Paradox within the twin paradox

[ghwellsjr]

Ya sorry. I meant pulse, not beam.

OK, so I will change your question to be:

"Do you not believe that it is possible for two pulses of light to become fixed at a distance between one another? If so, please explain why it is not theoretically possible."

The only way I can understand this question is if we think of two light sources a fixed distance apart, emitting two pulses of light in the same direction, and they travel forever at that same fixed distance apart. But they don't "become fixed at a distance between one another" some time after they were emitted, they started off that way and they remain that way forever (unless they hit something which ends their existence and then I wouldn't say they were any distance apart).

And all of this has nothing to do with any frames of reference, it's just physics.

So please help me understand your concepts of light and being fixed and how that relates to reference frames. It's very important if we are going to make progress.

 

[teachmemore]

Your concept of a reference frame is all mixed up. It is not something physical. It is a co-ordinate system that we use to describe and analyze situations. All inertial reference frames extend in all directions throughout all of space and go from eternity past to eternity future. It doesn't make sense to ask about frames that are in contact because all frames cover all space.

You can take the situation involving all of your objects described in one frame and use the Lorentz Transformation to see what it would look like in another frame but there is no sense in which the frames come in contact.

Now if you want to talk about exchanging information between two observers that are traveling with respect to one another, there's no problem with that, as long as you limit the speed of information exchange to c. You don't have to actually build such a device, or be concerned about the practicality of such a device, all you have to do is say that observer A sends a coded light beam to observer B and it travels at the speed of light in the reference frame that you have defined the motions of the two observers. You don't have to get more complicated than that.

ah. thank you, but the coded light pulse would not be able to relay all the information that my device would because the observer A does not know the distance between himself and observer B from observer B's frame of reference. Observer A only knows the distance as he observes it to be from his own reference frame and when the pulse reaches observer B, observer B has no way to find out at what distance observer A was from him in his own reference frame. The device gets around this.

 

[teachmemore]

OK, so I will change your question to be:

"Do you not believe that it is possible for two pulses of light to become fixed at a distance between one another? If so, please explain why it is not theoretically possible."

The only way I can understand this question is if we think of two light sources a fixed distance apart, emitting two pulses of light in the same direction, and they travel forever at that same fixed distance apart. But they don't "become fixed at a distance between one another" some time after they were emitted, they started off that way and they remain that way forever (unless they hit something which ends their existence and then I wouldn't say they were any distance apart).

And all of this has nothing to do with any frames of reference, it's just physics.

So please help me understand your concepts of light and being fixed and how that relates to reference frames. It's very important if we are going to make progress.

Now add to this the condition that the two light pulses do not begin at this fixed distance.

Edit: oh geeze, sorry. scratch that. The condition is that the two light pulses begin at a fixed distance, but then become an unfixed distance, and then return to a fixed distance. how is that? ;-)

Edit: although, I suppose it could be said that they do not begin at a fixed distance from one another, because they are moving away from one another from the instant they start. The distance between them is always changing from the beginning until they become "trapped" in the same frame of reference.

 

[teachmemore]

if i understand you correctly, the purpose of this hypothetical device is to help synchronize the two clocks A and B?
if so, it doesn't really provide a solution, and to explain why i need to first see if we agree about relative simultaneity. what i say is that if from A's point of view two events happen at the same time, then from B's point of view they will happen at different times (and vice-versa).
so if the two events are A's counter starting and B's counter starting, then it is impossible that both A and B will think that those events happened at the same time.
either from B's point of view it will happen at the same time and from A's view B's clock started sooner, or the other way around.
do you agree with that?

Ah, maybe here is where it will clear up the difference in our understanding.

If the two clocks are in contact with one another, even though they are in different frames of reference; when B perceives that they start at the same time, A will also perceive that the two clocks start at the same time. It is the only case when relativity can be simultaneous. If there is any distance between them, then relativity cannot be simultaneous and the mere fact that from one frame the events are simultaneous tells us that from the other reference frame, they are not.

If you can tell me why I'm wrong here, then maybe I will finally understand!

Edit: So to summarize what I have said here. I agree with you for all cases except the case where the two clocks are in contact with one another. If the two clocks are in contact with one another, then the exchange is instantaneous and they are both immediately aware of one another's event.

 

[JesseM]

Ah, maybe here is where it will clear up the difference in our understanding.

If the two clocks are in contact with one another, even though they are in different frames of reference; when B perceives that they start at the same time, A will also perceive that the two clocks start at the same time. It is the only case when relativity can be simultaneous. If there is any distance between them, then relativity cannot be simultaneous and the mere fact that from one frame the events are simultaneous tells us that from the other reference frame, they are not.

Yes, that's exactly right. Disagreements between frames about simultaneity only occur for events that happen at different spatial locations, if two events happen at the same time and the same position in one frame, then all other frames agree those events happened at the same time and position.

 

[cavalier3024]

Ah, maybe here is where it will clear up the difference in our understanding.

If the two clocks are in contact with one another, even though they are in different frames of reference; when B perceives that they start at the same time, A will also perceive that the two clocks start at the same time. It is the only case when relativity can be simultaneous. If there is any distance between them, then relativity cannot be simultaneous and the mere fact that from one frame the events are simultaneous tells us that from the other reference frame, they are not.

If you can tell me why I'm wrong here, then maybe I will finally understand!

Edit: So to summarize what I have said here. I agree with you for all cases except the case where the two clocks are in contact with one another. If the two clocks are in contact with one another, then the exchange is instantaneous and they are both immediately aware of one another's event.

yep. that's true, but still, only one can happen - either the two clocks are at the same location when they start, or they are at the same location some time after they start, when they measure each other. in either case there is a point in time where the two clocks are at different locations, and at that point my argument is valid.

so if what you say is that the two clocks start at the same location then yes, they will start simultaneously. but let's say that after some time they measure each other. each will see that the other is 'younger' (it's counter ticked less times) but that's not a paradox.. its a basic relativity consequence - when a moving object looks towards the direction opposite to its speed, the further away he looks the more backwards in time he will see.
so in this case they will both see each other's clocks tick slower and they will both see each other as younger..

 

[JesseM]

so if what you say is that the two clocks start at the same location then yes, they will start simultaneously. but let's say that after some time they measure each other. each will see that the other is 'younger' (it's counter ticked less times) but that's not a paradox.. its a basic relativity consequence - when a moving object looks towards the direction opposite to its speed, the further away he looks the more backwards in time he will see.

You seem to be saying that the fact that each measures the other to be younger is just a consequence of the fact that light takes some time to travel between them so they are seeing the other as they were in the past, but that's not correct! Time dilation is what remains after you correct for light delays. For example, suppose you and I are moving apart at 0.6c, and we both started from the same location when we were both aged 30. Also suppose I have a ruler at rest relative to me, and I am at the x=0 light years on my ruler. Then when I am 46 years old, if I look through my telescope I see an image of you next to the the x=6 light-year mark on my ruler, at age 38. Since you emitted this light when you were 6 light-years away from my position, I can calculate that you "really" emitted that light 6 years earlier, when I was only 40. Still, that means that even when I account for the light delay, I'm still left with the conclusion that you were age 38 when I was 40, so that's the "real" time dilation in my frame. And this effect is also completely symmetrical as long as we are both moving inertially: if you had your own ruler at rest relative to yourself and you were at the x=0 mark on your ruler, then when you were 46 you'd be seeing the light from the event of me turning 38 and passing the x=6 mark on your ruler, so you'd conclude that in your frame I was turning 38 when you were turning 40.

 

[cavalier3024]

You seem to be saying that the fact that each measures the other to be younger is just a consequence of the fact that light takes some time to travel between them so they are seeing the other as they were in the past, but that's not correct! Time dilation is what remains after you correct for light delays. For example, suppose you and I are moving apart at 0.6c, and we both started from the same location when we were both aged 30. Also suppose I have a ruler at rest relative to me, and I am at the x=0 light years on my ruler. Then when I am 46 years old, if I look through my telescope I see an image of you next to the the x=6 light-year mark on my ruler, at age 38. Since you emitted this light when you were 6 light-years away from my position, I can calculate that you "really" emitted that light 6 years earlier, when I was only 40. Still, that means that even when I account for the light delay, I'm still left with the conclusion that you were age 38 when I was 40, so that's the "real" time dilation in my frame. And this effect is also completely symmetrical as long as we are both moving inertially: if you had your own ruler at rest relative to yourself and you were at the x=0 mark on your ruler, then when you were 46 you'd be seeing the light from the event of me turning 38 and passing the x=6 mark on your ruler, so you'd conclude that in your frame I was turning 38 when you were turning 40.

exactly! what i tried to say is that: the fact that each measures the other to be younger is NOT just a consequence of the fact that light takes some time to travel between them. i don't know why you thought that i thought the opposite.. but anyway thanks for clarifying my case..

 

[JesseM]

exactly! what i tried to say is that: the fact that each measures the other to be younger is NOT just a consequence of the fact that light takes some time to travel between them. i don't know why you thought that i thought the opposite.. but anyway thanks for clarifying my case..

OK, I thought you were saying it was just an optical effect because of the statement "when a moving object looks towards the direction opposite to its speed, the further away he looks the more backwards in time he will see." That's true in terms of what's seen visually (for example, if I look at two stars 10 and 20 light-years away and at rest in my frame, then the image of the one 20 light-years away is what the star looked like 20 years 'backwards in time' in my frame, further back than the image of the one 10 light-years away), but after you correct for light delays I don't think it makes sense to say "the further away he looks the more backward in time he will see", but time dilation still exists when you correct for light delays with an object in motion relative to you. As long as you agree with that, no problem!

 

[cavalier3024]

after you correct for light delays I don't think it makes sense to say "the further away he looks the more backward in time he will see"

why not? let's ignore the delay caused by light. even now, two observers moving relative to each other will 'see' the same event happen at different times (lets assume that both observers are currently at the same position and that the event occurs at a different position). and the difference between this times will be bigger the farther away the event is from the observers. therefore, the moving observer will 'see' farther into the past the farther away he looks, relative to the other observer (if he looks at a direction opposite to his movement. if he will look forward he will see into the future, relative to the other observer)
thats why relativity of simultaneity is possible..

its kinda hard to explain it in words but from lorentz transform u get http://upload.wikimedia.org/math/c/0/f/c0ff5f091774a86621f711d11e7c0068.png" . so what i was saying is that the bigger x is (the farther away you look) the bigger the time difference (the more you will see into the past, relatively)

 

[JesseM]

why not? let's ignore the delay caused by light. even now, two observers moving relative to each other will 'see' the same event happen at different times (lets assume that both observers are currently at the same position and that the event occurs at a different position). and the difference between this times will be bigger the farther away the event is from the observers.

The difference is greater, yes, but it's not the case that this always means the event is further in the past the greater the distance of the observers, it might be further in the future for at least one of the observers.

 

[teachmemore]

People keep trying to explain this with a 'delay', claiming 'relativity of simultaneity', which shows that they are missing it.

It is easy to setup a thought experiment where simultaneity can be ensured. In this case I used a device which is in the frame of reference of counter B, when it comes in contact with counter A.

This device could potentially record a variety of information from A, and at the time within its own reference frame that corresponds to the simultaneous instant that counter A resets itself. This information can then be transmitted to counter B. Counter B can then do the calculation in order to set itself to the time it 'would' have counted to had it simultaneously reset itself with counter A. Counter B could also find out other details about Counter A, using information such as A's relative speed to B (of course, an agreement could have been made between the two counters at some point in history that this whole experiment would be done at a specific relative speed to one another, and this could have been arranged.)

If you are following me so far, then maybe you will be able to help me understand.

I'm going to simplify my thought experiment slightly.

Firstly, counter A and B are relative, so we make no assumptions about who is stationary and who is traveling from a "god's" view.

When counter B receives its signal, it will be able to do a relativistic calculation to find out the precise count that counter A has reached at the instant in its own reference frame that is simultaneous to the instant in B's reference frame for which it received the signal.

Counter B can now be assured that it has set itself to the same count in its own reference frame that A has reached in its own reference frame at that simultaneous instant. Counter A is not aware of this information, but it is still a fact for counter A of course.

When the two counters meet, Counter B reads A's counter.

Now here is the paradox: The counters were matched in there own respective frames of reference at a simultaneous instant in time, so they should still match counts when they meet on another, due to the symmetry between them. But, if this is the case, counter B will get a surprise, because from the time it set itself to match counter A in A's own reference frame, counter A has been moving slow for counter B.

This thought experiment is meant to rule out the relativity of simultaneity and to be independent of any "godly" knowledge of exact velocities of the two counters. It has also been designed to rule out any acceleration.

I should mention, the device was probably not required at all, just an ability for counter B to do special relativity calculations. The exact distance from A to B in A's reference frame and the relative velocity between A and B could be agreed upon ahead of time before A goes off into space to start the experiment. A could easily determine these factors by monitoring acceleration and time as it got itself into position to start the experiment. In this way counter B could be previously endowed with such knowledge.

Edit: This paradox would imply that absolute knowledge of frames of reference does exist, due to an impossibility of symmetry between to frames of reference in the universe, we can just not attain such absolute knowledge through special relativity calculations alone. For instance, if this experiment was carried out in space, we could then compare clocks and determine the actual absolute values for Earth's frame of reference relative to the rest of space.

Edit: Also, it follows that IF absolute frames of reference exist in space, symmetry between two frames of reference would be impossible; meaning, you can't just pick and choose, if someone is going faster than you, he will age slower and you would need to know if he is going faster before you could know whether he is aging slower.

Edit: Of course it also follows that if all reference frames are symmetrical, then there are no absolute frames of reference in the universe and that it is impossible to be in an absolutely stationary frame.

 

[JesseM]

People keep trying to explain this with a 'delay', claiming 'relativity of simultaneity', which shows that they are missing it.

It is easy to setup a thought experiment where simultaneity can be ensured. In this case I used a device which is in the frame of reference of counter B, when it comes in contact with counter A.

This device could potentially record a variety of information from A, and at the time within its own reference frame that corresponds to the simultaneous instant that counter A resets itself. This information can then be transmitted to counter B. Counter B can then do the calculation in order to set itself to the time it 'would' have counted to had it simultaneously reset itself with counter A.

"Simultaneously" in whose frame? Of course you could design things so that B will reset itself to read some time T "simultaneously" with A reading T as simultaneity is defined in B's rest frame, but in that case A and B will not read T simultaneously in A's rest frame. Likewise you could design things so they both read T simultaneously in A's rest frame, but then they wouldn't read T simultaneously in B's rest frame. So simultaneity hasn't been "ensured" in any absolute sense, only relative to some particular choice of reference frame. And since A and B are moving relative to one another, even if they both read T simultaneously in one of these frames, after that they will start to get progressively more out-of-sync because one is ticking slower than the other.

I'm going to simplify my thought experiment slightly.

Firstly, counter A and B are relative, so we make no assumptions about who is stationary and who is traveling from a "god's" view.

When counter B receives its signal, it will be able to do a relativistic calculation to find out the precise count that counter A has reached at the instant in its own reference frame that is simultaneous to the instant in B's reference frame for which it received the signal.

This is confusing, does "in its own reference frame" refer to A's frame or B's? Certainly when B receives the signal, it could figure out what count A has reached simultaneously with B receiving the signal using the definition of simultaneity in B's frame, or it could figure out what count A has reached simultaneously with B receiving the signal using the definition of simultaneity in A's frame. Which one are you talking about here, if either?

Counter B can now be assured that it has set itself to the same count in its own reference frame that A has reached in its own reference frame at that simultaneous instant. Counter A is not aware of this information, but it is still a fact for counter A of course.

I don't get what you mean here. If B set sets itself to time T such that the event of B reading T is simultaneous with the event of A reading T in A's frame, then they do not read T simultaneously in B's frame. Likewise if B sets itself so that they both read T simultaneously in B's frame, they do not read T simultaneously in A's frame. You seem to be suggesting it could be a "fact" in both frames that they are both at the same count T the moment after B receives the first signal and resets itself, but that's impossible.

 

[ghwellsjr]

Teachmemore, when describing two light pulses, you finally settled on:

I suppose it could be said that they do not begin at a fixed distance from one another, because they are moving away from one another from the instant they start. The distance between them is always changing from the beginning until they become "trapped" in the same frame of reference.

Are you thinking that the faster something travels, the slower time goes for that thing and does this have anything to do with your statement that these two light pulses traveling in opposite directions from a common starting point eventually 'become "trapped" in the same frame of reference'?

 

[ghwellsjr]

Firstly, counter A and B are relative, so we make no assumptions about who is stationary and who is traveling from a "god's" view.

Teachmemore, I have been trying to get you to use just one frame of reference and now you want to use no frame of reference. When describing these thought experiments, you get to play "god". In fact you must play "god". You must define counter A's position and velocity starting from time zero and you must define counter B's position and velocity starting from time zero in a single frame of reference. You must define when the counters get reset as defined by the one frame of reference. You cannot talk about the reference frame for A and a different reference frame for B as if each reference frame applies only to one counter. All reference frames apply to all objects all the time.

I explained all this and gave you an example back in post #31. Please study and understand this, and follow its advice before you try to explain another thought experiment. If you don't understand it, please ask for clarification:

You have a misunderstanding of what a reference frame is. It is a co-ordinate system with time added. Think of the graphs you made in school. You drew a horizontal line and a vertical line. At the intersection you wrote the number zero to specify the origin. Then you put positive numbers off to the right and upwards and negative numbers off to the left and downwards. Even though the numbers stopped because of the limited size of your paper, you knew that they really extended all the way to infinity in all directions. Then if you wanted to describe two objects, you specified their locations in terms of the x,y co-ordinates and you could calculate their distance apart or whatever.

Now what if someone told you that they were going to make two graphs on two separate pieces of paper and put one object in one of the graphs and another object in the other graph and then they started moving the pieces of paper around trying to explain how the two objects in the two separate graphs moved in relation to one another. Wouldn't you say they were mixed up?

That's what you are doing in your thought experiment. Instead, you must start with one and only one reference frame. That's like your graph. You can say that one counter is stationary at the origin and the other one is one light year off to the left (for example) and traveling at 0.99995c toward the origin and at time zero both counters are set to zero and they send out a signal once per minute. Then we can analyze what will happen approximately one year later when the traveling counter reaches the stationary counter. They will have different counts, the traveling one will have a much lower count on it.

Just think about something. When this scenario starts, neither counter will have any knowledge of what the other counter is doing. It will take almost one year for the signals coming from the traveling counter to reach the stationary counter and then within less than half an hour, all the signals will arrive in a burst and then the counter will arrive. At a speed of 0.99995c, the traveling counter's time has slowed to 1% of normal. Since there are 525600 minutes in a year, this counter will have only sent out 5256 signals during his entire trip but they will arrive during the last 26.28 minutes (1-0.99995 or 0.00005 times the number of minutes in a year). That's a rate of 200 signals per minute as measured by the stationary clock.

On the other hand, the traveling counter will not see anything from the stationary counter until after about half a year. Then it will start seeing signals coming in two per minute (they are traveling towards him at the speed of light and he is traveling toward them at almost the speed of light, which approximately doubles the rate at which he receives the signals), except since his clock is running at 1% of normal, he will think they are coming in at 100 times that rate which is 200 per minute. And by the time he gets to the stationary clock, he will have received almost a year's worth of signals sent out at one per minute or very nearly 525600.

Now if you want to analyze the same scenario in another reference frame, the one in which the traveling counter is stationary, you have to correctly transform everything in the first frame to the second frame. You just can't say that the two counters will be zero at the start at the same time because that would be a different scenario.

 

[ghwellsjr]

why not? let's ignore the delay caused by light. even now, two observers moving relative to each other will 'see' the same event happen at different times (lets assume that both observers are currently at the same position and that the event occurs at a different position). and the difference between this times will be bigger the farther away the event is from the observers. therefore, the moving observer will 'see' farther into the past the farther away he looks, relative to the other observer (if he looks at a direction opposite to his movement. if he will look forward he will see into the future, relative to the other observer)
thats why relativity of simultaneity is possible..

its kinda hard to explain it in words but from lorentz transform u get http://upload.wikimedia.org/math/c/0/f/c0ff5f091774a86621f711d11e7c0068.png" . so what i was saying is that the bigger x is (the farther away you look) the bigger the time difference (the more you will see into the past, relatively)

What do you mean by "if he looks at a direction opposite to his movement. if he will look forward he will see into the future, relative to the other observer"?

 

[cavalier3024]

What do you mean by "if he looks at a direction opposite to his movement. if he will look forward he will see into the future, relative to the other observer"?

what i meant is that if he will look towards the direction opposite to his movement, the farther away he will look the more he will see into the past. but if he will look forward, towards where he is moving, the farther he will look the more into the future he will see. (all the times of course are relative to the observer that he is moving relatively to). this is why simultaneity isn't the same at two moving reference frames.

if you ask why this is happening then there are many ways to show it in math or in graphs but honestly i have know idea why its like that..

It is easy to setup a thought experiment where simultaneity can be ensured.

what do you mean by simultaneity? let's say that both clocks can somehow magically know everything about each other without the need to send signals.
do you claim that both A and B will see each other starting at the same time?

 

[ghwellsjr]

what i meant is that if he will look towards the direction opposite to his movement, the farther away he will look the more he will see into the past. but if he will look forward, towards where he is moving, the farther he will look the more into the future he will see. (all the times of course are relative to the observer that he is moving relatively to). this is why simultaneity isn't the same at two moving reference frames.

if you ask why this is happening then there are many ways to show it in math or in graphs but honestly i have know idea why its like that..

Everybody always and only "looks into the past" no matter what direction they look in. I have no idea why you think this way.

 

[JesseM]

Everybody always and only "looks into the past" no matter what direction they look in. I have no idea why you think this way.

cavalier3024 explained earlier that this idea was about simultaneity, not optical appearances. If I pass you at relativistic speed, events that in my frame are simultaneous with me passing you might be either in the past or future from your frame's perspective, depending on the direction.

 

[cavalier3024]

Everybody always and only "looks into the past" no matter what direction they look in. I have no idea why you think this way.

not after you correct the light delay..
as i said before - take a look at http://upload.wikimedia.org/math/c/0/f/c0ff5f091774a86621f711d11e7c0068.png" (t is the time in one reference frame, and t' is the time in a frame moving at speed V relative to it).
now let's look at a situation where a certain event happens at t=0 and position x. an observer in the first frame will obviously 'see' (after correcting light delay) this event happen at t=0, no matter what x is. the observer at the moving frame however will 'see' this event at time t' that depends on x according to the given formula. now if V and X have a different sign (the observer is looking opposite to his movement direction) then t' will be positive, meaning that he will 'see' this event happen later then the first observer, so technically you may say he is looking into the past. now if X and V have the same sign (the observer is looking towards his movement direction) then t' will be negative meaning that he 'sees' this event happen before the first observer, technically meaning he is looking to the future.

 

[teachmemore]

"Simultaneously" in whose frame? Of course you could design things so that B will reset itself to read some time T "simultaneously" with A reading T as simultaneity is defined in B's rest frame, but in that case A and B will not read T simultaneously in A's rest frame. Likewise you could design things so they both read T simultaneously in A's rest frame, but then they wouldn't read T simultaneously in B's rest frame. So simultaneity hasn't been "ensured" in any absolute sense, only relative to some particular choice of reference frame. And since A and B are moving relative to one another, even if they both read T simultaneously in one of these frames, after that they will start to get progressively more out-of-sync because one is ticking slower than the other.

This is confusing, does "in its own reference frame" refer to A's frame or B's? Certainly when B receives the signal, it could figure out what count A has reached simultaneously with B receiving the signal using the definition of simultaneity in B's frame, or it could figure out what count A has reached simultaneously with B receiving the signal using the definition of simultaneity in A's frame. Which one are you talking about here, if either?

I don't get what you mean here. If B set sets itself to time T such that the event of B reading T is simultaneous with the event of A reading T in A's frame, then they do not read T simultaneously in B's frame. Likewise if B sets itself so that they both read T simultaneously in B's frame, they do not read T simultaneously in A's frame. You seem to be suggesting it could be a "fact" in both frames that they are both at the same count T the moment after B receives the first signal and resets itself, but that's impossible.

There is a problem with what you have said here.

Firstly, the answer to your question about what 'count' B is setting itself to was already explained here:

When counter B receives its signal, it will be able to do a relativistic calculation to find out the precise count that counter A has reached at the instant in its own reference frame that is simultaneous to the instant in B's reference frame for which it received the signal.

In other words, B is using the definition of simultaneity in A's reference frame to set itself.

Now, here is the problem with what you have said:

And since A and B are moving relative to one another, even if they both read T simultaneously in one of these frames, after that they will start to get progressively more out-of-sync because one is ticking slower than the other.

By making this statement, you are only repeating what I have already said. The fact that they get "progressively more out-of-sync" is the paradox!

Are you following me?

1. B calculates what the count is for A in A's reference frame.
2. B perceives A as going slower than itself.
3. When A reaches B, B sees that A is the same as itself.

Wait! no. What if A is not the same as B when they meet?

If A is not the same as B when they meet, then how could they be symmetrically relative?

That is the paradox.

I don't get what you mean here. If B set sets itself to time T such that the event of B reading T is simultaneous with the event of A reading T in A's frame, then they do not read T simultaneously in B's frame.

Nothing is being "read". B is doing a calculation, which is not the same as "reading". "Reading" implies relativistic transformations brought about by spatial differences. Calculations are not dependent on location. Imagine that A and B are touching in space and that B is reading from A instantly. This would be the equivalent. Remember what you told me?

Yes, that's exactly right. Disagreements between frames about simultaneity only occur for events that happen at different spatial locations, if two events happen at the same time and the same position in one frame, then all other frames agree those events happened at the same time and position.

To understand why doing a relativistic calculation from B's frame is the same as if they were touching and exchanging the same information derived from the calculation, all you have to do is realize that the calculation is being done in B's frame, so the information that B derives about A's frame is not dependent on space or time differences (assuming the calculation is done instantly for B in B's frame. If the calculation is not done instantly, the same argument applies, but just with a slight deviation).

You seem to be suggesting it could be a "fact" in both frames that they are both at the same count T the moment after B receives the first signal and resets itself

You are making an absolute statement. What "moment" are you referring to? The moment in A's frame or the moment in B's frame? I thought you didn't believe in absolutes? ;-) Joking.

 

[teachmemore]

Teachmemore, when describing two light pulses, you finally settled on:

Are you thinking that the faster something travels, the slower time goes for that thing and does this have anything to do with your statement that these two light pulses traveling in opposite directions from a common starting point eventually 'become "trapped" in the same frame of reference'?

I'm not sure what you mean here.

In what you refer to the "what I finally settled on", all I was doing was defining "fixed distance".

The answer to your second question is no.

 

[teachmemore]

Teachmemore, I have been trying to get you to use just one frame of reference and now you want to use no frame of reference. When describing these thought experiments, you get to play "god". In fact you must play "god". You must define counter A's position and velocity starting from time zero and you must define counter B's position and velocity starting from time zero in a single frame of reference. You must define when the counters get reset as defined by the one frame of reference. You cannot talk about the reference frame for A and a different reference frame for B as if each reference frame applies only to one counter. All reference frames apply to all objects all the time.

I explained all this and gave you an example back in post #31. Please study and understand this, and follow its advice before you try to explain another thought experiment. If you don't understand it, please ask for clarification:

You completely mis-interpreted the statement for which you just quoted.

I understand your post #31 quite clearly. If I had any issue with it, I would not hide that from you ;-)

If you want to understand what I mean by "independent of any godly knowledge", please refer to posts #74 or read through the entire discussion on page 5 of this thread.

 

[JesseM]

There is a problem with what you have said here.

Firstly, the answer to your question about what 'count' B is setting itself to was already explained here:

When counter B receives its signal, it will be able to do a relativistic calculation to find out the precise count that counter A has reached at the instant in its own reference frame that is simultaneous to the instant in B's reference frame for which it received the signal.

In other words, B is using the definition of simultaneity in A's reference frame to set itself.

As I said, "in its own reference frame" wasn't clear about whether it was referring to A or B which was why the quoted explanation was unclear. But OK, B is using A's definition of simultaneity to set itself. Let me make sure I understand this correctly with a numerical example. Suppose in B's rest frame, B is at rest at position x=0, while A is moving at 0.6c in the +x direction and crosses A's path at t=0. Also suppose that at t=-100 in this frame, A's own clock reads t_A=-80, and it's running slow by a factor of $$\sqrt{1 - 0.6c^2/c^2}$$ = 0.8 so 100 seconds later at t=0 A's clock will have elapsed a time of 80 seconds, so it'll read t_A=0 when A meets B (though because of the way B sets itself, B will not read t_B=0 at t=0 in its own frame, instead it'll read t=-10 as we'll see below).

Now suppose at t=-100 in this frame when A reads t_A=-80, it sends its first signal towards B, and since it's 60 light-seconds away from B at this moment, the signal will reach B 60 seconds later at t=-40. If the event of the signal reaching B happens at coordinates x=0, t=-40 in B's frame, then we can use the Lorentz transformation to figure out the position and time of this event in A's frame:

x' = gamma*(x - vt)
t' = gamma*(t - vx/c²)

With gamma = $$\frac{1}{\sqrt{1 - v^2/c^2}}$$, in this case 1/0.8 = 1.25.

So, for x=0, t=-40 in B's frame, the corresponding x' and t' coordinates in A's frame would be:

x' = 1.25*(-0.8*-40) = 40
t' = 1.25*(-40) = -50

So, the event of A's first signal reaching B happens at a time coordinate of t'=-50 in A's frame, and A's clock is running in sync with the time coordinate in his frame, so if B wants to set itself to read the same time as A in A's frame, then B would calculate that it must set itself to a time of t_B=-50 seconds at that moment. Is this what you had in mind?

Note that if this is right, then in B's own frame it does not read the same time as A when it resets itself--if B sets itself to read t_B=-50 seconds at t=-40 in its own frame, at this moment in its own frame A actually reads t_A=-32 (A's clock is running slow by a factor of 0.8 in this frame, so if A reads t_A=-32 at t=-40 in this frame, that ensures that 40 seconds later at t=0 in this frame, A will read t_A=-32 + 0.8*40 = 0 as originally assumed). It is only in A's frame that the event of B setting itself to read t_B=-50 is simultaneous with the event of A reading t_A=-50.

Now, here is the problem with what you have said:

And since A and B are moving relative to one another, even if they both read T simultaneously in one of these frames, after that they will start to get progressively more out-of-sync because one is ticking slower than the other.

By making this statement, you are only repeating what I have already said. The fact that they get "progressively more out-of-sync" is the paradox!

When I said "even if they both read T simultaneously in one of these frames", the subsequent comment about them getting progressively more out-of-sync was only meant to apply to the one frame where they were both initially reading the same time. In my example above, they both read a time of -50 seconds simultaneously in A's frame, so they get progressively more out-of-sync in that frame. But in B's frame, when B sets itself to read t_B=-50 this is simultaneous with the event of A reading t_A=-32, so in this frame A starts out ahead of B, which means that if A is running slow in this frame by a factor of 0.8 that actually means they get closer to being synchronized as time goes on! Although B's time will not have quite caught up with A's time when they meet, the gap will have narrowed, with A reading t_A=0 and B reading t_B=-10. And 50 seconds later in B's frame, B will read t_B=-10 + 50 = 40, while A is running at 0.8 the normal rate in this frame so it reads t_B= 0.8*50 = 40, so this is the time when they have momentarily become perfectly synchronized in this frame.

Are you following me?

1. B calculates what the count is for A in A's reference frame.
2. B perceives A as going slower than itself.
3. When A reaches B, B sees that A is the same as itself.

Wait! no. What if A is not the same as B when they meet?

They won't be the same, not if B set itself so that its time was the same as A's time in A's rest frame, that must mean that B will be behind when they meet since in A's frame B was initially set to the same time as A but was running slow thereafter. You can see this is true in my example above, where at t=-40 in B's frame it must set itself to t_B=-50 in order for it to be synchronized with A in A's frame at that moment, then when A and B meet, A will read t_A=0 while B will read t_B=-10.

If A is not the same as B when they meet, then how could they be symmetrically relative?

What do you mean by "symmetrically relative"? Their rates are symmetrically relative in the sense that A is running slow in B's frame while B is running slow in A's frame. In A's frame, at t'=-50 seconds A reads t_A=-50 seconds and at this moment B resets itself to read t_B=-50 seconds, then they meet 50 seconds later, and A has elapsed a full 50 seconds in this time but B is running slow by a factor of 0.8 so it has only elapsed a time of 0.8*50 = 40 seconds, meaning A will read t_A = -50 + 50 = 0 while B will read t_B = -50 + 40 = -10. Meanwhile, in B's frame, at t=-40 seconds A reads t_A=-32 and at this moment B sets itself to read t_B=-50, then 40 seconds later when they meet, B has elapsed a full 40 seconds so it reads t_B=-50+40=-10, while A is running slow by a factor of 0.8 so it's only elapsed a time of 40*0.8=32 seconds, so it reads a time of t_A=-32+32=0. You can see that both frames agree that A reads t_A=0 when they meet and B reads t_B=-10 when they meet, but the two frames disagree on whether A or B has elapsed more time since the moment B received A's first signal and reset itself to read t_B=-50 seconds.

I don't get what you mean here. If B set sets itself to time T such that the event of B reading T is simultaneous with the event of A reading T in A's frame, then they do not read T simultaneously in B's frame.

Nothing is being "read". B is doing a calculation, which is not the same as "reading". "Reading" implies relativistic transformations brought about by spatial differences.

Huh? How does it "imply" that? By "read" I just meant the normal idea of asking what a clock "reads" at a given moment, i.e. what time the clock is displaying on its clock face (or digital readout or whatever) at any given coordinate time. In my example above A "reads" a time of t_A=0 at t=0 in B's frame, and it is running slow by a factor of 0.8, so at t=-10 A "reads" t_A=-8, at t=-20 A "reads" t_A=-16, etc. etc. Unless English is not your native language, I assume you have heard people talk about "reading time" on a clock.

To understand why doing a relativistic calculation from B's frame

What "relativistic calculation", the one to reset the clock? You said before this calculation was done so their times would match in A's frame, not B's frame: "B is using the definition of simultaneity in A's reference frame to set itself." That was the assumption I was using in my calculation above, hopefully you're not changing the assumption here.

is the same as if they were touching and exchanging the same information derived from the calculation,

"The same" in what sense? It obviously is not the same the important sense that if B resets itself so that its reading at that moment matches A's reading in A's frame, then if B does this when they are far apart its reading will not match A's from the perspective of B's own frame, whereas if B resets itself this whey when they are touching then all frames will agree they show the same time at that moment.

You seem to be suggesting it could be a "fact" in both frames that they are both at the same count T the moment after B receives the first signal and resets itself

You are making an absolute statement. What "moment" are you referring to?

Sorry if it was confusing, but I was saying that the corresponding version of that statement couldn't be true in "both frames". In other words, it's impossible that it could be true in frame A that both clocks are "at the same count T the moment after B receives the first signal and resets itself" (with 'moment' defined here in terms of A's definition of simultaneity) and be true in frame B that both clocks are "at the same count T the moment after B receives the first signal and resets itself" (with 'moment' defined here in terms of B's definition of simultaneity). One or the other of those statements could be true, but they can't both be true in the same physical scenario.