# A question about a thought experiment in space

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.

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## Answers and Replies

russ_watters
Mentor
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)?

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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.

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.

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.

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
Mentor
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.

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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"?

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.

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
Mentor
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.

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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. Lets 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. Lets 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
Mentor
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.

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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.

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.
marlos jacob said:
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?

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?

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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.

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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

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• a thought experiment in space2.doc
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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 dont 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.

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.

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.

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

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 ?

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.

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.

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.
matheinste said:
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.

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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.