Discussion:
The rigid rod paradox with 8 clocks.
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Nicolaas Vroom
2019-11-13 16:26:25 UTC
Permalink
Consider a rod with 8 clocks, equally spaced, a distance l apart.
The rod is considered at rest. This implies that the speed of light c
in all directions is the same. We call this rule 1.
The clocks are numbered from #1 to #8.
The strategy is to perform a certain number of experiments.

The first experiment is called clock synchronisation.
Halfway in between clock #4 and #5, there is a light source which emits a
reset signal. The setup is such that the length lightpath to each clock is
the same. Using rule 1 the light signal will reach all the clocks simultaneous.
This is important because all the clocks at any moment will all show the same
count.

The second experiment starts with making an exact copy of rod #1. Also
attached to each clock there is an engine which can be fired with a standard
burst in either the forward or backward direction. Each clock also has
an observer.
The second experiment consists that each observer on rod #2 fires his engine
with a standard burst in the same direction when his clock is reset.
This burst will give the rod a certain speed v.
The now moving observers will perform the next tasks when they reach the next
clock at rest: They will write down the reading of the clock at rest and
the reading of their own moving clock.
This is the result:
They are the same for all observers. The number of counts of the moving
clocks is less than the number of counts of the clocks at rest.
This is not so strange because it means that the physical forces which
influence the behaviour of each clock are identical. Specific what this
means is that all the moving clocks stay synchronised. This is rule 2.
You can repeat this experiment, but still, rule 2 applies.

Experiment 3 is almost identical to experiment 2. That means all
the engines are fired after the reset signal is received.
This defines the starting condition of experiment 3. The starting
condition of experiment 3 is a moving rod with the speed v.
Experiment 3 involves that a certain moment the light signal between
clock #4 and #5 of the moving rod issues a reset signal.
Like before the moving observers write down the results when they reach
the next clock at rest. This is the result:
All the observers write down the same number of counts for the clocks
at rest. For the moving clocks, the results are different. The clock
in front will have the lowest count. The clock at the back the highest
count. Physical the clock in the back is reset the first.

Experiment 4 is the same as Experiment 3 with the difference that
we again make an exact copy of rod #2 before the reset signal is issued.
This is rod #3.
The extra complication is that in experiment 4 both the moving rod #2
and #3 receive the same reset signal.
The next complication is that when each of the clocks of rod #3 receives a
reset signal also the engine is fired in the same direction as experiment #2.
However, this will also give a physical complication, because the engine
in the back will start first and in front of the latest. As such the physical
forces will try to compress the rod.
The opposite case is also possible. That means physical forces will try to
expand the length of the physical rod.

Experiment 2 belongs to what you can call a symmetrical experiment.
This is the case if you start from a state at rest than in either direction
the results are the same i.e. how higher the speed how slower the
moving clock ticks.
What is also the case, after reaching a certain speed and the speed is
decreased the clock starts to run faster until the speed reaches zero.

Experiment 4 belongs to what you can call an asymmetrical experiment.
This is the case when the starting condition involves a moving rod.
In that case when a clock receives a burst in the same direction as the
original speed the clock will start to run slower.
In the opposite direction, it is first faster and then slower.

These results are maybe different as what is expected.
They challenge the concept of what means at rest.

For much more detail read this:
http://users.telenet.be/nicvroom/Article_Review_Moving%20Bodies_Appendix2.htm

Nicolaas Vroom
r***@gmail.com
2019-11-14 20:51:24 UTC
Permalink
Post by Nicolaas Vroom
Consider a rod with 8 clocks, equally spaced, a distance l apart.
The rod is considered at rest. This implies that the speed of light c=20
in all directions is the same. We call this rule 1.=20
The clocks are numbered from #1 to #8.
The strategy is to perform a certain number of experiments.
=20
The first experiment is called clock synchronisation.=20
Halfway in between clock #4 and #5, there is a light source which emits a=
=20
Post by Nicolaas Vroom
reset signal. The setup is such that the length lightpath to each clock i=
s
Post by Nicolaas Vroom
the same. Using rule 1 the light signal will reach all the clocks simulta=
neous.
Post by Nicolaas Vroom
This is important because all the clocks at any moment will all show the =
same=20
Post by Nicolaas Vroom
count.
As stated this implies the clocks are in a circle. If they were in a strai=
ght line they could not all be the same distance from the source of the lig=
ht pulse. However I accept that the clocks can be synchronized with the re=
sult you state by using a slightly more elaborate procedure.
Post by Nicolaas Vroom
=20
The second experiment starts with making an exact copy of rod #1. Also
attached to each clock there is an engine which can be fired with a stand=
ard
Post by Nicolaas Vroom
burst in either the forward or backward direction. Each clock also has
an observer.=20
The second experiment consists that each observer on rod #2 fires his eng=
ine
Post by Nicolaas Vroom
with a standard burst in the same direction when his clock is reset.=20
This burst will give the rod a certain speed v.
The now moving observers will perform the next tasks when they reach the =
next
Post by Nicolaas Vroom
clock at rest: They will write down the reading of the clock at rest and
the reading of their own moving clock.=20
This is the result:=20
They are the same for all observers. The number of counts of the moving
clocks is less than the number of counts of the clocks at rest.
This is not so strange because it means that the physical forces which
influence the behaviour of each clock are identical. Specific what this
means is that all the moving clocks stay synchronised. This is rule 2.
You can repeat this experiment, but still, rule 2 applies.=20
This is where you are misunderstanding SR. You are applying an impulse to =
each clock/observer on rod #2 and the result is that each clock/observer is=
instantaneously accelerated to some speed, and they all start at the same =
time in this frame. Also, in this frame, they remain the same distance apa=
rt and synchronized as you say, BUT ONLY IN THIS FRAME. In the frame of th=
e observers on the accelerated clocks the other clocks are no longer synchr=
onized, and the spacing between the clocks on rod #2 is no longer the same =
as on rod #1.
Post by Nicolaas Vroom
=20
Experiment 3 is almost identical to experiment 2. That means all
the engines are fired after the reset signal is received.
This defines the starting condition of experiment 3. The starting
condition of experiment 3 is a moving rod with the speed v.
Experiment 3 involves that a certain moment the light signal between
clock #4 and #5 of the moving rod issues a reset signal.=20
Like before the moving observers write down the results when they reach
All the observers write down the same number of counts for the clocks
at rest. For the moving clocks, the results are different. The clock
in front will have the lowest count. The clock at the back the highest
count. Physical the clock in the back is reset the first.=20
This experiment is not so clear what you intend. I believe you are saying =
that rod #2 is initially moving as speed v and that when alongside rod #1 t=
hat the engines are fired "simultaneously" to cause rod #2 to stop alongsid=
e rod #1. It is important to be clear what you mean by simultaneous, i.e. =
in what frame are they simultaneous? If in the rod #1 frame, and if rod #2=
was accelerated as in experiment #1, then yes, all the clocks will show th=
e same time differences. But be aware that the engines are not being fired=
simultaneously in the moving rod #2 frame. Nor is the initial spacing bet=
ween the clocks on rod #2 in the moving frame the same as on rod #1 measure=
d in its rest frame.
Post by Nicolaas Vroom
=20
Experiment 4 is the same as Experiment 3 with the difference that
we again make an exact copy of rod #2 before the reset signal is issued.
This is rod #3.=20
The extra complication is that in experiment 4 both the moving rod #2=20
and #3 receive the same reset signal.=20
The next complication is that when each of the clocks of rod #3 receives =
a=20
Post by Nicolaas Vroom
reset signal also the engine is fired in the same direction as experiment=
#2.
Post by Nicolaas Vroom
However, this will also give a physical complication, because the engine
in the back will start first and in front of the latest. As such the phys=
ical
Post by Nicolaas Vroom
forces will try to compress the rod.=20
You don't mention, and maybe are not aware, that in experiment 1 that rod #=
2 will be stretched, under tension, immediately after firing the engines th=
at accelerated the rod.
Post by Nicolaas Vroom
The opposite case is also possible. That means physical forces will try t=
o
Post by Nicolaas Vroom
expand the length of the physical rod.
=20
Experiment 2 belongs to what you can call a symmetrical experiment.
This is the case if you start from a state at rest than in either directi=
on=20
Post by Nicolaas Vroom
the results are the same i.e. how higher the speed how slower the=20
moving clock ticks.
What is also the case, after reaching a certain speed and the speed is=20
decreased the clock starts to run faster until the speed reaches zero.
I'm not following what you are intending to say here. If you are seeing so=
me inconsistency, that is because you are not appreciating all the effects =
of the accelerations and SR. Rod #2 will be stretched immediately after th=
e impulse and, in the frame moving with rod #2, the clocks will no longer b=
e synchronized IN THAT FRAME. They will remain synchronized in the rest fr=
ame of rod #1 however, and will continue to be spaced the same distance, un=
til the stiffness of rod #2 pulls the clocks back to their "normal" spacing=
. There are three effects: time dilation, change in synchonization and ch=
ange in spacing. If you correctly understand all of these there is no para=
dox or inconsistency.
Post by Nicolaas Vroom
=20
Experiment 4 belongs to what you can call an asymmetrical experiment.
This is the case when the starting condition involves a moving rod.
In that case when a clock receives a burst in the same direction as the
original speed the clock will start to run slower.=20
In the opposite direction, it is first faster and then slower.
=20
These results are maybe different as what is expected.=20
They challenge the concept of what means at rest.
=20
http://users.telenet.be/nicvroom/Article_Review_Moving%20Bodies_Appendix2=
.htm
Post by Nicolaas Vroom
=20
Nicolaas Vroom
Nicolaas Vroom
2019-11-17 09:35:31 UTC
Permalink
Post by r***@gmail.com
The setup is such that the length lightpath to each clock is the same.
Using rule 1 the light signal will reach all the clocks simultaneous.
This is important because all the clocks at any moment will all show the
same count.
As stated this implies the clocks are in a circle.
If they were in a straight line they could not all be the same distance
from the source of the light pulse.
However I accept that the clocks can be synchronized with the result
you state by using a slightly more elaborate procedure.
Please select the link mentioned at the bottom of this posting.
Part of that document (Appendix 2) are two pictures.
For Picture 1 select:
Loading Image...
For Picture 2 select:
Loading Image...
Picture 1 shows clock synchronization using clocks at rest.
Pucture 2 shows clock synchronization using moving clocks.
Post by r***@gmail.com
The number of counts of the moving
clocks is less than the number of counts of the clocks at rest.
This is not so strange because it means that the physical forces
which influence the behaviour of each clock are identical.
This is where you are misunderstanding SR.
You are applying an impulse to each clock/observer on rod #2 and
the result is that each clock/observer is instantaneously accelerated
to some speed, and they all start at the same time in this frame.
The whole idea is to do the experiment in small increments.
See for example Picture 3 in document "Appendix 2"
The experiment starts with clock synchronization.
Thereafter, all the clock show the same count and run at the same rate.
The first that happens is that at a certain count all the engines are
fired.
Thereafter, the whole rod gets a certain speed, all the clocks start
to run slower, but all the moving clocks stay synchronised.
This can be repeated, meaning the engines are again fired at a certain
count.
Thereafter, the speed of the rod increases, all the clocks start to
run even slower, but all the moving clocks stay synchronised.
Post by r***@gmail.com
Also, in this frame, they remain the same distance apart and
synchronized as you say, BUT ONLY IN THIS FRAME.
In the frame of the observers on the accelerated clocks the other
clocks are no longer synchronized, and the spacing between the clocks
on rod #2 is no longer the same as on rod #1.
All the moving clocks stay synchronised (as a set) and the clocks at rest
(as a set) stay synchronised. Their ticking rate can be different.
Consider a moving rod which 8 clocks #1 to #8 (#1 is left).
The moving is towards the right.
Consider 20 clocks at rest marked #1 to #20.
Consider that moving clock #8 (in front) meets clock at rest #18.
The clock readings are 40 counts and 50 counts.
When that is the case also moving clock #1 will meet clock at rest #11
and the clock readings are 40 and 50.
This type of behaviour will be observed after the first time when the
engines are started, but also after the second time etc etc.
What this means is that the distance between clock #1 and clock # 8
for both rods stays the same.
Post by r***@gmail.com
Experiment 3 is almost identical to experiment 2. That means all
the engines are fired after the reset signal is received.
This defines the starting condition of experiment 3. The starting
condition of experiment 3 is a moving rod with the speed v.
Experiment 3 involves that a certain moment the light signal between
clock #4 and #5 of the moving rod issues a reset signal.
This experiment is not so clear what you intend.
I believe you are saying that rod #2 is initially moving as speed v
and that when alongside rod #1 that the engines are fired "simultaneously"
to cause rod #2 to stop alongside rod #1.
It is important to be clear what you mean by simultaneous, i.e. in
what frame are they simultaneous?
If in the rod #1 frame, and if rod #2 was accelerated as in experiment #1,
then yes, all the clocks will show the same time differences.
But be aware that the engines are not being fired simultaneously
in the moving rod #2 frame.
Nor is the initial spacing between the clocks on rod #2 in the moving
frame the same as on rod #1 measured in its rest frame.
The whole idea behind experiment 3 is first to perform experiment 1
and then experiment 2. Now you have a moving rod, called rod #3.
You repeat this, now you have rod #4.
On both moving rods you perform clock synchronisation.
This is shown in picture 2 (bottom part).
On clock #4 after this reset set is reveived you fire the engines.
This is shown in picture 2 by means of the green and red lines
for different speeds.
What picture 2 also shows (top part) that there is length contraction
involved.
Post by r***@gmail.com
Experiment 4 is the same as Experiment 3 with the difference that
we again make an exact copy of rod #2 before the reset signal is issued.
This is rod #3.
However, this will also give a physical complication, because the engine
in the back will start first and in front of the latest.
As such the physical forces will try to compress the rod.
You don't mention, and maybe are not aware, that in experiment 1 that rod #2
will be stretched, under tension, immediately after firing the engines
that accelerated the rod.
I'm fully aware of this. That is why in experiment 2 the more clocks
and engines there are the better. The result is that the whole rod, as one
object, will start to move (and move) simultaneously.
In experiment 4 this is not the case.
The back will start to move first and the front the latest.
Post by r***@gmail.com
Experiment 2 belongs to what you can call a symmetrical experiment.
This is the case if you start from a state at rest than in either
direction the results are the same i.e. how higher the speed how slower
the moving clock ticks.
What is also the case, after reaching a certain speed and the speed
is decreased the clock starts to run faster until the speed reaches zero.
I'm not following what you are intending to say here.
My advice is to study Picture 3 of Appendix 2.
Post by r***@gmail.com
Experiment 4 belongs to what you can call an asymmetrical experiment.
These results are maybe different as what is expected.
They challenge the concept of what means at rest.
Appendix 2:
http://users.telenet.be/nicvroom/Article_Review_Moving%20Bodies_Appendix2.htm

Thanks for raising questions.

Nicolaas Vroom

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