In another thread 2inquisitive posed a possible paradox for relativist. As an excersise for our forum readers and posters I wanted to post this light clock for discussion. I make no claim to exposing a paradox or flaw in relativity as the discussion it self will show this either way. So just to get the issue totally clear I have drawn this diagram.

Here is the clock/rocket ship described in schematic.

<img src=http://www.paygency.com/lightclock1.jpg>
Edited with new diagram
Preamble:

The clocks [inside our rocket] iare structured in a way that an optic fibre cross is formed with it's arms at a fixed distance of 299792 kms..[for ease of math]

In the centre of this cross is a common light source. [colored red] A pulse of 0.5 seconds is emmitted from this common source every 10 seconds.

The clocks are built and calibrated on earth in a warehouse and was stationary at the time of calibration.

As expected each arm records a time of 2 second for a round trip at the rate of 299792ks per second. so:

A = 2 second
B = 2 second
C = 2 second

Now what I wish to discuss exactly what relativity predicts would be the time rate for all out three arms A,B and C, if our clocks are travelling at 0.8'c' in the vector as shown.

Time
A= ?
B= ?
C= ?

The reason for using optical fibre is to fix the space that the light has to pass through. It is true that the velocity figure would be less than 'c' due to the fibres density but this issue is superfluous to the task. As it just changes the initial figures and not the relativity.

We can assume that at the end of our arms is a mirror and our clocks determine time for light to travel by comparing outgoing light with incoming light.
Possibly at a later stage in the threads progress we will link the ends of the arms so that the light travels diagnally from their ends but this is later.....

If there are questions about our clock then maybe we should resolve these before getting bogged down in the detail.

Care to discuss: :)

We can assume that at the end of our arms is a light sensitive sensor that is linked to the source to give us the time delay. [it could have been designed using a two way system - mirrors..... but this is unecessary I think]

It is necessary to know how the link works. A simple signal wire would be sufficient.

Is this four clocks, or one?
Is the clock/s located in the middle, or at the ends of the arms?

Pete, what do youthink would be best here?
I would suggest a total of three clocks on the assumption that the right hand arm is identicle to the left....
Clock A, B, and C,

The clocks measure the time it takes the light to "travel" to the end of each the appropriate arm.

Clock A = time for Arm A etc.....

possibly it woudl be best to work in 4 mirrors and use the center of the cross to house our 4 clocks [or three if you like]

I'll redraw the diagram to allow for a two way measurement

The clocks measure the time it takes the light to "travel" to the end of each the appropriate arm.
How do the clocks know when the light left the centre?

ok pete ...put the clocks in the middle of the cross and make it a return trip of 2 *299792kms and pulse the comon light source for 0.5 seconds

Pete, what do youthink would be best here?
I would suggest a total of three clocks on the assumption that the right hand arm is identicle to the left....
Clock A, B, and C,

The clocks measure the time it takes the light to "travel" to the end of each the appropriate arm.



Clock A = time for Arm A etc.....

possibly it woudl be best to work in 4 mirrors and use the center of the cross to house our 4 clocks [or three if you like]

I'll redraw the diagram to allow for a two way measurement

Yes, because without having the light returning to the source point you do not have a proper light clock.

Is the length of the pulse important, or can we make it an arbitrarily short flash?
(I'm wondering how the 0.5 seconds is timed).

Now...
Now what I wish to discuss exactly what relativity predicts would be the time rate for all out three arns A,B and C, if our clock is travelling at 0.8'c' in the vector as shown.

Just to confirm - you're asking how often each clock ticks in a reference frame in which the clock is moving at 0.8c, right?

A: 3.33s (3 seconds out, 1/3 seconds back)
B: 3.33s (1/3 seconds out, 3 seconds back)
C: 3.33s (5/3 seconds each way)

ok...guys I have redrawn it to show a return trip of 2*299792 kms.
I think I've worded the experiment better as well....you'll need to go back to the first post....sorry or I can repost the diagram......

Any other suggestions?

I'm not sure if it's relevant or not, but I'm assuming light travels through an optical fibre as if it were travelling through an evacuated tube.

Is the length of the pulse important, or can we make it an arbitrarily short flash?
(I'm wondering how the 0.5 seconds is timed).

No ...not important and just used to give reality to it....same with the 10 second frequency of the pulse
A: 3.33s (3 seconds out, 1/3 seconds back)
B: 3.33s (1/3 seconds out, 3 seconds back)
C: 3.33s (5/3 seconds each way)

let me see if I understand your figures [just to be sure]
all clocks have the same time of 3.33 seconds for the round trip?

am I correct?

If I am

what are the lengths of the arms at this velocity.

length
A=?
B=?
C=?

I'm not sure if it's relevant or not, but I'm assuming light travels through an optical fibre as if it were travelling through an evacuated tube.
I think for the sake of ease that would be a fair assumption.

pete, maybe you can tell me why the time is longer going out and shorter coming in for arm A?

the diagram again:
<img src=http://www.paygency.com/lightclock1.jpg>

QQ,

I will only note that the responses you have gotten have not keyed in on your original post which was one way timing, that "How do you time that interval?", "and you need to synchronize the clocks between the start of the pulse and the receipt of the pulse at the end"; yada, yada, yada".

Had I posted this I can assure you it would have been attacked in that manner. :D

I guess the same could be said for "how do get a clock to go at 0.8'c'?" as well.....

Pete and James, I just thought of another approach that might be interesting to see the results of:

We have a perfect circle of optic fibre whose circumference is 299792kms.
Only one pulsed light source and only one clock that times the velocity as the light leaves to the right and returns from the left.

<img src=http://www.paygency.com/lightclockcircular.jpg>

The clock is calibrated on earth and the time for travel of the light is 1 second.

What would be the time delay at

a] 0.6 'c'
b] 0.8 'c'

I guess the same could be said for "how do get a clock to go at 0.8'c'?" as well.....
QQ: In order to have all your clocks start at the same instant exactly, you build a physical unit with mechanical switches located at the five points that the reflectors and source of the light pulses are located. Your ships passes by this frame and he mechanical switches trip each clock at the same time. You can also design the reflectors to reflect the first photon only and ignore trailing photons that trailed after the front running photon when emitted.

Someone may want to claim that the frame will contract in the direction of motion. This is no problem in synchronization of the clocks, or measurements later, as you have another mechanical switch triggering unit located ahead of he first. As the ship moves past this 2nd frame the current times on the clocks can all be read and compared, as well as the distances between the midpoint source and the various clockand reflector locations on the frame and this data can be radioed to the moving ship to provide calibration corrections.

I guess the same could be said for "how do get a clock to go at 0.8'c'?" as well.....

QQ: Minimizing motion effects on the measurement process.
You may also send your pulses that are time coded. The pulse itself carries the time the pulses were sent from the sources as well as the return time which are also time coded. Secondly, the arrival of every pulse at the reflector/detector is recorded at that time and place. Each measuring location is an information source center containing reflector, detector and computer data base that can all be linked to a main frame computer located anywhere on the ship. As all data is time coded there is no need to consider any motion effects on the data or the data acquisition after the data is stored as all data is time marked, or time coded.

given the length of the arms 299792kms and the possible microscopic size of the clocks I would not consider synchronisation to be a huge issue...I am sure with a little thought all clocks could be synchronised to the common light pulse anyway.
What is puzzling me is the figures Pete has arrived at:
A: 3.33s (3 seconds out, 1/3 seconds back)
B: 3.33s (1/3 seconds out, 3 seconds back)
C: 3.33s (5/3 seconds each way)

Arm A has 3 seconds outward and 1/3 second back....and this stikes me as interesting. It does initially suggest that velocity is determined by vector...but I am sure that this is not the case.....or is it Pete.
Pete are these figures a mistake?

What I was attempting to discover with this experiment:

How length contraction effects the arms and thus dillated time rates.

A = 299792km = 2 seconds
B = 299792km = 2 seconds
C = 299792km = 2 seconds

At 0.8 'c'

A = ???????km = ?? dilated seconds
B = ???????km = ?? dilated seconds
C = ???????km = ?? dilated seconds

The interesting thing with this light clock , if I am not mistaken, is that once it is calibrated as 1 second for the circumference of 299792kms as the clock contracts and dilates we always nknow by how much on board the clock so an absolute time reference is always available.
<img src=http://www.paygency.com/lightclockcircular.jpg>

so as an abstraction:

Light clock travels at 0.8 c
time delay = 3.33 seconds

means we are dilated by 2.33 seconds

On board the clock we can infer a velocity of the ship simply by it's dilated time. [ a speedometer if you like ]
and no matter how many ships are in transit they will all share the same absolute time reference.

Ship A = dil time 2.25 seconds = earth time = 1 second = velocity [a]
Ship B = dil time 3.33 seconds = earth time = 1 second = velocity [b]
Ship C = dil time 5.45 seconds = earth time = 1 second = velocity [c]

So if we didn't have absolute time to start with we certainly have now created absolute time using earth as our universal preferred frame.

pete, maybe you can tell me why the time is longer going out and shorter coming in for arm A?

the diagram again:
<img src=http://www.paygency.com/lightclock1.jpg>

For an observer for which this system is moving at a relative speed of .8c, light moves in the same direction at 1c(relative to this observer.) The result is that he sees the light moving at 1c as "chasing" the mirror which is moving at .8c. By the time the light travels a distance equal to the distance between the light source and the mirror, the mirror has moved to a new position. the light continues to travel unitl it finally catches up to the mirror. the time it will take will be equal to D/(c-v), Where D is the length of Arm A as measured by the Observer.

Once the mirror is reached and the pulse reflected, the reflected pulse and the light source will be moving in opposite directions on a collision course, the two will meet in less time than it would have taken the light to traverse the length of Arm A, because the lightsource is rushing to meet it. The time it will take will be equal to D/(c+v).

We know that the pulses all return to the center of the device at the same time in the frame of the device. This means that they all have to return to the center at the same time as seen by our observer's frame. Anything else would lead to contradictions.

Since we can show that the light traveling along arm A travels a longer distance according to our observer, (the light not only travels the length of Arm C, but also in the direction of arm A , and thus travels a diagonal longer than arm A) and that in the observer's frame the light must have a velocity of c, the light takes longer to return to the center in his frame than in the devices frame (where the light only travels the length of the Arm) .

This difference is expressed as T1 = T0/sqrt(1-v²/c²) where T0 is the time as measured in the devices frame.

Going back to arm A, we know how long it takes for the light to travel back and forth, namely D/(c-v)+D/(c+v). This must equal T1, so

T0/sqrt(1-v²/c²) = D/(c-v)+D/(c+v)

T0/sqrt(1-v²/c²) = D(1/(c-v)+1/(c+v))

T0/sqrt(1-v²/c²) = D((c+v)/(c-v)(c+v)+(c-v)/(c+v)(c-v))

T0/sqrt(1-v²/c²) = D(((c+v)+(c-v))/((c-v)(c+v)))

T0/sqrt(1-v²/c²) = D((c+v+c-v)/((c-v)(c+v)))

T0/sqrt(1-v²/c²) = D(c+c)/((c-v)(c+v)))

T0/sqrt(1-v²/c²) = D2c/((c-v)(c+v)))

T0/sqrt(1-v²/c²) = D2c/((c-v)(c+v)))

T0/sqrt(1-v²/c²) = D2c/(c²-cv+cv-v²)

T0/sqrt(1-v²/c²) = D2c/(c²-v²)

T0/sqrt(1-v²/c²) = D2c/(c²(1-v²/c²))

T0/sqrt(1-v²/c²) = D2/(c(1-v²/c²))

T0/sqrt(1-v²/c²) = 2D/(c(1-v²/c²))

T0c(1-v²/c²)/sqrt(1-v²/c²) = 2D

T0c sqrt(1-v²/c²) = 2D

T0c/2 * sqrt(1-v²/c²) = D


Now in the frame of the device, the length of the time it takes the pulse to travel the length of arm A is

Ta = D0/c

where D0 is the length of Arm A as measured by the devices frame. The time to make the whole trip out and back is equal to:

T0 = 2D0/c Thus:

(2D0/c)c/2 * sqrt(1-v²/c²) = D

The c's and 2's cancel out,and

D0 * sqrt(1-v²/c²) = D

And we have the length contraction formula which shows that arm A is shorter as measured by our observer than it is as measured by the device.

The same is true for arm B in which the out bound and return trip time are simply interchanged.

The interesting thing with this light clock , if I am not mistaken, is that once it is calibrated as 1 second for the circumference of 299792kms as the clock contracts and dilates we always nknow by how much on board the clock so an absolute time reference is always available.
<img src=http://www.paygency.com/lightclockcircular.jpg>

so as an abstraction:

Light clock travels at 0.8 c
time delay = 3.33 seconds

means we are dilated by 2.33 seconds

On board the clock we can infer a velocity of the ship simply by it's dilated time. [ a speedometer if you like ]
and no matter how many ships are in transit they will all share the same absolute time reference.

Ship A = dil time 2.25 seconds = earth time = 1 second = velocity [a]
Ship B = dil time 3.33 seconds = earth time = 1 second = velocity [b]
Ship C = dil time 5.45 seconds = earth time = 1 second = velocity [c]

So if we didn't have absolute time to start with we certainly have now created absolute time using earth as our universal preferred frame.


According to the device, time is neither dilated nor is there length contraction of the device, thus the device by its own clock will always measure one second for the light to make the circuit. Place this on a Ship and no matter how fast the ship moves in relation to the Earth, the ship will still measure the time of the light making the circuit as one second. So, no, you haven't come upon with a way to measure absolute time.

No ...not important and just used to give reality to it....same with the 10 second frequency of the pulse
I missed that 10 second frequency... it's probably better to leave that out as wel (10 seconds in what frame? Measured how?), and just have each pulse triggered by the return of the last one.
all clocks have the same time of 3.33 seconds for the round trip?
Yes
what are the lengths of the arms at this velocity.
A=0.6 light-seconds (179875km)
B=0.6 light-seconds (179875km)
C= 1 light-second (299792km)

pete, maybe you can tell me why the time is longer going out and shorter coming in for arm A?
Outward bound
A pulse leaves the center heading out on Arm A (which is 0.6 light-seconds long) at time zero.
The whole device is moving in the same direction as this pulse at 0.8c.

After 1 second:
The pulse has travelled 1 light-second.
The device has travelled 0.8 light-seconds.
The pulse is now 0.2 light-seconds away from the centre.

After 2 seconds:
The pulse has travelled 2 light-seconds.
The device has travelled 1.6 light-seconds.
The pulse is now 0.4 light-seconds away from the centre.

After 3 seconds:
The pulse has travelled 3 light-seconds.
The device has travelled 2.4 light-seconds.
The pulse is now 0.6 light-seconds away from the centre, and has reached the end of Arm A.


Now, heading back:
The pulse leaves the end of Arm A heading back to the centre.
The whole device is moving in the opposite direction as the pulse at 0.8c.

After 1 seconds:
The pulse has travelled 1 light-seconds.
The device has travelled 0.8 light-seconds.
The pulse is now 1.8 light-seconds away from the end of Arm A... but that's three times further than it needed to go to reach the centre. The pulse must have reached the centre after 1/3 of a second.



Before you return the obvious objection, please think carefully about what "the same speed in all reference frames" means, and what it doesn't mean. Don't fall into a common conceptual trap.

On board the clock we can infer a velocity of the ship simply by it's dilated time.
No.

On board the clock (ie in the clock's rest frame), the travel time is 1 second each way on each arm.

Read what I said carefully:
Just to confirm - you're asking how often each clock ticks in a reference frame in which the clock is moving at 0.8c, right?

A: 3.33s (3 seconds out, 1/3 seconds back)
B: 3.33s (1/3 seconds out, 3 seconds back)
C: 3.33s (5/3 seconds each way)

The dilated times are the travel times in a frame in which the clock is moving at 0.8c. Not the clock frame. In the clock frame, the clock is stationary. That's the whole point of reference frames.

We have a perfect circle of optic fibre whose circumference is 299792kms.
Only one pulsed light source and only one clock that times the velocity as the light leaves to the right and returns from the left.

The clock is calibrated on earth and the time for travel of the light is 1 second.

What would be the time delay at

a] 0.6 'c'
b] 0.8 'c'
On board the clock, the time delay is always the same (as Janus said) - 1 second.

In a reference frame in which the clock is moving at 0.8c, it is tricky to work out directly... the light follows an oddly shaped path, which must be integrated to get the path length. I don't know if I'm up to it.

given the length of the arms 299792kms and the possible microscopic size of the clocks I would not consider synchronisation to be a huge issue...I am sure with a little thought all clocks could be synchronised to the common light pulse anyway.
What is puzzling me is the figures Pete has arrived at:


Arm A has 3 seconds outward and 1/3 second back....and this stikes me as interesting. It does initially suggest that velocity is determined by vector...but I am sure that this is not the case.....or is it Pete.
Pete are these figures a mistake?

What I was attempting to discover with this experiment:

How length contraction effects the arms and thus dillated time rates.

A = 299792km = 2 seconds
B = 299792km = 2 seconds
C = 299792km = 2 seconds

At 0.8 'c'

A = ???????km = ?? dilated seconds
B = ???????km = ?? dilated seconds
C = ???????km = ?? dilated seconds

To eliminate the time dilation conundrum simply have your optical fiber squashed down into a straight line and oriented transverse to the direction of motion (move perpendicular to the line). The round trip times will always be the calibrrated standard of one second. In fact all moving frames can eliminate time dilation by using a transverse pulse/echo system where the pulse and echo are transverse to the motion of the frame. In this manner another optiocal fiber strung in a compressed line with axis parallel to the motion the round trip times of the transverse and parallel pulses can be monitored in real time and hence, real velocity can be determined, that is absolutely real velocity may be determiend.

Janus58, Pete and Geist, Thanks for your posts,

Janus and Pete, I see why the difference in travel times thanks you for explaining that.

I also understand that according to an on board observer ALL clocks will continue to show 2 seconds and only be seen as dilated by an Earth observer frame.
This is correct yes?

from earth frame 3.33 seconds have elapsed but from the clocks frame only 2 seconds have elapsed.......correct?

The second version of the clock [circle clock] proves this point quite well I think yes?

in that the circuit would always be seen to take 1 second on board the clock but appear dilated to an Earth frame.

Geist, I think the problem is no matter the angle or vector we will always end up in a single reference frame that being the clock unless the clock is divided into to reference frames meaning that half the clock is left back on stationary earth and the other half goes flying off somewhere....

The reason I placed the common source at the center of the cross of optic fibre was to see if all delays were identicle and this has been proved in math at least....
It clearly shows the clock conundrum as dealt with by SR.

It shows how within the frame the lengths and times are recipricating.

The circle clock shows this even clearer if I am not mistaken.

Janus58, Pete and Geist, Thanks for your posts,

Janus and Pete, I see why the difference in travel times thanks you for explaining that.

I also understand that according to an on board observer ALL clocks will continue to show 2 seconds and only be seen as dilated by an Earth observer frame.
This is correct yes?

from earth frame 3.33 seconds have elapsed but from the clocks frame only 2 seconds have elapsed.......correct?

The second version of the clock [circle clock] proves this point quite well I think yes?

Yep, all correct, as far as my understanding goes.
I personally like the straight lines of the first clock better than the circle clock (because with the straight lines I can work things out in more than one way to check), but that's just me :)

An interesting aside that you guys might like to comment on and correct me if I am wrong in extending the logic....
If we accelerate our 'cross' clock to 'c' the times in our clock will remain 2 seconds but from earths perspective the clocks are infinitely slowed. or eternal and the clock is contracted to a state of 2 dimensions.

Thus this would prove light to be a 2 dimensional object that has no distance to travel.......funny that hey?

if our clock contracted to 2 dimensions (zero vector dimension) then would it not be at it's entire journey simultaneously or to put it another way be at both source and destination simultaneously?

Yes... but! (there's always a but!)

Infinities (singularities) in mathematical models should be treated with caution. They might not necessarily have a meaningful relationship with the physical reality that they're modelling.

I think that these sort of extremes are fun to explore, but be careful to treat your conclusions with a healthy degree of skepticism.

A = B = C = 2sec

pete why do we not trust the math even if we are dealing with infinite or near infinite results..... an example would be pi I guess.....or the curve on a graph that leads too and starts from zero.....etc
The question is really if we can't trust the math at extremes then is there a problem with the math or our ability to conceptualise infinities?

Hi QQ,
The collapse of a star into black hole is a good example, I think.

(I'm stepping out of my comfort zone here, so treat what I say here with extra suspicion):

When you use the mathematical model of GR together with quantum physics and astrophysics to determine what happens to a star as it collapses down to its critical circumference and beyond, you find something strange - as the star gets close to the critical circumference, the collapse slows down, and slows down, and never ever quite reaches that critical circumference. It seems that it takes an infinite time for the star to actually collapse into a black hole.

However, that infinity is due to a singularity at the event horizon when the calculations are performed in a reference frame far from the hole.
When you use the reference frame of the star's surface, you find that the collapse proceeds at an ever increasing rate. The collapse does not freeze at the event horizon after all.

This is interesting Pete, you may recall I was theorising a while back about governed singularities or more precisely self governed singularities. What I put forward was teh notion that if a pinhole to absolute nothing [our singularity[ occurred space time would form a plug thus mass is formed. compressed space time as the universe applies pressure on the hole.

Simply put if you have an absolute vacuum the universe will try to fill the void. Space time compresses around the singularity automatically forming a 3 dimensional plug that governs the singularity. With this in mind I would suggest that a black hole could only exist within something as a singularity because if it wasn't the universe would disappear into the hole quick smart.
Now I am just talking with very limited knowledge of pressure systems I learned as an A grade motor mechanic years ago. But it seems to make sense to me.

When you say that within the horizon the collapse continues this makes sense because the horizon that is slowing to it's stop that it never gets to will eventually form matter as space collapses around the event horizon and the center of the hole disappears in to a singularity. [nothingness]

Essentially this means that a 'naked' black hole can not exist, not in this universe any way as the universe's ongoing existent is testament to this fact.