# Why nothing can go faster than speed of light.

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One clock would appear to be going either slow or fast.

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One clock would appear to be going either slow or fast.

That would only make sense if the clocks were synchronized to begin with and that they had the same tick rate (calibrated at least).

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That would only make sense if the clocks were synchronized to begin with and that they had the same tick rate (calibrated at least).

In any thought experiment like this I think one should assume that all clocks are synchronised in the same inertial frame prior to the experiment, or else it would be pointless.

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You described more than one frame. You talk of them moving slower or faster, which only makes sense if it's relative to another frame. Frames going the same velocity are the same frame.

I was going to see if a similar method would work for other frames but just picked on the multiple clocks in the same frame situation first. Janus had 3 clocks on the craft, and multiple on the exterior, that were synchronized but I can't recall him saying how they were synchronised or calibrated.

Would this method be suitable? #71

"How to synchronize their clocks.

Return time = time period between start of outgoing signal from A and the reflected return signal.

1. If two IFoR/clocks are moving at the same rate (in the same frame) and light reflected off one clock will return in the same time period to the sender/central clock.

step 1. send a signal with the time plus half the return time, and get second person (B) to set clock to this time."

In any thought experiment like this I think one should assume that all clocks are synchronised in the same inertial frame prior to the experiment, or else it would be pointless.

This maybe so, but sometimes this can't be done. This would depend on how the thought experiment was designed.

Edited by Robittybob1
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I was going to see if a similar method would work for other frames but just picked on the multiple clocks in the same frame situation first. Janus had 3 clocks on the craft, and multiple on the exterior, that were synchronized but I can't recall him saying how they were synchronised or calibrated.

"If one IFoR is moving away slower the light reflected off one will return in shorter time period to the sender.

If one IFoR is moving away faster the light reflected off one will return in a longer time period to the sender."

Would this method be suitable? #71

"How to synchronize their clocks.

Return time = time period between start of outgoing signal from A and the reflected return signal.

1. If two IFoR/clocks are moving at the same rate (in the same frame) and light reflected off one clock will return in the same time period to the sender/central clock.

step 1. send a signal with the time plus half the return time, and get second person (B) to set clock to this time."

For a single frame (which one takes to be at rest), compensating for the light travel time is known as the Einstein synchronization method. It's described in his 1905 relativity paper.

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"If one IFoR is moving away slower the light reflected off one will return in shorter time period to the sender.

If one IFoR is moving away faster the light reflected off one will return in a longer time period to the sender."

For a single frame (which one takes to be at rest), compensating for the light travel time is known as the Einstein synchronization method. It's described in his 1905 relativity paper.

True there were 3 situations that needed to be worked on (no motion, slower, faster) but I only tried to find a solution for one of them (the no relative motion situation) so far.

OK I need to look at Einstein's synchronization method. Thanks.

IFoR = inertial frames of reference.

Situation 1. If two IFoR are moving at the same rate light reflected off one will return in the same time period to the sender.

Situation 2 If one IFoR is moving away slower the light reflected off one will return in shorter time period to the sender.

Situation 3 If one IFoR is moving away faster the light reflected off one will return in a longer time period to the sender.

How to synchronize their clocks.

Return time = time period between start of outgoing signal from A and the reflected return signal.

Situation 1. If two IFoR are moving at the same rate light reflected off one will return in the same time period to the sender.

step 1. send a signal with the time plus half the return time, and get second person (B) to set clock to this time.

I have now may it clear there are 3 situations.

Edited by Robittybob1
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Relative to each other. All motion is relative to the sender frame, so if a signal returns at same interval the two IFoR are going at the same relative velocity. If the time period is shorter the distance between them is getting shorter or longer the frames are not going the same velocity.

I'm surprised that you question this?

That doesn't really make any sense. Your seen to be implying absolute velocities. Or you have an implied third frame but you say you don't...

Try naming your frames and explaining in what frame each velocity is measured in.

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That doesn't really make any sense. Your seen to be implying absolute velocities. Or you have an implied third frame but you say you don't...

Try naming your frames and explaining in what frame each velocity is measured in.

I understand there is no preferred frame, and all motion is relative to some other frame.

So the sender feels he is stationary i.e things go past his frame not the other way around, even if it seems logically incorrect e.g. the platform moves past the train to an observer on the train.

So all clocks are to be synchronized to the sender's frame for he has the equipment to send out and receive the signal. The other frames have clocks, mirrors on the clock and device to receive instantaneous messages and can adjust their clocks instantaneously to S1 being the sender's clock (the reference clock).

So sender's frame call it S and all the clocks on the S frame are synchronized one at a time against S1 firstly S2, S3 ...Sn

I have been trying to find someone doing a video description of Einstein's synchronization (ES) and as far as I can tell it was the same as I was using but only someone familiar with the ES method can confirm that.

So I was thinking how does S1 work out whether the other clocks are on it own frame or on different frames, and are those different frames coming closer or moving away, so S1 uses the light beam and measures the return time, if on two tests the return time decreases the other clock is on a frame moving toward S1.

If on two tests the return time increases the other clock is on a frame moving away from S1.

If on two tests the return time is equal the other clock is on the same frame as S1.

"so if a signal returns at same interval the two IFoR are going at the same relative velocity."

@Klaynos - that was not said properly "the same relative velocity" should have been "zero relative velocity". To the observer it would appear the other clock is not moving.

If there was another (external) observer on another IFoR it would appear as they both had "the same relative velocity".

Edited by Robittybob1
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I guess I wasn't clear. Setting them to tick together would be like matching your clock with one in a sped up or slowed down film.

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I guess I wasn't clear. Setting them to tick together would be like matching your clock with one in a sped up or slowed down film.

That is the consequence for there is no universal time standard. Is that what you mean? Whoever does the measuring sets the time to their frame. The person doing the measuring doesn't feel slowed or sped up.

To me you still aren't that clear.

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I understand there is no preferred frame, and all motion is relative to some other frame.

So the sender feels he is stationary i.e things go past his frame not the other way around, even if it seems logically incorrect e.g. the platform moves past the train to an observer on the train.

So all clocks are to be synchronized to the sender's frame for he has the equipment to send out and receive the signal. The other frames have clocks, mirrors on the clock and device to receive instantaneous messages and can adjust their clocks instantaneously to S1 being the sender's clock (the reference clock).

So sender's frame call it S and all the clocks on the S frame are synchronized one at a time against S1 firstly S2, S3 ...Sn

I have been trying to find someone doing a video description of Einstein's synchronization (ES) and as far as I can tell it was the same as I was using but only someone familiar with the ES method can confirm that.

So I was thinking how does S1 work out whether the other clocks are on it own frame or on different frames, and are those different frames coming closer or moving away, so S1 uses the light beam and measures the return time, if on two tests the return time decreases the other clock is on a frame moving toward S1.

If on two tests the return time increases the other clock is on a frame moving away from S1.

If on two tests the return time is equal the other clock is on the same frame as S1.

"so if a signal returns at same interval the two IFoR are going at the same relative velocity."

@Klaynos - that was not said properly "the same relative velocity" should have been "zero relative velocity". To the observer it would appear the other clock is not moving.

If there was another (external) observer on another IFoR it would appear as they both had "the same relative velocity".

I am not thinking in terms of time dilation occurring on the frames with relative motion, we are just looking at the situation in the idealised form where all distances are measured in the x dimension. When we look at how distant some object is, the angle of sight to its direction of travel doesn't change (coming straight at you). Later on we might need to account for this time dilation effect. For all we are trying to do at this stage is to synchronise every frame's clocks to begin with and later we will see if their time is dilated.

Since we can only do one clock at a time will the clocks stay in sync during the time taken till we get to the point of saying "job completed"?

Houston we have a problem!

Edited by Robittybob1
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I understand there is no preferred frame, and all motion is relative to some other frame.

So the sender feels he is stationary i.e things go past his frame not the other way around, even if it seems logically incorrect e.g. the platform moves past the train to an observer on the train.

So all clocks are to be synchronized to the sender's frame for he has the equipment to send out and receive the signal. The other frames have clocks, mirrors on the clock and device to receive instantaneous messages and can adjust their clocks instantaneously to S1 being the sender's clock (the reference clock).

So sender's frame call it S and all the clocks on the S frame are synchronized one at a time against S1 firstly S2, S3 ...Sn

I have been trying to find someone doing a video description of Einstein's synchronization (ES) and as far as I can tell it was the same as I was using but only someone familiar with the ES method can confirm that.

So I was thinking how does S1 work out whether the other clocks are on it own frame or on different frames, and are those different frames coming closer or moving away, so S1 uses the light beam and measures the return time, if on two tests the return time decreases the other clock is on a frame moving toward S1.

If on two tests the return time increases the other clock is on a frame moving away from S1.

If on two tests the return time is equal the other clock is on the same frame as S1.

"so if a signal returns at same interval the two IFoR are going at the same relative velocity."

@Klaynos - that was not said properly "the same relative velocity" should have been "zero relative velocity". To the observer it would appear the other clock is not moving.

If there was another (external) observer on another IFoR it would appear as they both had "the same relative velocity".

I am not thinking in terms of time dilation occurring on the frames with relative motion, we are just looking at the situation in the idealised form where all distances are measured in the x dimension. When we look at how distant some object is, the angle of sight to its direction of travel doesn't change (coming straight at you). Later on we might need to account for this time dilation effect. For all we are trying to do at this stage is to synchronise every frame's clocks to begin with and later we will see if their time is dilated.

Since we can only do one clock at a time will the clocks stay in sync during the time taken till we get to the point of saying "job completed"?

Houston we have a problem!

!

Moderator Note

In observing so much of the confusion rampant in many of the threads you're engaged in, Robittybob1, I find your propensity to unnecessarily quote yourself to be very misleading, especially when so many of your conclusions are conjecture, or just misunderstood, or just wrong. It occurs that perhaps using yourself as a source for supporting your arguments is counterproductive. It's also frustrating to have to re-read something to no purpose. It's also frustrating because it makes it seem like you're only listening to yourself instead of responding to the replies of others.

Take this on board or not. I'm concerned that so many members are either calling for clarity regarding your posts, or trying to explain their own responses to you. The signal to noise ratio is getting out of hand, imo.

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Was there some error in what I have said above? I am trying to be 100% accurate in what I post. If there was a mistake please discuss the issue on the forum.

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There is one more important aspect of SR that you are missing: The Relativity of simultaneity.

Let's assume that our spaceship has three clocks, one in the front, one in the back and one in the middle, and according to the ship, they are all in sync with each other. The ship is long (one light sec long as measured by the ship). A laser in fired from the middle clock towards both the front and back and reflected back to the middle. The time on the front and rear clocks upon arrival is noted, along with the time on the middle clock upon the light's return.

The laser's leave the middle clock when it reads 0, strikes both of the other clocks when they read 0.5 and returns to the middle clock when it reads 1 sec. In the ship this is explained by the fact that light travels at c and the round trip distance is 1 light sec.

Now consider what happens to each laser according to an observer watching the ship traveling at 0.5 c. One laser leaves the middle clock when it reads 0, and heads toward the front of the ship. The ship is length contracted, so the front for the ship is 0.433 light sec away, and due to the relative difference between the ship's velocity and the speed of light, arrives at the front of the ship after .866 sec. it reflects back towards the middle. Because the light is moving in the opposite direction from the ship, it will take 0.2887 sec to meet up with the middle clock for a total round trip time of 1.1547 sec. Due to time dilation,the middle clock ticks off 1 sec in this time.

.....

@Janus - Can you help me here please? I was reading http://galileoandeinstein.physics.virginia.edu/lectures/synchronizing.html and it sounded very much like what you descibed without going into too much detail.

My thought was if we had one central clock and 2 clocks at each end. We synchronize the first two using the method described

"the clocks are synchronized, assuming they are both accurate, is to start them together. How can we do that? We could, for example, attach a photocell to each clock, so when a flash of light reaches the clock, it begins running. If, then, we place a flashbulb at the midpoint of the line joining the two clocks, and flash it, the light flash will take the same time to reach the two clocks, so they will start at the same time, and therefore be synchronized."

So we synchronise the first two end clocks (have only 1 uncovered), and then accelerate the craft to half light speed. Do the synchronised clocks stay synchronised? How can we tell if they are synchronised?

When at half light speed if we were to repeat our experiment on the remaining two end clocks and record the time it happened on the already synchronised clocks would they both record the event, on a readout, as happening at same time?

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Try to explain why nothing can go faster than speed of light without using the term relativistic mass.

Can it be done?

well, you cannot fit something so easily into others brains without any solid proof that this particle is moving faster then light which is quite impossible since if you try to measure something faster then light then how will you see it?

no matter how much you try you cannot observe it since you need light to observe something.

for example,

if you send a particle you know faster then light about a light year away and maybe it do reach the distance before light but.... you will only be able to observe it only when the light from it comes back which will obviously take 1 year and if you add the processing time in it then it will obviously give result which shows that it was slower then light or equal to speed of light.

well, this is my theory to agree or not is all upto you.

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@Janus - Can you help me here please? I was reading http://galileoandeinstein.physics.virginia.edu/lectures/synchronizing.html and it sounded very much like what you descibed without going into too much detail.

My thought was if we had one central clock and 2 clocks at each end. We synchronize the first two using the method described

So we synchronise the first two end clocks (have only 1 uncovered), and then accelerate the craft to half light speed. Do the synchronised clocks stay synchronised? How can we tell if they are synchronised?

When at half light speed if we were to repeat our experiment on the remaining two end clocks and record the time it happened on the already synchronised clocks would they both record the event, on a readout, as happening at same time?

Now is not the time to get into what happens to separated clocks when you accelerate them, as you still have not come to grips with the behavior of clocks with relative motion is inertial frame.

I think you are getting too tied up in the minutiae of how the clocks are brought into sync when all that is needed is to establish that they are in sync.

A simple scheme would be such: you have a long string of clocks that are spaced at equal intervals. Each clock has an observer. The observer's note the times they read on the other clocks. All they are concerned with is whether or not clock equal distances from them show equal times. For instance, that the pair of clocks immediately to either side of their clock read the same, and that the pair of clocks one step further out read the same, etc. If this is true for every observer on every clock, then we can say that these clocks are all in sync in this frame. Again, it doesn't matter how we originally got the clocks in sync, only that we have established that they are in sync.

So let's say that we have set up such a system of clocks, and we now set up another set of clocks, again equally spaced, but this time they are moving with respect to the first line of clocks. Further, let's say that we use the first line of clock to set the second line of clock. The spacing between the second set of clocks will be such that they, at moments, line up with the first set of clocks. We we further stipulate that the when two clocks in the two sets pass each other, they read the same time. thus we get the following situation.

(note that the animation runs for a bit and then resets.)

So far, so good.

But what does this arrangement look like if we switch to the upper set of clocks so that the they are considered stationary and the bottom line is moving right to left?

First you have to consider the fact that in the above animation, the top line of clocks is in motion, and as such the rate we see them run at is a time dilated rate and the distances between them are length contracted distances. So in the frame where they are not moving, the clocks tick faster and the distances between them are greater and this how they need to be shown in this frame.

Also, since the bottom line of clocks is now in motion, the distances are length contracted and the clock rates time dilated, so in the new view they will be closer together and tick slower.

the result is the following;

(another point I need to make is that in neither of these animations did I apply length contraction to the clocks themselves, even though in reality they would have the same length contraction as the distances between them. I did it this way because, for one, it has no bearing on the ideas being presented and for another it makes the clock reading comparisons easier)

Note that just like in the first animation, whenever a clock in the top line passes a clock in the lower line, both clocks read the same time. This must remain true because this is the condition we used to set up the arrangement shown in the first animation.

However to maintain this condition, due to the fact that the tick rates and distances between the clocks in the two lines differ, The clocks in neither line are can be in sync. Clocks in one direction have to read ahead and clocks in the other read behind. The only time two clocks read the same is when it is a pair of clocks passing each other.

If we freeze the animation at a moment when two clocks passing each other read 12:00, you can see that none of the other clocks read 12:00

Even though with our initial set up where the lower set of clocks were considered stationary, all the clocks read the same time at all moments.

Events that are considered simultaneous when you are rest with respect to the lower line of clocks are not simultaneous when you are at rest with the upper line. This is the Relativity of Simultaneity, and is a very important part of Relativity. Until you can wrap your mind around and accept it, you will not be able to understand Relativity.

Edited by Janus
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Now is not the time to get into what happens to separated clocks when you accelerate them, as you still have not come to grips with the behavior of clocks with relative motion is inertial frame.

I think you are getting too tied up in the minutiae of how the clocks are brought into sync when all that is needed is to establish that they are in sync.

A simple scheme would be such: you have a long string of clocks that are spaced at equal intervals. Each clock has an observer. The observer's note the times they read on the other clocks. All they are concerned with is whether or not clock equal distances from them show equal times. For instance, that the pair of clocks immediately to either side of their clock read the same, and that the pair of clocks one step further out read the same, etc. If this is true for every observer on every clock, then we can say that these clocks are all in sync in this frame. Again, it doesn't matter how we originally got the clocks in sync, only that we have established that they are in sync.

So let's say that we have set up such a system of clocks, and we now set up another set of clocks, again equally spaced, but this time they are moving with respect to the first line of clocks. Further, let's say that we use the first line of clock to set the second line of clock. The spacing between the second set of clocks will be such that they, at moments, line up with the first set of clocks. We we further stipulate that the when two clocks in the two sets pass each other, they read the same time. thus we get the following situation.

(note that the animation runs for a bit and then resets.)

So far, so good.

But what does this arrangement look like if we switch to the upper set of clocks so that the they are considered stationary and the bottom line is moving right to left?

First you have to consider the fact that in the above animation, the top line of clocks is in motion, and as such the rate we see them run at is a time dilated rate and the distances between them are length contracted d

My initial thought to this is that it is a physical impossibility. Is this physically possible? Do they check the neighbouring clocks by sight? Seeing involves the speed of light so you see something as it was and not as it is.

The two clocks either side of me could be reading the same time.

One of these two clocks is also read by you and the two clocks either side of you could be reading the same time.

So this could be extended indefinitely.

Logically it can be done.

Now I have to get my head around accepting that all those clocks will reading the same time. OK yes they are.

But what does this arrangement look like if we switch to the upper set of clocks so that the they are considered stationary and the bottom line is moving right to left?

That was my question to you wasn't it. If clocks are synchronised in one frame and then accelerated will they still be synchronised?

If I turned your animation over the top row becomes the bottom and the animation will work just as well.

But if I was to slow the top row and accelerate the bottom row right to left? Look I'd stabbing in the dark.

They won't be in sync top and bottom any longer. But along the line they could still be in sync.

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well, you cannot fit something so easily into others brains without any solid proof that this particle is moving faster then light which is quite impossible since if you try to measure something faster then light then how will you see it?

no matter how much you try you cannot observe it since you need light to observe something.

for example,

if you send a particle you know faster then light about a light year away and maybe it do reach the distance before light but.... you will only be able to observe it only when the light from it comes back which will obviously take 1 year and if you add the processing time in it then it will obviously give result which shows that it was slower then light or equal to speed of light.

well, this is my theory to agree or not is all upto you.

Your argument is completely off base. Let's say you send a object at 0.999c to a planet 1 light year away. You watch for the flash it creates when it hits. Since it will take a bit over a year for the object to reach the planet, and then another year for the light to return to you, you will see the light a bit more than 2 years after you sent it.

However, let's say you were able to send the object at 2 times the speed of light. It will take 1/3 year to get there, the light will still take 1 year to reach you, and you will see the flash 1 1/2 years after you sent the object. The very fact that the light of the flash arrived in less than two years is a measurement that the object took less than 1 year to reach the planet and thus traveled faster than light.

if you were watching the first (.99c) object the whole way on its trip, you would see it make the trip in a bit over 2 years. If you watched the 2c object make the trip, you would see it make the trip in 1.5 years. Again, the fact that you see the image of the object make the trip in less than 2 yrs means you are watching an object that is traveling faster than light. In other words, the idea that you can't measure the speed of an object traveling faster than light with light is bogus.

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well, this is my theory to agree or not is all upto you.

!

Moderator Note

STOP interjecting your own made up ideas into mainstream threads. When someone is trying to determine our best current explanation, we don't need unsupported guesses, we need evidence and science. You've been told this many times before, so have a warning point too.

Do NOT respond to this action in thread. Report it if you object. That's what someone else did to your post.

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@Janus - what I have just noticed is that if you number all the clocks in one frame , 1,2,3, .....n and the first person sitting in front of clock 1 would have only one clock to check that is clock 2. The second person would check clocks 1 & 3 and the third person would check 2 & 4 so in effect you could have two series of clocks the odd numbered ones showing a different time to the even numbered ones and no one would know.

Edited by Robittybob1
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@Janus - what I have just noticed is that if you number all the clocks in one frame , 1,2,3, .....n and the first person sitting in front of clock 1 would have only one clock to check that is clock 2. The second person would check clocks 1 & 3 and the third person would check 2 & 4 so in effect you could have two series of clocks the odd numbered ones showing a different time to the even numbered ones and no one would know.

All clocks are checked by every motionless observer.

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All clocks are checked by every motionless observer.

Each observer is motionless with his own inertial frame, but who said each observer has to read all the clocks at once?

So even if the observer 28 checked clocks 27 & 29, then 26 & 30, then 25 & 31 ..... he is always checking an even numbered clock against another even numbered clock, the same with the odd ones. The odds and the evens could be out of sync and you can't tell.

Quote Janus

All they are concerned with is whether or not clock equal distances from them show equal times. For instance, that the pair of clocks immediately to either side of their clock read the same, and that the pair of clocks one step further out read the same, etc. If this is true for every observer on every clock, then we can say that these clocks are all in sync in this frame.

The observer is only comparing equidistance clocks, in other words an odd with an odd and even with even.

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@Janus - what I have just noticed is that if you number all the clocks in one frame , 1,2,3, .....n and the first person sitting in front of clock 1 would have only one clock to check that is clock 2. The second person would check clocks 1 & 3 and the third person would check 2 & 4 so in effect you could have two series of clocks the odd numbered ones showing a different time to the even numbered ones and no one would know.

No, each observer is able to read any of the other clocks, so observer 6 would be checking the pairs 5&7, 4&8, 3&9, 2&10, etc. observer 7 is checking pairs 6&8, 5&9, 4&10, 3&11 etc. So there is cross-checking

My initial thought to this is that it is a physical impossibility. Is this physically possible? Do they check the neighbouring clocks by sight? Seeing involves the speed of light so you see something as it was and not as it is.

The two clocks either side of me could be reading the same time.

One of these two clocks is also read by you and the two clocks either side of you could be reading the same time.

So this could be extended indefinitely.

Logically it can be done.

Now I have to get my head around accepting that all those clocks will reading the same time. OK yes they are.

It's perfectly possible to set things up this way. All each clock has to do is confirm that when a clock in the other row passes it, they read the same time. Since the distance between the two lines of clocks can be effectively zero, there is no propagation delay involved. Even if an observer was checking clocks further down the line, it doesn't factor in. If I'm at clock 1 looking at clock 100 in my line, all I have to do is verify that when a clock in the other line is adjacent to clock 100 it reads the same as clock 100 in my line. Since the distance to the two clocks is the same, the light propagation delay doesn't factor in.

But what does this arrangement look like if we switch to the upper set of clocks so that the they are considered stationary and the bottom line is moving right to left?

That was my question to you wasn't it. If clocks are synchronised in one frame and then accelerated will they still be synchronised?

No, it is not the same question. There is no acceleration involved in my scenario, I'm just showing how the events occur as measured from each of the two frame which are in inertial motion with respect to each other. Dealing with acceleration of either set of clock involves a whole new set of complications that goes beyond the principles we are trying to discuss here. (For example, in the accelerating frame, the clocks in that frame all run at different rates even as measured from that frame.) So, as I said before, you are not ready to deal with acceleration until you are perfectly comfortable inertial frames.

If I turned your animation over the top row becomes the bottom and the animation will work just as well.

No. it doesn't work that way. Let's assume that in animation 1, the distance between the clocks is 0.6 light hrs and the upper row of clocks is moving at 0.6 c. The bottom row of clocks is at rest in this animation, so this is the proper distance between them. The upper row is moving, so this is the length contracted distance between them and not the distance someone moving with these clocks would measure between the clocks. The length contraction factor at 0.6c is 0.8, so this means that in the rest frame of the upper row, these clocks are 0.75 light sec apart. 0.75 x 0.8 = 0.6, which is the distance rest frame of the lower clocks measures as the distance. In addition, the tick rate of the upper clocks as measured by the lower clock, is a time dilated rate, This means that the clocks in the upper row have to be rigged to run fast, so that when measured from the lower row frame, they run at the same rate as the lower clocks. In other words, if you took a clocks from the upper and lower rows and put them at rest with respect to each other the clock from the upper row would run 1.25 times faster than the one from the lower row.

Further. When measured from the rest frame of the upper row, it is the distance between the lower row clocks that is contracted and the tick rate that is slow. Since the proper distance between the clocks is 0.6 light hours, the distance as measured by the upper row frame will be 0.48 light hours. The clocks will also tick 0.8 as fast. So when the two rows of clocks are compared by someone at rest with respect to the upper row of clocks, their clocks will be 0.75 light hrs apart, compared to the 0.48 light hrs for the lower clocks, and the lower clocks will tick 0.64 as fast as the upper clocks.

This is what I showed in the second animation.

To get things as seen from the upper row to look like the first animation. You would have to adjust the distance distance between and the tick rate for the clock in one row or the other. Now, if we for instance, decrease the distance between the upper clocks and slow them down while simultaneously speeding up the lower clocks and increasing the distance between them can duplicate the first animation with the top row at rest and the lower row moving right to left, but now if we switch to the lower clock rest frame we find clocks in the upper row being closer together and ticking slower than the clocks in the lower row.

There is no way to arrange things so that the rest frames of both sets of clocks will say that the distance between the clocks in the two rows are the same or that the clocks in the two rows tick at the same rate. You can arrange it so that it true in one frame or the other, but not both. One one thing both rows do agree upon is that when any upper clock and lower clock pass each other, they read the same time.

But if I was to slow the top row and accelerate the bottom row right to left? Look I'd stabbing in the dark.

They won't be in sync top and bottom any longer. But along the line they could still be in sync.

Accelerating the clocks throws a monkey wrench into the works, which as I've already stated, you are not in a position to deal with yet. (accelerating a row of clocks that are sync in an inertial frame will put them out of sync even according to the clocks themselves.)

I'll repeat: Forget about acceleration until you after you have come to terms with purely inertial motion.

Edited by Janus
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Each observer is motionless with his own inertial frame, but who said each observer has to read all the clocks at once?

So even if the observer 28 checked clocks 27 & 29, then 26 & 30, then 25 & 31 ..... he is always checking an even numbered clock against another even numbered clock, the same with the odd ones. The odds and the evens could be out of sync and you can't tell.

Any observer has to read relatively simultaneous indications of the clocks.

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@Janus - what I have just noticed is that if you number all the clocks in one frame , 1,2,3, .....n and the first person sitting in front of clock 1 would have only one clock to check that is clock 2. The second person would check clocks 1 & 3 and the third person would check 2 & 4 so in effect you could have two series of clocks the odd numbered ones showing a different time to the even numbered ones and no one would know.

No, each observer is able to read any of the other clocks, so observer 6 would be checking the pairs 5&7, 4&8, 3&9, 2&10, etc. observer 7 is checking pairs 6&8, 5&9, 4&10, 3&11 etc. So there is cross-checking

Even in your answer pairs of clocks they are even with even, odd with odd.

Any observer has to read relatively simultaneous indications of the clocks.

And how does he do that?

We seem to have confused the quoting system.

@ Janus - Even in your answer your pairs of clocks they are even with even, odd with odd.

Edited by Robittybob1

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