[Jaaanosik]

Try this:

turnaround.jpg (60.55 kB . 310x458 - viewed 2822 times)

Halc,
Do you claim this is a correct deceleration/acceleration?
The back of the train car deceleration/acceleration does not match what platform observers see due to the simultaneity.
The front of the train car has instantaneous change of frames.
What is this space-time diagram?
Jano

[Jaaanosik]

The Triplets Challenge
Here is a very smart book:





Now the Triplets Paradox.



The left triplet travels left, the right triplet travels right, the third one does not travel.
I challenge every relativist out there to show us a solution what time will be displayed on the left triplet clock and on the right triplet clock when all triplets meet again!
Jano

[Halc]

In the returning twin's ct'' frame, he's 4.57 seconds old.

where did you get that number???
The 4s'' in the ct'' frame are 28s' in the ct' frame.

Yes, but nobody at the reunion has reason to use the ct' frame.  I said what frames I used, which are the respective frames of each of the two twins comparing notes at the reunion.

Do you claim this is a correct deceleration/acceleration?

It is a correct illustration of a very rapid turnaround with negligible time to reverse the front.  It is not instant on the right, but higher G than the resolution of the picture.

The back of the train car deceleration/acceleration does not match what platform observers see due to the simultaneity.

Of course the observations don't match.  The people on the platform don't see it until the light from the maneuver gets to them. They're not present with the observers in the train.  They measure the duration as being longer than the guy in the back of the train.

The left triplet travels left, the right triplet travels right, the third one does not travel.
I challenge every relativist out there to show us a solution what time will be displayed on the left triplet clock and on the right triplet clock when all triplets meet again!

Obviously the same time, due to symmetry.
Assuming the middle clock reads 1 (year? The bits you quote don't indicate units) at the end, it appears that both traveling clocks will read about 0.875 at reunion.  What's supposed to be even remotely different about that situation compared to the first one with only one traveler?

[Jaaanosik]

Yes, but nobody at the reunion has reason to use the ct' frame.  I said what frames I used, which are the respective frames of each of the two twins comparing notes at the reunion.
It is a correct illustration of a very rapid turnaround with negligible time to reverse the front.  It is not instant on the right, but higher G than the resolution of the picture.

The back of the train car deceleration/acceleration does not match what platform observers see due to the simultaneity.

Of course the observations don't match.  The people on the platform don't see it until the light from the maneuver gets to them. They're not present with the observers in the train.  They measure the duration as being longer than the guy in the back of the train.

There is a grid of inertial observers on the platform.
Their observation does not match your space-time diagram.

The left triplet travels left, the right triplet travels right, the third one does not travel.
I challenge every relativist out there to show us a solution what time will be displayed on the left triplet clock and on the right triplet clock when all triplets meet again!

Obviously the same time, due to symmetry.
Assuming the middle clock reads 1 (year? The bits you quote don't indicate units) at the end, it appears that both traveling clocks will read about 0.875 at reunion.  What's supposed to be even remotely different about that situation compared to the first one with only one traveler?

The Relativity says: "When two observers move in relation to each other then they cannot have the same proper time on their clocks."
That's it, that's the LT.

Does the left triplet move in relation to the right triplet?
If yes then they cannot have the same time on their clocks!!!
This is the Relativity!!!

If they end up with the same time on their clocks as per the middle triplet then the middle triplet reference frame is preferred one!
Jano

[Halc]

There is a grid of inertial observers on the platform.
Their observation does not match your space-time diagram.

The spacetime diagram is in the frame of the platform, so by definition, the 'grid' observers on said platform must witness the events that occur at their respective locations.  Translation: your statement above is self contradictory.

If you say my placement of events/worldlines is wrong, that's a different assertion, but you didn't say that this time.  You seem to say something different each time, but consistent in being wrong at least.

I drew a high-acceleration reverse of a million km rigid object moving at .866c, centered at t=4 platform time. In that regard, the diagram is exactly correct.  Your diagrams depict a non-rigid turnaround since the proper length of the object is often too long or too short. That doesn't make it wrong, it just means your train needs to be rubber in order to survive what you're doing to it.

The Relativity says: "When two observers move in relation to each other then they cannot have the same proper time on their clocks."

Relativity theory says no such thing (assuming "the relativity" means the theory).

[Jaaanosik]

There is a grid of inertial observers on the platform.
Their observation does not match your space-time diagram.

The spacetime diagram is in the frame of the platform, so by definition, the 'grid' observers on said platform must witness the events that occur at their respective locations.  Translation: your statement above is self contradictory.

If you say my placement of events/worldlines is wrong, that's a different assertion, but you didn't say that this time.  You seem to say something different each time, but consistent in being wrong at least.

I drew a high-acceleration reverse of a million km rigid object moving at .866c, centered at t=4 platform time. In that regard, the diagram is exactly correct.  Your diagrams depict a non-rigid turnaround since the proper length of the object is often too long or too short. That doesn't make it wrong, it just means your train needs to be rubber in order to survive what you're doing to it.

Halc,
The deceleration starts at 6s for the platform frame.
This is not the case in your space-time diagram.
The back observer starts to decelerate at 0s in your diagram.

The Relativity says: "When two observers move in relation to each other then they cannot have the same proper time on their clocks."

Relativity theory says no such thing (assuming "the relativity" means the theory).

What is the Twin paradox then?
How can two observers have the same time?
Please, explain how Twin paradox is wrong,
Jano

[Halc]

The deceleration starts at 6s for the platform frame.

Yours does, which is why your object gets crushed at first.

This is not the case in your space-time diagram.
The back observer starts to decelerate at 0s in your diagram.

At 2.3 seconds actually.  It's inertial up to that point.  That event is simultaneous with the front of the train in the train frame, when the entire train begins to decelerate all at the same time.

Please, explain how Twin paradox is wrong,

I think it's wrong to call it a paradox. I didn't assert that the twins scenario is wrong. I said your assertion above is wrong.

[Jaaanosik]

The deceleration starts at 6s for the platform frame.

Yours does, which is why your object gets crushed at first.

This is not the case in your space-time diagram.
The back observer starts to decelerate at 0s in your diagram.

At 2.3 seconds actually.  It's inertial up to that point.  That event is simultaneous with the front of the train in the train frame, when the entire train begins to decelerate all at the same time.

Please, explain how Twin paradox is wrong,

I think it's wrong to call it a paradox. I didn't assert that the twins scenario is wrong. I said your assertion above is wrong.

Halc,
what is the position of the front of the train car for x=0.2598cs and t=2.3s in the train frame coordinates?


The Triplets challenge - is the right triplet going to say that the left triplet proper time is the same as his/her (the right triplet)?
If the proper time is not the same then the middle triplet reference frame analysis is useless, essentially, we do not have physics.
If the left triplet proper time is observed/analyzed as to be same as the proper time of the right triplet from the right triplet frame then we need to see those claims/analysis because according to MCF (momentarily co-moving inertial reference frame) the left frame is moving and the Lorentz Transformation does not allow for the proper time to be the same.
MCF analysis is an approximation for the accelerated frames time dilation analysis.

The claim is:
- if the left time is not the same from the right frame point of view ... then we do not have physics because it contradicts the middle triplet's frame.
- if the left time is the same as the right frame time ... then we do not have the Lorentz Transformation, we do not have the Relativity

I am just talking about the thought experiment analysis but if this would be a real life experiment then either result contradicts the Relativity,
Jano

[Halc]

what is the position of the front of the train car for x=0.2598cs and t=2.3s in the train frame coordinates?

I didn't do any calculations to 4 digits of precision.  The yellow lines show lines of simultaneity along the changing frame of the train as it turns around.  You can see the line from that event being simultaneous with the short duration at t'=4 x'=0 measured at the front of the train.

The Triplets challenge - is the right triplet going to say that the left triplet proper time is the same as his/her (the right triplet)?

Proper time is frame independent, so all observers will agree that the proper times of the left and right worldlines between events A and B are identical.

If the proper time is not the same then the middle triplet reference frame analysis is useless, essentially, we do not have physics.
If the left triplet proper time is observed/analyzed as to be same as the proper time of the right triplet from the right triplet frame then we need to see those claims/analysis because according to MCF (momentarily co-moving inertial reference frame) the left frame is moving and the Lorentz Transformation does not allow for the proper time to be the same.

Maybe you should not only read, but actually comprehend that book you linked.

The claim is:
- if the left time is not the same from the right frame point of view ... then we do not have physics because it contradicts the middle triplet's frame.
- if the left time is the same as the right frame time ... then we do not have the Lorentz Transformation, we do not have the Relativity

Have fun justifying these claims.

[Jaaanosik]

what is the position of the front of the train car for x=0.2598cs and t=2.3s in the train frame coordinates?

I didn't do any calculations to 4 digits of precision.  The yellow lines show lines of simultaneity along the changing frame of the train as it turns around.  You can see the line from that event being simultaneous with the short duration at t'=4 x'=0 measured at the front of the train.

The Triplets challenge - is the right triplet going to say that the left triplet proper time is the same as his/her (the right triplet)?

Proper time is frame independent, so all observers will agree that the proper times of the left and right worldlines between events A and B are identical.

If the proper time is not the same then the middle triplet reference frame analysis is useless, essentially, we do not have physics.
If the left triplet proper time is observed/analyzed as to be same as the proper time of the right triplet from the right triplet frame then we need to see those claims/analysis because according to MCF (momentarily co-moving inertial reference frame) the left frame is moving and the Lorentz Transformation does not allow for the proper time to be the same.

Maybe you should not only read, but actually comprehend that book you linked.

The claim is:
- if the left time is not the same from the right frame point of view ... then we do not have physics because it contradicts the middle triplet's frame.
- if the left time is the same as the right frame time ... then we do not have the Lorentz Transformation, we do not have the Relativity

Have fun justifying these claims.

Halc,
using this: http://www.trell.org/div/minkowski.html


The time on the train car is 4.15s'.
The train starts to decelerate at 2.3s according to your calculations.
The front of the train car is already past the turning point for the platform 2.3s.
What are you trying to show us with your spacetime diagram?
Jano

[Halc]

using this: http://www.trell.org/div/minkowski.html

Cool tool.

The time on the train car is 4.15s'.
The train starts to decelerate at 2.3s according to your calculations.

So much for me just eyeballing it.
It appears that my acceleration begins more at t=2.13, not 2.3, but nowhere near the 6 second mark where you put it.
That means it also ends around t=13.87

I said the object was a million km long, so that should make d' always equal to -3.3356 or so.

[Jaaanosik]

using this: http://www.trell.org/div/minkowski.html

Cool tool.

The time on the train car is 4.15s'.
The train starts to decelerate at 2.3s according to your calculations.

So much for me just eyeballing it.
It appears that my acceleration begins more at t=2.13, not 2.3, but nowhere near the 6 second mark where you put it.
That means it also ends around t=13.87

I said the object was a million km long, so that should make d' always equal to -3.3356 or so.

Halc,
Your t=2.13s will be more than t'=4s' because anything above 2s is above 4s'. This is the Lorentz transformation.
The deceleration is arbitrary it can last longer or shorter.
I said from 3s' to 4s'. There is nothing wrong with that.
But there is the simultaneity problem. It is 0s for the back and 6s for the front,
... or it is 6s from the platform point of view.

Let us assume that the train would be a spaceship with engines at the front.
The back of the spaceship starts to decelerate at 0s and the engine is ignited at 6s.
Where is causality in this?
This does not make any sense,
Jano

[Halc]

Your t=2.13s will be more than t'=4s' because anything above 2s is above 4s'. This is the Lorentz transformation.

You have no idea what you're doing.  The front is well above the 2s mark, and is not above 4s', so your statement is obviously false.  It all depends on how far down x' you want to go, and -3.3356 isn't enough to push it below the 2s mark at that speed.

But there is the simultaneity problem. It is 0s for the back and 6s for the front,
... or it is 6s from the platform point of view.

Your problem is that you're doing things to the train in the platform frame.  Do it in the train frame like any train controller would.  Assuming the clocks in the train are initially in sync while inertial, all points in the train begin to accelerate left at t'=4.  Those clocks will run at different speeds relative to the train while accelerating, so they'll not be in sync anymore one the inertial phase is over.  This is consistent with SR and also with the equivalence principle.

Let us assume that the train would be a spaceship with engines at the front.

The engines need to be everywhere, else any changes in thrust cannot be felt elsewhere in the ship an sooner than the speed of sound will allow, and there can be no material that transmits pressure waves faster than light.  In other words, it is physically impossible to accelerate a ship from thrust at one end without introducing strain (deformation) to the ship.  So since I am doing this as a born-rigid object, the thrust must be applied everywhere to maintain that constant proper length.

The back of the spaceship starts to decelerate at 0s and the engine is ignited at 6s.
Where is causality in this?

Exactly so. This is why you can't thrust from one point.  Causality says the ship must bend if you do it that way.

[Jaaanosik]

Your t=2.13s will be more than t'=4s' because anything above 2s is above 4s'. This is the Lorentz transformation.

You have no idea what you're doing.  The front is well above the 2s mark, and is not above 4s', so your statement is obviously false.  It all depends on how far down x' you want to go, and -3.3356 isn't enough to push it below the 2s mark at that speed.
...

Halc,
let us just talk about the first part of your post:



As you can see 2s in the platform frame is 4s' in the train frame.
Where did you get the 3.3356cs' when the train car length is L'=3.4641cs'?
Jano

[Halc]

You have no idea what you're doing.  The front is well above the 2s mark, and is not above 4s', so your statement is obviously false.  It all depends on how far down x' you want to go, and -3.3356 isn't enough to push it below the 2s mark at that speed.
...

Halc,
let us just talk about the first part of your post

OK.  The first part of my post was saying you have no idea what you're doing.  To illustrate:

As you can see 2s in the platform frame is 4s' in the train frame.

A time in one frame cannot correspond to a specific time in another.  An event does, but the time 2s on the platform intersects every time in the train frame, depending on location.  This is a good example of you showing that you don't know what you're doing.

Where did you get the 3.3356cs' when the train car length is L'=3.4641cs'?

I was guessing a million km, something I've said in multiple posts.  Your pictures never said a different figure.  Yes, if you extend the train to 3.464, it will need to begin acceleration left immediately as it fully passes the x=0 mark on the platform in two seconds.

[Jaaanosik]

Halc,
just check my posts #106 and #107. I never changed the initial point of the discussion.

I know what I am talking about.
If the back of the train car is at platform x=0cs and t=2s then L' () - the proper length to the right, the simultaneity line of the train car, will end up at 4s'.
That's what I am saying.
Your deceleration after 2s does not make sense for the original setup.
The length units of measure were cs and not km at the beginning.

Even if we consider the 'born-rigid object' and the deceleration everywhere there is a problem.
The train car will start to stretch from the back side of the train car for the platform observer.
Remember the post where I showed ct, ct' and ct'' frames?
Here is what is happening.

The train car is decelerating in ct frame and stretching out from the back, the left side.
The train car is accelerating in ct' frame and shrinking from the front, the right side.
A total chaos and disagreement between the frames on what is going on.
The observers disagree on the acceleration because of different simultaneity.
Jano

[Halc]

Halc,
just check my posts #106 and #107. I never changed the initial point of the discussion.

That did not involve the current turnaround scenario, so I didn't know the length came from there.
The picture there pictures a train car contracted by a factor of 4 despite a gamma of only 2. All very deceptive.

I know what I am talking about.

I suspect you do, but that makes you trolling: deliberate misrepresentation.  Your posts in the energy/momentum conservation thread bear this out.

Even if we consider the 'born-rigid object' and the deceleration everywhere there is a problem.
The train car will start to stretch from the back side of the train car for the platform observer.

Exactly as it needs to.  It length is going to change from 1.73 to twice that when it matches the platform speed.

[Jaaanosik]

Halc,
just check my posts #106 and #107. I never changed the initial point of the discussion.

That did not involve the current turnaround scenario, so I didn't know the length came from there.
The picture there pictures a train car contracted by a factor of 4 despite a gamma of only 2. All very deceptive.

I know what I am talking about.

I suspect you do, but that makes you trolling: deliberate misrepresentation.  Your posts in the energy/momentum conservation thread bear this out.

Even if we consider the 'born-rigid object' and the deceleration everywhere there is a problem.
The train car will start to stretch from the back side of the train car for the platform observer.

Exactly as it needs to.  It length is going to change from 1.73 to twice that when it matches the platform speed.

Halc,
This is a disagreement on physics!!!
The platform observer sees the deceleration at the back but the front does not have any deceleration.
The train car is not a rigid body. It is stretching out.
Are you saying that the train car is a preferred frame?
Jano

[Jaaanosik]

...
I suspect you do, but that makes you trolling: deliberate misrepresentation.  Your posts in the energy/momentum conservation thread bear this out.
...

Halc,
I am not trolling!
I started my argument with pure, clean SR logic.
The end result is a logical fallacy.
Pointing out the truth is NOT trolling!
Jano