Twin paradox without accelerative frame.

[Trojan666ru]

I've asked this question a few times. This time i have tiny change in the question. Instead of people I'm using stopwatch & i have removed the accelerative frame.

If you have any doubts I'll clear it but do not skip the topic.

All i want is an answer that can be agreeable for all observers.

Imagine one observer and two apparatus.

Observer A who is coordinator and two boxes box 1 & box 2 which contain inactive stopwatches and automatic propulsion system which is motion dependent.
Observer A sets the two boxes away from each other equidistant from him. It forms an equilateral triangle. The distance between box 1 & 2 is 4 Lightyears.
The coordinator first send some laser signals towards each box to confirm (to analyse in future) they were equal distance away.
Then he sends another laser pulses to both of the boxes to activate their propulsion system and they starts to move towards each other. Both boxes accelerates and stop automatically when they reach a velocity .5c

Then the coordinator sends another pulse to activate their stopwatches.

So now we have two boxes moving towards each other with same velocity. No acceleration included.
Suppose if there were observers in each box, they thinks they are at rest and the other one is moving towards them at .99c. Therefore they thinks the others stopwatch has run slower.

In this case whose watch will have run longer when they meet each other?

 

[HallsofIvy]

By the symmetry of the situation, their stop watches will be exactly the same. Why do you ask?

 

[Demystifier]

... Suppose if there were observers in each box, they thinks they are at rest and the other one is moving towards them at .99c. Therefore they thinks the others stopwatch has run slower.

In this case whose watch will have run longer when they meet each other?

Good one!

Since this is a completely symmetrical situation, both observers will agree that the two clocks show the same time when they meet. Yet, each of the observers will think that his clock is faster. But how is that possible? The catch is that:
1) The two observers will agree that the two clocks do not show the same time BEFORE they meet.
2) Before the meeting they will not agree whick clock shows larger time and which clock shows smaller time.

 

[Trojan666ru]

By the symmetry of the situation, their stop watches will be exactly the same. Why do you ask?

The symmetry is only from the 3rd observers point of view. Observers in box do not realize they are moving so they thinks others clock has become slower

 

[Trojan666ru]

How do this situation become symmetrical? because both are sharing half light speed?
In each persons frame they think they are at rest because they are in an inertial frame so they can only perceive the others as moving.
In this case the box 2 will have nearly the speed of light from box 1’s point of view or not?

 

[ZapperZ]

Actually, the problem here is the definition of "meet". Are you simply defining it when they pass next to each other without stopping, or are you saying they then stop and be at rest in some reference frame? The latter requires a more explicit description of the deceleration of each one of them.

Zz.

 

[Trojan666ru]

They are not stopping. They passes by but at the point of intersection they send each others stopwatch data via laser for analysing

 

[ZapperZ]

They are not stopping. They passes by but at the point of intersection they send each others stopwatch data via laser for analysing

Then I don't see what the problem here is, or why you are going to such an extent with this example. Everyone in his/her own frame will see the clock in another moving frame as being slower. This is plainly illustrated in the S-S' frame example without having to resort to a 3rd observer. Someone in S will see S' clock slower, and someone in S' will see S clock slower. It is very democratic.

Zz.

 

[PeterDonis]

In this case the box 2 will have nearly the speed of light from box 1’s point of view or not?

Yes, it will. But from box 1's point of view, box 2 will have started moving a long time *before* box 1's stopwatch starts, and box 2's stopwatch will have been running during all that time. So from box 1's point of view, box 2's stopwatch runs slower but also starts running sooner, and the two effects just cancel out so that box 2's stopwatch reads the same as box 1's when they meet.

A good general rule in all these types of problems is to make sure you're taking into account relativity of simultaneity.

 

[Dale]

How do this situation become symmetrical?

It became symmetrical because of how you set up the problem. If you start with A and rotate by 180 degrees then you get B. Since the laws of physics are invariant under rotations the situation is symmetric.

 

[Trojan666ru]

Yes, it will. But from box 1's point of view, box 2 will have started moving a long time *before* box 1's stopwatch starts, and box 2's stopwatch will have been running during all that time. So from box 1's point of view, box 2's stopwatch runs slower but also starts running sooner, and the two effects just cancel out so that box 2's stopwatch reads the same as box 1's when they meet.

A good general rule in all these types of problems is to make sure you're taking into account relativity of simultaneity.

Let me ask you a thing
Apply the same circumstance and imagine the box 1 is not moving and box 2 is moving with a uniform velocity towards box 1. The speed is .99c
We have the similar situation in diferent way
Now how do you distinguish between the situations? I mean how can you say which one is symmetrical and which one is not?

 

[.Scott]

I believe the actual closing rate between the two boxes, each traveling 0.5c relative to the observer, is 0.80c.
Each box would see the clock in the other box ticking at 36 seconds/minute.
The unboxed observer would see each of the clocks ticking at about 52 seconds per minute.

As PeterDonis explained, only the unboxed observer sees the clocks activating simultaneously.

 

[Trojan666ru]

I believe the actual closing rate between the two boxes, each traveling 0.5c relative to the observer, is 0.80c.
Each box would see the clock in the other box ticking at 36 seconds/minute.
The unboxed observer would see each of the clocks ticking at about 52 seconds per minute.

As PeterDonis explained, only the unboxed observer sees the clocks activating simultaneously.

Agreed only unboxed observer sees time ticking simultaniously. The moving observers sees others clock started later ie each observer perceives others clock has slower working rate. So both are true but when they intersect they both will dasagree on their recordings

 

[Ibix]

You are correct that the situation is symmetrical in one frame only. In that frame, then, the analysis is straightforward and the stopwatches must agree.

Therefore, the stopwatches must agree at the meeting as analysed in any other frame. As you pointed out, from the frame of either stopwatch, the other is ticking more slowly. Therefore, to get to the same time at the meeting the moving stopwatch must have started first.

Indeed, this is what the Lorentz transforms show. The watches only start simultaneously in the frame where they tick at the same rate. In all other frames, the slower ticking one starts earlier, so the times are always equal when they meet.

You need to let go of the notion of a universal "now" if you want an intutive feel for SR. The distinction between "simultaneous for me" and "simultaneous for you" keeps tripping you up.p

 

[PeterDonis]

when they intersect they both will dasagree on their recordings

I'm not sure "disagree" is the right term. Certainly their recordings will be different, but that's because they are in relative motion. They will agree on the direct observable, which is that their clock readings are the same when they meet. It's true that they will "disagree" on which stopwatch started first, but that's not a direct observable.

 

[Trojan666ru]

Therefore, the stopwatches must agree at the meeting as analysed in any other frame. As you pointed out, from the frame of either stopwatch, the other is ticking more slowly. Therefore, to get to the same time at the meeting the moving stopwatch must have started first.

If it is like that then ill have to some additional setup to disqualify your prediction
At the beginning the unboxed observer sends light pulses to start their stopwatches. Here the sent light pulse is reflected towards each other. At the exact middle point of their distance, the light pulses meet together. There is an instrument kept at the midde (by the observer) it receieves the signal and analyses wheather it was meeted simultaneously. If not the lagging signal sender will be destroyed. But in the unboxed observers frame, there will be no problem at all and the experiment goes well

 

[ghwellsjr]

I've asked this question a few times. This time i have tiny change in the question. Instead of people I'm using stopwatch & i have removed the accelerative frame.

If you have any doubts I'll clear it but do not skip the topic.

All i want is an answer that can be agreeable for all observers.

Imagine one observer and two apparatus.

Observer A who is coordinator and two boxes box 1 & box 2 which contain inactive stopwatches and automatic propulsion system which is motion dependent.
Observer A sets the two boxes away from each other equidistant from him. It forms an equilateral triangle. The distance between box 1 & 2 is 4 Lightyears.

Let me modify your scenario slightly, Instead of it forming an equilateral triangle, let's make them all be along a straight line. Otherwise, we have a confusing situation when later on you say they move towards each other. This change won't make any difference to the point of your scenario.

The coordinator first send some laser signals towards each box to confirm (to analyse in future) they were equal distance away.
Then he sends another laser pulses to both of the boxes to activate their propulsion system and they starts to move towards each other. Both boxes accelerates and stop automatically when they reach a velocity .5c
Then the coordinator sends another pulse to activate their stopwatches.

Let me make another slight modification to your scenario. Instead of three separate sets of pulses, we only need one and we will assume that the boxes can accelerate instantaneously, OK? It won't make any difference.

So now we have two boxes moving towards each other with same velocity. No acceleration included.
Suppose if there were observers in each box, they thinks they are at rest and the other one is moving towards them at .99c. Therefore they thinks the others stopwatch has run slower.

They won't think the other one is moving towards them at .99c. The correct number is 0.8c but even then it is a mixed story. There are two ways to look at the situation. If you want to know what they will think based on what they actually see and can directly measure, they will see the other observer moving towards them at 0.5c for most of the trip and then near the end they will see the other observer start to accelerate toward them at 0.8c. It is at this point that they see the other ones stopwatch start ticking.

The other way to look at the situation is from the standpoint of their individual rest frames. In that case, the other observer is only moving at one speed 0.8c but the other observer started moving considerably earlier than they did and their stopwatch started ticking considerably earlier than their own did.

In this case whose watch will have run longer when they meet each other?

As has been mentioned by others, their stopwatches will have the same reading when they pass each other.

In order to make these analyses more clear, I have created some spacetime diagrams. First is the rest frame of the coordinator, shown as the thick blue line. I take up the scenario after the two boxes in red and green have been moved to a distance of 4 light-years apart. The dots represent 1-year increments of time for each observer:

At the Coordinate Time of 1 year, the coordinator sends a blue signal to both remote boxes to start their instant accelerations towards each other and to start their stopwatches.

As you can see, they each see the other ones stopwatch sitting at zero for over two years. Then they see the other one start accelerating towards them and then they see the other ones stopwatch start running at three times the rate of their own. When they finally meet, their own clocks have advanced by about three and a half years and they have seen the other ones clock advance by the same amount of time. I think this is very clear. What do you think?

Now I will transform the scenario into the rest frame of the red box after it has accelerated to 0.5c:

As you can see, the previous analysis of what they each can actually see is maintained in this space time diagram, but in addition, we can see the other analysis where the green box started accelerating way before the red box did and that's why both of them can accumulate the same amount of time (3.5 years) during their trips, but only the green box is Time Dilated in red's rest frame.

And we can transform to green's rest frame and see that the prior analysis of what each observer sees still holds true but in addition, it is only the red observer that is Time Dilated in green's rest frame:

Does this make perfect sense to you? Any questions?

 

[PeterDonis]

There is an instrument kept at the midde (by the observer) it receieves the signal and analyses wheather it was meeted simultaneously.

Simultaneity is relative. As you have specified the scenario, the instrument in the middle will indeed receive the return signals simultaneously, indicating that the stopwatches started simultaneously *in the observer's frame*. But by relativity of simultaneity, that means the stopwatches do *not* start simultaneously in either of the "moving" frames (box 1's rest frame while the boxes are approaching each other, or box 2's rest frame while the boxes are approaching each other).

The spacetime diagrams posted by ghwellsjr make all this visually obvious.

 

[Trojan666ru]

Simultaneity is relative. As you have specified the scenario, the instrument in the middle will indeed receive the return signals simultaneously, indicating that the stopwatches started simultaneously *in the observer's frame*. But by relativity of simultaneity, that means the stopwatches do *not* start simultaneously in either of the "moving" frames (box 1's rest frame while the boxes are approaching each other, or box 2's rest frame while the boxes are approaching each other

If the clocks do not start simultaneously in the moving frame, then that means the simultaneity analyser analyses the reflected signal has lagged and thereby destroys the moving observer, is that right?

 

[ghwellsjr]

If it is like that then ill have to some additional setup to disqualify your prediction
At the beginning the unboxed observer sends light pulses to start their stopwatches. Here the sent light pulse is reflected towards each other. At the exact middle point of their distance, the light pulses meet together. There is an instrument kept at the midde (by the observer) it receieves the signal and analyses wheather it was meeted simultaneously. If not the lagging signal sender will be destroyed. But in the unboxed observers frame, there will be no problem at all and the experiment goes well

Yes, as PeterDonis has pointed out, the blue coordinator will see the first signals sent out by the red and green boxes (at the point of their accelerations and when their stopwatches are started) at the same time in all three frames. In fact, the blue coordinator will see each pair of the signals sent by the red and green boxes as they pass by his location. In other words, the 1-, 2-, and 3-year signals pass by him simultaneously as well as the two boxes pass by him simultaneously.

 

[PeterDonis]

If the clocks do not start simultaneously in the moving frame, then that means the simultaneity analyser analyses the reflected signal has lagged and thereby destroys the moving observer, is that right?

No, because as you specified the scenario, the simultaneity analyser is at rest in the observer's frame (the one in which the motion of both boxes is symmetrical), so it is detecting simultaneity in that frame. The stopwatches start simultaneously in that frame, so that's what the simultaneity analyser will detect.

 

[ghwellsjr]

If the clocks do not start simultaneously in the moving frame, then that means the simultaneity analyser analyses the reflected signal has lagged and thereby destroys the moving observer, is that right?

No, that's not right. Look at the diagrams. Aren't they abundantly clear to you?

 

[Trojan666ru]

I haven't accepted those diagrams at the moment i saw the word "acceleration" because its a nice paradox solver. Thats why i insisted to remove any accerative frame
I insist i don't need accelaration, that's the way i made my paradox

 

[PeterDonis]

I haven't accepted those diagrams at the moment i saw the word "acceleration" because its a nice paradox solver. Thats why i insisted to remove any accerative frame

The diagrams don't use any accelerated frames. They do show each of the stopwatch boxes being accelerated to 0.5c relative to the observer's frame, but all the frames used are inertial frames.

 

[ghwellsjr]

I haven't accepted those diagrams at the moment i saw the word "acceleration" because its a nice paradox solver. Thats why i insisted to remove any accerative frame
I insist i don't need accelaration, that's the way i made my paradox

You should read your first post. You said that the two boxes started to accelerate when they received the second signal from the coordinator. I just compressed the start of acceleration, the ending of acceleration, and the starting of the stopwatches into one event, just to make my diagrams easier to draw.

Furthermore, all three of my diagrams are Inertial Reference Frames, they are not accelerated frames. If you truly want to construct a scenario with no accelerations, that is easy to do, but it's not what you asked for so please don't reject my diagrams because I did exactly what you asked for. Please note that you did not specify how much time there was between the three pulses sent out by the coordinator so I took the liberty of making those times equal zero. If you don't like the way I presented the diagrams, you have to be more explicit in how you specify your scenario so that I can make diagrams that conform to your wish. I can't read your mind.

 

[ghwellsjr]

OK, I have changed your scenario so that there truly is no acceleration anywhere. The two boxes come out of infinity and have been inertial the whole way as they approach the coordinator. At some point in time, the coordinator sends out a pulse to each box to start their stopwatches. Now here are the three spacetime diagrams without further explanation (the explanation in post #17 works here just as well):

Now is it all perfectly clear to you? No objections?

 

[Ibix]

If it is like that then ill have to some additional setup to disqualify your prediction
At the beginning the unboxed observer sends light pulses to start their stopwatches. Here the sent light pulse is reflected towards each other. At the exact middle point of their distance, the light pulses meet together. There is an instrument kept at the midde (by the observer) it receieves the signal and analyses wheather it was meeted simultaneously. If not the lagging signal sender will be destroyed. But in the unboxed observers frame, there will be no problem at all and the experiment goes well

The distances from the stopwatches to the simulaneity detector are only equal in the symmetric frame. So although everyone must agree that the pulses leave your coordinator simultaneously and arrive at the detector simultaneously, they don't have to agree that they bounced off the stopwatches simultaneously.

Why don't you do the maths yourself? Or draw George's spacetime diagrams? Either will resolve your misunderstandings.

 

[ghwellsjr]

I reread your first post so I need to clarify some issues. As is usual in scenarios like this, we will assume that the coordinator has synchronized his clock to the Coordinate Time in his rest frame so that his first dot is at his Proper Time of zero and his second dot is at his Proper Time of 1 year and so on.

You wanted the first pulse sent by the coordinator for later analysis to verify that the two boxes were equal distance away from him. You will note that the blue signal sent out by the coordinator at his Proper Time of 1 year serves this purpose. He receives the signals sent from both boxes that were sent at the moment they received his signal at his Proper Time of 5 years. This is referred to as a Radar Signal. He determines the two distances by taking the difference in the sent and received times and dividing by 2 and assuming that the distance is how far light would travel in that amount of time. So we take 5-1 which equals 4 and divide by 2 to get 2 light-years. Note that this works in all Inertial Reference Frames, that is, he gets the same answer in all IRF's. Furthermore, he takes the average of the sent and received signals and assumes that the radar reflection occurred at that time. Since the average of 1 and 5 is 3, he assumes that the reflection occurred at his Proper Time of 3 years. These numbers conform to his rest frame.

Now your second pulse is what started the acceleration but since you decided that you don't want any acceleration, there is no need for the second pulse.

The third pulse is what starts the stopwatches but since there is no acceleration and no time for the boxes to reach 0.5c (because they are inertial and have always been traveling at 0.5c with respect to the coordinator or according to the rest frame of the coordinator), then there is no reason to postpone the third pulse from the first pulse so I have used the one pulse to both serve as a radar measurement of distance and to start the stopwatches.

Is this all agreeable to you?

 

[Trojan666ru]

Well your spacetime diagram is a good example that there's no paradox but logic still contradicts.
The root of the solution starts from the simultaneity. You have drawn that diagram assuming the moving box receives the signal way before the observer receives the signal
So i decided to solve it from the simultaneity part
From the unboxed (blue line) observers frame each travelers receive the signals simultaneously.
There's a setup here
After receiving the signals, it reflects towards a beam joiner which is placed at the centre and that beam is sent to the box1’ interferometer (Red box) . Since according to the unboxed observers frame of reference, they must meet at the beam joiner and hit the interferometer simultaneously, therefore there won't be an interference pattern.
But when you apply simultaneity, from the red box’ point of view, the other one (green box) has already received the signal and therefore the light beams won't reflect simultaneously and will definitely form an interference pattern.
Solve this first.
If you can draw this in spacetime diagram, then it means it contradicts with logic

 

[PAllen]

Well your spacetime diagram is a good example that there's no paradox but logic still contradicts.
The root of the solution starts from the simultaneity. You have drawn that diagram assuming the moving box receives the signal way before the observer receives the signal
So i decided to solve it from the simultaneity part
From the unboxed (blue line) observers frame each travelers receive the signals simultaneously.
There's a setup here
After receiving the signals, it reflects towards a beam joiner which is placed at the centre and that beam is sent to the box1’ interferometer (Red box) . Since according to the unboxed observers frame of reference, they must meet at the beam joiner and hit the interferometer simultaneously, therefore there won't be an interference pattern.
But when you apply simultaneity, from the red box’ point of view, the other one (green box) has already received the signal and therefore the light beams won't reflect simultaneously and will definitely form an interference pattern.
Solve this first.

Ibix already answered this. In all frames, the signals leave simultaneously (trivial because they are colocated), and arrive together simultaneous (same reason), but the reflections off the boxes are simultaneous only in the unboxed frame. In one of the box frames, the signal reaches the other box first, but then has longer to go to meet the other signal; the other signal reaches given box later, but then has less to go to meet the other reflected signal. Thus everyone agrees the reception will be simultaneous, but only the unboxed observer finds the reflection events to be simultaneous.

 

[ghwellsjr]

Well your spacetime diagram is a good example that there's no paradox but logic still contradicts.
The root of the solution starts from the simultaneity. You have drawn that diagram assuming the moving box receives the signal way before the observer receives the signal

In order to avoid confusion, I have been using the terminology that you specified in your first post. I don't know what you mean by observer. Originally, you said there was going to be just "one observer and two apparatus". Then you called that one observer the coordinator. That's the terminology that I have consistently used and I have further identified the coordinator as being blue in my drawings. You also identified the "two apparatus" as two boxes which I have identified as red and green. But then you introduce two more observers to go with the two boxes. Now we have three observers and I have no idea what you mean by the statement:

"You have drawn that diagram assuming the moving box receives the signal way before the observer receives the signal"

Each moving box receives only one signal in my diagrams and that signal is sent by the coordinator so who is the observer that you are referring to in the above quote that also receives the signal?

If you want to make charges that "logic still contradicts", you really should present your case in a logical manner and not force us to read your mind. I, for one, can't do that.

So i decided to solve it from the simultaneity part
From the unboxed (blue line) observers frame each travelers receive the signals simultaneously.
There's a setup here
After receiving the signals, it reflects towards a beam joiner which is placed at the centre and that beam is sent to the box1’ interferometer (Red box) .

I have no idea what you have in mind. If I just read what you wrote, you are describing a scenario that looks like this, correct?

Since according to the unboxed observers frame of reference, they must meet at the beam joiner and hit the interferometer simultaneously, therefore there won't be an interference pattern.
But when you apply simultaneity, from the red box’ point of view, the other one (green box) has already received the signal and therefore the light beams won't reflect simultaneously and will definitely form an interference pattern.
Solve this first.
If you can draw this in spacetime diagram, then it means it contradicts with logic

You have to first tell me if I drew the spacetime diagram according to what you had in mind. I think I drew it according to my best understanding of your description.

 

[PeterDonis]

But when you apply simultaneity, from the red box’ point of view, the other one (green box) has already received the signal and therefore the light beams won't reflect simultaneously and will definitely form an interference pattern.

What does "apply simultaneity" mean? The only meaning I can come up with is "apply the Lorentz transformation", but if I interpret it that way, your statement is false: the beam joiner receiving both return beams at the same event is Lorentz invariant, so it's true in every frame. Put another way, whether or not the return beams form an interference pattern is Lorentz invariant, since it's a direct observable; so if a pattern does not form in the beam joiner's frame--which it doesn't, according to your own hypothesis, the way you have specified the scenario--then it doesn't in any frame. (And note that when this type of experiment is done in real life, it comes out as I just said: whether or not an interference pattern forms is indeed Lorentz invariant, the same in all frames.)

If you mean something else by "apply simultaneity", you're going to need to tell us what it is--and then explain how it can possibly be true since what you are deducing from it conflicts with the observed Lorentz invariance in actual experiments.

 

[Ibix]

What does "apply simultaneity" mean?

Peter: As I (indirectly) observed earlier, I interpret the OP's arguments as implying that (s)he assumes absolute simultaneity. That means that I suspect that the OP thinks that, since the pulses of light reflect simultaneously off the stopwatches in the "unboxed" co-ordinator's frame (blue in George's diagrams), they do so in every frame. That reasoning does indeed lead to a contradiction - if you assert that the pulses leave the stopwatches simultaneously ("apply simultaneity") in any other frame (e.g. the red one), they can't possibly reach the detector simultaneously. That we keep saying that the reflection events are simultaneous in some frames and not others is what the OP says "contradicts logic" in #29.

OP: Am I right?

 

[PeterDonis]

That we keep saying that the reflection events are simultaneous in some frames and not others...

Just to clarify: they are simultaneous in *exactly one* frame, the rest frame of the "joiner" and the "observer". That's true of any pair of spacelike separated events in SR: there is exactly one frame in which they are simultaneous.

...is what the OP says "contradicts logic" in #29.

If this is the case, then actual experiments also "contradict logic" by the OP's criterion. You can't change the actual experiments, they are what they are; so that means the OP's criterion needs to change.

 

[Ibix]

Just to clarify: they are simultaneous in *exactly one* frame, the rest frame of the "joiner" and the "observer".

True - poor proofreading on my part, I'm afraid.

If this is the case, then actual experiments also "contradict logic" by the OP's criterion.

Indeed. My last post is an attempt to draw out the theoretical basis behind the OP's thinking, not to make any claim that said basis is correct. I don't feel like an infraction today...