If Two Spaceships Close at 1.4c...?

[LightHeavyW8]

Although I have not as yet determined if time passes more slowly or more quickly after retirement, it has allowed me to explore subjects heretofore archived, such as Relativity. If the speed of light in a vacuum, c, is the same for every observer, does it follow that no information may be exchanged among observers at a speed greater than c? Quantum tunneling experiments appear to have succeeded in this regard. But how about space travelers?

 

First, consider a deaf observer and a blind "observer", and two high-powered rifles fired at them in unison from a distance. If the blind "observer" wishes to survive, he would do well to put his hand on the deaf observer's shoulder and duck when the deaf observer does, since the muzzle flashes will be seen before the bullets arrive, but the muzzle blasts will be heard just a little too late. Next, consider the Wikipedia entry for particle "closing speeds" below:

 

(From http://en.wikipedia....#Closing_speeds) -

 

"Closing speeds

An observer may conclude that two objects are moving faster than the speed of light relative to each other, by adding their velocities according to the principle of Galilean relativity.

 

For example, two fast-moving particles approaching each other from opposite sides of a particle accelerator will appear to be moving at slightly less than twice the speed of light, relative to each other, from the point of view of an observer standing at rest relative to the accelerator. This correctly reflects the rate at which the distance between the two particles is decreasing, from the observer's point of view and is called the closing speed. However, it is not the same as the velocity of one of the particles as would be measured by a hypothetical fast-moving observer traveling alongside the other particle. To obtain this, the calculation must be done according to the principle of special relativity. If the two particles are moving at velocities v and −v, then this relative velocity (again in units of the speed of light c) ... will always turn out to be less than the speed of light, regardless of the velocities of the two particles."

 

It seems that the particles collide and annihilate each other when an observer at rest relative to the accelerator thinks they will, and he thinks they are approaching each other at just under 2c - the poor "hypothetical fast-moving observer traveling alongside the other particle" never knew what hit him/them! Or, is he still merrily brewing tea in his own space-time continuum, somehow oblivious to the fact that we observed him/them to be annihilated?

 

Now, let's scale this up a bit. Draw the shortest possible straight line x-y between the Earth and Moon, and place a target at the midpoint. Have two opposing spaceships approach this target on a collision course a-b which is perpendicular to x-y, each one closing on the target at a speed, relative to the target, of .7c. Shine two laser beams from Earth so they intersect a-b at the minimum distance from the target to allow the travelers in each spaceship time to eject at their closing speed of 1.4c as seen by Earth observers, provided they notice when they pass by the beams. If they don't notice, there is a backup system. When Spaceship A crosses the beam, it illuminates its own laser beam aimed at a (broad spectrum) light detector alarm in Spaceship B, and Spaceship B instead fires a proton accelerator at .99c, (relative to Spaceship B) toward the approaching Spaceship A, which just happens to have a proton detector alarm on board. It seems to me that if the travelers fail to notice when they pass by the beams, the poor ("blind") traveler in Spaceship B will never know what hit him because he cannot receive light information from A at a speed greater than c, while the traveler in Spaceship A will be alerted just in time to yell "OH, SH_T!!!". Now, the important questions - Who conveyed information to whom, and when? Did they not collide at 1.4c (according to Earth observers), imparting "information" to both parties? Did not traveler A receive "information" about the impending collision even faster, at 2.39c? Do travelers A and/or B still merrily exist in their own space-time continua even AFTER observers on Earth see them annihilated?

[Spyman]

It seems that the particles collide and annihilate each other when an observer at rest relative to the accelerator thinks they will, and he thinks they are approaching each other at just under 2c - the poor "hypothetical fast-moving observer traveling alongside the other particle" never knew what hit him/them! Or, is he still merrily brewing tea in his own space-time continuum, somehow oblivious to the fact that we observed him/them to be annihilated?

The observer at rest is not in a privileged frame and as such the hypothetical fast-moving observer traveling alongside one of the particles is also equally able to from his observations predict when and where the collision will take place.

[Sisyphus]

 

It seems that the particles collide and annihilate each other when an observer at rest relative to the accelerator thinks they will, and he thinks they are approaching each other at just under 2c - the poor "hypothetical fast-moving observer traveling alongside the other particle" never knew what hit him/them! Or, is he still merrily brewing tea in his own space-time continuum, somehow oblivious to the fact that we observed him/them to be annihilated?

 

No, he will see it coming. From the traveler's perspective, the other particle is approaching at less than C. And, naturally, the light from that particle is approaching at C. Thus, the light gets there first, and he sees it before it hits him.

 

This applies to the scaled up scenario as well. The closing speed of the two spaceships in the Earth's rest frame doesn't matter. In the frame of either spaceship, the other will be approaching at less than C, and so will see the light signal (or the proton-at-just-less-than-C signal) before they collide.

[LightHeavyW8]

No, he will see it coming. From the traveler's perspective, the other particle is approaching at less than C. And, naturally, the light from that particle is approaching at C. Thus, the light gets there first, and he sees it before it hits him.

 

This applies to the scaled up scenario as well. The closing speed of the two spaceships in the Earth's rest frame doesn't matter. In the frame of either spaceship, the other will be approaching at less than C, and so will see the light signal (or the proton-at-just-less-than-C signal) before they collide.

 

But if light from A cannot reach B at a speed > c (Einstein's postulate), how can light from A arrive at B before A itself does? (Unless you add c to the velocity of A, which means that c is NOT constant.) And since B is actually approaching A at 1.4c, from an Earth observer's perspective, and the proton stream B fires at A is fired at .99c relative to B itself, what is to prevent that proton stream from arriving at A before the collision, which in turn happens before B can receive the light from A? This sequence follows from the behavior observed in particle accelerators (and most likely cosmic rays, which is probably what A thinks is coming from B), or so it seems to me. My intent is to try to distinguish the OBSERVATION, or not, of an event from the EVENT ITSELF. All parties WILL agree on the instant my experiment ends, though - won't they? I appreciate and thank you for all for your responses, btw.

Edited January 20, 2011 by LightHeavyW8

[losfomot]

But if light from A cannot reach B at a speed > c (Einstein's postulate), how can light from A arrive at B before A itself does? (Unless you add c to the velocity of A, which means that c is NOT constant.) And since B is actually approaching A at 1.4c, from an Earth observer's perspective, and the proton stream B fires at A is fired at .99c relative to B itself, what is to prevent that proton stream from arriving at A before the collision, which in turn happens before B can receive the light from A?

 

You have made the scenario a bit too complicated to respond to all of it... but maybe this will help...

 

The Earth observer sees two spaceships approach each other, each with a velocity of .7c, so the 'closing speed', as you've stated is 1.4c.

 

If one of the ships, B, fires a 'proton stream' toward the other ship at .99c relative to B... what will the Earth observer see (if the Earth observer could 'see' the proton stream)? The Earth observer would see a proton stream fired toward A from B at a little over .99c...

 

The Earth observer would see the proton stream moving away from B at just under .3c

 

The Earth observer would see the proton stream and A moving toward each other at a 'closing speed' of just under 1.7c

 

 

The Earth observer would not see the proton stream moving toward A at .7c (ship B velocity) + .99c (proton stream velocity) + .7c (ship A velocity) = 2.39c

Edited January 20, 2011 by losfomot

[Sisyphus]

But if light from A cannot reach B at a speed > c (Einstein's postulate), how can light from A arrive at B before A itself does? (Unless you add c to the velocity of A, which means that c is NOT constant.) And since B is actually approaching A at 1.4c, from an Earth observer's perspective, and the proton stream B fires at A is fired at .99c relative to B itself, what is to prevent that proton stream from arriving at A before the collision, which in turn happens before B can receive the light from A? This sequence follows from the behavior observed in particle accelerators (and most likely cosmic rays, which is probably what A thinks is coming from B), or so it seems to me. My intent is to try to distinguish the OBSERVATION, or not, of an event from the EVENT ITSELF. All parties WILL agree on the instant my experiment ends, though - won't they? I appreciate and thank you for all for your responses, btw.

 

I'm not quite sure where the misunderstanding is, so I'll just say a bunch of stuff that might help:

 

Remember that for each observer, light is always traveling at C relative to himself, and that all objects are moving at less than C relative to himself. Every observer is at rest in his own reference frame. There is no such thing as "moving at 0.7C." There is only "moving at 0.7C relative to the Earth," for example. It is impossible for an object to be approaching you at >C in your own rest frame.

 

The two cases of the light signal and the proton signal will be pretty much the same, just with the proton signal moving slightly slower. The proton will be moving at greater than 0.99C but less than C relative to the Earth. Relative to A, it will be moving even faster, but still less than C.

 

As for all parties agreeing, any event that occurs will occur in all reference frames. However, they won't agree on when and where those events occur or even necessarily what order they occur in.

[LightHeavyW8]

You have made the scenario a bit too complicated to respond to all of it... but maybe this will help...

 

The Earth observer sees two spaceships approach each other, each with a velocity of .7c, so the 'closing speed', as you've stated is 1.4c.

 

If one of the ships, B, fires a 'proton stream' toward the other ship at .99c relative to B... what will the Earth observer see (if the Earth observer could 'see' the proton stream)? The Earth observer would see a proton stream fired toward A from B at a little over .99c...

 

The Earth observer would see the proton stream moving away from B at just under .3c

 

The Earth observer would see the proton stream and A moving toward each other at a 'closing speed' of just under 1.7c

 

 

The Earth observer would not see the proton stream moving toward A at .7c (ship B velocity) + .99c (proton stream velocity) + .7c (ship A velocity) = 2.39c

 

But wiki's definition of "closing speed" allows the Earth observer to see A and B close at 1.4c, (or open at 1.4c if they miss each other) so why would the proton stream from B appear to be so much slower for the Earth observer than for B? Why wouldn't Earth see the proton stream precede B by adding the velocities just as we do for A + B? What if we remove the "appear to" component thusly - When the proton stream from B hits the proton detector in A, A flashes a laser pulse to Earth, and when B receives the original light signal from A, B will also send a laser pulse to Earth. I know the Earth observer will have to rely on light now, but I maintain Earth will see the laser pulse from A first, then the collision, then the laser pulse from B. This follows from the known behavior of particle accelerators, and assumes c is a constant for all inertial observers. Of course, if c is NOT constant, all bets are off - including SR and GR, no?

[Sisyphus]

But wiki's definition of "closing speed" allows the Earth observer to see A and B close at 1.4c, (or open at 1.4c if they miss each other) so why would the proton stream from B appear to be so much slower for the Earth observer than for B? Why wouldn't Earth see the proton stream precede B by adding the velocities just as we do for A + B? What if we remove the "appear to" component thusly - When the proton stream from B hits the proton detector in A, A flashes a laser pulse to Earth, and when B receives the original light signal from A, B will also send a laser pulse to Earth. I know the Earth observer will have to rely on light now, but I maintain Earth will see the laser pulse from A first, then the collision, then the laser pulse from B. This follows from the known behavior of particle accelerators, and assumes c is a constant for all inertial observers. Of course, if c is NOT constant, all bets are off - including SR and GR, no?

 

First of all, "appear to" doesn't matter - just assume that the Earth observer can see everything as it is. Second, velocities don't add like that - in the reference frame of the observer on Earth, nothing will ever be moving at greater than C.

 

Check this out:

 

http://en.wikipedia.org/wiki/Velocity-addition_formula

 

I'm not really following why you're having a problem with the proton stream but not with the light signal, since the situations are essentially the same. In the reference frame of A, the light signal will move away from A at C. In the Earth reference frame, the light signal will move away from A at 0.3C. This is possible because time and distance are not the same between reference frames.

[LightHeavyW8]

I'm not quite sure where the misunderstanding is, so I'll just say a bunch of stuff that might help:

 

Remember that for each observer, light is always traveling at C relative to himself, and that all objects are moving at less than C relative to himself. Every observer is at rest in his own reference frame. There is no such thing as "moving at 0.7C." There is only "moving at 0.7C relative to the Earth," for example. It is impossible for an object to be approaching you at >C in your own rest frame.

 

The two cases of the light signal and the proton signal will be pretty much the same, just with the proton signal moving slightly slower. The proton will be moving at greater than 0.99C but less than C relative to the Earth. Relative to A, it will be moving even faster, but still less than C.

 

As for all parties agreeing, any event that occurs will occur in all reference frames. However, they won't agree on when and where those events occur or even necessarily what order they occur in.

 

In my example, A is moving at .7c relative to a stationary target, and B is moving at .7c relative to the same target, coming from the opposite direction. My original question is - "can information be exchanged among observers at > c?" I wanted to scale up an observed example where speeds > c occur, i.e., closing speeds in particle accelerators. I chose a collision so there would at least be agreement, albeit posthumously, about when the experiment ended. AND who would or would not be surprised...

 

First of all, "appear to" doesn't matter - just assume that the Earth observer can see everything as it is. Second, velocities don't add like that - in the reference frame of the observer on Earth, nothing will ever be moving at greater than C.

 

Check this out:

 

http://en.wikipedia....ddition_formula

 

I'm not really following why you're having a problem with the proton stream but not with the light signal, since the situations are essentially the same. In the reference frame of A, the light signal will move away from A at C. In the Earth reference frame, the light signal will move away from A at 0.3C. This is possible because time and distance are not the same between reference frames.

 

I tried to pose the problem STARTING with the closing speeds of particles in accelerators, which can exceed c, and scaling this up to space travellers. I claim that B, who relies on light for his information, will be surprised by his collision but A, who relies on a proton stream, will not.

[Spyman]

Different observers in different frames will view the world differently, they will measure distances and durations to be different relative each other.

 

You need to read & learn about Length Contraction and Time Dilation.

Edited January 21, 2011 by Spyman

[LightHeavyW8]

<br />Different observers in different frames will view the world differently, they will measure distances and durations to be different relative each other.<br /><br />You need to read & learn about <a href='http://en.wikipedia.org/wiki/Length_contraction' class='bbc_url' title='External link' rel='nofollow external'>Length Contraction</a> and <a href='http://en.wikipedia.org/wiki/Time_dilation' class='bbc_url' title='External link' rel='nofollow external'>Time Dilation</a>.<br />

 

While I have read about them, I do not claim to be an expert. I have tried to pose my question and experiment around the known real-world behavior of particles which can be made to close at > c in accelerator/colliders, and that question is "can information be exchanged among observers at > c?". So far, no response has adequately supported or refuted this question, imo. In framing my experiment, I have tried to distinguish between the observation of an event and the event itself. Many discussions concerning relativity seem to regurgitate "what the observer sees" to excess and retreat into realms that are unprovable or unfalsifiable, so I included an indisputable and identical termination of the experiment that all parties must agree upon - BUT WHO WAS SURPRISED? How can light precede the 1.4 c closing speed of A and B if Einstein's Postulate says that c must be the same for all inertial observers?

[swansont]

How can light precede the 1.4 c closing speed of A and B if Einstein's Postulate says that c must be the same for all inertial observers?

 

Because according to the two people involved, the closing speed is not 1.4 c.

[Spyman]

While I have read about them, I do not claim to be an expert. I have tried to pose my question and experiment around the known real-world behavior of particles which can be made to close at > c in accelerator/colliders, and that question is "can information be exchanged among observers at > c?". So far, no response has adequately supported or refuted this question, imo. In framing my experiment, I have tried to distinguish between the observation of an event and the event itself. Many discussions concerning relativity seem to regurgitate "what the observer sees" to excess and retreat into realms that are unprovable or unfalsifiable, so I included an indisputable and identical termination of the experiment that all parties must agree upon - BUT WHO WAS SURPRISED? How can light precede the 1.4 c closing speed of A and B if Einstein's Postulate says that c must be the same for all inertial observers?

I think it has been explained for you several times by several posters already, but I will repeat it once more: an observer in either one of the spaceships will NOT measure the closing speed to be 1.4 c since they are in another frame of reference and both space and time appear different for them relative observers on Earth, in their view the spaceships will be closing at speeds less than c. The observer using photons will have slightly newer information than the one using matter but both will be able to observe when and where the collision will take place before it happens.

[LightHeavyW8]

...an observer in either one of the spaceships will NOT measure the closing speed to be 1.4 c since they are in another frame of reference and both space and time appear different for them relative observers on Earth, in their view the spaceships will be closing at speeds less than c.

 

I do not dispute this - I only claim they will be totally surprised by their collision at 1.4 c as observed from Earth, if they rely upon light for their information.

 

The observer using photons will have slightly newer information than the one using matter but both will be able to observe when and where the collision will take place before it happens.

 

Imho, this statement is inconsistent with the behavior observed by particles which can be made to close at > c in accelerator/colliders. An answer that requires a separate space-time continuum for every particle is somehow less than satisfying for me, stuck as I am in the same one as you. I still appreciate and thank you for your response!

 

<br />Because according to the two people involved, the closing speed is not 1.4 c.<br />

 

This is why I tried to distinguish between OBSERVING an event and the EVENT ITSELF. Like my poor "blind" observer who would duck too late if he waits until he hears the rifle fire, how do you know that the same problem cannot occur for Space Traveller B, if he relies upon light which is limited in speed to c? Particle closing speeds > c are not so limited.

[Sisyphus]

I do not dispute this - I only claim they will be totally surprised by their collision at 1.4 c as observed from Earth, if they rely upon light for their information.

 

Why would they be? A signal at C (or just less than C) is sent from the other ship approaching at less than C. The signals get there first. No surprise.

 

Imho, this statement is inconsistent with the behavior observed by particles which can be made to close at > c in accelerator/colliders. An answer that requires a separate space-time continuum for every particle is somehow less than satisfying for me, stuck as I am in the same one as you. I still appreciate and thank you for your response!

 

What do you mean by a separate space time continuum? If you mean that velocity, time, and distance are different in different reference frames, then it doesn't matter if it's satisfying, because it's demonstrably true.

 

This is why I tried to distinguish between OBSERVING an event and the EVENT ITSELF. Like my poor "blind" observer who would duck too late if he waits until he hears the rifle fire, how do you know that the same problem cannot occur for Space Traveller B, if he relies upon light which is limited in speed to c? Particle closing speeds > c are not so limited.

 

The closing speed is not greater than C in the reference frame of the ship. It is greater in the reference frame of Earth, perhaps, but in that reference frame the closing speed between the ship and the light signal is even greater. 1.7C vs. 1.4C.

 

EDIT: Here, I drew a diagram showing the situation in Earth's reference frame and Ship A's reference frame. Ship B's reference frame will look almost like the mirror image of Ship A's, except that the proton signal will only be going at 0.99C.

 

 

As you can see, nowhere is there a velocity greater than C, and both signals are faster than the ships that sent them. Neither reference frame is the "true" reference frame, because there is no such thing. They're both real. Neither one is just "appearances."

[swansont]

I do not dispute this - I only claim they will be totally surprised by their collision at 1.4 c as observed from Earth, if they rely upon light for their information.

 

 

In that view the closing speed of light is 2c. So, no problem.

[LightHeavyW8]

The closing speed is not greater than C in the reference frame of the ship. It is greater in the reference frame of Earth, perhaps, but in that reference frame the closing speed between the ship and the light signal is even greater. 1.7C vs. 1.4C.

 

You now have light moving at 1.7 c - would Einstein approve?

 

What do you mean by a separate space time continuum? If you mean that velocity, time, and distance are different in different reference frames, then it doesn't matter if it's satisfying, because it's demonstrably true.

 

Demonstrably? Please, do!

 

From Wikipedia: The speed of light, usually denoted by c, is a physical constant important in many areas of physics. Light and all other forms of electromagnetic radiation always travel at this speed in empty space (vacuum), regardless of the motion of the source or the inertial frame of reference of the observer. Therefore light CANNOT get from A to B faster than c, but A and B themselves CAN.

 

Ship B's reference frame will look almost like the mirror image of Ship A's, except that the proton signal will only be going at 0.99C.

 

Here is where your numbers depart from what is demonstrably so, i.e., particles that can be made to close and collide at up ro 1.99c. And if they miss each other, they are opening at 1.99c. If particles can do this, why would a proton shot ahead of B at .99c relative to B not be added to his .7c, i.e., 1.69c? And again, to separate "what the observer sees" from what ultimately and synchronously happens, rely upon A's proton detector, which is demonstrably NOT limited to c.

Edited January 21, 2011 by LightHeavyW8

[losfomot]

Here is where your numbers depart from what is demonstrably so, i.e., particles that can be made to close and collide at up ro 1.99c. And if they miss each other, they are opening at 1.99c. If particles can do this, why would a proton shot ahead of B at .99c relative to B not be added to his .7c, i.e., 1.69c?

 

because you are talking about what the Earth observer sees. A 'closing speed' is deduced or calculated by the Earth observer... it is not measured. However you must measure the two speeds that you use to calculate a closing speed. EO (earth observer) measures 0.7c as the speed of each spaceship... so EO can use those measured speeds to calculate a closing speed between the two (1.4c). EO cannot just assume that .99c should be added on top of that as the speed of the proton stream. If EO wants to use the speed of the proton stream in any of his calculations, EO must measure the speed of the proton stream seperately. If EO does this they will find that the speed of the proton stream is just over.99c

 

If particles can do this, why would a proton shot ahead of B at .99c relative to B not be added to his .7c, i.e., 1.69c?

 

Velocity is relative.

 

What exactly are you asking for in this question? ie relative to which observer?

In B's frame of reference, B is standing still and the proton stream is moving away from B at .99c

 

In Earth's frame of reference, B is moving at .7c, and the proton stream is moving at just over .99c... the EO sees the proton stream moving away from B at an 'opening speed' of just under .3c

 

Like you've stated before, closing speeds cannot be greater than 2c... the reason is that you cannot arbitrarily add speeds together like you are doing with the ships and proton stream. A 'closing speed' is calculated, but the two speeds used in such a calculation must be based on measurement. Since nothing can travel faster than c, even a 'closing speed' cannot be greater than 2c.

[Sisyphus]

You now have light moving at 1.7 c - would Einstein approve?

 

No, I have it moving at exactly C, like always. The other ship, however, is moving at 0.7C in the Earth's frame towards it, for a total "closing speed" of 1.7C. Note that this is the closing speed in the Earth's rest frame, not on either ship. This is different in each frame:

 

On sending ship, the signal is going C ahead relative to the sender, while the other ship is approaching at 0.94C, for a total of 1.94C.

 

On Earth, the signal is going C ahead (and 0.3C relative to the sender), while the other ship is approaching at 0.7C, for a total of 1.7C.

 

On the receiving ship, the signal is approaching at C (0.06C relative to the sender), and of course they aren't moving relative to themselves, so the total is exactly C.

 

Demonstrably? Please, do!

 

From Wikipedia: The speed of light, usually denoted by c, is a physical constant important in many areas of physics. Light and all other forms of electromagnetic radiation always travel at this speed in empty space (vacuum), regardless of the motion of the source or the inertial frame of reference of the observer. Therefore light CANNOT get from A to B faster than c, but A and B themselves CAN.

 

No, neither can. Or rather, light moves at C. Particles move at less than C. You can add up the speed of light and the speed of a particle and get more than C, or you can add up the speed of two particles and get more than C.

 

Here is where your numbers depart from what is demonstrably so, i.e., particles that can be made to close and collide at up ro 1.99c. And if they miss each other, they are opening at 1.99c. If particles can do this, why would a proton shot ahead of B at .99c relative to B not be added to his .7c, i.e., 1.69c? And again, to separate "what the observer sees" from what ultimately and synchronously happens, rely upon A's proton detector, which is demonstrably NOT limited to c.

 

Did you look at the diagram? Which part of it do you not understand? The fact that there are different frames of reference? The diagram is not just a picture of what each observer sees, it is a picture of what actually happens. The closing speed between the two ships is 1.4C in the Earth's frame, though neither ship going faster than C. In the ship's frame, the closing speed is only 0.94C. Velocities depend on frame of reference. That's just how the universe works.

[LightHeavyW8]

At the risk of repeating myself, my original question is - "can information be exchanged among observers at > c?" I wanted to scale up an observed example where speeds > c occur, i.e., closing speeds in particle accelerators. I chose a collision so there would at least be agreement, albeit posthumously, about when the experiment ended... Perhaps another way to put it is, can nothing truly travel faster than c, or are we just unable to tell?

[Sisyphus]

Nothing can truly travel faster than C.

[swansont]

From Wikipedia: The speed of light, usually denoted by c, is a physical constant important in many areas of physics. Light and all other forms of electromagnetic radiation always travel at this speed in empty space (vacuum), regardless of the motion of the source or the inertial frame of reference of the observer. Therefore light CANNOT get from A to B faster than c, but A and B themselves CAN.

I can't find the bolded part on the wikipedia page

http://en.wikipedia.org/wiki/Speed_of_light

[LightHeavyW8]

Nothing can truly travel faster than C.

 

Except light emanating from A, apparently...

[Sisyphus]

Except light emanating from A, apparently...

 

No, it's moving at exactly C.

[swansont]

At the risk of repeating myself, my original question is - "can information be exchanged among observers at > c?" I wanted to scale up an observed example where speeds > c occur, i.e., closing speeds in particle accelerators. I chose a collision so there would at least be agreement, albeit posthumously, about when the experiment ended... Perhaps another way to put it is, can nothing truly travel faster than c, or are we just unable to tell?

 

The answer to your question is no. Your first two sentences are inconsistent with each other; the first mentions "exchanged among observers" and the second is from the example of a third, uninvolved, observer.