Twin paradox without accelerative frame.

[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...

 

[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?