A question about a thought experiment in space

[marlos jacob]

Attached to this post is a thought experiment in space. It is only one page (38kb) in microsoft Word. At the end of the page is my question, and I hope some of you may help me about it. Thank you in advance.

 

[russ_watters]

All that thought experiment shows is whether the ship is at rest wrt the two buoys. How would you know if the buoys are at an absolute state of rest (assuming such a state existed)?

 

[marlos jacob]

All that thought experiment shows is whether the ship is at rest wrt the two buoys. How would you know if the buoys are at an absolute state of rest (assuming such a state existed)?

I am not understanding your above reply. Indeed, the experiment (Attached to the thread) does not mention any buoy at all. The experiment only considers one spaceship (ABC), with three apparatus and one observer inside it. Sorry, but I think there has been some misunderstanding. Can you please look again at the experiment? Any way, thanks for your attention.

 

[Mentz114]

An observer in the spaceship will only ever see T_ab = T_cb. The velocity of light will always be same inside the ship irrespective of any motion.

That is the first principle of special relativity and appears to be confirmed by experiment.

Oh, I should also add that there is no such thing as a state of absolue rest.

 

[marlos jacob]

An observer in the spaceship will only ever see T_ab = T_cb. The velocity of light will always be same inside the ship irrespective of any motion.

That is the first principle of special relativity and appears to be confirmed by experiment.

Oh, I should also add that there is no such thing as a state of absolue rest.

The experiment does not deny that the velocity of the light pulse is the same and constant inside the spaceship. Just because of that, if the spaceship is moving to the right, the pulse from C will strike B before the pulse from A does it, and Tc<Ta. In the same way, if the spaceship is moving to the left the observer will find Tc>Ta. This seems to be obvious. See for example, figure 3-1, at page 63, of the book SPACETIME PHYSICS, of Professors Taylor and Wheeler, edited by W. H. FREEMAN AND COMPANY (1997). It is also recognized by Professor Paul A Tipler, in his book "Physics for Scientists and Engineers", Fourth Edition, Volume 3. And finally it is just the old story of Einstein´s long train, whose front and back ends were striken by two lightnings. It becomes obvious that Ta = Tc will only happen if the spaceship is not moving. But not moving with respect to what? The only possible answer is that it is not moving with respect to spacetime. In other words. It must be in absolute rest in space-time. Also the experiment itself seems to be one capable of distinguishing physically if a body is moving at v=constant or if it is at rest.

 

[Mentz114]

No, you're wrong and so are your sources ( unless you're misinterpreting them).

I don't want to argue about this - there is no absolute rest frame.

Imagine an expanding sphere. Every point on the surface has the same relationship to other points so you cannot distinguish betwen them.
How can you define an absolute frame in this case except arbitrarly, which means it is not absolute.

There is no absolute rest frame.

I'm outa here.

 

[russ_watters]

I am not understanding your above reply. Indeed, the experiment (Attached to the thread) does not mention any buoy at all. The experiment only considers one spaceship (ABC), with three apparatus and one observer inside it. Sorry, but I think there has been some misunderstanding. Can you please look again at the experiment? Any way, thanks for your attention.

I'm calling the two objects on the ends "buoys" for simplicity - that is essentially what they are (signaling devices for determining your position/motion).

But not moving with respect to what? The only possible answer is that it is not moving with respect to spacetime.

Why is that the only possible answer? You are measuring the speed with respect to the two objects at the ends. So you are only saying the other object is not moving with respect to them. In order to know it is not moving wrt the other two objects, you would need to know that they are stationary.

You're going to need to figure this out quickly - I don't want to seem short here, but this is a pretty simple concept and we don't entertain crackpottery here.

Edit: BTW, there is no need to do a thought experiment here. We already have a system that does this - it is called GPS. But instead of each GPS receiver interrogating the satellites, the satellites are synchronized by a ground station and constantly send out a time signal. The receiver gets the time signal and calculates the location of the receiver wrt the satellites.

 

[JesseM]

Attached to this post is a thought experiment in space. It is only one page (38kb) in microsoft Word. At the end of the page is my question, and I hope some of you may help me about it. Thank you in advance.

Marlos, you are ignoring the fact that different reference frames disagree about whether the pulse from A was emitted "at the same instant" as the pulse from C, and that this disagreement about simultaneity ensures that every frame agrees both pulses reached B at the same moment and that both pulses were traveling at the same speed, regardless of the speed they see the ship moving (for example, a frame that observe the ship moving to the right will observe that A emitted its pulse before C, while a frame that observes the ship moving to the left will observe that C emitted its pulse before A). Are you familiar with the concept of the "relativity of simultaneity"?

 

[marlos jacob]

You are measuring the speed with respect to the two objects at the ends. So you are only saying the other object is not moving with respect to them. In order to know it is not moving wrt the other two objects, you would need to know that they are stationary.

You're going to need to figure this out quickly - I don't want to seem short here, but this is a pretty simple concept and we don't entertain crackpottery here.

Mr. Russ Watters. 1. The experiment is not measuring any speed among the objects in the spaceship. All of them, A and C (the emitters), and B (the receiver), are fixed inside the spaceship and their relative speed is none. 2. The experiment, due to its purpose, says nothing about other objects. 3. The name of this forum contains two times the world HELP. And all I asked for was help to clear my doubts, specially the help of people like you, who certainly has a strong knowledge on Physics. 3. What I expected then, was that people like you, would, without any intellectual prejudice, make an attentive appreciation of the experiment, and with reasonable arguments to accept or reject it, in adherence to the purposes and rules of this phorum. Instead, I received from you the the stinking smell of dogmatism, and the sick desire of imposing humiliation (English is not my native language, but I discovered in the Dictionary the meaning of the word Crackpottery).I am not a specialist on Relativity. I am only trying to understand it, asking help to clear my doubts.

 

[marlos jacob]

OK, Jesse. You are right about how other observers will see the events inside the spaceship. But remember that the experiment, to fulfil its purpose, need not to consider other observers. Just look at each possibility the experiment generates, and try to judge if the correspondent conclusions of the observer are right or wrong. Then you can decide about its validity. Thanks.

 

[JesseM]

OK, Jesse. You are right about how other observers will see the events inside the spaceship. But remember that the experiment, to fulfil its purpose, need not to consider other observers. Just look at each possibility the experiment generates, and try to judge if the correspondent conclusions of the observer are right or wrong.

Well, if the observer concludes that in his own frame, the two signals were emitted at the same time and traveled at the same speed, then he's right. But if he concludes that his point of view is "right" in some absolute sense, while other frames are "wrong" in an absolute sense, then you need to explain why you think the point of view of other frames is not equally valid.

Also, I don't understand why you think it matters whether an observer is "inside the spaceship" or not, all that matters is the observer's speed relative to the ship. If an observer is driving a car at a constant speed through the ship, then her rest frame will be different than the ship's frame, and in her rest frame she will conclude that the two light signals from A and C were emitted at slightly different times.

 

[russ_watters]

Mr. Russ Watters. 1. The experiment is not measuring any speed among the objects in the spaceship. All of them, A and C (the emitters), and B (the receiver), are fixed inside the spaceship and their relative speed is none.

Ahh, I misunderstood that part. Sorry. (a better drawing would help)

Ok, so this just becomes a Michelson-Morley type experiment hoping to show an anisotropy of the speed of light? If the objects are all at rest wrt each other, then there is nothing happening here, not even a relativity of simultenaity issue. The objects A and C can be triggered either by object B or be synchronized separately and constantly fire time-coded signals at object B. Since the speed of light is constant regardless of the motion (but not acceleration) of the ship, the signals always travel the distance in the same time and the operator of the ship would always think he was stationary.

You may be envisioning a situation such as a boat traveling against a current or even ultrasonic sensors with sound travleing with/against the wind. Such setups would show motion wrt to the fluid medium because things like the speed of sound are not constant across different reference frames.

 

[marlos jacob]

Ahh, I misunderstood that part. Sorry. (a better drawing would help)

Ok, so this just becomes a Michelson-Morley type experiment hoping to show an anisotropy of the speed of light? If the objects are all at rest wrt each other, then there is nothing happening here, not even a relativity of simultenaity issue. The objects A and C can be triggered either by object B or be synchronized separately and constantly fire time-coded signals at object B. Since the speed of light is constant regardless of the motion (but not acceleration) of the ship, the signals always travel the distance in the same time and the operator of the ship would always think he was stationary.

Thanks for your attention, and I would say that now you got the real conception of the experiment. Let's put it in a little diferent way. The sapaceship is ABC, as before. B is in the center and the emitters A and C are at the opposite ends of the spaceship. The movement of ABC, if any, can only be along the X axis. A and C emit a pulse to B, simultaneously. If the spaceship is eventually moving to the right, the pulse from C will travel a certain distance Dc<(CB=AB), to reach B (because B is aproaching the pulse, due to his movement to the right, with the spaceship, just in the direction of point C, from where the pulse departed). On the other hand, the pulse coming from A, will have to travel a bigger distance Da>(AB=CB) to reach B (because B is moving away from the pulse, due to his movement to the right, with the spaceship). Being Da>Db, and because c is constant, Ta=Da/c > Tc=Dc/c.

Now, if the ship is eventually moving to the left, receiver B will report Ta<Tc. This seems clear to me, and is exactly what happens in the story of the Einstein`s long train, where an observer, in the middle of the train, must see the light coming from the lightning that striked the front end of the train, before he sees the light coming from the back end of the train,etc, etc. In consequence of the above, my reasoning is that, if the receiver B eventually reports Ta=Tc, the spaceship cannot be moving to the right, neither to the left, and consequently it is not moving, and it should be, I suppose, in a kind of Absolute Reference Frame. And that any spaceship where Ta=Tc, would be in the same situation.

Additionally, as you know, it is said and accepted that “there is no physical experiment through which one can say if a body is moving uniformly or if it is at rest”. Accordingly to the above reasonings, this experiment can do it (indeed, if the observer finds Ta<>Tc, the body (spaceship) is moving. If Ta=Tc, the body is not moving).

Finally I should tell that I cannot really see where the above reasoning hurts the Theory of Relativity, just because they contain nothing against the two basic principle of that so solid theory: 1. The velocity of light is the same for all observers in inertial frames; 2. The laws of physics remain the same for all those observers.

Thanks again for being tolerant with those reasonings which may very well prove to be only silliness, fruit of my lack of better knowledge. That is my big doubt.

 

[JesseM]

Thanks for your attention, and I would say that now you got the real conception of the experiment. Let's put it in a little diferent way. The sapaceship is ABC, as before. B is in the center and the emitters A and C are at the opposite ends of the spaceship. The movement of ABC, if any, can only be along the X axis. A and C emit a pulse to B, simultaneously. If the spaceship is eventually moving to the right

But there is no absolute truth about whether the spaceship is moving to the right in relativity, it just depends on what frame you choose. If the light signals from A and C are emitted simultaneously in the ship's rest frame, then obviously in this frame the ship is not moving at all, so the light signals reach B at the same time. And if this is viewed by some other frame where the ship is moving to the right, then in this frame the signals from A and C are not emitted simultaneously, instead the signal from A is emitted prior to the signal from C, in such a way that both signals reach B at the same time. So this example doesn't allow you to judge that one frame's view is "more correct" than any other.

You could also imagine what would happen if the signals from A and C were emitted simultaneously in a frame where the ship was moving to the right. In this case, the signal from C would reach B before the signal from A. But this doesn't indicate that the ship is moving in an absolute sense...in the rest frame of the ship, it's easy to explain the same observations, since in this frame the signal from C was emitted before the signal from A, so of course the signal from C reached B first. Again, this example doesn't lead you to prefer one frame's perspective over another in any absolute sense.

 

[russ_watters]

Finally I should tell that I cannot really see where the above reasoning hurts the Theory of Relativity, just because they contain nothing against the two basic principle of that so solid theory: 1. The velocity of light is the same for all observers in inertial frames; 2. The laws of physics remain the same for all those observers.

What you have wrong is thinking that the distances are different when the spaceship is moving. If the apparatus is physically fixed, it is easy to see that the distances are fixed. You can measure it with a tape measure! Since you are saying that the distances depend on the speed of the ship, that's a violation of #2.

Since the pieces of the apparatus are fixed with respect to each other, if the time for the signal to transit varies, the speed at which they transit (the speed of light) must very as well. So that would violate #1.

Btw, those are reversed... #2 is the first postulate and predates Einstein's version of relativity. How you said it wasn't even true for Galileo's version. It would probably help for this problem to get them in order since you need to start with what you know: that due to the principle of relativity, the length of your apparatus is fixed in your reference frame. This was the starting point for the MM experiment and the reason they postulated that the speed of light was frame dependent.

It may also help you to apply some arbitrary math here. How would you actually do the calculations? What does the speed/distance/time equation look like? Since the two pieces of information you measure are the distance between the parts of the apparatus and the timing of the signals, it is s=d/t...making s (which is the speed of light) vary.

 

[marlos jacob]

But there is no absolute truth about whether the spaceship is moving to the right in relativity, it just depends on what frame you choose. If the light signals from A and C are emitted simultaneously in the ship's rest frame, then obviously in this frame the ship is not moving at all, so the light signals reach B at the same time. And if this is viewed by some other frame where the ship is moving to the right, then in this frame the signals from A and C are not emitted simultaneously, instead the signal from A is emitted prior to the signal from C, in such a way that both signals reach B at the same time. So this example doesn't allow you to judge that one frame's view is "more correct" than any other.

Jesse. I think that we are discussing exactly the core of the experiment, and what you say is in perfect adherence to the Theory. I accept that the spaceship is at rest in its reference frame. What I am doing, which seems to be somewhat “not accepted”, is to forget that convention of reference frames, and imagining the spaceship on space, alone, full closed, with no windows, and imagining a way for the observer in it to discover if the spaceship, with himself and his apparatus, is moving uniformly or not. For that, the observer only can and needs to cope with the spaceship itself, and its apparatus A, B and C. He even don´t need to know about the concept of reference frames. Also, I am treating the problem only using classical kinematics, with the velocity of light constant. Looking at the problem that way, the experiment seems to be valid, because if the spaceship is moving to the left or to right, B wil register Ta<>Tc. And this happens because the velocity of the pulse inside de spaceship is independent of the eventual real movement of the spaceship itself in space.

To make it clearer maybe I will need to make a diagram and demonstrative yet simple kinematic calculus, but I am afraid the diagram, being a little complicated, will be deformed because I have to do it in Microsoft Word, accordingly to the rules of the forum.

 

[marlos jacob]

What you have wrong is thinking that the distances are different when the spaceship is moving. If the apparatus is physically fixed, it is easy to see that the distances are fixed. You can measure it with a tape measure! Since you are saying that the distances depend on the speed of the ship, that's a violation of #2.

Since the pieces of the apparatus are fixed with respect to each other, if the time for the signal to transit varies, the speed at which they transit (the speed of light) must very as well. So that would violate #1.

Mr Russ. I think that we are discussing exactly the core of the experiment, and what you say is in perfect adherence to the Theory. What I am doing, which seems to be somewhat “not accepted”, is to forget that convention of reference frames, and imagining the spaceship on space, alone, full closed, with no windows, and imagining a way for the observer in it to discover if the spaceship, with himself and his apparatus, is moving uniformly or not. For that, the observer only can, and only needs to cope with the spaceship itself, and its apparatus A, B and C. He even don´t need to know about the concept of reference frames. Also, I am treating the problem only using classical kinematics, with the velocity of light constant. Looking at the problem that way, the experiment seems to be valid, because if the spaceship is moving to the left or to right, B wil register Ta<>Tc. And this happens because the velocity of the pulse inside de spaceship is independent of the eventual real movement of the spaceship itself in space.

To make it clearer maybe I will need to make a diagram and demonstrative yet simple kinematic calculus, but I am afraid the diagram, being a little complicated, will be deformed because I have to do it in Microsoft Word, accordingly to the rules of the forum.

 

[JesseM]

Jesse. I think that we are discussing exactly the core of the experiment, and what you say is in perfect adherence to the Theory. I accept that the spaceship is at rest in its reference frame. What I am doing, which seems to be somewhat “not accepted”, is to forget that convention of reference frames, and imagining the spaceship on space, alone, full closed, with no windows, and imagining a way for the observer in it to discover if the spaceship, with himself and his apparatus, is moving uniformly or not.

But how? Do you understand that if the observer sets the clocks at A and C so that they are synchronized in his rest frame, then if the two signals are emitted at the same time according to those clocks, the signals will both reach B at the same moment? Do you understand that all frames would agree about this prediction, regardless of whether the ship is moving in that frame or not? Remember, if the ship sets the clocks so they are synchronized in the ship's rest frame, then another frame will see the clocks as out-of-sync, and thus see the signals emitted at different times, in just the right way so that both signals will be predicted to arrive at B at the same moment given the assumption that both signals travel at c in this other frame.

Also, I am treating the problem only using classical kinematics, with the velocity of light constant. Looking at the problem that way, the experiment seems to be valid, because if the spaceship is moving to the left or to right, B wil register Ta<>Tc.

Yes, in a frame where the ship is moving, the two signals will take a different amount of time between being emitted and reaching B, due to the motion of the ship; this is just as true in relativity as it is in classical kinematics. But this will be exactly compensated for by the fact that the different signals were emitted at different moments in this frame, so both signals are predicted to reach B at the same moment (in all frames). Again, this is assuming that the observer on the ship had clocks at the front and back which were synchronized in his rest frame, and that A and C emitted their signals when the clock at each end showed the same reading.

If you don't want the observer to pick the moment the signals are sent using clocks which have been synchronized according to the Einstein synchronization procedure (where two clocks are judged to be 'synchronized' in their mutual rest frame if they both show the same reading at the moment they are hit by light from a source turned on at the midpoint of the two clocks), then how do you want the observer on the ship to pick the moment the two signals are sent, exactly?

 

[marlos jacob]

What you have wrong is thinking that the distances are different when the spaceship is moving. If the apparatus is physically fixed, it is easy to see that the distances are fixed. You can measure it with a tape measure! Since you are saying that the distances depend on the speed of the ship, that's a violation of #2.

Since the pieces of the apparatus are fixed with respect to each other, if the time for the signal to transit varies, the speed at which they transit (the speed of light) must very as well. So that would violate #1.

Mr Russ. I think that we are discussing exactly the core of the experiment, and what you say is in perfect adherence to the Theory. What I am doing, which seems to be somewhat “not accepted”, is to forget that convention of reference frames, and imagining the spaceship on space, alone, full closed, with no windows, and imagining a way for the observer in it to discover if the spaceship, with himself and his apparatus, is moving uniformly or not. For that, the observer only can, and only needs to cope with the spaceship itself, and its apparatus A, B and C. He even don´t need to know about the concept of reference frames. Also, I am treating the problem only using classical kinematics, with the velocity of light constant. Looking at the problem that way, the experiment seems to be valid, because if the spaceship is moving to the left or to right, B wil register Ta<>Tc. And this happens because the velocity of the pulse inside de spaceship is independent of the eventual real movement of the spaceship itself in space.

To make it clearer maybe I will need to make a diagram and demonstrative yet simple kinematic calculus, but I am afraid the diagram, being a little complicated, will be deformed because I have to do it in Microsoft Word, accordingly to the rules of the forum.

 

[JesseM]

What I am doing, which seems to be somewhat “not accepted”, is to forget that convention of reference frames, and imagining the spaceship on space, alone, full closed, with no windows, and imagining a way for the observer in it to discover if the spaceship, with himself and his apparatus, is moving uniformly or not.

You say that you want to "forget that convention of reference frames", but your experiment depends on the idea that the signals from A and C are emitted "at the same time"...so unless you specify which frame's definition of simultaneity you're using, or else specify a physical procedure for A and C to decide when to emit their signals (like my suggestion earlier of having clocks at each end of the ship which have been synchronizing using the Einstein synchronization convention, which involves setting each clock based on the moment they're hit by light from a source at their midpoint), then your experiment is simply ill-defined.

So again, please answer the question from my last post: how do you want the observer on the ship to pick when each of the two signal is sent?

 

[matheinste]

Light pulses reaching the mid point, B, of the spaceship at different times means that the light pulses cannot have been emitted at the same time in the frame of the ship.
If light pulses are emitted simultaneously at both ends of the spaceship the light will meet at the mid point between the emitters NO MATTER WHAT THE INERTIAL MOTION OF THE SHIP IS.

 

[marlos jacob]

But there is no absolute truth about whether the spaceship is moving to the right in relativity, it just depends on what frame you choose. If the light signals from A and C are emitted simultaneously in the ship's rest frame, then obviously in this frame the ship is not moving at all, so the light signals reach B at the same time. And if this is viewed by some other frame where the ship is moving to the right, then in this frame the signals from A and C are not emitted simultaneously, instead the signal from A is emitted prior to the signal from C, in such a way that both signals reach B at the same time. So this example doesn't allow you to judge that one frame's view is "more correct" than any other.

Dear Mr Jesse . Thank you for your genuine desire to help. As I told you before, I decided to accept the risk of sending you the graphics, where I try to picture to you exactly what I imagine is happening on the spaceship ABC. I had to append it because if I put it here the graphics will be distorted. Please can you take a look on it? If it is distorted can you send me your email so that I can manage to send it to you without distortion? My email is
marlosjacob@hotmail.com

 

[marlos jacob]

If light pulses are emitted simultaneously at both ends of the spaceship the light will meet at the mid point between the emitters NO MATTER WHAT THE INERTIAL MOTION OF THE SHIP IS.

Dear Mr Matheinste,
That is the point that I have dificulty in accepting. Remember the old story of the Einstein long train which was striken by two lightinings at the same time, at its both ends? How one can deny that, considering that the train is moving, the light of the lightenings will arrive at different times in the center of the train?

Anyway, today I posted a thread in response to Mr JesseM about the subject we are discussing. I kindly ask you to see it, because certainly it will make clearer my way of seeing what is really happening on the spaceship, wtihout regard to things like frames of reference. Nothing against them, but we don't need them to approach this subject, as you will see.
Thank you for your attention.

Marlos Jacob

 

[JesseM]

Dear Mr Jesse . Thank you for your genuine desire to help. As I told you before, I decided to accept the risk of sending you the graphics, where I try to picture to you exactly what I imagine is happening on the spaceship ABC. I had to append it because if I put it here the graphics will be distorted. Please can you take a look on it? If it is distorted can you send me your email so that I can manage to send it to you without distortion? My email is
marlosjacob@hotmail.com

Your diagrams are fine, but they shed no light on the question I keep asking you about, namely, what do you mean when you say two different events (specifically the emission of a light pulse from A and the emission of a light pulse from C) happen at the "same time"? Surely you'd agree that if the ship is moving to the left, but the pulse from C was emitted before the pulse from A, then even though the pulse from C would take longer to reach B (since B was moving away from it), the fact that the pulse from C had a "head start" might allow it to reach B at the same moment or even before the pulse from A reached B.

Again, imagine that you have an experimenter sitting at A who pushes a button on an device there to send a pulse towards B, and another experimenter at C who pushes a button on his own device to send a pulse towards B. What procedure do you want to give them to decide when to push their own buttons? For example, you could give each one their own clock, and say they should each push the button when the clock shows a certain time, like 4:00. But then you have to explain how you want to make sure the two clocks are "synchronized", and if you use the Einstein synchronization procedure which I described earlier, this will guarantee that if both experimenters push their buttons at the "same time" according to their clocks, then the two light pulses will reach B simultaneously. But if you don't want to synchronize the clocks this way, or if you don't even want the two experimenters to use clocks to decide when to push their buttons in the first place, the you need to specify a procedure for deciding how each experimenter chooses the moment to send their signal.

 

[matheinste]

Hello Marlos Jacob. I understood your question when I first saw it. Thanks for the diagram but your verbal description was quite clear. You are not alone. I too had this misunderstanding before it became cklear. The error is no less fundamental than a misinterpretation of a consequence of the basic postulate that the speed of light is the same for all observers. A clear explanation is simple but needs careful reasoning and takes many words. I will give a short version but if this does not make the answer clear I will gladly give a fuller version.
Think of an emitter emitting a light pulse. The light moves outwards from the emitter ( with an observer at the same point if you wish ) who is therefore at the centre of an expanding sphere of light. Here is the crucial point. THE EMITTER, IF YOU LIKE WITH AN OBSERVER AT THE SAME POINT, REMAINS AT THE CENTRE OF THIS SPHERE NO MATTER WHAT CONSTANT VELOCITY THE EMITTER HAS. The mistake you make is to imagine that the emitter/observer leaves the point of emission behind in space. It in effect travels with it/him.

So the emission points at the ends of your ship carry with them, and remain central to, the expanding light spheres. And so the centre of these light spheres remain at equal distances from B at the centre of the ship and so if emitted simultaneously, by whatever means, in the ship's frame, the light pulses meet at B NO MATTER WHAT THE INERTIAL VELOCITY OF THE SHIP IS.
Sorry to mention reference frames but it is essential to define simultaneity.

Please ask for a longer explanation of the train scenario if needed. I hope this very non technical reasoning has helped.

 

[Mentz114]

the experiment seems to be valid, because if the spaceship is moving to the left or to right, B will register Ta<>Tc.

Until you can see that this is wrong and Ta=Tc ALWAYS you're in trouble.

 

[JesseM]

THE EMITTER, IF YOU LIKE WITH AN OBSERVER AT THE SAME POINT, REMAINS AT THE CENTRE OF THIS SPHERE NO MATTER WHAT CONSTANT VELOCITY THE EMITTER HAS. The mistake you make is to imagine that the emitter/observer leaves the point of emission behind in space. It in effect travels with it/him.

This would only be true in the rest frame of the emitter. In a frame where the emitter is moving, the emitter will not remain at the center of the expanding light sphere.

 

[JesseM]

Until you can see that this is wrong and Ta=Tc ALWAYS you're in trouble.

If Ta represents the time for the light to go from A to B, and Tc represents the time for the light to go from C to B, then it is not true that Ta=Tc always; in a frame where the ship is moving, they are different. However, as long as the signals from A and C are emitted simultaneously in the ship's rest frame, other frames will see the signals emitted at different times, in such a way that both signals still reach B at the same moment. For example, in a frame where the ship is moving to the left at 0.25c, Tc will be 1.333... seconds while Ta will be only 0.8 seconds; but if the two signals were emitted simultaneously in the ship's rest frame, then in this frame the signal from C will be emitted 0.5333... seconds before the signal from A, so that both signals reach B simultaneously.

 

[Mentz114]

Yes, I should have qualified my remark with 'in the spaceship'.

 

[Mentz114]

This would only be true in the rest frame of the emitter. In a frame where the emitter is moving, the emitter will not remain at the center of the expanding light sphere.

Would the emitter then be catching up on the light going in his direction ?

 

[matheinste]

This would only be true in the rest frame of the emitter. In a frame where the emitter is moving, the emitter will not remain at the center of the expanding light sphere.

Again I will keep it brief. Any emitter is always in its own rest frame ( as is any other object ) and so every emitter is always central to its own expanding light sphere.

In the context of the spaceship question all other frames but those of the spaceship can be ignored providing simultaneity of emission is defined to be in this frame.

 

[Mentz114]

Any emitter is always in its own rest frame ( as is any other object ) and so every emitter is always central to its own expanding light sphere.

Thank you, Math, if only this had been said at the beginning. This picture encapsulates the fact that the speed of light is not relative.

 

[yogi]

The situation is actually the analog of a method Einstein suggested in 1916 of synchronizing to clocks from a source midway between them.

For the person who started this thread, take solice in the fact that SR is counter intuitive - it is difficult to get it into ones thinking that if a source (or a receiver) midway between two clocks will generate simultaneous signals when it is at rest - and also when its velocity has been changed to some value different - but that is what Einstein imposes upon the real world. But every answer given in this thread and all the thousands of other threads that have discussed the problem are based upon a key part of the SR that in fact has never been measured - namely that the one way velocity of light in free space is isotropic. So if you accept SR you are forced to conclude that the one way velocity between the endpoints and the center is independent of the velocity of the capsule. If you do not accept that - that is not a crime and not something that deserves condemnation or ridicule - If that one aspect of the special theory is every proved to be untrue, it would have no effect upon all of the successes of the theory -

 

[JesseM]

Again I will keep it brief. Any emitter is always in its own rest frame ( as is any other object ) and so every emitter is always central to its own expanding light sphere.

I would say an object isn't really "in" one frame or another, a frame is just a coordinate system used to keep track of any objects you choose. But I think we're agreed that your statement about the emitter remaining at the center of the expanding light sphere is true in the emitter's rest frame.

In the context of the spaceship question all other frames but those of the spaceship can be ignored providing simultaneity of emission is defined to be in this frame.

If you assume that the emissions are simultaneous in the ship's rest frame, I agree, you can just look at the ship's rest frame to see the light beams will arrive at B at the same moment (although you are also free to translate things into a different frame if you choose). But the point here is that marlos jacob seems unclear on the idea that "the emissions happen simultaneously" can only be true in a single reference frame, and in fact he has not specifically stated which frame this is supposed to be true of. He seems to think that there is some objective frame-independent sense in which they can be said to have happened simultaneously, so that if the ship is moving and they happen simultaneously, they will reach B at different times and this will mean the ship is "objectively" in motion. I'm trying to make clear to him that there is no frame-independent way of defining simultaneity, and that he has to specify what procedure is being used to decide the timing of the two signal-emissions, which will determine whether the signals arrive at B at the same moment or not.

 

[matheinste]

In reply to JesseM post 34. Greetings.

An object can only be in its own rest frame. This frame may of course be moving relative to other frames.

With regard to simultaneity. If two objects have zero relative velocity, such as the emitters in the spaceship, and if they emit a light pulse, and if these light pulses meet halfway between the emitters, we will regard the emissions as simultaneous. NOTE We have said nothing about the ships motion in this definition or mentioned any other moving frames or co-ordinates of any sort.
We have however accepted without proof that the speed of light is the same at all points in space.

We assume that we have some mechanism that can be adjusted to fire the emitters so that these requirements are met. We will also assume that this mechanism is not affected by the motion of the ship.

We both agree that emitters in their own rest frame remain central to their emitted spheres of light ( we differ about this being true as seen from other frames but in this case it does not matter ) The emitters are both in the same frame ( of course their commom rest frame ) AND REMAIN SO WHATEVER THE MOTION OF THE SHIP and so the light fronts must meet halfway between them providing the firing mechanism we use is not affected by such motion. We must agree that under ALL inertial motion of the ship that as long as the emitters have zero relative velocity the controlled emitted light fronts will meet at the same point which we have defined, halfway.

This I hope answers the ORIGINAL question. You cannot detect whether the ship is moving or at rest relative to anything else by carrying out the experiment described.

This is not a rigorous argument and many assumptions are made and many other non trivial queries could be raised ( and answered ), but the core of the argument is sound and the idea was only to answer the original question.

Thankyou for reading this and I hope we are nearer agreement than we were. Best wishes Matheinste.

 

[JesseM]

An object can only be in its own rest frame. This frame may of course be moving relative to other frames.

What does it mean for an object to be "in" one frame but not another? What is the physical meaning of this statement? Again, a frame is just a coordinate system. You could say "an object is defined to be 'in' a frame if it is at rest in that frame" and then your statement would be true by definition, but I don't think this is standard terminology in relativity.

With regard to simultaneity. If two objects have zero relative velocity, such as the emitters in the spaceship, and if they emit a light pulse, and if these light pulses meet halfway between the emitters, we will regard the emissions as simultaneous.

Yes, this is a version of the Einstein synchronization convention. I gave another version in post #18 when I said "two clocks are judged to be 'synchronized' in their mutual rest frame if they both show the same reading at the moment they are hit by light from a source turned on at the midpoint of the two clocks". The problem is not that I don't understand how simultaneity works in relativity, the problem is that marlos jacob doesn't seem to understand it, and doesn't seem to understand that you need to specify a particular synchronization convention before you can say that the signals from A and C were emitted at the "same time".

We both agree that emitters in their own rest frame remain central to their emitted spheres of light ( we differ about this being true as seen from other frames but in this case it does not matter )

When you say "we differ", do you mean you think that in a frame where the emitter is moving, it will still be observed to remain at the center of the expanding light sphere? Suppose the emitter is moving along the x-axis at 0.5c, and as it passes the origin at t=0 it emits a flash of light in all directions. at t=1 seconds it will be at position x=0.5 light-seconds in this frame...are you saying that you think that in this frame, the right edge of the light sphere would be at x=1.5 light-seconds, and the left edge would be at -0.5 light-seconds, so that both edges are 1 light-second from the emitter at this moment? This would violate the second postulate of SR which says that the speed of light must be c in all frames...for this frame to see the light moving at c, if the flash was emitted at x=0, then after 1 second the right edge must be at x=1 light-second and the left edge must be at x=-1 light seconds. So in this frame, since the emitter is at x=0.5 light-seconds at this moment, it is 1.5 light-seconds away from the left edge and 0.5 light-seconds away from the right edge of the expanding light sphere.

The emitters are both in the same frame ( of course their commom rest frame ) AND REMAIN SO WHATEVER THE MOTION OF THE SHIP and so the light fronts must meet halfway between them providing the firing mechanism we use is not affected by such motion. We must agree that under ALL inertial motion of the ship that as long as the emitters have zero relative velocity the controlled emitted light fronts will meet at the same point which we have defined, halfway.

This is of course true as long as both emitters sent the light simultaneously in their mutual rest frame (the ship's rest frame). But again, the problem is that marlos jacob did not specify that they sent the light "at the same time" in the ship's rest frame, he didn't specify a frame at all, and he doesn't seem to understand that there is no objective frame-invariant procedure for defining what it means for two emitters at different locations to send out light "at the same time". Without specifying which frame's definition of simultaneity he's using, or without a physical procedure for deciding when each of the two signals are sent, his scenario is simply ill-defined.

 

[pervect]

I haven't read this very long thread in detail, but I do have a comment.

If a object A, regarded as "stationary" emits a light signal, and object B, regarding as "moving" both emit a signal at the same location in space and time, there will not be two different wavefronts. There will only be one wavefront - i.e. if object A emits a radio signal, and object B emits a light signal, a receiver will receive the radio signal at the same time a photocell detects the light flash. (This assumes that the doppler shift is low enough that radio waves aren't converted into light, or vica-versa).

There is experimental evidence for this from astronomy from the measurement of binary stars. Light emitted from a star moving towards us does not arrive earlier than light emitted from a star moving away from us.

Thus Observer A will see the emitted as circular, using his defintion of simultaneity, regardless of the motion of the source of the emission. Observer B, using a different defintion of simultaneity than observer A, will also see this wavefront as being circular.

 

[robphy]

If a object A, regarded as "stationary" emits a light signal, and object B, regarding as "moving" both emit a signal at the same location in space and time, there will not be two different wavefronts. There will only be one wavefront - i.e. if object A emits a radio signal, and object B emits a light signal, a receiver will receive the radio signal at the same time a photocell detects the light flash.

(color-emphasis mine)


I would say there is only one light-cone [with vertex at the emission event]. However, because these two inertial observers are in relative motion, each will slice up that light-cone with a different set of parallel spatial sections... i.e., different sets of events comprising their wavefronts. For each inertial [source] observer, however, the events on his wavefront are spatially-equidistant from him. That is, as you said,

Observer A will see the emitted as circular, using his defintion of simultaneity, regardless of the motion of the source of the emission. Observer B, using a different defintion of simultaneity than observer A, will also see this wavefront as being circular.

 

[pervect]

(color-emphasis mine)


I would say there is only one light-cone [with vertex at the emission event]. However, because these two inertial observers are in relative motion, each will slice up that light-cone with a different set of parallel spatial sections... i.e., different sets of events comprising their wavefronts. For each inertial [source] observer, however, the events on his wavefront are spatially-equidistant from him. That is, as you said,

Right, that is probably a better terminology. The important point is that at any event (location in space and time) there will be a radio signal from A if and only if there is also a light signal from B, even though A and B were moving at different speeds when the signal was emitted.

Different observers parameterize this unique light cone differently, because they have different ideas of what events are simultaneous.

 

[matheinste]

Hi Jesses M. I was not questioning your understanding of the subject just trying to comment generally on the questions arising from the original question.

Secondly regarding any difference in meaning on the subject of emitters remaining central to expanding spheres of light, if when you say " a frame in which an emitter is moving " - you mean a frame which an emitter is moving relative to, then we agree. An observer in a frame moving relative to the emitter will not see the emitter remaining central to the light sphere.Any misunderstanding was probably caused by my not being perfectly clear as to what I meant and I apologise for this. I know that what you say is correct regarding this and there is perhaps a mis-use of phraseology on my part.

I know that we both agree on the answer to the original post that because of a lack of universal rest frame we cannot tell from the proposed experiment even if perfectly defined whether the ship is in motion relative to anything

Bye.

 

[marlos jacob]

Your diagrams are fine, but they shed no light on the question I keep asking you about, namely, what do you mean when you say two different events (specifically the emission of a light pulse from A and the emission of a light pulse from C) happen at the "same time"?

Dear Mr Jesse.
I just have to suppose that the two pulses, from A and from C, depart at the same T=0, on the spaceship, toward B. And this reasonable and acceptable in this thought experiment. Any acceptable and correct method to make the synchronization of the clocks on A and C is valid for the purpose of the experiment.

 

[JesseM]

Dear Mr Jesse.
I just have to suppose that the two pulses, from A and from C, depart at the same T=0, on the spaceship, toward B.

When you say "on the spaceship", do you mean according to the definition of simultaneity used by the rest frame of the ship? (someone could be 'on the spaceship' in the sense of being inside it but have a different rest frame than the ship, like if they were walking from one end of the ship to another) If so, then using this definition of simultaneity ensures that the two light pulses always reach B at the same moment, regardless of the motion of the ship. Again, if you choose a frame where the ship is moving (as you seemed to do in the diagrams you provided), then if you defined the scenario so that the pulses were emitted simultaneously in the ship's rest frame, that means that in this new frame the pulses were not emitted simultaneously, and were in fact emitted with just the right time offset so that they reach B at the same moment even though it took longer for the pulse to go from A to B than it took for the other pulse to go from C to B (or vice versa). What part of this are you having trouble understanding? Would you like to see a numerical example?

 

[matheinste]

Hello everyone.

Surely if A-B and C-B are both x units apart then if c is the same for both then the time taken for light to travel the distance x between them is the same under all circumstances ( any movement concerned is assumed to be inertial ).

If emissions in the spasceship frame are simultaneous then they are not simultaneous in any other frame relative to which the ship is moving and so will not meet at B as long as B continues to be defined as the mid point of the ship and not some point in space where the mid point of the ship was at emission time. The mid point of the ship and where this point was ( at emission time ) if the ship is in motion are of course not the same point and perhaps this is somehow the source of the misunderstanding on the part of Marlos Jacob.

 

[JesseM]

Surely if A-B and C-B are both x units apart then if c is the same for both then the time taken for light to travel the distance x between them is the same under all circumstances ( any movement concerned is assumed to be inertial ).

In the ship's rest frame, yes. But in any frame where the ship is moving along the A-B-C axis, the point B (the midpoint of the ship, which is moving in any frame where the ship is moving) will be moving away from the point where one pulse was emitted, and towards the point where the other was emitted, so if both pulses move at c in this frame, this frame must measure one pulse to take longer to reach B than the other. But of course, if the pulses were emitted simultaneously in the ship's frame, in this frame they were emitted at different times, with the difference in time being just the right amount to ensure both pulses reach B at the same moment.

If emissions in the spasceship frame are simultaneous then they are not simultaneous in any other frame relative to which the ship is moving and so will not meet at B as long as B continues to be defined as the mid point of the ship and not some point in space where the mid point of the ship was at emission time.

Unless I'm misunderstanding, you seem to have it backwards--If the emissions were simultaneous in the ship's frame, then that guarantees that they do both reach B at the same moment, if B is the midpoint of the ship. Again, in a frame where the ship is moving, one pulse has to travel further to reach B than the other because B is moving away from the point in space where one pulse was emitted and towards the point where the other was emitted, but the pulse that has to travel further will also have been emitted earlier in this frame, by just the right amount so that they both reach B at the same moment. Different frames can't disagree about local events, so if the pulses reach B at the same moment in the ship's rest frame, they must do so in every other frame too.

 

[matheinste]

Hello again JesseM.

Yes all observers must see the light fronts meet at what IS STILL B ( mid way between A and C ) in the ship's frame but not at what WAS B in other frames because the pulses are not simultaneous in the other frames and so it is not required that the fronts meet halfway between where the points A and C ( on the ship ) WERE ( in a frane moving relative to the ship or vice versa ) when the simultanous emissions happened in the ship's frame. Perhaps it woulld be clearer if the points were lablelled A*, B* and C* in another frame to emphasise this

I am explaining it exactly as it is explained in the many textbooks which use the rail train scenario as an example of non absolute simultaneity. I am sure we both mean the same but express things differently. I think the confusion may be that the points of emission and therefore their mid point are different in the two frames. this is fundamental to the whole argument.

Goodbye and good thinking. My brain hurts.

 

[JesseM]

Yes all observers must see the light fronts meet at what IS STILL B ( mid way between A and C ) in the ship's frame but not at what WAS B in other frames because the pulses are not simultaneous in the other frames and so it is not required that the fronts meet halfway between where the points A and C ( on the ship ) WERE ( in a frane moving relative to the ship or vice versa ) when the simultanous emissions happened in the ship's frame. Perhaps it woulld be clearer if the points were lablelled A*, B* and C* in another frame to emphasise this

Ah, I see what you mean. Yes, I agree the pulses don't have to meet at the position coordinate that's midway between the position coordinates where each pulse was emitted in that frame. But I was confused because it seemed like you were specifically saying that you still were using "B" as a label for the midpoint of the ship, when you said the two pulses "will not meet at B as long as B continues to be defined as the mid point of the ship and not some point in space where the mid point of the ship was at emission time." If B is defined as the midpoint of the ship, and the two pulses are emitted simultaneously in the ship's frame, you agree that the pulses will meet at B in all frames, right?

 

[matheinste]

Hello JesseM. I am pleased that we agree.

I hope Marlos Jacob is happy with the answers. If not would he let the forum know. As Metz said it is essential that he grasps this point if he wishes to progress further.

 

[marlos jacob]

Hello JesseM. I am pleased that we agree.

I hope Marlos Jacob is happy with the answers. If not would he let the forum know. As Metz said it is essential that he grasps this point if he wishes to progress further.

Dear Mr JesseM. and Mr Matheinste.
No, I am unfortunately not happy. And here are my reasons:

The explanations below are directed to you and to all our friends who have helped me, working in the question that originated this thread.

The Thought Experiment in this thread is very simple, and so it should be, because I am not a highly educated person in the Theory of Relativity. I only admire it and study it like many others amateurs. I developed that Experiment because I yet cannot accept the postulate that “there is not a physical experiment capable of detecting if a body is moving uniformly or if it is at rest”. The objective is to prove that this Experiment is possible. I yet think that my initial post of this thread, made more clear through the diagrams I appended to my reply #22, are sufficient to demonstrate it.

It continues clear to me that, for understanding and evaluating the Experiment, one needs not to know Relativity, or what a frame of reference is. It does not matter if observers outside the spaceship, and moving or not in relation to it, see the pulses from A and C departing at the same time or not; it does not matter if those observers see the pulses arriving at B at the same time or not. The Experiment does not need other observers to be consistent and valid. Also, it is not necessary for me to specify how the experiment manages to get the pulses departing simultaneously. We all know that there are more than one way to do this. All of us also know that to achieve this simultaneity inside the ship is something possible. So let's take it for granted. Also one need not to have an operator at each emitter (A and C) to perform this task. Also, it is not necessary to use frames of reference. This is only a concept, created by man, and is not part of nature. All those considerations can complicate things and do not help to prove if the Experiment is valid or not.

All that one needs to evaluate this Experiment is:
1. To know that the velocity of light is constant and equal to c, relative to anybody in space, independently of the velocity of this body;
2. To know that a pulse of light is different of a bullet, because the velocity of the pulse of light, after it leaves its source, suffers no influence of the velocity of that source (it is always c). As we know, a bullet performs differently ( its velocity, after it leaves its source, is clearly dependent of the velocity of that source). And one needs to know that this difference between them was the basic reason why I choose the pulse, and not the bullet, to design the Experiment;
5. To know only the basics of Kinematics.

Well, forgetting reference frames, and fixing our attention in the diagrams of my reply #22, I think that nobody, up to now, did refute that, for the observer in the spaceship, if he finds Ta>Tc, he can be confident that the spaceship is moving to right in space, with some positive velocity say +V; also nobody refuted that if he finds Ta<Tc, he can be confident that the spaceship is moving to the left in space with some negative velocity, say -V. In conclusion, if he finds Ta<>Tc, the spaceship is moving. And this becomes obvious because point B will be moving away or towards the points (A and C) from where the light pulses were emitted. Otherwise, or if he finds Ta=Tc, he is obliged, inclusive by logic, to be confident that the spaceship cannot be moving. This is pure kinematics and logics and I see no reason to complicate those simple facts. I ALSO CANNOT SEE WHY THOSE PHYSICAL FACTS CAN BE DENIED, OR MODIFIED IN THEIR INHERENT REALITY, ONLY BECAUSE SOMEBODY DECIDES TO ASSOCIATE A FRAME OF REFERENCE TO THE SPACESHIP. I mean that, if the spaceship is detected by this experiment to be moving, the fact that somebody decides to associate a restframe to it, cannot change the fact that it is moving, just because it was found that Ta<>Tc. If this mental operation (of associating a restframe to the spaceship) changes the observed and measured reality, then I only can suppose that this operation is not correct.

No matter if other bodies are moving with respect to our spaceship. The important thing is that, if the observer finds Ta=Tc, the spaceship ABC is not moving, and consequently its velocity has to be V=0. And any other spaceship DEF, in space, where the performance of the experiment yields Td=Tf, i.e., spaceship with V=0, also is not moving, and an observer in it will see the pulses from A and C to B, departing simultaneously and arriving simultaneously at B.

No matter if other observers, in others spaceships XYZ, where the performance of the experiment yields Tx<>Tz, i.e., spaceships with V<>0, sees the pulses of light departing from A or C at different times (not simultaneously); no matter if they see the pulses arriving at different or equal times at B. This is only a consequence of their having a velocity V<>0, in space, and all those different views are predicted and are in accordance to the Relativity Theory.

And now, it seems to me, we are at the crucial question. Up to now we realized that our observer could find, through the measurements provided by receptor B, that the spaceship could be detected eventually moving with some velocity V<>0, and could be detected eventually not moving, i..e, having a velocity V=0. The crucial question is: THOSE VELOCITIES V, ARE IN RELATION TO WHAT?

I think that the only reasonable answer is to say that V occurs in relation to spacetime. And one only can conclude that, if he finds V=0, the spaceship is in a special state of rest in spacetime. In absolute rest. And, as far as I can see, this does not invalidate the Theory of Relativity.

On the other hand, this peculiar situation, makes me to remember, that something similar happened when the famous Maxwell Equations, provided a certain constant value for the velocity of light: “c”. But those equations did not say nothing about TO WHAT REFERENCE FRAME this velocity was referred to. And Einstein concluded that this velocity was true in relation to anybody in the universe. And he was right. (Please no possible comparison between he and me is involved here: HE WAS CONFIDENT, I AM IN DOUBT; HE WAS A GENIOUS, I ONLY WOULD LOVE TO BE ONE ).

Someone could say that V=0 can never be found in the experiment. But this is not true. To demonstrate it, we can suppose that our spaceship, now supposed to be moving to the right with some velocity V, and far distant from any matter concentration, has a special rocket that, when ignited, has sufficient force to decelerate it. If we keep the rocket functioning by a sufficient long time, it will necessarily be able of reversing the velocity V of the spaceship, to –V. In this process, there will be a certain moment, or a certain lapse of time, during which the velocity of the spaceship is necessarily zero (V=0). In fact, how could any spaceship, moving to right along the X axis, to start moving to the left, without stopping? Of course it has to stop. Aditionally, the observer can keep the spaceship stopped only by turning off the rocket, if he does it exactly when receptor B indicates Ta=Tc. And, if when this is happening, i.e., while V=0, the observer emits the pulses, then, and only then, they will reach B simultaneously.

This is how I am seeing this experiment. It seems to prove that we can produce an absolute reference frame in space, and we can do it, for example, in a point between the Earth and the moon, where gravity is weak, just by managing a spaceship with sufficient powerful rockets to counter-balance the gravity forces, in such a way that receptor B, keeps continuously reading Ta=Tc. Or, in a tridimensional situation, reading TaX=TcX, TaY=TcY, and TaZ=TcZ. I mean, the three components of the velocity V of the spaceship, being zero, along the three axis X, Y and Z.

If you are an observer on that spaceship, you would see the earth, the moon, the planets and the sun, as well as galaxies, to run away from you, in directions that you can determine, and in velocities that you can measure with respect to this frame ( I personally would like to know which would be the velocity of the Earth with respect to that frame. Certainly it will not be 30 km/sec). On the other hand, with the time running, it is quite probable that you would see some other celestial bodies (stars, galaxies, etc), approaching you, also in velocities and directions that you could measure in absolute terms. Of course, to keep the spaceship in the same place, in the spacetime, it has to have sufficient combustible to the rockets, and automatic mechanism to ignite the appropriate rockets every time that the approaching of some celestial body brings new gravity forces to be counter-balanced by the rockets in such a way to keep the spaceship with Vx=Vy=Vz=0, i.e., to keep it in the same place, i.e., to keep it not moving. Do we have sufficient technology and resources to make it by one year ? Or by only 15 days? Or by half an hour?.

If this thought experiment is consistent and valid, it would be worthwhile to try to make such spaceship.

But we must remember that to have the absolute reference frame, it is not sufficient only to counter-balance gravity. It is necessary, also, to manage the rocket so that you measure Ta=Tc.

Thank you for being tolerant with my dreams, and excuse-me for talking so much this time.

I would like to hear from you.

 

[matheinste]

Hello again Marlos Jacob. please don't feel that you are wasting my time. I feel it important that we should all seek the truth. For now I have only one comment bit if requested will give a much more lengthy reasoning as far as possible not referring to frames or other observers outside the ship.

You say--
"" I think that the only reasonable answer is to say that V occurs in relation to spacetime. And one only can conclude that, if he finds V=0, the spaceship is in a special state of rest in spacetime. In absolute rest. And, as far as I can see, this does not invalidate the Theory of Relativity--""

This contradicts the basic postulate of Relativity which denies an absolute or special frame of rest. If there was such a frame then your reasoning WOULD be valid and motion relative to this preferred frame could be detected.

Goodbye for now Matheinste.

 

[JesseM]

It continues clear to me that, for understanding and evaluating the Experiment, one needs not to know Relativity, or what a frame of reference is.

Well, you at least need a procedure for "synchronizing" the clocks at A and C. And if you want to predict what will actually happen when the signals are sent, you need to use relativity to predict it.

It does not matter if observers outside the spaceship, and moving or not in relation to it, see the pulses from A and C departing at the same time or not; it does not matter if those observers see the pulses arriving at B at the same time or not.

But it was you who introduced the idea that the spaceship might be "moving"--your diagrams illustrate it moving the the left. Obviously the ship is not moving in its own rest frame, so these diagrams only make sense as the perspective of an observer or frame that the ship is moving relative to.

In any case, without bringing in the concept of reference frames, I can tell you right now what the result of the experiment will be--if you use the Einstein synchronization procedure to synchronize the clocks at A and C, then the pulses are sent from A and C at the "same time" on the clocks, then it is guaranteed that the two pulses will always meet at B at the same time--there is no way it can be otherwise.

Remember the Einstein synchronization procedure is itself based on light. One way of stating the Einstein synchronization procedure would be to set off a flash at B, and then set the clocks at A and C to read the same time (say, 12:00) at the moment the light from the flash hits them. So isn't it obvious that if you "synchronize" the clocks at A and C in this way, and then A and C send return pulses back to B at the same time according to these clocks, the return pulses will meet B simultaneously? After all, if you take a film of the light traveling from B to A and C, with each clock reading 12:00 when the light hits them, it will look precisely like a backwards version of a film of A and C sending light towards B when they read 12:00, and the light converging on B at the same moment.

If we actually figure out the coordinates of all these events in some frame where the ship is in motion, we can see how this works explicitly. Suppose we are in a frame where the ship is traveling at 0.5c to the left, and in this frame the distance between A and B is 6 light-seconds, as is the distance between B and C. Suppose that at t=0 seconds in this frame, the coordinates of all three points are:

A is at x=50 light-seconds
B is at x=56 light-seconds
C is at x=62 light seconds

Also suppose that at t=0 seconds, B sends out a flash of light, in order to synchronize the clocks at A and C. After 4 seconds, the pulse moving in the direction of C will have traveled 4 light-seconds to the right, so it'll now be at x=56+4=60 light-seconds; meanwhile, since the ship is moving at 0.5c it'll have moved 2 light-seconds to the left, so C will be at x=62-2=60 light seconds. So, at t=4 seconds in this frame, the light pulse from B reaches C, and C sets its clock to read 12:00 (or whatever time you like).

Then at t=12 seconds in this frame, the light pulse moving in the direction of A has moved 12-light seconds to the left, so it'll be at x=56-12=44 light-seconds. Meanwhile, since the ship is moving at 0.5c to the left, A will have moved 6 light-seconds to the left in this time, so it'll be at x=50-6=44 light-seconds. So, at t=12 seconds, the light pulse from B reaches A, and A sets its clock to read 12:00. Remember, although A and C set their clocks to 12:00 at different times in this frame--C at t=4 seconds, and A at t=12 seconds--according to the Einstein synchronization procedure these clocks are defined to be "synchronized" in the rest frame of the ship.

Now, suppose that A and C have both decided that they will send return pulses back to B when their own clocks read 12:00. So at t=4 seconds, C sends a pulse back to the left, in the direction of B. At the moment the pulse is sent, C is at x=60 light-seconds, and B is at x=54 light seconds. Then 12 seconds later, at t=16 seconds, the light pulse has moved 12 light-seconds to the left, so it's now at x=60-12=48 light-seconds, while B has moved 6 light-seconds to the left, so it's now at 54-6=48 light-seconds. So, this is the time when the pulse from C reaches B in this frame, at t=16 seconds.

Meanwhile, at t=12 seconds, A's own clock read 12:00, so it sent a light pulse to the right, in the direction of B. At the moment the pulse was sent, A was at x=44 light-seconds, while B was at x=50 light-seconds. 4 seconds later, at t=16 seconds, the light pulse moved 4 light-seconds to the right of B, and was at x=44+4=48 light-seconds; meanwhile B had moved 2 light-seconds to the left, and was not at x=50-2=48 light-seconds. So, again we find that the light pulse from C reached B at t=16 seconds, just like the light pulse from A. Even though in this frame the signals from A and C were sent at different times, they both reached B at the same time. This was guaranteed by the fact that the clocks at A and C were themselves synchronized using light-signals from B, using the Einstein synchronization procedure.

If you still disagree that the light signals from A and C are guaranteed to reach B at the same moment as long as clocks at A and C are synchronized using the Einstein synchronization procedure, please take the time to make sure you follow this example and see how it works. It might help to draw diagrams of the position of the ship at each of the following 4 times in this frame:

t=0 seconds: synchronization signals sent from B to A and C
A at x=50 ls
B at x=56 ls
C at x=62 ls

t=4 seconds: synchronization signal from B reaches C, C sets clock to 12:00, and sends return signal to B
A at x=48 ls
B at x=54 ls
C at x=60 ls

t=12 seconds: synchronization signal from B reaches A, A sets clock to 12:00, and sends return signal to B
A at x=44 ls
B at x=50 ls
C at x=56 ls

t=16 seconds: return signals from A and C, sent when clocks at A and C both read 12:00, converge at B
A at x=42 ls
B at x=48 ls
C at x=54 ls

Do you disagree with any of the numbers I've given here? Do you disagree that even though the ship is moving in this frame, because the clocks at A and C were "synchronized" according to the Einstein synchronization procedure which involved sending signals from B, this ensures that the return signals from A and C to B both reach B at the same moment? Again, please make sure you follow this example and agree with all the numbers if you still are not convinced that signals from A and C always reach B at the same moment regardless of the motion of the ship.

Also, it is not necessary for me to specify how the experiment manages to get the pulses departing simultaneously.

It is necessary for you to understand that there is no universal definition of what it means for the pulses to depart simultaneously. If the pulses depart simultaneously in the frame where the ship is at rest, the pulses depart non-simultaneously in any frame where the ship is moving--do you agree? And do you agree that as long as the pulses are sent "at the same time" according to clocks at A and C which are synchronized in the rest frame of the ship (meaning they have been synchronized using a pulse from B to A and C, with both set to the same time when the light hits them), then this guarantees that the pulses will both reach B at the same time, no matter how the ship is moving? There is no possible way that the clocks at A and C could have been synchronized using the Einstein synchronization procedure and yet the pulses could fail to reach B at the same time!

Also, it is not necessary to use frames of reference. This is only a concept, created by man, and is not part of nature.

But neither are the words "at the same time" a part of nature. If there are two events happening at different locations, like a pulse sent from A and a pulse sent from C, the only way you can use the words "same time" or "different time" is if you either have a reference frame which assigns time-coordinates to the two events, or if you have a physical procedure for synchronizing clocks which were right next to each event when they happened. Both of these ideas are also "created by man"! Nature doesn't have a single correct answer to whether the events "really" happened at the same time, any more than Nature has a single correct answer to whether two objects in space "really" have the same x-coordinate or a different x-coordinate.

All that one needs to evaluate this Experiment is:
1. To know that the velocity of light is constant and equal to c, relative to anybody in space, independently of the velocity of this body;

And how exactly do you think it's possible to make sense of "velocity" without a coordinate system? To measure an object's one-way speed, I need two clocks which I have "synchronized" according to some procedure and which lie a fixed distance D apart, and then I note the time t1 that it passes the first clock and the time t2 it passes the second, and then I calculate distance/time, i.e. D/(t2-t1). Without a way to define whether two clocks at different locations are "synchronized" or not, the notion of "speed" makes no sense whatsoever! And again, Nature has no single definite answer to whether two clocks are synchronized or not, you can only say whether they are synchronized in one frame or another.

5. To know only the basics of Kinematics.

Well, look over my numerical example above, and see if it fits with your understanding of kinematics.

Well, forgetting reference frames, and fixing our attention in the diagrams of my reply #22, I think that nobody, up to now, did refute that, for the observer in the spaceship, if he finds Ta>Tc

The observer on the ship can never find that Ta is different than Tc, not if his clocks were synchronized using the Einstein synchronization procedure, which is itself based on the assumption that light takes the same amount of time to go from B to A that it takes to go from B to C! That's the whole basis for this form of synchronization--you set off pulses going in both directions from B at a single moment, and then you set clocks at A and C to read the same time at the moment the pulses reach them. If you think that there is any way possible that an observer could synchronize clocks at A and C using this procedure, yet not find that according to these clocks light takes the same amount of time to go from A to B as it takes from C to B, then you really need to think about it more carefully, because the procedure itself guarantees that the measured time (again, according to the clocks 'synchronized' using the procedure) must be identical.

I ALSO CANNOT SEE WHY THOSE PHYSICAL FACTS CAN BE DENIED, OR MODIFIED IN THEIR INHERENT REALITY, ONLY BECAUSE SOMEBODY DECIDES TO ASSOCIATE A FRAME OF REFERENCE TO THE SPACESHIP.

There are no "physical facts" which are modified by using a frame of reference; you're just wrong about what the physical facts are in the first place. Again, if the clocks at A and C are synchronized using the Einstein synchronization procedure, it is absolutely impossible that pulses sent "at the same time" according to these clocks would fail to meet at B at the same time (assuming the ship does not accelerate, of course).