't matter. The clocks have stopped when they hit the wall.

Thomas

[Thomas Smid]

On Feb 16, 8:51 am, Jos Bergervoet <jos.r.bergerv...@gmail.com> wrote:

> On Feb 15, 12:56 am, Thomas Smid <thomas.s...@gmail.com> wrote:

> > Saying that 'The standard meaning
> > of this phrase necessarily includes an inertial frame' is circular
> > reasoning if you want to support your initial statement that 'there is
> > nothing "absolute" about this' (because if you define 'in synch' as a
> > frame dependent concept, then it can obviously not be absolute).
>
> Do not accuse Thomas Smid. *Others* have defined it in that way
> since it is the standard meaning..

I did not accuse. I just pointed out that Tom Robert's argument
(whether you call it standard meaning or not) effectively amounts to
saying that the clock synchronization is not absolute because it is
relative. This is obviously a logical tautology, so the footnote does
not do anything to support the claim that 'there is nothing absolute
about this'.

>
> > Assume that the
> > clocks hit a brick wall after some distance x (and -x)
>
> The distance is dependent on the movement of the observer.
> Suppose the observer moves along with one of the clocks,
> then he will say that clock has travelled no distance at all
> when hitting your wall! (While the other certainly has.)

But suppose a second observer is moving with the other clock. The two
observers couldn't possibly agree about the outcome of the experiment,
leading to a logical paradox, as certainly the clocks will show a
definitive time each when they are examined after impact with the
brick walls.

Thomas

[holog]

>
> But let's spin my example of the two clocks moving with the same
> velocity into opposite directions a bit further. Assume that the
> clocks hit a brick wall after some distance x (and -x) and that this
> stops both clocks at the corresponding times. The question whether the
> clocks were 'in synch' or not can then simply be answered by picking
> the clocks up from the foot of the brick wall and comparing the times
> they show. And this will be a unique and absolute result, independent
> of any external observers and their state of motion.
>
> Thomas

get a timex , it takes a licking and keeps on ticking, mechanical
devices are not a good analogy, anything mechanical that moves will be
affected, and there is no bringing it back, two separate clocks will
be different

holog

[holog]

> But let's spin my example of the two clocks moving with the same
> velocity into opposite directions a bit further. Assume that the
> clocks hit a brick wall after some distance x (and -x) and that this
> stops both clocks at the corresponding times. The question whether the
> clocks were 'in synch' or not can then simply be answered by picking
> the clocks up from the foot of the brick wall and comparing the times
> they show. And this will be a unique and absolute result, independent
> of any external observers and their state of motion.
>
> Thomas

the mere "picking the clock up" will change its "time" ,
acceleration slows matter

holog

[Tom Roberts]

On 2/15/12 2/15/12 4:23 PM, harald wrote:
> "Tom Roberts"<tjro...@sbcglobal.net> wrote in message
> news:ROednWhKqd1...@giganews.com...

>> But in EVERY case, including the exceptions of the footnote and your
>> example, the synchronization is valid only in a single inertial frame. If
>> the clocks aren't at rest in that frame, then they must be moving at the
>> same speed relative to that frame, or they cannot remain in synch in that
>> frame.
>
> That is still imprecise: clocks that are moving at a speed relative to
> an inertial frame only remain in sync with clocks that are at rest in
> that frame if their rate is adjusted.

I was discussing standard clocks, synchronized with each other (not with
coordinate clocks of some inertial frame).

> And apart of imperfections, clocks
> in circular orbit can remain continuously in synch with clocks on earth
> as well as with other clocks in a different circular orbit if their
> rates have been properly adjusted, such that they all remain in synch in
> the ECI frame. There is nothing impossible about that.

No, but that is a different meaning of "clock" than I used. The lack of
precision in the English language is often a poor match to the needs of the
technical vocabulary of physics. A "modified clock" is not the same as "a
clock", and your objection hinges on not distinguishing between them.

After your circumlocutions are understood, and also mine, my main point here
remains: this "synchronization" among "clocks" is valid ONLY in a single
inertial frame (regardless of the details of the words in quotes). As best we

[harald]

"Thomas Smid" <thoma...@gmail.com> wrote in message
news:a6480be6-877e-408c...@i2g2000vbv.googlegroups.com...

> On Feb 16, 8:51 am, Jos Bergervoet <jos.r.bergerv...@gmail.com> wrote:
>> On Feb 15, 12:56 am, Thomas Smid <thomas.s...@gmail.com> wrote:
>

[..]

>> > Assume that the
>> > clocks hit a brick wall after some distance x (and -x)
>>
>> The distance is dependent on the movement of the observer.
>> Suppose the observer moves along with one of the clocks,
>> then he will say that clock has travelled no distance at all
>> when hitting your wall! (While the other certainly has.)
>
> But suppose a second observer is moving with the other clock. The two
> observers couldn't possibly agree about the outcome of the experiment,
> leading to a logical paradox, as certainly the clocks will show a
> definitive time each when they are examined after impact with the
> brick walls.

I suppose that you mean a second observer who sets up his/her own
system, using the standard clock synchronization. That observer will
thus use a different clock synchronization than the first one. When
calculating it with SR, the two observers *will* agree on the outcome:
SR takes care of that. The calculations for one clock moving and the
other not, as well as for the two clocks going in different directions,
are not the same with respect to such a moving system.

Harald

[Thomas Smid]

On Feb 16, 10:31 pm, Tom Roberts <tjroberts...@sbcglobal.net> wrote:

> Perhaps, depending on the meanings of your words. If both clocks stopped
> ticking but could still be read, then yes (but the two objects you
> called "clocks" ceased to be clocks when they stopped ticking). If they
> keep ticking, then no.

> Does your "this stops both clocks" mean they stopped
> moving or stopped ticking?

Obviously I was implying that the clocks stop on impact. This is how
you normally measure the duration of events: you start the clock at
the beginning of the event and you stop the clock at the end. The
reading gives you the duration, assuming obviously that the clock runs
in sync with some time standard. Reversely, if you know the duration
of the event in advance, then this gives you the opportunity to check
whether the clock actually runs in synch with the time standard. The
latter scenario is indeed what my example is addressing: you know the
distance the clock travels and you know the velocity, so you know the
time it should show when it hits the wall. And whether the time it
shows uppn subsequent examination agrees with your expection in this
sense or not, this won't be affected by what some 'observer' thinks it
should show. So there really is no room for any disagreement about the
result here. The two clocks are either in synch or they aren't.

Thomas

[Thomas Smid]

On Feb 18, 9:18 am, "harald" <h...@swissonline.ch> wrote:
> "Thomas Smid" <thomas.s...@gmail.com> wrote in message

> > But suppose a second observer is moving with the other >clock. The two
> > observers couldn't possibly agree about the outcome of the >experiment,
> > leading to a logical paradox, as certainly the clocks will >show a
> > definitive time each when they are examined after impact >with the
> > brick walls.

> I suppose that you mean a second observer who sets up his/her
> own system, using the standard clock synchronization. That
> observer will thus use a different clock synchronization than
> the first one. When calculating it with SR, the two observers
> *will* agree on the outcome:

As far as the clock rates are concerned, an observer moving with clock
A and an observer moving with clock B could only agree about it if
they conclude their clock rates are identical (i.e. in synch). Only
this would be consistent with the fact that the clocks will show
identical times after they have been stopped by the brick wall
(assuming the clocks are physically identical).

Thomas

[harald]

"Thomas Smid" <thoma...@gmail.com> wrote in message

news:29b03ed4-9fcb-4b35...@t24g2000yqj.googlegroups.com...

That is not a logical paradox but basic SR; what you present is a typical SR
calculation exercise.

According to SR they will agree about the outcome of such experiments;
moreover, they will agree to disagree about the clock rates. According to
the second observer, the one clock will be moving fast, while the other
clock will be at rest. SR predicts different clock rates (time dilation) and
relative velocities (velocity "addition" rule), and these effects compensate
each other for the outcome of such experiments.

Harald

[Tom Roberts]

When you use words so loosely, and in disagreement with their standard
meanings in physics, it's no wonder you are confused (and probably don't
even realize it).

Look at your first sentence, then look at the last sentence of mine that
you quoted -- then consider why you did not bother to make the VERY
IMPORTANT distinction I did. With such a cavalier attitude towards
precision in thought and word, you will not be able to understand the
subtleties of modern physics until you GREATLY improve your approach and
become MUCH more careful and precise.

Let me discuss just your last two sentences. Yes, there is no room for
disagreement here: after they stop ticking, the objects you call
"clocks" are not clocks in the sense the word is used in physics; they
are no more useful than paintings of clocks. There is no disagreement
about PAINTINGS OF CLOCKS THAT SHOW THEIR HANDS TO POINT THE SAME. But
for clocks (not paintings), there is no "absolute" statement like "they
are either in synch or they aren't", because different observers can
legitimately differ (because they compare them differently, using the
different definitions of simultaneity in their different rest frames).

Tom Roberts

[Thomas Smid]

On Feb 20, 1:39 pm, "harald" <h...@swissonline.ch> wrote:
> "Thomas Smid" <thomas.s...@gmail.com> wrote in message

> > As far as the clock rates are concerned, an observer moving with clock
> > A and an observer moving with clock B could only agree about it if
> > they conclude their clock rates are identical (i.e. in synch). Only
> > this would be consistent with the fact that the clocks will show
> > identical times after they have been stopped by the brick wall
> > (assuming the clocks are physically identical).
>
> > Thomas
>

> That is not a logical paradox but basic SR; what you present is a typical SR
> calculation exercise.
>
> According to SR they will agree about the outcome of such experiments;
> moreover, they will agree to disagree about the clock rates. According to
> the second observer, the one clock will be moving fast, while the other
> clock will be at rest. SR predicts different clock rates (time dilation) and
> relative velocities (velocity "addition" rule), and these effects compensate
> each other for the outcome of such experiments.
>
> Harald

Let's make the example a bit clearer, to have a better definition of
the relevant quantities here: assume instead of the brick wall that
the clocks (moving again with the same velocity v in opposite
directions in the lab frame) register the corresponding times when
touching a series of mechanical contacts mounted equidistantly at
distances dx along their direction of travel. Both clocks will
therefore touch the contacts in time increments of dt=dx/v, and this
is what the registered time series in both clocks should show provided
they are physically identical. So on what basis (in terms of
experimental evidence) could one then possibly disagree about the
clock rates, considering that both registered time-series match each
other entry for entry?

Thomas

[harald]

"Thomas Smid" <thoma...@gmail.com> wrote in message

news:82521744-5c48-43e3...@t30g2000vbx.googlegroups.com...

Not so according to SR. As I mentioned, your account completely neglects
time dilation and velocity transformation. It also doesn't account for
length contraction. Thus, the following distinctions are missing:

- which reference system (S, S', S'', ...)
- which clock (C1, C2)
- the corresponding distance measurement (dx, dx', dx'', ...)
- the corresponding velocity (v1, v2, v1', v2', v1'', v2'', ...)

We have the choice out of 3 convenient reference systems (out of an infinite
number):
- S: co-moving with the the contacts and/or brick walls (the "lab frame")
- S': co-moving with clock 1
- S'': co-moving with clock 2

> So on what basis (in terms of
> experimental evidence) could one then possibly disagree about the
> clock rates, considering that both registered time-series match each
> other entry for entry?

The experimental evidence agrees with SR insofar as gravitational effects
don't play a role: very fast moving atomic clocks tick measurably slower.
Please take your pick of S, S' or S'' and redo your analysis separately for
each clock and with max. precision.

Harald

[Thomas]

Words are not enough in science. It also required to substantiate then
through hard facts i.e. experimental results (in these case actual
clock readings). It is here were any disagreement about the clock
rates would become paradoxical. See also my previous post (reply to
Harald).

Thomas

[Thomas Smid]

On Feb 24, 8:49=A0am, "harald" <h...@swissonline.ch> wrote:
> "Thomas Smid" <thomas.s...@gmail.com> wrote in message

> > Let's make the example a bit clearer, to have a better definition of
> > the relevant quantities here: assume instead of the brick wall that
> > the clocks (moving again with the same velocity v in opposite
> > directions in the lab frame) register the corresponding times when
> > touching a series of mechanical contacts mounted equidistantly at
> > distances dx along their direction of travel. Both clocks will

> > therefore touch the contacts in time increments of dt=3Ddx/v, and this

> > is what the registered time series in both clocks should show provided
> > they are physically identical.
>

> > So on what basis (in terms of
> > experimental evidence) could one then possibly disagree about the
> > clock rates, considering that both registered time-series match each
> > other entry for entry?


> The experimental evidence agrees with SR insofar as gravitational effects
> don't play a role: very fast moving atomic clocks tick measurably slower.

> Please take your pick of S, S' or S'' and redo your analysis separately f=

or
> each clock and with max. precision.

Experimental evidence where both clocks show the same reading at the
end (as in my thought experiment)? You have to explain to me how this
is evidence for different clock rates in the two clocks (or how this
evidence could be obtained from the data under this constraint).

Thomas

[harald]

On Feb 25, 5:32=A0pm, Thomas Smid <thomas.s...@gmail.com> wrote:

> On Feb 24, 8:49=3DA0am, "harald" <h...@swissonline.ch> wrote:
>
>> > Let's make the example a bit clearer, to have a better definition of
>> > the relevant quantities here: assume instead of the brick wall that
>> > the clocks (moving again with the same velocity v in opposite
>> > directions in the lab frame) register the corresponding times when
>> > touching a series of mechanical contacts mounted equidistantly at
>> > distances dx along their direction of travel. Both clocks will
>> > therefore touch the contacts in time increments of dt=3Ddx/v, and this
>> > is what the registered time series in both clocks should show provided
>> > they are physically identical.
>>
>> > So on what basis (in terms of
>> > experimental evidence) could one then possibly disagree about the
>> > clock rates, considering that both registered time-series match each
>> > other entry for entry?
>

>> The experimental evidence agrees with SR insofar as gravitational effect=
s
>> don't play a role: very fast moving atomic clocks tick measurably slower=

>> Please take your pick of S, S' or S'' and redo your analysis separately

>> for each clock and with max. precision.
>> [..]

>
> Experimental evidence where both clocks show the same reading at the
> end (as in my thought experiment)? You have to explain to me how this
> is evidence for different clock rates in the two clocks (or how this
> evidence could be obtained from the data under this constraint).
>
> Thomas

Your conclusion lacks again the necessary precision: according to S' and
S'' the clock rates will differ from each other, while according to S
the clock rates will be equal to each other, with time increments that
differ from dt=3Ddx/v (I notice that you didn't do the analysis that
would have told you this).

While your imagined setup has probably not been tested (that would be
very lucky), the experimental support for SR is summarized here:

http://math.ucr.edu/home/baez/physics/Relativity/SR/experiments.html

Harald

[Justintruth]

> Yes, A and B can use that determine the correct match-up of the sequence of
> particles, and can use this to synchronize their clocks. But it would be MUCH
> easier to modulate the particles being sent to both places and use that to match
> them up.
>

I don't think you can "modulate the particles being sent to both
places and use that to match them up". Here's why:

Assume I modulate an incrementing counter on a wave and send it in two
directions at the same time. Say one goes to Alice and one to Bob. Now
Alice and Bob know the distance to the emitter and they divide by c to
find the time that a given count was sent and then set their clocks
accordingly. Thereafter there will be a fixed difference between what
they observe on their clock and what they receive in the signal. Note
that they cannot just set the time to what they receive. They must
back out via calculation when what they receive was transmitted in
order to synchronize.

Does that establish an absolute frame? No because in the second frame
the same procedure is used to synchronize its clocks and when it does
and then records the receipts it will disagree that the time of
receipts by Alice and Bob of a given count is the same. How could it
be as let's say Alice is moving toward where the count was transmitted
and Bob away and so the light will have to travel a shorter distance
to Alice and a longer one to Bob.

Now look at the procedure I am proposing. We place at the center a
generator of entangled states and send them to Alice an Bob every
second. Alice measure the state one way until a given time on her
clock and then "turns the handle" of let's say a polarizer and then
sends the entire record of her measurements to Bob any way she likes.
Bob the compares Alice's measurements with his own and sets his clock.

Now look at that from another frame of reference. Remember that the
effect of turning the handle takes place instantaneously. So the
correlations will change for all observers independent of frame.
Therefore there is no need to "back out via calculation" when the
package was sent. In fact you could store it for a year and not let me
know and I still could correlate the clocks. This constitutes a frame
of reference that can be defined to be the "absolute" frame or the
"frame of the ether" if you prefer.

What is wrong with my thinking?

[Justintruth]

> Not sure if this is what you have in mind, but there is no issue whatsoever
> about clocks that slide next to each other; they can always be compared.
> However, different systems disagree on what is the "correct"
> synchronization.
>

Actually they can be compared only at one point in time as they then
are no longer co-located at any other. The clock at my zero (space)
can be compared to your clock at your zero (space) for one tick but
the second tick will have your clock at your zero (space) being
compared to a clock of mine that is not at my zero (space) . You
cannot therefore establish an interval without being able to
synchronize distant clocks.

But that is not what I had in mind.

Einstein's theory is based on the following fact: If I take a light
wave and send it out to a distant point and reflect it back to the
origin I can know (observe) the round trip time of the light but I
cannot know (observe) whether this time is evenly divided into an
outbound period of half the duration and an inbound period of half the
duration. If I could know (observe) this then I could use it to
determine whether two events were simultaneous absolutely. Because I
cannot know (observe) this each frame is free to assume that relative
to it the time period is evenly divided. Once I do that then the
frames disagree on simultaneity and the theory evolves. It is
important to realize that the constancy of the speed of light is an
assumption in Einstein's theory because there is no way to observe the
one way speed. If you could then his theory would not work as you
would use that observation as a basis to determine which frame was
"correct".

My point is that there may be a way to make such an observation.

It appears to me that Bell's theorem will allow us to establish that
frame of reference for which a change in the state at one point is
simultaneous with observations at another point. I had thought that
the fact that Bell's theorem requires correlation and correlation
requires information to be transmitted by a signal that travels as
fast or slower than light meant that Bell's theorem could not be used
to synchronize the clocks. "What I have in mind" is that that "fact"
may be wrong. I think I have a procedure that might synchronize two
clocks that does not depend on assuming anything about an intervening
signal but that can be based on an observed change in the correlation.
You could mail the results and the mailman could stop and have a
cheeseburger on the way without you knowing and it would be no
problem. It does not depend on knowing anything about the speed of any
signal.

Actually I do not think I am right. I am probably doing something
wrong but.... it seems to work... that is my problem.

If it is true, and I am not wrong, then a new Michelson Morly like
experiment could be done that determines the motion through the ether
by seeing what the arrival time stamps were at distant clocks
synchronized by correlating measurements of coupled quantum states.
The "ether" would be the frame of reference in which the change in
correlation of measurements at one site (Alice) is simultaneous (by
definition) with a measurement taken at the other site (Bob). As long
as both sites keep records it appears that the synchronization does
not require any information on how fast or slow the data from each
site is exchanged.

If one frame of reference says that two events are simultaneous based
on Einstein's synchronization procedure then we could check whether
the correlations change at that time. If they do then that is the
absolute frame and the other frame of reference would be "wrong" as
"simultaneous" would not have an absolute meaning. The measurement can
be done with an arbitrarily large number of locally simultaneous
samples at each time (say every second) so as to produce each time at
one location a time for that "tick". The distant location would then
correlate its own measurements of those samples with those sent to it
from the other site and define time "0" as the time when the amount of
correlation between the events changed.

> It is unclear to me how you simulate an alternative procedure with your
> sliding number sets. Perhaps you can make a sketch of what you have in mind?

Sure. Here is the procedure. Take a spread sheet and use the random
number function to produce a set of ones and zeros down a column. Now
take a second column and for half of it enter random ones and zeros
and for the second half of it enter a copy of the first column. Now
assign to the first column a set of times. For convenience mark the
row that the second column changes from random numbers to a copy as
"0" and mark previous rows -1,-2,-3.... and subsequent rows 1,2,3....
This "zero" could be the time that a polarizer handle is turned for
example.

Here is the formula for column 1 and the first half of column 2:
=IF(RAND()>=0.5,1,0). Here is the formula for the second half of
column 2: =IF(B2=1,1,0)

Note that there is no way to tell where the formula changes by looking
only at column 2. You need to look at the correlation between column
one and column 2 not just the column 2 data which is random. In other
words a copy of a random sequence is itself random but the correlation
between a copy of a random sequence and another random sequence are
not the same. Two different random sequences correlate 50% but a copy
correlates 100%. That means that you must send the data from one site
to another at sub-luminary speed and I had originally thought that
that meant that you could not do a synchronization. That now appears
wrong to me.

So now here is the procedure. Assume you only have both columns of
ones and zeros. You do not have the times. Can you find the point at
which the correlation changes? What I did was take the second column
and place the first column offset by 10 rows, then 9 rows, then 8 rows
etc. Then just look at the correlations. (By correlate I just mean
that if they are equal they are correlated) Only when the columns are
lined up with zero row offset relative to the original alignment do
you get a correlation that changes from 50% to 100%. So you can know
from the data how to align it and then once its aligned its obvious
where the correlation goes from 50% to 100% and you call that 0.
(Admittedly there is some chance of correlation by accident but you
can make that arbitrarily small by using an arbitrary number of
columns. So you have now synchronized the clocks without needing to
measure the travel time of any signal.

> In particular how do you quantify (in an "absolute" way) your statement:
> "B could just lay the data next to his own WHEN he gets it and see WHERE
> they changed correlation."

As described above.

>
> > So if two frames of reference moving relative to one another disagree
> > on the present can't I use this procedure to establish a time on the
> > clock that is independent of frame of reference to disambiguate?
>
> They don't necessarily disagree on the present but on distant simultaneity
> along one direction.

If we disagree on what is simultaneous with what, we disagree with
what is at a "present" moment. That is all I meant. What is
simultaneous with me now is the present. If we disagree on
simultaneity we will disagree on what state of the universe was
"present" at a given "time".

Imagine two distant flashing lights that change from red to blue.
Imagine an observer in the middle who notes that both are red at the
same time and both are blue at the same time. For an observer moving
relative to him the lights may never be red at the same time but
rather the statement that whenever one is blue the other is red may be
true relative to him. So they disagree on what constitutes the
present. One will say it has two lights of the same color and another
will say no. That is what I meant.

>
> > What am I doing wrong?
>
> Perhaps you misunderstand the problem?

I actually believe that that is probable. However it does not help me
understand how I have misunderstood it. Can you (or anyone) see how?
If so let me know and thanks in advance.

[Norbert Dragon]

* Justintruth wrtie:

> Actually they can be compared only at one point in time as they then
> are no longer co-located at any other.

Proof by lack of imagination.

A referee, who is always in the middle of uniformly moving clocks
can easily see whether they are equal. He sees both receding clocks
redshifted but the redshift is the same.

Fig. 2.2

http://www.itp.uni-hannover.de/~dragon/rel_e.pdf

> Einstein's theory is based on the following fact: If I take a light
> wave and send it out to a distant point and reflect it back to the
> origin I can know (observe) the round trip time of the light but I
> cannot know (observe) whether this time is evenly divided into an
> outbound period of half the duration and an inbound period of half the
> duration.

Relativity is not based on the problem which you mention. The problem
is easily solved by a referee in the middle of the emitter and
reflector. He can see that light to and fro takes the same time.

Fig. 1.5

http://www.itp.uni-hannover.de/~dragon/rel_e.pdf

See also page 10, which shows that isotropic propagation of light in the
vacuum is not a mere convention: it leads to verifiable consequences.

> It appears to me that Bell's theorem will allow us to establish that
> frame of reference for which a change in the state at one point is
> simultaneous with observations at another point.

No. It is not the state which changes but the knowledge about the state.

chapter 1.4

http://www.itp.uni-hannover.de/~dragon/rel_e.pdf

--
Superstition brings bad luck.

www.itp.uni-hannover.de/~dragon

[harald]

"Norbert Dragon" <dra...@itp.uni-hannover.de> wrote in message
news:4f5d003d$0$7611$9b4e...@newsspool1.arcor-online.net...
[..]

> Relativity is not based on the problem which you mention. The problem
> is easily solved by a referee in the middle of the emitter and
> reflector. He can see that light to and fro takes the same time.

[...]

According to SR the point of view of that observer is not privileged as the
"true" point of view: he can't claim that the others are "wrong", based on
his measurements.
See my earlier reply of 2 March: According to S' and S'' the clock rates

will differ from each other, while according to S the clock rates will be

equal to each other.

[harald]

"Justintruth" <justi...@hotmail.com> wrote in message
news:d4d681ac-0ae3-49dd...@j11g2000yqj.googlegroups.com...

[..]

> But that is not what I had in mind.
>
> Einstein's theory is based on the following fact: If I take a light
> wave and send it out to a distant point and reflect it back to the
> origin I can know (observe) the round trip time of the light but I
> cannot know (observe) whether this time is evenly divided into an
> outbound period of half the duration and an inbound period of half the
> duration. If I could know (observe) this then I could use it to
> determine whether two events were simultaneous absolutely. Because I
> cannot know (observe) this each frame is free to assume that relative
> to it the time period is evenly divided. Once I do that then the
> frames disagree on simultaneity and the theory evolves. It is
> important to realize that the constancy of the speed of light is an
> assumption in Einstein's theory because there is no way to observe the
> one way speed. If you could then his theory would not work as you
> would use that observation as a basis to determine which frame was
> "correct".

Right. It could be that you have not yet fully understood its consequences.
For example, if you put two polarizers in parallel on each end of a stiff
axle that you make to rotate, polarized light from a light source in the
middle going in both directions will only pass them 50% at those times that
it is polarized the same as the polarizers when reaching them. That is the
description with respect to the lab frame S.

These events must of course be the same as described in a reference system
S' in which the system is moving with the axle along X, because both
polarizers are *not* parallel according to S' - according to S', the
rotating axle is twisted! Indeed, in S' those two events are not
simultaneous.

> My point is that there may be a way to make such an observation.
>
> It appears to me that Bell's theorem will allow us to establish that
> frame of reference for which a change in the state at one point is
> simultaneous with observations at another point. I had thought that
> the fact that Bell's theorem requires correlation and correlation
> requires information to be transmitted by a signal that travels as
> fast or slower than light meant that Bell's theorem could not be used
> to synchronize the clocks. "What I have in mind" is that that "fact"
> may be wrong. I think I have a procedure that might synchronize two
> clocks that does not depend on assuming anything about an intervening
> signal but that can be based on an observed change in the correlation.
> You could mail the results and the mailman could stop and have a
> cheeseburger on the way without you knowing and it would be no
> problem. It does not depend on knowing anything about the speed of any
> signal.

[..]

> Note that there is no way to tell where the formula changes by looking
> only at column 2. You need to look at the correlation between column
> one and column 2 not just the column 2 data which is random. In other
> words a copy of a random sequence is itself random but the correlation
> between a copy of a random sequence and another random sequence are
> not the same. Two different random sequences correlate 50% but a copy
> correlates 100%. That means that you must send the data from one site
> to another at sub-luminary speed and I had originally thought that
> that meant that you could not do a synchronization. That now appears
> wrong to me.
>
> So now here is the procedure. Assume you only have both columns of
> ones and zeros. You do not have the times. Can you find the point at

> which the correlation changes? [..]

I don't (and won't) have the time to redo your simulation, but it sounds as
if you try to establish something in QM that resembles my abovementioned
relativistic scenario (which is why I presented it).
Also there one could make the polarizers rotate independently; and the times
that the maximum amount of light comes through, merely corresponds to
synchronisation in the rest frame of the set-up.

Does that help?

[..]

Regards,
Harald

[Tom Roberts]

On 3/10/12 3/10/12 - 9:07 AM, Justintruth wrote:
> [...]

Your claims fail because the correlation begins with a SPECIFIC PAIR of
particles, corresponding to the pair at which Alice turned her handle. Bob will
see the correlation begin when the CORRESPONDING PARTICLE REACHES HIM, not
"instantaneously". Yes, in their mutual inertial frame this occurs at the same
time, but that is unique to this frame -- there is no aspect of "absolute
synchronization" here.

This is no different from the case where a single pair is sent,
corresponding to the first pair after Alice turns her handle.
This is, of course, one of Einstein's synchronization methods.


Tom Roberts

[Norbert Dragon]

* harald writes:

>* Norbert Dragon wrote

>> Relativity is not based on the problem which you mention. The problem
>> is easily solved by a referee in the middle of the emitter and
>> reflector. He can see that light to and fro takes the same time.

> According to SR the point of view of that observer is not privileged as the
> "true" point of view: he can't claim that the others are "wrong", based on
> his measurements.

Relativity is a physical theory not a judicial theory.

It is irrelevant, whether you claim this or that observer to be right.
What counts is what he observes.

Obviously, given two observers, a third one in their middle
differs from all other observers who are not in their middle.

In Euclidian geometry, all straight lines are equally "valid",
but given two intersecting lines the one which bisects their angle
is special.

[Jos Bergervoet]

On Mar 15, 9:07=A0am, Norbert Dragon <dra...@itp.uni-hannover.de> wrote:
> * =A0harald writes:
...
> > According to SR the point of view of that observer is not privileged as=
the
> > "true" point of view: he can't claim that the others are "wrong", based=

on
> > his measurements.
>
> Relativity is a physical theory not a judicial theory.
>
> It is irrelevant, whether you claim this or that observer to be right.
> What counts is what he observes.
>
> Obviously, given two observers, a third one in their middle
> differs from all other observers who are not in their middle.
>
> In Euclidian geometry, all straight lines are equally "valid",
> but given two intersecting lines the one which bisects their angle
> is special.

For two infinite lines there are two lines bisecting the angle
(could it be that they are perpendicular to each other?!) Both
might equally well claim that they are special, I think..

Now returning to the subject: for the two observers there is
indeed only one special third reference frame in the middle.
This could serve to define synchronicity of the two clocks.
If the third-frame observer sees them ticking at the same
rate (not necessarily at the rate of its own clocks!) then we
could define them as synchronized. This would seem to be
an extension of the standard definition, which only allows
for clocks at rest r.t. each other to be called synchronized.

But is this a wise extension? Would "being synchronized"
be an equivalence relation in this way, i.e. would it have
reflexivity, symmetry, and transitivity? (Well, the first two
by construction, I'd say.. But try the third!)

--
Jos

[Norbert Dragon]

* Jos Bergervoet writes:

> For the two observers there is

> indeed only one special third reference frame in the middle.
> This could serve to define synchronicity of the two clocks.
> If the third-frame observer sees them ticking at the same
> rate (not necessarily at the rate of its own clocks!) then we
> could define them as synchronized. This would seem to be
> an extension of the standard definition, which only allows
> for clocks at rest r.t. each other to be called synchronized.

> But is this a wise extension? Would "being synchronized"
> be an equivalence relation in this way, i.e. would it have
> reflexivity, symmetry, and transitivity? (Well, the first two
> by construction, I'd say.. But try the third!)

No, you should not call two clocks synchronized, if they run equally
fast for an observer in their middle, but equal. If two clocks are
equal to a third, then they are equal to each other. Equality of
clocks which move relative to each other is transitive.

pages 21,22

http://www.itp.uni-hannover.de/~dragon/rel_e.pdf

[Justintruth]

On Mar 13, 6:25=A0pm, "harald" <h...@swissonline.ch> wrote:
> "Justintruth" <justintr...@hotmail.com> wrote in message

>
> news:d4d681ac-0ae3-49dd...@j11g2000yqj.googlegroups.com...
>
> [..]
>
>
>
> > But that is not what I had in mind.
>
> > Einstein's theory is based on the following fact: If I take a light
> > wave and send it out to a distant point and reflect it back to the
> > origin I can know (observe) the round trip time of the light but I
> > cannot know (observe) whether this time is evenly divided into an
> > outbound period of half the duration and an inbound period of half the
> > duration. If I could know (observe) this then I could use it to
> > determine whether two events were simultaneous absolutely. Because I
> > cannot know (observe) this each frame is free to assume that relative
> > to it the time period is evenly divided. Once I do that then the
> > frames disagree on simultaneity and the theory evolves. It is
> > important to realize that the constancy of the speed of light is an
> > assumption in Einstein's theory because there is no way to observe the
> > one way speed. If you could then his theory would not work as you
> > would use that observation as a basis to determine which frame was
> > "correct".
>

> Right. It could be that you have not yet fully understood its consequence=

s.
> For example, if you put two polarizers in parallel on each end of a stiff
> axle that you make to rotate, polarized light from a light source in the

> middle going in both directions will only pass them 50% at those times th=
at
> it is polarized the same as the polarizers when reaching them. That is th=

e
> description with respect to the lab frame S.
>

> These events must of course be the same as described in a reference syste=

> I don't (and won't) have the time to redo your simulation, but it sounds =

as
> if you try to establish something in QM that resembles my abovementioned
> relativistic scenario (which is why I presented it).

> Also there one could make the polarizers rotate independently; and the ti=

mes
> that the maximum amount of light comes through, merely corresponds to
> synchronisation in the rest frame of the set-up.
>
> Does that help?
>
> [..]
>
> Regards,
> Harald

It certainly does help. I never thought through that scenario. I will
try to find time to see if it is the cause of my misunderstanding.

[Justintruth]

I don't think that what you're saying works as it no longer matters
when the corresponding particle reaches Bob. You do not have to do the
synchronization at any particular time. That is the whole point.The
procedures is independent of any hypothesis about the transit time. It
is only based on the ability to count particles and to slide the
records next to each other to see where the change in correspondence
occurred. All that matters is whether Bob will be able to set his
clock so that it reads what Alice's clock reads. I think that he can
because he has the series of results at his end and he can be sent the
series of results form the other end and he can lay them side by side
and then slide them until the correlations match the expected change.
He can then see where the change occurred (which particle it occurred
on). He can then label that time with the same label as in Alices
data.

Can you show me how "when the particle reaches him" is used in the
procedure I am suggesting? Rather the procedure I am suggesting
defines "when the particle reaches him".

Not do divert my own post... but assuming that two bell experiments
are conducted in two inertial frames of reference moving relative to
one another along an axis along which the Alices are separated from
the Bobs what do you think will happen with the times that the
correlation changes in each frame? Will the experiment in each frame
show that their own experiment shows simultaneous turning of the
handle and change in correlation or will there be some frame of
reference for which all experiments show a simultaneous result?

My guess is that it will be simultaneous relative to the frame of the
experiment. So if we did it in Jan and Jun the change in correlation
would be simultaneous with the turning of the handle in both
experiments. But what will happen if Bob and Alice are moving apart
from one another?! We can still do the experiment but which frame of
reference, Bob's or Alice's will be the one in which the effect is
simultaneous!

My current WAG (Wild Ass Guess) which I cannot justify in any way is
that if the experiment is done in either frame that it will result in
a simultaneous result in that frame. I currently guess that that means
you could use that to establish an absolute time but it would not be
fruitful because if you did you have the problem of all of the laws of
electromagnetism being different in both frames. It would be better to
leave the definition of time alone like it is now according to
Einstein. But I have no idea what will happen if Bob is moving
relative to Alice.

If these things are as I "guess" (I hesitate to call it even that)
then I predict that if you do a Bell experiment in a frame of
reference moving relative to you you will find that the correlation
change does not occur simultaneously with the turning of the handle.
It could be before and it could be after.

If someone offered me a bet as to whether I am confused at 100 to 1
odds I would bet I am confused. I just think I am missing something. I
will find it eventually I hope.

[Justintruth]

On Mar 12, 1:31 am, Norbert Dragon <dra...@itp.uni-hannover.de> wrote:
> * Justintruth wrtie:
>
> > Actually they can be compared only at one point in time as they then
> > are no longer co-located at any other.
>
> Proof by lack of imagination.
>
> A referee, who is always in the middle of uniformly moving clocks
> can easily see whether they are equal. He sees both receding clocks
> redshifted but the redshift is the same.

How does he know the time on the clocks from the equality of the
redshift?

What I did was respond to the fact that you cannot compare two clocks
moving relative to one another for more than one instant because if
the clock moves it requires a signal from that clock to travel once
the clocks are no longer co-located. Having the redshift will not
help.

>
> Fig. 2.2
>
> http://www.itp.uni-hannover.de/~dragon/rel_e.pdf
>
> > Einstein's theory is based on the following fact: If I take a light
> > wave and send it out to a distant point and reflect it back to the
> > origin I can know (observe) the round trip time of the light but I
> > cannot know (observe) whether this time is evenly divided into an
> > outbound period of half the duration and an inbound period of half the
> > duration.
>
> Relativity is not based on the problem which you mention. The problem
> is easily solved by a referee in the middle of the emitter and
> reflector. He can see that light to and fro takes the same time.

I don't think you are right. Here is what Einstein says:

Begin Quote:
"If an observer be stationed at A with a clock, he can estimate the
time of events occurring in the immediate neighbourhood of A, by
looking for the position of the hands of the clock, which are
synchronous with the event. If an observer be stationed at B with a
clock, ? we should add that the clock is of the same nature as the one
at A, ? he can estimate the time of events occurring about B. But
without further premises, it is not possible to compare, as far as
time is concerned, the events at B with the events at A. We have
hitherto an A-time, and a B-time, but no time common to A and B. This
last time (i.e., common time) can be defined, if we establish by
definition that the time which light requires in travelling from A to
B is equivalent to the time which light requires in travelling from B
to A. For example, a ray of light proceeds from A at A-time tA towards
B, arrives and is reflected from B at B-time tB, and returns to A at A-
time t'A . According to the definition, both clocks are synchronous,
if

tB - tA = t'A - tB.

End Quote

Why do you think that Einstein said that we must establish the fact of
the equality of the time of travel from A to B with the time of travel
of B to A by "definition"?

Here is why:

Here is what happens in your method: The light is emitted at T0. The
referee sees the arriving signal at some time T1. The light is
reflected at T2. He then sees the signal coming the other way at T3.
The light arrives back at the source at T4.

The referee knows T1 and T2. Now you want him from those numbers to
calculate the velocity of the light in either direction. How does he
do that?

The delta time from the emitter to the referee T01 is unknown because
T0 is unknown.
The delta time from the referee to the reflector T12 is unknown
because T2 is unknown.
The delta time from the reflector to the referee T23 is unknown
because T2 is unknown.
The delta time from the referee to the source T34 is unknown because
T4 is unknown.
The delta time between the arrivals at the referee are known: T13=T3-
T1.

So T12+T23=T3-T1.

The referee must establish that T01+T12=T23+T34 from T12+T23.

How will the referee do the math?

The whole theory is based on this fact because if you could know the
"actual" time it took then time would not be relative.

You can calculate the time if you assume that c is constant but you
cannot observe whether it is. If I am wrong then you should be able to
do the referee's calculation without further assumption. Can you?

>
> Fig. 1.5
>
> http://www.itp.uni-hannover.de/~dragon/rel_e.pdf
>
> See also page 10, which shows that isotropic propagation of light in the
> vacuum is not a mere convention: it leads to verifiable consequences.
>
> > It appears to me that Bell's theorem will allow us to establish that
> > frame of reference for which a change in the state at one point is
> > simultaneous with observations at another point.
>
> No. It is not the state which changes but the knowledge about the state.

Actually it is the state of the correlation that changes not our
knowledge about it.

>
> chapter 1.4
>
> http://www.itp.uni-hannover.de/~dragon/rel_e.pdf
>
> --
> Superstition brings bad luck.
>

Sure does!.... and thanks for the free pdf!

btw the logic in the first paragraph on page 10 seems to require the
assumption that light travels at speed c and therefore the result
cannot be used to check it.

[Justintruth]

On Mar 13, 6:25 pm, Tom Roberts <tjroberts...@sbcglobal.net> wrote:

I used to think that the fact that the coupled photons (or particles)
and the messages required to compare results were at the speed of
light or less meant that it was no problem for synchronization. I am
not sure now. The reason is that it seems that in these correlation
experiments you can determine synchronization independently of
measuring how fast the particles travel or how fast the messages are
exchanged about the measurement results. That is the point of what I
am saying. You can synchronize without knowing either. Several people
have responded in a way that makes me think there is something wrong
in the way I am communicating the problem. I know that both the
coupled particles and the messages required to compare the messages
can only be sent at light or sub light speed. I just think its
irrelevant.

So let's see if I can justify to you the fact that I don't need to
know the time of flight or speed of flight of either the message or
the coupled photon. If I can't justify it its wrong.

Normally we send a signal from Alice to Bob and based on knowledge of
the distance we divide by the speed of light, get the time of flight
and subtract it from the receipt time to know how long ago the
particle was emitted. I then set the clock to receipt time minus this
delta time. Clearly this depends on knowing the rate at which the
signal arrived.

But now lets assume that Alice turns her polarizer handle at some time
T0 relative to her local clock. After a while she sends Bob a letter
via US postal service that has a record of when on her clock she
turned the handle and a complete record of her measurements. Unknown
by Bob the letter carrier stops and has a hamburger and then delivers
the mail. Clearly Bob knows nothing about the time the letter took. He
then takes the list he gets and compares it against the list of his
own measurements. By sliding one list past another he gets a point
where the correlations are as they should be up to a point and then
change to what would be expected for a change in handle position. He
now has correlated the events. Because he has his own local time for
each of his measurements he can see the local time of the change. Call
it T1. Since Alice's records contain the time relative to Alice's
clock that the handle was turned. Bob now has T1 and T0. He then picks
a time in the future to sync his clock. Let's say the time he picks is
T1 plus one day. Call this time T2. He sets his clock to T0, which he
got from Alice, plus T2-T1, the elapsed time since the turning
occurred. Finally Bob starts his clock at T2. He now has a
synchronized clock.

The problem is that Bob did not use at any time either the time the
messages took or the time of flight of the coupled particles. In fact
it is also independent of distance measurements also.

That is my problem in as much detail as I can think of. I was
surprised when I realized this. What is wrong with this procedure?

The only answer I can think of is that there is nothing wrong with
this procedure in the rest frame but that in a moving frame it is a
problem. For the life of me I can't think what as it does not seem to
be dependent on the relative velocity of Alice or Bob either.

On another subject your statement that it is relative to the inertial
frame of the apparatus is appreciated. But Tom, it seems to me that
there are more possibilities. Consider these two possibilities: One is
that the simultaneity occurs in the rest frame of the apparatus as you
said. The other is that it is simultaneous to some single frame - sort
of like an ether for bell correlation vs the notion that the
simultaneity of the change is for example relative to the emitter
frame. Do you know if there have been attempts to do Bell experiments
at different times of year to check?

One other subject - sorry but I infrequently get to discuss this with
someone who knows the theory as I am an amateur and no one around me
is interested... Do you know what happens if one end of the apparatus
is moving relative to the other? Which frame is it simultaneous
relative to? Is it the frame that "turns the polarizer handle", or the
frame that sees the change in amount of correlation, or the frame of
the emitter of the coupled photons.

One more subject: If it is as you say that the simultaneity is
relative to the rest frame of the apparatus that means that in some
frames the correlations change before the handle is turned. What are
we to make of the physics of that frame? What caused the change? An
event in the future? Clearly it is not the same as the physics of the
rest frame unless causality can go back in time. If I change the
position of Alice and Bob in a frame moving with respect to me the
correlation change first precedes and then lags the turning of the
handle? This may be the case but it seems fishy to me.

Thx in advance.

[Justintruth]

Don't you need to know that the clocks are reading the same values in
addition to ticking at the same rate in order to declare
synchronization?

Also you need a way to check that you are at the midpoint. How do you
determine that from your observations? You would need to
simultaneously mark both ends at distant places. How do you do that
without the ability to establish a frame as Einstein showed based on
an assumption about light. If it turns out you can't establish the
midpoint the argument is circular. Circular but interesting! Sort of
like a new paradox. But can one show that one cannot determine the mid-
point without aprior sync? Not sure.

[Justintruth]

I keep getting that the angle of the polarizers at emission and
reception as being the same in both frames.

The tangent of the angle can be seen as the the ratio of two lengths
perpendicular to the direction of motion. These lengths stay the same
on transformation. The axle shortens, the rate of rotation slows, and
in the case I looked at the receiver moves toward where the
transmitter transmitted. In the end the relative angles (difference
between transmit angle at transmit and receive angle at receive) are
the same at transmission and reception in both frames.

In what I am thinking of the polarizers are not rotating except for
the brief time when the handle is turned. This can be made short
enough to consider the angle a step function. So I guess it doesn't
help.

[Norbert Dragon]

* Justintruth schreibt:

>* Norbert Dragon wrote:

>> A referee, who is always in the middle of uniformly moving clocks
>> can easily see whether they are equal. He sees both receding clocks
>> redshifted but the redshift is the same.

> How does he know the time on the clocks from the equality of the
> redshift?

He reads the clocks as each child does: he looks.=20

The light from the clocks show their reading.

For blind observers the clocks can send time signals.

> Having the redshift will not help.

No, but a glance does.

> I don't think you are right. Here is what Einstein says:

> Why do you think that Einstein said that we must establish the fact of
> the equality of the time of travel from A to B with the time of travel
> of B to A by "definition"?

Because Poincar=E9 had made this imprudent remark.

To define equal clocks at rest to each other, let a referee juge who
is in their middle. They are equal if they show equal times to the
referee -- no differing definition would deserve the name equal.

With these equal clocks one can simply check that it takes light
the same time to and fro.

Poincar=E9 overlooked that what cannot be done with two observers,=20
may possibly be achieved by three.

> Here is why:

> Here is what happens in your method: The light is emitted at T0.=20

While both clocks show T_0 to the referee.

> The referee sees the arriving signal at some time T1.=20

The time when the referee sees the signal is irrelevant. It only
matters, that with the light from the clocks he sees the initial=20
reading T_0 on both clocks.

> The light is reflected at T2.=20

The referee confirms that both clocks read T_2 when the light is
reflected by the other observer.

Therefore he concludes that it takes light T_2 - T_1 to run from one
to the other observer.

> He then sees the signal coming the other way at T3.

Again: the referee does not need a clock. He only has to confirm, that
light which he sends to the observers always comes back in the same=20
instant. He does not have to know the time of this instant.

> The light arrives back at the source at T4.

The referee confirms that both clocks read T_4 when the reflected light
returns to the original observer.

Therefore he concludes that it takes light T_4 - T_2 to run back from=20
the other observer to the first.

Moreover he confirms T_4 - T_2 =3D T_2 - T_0 : It takes light the same
time to run to as to run fro.

> The whole theory is based on this fact because if you could know the
> "actual" time it took then time would not be relative.

You still have not grasped the idea, that one can read distant clocks
by looking at them.

That time is relative means: time relates to the clock which you read.=20
Equal clocks, for example clocks in relative motion, can show different
times to observers.=20

> You can calculate the time if you assume that c is constant but you
> cannot observe whether it is. If I am wrong then you should be able to
> do the referee's calculation without further assumption. Can you?

Yes. This is detailed in=20

http://www.itp.uni-hannover.de/~dragon/rel_e.pdf

> btw the logic in the first paragraph on page 10 seems to require the
> assumption that light travels at speed c and therefore the result
> cannot be used to check it.

The assumption that the velocity of light is the same in all directions
and at all places endows space with a Euclidean geometry, which, if you
take it for granted, leads to relations among the n(n-1) distances of
n > 4 points. The distances can be measured and checked for the=20
fulfilment of these theoretical relations. No conflicting observation
is known.

--=20

[Jos Bergervoet]

On Mar 15, 5:47=A0pm, Norbert Dragon <dra...@itp.uni-hannover.de> wrote:

Is there a definition of "equal" in your link?

1) Do you include the requirement of showing
the same reading to this observer in the middle?
Or is only showing the same rate (with possibly
an offset in their zero-point) sufficient?

2) And does "being equal" only apply to clocks with
their worldlines intercepting? (The referee in the
middle can be uniquely defined in more general
cases).

The relation is transitive only if you require equal clocks
to show the same rate, not the same reading! A simple
counterexample of three clocks moving over the same
line with speed -v, 0, +v, but with 3 different intercept
points shows that the same reading is not possible
for all three respective pairs, each pair observed by its
own referee in the middle.

And I think this was what some posters in this thread
were actually looking for. The answer is: this is not
possible! No equivalence relation is left if you require
equal reading seen by the respective referee in the
middle for each pair of clocks.

--
Jos

[Justintruth]

On Mar 13, 6:25 pm, Tom Roberts <tjroberts...@sbcglobal.net> wrote:

I finally got what you were saying. Please ignore my other posts.
Thanks. What made it work for me was to consider uneven path lengths
to Alice and Bob.

[Norbert Dragon]

* Justintruth writes:

>* Norbert Dragon wrote:

>> http://www.itp.uni-hannover.de/~dragon/rel_e.pdf

> thanks for the free pdf!

Wait a little then you can pay for it if you want.
It is in the process of been printed by Springer.

--
Superstition brings bad luck.

www.itp.uni-hannover.de/~dragon

[Norbert Dragon]

* Jos Bergervoet schreibt:

> Is there a definition of "equal" in your link?

The definition is given for uniformly moving clocks with
intercepting worldlines.

> 1) Do you include the requirement of showing
> the same reading to this observer in the middle?
> Or is only showing the same rate (with possibly
> an offset in their zero-point) sufficient?

For intercepting worldlines this does not make a difference.

> 2) And does "being equal" only apply to clocks with
> their worldlines intercepting? (The referee in the
> middle can be uniquely defined in more general
> cases).

> The relation is transitive only if you require equal clocks
> to show the same rate, not the same reading!

Your are right, though your "only if ..." is not the only possible
solution. One could also define equal clocks with nonintercepting
worldlines by the property that each of them is equal to a clock
at relative rest with intercepting worldline and that these
intercepting clocks are equal.

The problem is related the fact that each vector field defines an
orthogonal subspace of the tangent space which however usually is not
tangent to a submanifold such as t=constant. I was not aware that the
problem shows already up if one considers three wordlines.

[harald]

On Mar 16, 6:03=A0pm, Justintruth <justintr...@hotmail.com> wrote:
> I keep getting that the angle of the polarizers at emission and
> reception as being the same in both frames.
> The tangent of the angle can be seen as the the ratio of two lengths
> perpendicular to the direction of motion. These lengths stay the same
> on transformation. The axle shortens, the rate of rotation slows, and
> in the case I looked at the receiver moves toward where the
> transmitter transmitted. In the end the relative angles (difference
> between transmit angle at transmit and receive angle at receive) are
> the same at transmission and reception in both frames.

That is erroneous. The calculation that leads to the correct result is
a bit complex (a factor 2 may be missing after a too hasty
calculation); however, it's easy to see that due to the rotation of
the axle, surface elements on the axle have a velocity that is at an
angle relative to that of the system. As a result those elements must
also be Lorentz contracted at an angle relative to the motion of the
system, so that the axle must be twisted. Several publications exist
in the literature about the calculation, for example:

http://www.opticsinfobase.org/josa/abstract.cfm?URI=3Djosa-29-11-472

Note that the only thing that matters is the angles of the polarizers
at reception; and even the point in time is not relevant, as the twist
is constant at constant rotation speed.

> In what I am thinking of the polarizers are not rotating except for
> the brief time when the handle is turned. This can be made short
> enough to consider the angle a step function.

That is not important: the rotating axle is just a striking example of
the fact that distant simultaneity is assessed differently by moving
independent reference systems, such that exactly the same detection
events are predicted.

> So I guess it doesn't help.

I now guess that it likely is the main issue. :-)

Regards,
Harald

[Jos Bergervoet]

On Mar 18, 12:18=A0pm, Norbert Dragon <dra...@itp.uni-hannover.de>
wrote:
> * =A0Jos Bergervoet schreibt:

>
> > Is there a definition of "equal" in your link?
>
> The definition is given for uniformly moving clocks with
> intercepting worldlines.

But what is the definition? Where in the
document is it?

>> 1) Do you include the requirement of showing
>> the same reading to this observer in the middle?
>> Or is only showing the same rate (with possibly
>> an offset in their zero-point) sufficient?
>
> For intercepting worldlines this does not make a difference.

Of course it does: clocks observed with the same
rate but with a fixed time difference will be called
"equal" or "not equal" depending on your choice.

And the relation is only transitive for one choice!

...

>> The relation is transitive only if you require equal clocks
>> to show the same rate, not the same reading!
>
> Your are right, though your "only if ..." is not the only
> possible solution. One could also define equal clocks with
> nonintercepting worldlines

Yes, but it changes nothing. Transitivity *already*
fails for clocks with intercepting wordlines, if the
definition of "equal" includes "equal observed
readings".

In your Fig. 2.3 you could get this situation by
*not* taking the interception points all equal to
O in the drawing, but allowing 3 different points,
as will in general be necessary.

If being equal is a property of two clocks only,
the definition cannot include a special point in
space-time where the interception of worldlines
should be. So if you require interception, you
have to allow different interception points for all
three pairs. Fig. 2.3 is only a special case of
three pairwise equal clocks.

--
Jos

[Norbert Dragon]

* Jos Bergervoet writes:

>* Norbert Dragon wrote:

>> The definition is given for uniformly moving clocks with
>> intercepting worldlines.

> But what is the definition? Where in the
> document is it?

page 20

http://www.itp.uni-hannover.de/~dragon/rel_e.pdf

"Both clocks run the same if they show the referee equal times.
This is the geometric definition of equal lengths on straight=20
worldlines of moving observers."

I admit that the definition gives only a sufficient condition
for beeing equal and that, at least on second thought, one should call
clocks also equal, if their reading differs only by an offset.

> In your Fig. 2.3 you could get this situation by
> *not* taking the interception points all equal to
> O in the drawing, but allowing 3 different points,
> as will in general be necessary.

I had assumed that we speak of three clocks with a common
intersection point. For them equality means that they show=20
referees the same time.=20

But it seems now that I get your point:

If the clocks do not intercept or if they intercept in different
events, then they are related by Poincar=E9-transformations and
only show equal time differences.

> If being equal is a property of two clocks only,
> the definition cannot include a special point in
> space-time where the interception of worldlines
> should be. So if you require interception, you
> have to allow different interception points for all
> three pairs. Fig. 2.3 is only a special case of
> three pairwise equal clocks.

I have to consider how to account for your arguments in my text.=20
To be as simple as possible I consider worldlines which intercept in a
common event and for brevity I prefer to speak of times rather than=20
time differences.=20

But may be this violates Einstein's guideline:

As simple as possible, but not simpler.

--=20

[Hendrik van Hees]

On 20/03/12 09:21, Norbert Dragon wrote:
> * Jos Bergervoet writes:
>
>> * Norbert Dragon wrote:
>
>>> The definition is given for uniformly moving clocks with
>>> intercepting worldlines.
>
>> But what is the definition? Where in the
>> document is it?
>
> page 20
>
> http://www.itp.uni-hannover.de/~dragon/rel_e.pdf
>
> "Both clocks run the same if they show the referee equal times.
> This is the geometric definition of equal lengths on straight

> worldlines of moving observers."
>
> I admit that the definition gives only a sufficient condition
> for beeing equal and that, at least on second thought, one should call
> clocks also equal, if their reading differs only by an offset.
>
>> In your Fig. 2.3 you could get this situation by
>> *not* taking the interception points all equal to
>> O in the drawing, but allowing 3 different points,
>> as will in general be necessary.
>
> I had assumed that we speak of three clocks with a common
> intersection point. For them equality means that they show

> referees the same time.

>
> But it seems now that I get your point:
>
> If the clocks do not intercept or if they intercept in different

> events, then they are related by Poincaré-transformations and

> only show equal time differences.
>
>> If being equal is a property of two clocks only,
>> the definition cannot include a special point in
>> space-time where the interception of worldlines
>> should be. So if you require interception, you
>> have to allow different interception points for all
>> three pairs. Fig. 2.3 is only a special case of
>> three pairwise equal clocks.
>
> I have to consider how to account for your arguments in my text.

> To be as simple as possible I consider worldlines which intercept in a
> common event and for brevity I prefer to speak of times rather than

> time differences.

>
> But may be this violates Einstein's guideline:
>
> As simple as possible, but not simpler.
>
> --

> Superstition brings bad luck.
>
> www.itp.uni-hannover.de/~dragon

I think the usual way in SRT is to define a standard clock as a
periodically functioning device which gives a well-defined tick rate to
an locally inertial observer, e.g., the atomic clock of the NIST, PTB,
etc. Of course, only time intervals are comparable in this case.

If the clock's rest frame is an inertial frame then it's even possible
to globally synchronize it with all clocks in this frame (up to an
arbitrary constant shift, which again is irrelevant for time intervals).
The readings of such clocks for any other uniformly moving observer with
respect to the clocks' inertial restframe relative to these observers'
own clocks (at rest relative to these observers) is then simply given by
the appropriate Poincare transformation (including possible constant
time shifts between clocks).

The relative readings of time intervals of a generally moving clock,
including acceleration, relative to a clock of an inertial frame is
given by the proper time of the moving clock.

In general relativity, you can synchronize clocks at different
space-time points and in relative motion wrt. each other only locally.
For a general space-time there's no possibility to synchronize two
clocks at far (temporal and spatial) distances at all.

--
Hendrik van Hees
Frankfurt Institute of Advanced Studies
D-60438 Frankfurt am Main
http://fias.uni-frankfurt.de/~hees/