Is the frame dragging part of the Special Relativity?

[SDG]

Originally Posted by HansMustermann

Not sure you understand what a black hole is. A black hole has an event horizon. Not every event horizon is a black hole or around a black hole.

Basically just like a bird has wings, but not everything that has wings is a bird. I just saw four wings on a helicopter, for example.

Additionally, unless I'm reading your graph wrong, there isn't even an event horizon there.

An event horizon ONLY happens when there's acceleration involved. Gravity IS acceleration, or at least equivalent, which is why a black hole has one. An accelerating body would also 'see' one behind it, although it would have to be accelerating really hard, and it's not really there for anyone else. Some guy standing on the ground next to the accelerating train would not actually see any event horizon there.

So basically not only there is no black hole there in any case, but if it's a constant speed motion (which seems to be the case in your graph) there isn't an event horizon at all either.

Except there is no motion, now flow of the time at the event A while there is motion and time flow from B to C at the back.
How do you explain that?

[Ziggurat]

Originally Posted by SDG

Two triplets travel to the right.
Do ct and ct' triplets agree on the ct'' time when they meet at t=14s?

Precision in language is important here. the unprimed and the single primed triplets will not have experienced the same proper time as as each other or as the double primed twin. They can still each calculate each other's proper time, and (if they do it right) they will all agree on that.

Quote:

The relative speed between ct' and ct'' is the same as between ct and ct'.

Yes. But the unprimed twin is not equivalent to the double primed twin from the perspective of the single primed twin. Why not? Because he accelerated, and he accelerated in a specific direction. That breaks the symmetry he sees between the single primed and double primed twins.

There is no way to go from your diagram to any of my three diagrams via a Lorenz transformation. But all three of my diagrams can be changed into each other by a Lorenz transformation.

[SDG]

Originally Posted by Ziggurat

Precision in language is important here. the unprimed and the single primed triplets will not have experienced the same proper time as as each other or as the double primed twin. They can still each calculate each other's proper time, and (if they do it right) they will all agree on that.
...

OK, what time there is going to be on ct'' clock when they meet at t=14s?

[Ziggurat]

Originally Posted by SDG

Is there an 'event horizon/black hole' at event A here?

There are no black holes in special relativity. There are only black holes in general relativity. Conceptually, event horizons exist in special relativity, but only for perpetually accelerating observers. If you don't accelerate forever, then there is no event horizon. So as a practical matter, they don't actually exist in special relativity.

But even on a conceptual level, I don't see any acceleration in your diagram. So how would there be any event horizon?

Quote:

This all goes back to what time dilation is and how it is reciprocal.

None of this is necessary. All you need to do is test the metric. I gave you the metric. It's pretty easy to show that proper time is invariant under Lorenz transformations. It's easy to see that the metric measures the proper time of a worldline. It's pretty easy to see that a straight line is the longest length for time-like separation in this metric. And it's pretty easy to see that the earthbound twin has a straight world line, while the traveling twin has a curved world line.

There is no paradox. There never was. There is only a lack of understanding. If you can accept that your understanding was wrong, then you might be able to reach a better understanding. If you're stuck on trying to prove everyone else wrong, you will never make progress, because we aren't.

[Ziggurat]

Originally Posted by SDG

OK, what time there is going to be on ct'' clock when they meet at t=14s?

Your diagram already has the answer.

What you're confused about is you don't understand how to view the proper time of the double primed triplet from the point of view of the single primed triplet. But the single primed triplet isn't an inertial observer. Unless you want to deal with his acceleration, then you either have pick either his outbound or his incoming reference frames, or you have to do it piecemeal. Either way, though, you have to contend with the fact that from the primed triplet's outgoing AND incoming reference frames, the primed triplet and the double primed triplet DO NOT turn around at the same time. That screws up your attempt to compare the unprimed triplet's view of the single primed triple with the single primed triplet's view of the double primed triplet. They aren't the same. Acceleration changes the situation. You don't need to know what the acceleration was, but you cannot simply ignore that it exists. That's simply wrong.

[HansMustermann]

Originally Posted by SDG

Except there is no motion, now flow of the time at the event A while there is motion and time flow from B to C at the back.
How do you explain that?

I'm afraid you've lost me again. You'll have to explain more detailed exactly what you mean. There's only so much I can deduce from a graph, and there's no telling if I'm even deducing the same thing you had in mind.

For a start, which is the primed frame? The one at the front, the one at the back, one of the guys at the station, or what?

Second, I'm still not sure what do you mean no motion and no time flow in any of those situations. If I even understand what they are.

I mean, the front and back observers don't move with respect to each other, because basically that's how you defined the problem. The back one is at a fixed coordinate, so yeah, the speed is zero with regards to the other guy.

The "no time flow" is more difficult for me to understand. If you take the two measurements synchronously (bearing in mind what that means in relativity), then sure, they both happen at the same moment in that reference frame. But that just means they're at the same time, not that time generally stops flowing.

Short of actually travelling at c, there is no way to make time stop flowing. Or more accurately said, you can't stop yourself from moving along the time axis.

But anyway, sure, you took two measurements in two different points, at the same moment in time. Well, at the same moment in the reference frame of the guy at the front of the train. I don't see any particular significance of that. Sure, it can happen that two different events happen at the same t, just like they could happen at the same x at different times.

The same events however, can occupy wildly different coordinates and times in some other frame of reference. It's the same thing I was trying to illustrate with that spaceship and hangar example on page 1. Two events that have the same t coordinate in one frame, can have very different times in another frame.

It's just the normal Lorentz transformations, really: t' = γ(t - vx/c²). Even when t in one coordinate system is the same for two events happening at coordinates x₁ and x₂, when v is not zero, and the coordinates are different, you obviously get different t'₁ and t'₂ for the same t.

Not sure how a black hole even enters the equation there in any way.

[SDG]

Originally Posted by Ziggurat

Your diagram already has the answer.

What you're confused about is you don't understand how to view the proper time of the double primed triplet from the point of view of the single primed triplet. But the single primed triplet isn't an inertial observer. Unless you want to deal with his acceleration, then you either have pick either his outbound or his incoming reference frames, or you have to do it piecemeal. Either way, though, you have to contend with the fact that from the primed triplet's outgoing AND incoming reference frames, the primed triplet and the double primed triplet DO NOT turn around at the same time. That screws up your attempt to compare the unprimed triplet's view of the single primed triple with the single primed triplet's view of the double primed triplet. They aren't the same. Acceleration changes the situation. You don't need to know what the acceleration was, but you cannot simply ignore that it exists. That's simply wrong.

All three triplets have to be in the same frame when ct' and ct'' stop at their respective return points.
This is right in the middle. There is no issue with the simultaneity. The simultaneity is the horizontal line.
This is why I insist on the acceleration because the acceleration has a stop in the middle and all triplets have to be in the same frame.
Your argument does not hold.

[theprestige]

I like how a seemingly innocent question about frame-dragging turned into a stalking horse for denying the resolution of the twin paradox and dismissing SR altogether.

[SDG]

Originally Posted by Ziggurat

There are no black holes in special relativity. There are only black holes in general relativity. Conceptually, event horizons exist in special relativity, but only for perpetually accelerating observers. If you don't accelerate forever, then there is no event horizon. So as a practical matter, they don't actually exist in special relativity.

But even on a conceptual level, I don't see any acceleration in your diagram. So how would there be any event horizon?



None of this is necessary. All you need to do is test the metric. I gave you the metric. It's pretty easy to show that proper time is invariant under Lorenz transformations. It's easy to see that the metric measures the proper time of a worldline. It's pretty easy to see that a straight line is the longest length for time-like separation in this metric. And it's pretty easy to see that the earthbound twin has a straight world line, while the traveling twin has a curved world line.

There is no paradox. There never was. There is only a lack of understanding. If you can accept that your understanding was wrong, then you might be able to reach a better understanding. If you're stuck on trying to prove everyone else wrong, you will never make progress, because we aren't.

The back of the train is moving from the event B to the event C.
The front of the train is not moving at all and the time is not flying as well when the front is stuck the event A.
How can the back train observer to see something different when looking out compared to platform observer looking in.
Is the reality different based on who is looking?

[Ziggurat]

Originally Posted by SDG

https://i.imgur.com/9RSrQN7.png

All three triplets have to be in the same frame when ct' and ct'' stop at their respective return points.
This is right in the middle. There is no issue with the simultaneity. The simultaneity is the horizontal line.
This is why I insist on the acceleration because the acceleration has a stop in the middle and all triplets have to be in the same frame.
Your argument does not hold.

You clearly don’t even understand my argument.

Pick a reference frame. Everything is easier if you use an inertial frame, but it doesn’t need to be inertial. Do the calculations in that frame. Pick another frame. Do the calculations again. The results will agree, if you don’t screw up.

In no reference frame, inertial or otherwise, do any of your three triplets make equivalent world lines. In all frames, all three triplets are different.

[SDG]

Originally Posted by Ziggurat

You clearly don’t even understand my argument.

Pick a reference frame. Everything is easier if you use an inertial frame, but it doesn’t need to be inertial. Do the calculations in that frame. Pick another frame. Do the calculations again. The results will agree, if you don’t screw up.

In no reference frame, inertial or otherwise, do any of your three triplets make equivalent world lines. In all frames, all three triplets are different.

OK, please, have a look here:




Now have a look here:





Every acceleration has own P point.
What is tangent to the worldline at that point?
They all point up, all triplets are in the same frame.
If u(B+) is agreed upon at the top then there is u(P) in the middle.
I know and I agree there is a constant acceleration in P but the motion stopped.
The simultaneity is the horizontal line.

[SDG]

Originally Posted by HansMustermann

I'm afraid you've lost me again. You'll have to explain more detailed exactly what you mean. There's only so much I can deduce from a graph, and there's no telling if I'm even deducing the same thing you had in mind.

For a start, which is the primed frame? The one at the front, the one at the back, one of the guys at the station, or what?

Second, I'm still not sure what do you mean no motion and no time flow in any of those situations. If I even understand what they are.

I mean, the front and back observers don't move with respect to each other, because basically that's how you defined the problem. The back one is at a fixed coordinate, so yeah, the speed is zero with regards to the other guy.

The "no time flow" is more difficult for me to understand. If you take the two measurements synchronously (bearing in mind what that means in relativity), then sure, they both happen at the same moment in that reference frame. But that just means they're at the same time, not that time generally stops flowing.

Short of actually travelling at c, there is no way to make time stop flowing. Or more accurately said, you can't stop yourself from moving along the time axis.

But anyway, sure, you took two measurements in two different points, at the same moment in time. Well, at the same moment in the reference frame of the guy at the front of the train. I don't see any particular significance of that. Sure, it can happen that two different events happen at the same t, just like they could happen at the same x at different times.

The same events however, can occupy wildly different coordinates and times in some other frame of reference. It's the same thing I was trying to illustrate with that spaceship and hangar example on page 1. Two events that have the same t coordinate in one frame, can have very different times in another frame.

It's just the normal Lorentz transformations, really: t' = γ(t - vx/c²). Even when t in one coordinate system is the same for two events happening at coordinates x₁ and x₂, when v is not zero, and the coordinates are different, you obviously get different t'₁ and t'₂ for the same t.

Not sure how a black hole even enters the equation there in any way.

Hans,
slowly...
The Minkowski space-time diagram.
An event on the Minkowski space-time diagram is a point.
Let us assume there is a smallest tick possible, Planck time interval (10^-43s) added to the point.
The result is already another point in the diagram because the time on the worldline moved.
Does this make sense?

[Ziggurat]

Originally Posted by SDG

The back of the train is moving from the event B to the event C.
The front of the train is not moving at all and the time is not flying as well when the front is stuck the event A.

Why would the front of the train be stuck at A? It isn’t. If you think it’s stuck because it’s at an event horizon, then I will again point out that there is no acceleration in this problem. There is no event horizon for any observer in this problem. Certainly observers at the back of the train will not observe one at A.

Quote:

How can the back train observer to see something different when looking out compared to platform observer looking in.
Is the reality different based on who is looking?

Of course it is. The platform observer sees the platform as stationary and the train as moving. The passenger on the train sees the train as stationary and the platform as moving.

Some things are invariant, and don’t depend on the observer. But lots of things do. This isn’t unique to special relativity, it’s true in Newtonian mechanics too.

[Ziggurat]

Originally Posted by SDG

Every acceleration has own P point.
What is tangent to the worldline at that point?
They all point up, all triplets are in the same frame.
If u(B+) is agreed upon at the top then there is u(P) in the middle.
I know and I agree there is a constant acceleration in P but the motion stopped.
The simultaneity is the horizontal line.

So what? Are you under the impression that this means the clocks should sync up? It doesn’t mean that at all. In fact, it’s not even relevant what each of them thinks is simultaneous unless you want to work from their reference frames, but those frames aren’t inertial, and the amount of non inertiality varies between them. In no case is there any equivalence between any of them.

[SDG]

Originally Posted by Ziggurat

So what? Are you under the impression that this means the clocks should sync up? It doesn’t mean that at all. In fact, it’s not even relevant what each of them thinks is simultaneous unless you want to work from their reference frames, but those frames aren’t inertial, and the amount of non inertiality varies between them. In no case is there any equivalence between any of them.

Let us assume that ct' and ct'' undergo acceleration/deceleration and they stop at the middle point P, just the first leg.
What does the ct'' clock shows?

[Ziggurat]

Originally Posted by SDG

Let us assume that ct' and ct'' undergo acceleration/deceleration and they stop at the middle point P, just the first leg.
What does the ct'' clock shows?

Your diagram already shows the answer. Why do you have a problem with that answer?

[SDG]

Originally Posted by Ziggurat

Why would the front of the train be stuck at A? It isn’t. If you think it’s stuck because it’s at an event horizon, then I will again point out that there is no acceleration in this problem. There is no event horizon for any observer in this problem. Certainly observers at the back of the train will not observe one at A.



Of course it is. The platform observer sees the platform as stationary and the train as moving. The passenger on the train sees the train as stationary and the platform as moving.

Some things are invariant, and don’t depend on the observer. But lots of things do. This isn’t unique to special relativity, it’s true in Newtonian mechanics too.

I am going to be specific.
At event A (x=x'=0 and t=t'=0) the back train observer reads -6.9282cs from the platform frame.
Correct?
So the platform is moving in the train frame and the platform ruler is length contracted into the train frame.
Agreed?

[SDG]

Originally Posted by Ziggurat

Your diagram already shows the answer. Why do you have a problem with that answer?

So you agree it is ct''=1s''.
How is it possible when Lorentz factor is 2 between ct' and ct'' and ct' has 3.5s'?

[Robin]

Originally Posted by SDG

Based on comoving inertial frames we can extrapolate:

https://i.imgur.com/9RSrQN7.png

So based on textbooks the time dilation is being an effect of the speed and not the acceleration.
This statement is supported by the Lorentz factor that is based on the speed and not an acceleration,
SDG

Time dilation due to velocity is the wrong way round for the twin paradox, the paradox is because of the acceleration.

Your diagram represents an idealisation where smooth acceleration and deceleration are replaced by an instantaneous change in velocity.at three points.

Try drawing lines of simultaneity on.that diagram for the segments AP and then for PB.

You will see the time dilation goes the wrong way for the twin paradox, the stationary observer is aging more slowly than the moving observer.

But notice what happens at P, the stationary observer "ages" instantly and then continues to age more slowly than the moving observer.

Now try breaking that journey into.four segments instead of two.and draw your lines of simultaneity.

Then try it with eight segments.

The more segments you add the closer you get to the smooth acceleration. And your lines of simultaneity will show you how the twin paradox depends on acceleration.

Try it.

[SDG]

Originally Posted by Robin

Time dilation due to velocity is the wrong way round for the twin paradox, the paradox is because of the acceleration.

Your diagram represents an idealisation where smooth acceleration and deceleration are replaced by an instantaneous change in velocity.at three points.

Try drawing lines of simultaneity on.that diagram for the segments AP and then for PB.

You will see the time dilation goes the wrong way for the twin paradox, the stationary observer is aging more slowly than the moving observer.

But notice what happens at P, the stationary observer "ages" instantly and then continues to age more slowly than the moving observer.

Now try breaking that journey into.four segments instead of two.and draw your lines of simultaneity.

Then try it with eight segments.

The more segments you add the closer you get to the smooth acceleration. And your lines of simultaneity will show you how the twin paradox depends on acceleration.

Try it.

Robin,
you are responding to 'very old post'
I'll wait till you catch up.

[Robin]

Originally Posted by SDG

Robin,

you are responding to 'very old post'

I'll wait till you catch up.

The thread is only a day old.

And a faulty premise is a faulty premise no matter how long ago it was stated.

[Ziggurat]

Originally Posted by SDG

So you agree it is ct''=1s''.
How is it possible when Lorentz factor is 2 between ct' and ct'' and ct' has 3.5s'?

Sorry, I spoke too quickly. So much of your stuff is taken from textbooks, I assumed that picture was too, but it isn't.

Yes, it's 1 second. How is it possible that it's 1 second and not 1.75 seconds? Easy: you forgot that simultaneity is relative. If you transform to the outbound single primed frame, then the double primed frame will turn around before the single primed frame does. Go look at my diagrams again. Note that when you change reference frames, the time at which the turnaround happens changes too. The turnaround point is no longer halfway through the journey.

And your objection about them being simultaneous halfway through the turnaround is irrelevant. You either switched frames already (in which case you're no longer in a frame where gamma for the double prime is 2), or you're working in an accelerating reference frame. You can't just mix and match without accounting for that mixing and matching, and you aren't doing that.

Remember what I said earlier about accelerating reference frames introducing artificial stuff? In Newtonian mechanics, those are fictitious forces like centripetal force. In special relativity, it gets weirder. Time flows differently at a distance for special relativity in an accelerating frame. It can even flow backwards. It doesn't really flow backwards in any meaningful way, it's just a consequence of your bad choice of coordinates, but that's still what you get when you choose to use an accelerating reference frame.

[Ziggurat]

Originally Posted by SDG

I am going to be specific.
At event A (x=x'=0 and t=t'=0) the back train observer reads -6.9282cs from the platform frame.
Correct?

No, that's not correct. That's not even wrong. An event is a single point in spacetime. The back of the train isn't at x=0 t=0. It doesn't even make sense to say what something else reads at an event some distance away from it.

Now, you COULD rationally ask how far away from A some other event is. This is an invariant quantity. You could also ask how far away some world line is from A, in some specified reference frame. The answer is not invariant, and will depend on your choice of reference frames. These are different questions from each other, although there is some reference frame in which the answers will be the same. And they are also different questions from what you asked, which doesn't actually make sense.

Quote:

So the platform is moving in the train frame and the platform ruler is length contracted into the train frame.
Agreed?

In the train's frame, the platform is contracted, yes. And in the platform's frame, the train is contracted. What's your point? Do you imagine that this leads to a paradox as well? I promise you, it doesn't. You aren't the first person to think it does. The only question is, will you actually learn why it doesn't?

[SDG]

Originally Posted by Ziggurat

Sorry, I spoke too quickly. So much of your stuff is taken from textbooks, I assumed that picture was too, but it isn't.

Yes, it's 1 second. How is it possible that it's 1 second and not 1.75 seconds? Easy: you forgot that simultaneity is relative. If you transform to the outbound single primed frame, then the double primed frame will turn around before the single primed frame does. Go look at my diagrams again. Note that when you change reference frames, the time at which the turnaround happens changes too. The turnaround point is no longer halfway through the journey.

And your objection about them being simultaneous halfway through the turnaround is irrelevant. You either switched frames already (in which case you're no longer in a frame where gamma for the double prime is 2), or you're working in an accelerating reference frame. You can't just mix and match without accounting for that mixing and matching, and you aren't doing that.

Remember what I said earlier about accelerating reference frames introducing artificial stuff? In Newtonian mechanics, those are fictitious forces like centripetal force. In special relativity, it gets weirder. Time flows differently at a distance for special relativity in an accelerating frame. It can even flow backwards. It doesn't really flow backwards in any meaningful way, it's just a consequence of your bad choice of coordinates, but that's still what you get when you choose to use an accelerating reference frame.

The bold part is just wrong.
You either know how to draw the simultaneity line or not.
They stopped at respective P events. Why there is a time discrepancy?
Can you be reasonable?

Edit: There is no frame switching. The triplets are in the same frame at the beginning and midway through at P events/points.
Couple of them accelerate in between but what is the time at the P events when they are in the SAME frame?

[SDG]

Originally Posted by Ziggurat

No, that's not correct. That's not even wrong. An event is a single point in spacetime. The back of the train isn't at x=0 t=0. It doesn't even make sense to say what something else reads at an event some distance away from it.

Now, you COULD rationally ask how far away from A some other event is. This is an invariant quantity. You could also ask how far away some world line is from A, in some specified reference frame. The answer is not invariant, and will depend on your choice of reference frames. These are different questions from each other, although there is some reference frame in which the answers will be the same. And they are also different questions from what you asked, which doesn't actually make sense.



In the train's frame, the platform is contracted, yes. And in the platform's frame, the train is contracted. What's your point? Do you imagine that this leads to a paradox as well? I promise you, it doesn't. You aren't the first person to think it does. The only question is, will you actually learn why it doesn't?

There is the event A when the front of the train aligns with the origin of the platform frame.
Therefore x=x'=0 and t=t'=0.
There are two simultaneity lines going through the event A.
One for train frame and one for the platform.
We follow the train simultaneity line.
The train back observer at -3.4641cs' and at t'=0 sees -6.9282cs outside and time -6s.
The platform ruler is Lorentz contracted to the train frame.
To clarify the diagram for you.
Yes, it leads to contradiction. It leads to the twin paradox that is not resolved.
The twin paradox is not resolved, the triplet paradox is not resolved.
You cannot explain the times at P of ct' and ct'' triplets.

[SDG]

Originally Posted by Robin

The thread is only a day old.

And a faulty premise is a faulty premise no matter how long ago it was stated.

I'll wait till you catch up with the accelerations and the triplet paradox.

[Ziggurat]

Originally Posted by SDG

The bold part is just wrong.
You either know how to draw the simultaneity line or not.
They stopped at respective P events. Why there is a time discrepancy?
Can you be reasonable?

There's a clock difference. Clocks measure proper time, they do not measure coordinate time. The fact that the turnaround is simultaneous in a particular reference frame doesn't mean that different clocks will be synchronous.

Quote:

Edit: There is no frame switching. The triplets are in the same frame at the beginning and midway through at P events/points.

But not during. And it's during the trip that your double primed triplet has a gamma of 2 relative to your single primed triplet. You switched from the outbound frame of the single prime twin to the mid turnaround frame of the single prime twin. These are either different inertial frames, or the same non-inertial frame.

Quote:

Couple of them accelerate in between

Then they aren't inertial, and Lorenz time dilation doesn't suffice for the calculation. You can only do the calculation with Lorenz time dilation alone if you only work within inertial reference frames. And you aren't doing that.

[Robin]

Originally Posted by SDG

I'll wait till you catch up with the accelerations and the triplet paradox.

You seem to be the only one who hasn't caught up on the triplet "paradox"

[Robin]

Originally Posted by SDG

The bold part is just wrong.
You either know how to draw the simultaneity line or not.
They stopped at respective P events. Why there is a time discrepancy?
Can you be reasonable?

Edit: There is no frame switching. The triplets are in the same frame at the beginning and midway through at P events/points.
Couple of them accelerate in between but what is the time at the P events when they are in the SAME frame?

If the third triplet changes velocity instantaneously at P then by definition there is no "halfway through" the turnaround.

It simply skips T/2 in the original frame.

On the other hand if it changes direction in a short finite time then there is no discrepancy, each of the three will have the same time.

This is what I was pointing out to you before.

[HansMustermann]

Originally Posted by SDG

The back of the train is moving from the event B to the event C.
The front of the train is not moving at all and the time is not flying as well when the front is stuck the event A.

That doesn't even make any sense. An event is a single point in space-time. It doesn't mean you're "stuck" there in any meaningful sense. You took a measurement for where that point is at t=0, then you didn't take another one.

It's like saying I put a ball at the top of a ramp at exactly 8 AM. That's the event: the time and where the ball is. Maybe I did look where the ball is 1s later, or maybe I didn't, but in either case another measurement would be another event. But even if I didn't take another one, it just means I only know one event; it doesn't mean that the ball is stuck at the top of the ramp in any meaningful way.

Originally Posted by SDG

How can the back train observer to see something different when looking out compared to platform observer looking in.
Is the reality different based on who is looking?

Depending on what you mean by that, quite easily. That's kinda the whole point of the Lorentz transform.

[Pixel42]

Guys, we've already been through all this with SDG on Bjarne's "Relativity is about to fall apart" thread. He does not grasp the basic principles of Special Relativity and refuses to even try to do so, preferring to cling to his misunderstanding so he can claim to have found a problem with it.

[SDG]

Originally Posted by Ziggurat

There's a clock difference. Clocks measure proper time, they do not measure coordinate time. The fact that the turnaround is simultaneous in a particular reference frame doesn't mean that different clocks will be synchronous.



But not during. And it's during the trip that your double primed triplet has a gamma of 2 relative to your single primed triplet. You switched from the outbound frame of the single prime twin to the mid turnaround frame of the single prime twin. These are either different inertial frames, or the same non-inertial frame.



Then they aren't inertial, and Lorenz time dilation doesn't suffice for the calculation. You can only do the calculation with Lorenz time dilation alone if you only work within inertial reference frames. And you aren't doing that.

I'd like to address the acceleration. This is one of our previous posts regarding your triplet diagram.

Originally Posted by Ziggurat

Originally Posted by SDG

There is a contradiction.
The green triplet measures acceleration on his accelerometer in the first leg in diagram 1.
The green triplet DOES NOT measure any acceleration on his accelerometer in the first leg in diagram 3.
These are two distinct and different scenarios.
Your diagrams do not describe the same stuff.

No he doesn't. It's a straight line. There is no acceleration during the leg. Any acceleration before that leg can be made an arbitrarily small perturbation.

Yes, they do. They really do.

Your example, the green triple has to undergo miraculous instantaneous change of the frames at the event P (to keep 'P' in sync with my diagrams).
It means that the green triplet has to decelerate into the red triplet frame, do we agree on that?
This is the basis for my claim that there is a contradiction because in diagram 3 there is not deceleration to event/point P. Only acceleration from the event/point P.

Now you are talking about proper time measured by the clock.
I am glad you do that because we agree that ct'' observer can have only one value on his clock.
This means that we can extrapolate a red line (figure 2.9 below) that is equal to the proper time.



Question, is the red line OK to be used for time dilation calculation without acceleration, to simplify?
If we agree then it is logical to assume a real stop at the event/point P as well in spite of 'the jump' drawn currently by every physicist.
Agreed?

[SDG]

Originally Posted by Robin

If the third triplet changes velocity instantaneously at P then by definition there is no "halfway through" the turnaround.

It simply skips T/2 in the original frame.

On the other hand if it changes direction in a short finite time then there is no discrepancy, each of the three will have the same time.

This is what I was pointing out to you before.

Stop this nonsense.

[SDG]

Originally Posted by HansMustermann

That doesn't even make any sense. An event is a single point in space-time. It doesn't mean you're "stuck" there in any meaningful sense. You took a measurement for where that point is at t=0, then you didn't take another one.

It's like saying I put a ball at the top of a ramp at exactly 8 AM. That's the event: the time and where the ball is. Maybe I did look where the ball is 1s later, or maybe I didn't, but in either case another measurement would be another event. But even if I didn't take another one, it just means I only know one event; it doesn't mean that the ball is stuck at the top of the ramp in any meaningful way.



Depending on what you mean by that, quite easily. That's kinda the whole point of the Lorentz transform.

There is an event A and there is a simultaneity line.
The event A has 0s' on the train ct' clock at x'=0cs', the front of the train.
Every train observer along the simultaneity line has this time on his clock.
None of these observers can add a single Planck interval. Agreed?

I am going to assume yes, at the moment.
In this case the same applies to the platform observer.
The event A has 0s on the platform ct clock at the origin x=0cs.
Every platform observer along the simultaneity line has this time on his clock.
None of these observers can add a single Planck interval.

The events B and C were chosen based on the Lorentz transformation and the length contraction.
The problem is the events B and C are bi-located in the Minkowski diagram.
The respective back observers at their respective time 0 are not looking at each other.
Remember, they cannot add a simple one Planck time interval.

The Special Relativity is reciprocal.

[Ziggurat]

Originally Posted by SDG

Your example, the green triple has to undergo miraculous instantaneous change of the frames at the event P (to keep 'P' in sync with my diagrams).

No he doesn't. It should be sufficiently quick compared to the other timescales of the problem so that we can leave it out of the calculations, but it doesn't have to be instantaneous.

Quote:

It means that the green triplet has to decelerate into the red triplet frame, do we agree on that?
This is the basis for my claim that there is a contradiction because in diagram 3 there is not deceleration to event/point P. Only acceleration from the event/point P.

I can't make heads or tales of what you're saying. There is no P in my diagram. There is a P in your diagram, but your diagram isn't my diagram. And I have no idea what you're trying to convey by the distinction between "deceleration to" and "acceleration from". There is no real distinction between deceleration and acceleration. And since your P point is in the middle of continuous acceleration, it involves both acceleration to and acceleration from that point.

In short, you're babbling nonsense.

Quote:

Now you are talking about proper time measured by the clock.

Of course I am. Because that's what clocks measure: proper time. They don't measure coordinate time.

Quote:

Question, is the red line OK to be used for time dilation calculation without acceleration, to simplify?

I've been saying that all along.

Quote:

If we agree then it is logical to assume a real stop at the event/point P as well in spite of 'the jump' drawn currently by every physicist.
Agreed?

The traveler on the red path will be momentarily stationary with respect to the reference frame of the earthbound twin, yes. But you've invested this statement with meaning that it never had, and that you can't explain in anything approaching a coherent manner.

You might have heard the statement, "if everyone around you is a jerk, it's you". The same applies here: if everyone around you is wrong, it's you.

You aren't a genius. You aren't special. If you think that the entire edifice of modern physics is built on a fraud, well, no. You're the one who made the error, not everyone else.

[Ziggurat]

Originally Posted by SDG

There is an event A and there is a simultaneity line.

There is not "a" line of simultaneity intersecting A. There are an infinitude of such lines.

Quote:

The event A has 0s' on the train ct' clock at x'=0cs', the front of the train.
Every train observer along the simultaneity line has this time on his clock.

If they have synchronized their watches with each other within the train's reference frame, sure.

Quote:

The problem is the events B and C are bi-located in the Minkowski diagram.

What do you mean, "bi-located"? This is not standard terminology in relativity. Do you mean that they have the same spatial coordinates in the train's reference frame? Yes, they do. Why would that be a problem? It isn't.

Quote:

The respective back observers at their respective time 0 are not looking at each other.

I don't know what you mean by respective back observers. There's an observer at the back of the train. I don't recall you defining multiple observers on the platform. And what do you mean by they are not looking at each other? And why is that a problem?

Quote:

The Special Relativity is reciprocal.

When a situation is symmetric, yes. But often, an imagine symmetry isn't an actual symmetry. That's the case in the twin paradox. You may have inadvertently done that here as well, but your description isn't coherent enough to even make that determination.

[SDG]

Originally Posted by Ziggurat

No he doesn't. It should be sufficiently quick compared to the other timescales of the problem so that we can leave it out of the calculations, but it doesn't have to be instantaneous.



I can't make heads or tales of what you're saying. There is no P in my diagram. There is a P in your diagram, but your diagram isn't my diagram. And I have no idea what you're trying to convey by the distinction between "deceleration to" and "acceleration from". There is no real distinction between deceleration and acceleration. And since your P point is in the middle of continuous acceleration, it involves both acceleration to and acceleration from that point.

In short, you're babbling nonsense.



Of course I am. Because that's what clocks measure: proper time. They don't measure coordinate time.



I've been saying that all along.



The traveler on the red path will be momentarily stationary with respect to the reference frame of the earthbound twin, yes. But you've invested this statement with meaning that it never had, and that you can't explain in anything approaching a coherent manner.

You might have heard the statement, "if everyone around you is a jerk, it's you". The same applies here: if everyone around you is wrong, it's you.

You aren't a genius. You aren't special. If you think that the entire edifice of modern physics is built on a fraud, well, no. You're the one who made the error, not everyone else.

It appears you do not understand the motion.
Please, read and correct your mistakes.

[SDG]

Originally Posted by Ziggurat

There is not "a" line of simultaneity intersecting A. There are an infinitude of such lines.



If they have synchronized their watches with each other within the train's reference frame, sure.



What do you mean, "bi-located"? This is not standard terminology in relativity. Do you mean that they have the same spatial coordinates in the train's reference frame? Yes, they do. Why would that be a problem? It isn't.



I don't know what you mean by respective back observers. There's an observer at the back of the train. I don't recall you defining multiple observers on the platform. And what do you mean by they are not looking at each other? And why is that a problem?



When a situation is symmetric, yes. But often, an imagine symmetry isn't an actual symmetry. That's the case in the twin paradox. You may have inadvertently done that here as well, but your description isn't coherent enough to even make that determination.

The bold part.
They do not have the same space-time coordinates.
If the back train observer starts moving from the event B to event C it means we are going to add time across the train frame.
Not only the back but the front as well.
It means the initial alignment at the event A is not true any more.
It means going from B to C cannot be done without breaking the A.
This is the problem because C and A cannot happen for the platform observer without reading different numbers.
The paradox.

This leads to different coordinate times based on the different observers and there could be only one proper time along a worldline.

[Ziggurat]

Originally Posted by SDG

The bold part.
They do not have the same space-time coordinates.

I didn't say space-time coordinates. I said spatial coordinates. In other words, in that frame, those two events happen at the same place. Not the same time.

Quote:

If the back train observer starts moving from the event B to event C it means we are going to add time across the train frame.
Not only the back but the front as well.
It means the initial alignment at the event A is not true any more.

The front of the train doesn't stay at A. Not even if it crashes. Are you talking about a scenario where the train crashes? Is that what you're trying to get at? What happens to the back of the train if the front of the train stops? Because it you are, well, there's no paradox there, but there is something quite interesting which we could talk about. If not, then I have no idea what you're going on about.

Quote:

This leads to different coordinate times based on the different observers and there could be only one proper time along a worldline.

The front of the train doesn't have to trace out the same worldline as the back of the train. It doesn't have to have the same proper time. And proper time isn't coordinate time.

[Ziggurat]

Originally Posted by SDG

It appears you do not understand the motion.
Please, read and correct your mistakes.

I understand the diagram. I don't understand you. I cannot correct that. Either you can make yourself comprehensible, or you cannot.