Twin paradox without accelerative frame.

[Dale]

there's no paradox but logic still contradicts.

Nonsense. If the logic is contradictory then it is a paradox, essentially by definition. You cannot have it both ways.

Your difficulty understanding SR is neither unusual nor is it damaging to relativity. Everyone who has learned relativity has been where you are. You need to stop trying to disprove SR and instead try to learn it. Despite your current stage of learning, it is eminently logical and consistent.

 

[Trojan666ru]

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.

do you mean that even if the light hits the moving box earlier, it will take longer time to reflect off from that box in an adjusted way to hit the detector simultaneously?
Why is that so?
Is that because since the moving box has time dilated and therefore reflections also has to be dilated?

 

[Ibix]

No. Reflection is instantaneous here. But light has further to travel from one stopwatch to the simultaneity detector than from the other stopwatch in any frame except the co-ordinator's (blue frame). The extra distance for this leg balances out the extra time and the pulses arrive simultaneously.

This is obvious from George's diagrams or from the Lorentz tranformed co-ordinates of the events.

 

[ghwellsjr]

Trojan666ru, how about responding to my request for feedback on the diagram in post #31 that you asked me to draw for you? I can't solve your logic contradiction problem until you assure me that I understand what your problem is.

 

[Trojan666ru]

George

Your answers were correct according to my question
I understood your diagram. I'll summarise what i understood
In the red box’ rest frame, the green box has already received the signal and it reflected off too early but it got longer distance to travel to meet the simultaneity detector

But what i do not understand is, how did the proper distance between the simultaneity detector and the green box increased?
Spacetime diagram shows like that but, that's not how the actual experimental setup looks like

 

[PAllen]

George

Your answers were correct according to my question
I understood your diagram. I'll summarise what i understood
In the red box’ rest frame, the green box has already received the signal and it reflected off too early but it got longer distance to travel to meet the simultaneity detector

But what i do not understand is, how did the proper distance between the simultaneity detector and the green box increased?
Spacetime diagram shows like that but, that's not how the actual experimental setup looks like

Proper distance is invariant and applies to specific events which are spacelike separate (which means there exists a frame in which they are simultaneous). Events like sending a signal, reflecting it, joining with another signal - these are all causally connected and have timelike separation. Any statement about distance traveled is a frame variant coordinate quantity. Proper distance between such events is not and cannot be defined.

Distance between objects is also inherently frame variant because disagreement on simultaneity means different events on each object's history are compared for the purpose of measuring distance. Each frame picks different spacelike separated events as 'at the same time', thus each computes a different proper distance as the one that gives the distance between the objects.

 

[Trojan666ru]

I'm making some arrangements in distance from the above question to calculate reflected time

They are separated by 4 LY.
When they receive the first signal from the coordinator, they starts to accelerate and when both the boxes travel exact 1 LY (from the coordinator view) they will attain .5c. That instant is simultaneous for the coordinator. From that instant there will be only 2 LY distance between the boxes. As preplanned the coordinator has already sent another signal for activating the stopwatch at the exact instant it travels (green box) 1 LY. So these events are simultaneous in coordinator frame

From the rest frame of red box the green box is the one moving with .8c. So the signal which hits the green box has only 2LY distance between the boxes, so how much time does the signal takes to reach the red box?
Greater than 2 years, Lesser than 2 years or Exact 2 years?

 

[ghwellsjr]

George

Your answers were correct according to my question
I understood your diagram. I'll summarise what i understood
In the red box’ rest frame, the green box has already received the signal and it reflected off too early but it got longer distance to travel to meet the simultaneity detector

But what i do not understand is, how did the proper distance between the simultaneity detector and the green box increased?
Spacetime diagram shows like that but, that's not how the actual experimental setup looks like

I get the impression that my previous diagram was not correct:

Instead, I think this is really what you are describing:

I think you are saying that in this frame, there is no interference pattern because the two beams sent from the coordinator are symmetrical and simultaneous throughout their trips, including their simultaneous reflections and simultaneous arrival back at the coordinator but when we transform to the rest frame of either box (red or green), there will be an interference pattern because even though the two beams arrive at the coordinator simultaneously, they didn't reflect simultaneously and didn't follow simultaneous equal distant paths, is that correct?

So if I can show you that there is no interference pattern in both the red and green rest frames, then you will be convinced that there is no logical contradiction, correct?

 

[Trojan666ru]

But you have already shown me that there is no contradiction. Your spacetime diagram was perfect. But i just need an answer to my last post

 

[Trojan666ru]

Your second diagram is not right. The beam joiner is at the middle but the interference detector is placed only on the red box. It goes like this
The reflected light from green box and red box meets the beam joiner, then they combine and move towards the red box interferometre

 

[PeterDonis]

Your second diagram is not right. The beam joiner is at the middle but the interference detector is placed only on the red box. It goes like this
The reflected light from green box and red box meets the beam joiner, then they combine and move towards the red box interferometre

Doesn't that make the situation asymmetrical with respect to the joiner's frame? I thought you intended for the scenario to be symmetrical with respect to that frame.

Also, how are the beams "combined"?

 

[Trojan666ru]

The beam joiner is at the middle but the interference detector is placed only on the red box. It goes like this
The reflected light from green box and red box meets the beam joiner, then they combine and move towards the red box interferometre

All these are simultaneous from the blue box reference frame (coordinator frame)

When i said it won't be simultaneous in the red box reference frame (ie the combined rays won't hit the interferometre simultaneously), you said that it will be simultaneous for all observers with SP diagram
The beams are combined with beam joiner

That's what happened here

Everything was perfect until my #42nd post

 

[PeterDonis]

When i said it won't be simultaneous in the red box reference frame (ie the combined rays won't hit the interferometre simultaneously), you said that it will be simultaneous for all observers with SP diagram

I haven't said anything about when the combined rays hit the red box's interferometer, because up until now I didn't understand that that's where the interferometer was.

If your intended scenario is that the beam reflected by the red box meets the beam reflected by the green box at the joiner, and the beam reflected by the red box then gets reflected at the joiner so it's now moving along with the beam reflected by the green box (i.e., towards the red box), then yes, both beams will arrive at the red box interferometer at the same event, and that will be true in all reference frames.

However, the term "simultaneous" is not correct if used to describe what I just described, because both beams arrive at the red box interferometer at the same event. "Simultaneous" is a term applied to two distinct events which have the same time coordinate in some reference frame. It does not apply to a single event. The fact that both beams arrive at the red box interferometer at the *same*, single event is why they arrive together in all reference frames: a single event must be a single event in all reference frames.

A spacetime diagram of all this would look the same as the first diagram in ghwellsjr's post #43, except that both beams would move up and to the left from the "joiner" event to the "interferometer" event on the red box's worldline--i.e., both beams would overlap, travel on the *same* path through spacetime, between those two events.

 

[PeterDonis]

The beams are combined with beam joiner

So where do I buy a "beam joiner"? Is there one available on the Internet?

What I was asking was, what exactly does the "beam joiner" do, physically?

 

[Trojan666ru]

So if I can show you that there is no interference pattern in both the red and green rest frames, then you will be convinced that there is no logical contradiction, correct?

But i believe you have already shown it, don't you?

 

[ghwellsjr]

Your second diagram is not right. The beam joiner is at the middle but the interference detector is placed only on the red box. It goes like this
The reflected light from green box and red box meets the beam joiner, then they combine and move towards the red box interferometre

I guess my first diagram was almost correct, I just added a second interferometer by mistake. Here it is with only one interferometer and I made the joined green and red signals appear as yellow:

Is this correct?

If so, then is your concern that in the blue coordinator's rest frame, the interferometer will not detect an interference pattern but in the red box's rest frame it will? And if that is your concern, and I can show you that there will not be an interference pattern, will you agree that the logic does not contradict and we will be done with this thread?

 

[Trojan666ru]

So where do I buy a "beam joiner"? Is there one available on the Internet?

What I was asking was, what exactly does the "beam joiner" do, physically?

that beam joiner has no relavance. I used it only to combine beams. I can instead use a mirror. My aim is only to reflect both the beams and point it towards the red box interferometre

 

[PeterDonis]

that beam joiner has no relavance. I used it only to combine beams. I can instead use a mirror.

Well, it would have been clearer to just say "mirror" in the first place. But ok, now I understand, and my post #48 was correct in assuming a mirror.

 

[ghwellsjr]

So where do I buy a "beam joiner"? Is there one available on the Internet?

What I was asking was, what exactly does the "beam joiner" do, physically?

I thought it was just a half-silvered mirror placed so that the red beam reflects at 45 degrees in and 45 degrees out while the green beam passes right through (on the diagram--the beams are really coming in directly and reflecting or going through directly).

 

[Trojan666ru]

George: Exactly
But i won't stop the thread until i solve one more problem. But I'll post it after you prove it with SP diagram

 

[PeterDonis]

Here it is with only one interferometer and I made the joined green and red signals appear as yellow

Yep, this is exactly what I was describing in post #48.

(Btw, ghwellsjr, I know you've answered this before but it was a while ago and I don't remember what the answer was: what software do you use to generate these great spacetime diagrams?)

 

[ghwellsjr]

Yep, this is exactly what I was describing in post #48.

(Btw, ghwellsjr, I know you've answered this before but it was a while ago and I don't remember what the answer was: what software do you use to generate these great spacetime diagrams?)

I wrote my own application using LabVIEW by National Instruments. What I did in LabVIEW presumably could be done in any programming language but since I don't know any others (and don't want to learn any others), I did it in the one I have been using professionally for over two decades.

 

[Trojan666ru]

Your previous space time diagram was correct. Or do you want to post it again? Whatever I'm sure your spacetime diagram will look like this again

 

[ghwellsjr]

Your previous space time diagram was correct. Or do you want to post it again? Whatever I'm sure your spacetime diagram will look like this again

OK, we'll go with the acceleration version.

Here's the blue coordinator's rest frame showing how the interferometry would work. If it was real interferometry of visible light, it would be on too small a scale to distinguish any features so I have, in effect, used a one-year long light signal that has a period of three months. The five thin blue lines emanating from the blue coordinator represent the peaks of this very low frequency light:

As you can see, the light hitting the moving reflectors on the red and green boxes causes a Doppler "blue" shift in the frequency (higher) or wavelength (shorter) of the light but since it is symmetrical, the Doppler shift is the same for both boxes and is received by the Joiner coherently and passed on to the Interferometer in the red box with no interference (as you say), meaning the peaks of the reflections from both boxes arrive simultaneously at the Joiner and passed on to the Interferometer.

Now we transform to the rest frame of the red box:

As you can see, the signals leaving the blue coordinator start out Doppler shifted in opposite directions. Going to the red box, the signal is Doppler blue-shifted (higher frequency) and the signal going to the green box is Doppler red-shifted (lower frequency). But the signal bouncing off the red box doesn't get any further Doppler shifting, it has the same frequency as the sent signal. However, the signal bouncing off the green box gets a lot of Doppler red-shifting making it an even higher frequency than the signal reflecting off the red box. But this signal passes right through the Joiner without any more Doppler shifting while the signal from the red box gets Doppler red-shifted some more so that it ends up exactly in sync with the reflected signal from the green box. As a result, the interferometer in the red box detects no interference pattern, just as in the original rest frame.

Does this make perfect sense to you? Any questions or further concerns? Can you see that it is logically consistent and there is no contradiction?

 

[Trojan666ru]

We reached the same conclusion before the #42nd post. Now answer my #42 question?

 

[ghwellsjr]

We reached the same conclusion before the #42nd post. Now answer my #42 question?

OK, here's your post #42:

I'm making some arrangements in distance from the above question to calculate reflected time

They are separated by 4 LY.
When they receive the first signal from the coordinator, they starts to accelerate and when both the boxes travel exact 1 LY (from the coordinator view) they will attain .5c. That instant is simultaneous for the coordinator. From that instant there will be only 2 LY distance between the boxes. As preplanned the coordinator has already sent another signal for activating the stopwatch at the exact instant it travels (green box) 1 LY. So these events are simultaneous in coordinator frame

Here's a diagram depicting this scenario:

From the rest frame of red box the green box is the one moving with .8c.

Here's a diagram for the rest frame of the red box:

So the signal which hits the green box has only 2LY distance between the boxes, so how much time does the signal takes to reach the red box?
Greater than 2 years, Lesser than 2 years or Exact 2 years?

As you can see, in the blue coordinator's rest frame, it takes less than 2 years for the green signal to propagate from the green box to the red box but in the red box's rest frame, it takes more the 2 years.

I wasn't sure which frame you wanted the answer for so I gave you both answers.

Any more questions?

 

[Trojan666ru]

Now there's the problem!

As you said, in the blue box reference frame there is only 2Ly distance between the red box & green box at the instant it reaches .5c
So the reflected light will take only less than 2 years to reach the red box interferometer (in blue box frame)

In addition since the red box is moving with respect to the blue box, its time is dilated and therefore will receive the signal before its stopwatch ticks to 2 years

If the stopwatch & interferometer are a part of time bomb which is programmed not to explode if the signal hits there before 2 years, then from the blue box’ rest frame there won't be any explosion but from the red box’ rest frame, the signal has to travel more than 2Ly distance (as shown in your space-time diagram) and therefore it hits the stopwatch only after 2 years, therefore the bomb explodes!

 

[PAllen]

Now there's the problem!

As you said, in the blue box reference frame there is only 2Ly distance between the red box & green box at the instant it reaches .5c
So the reflected light will take only less than 2 years to reach the red box interferometer (in blue box frame)

In addition since the red box is moving with respect to the blue box, its time is dilated and therefore will receive the signal before its stopwatch ticks to 2 years

If the stopwatch & interferometer are a part of time bomb which is programmed not to explode if the signal hits there before 2 years, then from the blue box’ rest frame there won't be any explosion but from the red box’ rest frame, the signal has to travel more than 2Ly distance (as shown in your space-time diagram) and therefore it hits the stopwatch only after 2 years, therefore the bomb explodes!

The stopwatch starting is a specific event. A signal reaching the stopwatch is another event. The time a specific clock reads between two events on its history is invariant in all frames. You are confusing this with judgement about simultaneity, which has no effect on a specific clock. In both of gwellsjr's diagrams, look at the tick marks on the red world world line between when the stopwatch starts and when the signal arrives. It is less than 2 in both frames. This fact is independently computed in both frames. Thus both frames agree the bomb will not go off.

 

[ghwellsjr]

Now there's the problem!

I think you will see that there's no problem.

As you said, in the blue box reference frame there is only 2Ly distance between the red box & green box at the instant it reaches .5c
So the reflected light will take only less than 2 years to reach the red box interferometer (in blue box frame)

Correct.

In addition since the red box is moving with respect to the blue box, its time is dilated and therefore will receive the signal before its stopwatch ticks to 2 years

Correct, but again, note that the red box's stopwatch is only Time Dilated in the blue coordinator's rest frame (the first diagram). In the rest frame for the red box, it's own clock is not Time Dilated but the blue coordinator's clock is Time Dilated by the same amount as the red box's stopwatch was Time Dilated in the first frame. Can you see this difference in the two diagrams?

If the stopwatch & interferometer are a part of time bomb which is programmed not to explode if the signal hits there before 2 years, then from the blue box’ rest frame there won't be any explosion but from the red box’ rest frame, the signal has to travel more than 2Ly distance (as shown in your space-time diagram) and therefore it hits the stopwatch only after 2 years, therefore the bomb explodes!

In both frames, the green signal gets to the red stopwatch when it reads about 1.15 years. It's exactly the same in all frames so the bomb won't explode in any frame.

If you go back to my first post (#17) you will see that I said just before the first diagram, "The dots represent 1-year increments of time for each observer". As you can see, the first red dot representing zero years on the stopwatch occurs when the second blue signal arrives from the coordinator which starts the stopwatch ticking. (Actually, the stopwatch has been inactive and sitting at zero for the entire time along the thick red line prior to the first dot.) The next red dot represents one year on the stopwatch and a short time later, the green signal arrives from the green box. The last red dot is when the stopwatch gets to two years and it doesn't happen until after the red box passes both the blue coordinator and the green box. Isn't all this very clear in both diagrams?

It's important to realize that no frame changes what any observer can see or measure. The red box only knows the times that the signals arrive at its location, it has no awareness of when those signal were sent or when other signals arrive at other locations.

 

[PeterDonis]

from the red box’ rest frame, the signal has to travel more than 2Ly distance (as shown in your space-time diagram) and therefore it hits the stopwatch only after 2 years, therefore the bomb explodes!

You're forgetting relativity of simultaneity again. In the red box' rest frame, the signal leaves the green box *before* the red box' stopwatch starts--about a year before, looking at ghwellsjr's spacetime diagram. So the red box' stopwatch is only running for a little over a year when the signal arrives; and since it's the reading on the stopwatch that determines whether the bomb explodes, the bomb does not explode.

 

[ghwellsjr]

As you can see, in the blue coordinator's rest frame, it takes less than 2 years for the green signal to propagate from the green box to the red box but in the red box's rest frame, it takes more the 2 years.

I wasn't sure which frame you wanted the answer for so I gave you both answers.

What about this quote?

That quote was in response to these questions:

So the signal which hits the green box has only 2LY distance between the boxes, so how much time does the signal takes to reach the red box?
Greater than 2 years, Lesser than 2 years or Exact 2 years?

Those questions are not about the Proper Time on a single clock which was your criterion for the bomb, they are about the difference between two Coordinate Times which is different in each coordinate system we use. Note that the Proper Time on the green clock was zero when the signal started and it was 1.15 on the red clock when it was received (in both frames) but the Coordinate Times for those two events are totally unrelated. In the first frame, the Coordinate Times were about 7.15 and 8.5 (with a difference of about 1.35) and in the second frame they were about 7.7 and 10 (with a difference of about 2.3).

If you wanted to make the bomb explode according to the Coordinate Time in one frame and not the other, you would have to incorporate a switch in the bomb which would define which frame it was supposed to respond to and then, depending on the switch, it would either blow up or not in all frames.

 

[Trojan666ru]

In the beginning all three boxes were at rest relative to each other. It is also clear that from all three observers point of view the distances are well defined (ie 4Ly in total and in the middle a half silvered mirror)
But somehow from the red box’ point of view, the green box receives the signal from the coordinator earlier than the red box. Why is it so?

Is that because the coordinator sends the green signal earlier and red later?

Or

The distance between the coordinator and red box suddenly increased & the distance between the coordinator and green box decreased?

 

[Trojan666ru]

George: Probably the distance got suddenly changed, that's what spacetime diagram says, but how?

 

[Ibix]

The latter.

The two pulses are emitted at the same time in the same place. This is one event, so must be one event in all frames. They are received in the same place at the same time - again, one event, so one event in all frames.

However the reflections do not occur at the same place. Observers do not, in general, agree on the simultaneity or order of these events, nor their spatial locations.

In the red frame, the coordinator is moving towards the red box. The light pulse travels to the box, but by the time it gets there the blue co-ordinator is closer to the red box. Its return journey is shorter than its outbound one.

The green box is also coming towards the red box. For the other light pulse, then, the green box is closing the distance it has to cross. After reflection, though, the pulse is chasing after the blue box - it has extra distance to travel to catch it. That means that its outbound journey is shorter than its return journey.

If you require that lightspeed be constant, the only way to resolve this is to allow events to happen in different orders for different people, and different distances apart. The Lorentz transforms desribe the detail. This is Special Relativity.

If you require global simultaneity (that events that are simultaneous for one person are simultaneous for all), then you have to let lightspeed vary to cover the different distances in the same time. The Galilean transforms describe the details. This is Newtonian physics.

Both views are internally logically consistent. However, all experiments capable of differentiating the two agree with Special Relativity.

 

[Ibix]

George: Probably the distance got suddenly changed, that's what spacetime diagram says, but how?

You are thinking of distance as all there is to know about the separation of two points in space. This is not correct - remember that time is a dimension. The separation between two events in spacetime is the constant thing. It's called the interval:
$$s^2=(ct)^2-(x^2+y^2+z^2)$$

Wondering why the distance changes is analogous to wondering why people sitting round a table don't agree whether it's a long table or a wide table.

I suggest you google "block universe" for a better explanation.