On Wed, 10 Sep 2003 11:31:30 -0400, "kenseto" wrote:


"HenriWilson" wrote in message
.. .
On Tue, 9 Sep 2003 09:01:45 -0400, "kenseto" wrote:


"HenriWilson" wrote in message
.. .
On 8 Sep 2003 15:19:18 -0700, (Mark Szlazak) wrote:

(HenriWilson) wrote in message
...
You didn't understand the experment. See my other reply.

That maybe. Would you elaborate more on a small part of your test
set-up?
Since you say that the lasers could be replaced by rods, how about
just skipping the lasers for now and just using the ends of the
original rod and only two clocks. Please describe how you go about
setting up one clock and get it to react to one end of the rod and
then how you place the other clock a rods distance away from the first
and getting it to react to the other end of the rod. I'm still feeling
skeptical so maybe this might help with understanding the situation
better.

The experimental setup is quite simple.

Take two identical rods. Lay them side by side and mark adjacent points
at
each
end.

|__________________________|
|__________________________|

On the upper rod, fasten two clocks exactly at the end points. Both
clocks
have
photodetectors with fine slits or pinholes.

You forgot that the laser will diffract when it pass through the fine
slits.

Ken, the rods can be LY's in length and the slits nanometers from the
lasers.
Besides , both beams will behave identically. The path length is virtually
zero
anyway. The lasers are used purely for locations.

OK lets take 3 stationary clocks and they are not running and all are set at
zero. When the moving rod moves into position between the first two clocks
they are now running and they now absolutely synchroneous. Now the move rod
moves into position between the second and the third clock Now how do you
set the third clock so that it is synchroneous with the second
clock?....remember that the second clock is already running.

Ken Seto




See a full explanation in my latest thread. I think you will like this one.

Henri Wilson.

See my animations and physics book at:
http://www.users.bigpond.com/HeWn/index.htm

On 10 Sep 2003 11:24:52 -0700, (Mark Szlazak) wrote:

(HenriWilson) wrote in message . ..
... reading coincides with all other clocks as though communication between them is...

There is no doubt in my mind that this thought experiment refutes SR
completely.

It is possible to create an instantaneous universe.

I can hesitantly believe that you've shown how to globally sync clocks
no matter what the speed of their "frame", but it's all done with
signals that aren't instantaneous. If you sync/hit a pair at the same
time once then the fastest speed you can hit them again with the laser
beams is still the speed of light. So the speed of sending signals
will not be instantaneous. They'll both get the start and end of the
signal simultaneously but the signal will have duration. Maybe SR is
fine but the notions about relative simultaneity aren't.

No you hven't quite got it.
Terminate this thread and please continue in my latest.

Henri Wilson.

See my animations and physics book at:
http://www.users.bigpond.com/HeWn/index.htm

On Wed, 10 Sep 2003 15:19:53 +0200, "Paul B. Andersen"
wrote:


"HenriWilson" skrev i melding ...
On Tue, 9 Sep 2003 22:56:30 +0200, "Paul B. Andersen"
wrote:


"HenriWilson" skrev i melding
.. .

The experimental setup is quite simple.

Take two identical rods. Lay them side by side and mark adjacent points at each
end.

|__________________________|
|__________________________|

On the upper rod, fasten two clocks exactly at the end points. Both clocks have
photodetectors with fine slits or pinholes.

c__________________________c

On the lower rod, fix two lasers with their very fine beams pointing exactly at
slits on the clocks.

|__________________________|

These positions are finely adjusted while the two rods are at rest and adjacent
so that the beams are aligned exactly with the two slits.

The experiment requires that the lower rod is moved rapidly (wrt the upper one)
along its axis while both are parallel and in very close proximity (for
instance along a 'mini-mono-rail').

c__________________________c
|__________________________|-------------v-


Since no REAL PHYSICAL length change is experienced by the moving rod, the RH
laser should always be aligned exactly with the RH clock slit when the LH one
flashesinto the LH cell, irrespective of rod speed. (The time taken for the
laser beam to travel to the photocell can be ignored)

If both clock readings are empirically adjusted so that they always read the
same when flashes are received on subsequently repetitions of the procedure,
then it can only be assumed that the clocks are in some kind of 'absolute
synch'.

If anybody wonder what kind of "synchronization" this leads to
in the real world where the Lorentz transform apply, here it is:

Let the proper lengths of the rods be L.
The right clock will show a time (L/v)*(1 - sqrt(1-v^2/c^2))
less than a clock which is e-synched to the left clock..
If you measure the speed of light with those clocks, it will be anisotropic.
And you can make the anisotropy to be anything you want by
selecting the speed and direction of the synchronizing rod.

Paul as you know, the physical lengths of the rods DO NOT change with velocity.
Nor do physical clock rates.
The fact that ytjhey are observer\d to change makes no difference to this
procedure.

Anyway, if you insist on introducing your LTs, a simple way around that problem
is to accelerate the lower rod one way and the upper one by an identical amount
in the opposite direction.
Then, even if you apply your fictitious (observational) transforms, the clocks
will still end up absolutely synched.

The derivation above wasn't meant for you, Henry.
Your opinion about it is of no interest whatsoever,
because, as your naive assertion above demonstrates,
you have no idea of what the Lorentz transform leads to.
You don't understand it and are unable to use it.

please see my latest thread.


Isn't it strange that different sets of "absolutely synched" clocks
can give widely different values of the OWLS of the same light beam? :-)

Only if one is a closet aetherist - which is obviously what you are.


Shut up, Henry.
This is not for you, you understand nothing related to SR.

What would I want to go through life trying to understand something that is
completely wrong from the first paragraph.
I don't go in for self-delusion.

Thanks for confirming that you don't understand SR.

Please continue this discussion in my latest thread.


Note: if, as some SRians seem to think, the physical length of a rod DOES
change with velocity, then a simple way to eliminate such an effect is to
acccelerate both rods in opposite directions by the same amount. This does not
affect the synching procedure.

... as you so vividly demonstrates here. :-)

Paul, if both rods DID, by some magic, physically contract according to the
LT's, then they would surely both contract by the same amount if they received
identical accelerations in opposite directions. That is plainly obvious because
of the 'v^2' term.
Therefore when the LH laser is directly adjacent to the LH clock, the RH pair
will also be exactly lined up.

The following is not for you, Henry.
As we agree, you don't understand it anyway.

Let's find out how the clocks are synched by this approach.
Still according to SR, of course.
As observed in the stationary frame, both ends of the rods would line
up simultaneously. And if both clocks were set to the same value
at that instant, they would be synchronous in the stationary frame.
Nothing "absolute" about that, of course.
But in the rest frame of the clocks, the right clock would show
a time L*u/c^2 less than an e-synched clock, where u is the speed
of the clocks in the stationary frame, and L the proper length of the rods.

Hey wait a minute. I think you said the clock frame WAS the stationary frame.

The relative speed between the rods, v, is given by:
v = (u+u)/(1 + u^2/c^2)
u = (c^2/v)*(1 - sqrt(1 - v^2/c^2))
Thus the right clock will show a time (L/v)*(1 - sqrt(1-v^2/c^2))
less than a clock which is e-synched to the left clock.
The same as calculated above.
Of course the synchronization of the clocks is independent
on which frame the scenario is described in.

THIS is for you, Henry:
You say that you don't understand SR.
So why do you make a fool by yourself by asserting what
SR predicts?

What the hell are 'v' and 'u' above.
Please continue in my latest thread.


Paul




Henri Wilson.

See my animations and physics book at:
http://www.users.bigpond.com/HeWn/index.htm

"HenriWilson" wrote in message
...
On Wed, 10 Sep 2003 11:31:30 -0400, "kenseto" wrote:


"HenriWilson" wrote in message
.. .
On Tue, 9 Sep 2003 09:01:45 -0400, "kenseto"
wrote:


"HenriWilson" wrote in message
.. .
On 8 Sep 2003 15:19:18 -0700, (Mark Szlazak) wrote:

(HenriWilson) wrote in message
...
You didn't understand the experment. See my other reply.

That maybe. Would you elaborate more on a small part of your test
set-up?
Since you say that the lasers could be replaced by rods, how about
just skipping the lasers for now and just using the ends of the
original rod and only two clocks. Please describe how you go about
setting up one clock and get it to react to one end of the rod and
then how you place the other clock a rods distance away from the
first
and getting it to react to the other end of the rod. I'm still
feeling
skeptical so maybe this might help with understanding the situation
better.

The experimental setup is quite simple.

Take two identical rods. Lay them side by side and mark adjacent
points
at
each
end.

|__________________________|
|__________________________|

On the upper rod, fasten two clocks exactly at the end points. Both
clocks
have
photodetectors with fine slits or pinholes.

You forgot that the laser will diffract when it pass through the fine
slits.

Ken, the rods can be LY's in length and the slits nanometers from the
lasers.
Besides , both beams will behave identically. The path length is
virtually
zero
anyway. The lasers are used purely for locations.

OK lets take 3 stationary clocks and they are not running and all are set
at
zero. When the moving rod moves into position between the first two
clocks
they are now running and they now absolutely synchroneous. Now the move
rod
moves into position between the second and the third clock Now how do you
set the third clock so that it is synchroneous with the second
clock?....remember that the second clock is already running.

Ken Seto




See a full explanation in my latest thread. I think you will like this
one.

Don't direct me to see another thread. Answer my question please.

Kwen Seto

"HenriWilson" skrev i melding ...
On Wed, 10 Sep 2003 15:19:53 +0200, "Paul B. Andersen"
wrote:


"HenriWilson" skrev i melding ...
Paul, if both rods DID, by some magic, physically contract according to the
LT's, then they would surely both contract by the same amount if they received
identical accelerations in opposite directions.

Let's find out how the clocks are synched by this approach.
Still according to SR, of course.
As observed in the stationary frame, both ends of the rods would line
up simultaneously. And if both clocks were set to the same value
at that instant, they would be synchronous in the stationary frame.
Nothing "absolute" about that, of course.
But in the rest frame of the clocks, the right clock would show
a time L*u/c^2 less than an e-synched clock, where u is the speed
of the clocks in the stationary frame, and L the proper length of the rods.

Hey wait a minute. I think you said the clock frame WAS the stationary frame.

Read what I wrote, don't think it.

The relative speed between the rods, v, is given by:
v = (u+u)/(1 + u^2/c^2)
u = (c^2/v)*(1 - sqrt(1 - v^2/c^2))
Thus the right clock will show a time (L/v)*(1 - sqrt(1-v^2/c^2))
less than a clock which is e-synched to the left clock.
The same as calculated above.
Of course the synchronization of the clocks is independent
on which frame the scenario is described in.

What the hell are 'v' and 'u' above.

They are what they are clearly stated to be above.

Please continue in my latest thread.

Why would I do that?
I have shown how your "synch method" would synch clocks
according to SR. Twice.
It is correct, nothing to discuss.
It is not a matter of opinion what SR say.

I am not interested in what you may have to say about it,
because you don't understand SR, and are unable to say
what SR predicts. When you try, you are invariably wrong.

And you can't even read, as demonstrated above.

Paul

On Wed, 10 Sep 2003 11:31:30 -0400, "kenseto" wrote:


"HenriWilson" wrote in message
.. .
On Tue, 9 Sep 2003 09:01:45 -0400, "kenseto" wrote:


"HenriWilson" wrote in message
.. .
On 8 Sep 2003 15:19:18 -0700, (Mark Szlazak) wrote:

(HenriWilson) wrote in message
...
You didn't understand the experment. See my other reply.

That maybe. Would you elaborate more on a small part of your test
set-up?
Since you say that the lasers could be replaced by rods, how about
just skipping the lasers for now and just using the ends of the
original rod and only two clocks. Please describe how you go about
setting up one clock and get it to react to one end of the rod and
then how you place the other clock a rods distance away from the first
and getting it to react to the other end of the rod. I'm still feeling
skeptical so maybe this might help with understanding the situation
better.

The experimental setup is quite simple.

Take two identical rods. Lay them side by side and mark adjacent points
at
each
end.

|__________________________|
|__________________________|

On the upper rod, fasten two clocks exactly at the end points. Both
clocks
have
photodetectors with fine slits or pinholes.

You forgot that the laser will diffract when it pass through the fine
slits.

Ken, the rods can be LY's in length and the slits nanometers from the
lasers.
Besides , both beams will behave identically. The path length is virtually
zero
anyway. The lasers are used purely for locations.

OK lets take 3 stationary clocks and they are not running and all are set at
zero. When the moving rod moves into position between the first two clocks
they are now running and they now absolutely synchroneous. Now the move rod
moves into position between the second and the third clock Now how do you
set the third clock so that it is synchroneous with the second
clock?....remember that the second clock is already running.

Ken Seto


You just repeat the rod synchronising process using clocks 2 and 3 (or 1 and 3)

Henri Wilson.

See my animations and physics book at:
http://www.users.bigpond.com/HeWn/index.htm

(HenriWilson) wrote in message . ..
OK lets take 3 stationary clocks and they are not running and all are set at
zero. When the moving rod moves into position between the first two clocks
they are now running and they now absolutely synchroneous. Now the move rod
moves into position between the second and the third clock Now how do you
set the third clock so that it is synchroneous with the second
clock?....remember that the second clock is already running.

Ken Seto


You just repeat the rod synchronising process using clocks 2 and 3 (or 1 and 3)

Henri Wilson.

An eventually unworkable method would be to use a 3-clock rod then
4,5, etc.

My earlier concern seems similar to Ken Seto's. Even if you have
groups of clocks in sync, that doesn't increase the maximum speed at
which messages can be sent between them. SR allows local sync and
still doesn't say this implies faster than light signalling. If you
get a more global sync, still nothing has changed with regards to how
fast you can send messages (like this post) between these different
frames.

On Wed, 10 Sep 2003 23:35:00 +0200, "Paul B. Andersen"
wrote:


"HenriWilson" skrev i melding ...
On Wed, 10 Sep 2003 15:19:53 +0200, "Paul B. Andersen"
wrote:


"HenriWilson" skrev i melding ...
Paul, if both rods DID, by some magic, physically contract according to the
LT's, then they would surely both contract by the same amount if they received
identical accelerations in opposite directions.

Let's find out how the clocks are synched by this approach.
Still according to SR, of course.
As observed in the stationary frame, both ends of the rods would line
up simultaneously. And if both clocks were set to the same value
at that instant, they would be synchronous in the stationary frame.
Nothing "absolute" about that, of course.
But in the rest frame of the clocks, the right clock would show
a time L*u/c^2 less than an e-synched clock, where u is the speed
of the clocks in the stationary frame, and L the proper length of the rods.

Hey wait a minute. I think you said the clock frame WAS the stationary frame.

Read what I wrote, don't think it.

In what you wrote, you talk about the 'stationary frame' and the 'rest frame'.
They are both the clock frame.


The relative speed between the rods, v, is given by:
v = (u+u)/(1 + u^2/c^2)
u = (c^2/v)*(1 - sqrt(1 - v^2/c^2))
Thus the right clock will show a time (L/v)*(1 - sqrt(1-v^2/c^2))
less than a clock which is e-synched to the left clock.
The same as calculated above.
Of course the synchronization of the clocks is independent
on which frame the scenario is described in.

What the hell are 'v' and 'u' above.

They are what they are clearly stated to be above.

Please continue in my latest thread.

Why would I do that?
I have shown how your "synch method" would synch clocks
according to SR. Twice.
It is correct, nothing to discuss.
It is not a matter of opinion what SR say.

I am not interested in what you may have to say about it,
because you don't understand SR, and are unable to say
what SR predicts. When you try, you are invariably wrong.

And you can't even read, as demonstrated above.

I don't think you understood the experiment.

Paul



Henri Wilson.

See my animations and physics book at:
http://www.users.bigpond.com/HeWn/index.htm

On 11 Sep 2003 07:37:29 -0700, (Mark Szlazak) wrote:

(HenriWilson) wrote in message . ..
OK lets take 3 stationary clocks and they are not running and all are set at
zero. When the moving rod moves into position between the first two clocks
they are now running and they now absolutely synchroneous. Now the move rod
moves into position between the second and the third clock Now how do you
set the third clock so that it is synchroneous with the second
clock?....remember that the second clock is already running.

Ken Seto


You just repeat the rod synchronising process using clocks 2 and 3 (or 1 and 3)

Henri Wilson.

An eventually unworkable method would be to use a 3-clock rod then
4,5, etc.

My earlier concern seems similar to Ken Seto's. Even if you have
groups of clocks in sync, that doesn't increase the maximum speed at
which messages can be sent between them. SR allows local sync and
still doesn't say this implies faster than light signalling. If you
get a more global sync, still nothing has changed with regards to how
fast you can send messages (like this post) between these different
frames.

You don't have to send signals between them.

All experiments are carried out using the clocks (ie, instantly) and the
results collated afterwards.
Thus, human observers and light's finite travel time are eliminated from all
experiments.

Henri Wilson.

See my animations and physics book at:
http://www.users.bigpond.com/HeWn/index.htm

(HenriWilson) wrote in message . ..
You don't have to send signals between them.

All experiments are carried out using the clocks (ie, instantly) and the
results collated afterwards.
Thus, human observers and light's finite travel time are eliminated from all
experiments.

Yes, I agree with your method of syncing clocks but I'm referring to
it's implications for SR. One thing SR is about is light as the
fastest way signals can be sent and lights independence from source
speed. Nothing in your set-up shows instantaneous signalling. However,
it may provide a way to resolve some issues like the conventionality
of simultaneity (different from the relativity of simultaneity)
mentioned by Hans Riechenbak (sp?) and Einstein(?).