Is Faster than Light travel impossible?

[Doc Al]

Are you proposing that Faster than Light travel is possible then ?

I'm saying that the rate at which ships A and B approach each other as measured by frame C can be greater than the speed of light. That has nothing to do with any object or information traveling faster than light.

Are A and B both in the C Frame of reference ?

A and B are two ships traveling with different velocities with respect to frame C.

 

[A.T.]

Are you proposing that Faster than Light travel is possible then ?

Nothing is traveling faster than light here. Velocity is defined as change of space coordinate per time change. Change of distance to another moving object per time change is not velocity, and can be greater than c.

Ok, you got me really confused then, just when I thought I was beginning to get it.

Then forget what I wrote. I just wanted to point out what 'relative' means, and that not everything is relative in Relativity.

 

[jmallett]

Nothing is traveling faster than light here. Velocity is defined as change of space coordinate per time change. Change of distance to another moving object per time change is not velocity, and can be greater than c.

Then forget what I wrote. I just wanted to point out what 'relative' means, and that not everything is relative in Relativity.

Can I try my question then in a different way ?
I am in free space in my own frame of reference inside a box, sitting on a rod that has a light source mounted at one end of it. The light flashes and releases a photon. Am I correct in believing that I will see that photon traveling at the speed of light ? because that's what I was expecting.
I now open a door in the box and see that there are stars passing me by. Why do I now have to add the speed of the rod relative to the stars into my equation ? and if the speed is positive and opposed to the direction of the photon then doesn't that mean that the end of the rod is now closing in on the photon at the speed of the rod relative to the stars plus the speed of the photon (light).

Why does observing the stars make any difference ? Isn't that an argument that derives from the proposition of an ether ?

HELP !

 

[Doc Al]

Can I try my question then in a different way ?
I am in free space in my own frame of reference inside a box, sitting on a rod that has a light source mounted at one end of it. The light flashes and releases a photon. Am I correct in believing that I will see that photon traveling at the speed of light ? because that's what I was expecting.

You are correct.

I now open a door in the box and see that there are stars passing me by. Why do I now have to add the speed of the rod relative to the stars into my equation ? and if the speed is positive and opposed to the direction of the photon then doesn't that mean that the end of the rod is now closing in on the photon at the speed of the rod relative to the stars plus the speed of the photon (light).

You will measure the speed of any photon as traveling at the same speed with respect to you, regardless of your motion with respect to anything else. Only if the rod is moving with respect to you do you need to add or subtract its motion to find the closing speed of the photon and rod as seen by you.

If you are observing a rod that is moving with respect to you and there's a photon traveling along that rod, then the rate at which the photon and rod approach each other as seen by you can certainly be greater than the speed of light. (This is what Integral was explaining in post #26.)

 

[jmallett]

You are correct.

You will measure the speed of any photon as traveling at the same speed with respect to you, regardless of your motion with respect to anything else. Only if the rod is moving with respect to you do you need to add or subtract its motion to find the closing speed of the photon and rod as seen by you.

If you are observing a rod that is moving with respect to you and there's a photon traveling along that rod, then the rate at which the photon and rod approach each other as seen by you can certainly be greater than the speed of light. (This is what Integral was explaining in post #26.)

OK, great, thanks, I get that, and I am still with you, for the moment.
Now I go back to Einstein's 1905 paper and look at the last part of Para 2 of Section 1. Here he says that there are 2 clocks, one at each end of the rod and there is an observer at each clock, at least that's what I understand. These observers are not moving with respect to the rod, right ?

Next in about 2 very cryptic sentences he concludes that light will travel faster in one direction than in the other. In fact he does add/subtract the rod's motion even when the observer and the rod are not moving relative to each other by suddenly introducing the cosmos.

 

[Doc Al]

Now I go back to Einstein's 1905 paper and look at the last part of Para 2 of Section 1. Here he says that there are 2 clocks, one at each end of the rod and there is an observer at each clock, at least that's what I understand. These observers are not moving with respect to the rod, right ?

I think I'm looking at the passage that you are referring to. (It's not paragraph 2 of Section 1 in my version of "On the Electrodynamics of Moving Bodies"; more like the middle of Section 2.)

There are two sets of observers. The 'stationary' observers (who see the rod moving) and the 'moving' observers (who travel along with the rod).

Next in about 2 very cryptic sentences he concludes that light will travel faster in one direction than in the other. In fact he does add/subtract the rod's motion even when the observer and the rod are not moving relative to each other by suddenly introducing the cosmos.

I don't see any introduction of the cosmos, just some basic kinematics as viewed from the 'stationary' frame. (I wonder if we are looking at the same thing? )

 

[jmallett]

I think I'm looking at the passage that you are referring to. (It's not paragraph 2 of Section 1 in my version of "On the Electrodynamics of Moving Bodies"; more like the middle of Section 2.)

There are two sets of observers. The 'stationary' observers (who see the rod moving) and the 'moving' observers (who travel along with the rod).

I don't see any introduction of the cosmos, just some basic kinematics as viewed from the 'stationary' frame. (I wonder if we are looking at the same thing? )

Yes, we are looking at the same place in the paper. Yes, there are observers moving with the rod, then there are the stationary observers.

What are the stationary observers stationary relative to ? They are not stationary relative to the rod, they are stationary relative to an external frame of reference which I am equating to the cosmos in a general sense.

 

[Doc Al]

Yes, we are looking at the same place in the paper. Yes, there are observers moving with the rod, then there are the stationary observers.

OK.

What are the stationary observers stationary relative to ? They are not stationary relative to the rod, they are stationary relative to an external frame of reference which I am equating to the cosmos in a general sense.

Nothing quite so dramatic. Just imagine that some observers are in a space station floating in space and they are observing a long rod which is moving past them. All that matters is the relative motion between the space station and the rod. In order to specify who is who, we arbitrarily call the space station observers the 'stationary' frame, but this does not mean stationary in any absolute sense.

 

[jmallett]

OK.

Nothing quite so dramatic. Just imagine that some observers are in a space station floating in space and they are observing a long rod which is moving past them. All that matters is the relative motion between the space station and the rod. In order to specify who is who, we arbitrarily call the space station observers the 'stationary' frame, but this does not mean stationary in any absolute sense.

By the way, thanks very much for your help - it is really and truly appreciated.

So do these observers see what is happening on the rod instantaneously ? I think Einstein is saying that all of their clocks are synchronised, right ? so that would imply that they see the same thing at the same time, or else are at least easily able to make a correction based on time and distance, right ?

 

[Doc Al]

So do these observers see what is happening on the rod instantaneously ? I think Einstein is saying that all of their clocks are synchronised, right ? so that would imply that they see the same thing at the same time, or else are at least easily able to make a correction based on time and distance, right ?

I don't know what you mean by seeing things happen 'instantaneously'. The 'moving' clocks are synchronized in their own frame, as are the 'stationary' clocks. One point of Einstein's paper is to show that clock synchronization is frame dependent--the 'stationary' observers will see the 'moving' clocks as unsynchronized.

 

[heldervelez]

Light speed is independent of the movement of the source.
It means that the medium only (vacuum, aether, field, space, glass) determines the speed of propagation of light.
We measure the light speed as 'c', in our local referential, and the same occurs to the measures made in others referentials.
But the limiting value 'c' is a two way (closed loop) measure.
The one way light speed has an undefined value (AFAIK unmeasured), maybe/maybe_Not greater than 'c'.
May be is my answer to the OP question (ONE WAY speed, only)

 

[Doc Al]

Light speed is independent of the movement of the source.
It means that the medium only (vacuum, aether, field, space, glass) determines the speed of propagation of light.
We measure the light speed as 'c', in our local referential, and the same occurs to the measures made in others referentials.

Just a clarification: The speed of light in vacuum will be measured as 'c' in any local inertial frame. Since the speed of light slows in some media--such as glass--we'd need to add (relativistically, of course) the speed of the medium plus the speed of the light with respect to the medium to find its measured speed in some frame. (See Fizeau's experiments.)

But the limiting value 'c' is a two way (closed loop) measure.
The one way light speed has an undefined value (AFAIK unmeasured), maybe/maybe_Not greater than 'c'.
May be is my answer to the OP question (ONE WAY speed, only)

I don't think issues about one-way speed of light have anything to do with the OP's question, which has to do with understanding how a 'closing speed' > c does not violate relativity.

 

[jmallett]

I don't know what you mean by seeing things happen 'instantaneously'. The 'moving' clocks are synchronized in their own frame, as are the 'stationary' clocks. One point of Einstein's paper is to show that clock synchronization is frame dependent--the 'stationary' observers will see the 'moving' clocks as unsynchronized.

I had understood from the paper that all of the clocks were synchronized.

Why don't the stationary observers see the error that the moving observers made in the measurement of the length of the rod ?

 

[Doc Al]

I had understood from the paper that all of the clocks were synchronized.

You can't synchronize clocks that move with respect to each other. The moving clocks are synchronized with each other (in the moving frame) and the stationary clocks are also synchronized with each other (in the stationary frame).

Why don't the stationary observers see the error that the moving observers made in the measurement of the length of the rod ?

I wouldn't call it an error. The moving observers used standard methods to measure the length of the rod. The stationary observers see that the moving observers' clocks are unsynchronized and their measuring rods are shortened according to the stationary frame. The moving observers, in turn, see the 'stationary' observers' clocks as being unsynchronized. (From the viewpoint of what we are calling the 'moving' frame, it is the 'stationary' frame that is moving.)

 

[jmallett]

You can't synchronize clocks that move with respect to each other. The moving clocks are synchronized with each other (in the moving frame) and the stationary clocks are also synchronized with each other (in the stationary frame).


I wouldn't call it an error. The moving observers used standard methods to measure the length of the rod. The stationary observers see that the moving observers' clocks are unsynchronized and their measuring rods are shortened according to the stationary frame. The moving observers, in turn, see the 'stationary' observers' clocks as being unsynchronized. (From the viewpoint of what we are calling the 'moving' frame, it is the 'stationary' frame that is moving.)

OK, so I am going wrong somewhere, this is how I saw it as an error:

What the stationary observer sees is that the moving observers did not measure the actual length of the rod. Suppose there is a backdrop on the other side of the rod from the stationary observers (think of the rod going down a tunnel). The stationary observers see the photon emitted at the front of the rod against the backdrop, watch the rod move forward and the photon move "backward" until it reaches a coordinate where the end of the rod is going to be when it meets the photon. From the stationary observers point of view they definitely did not measure the rod's length, did they ? They measured a distance between two points that they observed and, if they had quick enough eyesight would see the front of the rod has already passed the reference point for the start of the measurement.

Now a stationary observer who knows the value of c and v can use simple math to determine the error in the measurement made my the moving observers.

The moving observers, because they don't recognize the stationary frame of reference, simply assume that the photon traveled the length of the rod. It didn't because by the time the photon reached the back end the back end had moved (they just didn't know it because they were completely absorbed by their own frame of reference).

What worries me is how does the photon and the end of the road move relative to each other at faster than the speed of light ? because when we think of either of those points being the "fixed" point and the other the moving one they are now seeing the other travel faster than the speed of light.

 

[Doc Al]

OK, so I am going wrong somewhere, this is how I saw it as an error:

What the stationary observer sees is that the moving observers did not measure the actual length of the rod. Suppose there is a backdrop on the other side of the rod from the stationary observers (think of the rod going down a tunnel). The stationary observers see the photon emitted at the front of the rod against the backdrop, watch the rod move forward and the photon move "backward" until it reaches a coordinate where the end of the rod is going to be when it meets the photon. From the stationary observers point of view they definitely did not measure the rod's length, did they ? They measured a distance between two points that they observed and, if they had quick enough eyesight would see the front of the rod has already passed the reference point for the start of the measurement.

From the 'moving' frame's viewpoint, the motion of the backdrop means nothing. They have a perfectly good and stationary frame in which to do their measurements--their own. Of course, being smart fellows they are well aware that the 'stationary' frame measures a different travel distance for the photon. But don't forget that according to the moving frame the clocks used to measure the time interval in the stationary frame are not even synchronized. So who made the 'error'?

Now a stationary observer who knows the value of c and v can use simple math to determine the error in the measurement made my the moving observers.

Again, there's no error. A stationary observer who knows the value of c and v and relativistic physics can translate measurements made in one frame into measurements made in another.

The moving observers, because they don't recognize the stationary frame of reference, simply assume that the photon traveled the length of the rod.

From their viewpoint--which is perfectly legitimate--the photon did travel the length of the rod!

It didn't because by the time the photon reached the back end the back end had moved (they just didn't know it because they were completely absorbed by their own frame of reference).

You still are hung up thinking that one reference frame is the 'correct' one. Either one is perfectly OK to use.

What worries me is how does the photon and the end of the road move relative to each other at faster than the speed of light ?

They don't! In relativity, 'relative speed' means the speed one thing has as measured in the frame of the other. From the frame of the rod, the photon moves at speed c. (What's interesting is that the photon moves at speed c with respect to both frames!)

because when we think of either of those points being the "fixed" point and the other the moving one they are now seeing the other travel faster than the speed of light.

No. Again, viewing things from the frame in which the rod is at rest, the photon moves at speed c, like usual. However, it's also true that viewed from the 'stationary' frame the rod is moving towards the incoming photon, thus the distance between them--according to the stationary frame--is closing at greater than the speed of light. So what?

 

[jmallett]

From the 'moving' frame's viewpoint, the motion of the backdrop means nothing. They have a perfectly good and stationary frame in which to do their measurements--their own. Of course, being smart fellows they are well aware that the 'stationary' frame measures a different travel distance for the photon. But don't forget that according to the moving frame the clocks used to measure the time interval in the stationary frame are not even synchronized. So who made the 'error'?


Again, there's no error. A stationary observer who knows the value of c and v and relativistic physics can translate measurements made in one frame into measurements made in another.


From their viewpoint--which is perfectly legitimate--the photon did travel the length of the rod!

You still are hung up thinking that one reference frame is the 'correct' one. Either one is perfectly OK to use.


They don't! In relativity, 'relative speed' means the speed one thing has as measured in the frame of the other. From the frame of the rod, the photon moves at speed c. (What's interesting is that the photon moves at speed c with respect to both frames!)

No. Again, viewing things from the frame in which the rod is at rest, the photon moves at speed c, like usual. However, it's also true that viewed from the 'stationary' frame the rod is moving towards the incoming photon, thus the distance between them--according to the stationary frame--is closing at greater than the speed of light. So what?

So I have no axe to grind, or pre-conceived notions, I am merely searching for the truth, and until I can understand this, on a personal basis, I would be required simply to take it on faith. I am hoping to achieve an understanding which reaches beyond faith, otherwise it's just a case of pick something at random and believe it. That's not my goal. I'd like to truly understand, and here is what I don't understand:
The math generated by Einstein is based on a moving observer who is denied the knowledge of a key fact - how fast am I travelling, what is my v.

OK, let's accept that for the moment. If I live in a purely relative world then I have no v in my own world and the equations postulated by Einstein should not contain v - my world could well be larger or smaller smaller than the world of someone traveling at a different speed, and since I am not aware of them then I don't know or care.

If I acknowledge I actually do possesses v then I must add it into my statement of measurement and knowing what it is becomes relevant to how large or small I am, and then, like Einstein I must use math that includes it.

Well let's try both in turn.
I live in a relative world. My v is not known to me, or relevant to me because of the frame or reference I live in. I am going to measure this rod in front of my and clearly, in my world, it has a length L. I no longer need to develop math which includes v, because that is not relevant/relative to my world and the equations I am going to use are not Einsteins.

Other World. I think I might have a v. The only way I can determine it's value is by finding something that actually does not have v and I am going to call that a stationary observer, for the sake of argument. Now I have a way to determine my v. I can simply ask him to watch me and to perform some math, and his math is going to use my v and determine my size. But in reality he doesn't have v=0. He simply has negative my v, which is a very special case and hardly a basis for generalization of the universe. I need more observers with a whole variety of v's.

In the first world I have no need of v and therefore an Einsteinian equation containing v is meaningless.
In the second world v is a very special case and using it in an equation is also meaningless because it will only work if, and when, I can find someone else with a velocity of -my(v)
So in this case why include v if it has no real meaning in this context either ?
Again I have problems justifying the existence of v in an Einsteinian equation.

The big headache, for me, comes when I eliminate v from the equations and now find that I am back at Newtonian math.

My v'ness would be simply a matter of faith, and that's not a step I am quite ready to take, especially since the whole of our modern physics is based on a requirement to believe in my own v'ness without being able to question it.

 

[A.T.]

The math generated by Einstein is based on a moving observer who is denied the knowledge of a key fact - how fast am I travelling, what is my v.

That is actually Galilean invariance, stated long before Einstein:
http://en.wikipedia.org/wiki/Galilean_invariance

In the first world I have no need of v and therefore an Einsteinian equation containing v is meaningless.

You are using 'v' for different things and confusing yourself. Just because you cannot determine an absolute value of your own velocity, doesn't mean that you cannot determine the relative velocity of some object relative to you. The letter 'v' is still quite useful to describe relative velocity.

Try to address more specific questions which have quantitative answers instead of juggling with terms and pseudo philosophical conclusions.

 

[cfrogue]

So I have no axe to grind, or pre-conceived notions, I am merely searching for the truth, and until I can understand this, on a personal basis, I would be required simply to take it on faith. I am hoping to achieve an understanding which reaches beyond faith, otherwise it's just a case of pick something at random and believe it. That's not my goal. I'd like to truly understand, and here is what I don't understand:
The math generated by Einstein is based on a moving observer who is denied the knowledge of a key fact - how fast am I travelling, what is my v.

This is not correct.

The math is generated based on the logic that how fast a frame is actually traveling is not logically decidable.

In order to apply your logic, you must produce this absolute v and then make your decisions.

You do not have a method to produce this v and thus, it is necessary to fix one frame and calculate the motion of other frames in a relative way.

 

[jmallett]

Try to address more specific questions which have quantitative answers instead of juggling with terms and pseudo philosophical conclusions.

Sorry if that's the impression I gave you, it was more of a fairly lame attempt at humor and irony because I recognize that the possession of v'ness is clearly stupid.

However having an absolute v appears to be an inherent assumption of Einstein's paper, based on the way that I am (currently) interpreting it.

He surely can't make that assumption.

 

[Doc Al]

However having an absolute v appears to be an inherent assumption of Einstein's paper, based on the way that I am (currently) interpreting it.

He surely can't make that assumption.

Of course he doesn't make such an assumption. I have no idea where you are getting that. All velocities are relative velocities.

 

[jtbell]

However having an absolute v appears to be an inherent assumption of Einstein's paper, based on the way that I am (currently) interpreting it.

Why do you think that? Can you point us to a specific place where he appears to use that assumption?

 

[jmallett]

Why do you think that? Can you point us to a specific place where he appears to use that assumption?

If there are 2 sets of observers neither can know they are stationary - they are just traveling relative to each other, right ?

In this case any perceived difference by either simply has to be the inverse of the other, right ?

When Einstein states "Observers moving with the moving rod would thus find that the two clocks were not synchronous, while observers in the stationary system would declare the clocks to be synchronous." he has to make an assumption of which one is stationary, other wise the statement would be:

observers moving in Frame(1) find that the clocks are not synchronous and observers in Frame(2) also find that the clocks are not synchronous.

 

[jmallett]

Of course he doesn't make such an assumption. I have no idea where you are getting that. All velocities are relative velocities.

If they are truly relative then one is simply the inverse of the other.
Einstein states they are not. He states that one observer sees the clocks as synchronous.

 

[Doc Al]

When Einstein states "Observers moving with the moving rod would thus find that the two clocks were not synchronous, while observers in the stationary system would declare the clocks to be synchronous." he has to make an assumption of which one is stationary,

The designations 'stationary' and 'moving' are arbitrary. (Any observer can choose to view himself as stationary.) In this example, the 'stationary' observers (say they are on a space station) are the ones observing the rod moving past them. The 'moving' observers are the ones moving with the rod.

The point is that whether or not a pair of clocks are synchronized depends on who is doing the observing. For a pair of clocks moving with the rod, the rod observers say they are synchronized but the other frame ('stationary') observers disagree.

Nothing here has anything to do with 'absolute' velocity. All that matters is the relative motion of observer and clock.

 

[Doc Al]

If they are truly relative then one is simply the inverse of the other.
Einstein states they are not. He states that one observer sees the clocks as synchronous.

The clocks are stationary in one particular frame (the one moving along with the rod), not in both. Observers in frames that are moving with respect to the clocks will see those clocks as unsynchronized.

But it certainly works both ways. A pair of synchronized clocks in the 'stationary' frame would be seen as unsynchronized by the observers moving with the rod.

 

[jmallett]

I will go back and study it again, then and think about it some more.

 

[jmallett]

The clocks are stationary in one particular frame (the one moving along with the rod), not in both. Observers in frames that are moving with respect to the clocks will see those clocks as unsynchronized.

But it certainly works both ways. A pair of synchronized clocks in the 'stationary' frame would be seen as unsynchronized by the observers moving with the rod.

So Einstein is saying there are observers at A and B, both located on the moving rod, right ? and that they see the clocks as being unsynchronized, right ?

I understand that to mean that they see that it takes light longer to go one way when compared with the other, right ? why don't they deduce, then, that they must be moving ?

 

[Doc Al]

So Einstein is saying there are observers at A and B, both located on the moving rod, right ? and that they see the clocks as being unsynchronized, right ?

No. The clocks attached to the rod are synchronized as far as the observers moving with the rod are concerned. (Other observers, who see the clocks as moving with respect to them, will see the clocks as unsynchronized.)

 

[jmallett]

No. The clocks attached to the rod are synchronized as far as the observers moving with the rod are concerned. (Other observers, who see the clocks as moving with respect to them, will see the clocks as unsynchronized.)

Einstein says:

Observers moving with the moving rod would thus find that the two
clocks were not synchronous, while observers in the stationary system would
declare the clocks to be synchronous.

There are three clocks in his experiment. One at A, one at B, and one "stationary" clock.

In his above statement which two clocks is he referring to ?

If he means the two clocks attached to the rod he is saying they are NOT synchronous for the moving observers. If he means one attached to the rod and one "stationary" clock then he is saying those ARE synchronized.
The problem is in in determining which two of the three clocks he is referring to, isn't it ?

 

[Doc Al]

Einstein says:

Observers moving with the moving rod would thus find that the two
clocks were not synchronous, while observers in the stationary system would
declare the clocks to be synchronous.

There are three clocks in his experiment. One at A, one at B, and one "stationary" clock.

In his above statement which two clocks is he referring to ?

The two clocks, at A and B, are in the 'stationary' frame. Thus they are synchronized in the stationary frame. But observers in the 'moving' frame (with the rod) view those clocks as unsynchronized.

 

[jmallett]

Einstein says:



If he means the two clocks attached to the rod he is saying they are NOT synchronous for the moving observers. If he means one attached to the rod and one "stationary" clock then he is saying those ARE synchronized.
The problem is in in determining which two of the three clocks he is referring to, isn't it ?

No, wait, I may have got that wrong. Let me try again. If he means the two clocks attached to the moving rod he is saying they are not synchronous. You are disagreeing with that, therefore the two clocks in qustion must be one on the moving rod and on "stationary" clock, and these are viewed as being not synchronized.

I think that is what you are saying, right ?

 

[jmallett]

No, wait, I may have got that wrong. Let me try again. If he means the two clocks attached to the moving rod he is saying they are not synchronous. You are disagreeing with that, therefore the two clocks in qustion must be one on the moving rod and on "stationary" clock, and these are viewed as being not synchronized.

I think that is what you are saying, right ?

I think first we must define which clocks are being considered. Let's assume there are three:
A, B (both on the moving rod) and S (stationary).

Which ones are synchronized and which are not ?

 

[Doc Al]

No, wait, I may have got that wrong. Let me try again. If he means the two clocks attached to the moving rod he is saying they are not synchronous. You are disagreeing with that, therefore the two clocks in qustion must be one on the moving rod and on "stationary" clock, and these are viewed as being not synchronized.

I think that is what you are saying, right ?

The two clocks are in the 'stationary' frame and thus are synchronized in that frame. Viewed from the frame of the moving rod, they are not synchronized.

(Rather than puzzle your way through Einstein's 1905 paper, I recommend you get one of the many pedagogical books explaining special relativity that have been written over the years. Here's a free one by Dan Styers: http://www.oberlin.edu/physics/dstyer/Einstein/SRBook.pdf" )

[I just noticed that JesseM recommended the same book back in post #16!]

 

[jmallett]

The two clocks are in the 'stationary' frame and thus are synchronized in that frame. Viewed from the frame of the moving rod, they are not synchronized.

(Rather than puzzle your way through Einstein's 1905 paper, I recommend you get one of the many pedagogical books explaining special relativity that have been written over the years. Here's a free one by Dan Styers: http://www.oberlin.edu/physics/dstyer/Einstein/SRBook.pdf" )

thanks, that's my homework for the weekend then, and I won't trouble you until I have studied it carefully, however I still don't quite understand the experiment.

There are 2 clocks in the stationary frame and one on the moving rod ? or is the moving rod considered the stationary frame, in which case the clocks are stationary to each other but moving relative to the stationary clock ?

Is there some wau n which these can be more specifically defined ? Let's say there3 is a clock A which has an observer a, a clock B, which has an observer b, and a stationary clock S, which has an observer s.

Which clocks are attached to the rod and (I had assumed A & B), which ones are defined as "stationary" ? I had assumed S. Are A & B in fact the "stationary" clocks because they are stationary relative to each other ?