Twin paradox without accelerative frame.

[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.

 

[ghwellsjr]

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)

Yes, that is all very true. However, you have to be careful with that term "point of view" because it sounds like we are referring to something each box can actually see. Instead, in this context it just means "the Inertial Reference Frame (IRF) in which the object is at rest". In order to determine what each box can actually see, you have to look at the thin lines coming from the other boxes and arriving at the box you are concerned with. Each box has to wait for the light signals to get to it. The farther away the source of the light, the longer it takes to get there.

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?

Since the red box has accelerated it has more than one IRF in which it is at rest so when we want to switch to the IRF in which the red box is at rest at the end of the scenario we have to use the Lorentz Transformation process to see what things look like in this second IRF. But if you do the same thing that I described for the first IRF (draw in light signal lines) you will discover that both IRF's show exactly the same phenomenon in terms of what the red box can actually see.

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

No, look at the thin lines coming out of the coordinator, in both diagrams, they depart at the same time.

Or

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

No, in the second IRF, the distances are Length Contracted by the same amount but it's because the propagation of light is c in all directions in any IRF (by SR's definition) and all the boxes are moving so the ones moving in the same direction as the light will take longer and the ones traveling against the light will take less itme.

 

[Sugdub]

... Any questions or further concerns? ...

According to your first diagram in #59 (the twins scenario being described in the IRF showing the coordinator at rest), the SR theory represents both travelers moving toward each other at a relative speed equal to c in that IRF (their initial distance is 4 light-years and they meet 4 years later).

Should each of them travel at 0.7c toward the coordinator (instead of 0.5c), would the SR theory indicate that their relative speed in respect to each other is equal to 1.4c in the coordinator's IRF?

More generally, is it correct to assume that the standard addition of vectors still holds for velocities provided their magnitudes relate to the same IRF?

 

[Dale]

According to your first diagram in #59 (the twins scenario being described in the IRF showing the coordinator at rest), the SR theory represents both travelers moving toward each other at a relative speed equal to c in that IRF (their initial distance is 4 light-years and they meet 4 years later).

Usually this speed is called a "closing speed" rather than a "relative speed", so as to distinguish it from the speed that each measures relative to themselves.

Should each of them travel at 0.7c toward the coordinator (instead of 0.5c), would the SR theory indicate that their relative speed in respect to each other is equal to 1.4c in the coordinator's IRF?

Yes, that would be a closing speed of 1.4 c.

More generally, is it correct to assume that the standard addition of vectors still holds for velocities provided their magnitudes relate to the same IRF?

Closing speed doesn't really have a direction, so it isn't a vector.

 

[jtbell]

I would say that if in frame A you have object B traveling with velocity $\vec v_{BA}$ and object C traveling with velocity $\vec v_{CA}$, then the separation velocity (in frame A) of object C from object B is just the difference $\vec v_{CA} - \vec v_{BA}$, which is a vector.

However, this separation velocity does not equal the the velocity of C in B's rest frame $\vec v_{CB}$ (which is what we normally mean by the relative velocity of C and B). For that, you need a vector version of the "relativistic velocity addition" formula, which I'm too lazy to look up at the moment.

 

[PAllen]

I would say that if in frame A you have object B traveling with velocity $\vec v_{BA}$ and object C traveling with velocity $\vec v_{CA}$, then the separation velocity (in frame A) of object C from object B is just the difference $\vec v_{CA} - \vec v_{BA}$, which is a vector.

However, this separation velocity does not equal the the velocity of C in B's rest frame $\vec v_{CB}$ (which is what we normally mean by the relative velocity of C and B). For that, you need a vector version of the "relativistic velocity addition" formula, which I'm too lazy to look up at the moment.

One formulation in, terms of 4-velocities, for the relative speed is:

v = || U2 - (U1 ∙ U2) U1 || / (U1 ∙ U2)

The spatial direction vector of U2 relative to U1 is then:

U2 - (U1 ∙ U2) U1

If this is computed in components in some inertial frame, the spatial vector will be expressed in axes parallel to that inertial frame, which is often what you want.

 

[Sugdub]

... Yes, that would be a closing speed of 1.4 c. ...

Thanks for your answer. I tried to extrapolate a diagram by Gwellsjr to address the first phase of the basic twins scenario, assuming that each traveler and the associated clock moves away from the coordinator at a relative speed of 0.7c (in respect to the coordinator's rest frame), whilst sending light pulses backwards at a definite frequency. I concluded that each traveler would still receive the light pulses propagating the “ticks” of the other traveler's clock, in spite of their “closing speed” (does this wording also hold if both travelers move away from each other?) equal to 1.4c in the coordinator's rest frame. This sounds non-intuitive. Is it correct? Actually on the diagram it seems that 2c is the absolute upper limit for this exchange of information taking place, isn't it?

 

[Trojan666ru]

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.

That's what i do not understand yet. Even before starting experiment (while everybody is at rest with respect to everyone) how can the green box receive the information for acceleration earlier than red box?
If i conduct this experiment in small scale, with a 100m distance, and if I'm sitting inside the red box, will i see the green box receiving the signal earlier than me?
What if we have no plan to accelerate the boxes? Still the green box receives the signal earlier?

 

[Dale]

I concluded that each traveler would still receive the light pulses propagating the “ticks” of the other traveler's clock, in spite of their “closing speed” ... equal to 1.4c in the coordinator's rest frame. This sounds non-intuitive. Is it correct?

Yes, it is correct. The 1.4 c closing speed is not the speed of anything, so it is not relevant to determining whether or not something can outrun a light signal.

Actually on the diagram it seems that 2c is the absolute upper limit for this exchange of information taking place, isn't it?

There is no exchange of information taking place at anything greater than c, but 2 c is the upper limit of closing speed.

 

[Dale]

(while everybody is at rest with respect to everyone) how can the green box receive the information for acceleration earlier than red box?

Don't forget that simultaneity is frame-dependent, which means that statements about simultaneity inherently have no meaning unless you also state the frame that simultaneity is judged relative to. So, the question you asked is inherently incomplete.

You should ask something more like "how can the green box receive the information for acceleration earlier than red box in a frame where they are all moving to the left?". Then the answer is "because the light has less distance to travel to reach the green box it takes less time".

If i conduct this experiment in small scale, with a 100m distance, and if I'm sitting inside the red box, will i see the green box receiving the signal earlier than me?

The distance scale is not relevant. If the green box moves towards the emitter then it will receive the signal earlier.

What if we have no plan to accelerate the boxes? Still the green box receives the signal earlier?

Future plans are also not relevant.

 

[PeterDonis]

Even before starting experiment (while everybody is at rest with respect to everyone) how can the green box receive the information for acceleration earlier than red box?

"Earlier" is frame-dependent. The green box receives the information earlier with respect to the frame in which the red box is at rest *after* it has accelerated. Changing frames changes the time coordinates of all events, not just the ones after a certain point.

If i conduct this experiment in small scale, with a 100m distance, and if I'm sitting inside the red box, will i see the green box receiving the signal earlier than me?

It depends on how you assign time coordinates to events on the green box's worldline. Since the events of the green box receiving the signal, and of the red box receiving the signal, are spacelike separated, their time ordering (which one happened "first") is not invariant; it's frame-dependent. So you can assign any time ordering you like and still be consistent with all observations.

The diagram ghwellsjr drew assumes that the red box assigns time coordinates to all events using the inertial frame in which it is at rest after it has accelerated. But there are other ways it could assign time coordinates to events that could give different answers.

The fundamental point you keep on missing is the one I said above, which I'll restate: The time ordering of spacelike separated events is not invariant. Therefore it is not a good thing to focus on if you're trying to understand what's happening. You should focus on things that are invariant, like the fact that the red clock and the green clock show the same reading when they meet.

 

[ghwellsjr]

Re-uploaded diagrams

That's what i do not understand yet. Even before starting experiment (while everybody is at rest with respect to everyone) how can the green box receive the information for acceleration earlier than red box?

You showed the rest frame for the red box after acceleration. Here's the rest frame for the red box before acceleration with the Proper Times on their stopwatches marked:

The issue of which box received the signal first depends on the frame because we are talking about simultaneity issues which are frame dependent. If you look at the third diagram in my first post (#17) you will see that the red box receives the signal for acceleration earlier than the green box. In the above frame, they both receive their respective signals at the same Coordinate Time of 3 years when their stopwatches are activated to start ticking from zero. But neither one can see that the other one received his signal at the same time as his own. Observers cannot see Coordinate Times for remote events.

But later on, each one can see the other one receiving the reflected signal or the image of the Proper Time on the other ones stopwatch. In the above diagram, can you see that the green signal that came from the green box when it started to accelerate (at its Proper Time of zero) arrives at the red box when the Proper Time on his stopwatch reached about 2.3 years? I hope the Proper Time markings have made this very clear.

The same thing applies for the green box watching the Proper Time on the red box's stopwatch. He also sees the red box's stopwatch start ticking when the Proper Time on his own stopwatch reaches 2.3 years.

I have added the same Proper Time markings on the diagram that you showed:

Can you see that the Proper Times on both stopwatches and the signals that go to and from them are identical in both frames even though the Coordinate Times and Coordinate distances are different? Both of these frames contain identical information. The boxes don't know that we have drawn two different diagrams with different coordinates. All they know is what they can see when, according to the Proper Times on their own stopwatches, the signals arrive at their own locations.

If i conduct this experiment in small scale, with a 100m distance, and if I'm sitting inside the red box, will i see the green box receiving the signal earlier than me?

No, just like in the above two diagrams, you won't see what is happening on the other box until later on. But if you go to the effort of sending radar signals (like you mentioned in your first post) and collecting observational data and you apply Einstein's convention that light travels at c in any Inertial Reference Frame, then you can construct a non-inertial rest frame for yourself and from that you can construct either of the above two inertial diagrams and then you will be able to see what we can see.

What if i have no plan to accelerate my boxes? Still the green box receives the signal earlier?

The issue of which is earlier is dependent on the frame you chose to answer the question in. But what they can see, measure and observe is the same in all frames. If you imagine in the above diagrams that the red and green lines continue in straight lines, then you will see the image of the red box starting its stopwatch when your own stopwatch reaches 4 years, but remember, in this case, your Proper Time is not dilated in the original defining IRF.

I'm working on another answer to your previous post which may help clarify your questions. I hope to post it shortly.

 

[Trojan666ru]

I can't see you drawings from #81 have you removed it?

 

[ghwellsjr]

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)

I answered this post already but it wasn't a very good answer.

Now I want to show how each observer/object can use radar signals and other observations to determine the answers to all your questions. The important point is that they each will follow the exact same procedure that you described in your very first post where the observer sends signals to the other objects and watches for the echo to get back to him, along with his observation of the time on the other objects' clocks so that he can verify how far away they are from him. And most importantly, the apply Einstein's convention that light travels in all directions at c. It is this one fact that makes the simultaneity issues different in each frame.

So for this purpose, we need to have clocks attached to each object (not just inactive stopwatches that are started at a particular time). We will assume that all three clocks are synchronized to the Coordinate Time in the defining Inertial Reference Frame (IRF) that applies when all three objects are in mutual rest. Here is a spacetime diagram for the defining IRF with Proper Time markings for the red and green boxes. The Proper Time for the blue coordinator matches the Coordinate Time so I didn't make special markings for him. I had to continue the scenario on both ends to enable the radar signals to work:

I have also marked the same signals that are shown in post #61, namely a pair sent by the blue coordinator at his 1-year mark and received by the boxes at their 3-year marks to start their accelerations toward the coordinator and another pair sent by the blue coordinator at his 6.2-year mark and received by the boxes at their 7-year marks when their accelerations end and when they start their stopwatches (which we are not concerned with now) and when they echo signals back toward the coordinator and the other box. When those red and green signals pass through the half silvered mirror located with the coordinator, I show them as yellow as they progress to the other box.

Now I want to go through a little exercise to show how an observer can use radar signals and other observations to construct his own reference frame in which he is at rest. I will start with the blue coordinator because he is already at rest in the defining IRF and the process will be much more obvious. Here is a diagram showing the important blue radar signals that the coordinator sends out and the return reflections coming from the red and green boxes. The coordinator has to be sending out radar signals much more often than I am showing them but I have selected just a few that will make our process easier to perform:

For each radar signal that is sent, the observer keeps track of when it was sent according to the Proper Time on his own clock. After he receives the reflection, he keeps track of the received time and the observed Proper Time on the target. He keeps a log of these three numbers for each radar signal. Then he calculates a distance and a time for each set of numbers. These calculations use Einstein's convention that the radar signal takes the same amount of time to reach the target as it takes for the reflection to get back to the sender. The distance is merely the received time minus the sent time divided by two and then multiplied by the speed of light which in our example is 1. The time at which this distance applies is the average of the sent and received times. This will enable him to make a diagram showing his results as a function of his Proper Time. He is doing this separately for each object.

Let's go through one example at the bottom of the diagram. At his time of -6 years, the coordinator sends out a radar signal (in both directions but we can focus on just one). At his time of -2 years, he gets the reflection signal. He takes the difference, which is 4 and divides by 2 to to get a distance of 2 light-years. He knows that the distance is to the left for the red box and to the right for the green box. He averages -6 and -2 to get -4 as the time the box was 2 LY away. I hope this is clear. Now he just repeats this over and over again and from the calculations, he can draw a diagram just like the one we have been looking at. As he gets to the portions where the boxes are accelerating, his log of sent and received times will include times with fractional parts so we have to interpolate between the dots. Here is his final diagram:

Now we want to graduate to the red box who is going to make a diagram showing the position of the coordinator and the position of the green box as a function of his own Proper Time (marked off by the red dots). We start with the same original defining IRF and draw in the red radar signals sent by the red box and going to the blue coordinator. The reflections are shown in blue as they come back to the red box:

Next, we have a similar diagram showing the red radar signal going to green box and the green reflections coming back to the red box:

And here is the diagram that the red box constructs from all the radar signals and observations of the Proper Times of the other two objects:

I have drawn in the same signals that were in the first diagram in this post so that we can see that the signals arrive at the same times.

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?

It's because the red box is applying Einstein's convention that light travels at c in his rest frame.

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

No, in all frames, the blue coordinator sends signals to both boxes at the same time.

Or

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

Not suddenly but gradually as you can see in the last diagram. But again, this is because the red box is applying Einstein's convention to his rest frame. It's not something that he can see or something that is intrinsic to nature. It's a man-made convention (the man being Einstein).

I said at the start that I didn't like my previous answer. I didn't realize at that time that the what you said about the two distances is true but now I realize that those two distances are different in the red box's rest frame. But, keep in mind that we could do this exercise all over again from the point of view of the green box and everything would be opposite. Then the red box would receive the signals from the coordinator earlier than the green box. It all has to do with Einstein's convention and the fact that the farther away an object is, the earlier you have to send a radar signal to it and the later you get its reflection compared to a closer object.

Well, I hope this helps. Keep asking if you have more questions. But please study this response in detail first.

 

[ghwellsjr]

I can't see you drawings from #81 have you removed it?

No, at least not on purpose. I don't know why this happens. I can see them on my webpage which means I did upload them but for some reason after doing all the edits, they don't appear for others on the final submission even though they appear for me.

Anyway, I have re-uploaded them so hopefully you can see them now.

 

[georgir]

With acceleration as slow as in these examples, an important detail may be missed about relativity.
Acceleration, or in other words changes of reference frame, mean changes in what one would call "present".

Unlike velocity, there is practically no limit on acceleration, so it can even be approximated to instantaneous. Then distances and remote clocks may actually be said to suddenly change, i.e. objects seem to jump through space and time. But of course it is not a real, physical change, just a change in point of view.

 

[ghwellsjr]

With acceleration as slow as in these examples, an important detail may be missed about relativity.
Acceleration, or in other words changes of reference frame, mean changes in what one would call "present".

Acceleration does not require a change in reference frame and doesn't have to mean the kind of instantaneous changes in what one would call "present" that you are talking about. You can do that if you want but since "present" at remote locations is fickle, you can do exactly what I did in post #83 where the acceleration is slow to the scenario in post #81 where the acceleration is instantaneous and you will see a point of view for the red box that is based on actual measurements that the red box can make and it won't have any of the abrupt changes that you refer to.

Unlike velocity, there is practically no limit on acceleration, so it can even be approximated to instantaneous. Then distances and remote clocks may actually be said to suddenly change, i.e. objects seem to jump through space and time. But of course it is not a real, physical change, just a change in point of view.

I invite you to make a diagram for the point of view you think is required by abrupt acceleration for the scenario in post #81 and then I invite you to make one based on the process I describe in post #83. Can you do that?

 

[georgir]

I've already had lectures on radar measurement by you or maybe by DaleSpam, thank you.
The fact that an observer's own past and future motion affects what he would consider the current state of remote objects makes this a pretty useless system for me.
You and DaleSpam are both free to continue using it of course, but try not to completely distract people interested in learning the subject from the much simpler and objective system of direct application of the Lorentz transforms for the momentarily comoving inertial reference frame of the observer.

 

[WannabeNewton]

...but try not to completely distract people interested in learning the subject from the much simpler and objective system of direct application of the Lorentz transforms for the momentarily comoving inertial reference frame of the observer.

You do realize that this only works locally (in general) right? Oh the irony.

 

[ghwellsjr]

I've already had lectures on radar measurement by you or maybe by DaleSpam, thank you.
The fact that an observer's own past and future motion affects what he would consider the current state of remote objects makes this a pretty useless system for me.
You and DaleSpam are both free to continue using it of course, but try not to completely distract people interested in learning the subject from the much simpler and objective system of direct application of the Lorentz transforms for the momentarily comoving inertial reference frame of the observer.

An Inertial Reference Frame (IRF) is based on an observer's own past and future motion--namely that it be inertial. Also, an IRF is exactly identical to using radar measurements. The Lorentz transform only works between two IRF's. So if you object to radar measurements, then you must also object to using a comoving IRF.

And if you think it is much simpler, then I challenge you to make a diagram showing the red box's comoving IRF for this scenario:

Or if you object to the slow acceleration, do it for this scenario:

​

 

[georgir]

No point in giving me homework, I know that gradual acceleration would be a hell of a math puzzle for me, and that instant acceleration results in disjointed worldlines, with repeated or missing times.
But the fact remains, radar measurements are no better in other aspects, namely objectivity.

If you have two observers, A at rest relative to a subject some distance away, and B in inertial motion retative to them, and you get observer C that "jumps ship" from A to B at the moment when the two cross, C would not agree with A about the distsance or velocity of the subject a certain time before the jump, nor with B a certain time after it, despite being in exactly the same spot observing exactly the same things as them at the specific moment. Such a system is not something you'd see in a physics textbook, ever.

 

[Nugatory]

If you have two observers, A at rest relative to a subject some distance away, and B in inertial motion retative to them, and you get observer C that "jumps ship" from A to B at the moment when the two cross, C would not agree with A about the distance or velocity of the subject a certain time before the jump, nor with B a certain time after it, despite being in exactly the same spot observing exactly the same things as them at the specific moment. Such a system is not something you'd see in a physics textbook, ever.

I don't know that I've ever seen it in a textbook, but it's certainly fair game for discussion during office hours - that's where I first encountered it. There's also at least one thread somewhere here that discusses this situation.

It's not so interesting in this simple example. If the velocity of the subject relative to A and to C is the same, and then C changes speed, then of course the velocity of the subject relative to C will change and therefore no longer be equal to the velocity of the subject relative to A, and that's no great surprise.

It is more interesting (at least an exercise in demonstrating the consistency of the Lorentz transforms) if you combine this frame-jumping with the traditional twin paradox. The subject is the stay-at-home twin, C is the traveling twin, and at the turnaround point he jumps from his outbound spaceship onto another ship that happens to be traveling in the opposite direction, inbound. Now consider how C reconciles his recollection of events with the ship's log that's been maintained by the crew of the incoming ship.

 

[georgir]

I know that gradual acceleration would be a hell of a math puzzle for me

And luckily I don't need to solve it anyway, as people have done it already. It's an interesting solution, with time rates dependant on distance from the observer, an apparent event horizon behind him and what not...

 

[georgir]

If the velocity of the subject relative to A and to C is the same, and then C changes speed, then of course the velocity of the subject relative to C will change.

But radar method will tell you that the velocity of the subject changed a while before C changed his, and then again a while after, and that's the wonky part that will never fly.

 

[Dale]

But radar method will tell you that the velocity of the subject changed a while before C changed his, and then again a while after, and that's the wonky part that will never fly.

Why not? The coordinates are non inertial so why shouldn't the coordinate velocity change before and after?

 

[georgir]

Why not? The coordinates are non inertial so why shouldn't the coordinate velocity change before and after?

Because describing the current the state of the world should not depend on what I did once upon a time, and even less so on what I will do in the future, just on my current state as an observer. And any other observer that is momentarily identical to me should agree with me.

 

[PAllen]

Because describing the current the state of the world should not depend on what I did once upon a time, and even less so on what I will do in the future, just on my current state as an observer. And any other observer that is momentarily identical to me should agree with me.

Why on Earth would I accept such propositions? The whole reason the Einstein definition of simultaneity makes so much sense for inertial frames is the particular past (and future - to receive a signal from what I think of as now). So two observers whose past and future are different have no reason whatsoever to agree on simultaneity just because their current instantaneous velocity is the same. Any operational synchronization procedure they use will come out different.

FYI, it is a historic fact that Einstein never used or endorsed your proposition for dealing with accelerated motion in SR. He did, in one paper, use radar coordinates. But mostly he just chose to not bother with a 'frame' for accelerated observer at all. Any observation by the accelerated observer can be derived most simply by using any single inertial frame over whole period of interest.

 

[WannabeNewton]

Because describing the current the state of the world should not depend on what I did once upon a time, and even less so on what I will do in the future, just on my current state as an observer. And any other observer that is momentarily identical to me should agree with me.

This is absolute nonsense. Not only does the formalism of SR, and Einstein simultaneity for non-inertial observers in particular, disagree with you but also you're making hand-wavy arguments in favor of your utterly flawed intuition.

 

[PeterDonis]

Because describing the current the state of the world

And right here is your problem: you think that "the current state of the world" is a well-defined concept. It isn't. The world does not have a unique "current state". You, at a particular event on your worldline, have a current state, but that's because you are a localized object. The different ways of assigning a "current state" to the rest of the world, at events spacelike separated from a particular event on your worldline, are arbitrary conventions; none of them make any difference to any physical predictions.

This, by the way, points out another flaw in your argument: you are assuming that a given observer is *forced* to use a certain simultaneity convention (which is what assigning a meaning to "the current state of the world" amounts to). That is a necessary premise for your further claim that two observers who are momentarily comoving must use the same convention, which is what your claim that "the current state of the world" must be the same for two momentarily comoving observers amounts to. This claim is false because the premise is false: any observer, in any state of motion, can choose whatever simultaneity convention he likes. Simultaneity conventions are just that: conventions, arbitrary choices that do not affect any physical predictions.

 

[WannabeNewton]

For some reason this issue keeps coming up even though it's really not that non-intuitive. The problem is some people seem to be stuck on Aristotelian notions of time and the "now". In Galilean relativity there is a prescribed partitioning of space-time into classes of "now" so there is no issue but in SR things aren't so simple. The only thing nature provides us with is light cones. Furthermore clocks only keep time at their locations. An accelerated clock and a momentarily comoving inertial clock will only be inherently equal at their common location (assuming the accelerated clock has been calibrated so as to prevent deviations from inertial clock behavior) because again clocks only keep time of local events. When you have clocks separated in space you need to adopt an arbitrary convention that prescribes how to synchronize clocks i.e. a convention that prescribes when two clock readings are simultaneous. For spatially separated inertial clocks there is a natural such clock synchronization convention and this is the Einstein synchronization ("natural" because it is directly related to orthogonality with respect to the world lines of the clocks as defined by $\eta_{\mu\nu}$). For non-inertial clocks there is no such natural synchronization convention.

I would really recommend reading "Relativity and Geometry"-Torretti. It's a very cheap book and it has detailed discussions of clock synchronization/simultaneity.

 

[pervect]

I agree totally with the first part of Peter's comment. However, I have a few remarks about the following:

This, by the way, points out another flaw in your argument: you are assuming that a given observer is *forced* to use a certain simultaneity convention (which is what assigning a meaning to "the current state of the world" amounts to). That is a necessary premise for your further claim that two observers who are momentarily comoving must use the same convention, which is what your claim that "the current state of the world" must be the same for two momentarily comoving observers amounts to. This claim is false because the premise is false: any observer, in any state of motion, can choose whatever simultaneity convention he likes. Simultaneity conventions are just that: conventions, arbitrary choices that do not affect any physical predictions.

This part of the argument I think needs a more lengthly treatment. There was a quote from Einstein on this point earlier - there is a unique simultaneity convention that's required to make Newtonian mechanics work. So while choice of simultaneity may be optional, if you have an IRF and you want to make Newtonian mechanics work, your choice of simultaneity conventions isn't arbitrary, you must choose the required option.

"Making Newtonian mechanics" work is a bit vague. The issue is that in order for the momentum p to be equal to m v, mass * velocity, one needs the correct simultaneity convention. While it is possible to find a Lagrangian to give one the correct conserved momentum and equations of motion,the usual formulation of Newtonian mechanics as taught in high school does in fact _require_ certain choices about the "convention" of simultaneity. As Einstein pointed out, the same choices that make Newtonian mechanics work (in the above sense) also make Maxwell's equations work - which in turn makes the speed of light in a vacuum equal to "c".

Einstein also demonstrated that the "required" simultaneity conventions generate different sets of "simultaneous" points for different observers to the same event O. So there isn't any "universal" notion of simultaneity that makes Newtonian mechanics work for everyone.