Average time/Actual time/Absolute time

[afterburner]

Could someone PLEASE help me out with the following scenario. Its a AAA clock. I hope I didn't miss anything.


We have a sphere 6km in diameter. We also have 21 devices. Each device being a computer, an atomic clock, a photon detector and a controlled light burst machine all in one.

At the centre of this sphere, we have one such device. Lets call it the mother device. On the inside, along the walls of the sphere, we have the rest of the devices - 20 devices, all spaced out with equal spaces between each other. Lets call these lesser devices.

So far, we have 20 lesser devices spaced out equally on the walls of the sphere, measuring the intervals between bursts, as well as the number of bursts received from the mother device at the centre of the sphere.

These lesser devices are programmed to know the number of bursts received and the intervals between bursts from the mother device, in a stationary reference frame. Each lesser device sends its own bursts to the mother device containing information on the number of bursts received and the intervals between bursts. All of the lesser devices are synchronized with the mother device.

When the entire AAA clock device is stationary, the bursts from the mother device are received simultaneously by each lesser device. Each lesser device then fires its own bursts based on how many bursts by the mother device were registered and the times between each burst. So basically when stationary, the lesser devices all fire their information simultaneously at the mother device. This will happen in such a way that the mother device will receive a signal from each lesser device every 1/10th of a second (for example).

When the entire AAA clock is not stationary, the bursts from the lesser devices are adjusted by the computer at each lesser device to fire in such a way, that the mother device will still receeive the signals at the same time from each lesser device - every 1/10th of a second.

Note that when stationary, the number of bursts and the intervals between bursts should be the same. When not stationary, the number of bursts and the intervals between bursts should be different for all 20 lesser detectors. Using this information, the lesser device computers figure out how to send their bursts (either faster or slower than "normal"), so that each lesser device's burst is received at the same time by the mother device (containing the varying number of bursts and the varying intervals between bursts).

When the information is received by the mother device, it is compared with that of the stationary reference frame, which is stored on the hard drive as a reference point. From this, the mother device spits out the average time and speed for the onboard personnel.


What I want to know is...

1. When we have this AAA clock moving at 0, 25, 50, 95, and 99 percent the speed of light, what will the average time for each increment of speed be?

2. Did I go wrong anywhere/miss something in my reasoning?

3. What do you think?


Thanks.

[Ken G]

The problem with this scenario is nothing is moving relative to anything else, if that's what you meant by "the entire AAA clock". Thus you always get the same answer, the computers never calculate anything differently, and it's impossible for the computers to infer any particular speed. Indeed, that impossibility is required by relativity. However, if you are imagining that the mother device is moving while the lesser devices stay put, you could have them infer the mother device's speed and correct their own signals accordingly such that the mother device receives them 1/10 of a second apart. Then, what is the question? You just seem to be talking about the Doppler formula, which you can easily Google.

[hhEb09'1]

There are some slight differences that would show up during acceleration, but you don't need a 6km clock to tell you that it's accelerating.

[Jeff Root]

What Ken and hh said. The speed makes no difference. All the
timings will be exactly the same as if the sphere were not moving.
That is why it is impossible to measure absolute motion.

-- Jeff, in Minneapolis

[Squashed]

Start the sphere rotating and then propel the sphere with a linear velocity in some direction and then you'll get some interesting results.

[Jeff Root]

Squashed,

The rotation will cause interesting results but the linear velocity
still will have no effect.

-- Jeff, in Minneapolis

[Squashed]

Squashed,

The rotation will cause interesting results but the linear velocity
still will have no effect.

-- Jeff, in Minneapolis

Jeff, I was thinking the combination of rotation with velocity would cause interesting effects. If the momentary rotational velocity of one clock added to the linear velocity then that clock would experience more SR than the clock 180 degrees around the sphere which subtracts its velocity from the linear velocity. Since the speed of light is the absolute speed limit then the addition of velocity would have less effect than the subtraction of velocity.

But then I guess the addition and subtraction would only be visible to an "outside the sphere" viewer - so you may very well be correct.

[afterburner]

okok...allow me to change the clock a little bit...

for ease...lets pretend its a circle with 8 lesser devices spaced out equally, and the mother device in the middle.

as i mentioned, all of the lesser devices are synchronized with each other, as well as the mother device.

This time, however, the lesser devices are going to be programmed to send their bursts in a consecutive manner, every second, one after another. So each device fires every 9th second, so that no two devices fire at the same time - in a circular manner. Each lesser device has its own unique wavelength. The mother device does not fire any signals this time around.

this goes on when the AAA clock is stationary, and the mother device stores the data as a reference point. each signal from the lesser device, when stationary, should arrive one second apart at the mother device. *1


When the ship is moving at lets say 75% the speed of light, all of the clocks onboard the ship/clock device will be dilated, but they still will be sending the signals one dilated second apart.

now, the signals from the lesser devices are going to be sent at the speed of light towards the mother device. However, the signals sent by the lesser devices "at the back" will be out of sync according to the synchronized mother device. that is to say, the signal will lag behind, since the ship is already moving at the 75% c. *2

the computer attached to the mother device therefore knows something is fishy, and adjusts the clock to the appropriate time. NOT the atomic clock, that is attached to the mother device, but the clock that spits out the time after calculations and comparisons between all lesser devices have been made.


the following is assuming that if you draw a circle 1 is at the bottom, 5 is at the top, 2 is one to the right of 1....the direction of travel is 5...

*1 when the ship is stationary the mother device will have the following data.
1-------2-------3-------4-------5-------6-------7-------8...




*2 when the ship is moving at 75% c...not exact..but appoximate

1-------2------3----4--5--6----7-------8...

or at higher speeds

5--4-6----3----7-------1-----2---------8...

at even higher speeds

5---5---5---5---4---6---4---6--3----7--------2----8--5---4---1--1-5--...


Do you see what im trying to get at here?

The computer will compare this to the stationary results and adgust the clock to the appropriate time. Once again, not the atomic clocks in the devices, but the one that spits out the time for the onboard people after comparison to the stationary data.


Or am I still getting something wrong?

[Ken G]

There is still a lot that is not clear. Are the lesser devices moving with the mother device, inside the "ship", or are they not? If they are, you will see nothing change at all in the onboard computers (except during periods of acceleration as already pointed out by Jeff Root). If they are not, you are just investigating the Doppler effect, and there is nothing unusual going on here. What is your question?

[Jeff Root]

When the ship is moving at lets say 75% the speed of light, all
of the clocks onboard the ship/clock device will be dilated, but
they still will be sending the signals one dilated second apart.

The clocks will all appear to be ticking slowly to observers
not moving with the ship. The time dilation is relative to
observers having motion different from the ship.

now, the signals from the lesser devices are going to be sent
at the speed of light towards the mother device. However, the
signals sent by the lesser devices "at the back" will be out of
sync according to the synchronized mother device. that is to
say, the signal will lag behind, since the ship is already
moving at the 75% c. *2

All the signals will arrive on the same schedule as they do
when the ship is not moving.

It is only an observer moving at a speed different from that of
the ship who would see the signals from the rear take longer to
reach the mother device than signals from the front of the ship.

Or am I still getting something wrong?

There is nothing for the computer to adjust. The computer does
not see any variation in the signals.

-- Jeff, in Minneapolis

[afterburner]

It is only an observer moving at a speed different from that of
the ship who would see the signals from the rear take longer to
reach the mother device than signals from the front of the ship.

So what you are saying is...
If we consider my tiger example with the cameras virtually moving at c, but change it the following way: two lasers, one at the back of the ship, one at the front - both firing at a detector in the middle....

What you're saying is...that the cameras would actually see the laser beam from the front of the ship firing to the back, reach the detector in the middle before the laser beam firing from the back of the ship...However, this would not be so for the observers inside the ship, as well as the detector...

Is that what you are saying?

[Ken G]

Yes, precisely. There is a disagreement on whether or not the tigers fired their lasers at the same time or not.

[Jeff Root]

It is only an observer moving at a speed different from that of
the ship who would see the signals from the rear take longer to
reach the mother device than signals from the front of the ship.

So what you are saying is...
If we consider my tiger example with the cameras virtually moving
at c, but change it the following way: two lasers, one at the back
of the ship, one at the front - both firing at a detector in the
middle....

What you're saying is...that the cameras would actually see the
laser beam from the front of the ship firing to the back, reach
the detector in the middle before the laser beam firing from the
back of the ship...However, this would not be so for the
observers inside the ship, as well as the detector...

Is that what you are saying?

Something like that.

Most likely, you have it set so that the two laser beams hit
the detector simultaneously. Dr. Foyle, aboard the ship, sees
the two lasers fire simultaneously, and a few nanoseconds later
they reach the detector simultaneously. Professor Fourmile,
watching the ship zip past at nearly the speed of light, sees
the rear laser fire first, and the beam is close to the detector
after travelling for nearly a microsecond before the front laser
fires, but just half a nanosecond later the two beams reach the
detector simultaneously, just as they do for Dr. Foyle.

-- Jeff, in Minneapolis

Tiger, tiger, burning bright
In the forests of the night
What immortal hand or eye
Can frame thy fearful symmetry?

[Ken G]

Yes, that's what I meant to say as well-- it's not that in one frame they reach at the same time and the other they don't, it's that in only one frame were they fired at the same time.

[afterburner]

But how can that be so? I'll describe the way I picture it and can you please tell me where I'm wrong.

Lets stick with the two lasers at opposite sides of the ship, these lasers will be the same at the lesser devices. The cameras outside will record everything. The mother device will be the thing that’s being fired at, and is exactly between the two lesser devices. Lasers are hard wired to fire every 10th second, as decided by the atomic clocks that are attached to each laser.

When the ship is stationary, we do a little test run. We synchronized the three devices, and make sure that the lesser devices fire on the same vibration of each atomic clock. Video: Atomic clocks show stationary, not dilated time. The two lasers fire at the same time, just as pre-programmed before the test run - on the same vibration (10th second). Onboard observer: Time feels like its going by as usual. The two lasers fire simultaneously, just as planned - on the same vibration. Mother device: Registered the two beams as both hitting at t=x

When the ship is moving at 75% c. Video: The atomic clocks are clearly running slower than stationary time, nevertheless all three are synchronized. The firing on the 10th second goes as planned - both lesser devices fire simultaneously, however, we see that the laser beam from the back of the ship is slowly creeping towards the mother device as compared to the beam from the front of the ship, which whizzed by and clearly hit the mother device first. Onboard observer: Time feels like its going by as usual. The two lasers fire simultaneously, just as planned - on the same vibration. Mother device: Registered the beam from the front of the ship as t=x, but the beam from the back of the ship as t=x+y

This is what how the scenario plays out in my head, apparently wrong.


If the cameras outside the ship, clearly see the two beams as they are moving through space, and the beam from the front of the ship clearly gets to the mother device firsts...I fail to see why the mother device would still register that the beams arrived at the same time, when the video clearly shows otherwise.

For the onboard observers...the atomic clocks are synchronized to fire at the same time, so they DO actually fire at the same time, just as seen on the cameras...the fact that the ship is going 75% c explains that the beam fired from the back will arrive after the beam fired from the front of the ship, since nothing can travel faster than light...

The beam fired from the back of the ship travels "with" the ship for a while, much like we see on the cameras. The beam from the front of the ship does not go through the same thing, much like we see on the cameras...

So why in the world would both beams from the lesser devices arrive at the same time, as registered by the mother device?

[Ken G]

When the ship is moving at 75% c. Video: The atomic clocks are clearly running slower than stationary time, nevertheless all three are synchronized.

Yes, this is what the video will show, but that's different from what the ship sees. As the ship accelerated, and it doesn't matter how long it took, the clocks that were synchronized will desynchronize. This must be true, it is the core of the twin paradox. Acceleration, coupled with separation, causes desynchronization. This is part of the "Lorentz transformation", and is the key forgotten element of special relativity (yes, it's pure special relativity). Almost every SR "paradox" comes from forgetting this, and this is another example.

For the onboard observers...the atomic clocks are synchronized to fire at the same time, so they DO actually fire at the same time, just as seen on the cameras...

No, the cameras are not moving with the ship, so cannot possibly agree with the ship on the synchronization of the lasers. What will happen is synchronization will be maintained for the cameras, but not onboard the ship. The clock in the front of the ship will instantaneously gain time relative to the clock in the back (if the acceleration is instantaneous), so the front laser will pop off some quick shots, arriving at the center ahead of the rear laser, in agreement with the findings of the stationary cameras for different reasons (the ones you described).

So why in the world would both beams from the lesser devices arrive at the same time, as registered by the mother device?

They wouldn't, the beam from the front would arrive first, as you expect. But it will be because of the desynchronization of the front clock, in the frame of the ship. This contradicts what Jeff Root said only because it depends on how you set it up. If you set it up so that the clocks are initially sychronized, and then accelerated without correcting the clocks in any way, what happens is what is being said now.

[Jeff Root]

I may be confused, but I disagree with Ken.

In order to describe the situation correctly, it is necessary to
discard some of afterburner's assumptions. Specifically, I get
rid of the assumption that the two laser beams reach the
central detector at different times.

It seems to me that once the lasers have been synchronized
before the trip starts, they will remain synchronized at any
speed, from the point of view of those aboard the ship. I'm
not quite sure what happens during acceleration, but as the
effect is fairly small, let's continue to ignore it.

Since the lasers are synchronized from the point of view of
those aboard the ship, the beams will continue to reach the
central detector simultaneously, and that will be true for all
observers. However, the lasers will be unsynchronized for
those not moving with the ship.

Don't you agree, Ken? Or am I confused?

-- Jeff, in Minneapolis

[Squashed]

... If you set it up so that the clocks are initially sychronized, and then accelerated without correcting the clocks in any way, what happens is what is being said now.

Wouldn't the acceleration be the same for the fore and aft "lesser" devices?

During the acceleration the two opposite lesser devices would de-synchronize but when the acceleration is completed the two would re-synchronize because both underwent the same acceleration - its just that the fore clock completed all its accleration first.

Now if the "mother" device flashes a signal to outside viewers whenever it receives a simultaneous signal from the two lesser devices - which is a positive indication of simultaneity - but if the outside viewers "see" the two lesser devices as de-synchronized then how can the outside viewers rectify what their eyes are observing: the two contradictory indications?

[Jeff Root]

Now if the "mother" device flashes a signal to outside viewers
whenever it receives a simultaneous signal from the two lesser
devices - which is a positive indication of simultaneity - but
if the outside viewers "see" the two lesser devices as
de-synchronized then how can the outside viewers rectify what
their eyes are observing: the two contradictory indications?

I don't think there is anything contradictory there. The two
lasers are synchronized from the point of view of people on the
ship, and it takes the same amount of time for the beams to
travel equal distances to the central detector. From the point
of view of someone not moving with the ship, the rear beam has
to travel a much greater distance than the front beam does, and
fires earlier, so the beams reach the detector simultaneously.

-- Jeff, in Minneapolis

[Ken G]

Wouldn't the acceleration be the same for the fore and aft "lesser" devices?

Yes.

During the acceleration the two opposite lesser devices would de-synchronize but when the acceleration is completed the two would re-synchronize because both underwent the same acceleration - its just that the fore clock completed all its accleration first.

No, the desynchronization is permanent, it does not go away when the acceleration stops. This is required to answer the twin paradox, by the way.

[Ken G]

It seems to me that once the lasers have been synchronized
before the trip starts, they will remain synchronized at any
speed, from the point of view of those aboard the ship.

It seems that way, but remember that in relativity, little is what it seems. The most commonly overlooked wrinkle in special relativity is that it is not just time dilation and length contraction (which are weird enough), to complete the Lorentz transformation there is also a term that says that whenever the speed changes, there is a permanent desynchronization of time that interacts with the acceleration such that the desynchronization is zero for two clocks in the same place, but is not zero if the clocks are separated, and indeed it increases in proportion to the separation. This effect is crucial for understanding the twin "paradox", for example, because without it, you end up concluding that time dilation should make each twin younger than the other. In the present example, it means that if the lesser devices are synchronized with respect to each other, in the frame of the people on the ship, prior to liftoff, then they will not be synchronized for those same people after the acceleration is complete. The best frame to analyze this phenomenon is always the non-accelerating frame, since there you really know what is going on, and be prepared for suprises in the accelerated frame, along the lines of what is causing Squashed some consternation. I think a key principle to remember is that if you are separated from a distant location, then the progress of time at that distant location can be thought of like a timeline in a history book. It has no direct connection to your sense of "now", but there is a useful convention for making that connection, defined as part of special relativity, and indeed that sense will change whenever you accelerate.

[afterburner]

Just to make it clear...I'm still a bit confused.

All three atomic clocks are synchronized when the ship is stationary for both the cameras, and the onboard observers. As the ship accelerates, the clock at the front of the ship will complete acceleration before the clock at the back, and therefore will be lagging behind slightly (compared to the other two clocks), firing it laser a little bit after the clock at the back of the ship. This explains why both beams reach the mother device at the same time for the people onboard the ship, keeping in mind that the mother device's time is somewwere in between the two lesser devices.

Why the outside cameras dont see that all three atomic clocks are not synchronized still puzzles me. Or is the above what the cameras see as well?

Also, does the effect of desynchronization that Ken G mentioned still hold for clocks that are perpendicularly alligned to the direction of travel, not one behind the other, such as in my example?

Thanks.

[Ken G]

Just to make it clear...I'm still a bit confused.

When in doubt, do all your calculations from the stationary frame, and just use those results to figure out what the accelerated frame will see.

This explains why both beams reach the mother device at the same time for the people onboard the ship, keeping in mind that the mother device's time is somewwere in between the two lesser devices.

No, both beams don't reach the mother device at the same time, that's what I have been saying. Jeff Root may have said that, but he was mistaken. Do the calculation in the stationary frame, and assume the acceleration that occurs is instantaneous. If there is first, prior to acceleration, a pulse from both lesser devices, and they reach the mother device at the same time, then we may imagine a second pulse emitted from each lesser device just after the acceleration (or at the same time as it, it matters not). The mother device then starts moving, and reaches the pulse from the front lesser device first. That's what the people on board will see as well. The lesser devices have desynchronized for the people on the ship, due to the acceleration. Time itself is flowing in a nonconstant way over the length of the ship during the acceleration (and this is all in the frame of the ship that I am speaking).

Why the outside cameras dont see that all three atomic clocks are not synchronized still puzzles me.

There outside cameras still think the atomic clocks are synchronized to each other because they are all doing the same thing. It is the people on the ship that see desynchronization, because it is necessary to apply a Lorentz transformation to enter their frame.

Also, does the effect of desynchronization that Ken G mentioned still hold for clocks that are perpendicularly alligned to the direction of travel, not one behind the other, such as in my example?

No, there is no lateral desynchronization. If there were, which sign would it have?

[afterburner]

Yet another clarification

We know what happens in a stationary reference frame.

During acceleration, time flows unevenly throughout the ship, which is the reason why all three clocks will no longer be synchronized.

After acceleration is complete, time flows smoothly everywhere on the ship, but all three clocks are still not synchronized because of the above. When the two lesser devices fire at the mother device, the beam from the front of the ship will arrive/register on the mother device before the beam coming from the back of the ship. So the mother device will know that the two beams that were supposed to arrive simultaneously, in fact arrived not simultaneously (as compared to a stationary reference frame data, which is used for comparison)

And if we extend this to the full circle/sphere device, the pulses from the lateral devices will still arrive simultaneously, much like they do at rest.



Now looking at the scenario from the video...
Completely lost. ... If we have the two atomic clocks showing time towards the window....and if the two atomic clocks are synchronized when at rest, and thats what the cameras see as well....why is it that after acceleration, the two clocks pointing towards the window will still show the same time, even though they are not synchronized...Shouldn’t the camera "pick up" the fact that time does not flow evenly throughout the ship during acceleration? and therefore show different times on all three devices?


Is this right?

[Ken G]

During acceleration, time flows unevenly throughout the ship, which is the reason why all three clocks will no longer be synchronized.

Yes, but this is only true in the ship's frame. In the stationary frame, the clocks maintain synchronization, as they all do exactly the same thing.

If we have the two atomic clocks showing time towards the window....and if the two atomic clocks are synchronized when at rest, and thats what the cameras see as well....why is it that after acceleration, the two clocks pointing towards the window will still show the same time, even though they are not synchronized...Shouldn’t the camera "pick up" the fact that time does not flow evenly throughout the ship during acceleration? and therefore show different times on all three devices?

Time only flows unevenly in the accelerated frame. There is no absolute time, so this uneven flow has no absolute quality.

[Jeff Root]

Ken,

Could you explain why clocks distributed along the length of the ship
appear to become de-synchronized during acceleration?

What happens if one synchronizes the clocks before launch, moves
them to their storage position at the center of the ship, accelerates
the ship to Really High Speed, then puts the clocks back in their
working positions?

-- Jeff, in Minneapolis

[hhEb09'1]

What happens if one synchronizes the clocks before launch, moves them to their storage position at the center of the ship, accelerates the ship to Really High Speed, then puts the clocks back in their working positions?

You'd think that the results would be the same? Moving the clocks involves an acceleration as well, as well as the relative movement.

[Ken G]

Could you explain why clocks distributed along the length of the ship
appear to become de-synchronized during acceleration?

I doubt I can explain why that's true, but to explain that it's true, you need to look up the "Lorentz transformation", and note how time and location get comingled when you transform from one frame to another (and that's essentially what you do when you accelerate the ship).

What happens if one synchronizes the clocks before launch, moves
them to their storage position at the center of the ship, accelerates
the ship to Really High Speed, then puts the clocks back in their
working positions?

That's an interesting question. They'll stay synchronized in that situation, in the rocket frame, and the desynchronization will appear in the stationary frame as you move them to their final locations. Excellent situations to contrast.

[Jeff Root]

What happens if one synchronizes the clocks before launch, moves
them to their storage position at the center of the ship, accelerates
the ship to Really High Speed, then puts the clocks back in their
working positions?

They'll stay synchronized in that situation, in the rocket frame,
and the desynchronization will appear in the stationary frame as
you move them to their final locations.

Why is it different from the original scenario in which the
clocks are distributed along the length of the ship?

-- Jeff, in Minneapolis

[Ken G]

Why shouldn't it be? It's a different situation. The motion of the clocks, if taken in the stationary frame for greater simplicity, is not the same in the two cases.