1. Assume the normal twins paradox. Let's call them O and O' where O is the stay at home twin.

The non-inertial twin will accelerate at "a" for time Ta through the 4 acceleration phases.

Also assume there is another observer O2.

Finally, when O and O' are at the same place and at the instant O' accelerates, a light pulse is emitted. Further, at that same instant the observer O2 will be at the same v that the O' twin will accelerate to and that observer is also co-located at the light pulse with O and O'.

What time will elapse on the clock of the stay at home twin while O' elapses Ta?

How will the center of the light sphere travel in the view of the accelerating twin from the frame of O to the frame of O2?

2.

3. Finally, when O and O' are at the same place and at the instant O' accelerates, a light pulse is emitted.
This notion of simultaneity is highly problematic - who determines "same place" and "same instant", and how do they do it ?

4. Originally Posted by Markus Hanke
Finally, when O and O' are at the same place and at the instant O' accelerates, a light pulse is emitted.
This notion of simultaneity is highly problematic - who determines "same place" and "same instant", and how do they do it ?
Under relativity, this is not a problem. Einstein assumed 2 frames in relative motion understand when their origins are common.

What you can't do under SR is determine equal simultaneity between 2 frames regarding the actions of am emitted light pulse.

5. Here is the correct answer for the time elapsed for the stay at home twin when the accelerating twin accelerates at a constant "a" for Ta seconds.
t = c/a sinh(a (Ta)/c).

physics/0411233] Differential aging from acceleration, an explicit formula

Now, what happens to the center of the light pulse in the view of the accelerating twin?

First, it starts out at the common location with the light pulse. Then, as the twins O' accelerates the center of the light sghere moves out in front of the accelerating twin.

Then, when O' joins the frame of O2, the center of the light sphere is located at O2 in the frame of O2, since that is the light emission point in that frame.

Therefore, as O' accelerates, the light sphere moves out in front of O' in its accelerating context.

6. Originally Posted by chinglu
Here is the correct answer for the time elapsed for the stay at home twin when the accelerating twin accelerates at a constant "a" for Ta seconds.
t = c/a sinh(a (Ta)/c).

physics/0411233] Differential aging from acceleration, an explicit formula

Now, what happens to the center of the light pulse in the view of the accelerating twin?
First, it starts out at the common location with the light pulse. Then, as the twins O' accelerates the center of the light sghere moves out in front of the accelerating twin.
Then, when O' joins the frame of O2, the center of the light sphere is located at O2 in the frame of O2, since that is the light emission point in that frame.
Therefore, as O' accelerates, the light sphere moves out in front of O' in its accelerating context.
Ok, I am not sure I really get what you are trying to do - is there is specific point/question to this scenario which we could formulate in a more general way ?

7. Chinglu has a long history of denying the basic postulates of Special Relativity. He seems unable to grasp the concept that the origin of any light remains at rest in relation to any (inertial) observer.

Good luck with this.

See:
Einstein's train experiment
a
nd
Proof there exists spherical light waves in the rest frame t

8. Originally Posted by SpeedFreek
Chinglu has a long history of denying the basic postulates of Special Relativity. He seems unable to grasp the concept that the origin of any light remains at rest in relation to any (inertial) observer.

Good luck with this.

See:
Einstein's train experiment
a
nd
Proof there exists spherical light waves in the rest frame t
Any comments to this chinglu ?

9. Originally Posted by Markus Hanke
Originally Posted by chinglu
Here is the correct answer for the time elapsed for the stay at home twin when the accelerating twin accelerates at a constant "a" for Ta seconds.
t = c/a sinh(a (Ta)/c).

physics/0411233] Differential aging from acceleration, an explicit formula

Now, what happens to the center of the light pulse in the view of the accelerating twin?
First, it starts out at the common location with the light pulse. Then, as the twins O' accelerates the center of the light sghere moves out in front of the accelerating twin.
Then, when O' joins the frame of O2, the center of the light sphere is located at O2 in the frame of O2, since that is the light emission point in that frame.
Therefore, as O' accelerates, the light sphere moves out in front of O' in its accelerating context.
Ok, I am not sure I really get what you are trying to do - is there is specific point/question to this scenario which we could formulate in a more general way ?
We are simply exploring the acceleration equations for the twin's paradox and all the consequences of the acceleration.

Now, do you agree after O' accelerates to the O2 frame and is at rest with the O2 frame, it must adopt the light emission point of O2 as the center of the light sphere?

Here is some additional mainstream logic.

10. Originally Posted by SpeedFreek
Chinglu has a long history of denying the basic postulates of Special Relativity. He seems unable to grasp the concept that the origin of any light remains at rest in relation to any (inertial) observer.

Good luck with this.

See:
Einstein's train experiment
a
nd
Proof there exists spherical light waves in the rest frame t
So what.

Nothing in there can be refuted.

But, in this thread, we are studying the twin's paradox in mathematical detail.

Did you want to do this or are you going to be a troll.

Now, if you feel there is something wrong with the math or logic, jump in. That is what learning and discussion is all about.

11. Originally Posted by Markus Hanke
Originally Posted by SpeedFreek
Chinglu has a long history of denying the basic postulates of Special Relativity. He seems unable to grasp the concept that the origin of any light remains at rest in relation to any (inertial) observer.

Good luck with this.

See:
Einstein's train experiment
a
nd
Proof there exists spherical light waves in the rest frame t
Any comments to this chinglu ?
What's to comment?

If you all would like to discuss those ideas later, I would be happy.

However, we are exploring the twin's paradox in extreme detail and we will not deviate from any truth of SR.

Did you have any comments on anything presented thus far for the twin's paradox?

12. Originally Posted by chinglu
Did you have any comments on anything presented thus far for the twin's paradox?
The coordinate speed of light is not constant in an accelerating frame.

13. Originally Posted by SpeedFreek
Originally Posted by chinglu
Did you have any comments on anything presented thus far for the twin's paradox?
The coordinate speed of light is not constant in an accelerating frame.
So what.

What does that have to do with the origin of the light sphere and how it moves during accelerating?

14. Originally Posted by chinglu
What does that have to do with the origin of the light sphere and how it moves during accelerating?
The coordinate speed of light gets progressively slower, along the axis of motion, the further in front of observer O' he cares to examine. There is a distance where the coordinate speed of light reaches zero, and there is a coordinate event horizon known as the Rindler horizon.

So, from the view of observer O' who is in an accelerating and thus non-inertial frame, the light pulse does not move out in front of him as it does for observer O2, who is in an inertial frame. For O', the coordinate speed of the light pulse ensures the correct distance to keep the origin centred on the observer.

For issues related to this, see Bells Spaceship Paradox and Rindler coordinates.

15. Originally Posted by SpeedFreek
Originally Posted by chinglu
What does that have to do with the origin of the light sphere and how it moves during accelerating?
The coordinate speed of light gets progressively slower, along the axis of motion, the further in front of observer O' he cares to examine. There is a distance where the coordinate speed of light reaches zero, and there is a coordinate event horizon known as the Rindler horizon.

So, from the view of observer O' who is in an accelerating and thus non-inertial frame, the light pulse does not move out in front of him as it does for observer O2, who is in an inertial frame. For O', the coordinate speed of the light pulse ensures the correct distance to keep the origin centred on the observer.

For issues related to this, see Bells Spaceship Paradox and Rindler coordinates.
Let's try to stay on task.

When O' enters the frame of O2, where is the center of the light sphere at that instant?

16. Originally Posted by chinglu
When O' enters the frame of O2, where is the center of the light sphere at that instant?
Frames O' and O2 are separated by distance, so "that instant" is not the same for O' as it is for O2. Perhaps they should use light, emitted after the acceleration has ended, to synchronise their clocks?

17. Originally Posted by chinglu
However, we are exploring the twin's paradox in extreme detail and we will not deviate from any truth of SR.
Did you have any comments on anything presented thus far for the twin's paradox?
I would have the following comments :

1. The twin paradox has been examined and considered to great detail by academics and students alike - Google it and you get 150,000 hits. Why do we need to do it again "in extreme detail" ? Please briefly explain.
2. Why can we not cut through the "extreme detail" and just go straight to the heart of the matter - just state clearly what your contention is on a general level, and we will address it then.

Generally speaking the twin paradox is resolved simply by recognizing that, amongst all possible worldlines connecting two events 1 and 2, the ones which are not accelerated take the longest proper time :

This has been shown to be true in the general case by von Laue and Langevin back in 1913. In my mind it is a waste of time to go through all this again "in extreme detail", unless of course you have a different agenda or contention which you have not yet mentioned.

18. Originally Posted by SpeedFreek
Originally Posted by chinglu
When O' enters the frame of O2, where is the center of the light sphere at that instant?
Frames O' and O2 are separated by distance, so "that instant" is not the same for O' as it is for O2. Perhaps they should use light, emitted after the acceleration has ended, to synchronise their clocks?
Yes, at the instant O' enters the frame of O2, they are in the same frame and instants are the same. Regardless, since O' is in the frame of O2, it must adopt the truths of that frame and the light emission in O2 is at O2, hence that is the center of the light sphere.

So, whereas, you accelerated away from the center of the light sphere, when the acceleration was complete, the light sphere ends up out in front of you.

19. Originally Posted by Markus Hanke
Originally Posted by chinglu
However, we are exploring the twin's paradox in extreme detail and we will not deviate from any truth of SR.
Did you have any comments on anything presented thus far for the twin's paradox?
I would have the following comments :

1. The twin paradox has been examined and considered to great detail by academics and students alike - Google it and you get 150,000 hits. Why do we need to do it again "in extreme detail" ? Please briefly explain.
2. Why can we not cut through the "extreme detail" and just go straight to the heart of the matter - just state clearly what your contention is on a general level, and we will address it then.

Generally speaking the twin paradox is resolved simply by recognizing that, amongst all possible worldlines connecting two events 1 and 2, the ones which are not accelerated take the longest proper time :

This has been shown to be true in the general case by von Laue and Langevin back in 1913. In my mind it is a waste of time to go through all this again "in extreme detail", unless of course you have a different agenda or contention which you have not yet mentioned.
Noone is having a problem with differential aging, did you bother to look at the link I posted?

What we are now discussing is the movement of the center of the light sphere during acceleration.

You may google that and find no hits.

Now, do you have a reference somewhere that the center of the light sphere moves during acceleration? That is what I mean by detail. We are considering everything.

20. Originally Posted by chinglu
What we are now discussing is the movement of the center of the light sphere during acceleration.
You mean the same again as you already did over on sciforums.com :

SR acceleration - Page 2

My question remains - why do we need to do it all over again ? Seems to me that your scenario was answered over there in some detail. Your insistence on doing it again here suggests to me that you aren't really interested in the answers given.

You may google that and find no hits.
How so ? If I search for "electromagnetic waves in accelerated frames" I am getting a number of hits. For example this :

[1110.5367] The electromagnetic field in accelerated frames

Seems to me that this is all very well worked out.

Now, do you have a reference somewhere that the center of the light sphere moves during acceleration?
This quote does not refere to any of my posts, so why do you ask me for references ?

On the other hand I have several issues with the proposed scenario :

1. The entire thing hinges on being able to determine "same place" and "same instant" for two separate frames, one of which is inertial, the other one accelerated. To the best of my understanding simultaneity is not defined for two such frames, except perhaps in a local sense. And if you define it locally, the measurements of simultaneity of the event will change as one observer accelerates, so the observers will not agree on the where the center is.
2. Once emitted, neither of the observers will be able to observe the wave front - so how are they supposed to determine where the center of the sphere is ?
3. In the accelerated frame the sphere ( suppose he could actually see it ) will be length contracted along the acceleration vector, thus will no longer be a sphere
4. In the accelerated frame space-time will be curved in accordance with the usual laws of GR, i.e. the sphere once again ceases to be a sphere, and thus the ( unobservable ) center point is allowed to shift without violating any principles of relativity. In fact, this shift is predicted and expected by the laws of GR.
5. The two observers ( the one at rest and the accelerated one ) do not have any way to compare their measurements about where the center of the sphere is. Even if they did, they use different rulers and clocks to perform those measurements.
6. If the accelerated twin goes through 4 phases of acceleration, he will end up with the same center point for the light sphere, because performing such transformations of the location of the center point successively in opposite directions will yield the unity matrix. What does change though is the length of the worldline for the accelerated twin as compared to the stationary one - but this is once again predicted and expected by GR due to his acceleration.

In short, everything here is exactly as it should be so far as relativity is concerned.

21. Originally Posted by Markus Hanke
Originally Posted by chinglu
What we are now discussing is the movement of the center of the light sphere during acceleration.
You mean the same again as you already did over on sciforums.com :

SR acceleration - Page 2

My question remains - why do we need to do it all over again ? Seems to me that your scenario was answered over there in some detail. Your insistence on doing it again here suggests to me that you aren't really interested in the answers given.

You may google that and find no hits.
How so ? If I search for "electromagnetic waves in accelerated frames" I am getting a number of hits. For example this :

[1110.5367] The electromagnetic field in accelerated frames

Seems to me that this is all very well worked out.

Now, do you have a reference somewhere that the center of the light sphere moves during acceleration?
This quote does not refere to any of my posts, so why do you ask me for references ?

On the other hand I have several issues with the proposed scenario :

1. The entire thing hinges on being able to determine "same place" and "same instant" for two separate frames, one of which is inertial, the other one accelerated. To the best of my understanding simultaneity is not defined for two such frames, except perhaps in a local sense. And if you define it locally, the measurements of simultaneity of the event will change as one observer accelerates, so the observers will not agree on the where the center is.
2. Once emitted, neither of the observers will be able to observe the wave front - so how are they supposed to determine where the center of the sphere is ?
3. In the accelerated frame the sphere ( suppose he could actually see it ) will be length contracted along the acceleration vector, thus will no longer be a sphere
4. In the accelerated frame space-time will be curved in accordance with the usual laws of GR, i.e. the sphere once again ceases to be a sphere, and thus the ( unobservable ) center point is allowed to shift without violating any principles of relativity. In fact, this shift is predicted and expected by the laws of GR.
5. The two observers ( the one at rest and the accelerated one ) do not have any way to compare their measurements about where the center of the sphere is. Even if they did, they use different rulers and clocks to perform those measurements.
6. If the accelerated twin goes through 4 phases of acceleration, he will end up with the same center point for the light sphere, because performing such transformations of the location of the center point successively in opposite directions will yield the unity matrix. What does change though is the length of the worldline for the accelerated twin as compared to the stationary one - but this is once again predicted and expected by GR due to his acceleration.

In short, everything here is exactly as it should be so far as relativity is concerned.
1) The consensus is that indeed the center of light sphere moves in the first stage of this twin's experiment. Then, we went on to stage 2 in which I am no long interested in talking with them.

1) Simultaneity is irrelevant. The issue is when O' is in the frame O2 does it adopt the center of the light sphere for the frame. The other forum understood this follows from SR. I suggest you read them further.
2) The center of the light sphere is frame dependent. That is really the nature of SR. Everything is not relative, the center if the light sphere is. I already posted a link from the mainstream helping you understand this. I will post it again.
3) This makes no difference to the problem. And, your reasoning applies to a solid sphere. But, at any co-moving inertial frame, the accelerated observer will see a constant speed of light in all directions. So, this does not follow from SR.
4) You mentioned a center shift of the light sphere, good you agree with my reasoning and understand I am 100% correct..
5) You need to make a decision as to whether the accelerated observer adopts the truths of an inertial frame in which it enters. If not, then it will disagree with the truths of the inertial frame and hence SR is in a contradiction. So, it is time you understood SR.I thought you said no further detail was needed?
6) We will get to this, but your answer is correct. When the 2 twins re-unite, they will both agree on that the location of the center of the light sphere and it is located at the stay at home twin O. But, the transformations of the light sphere center are needed in order to understand the final result. So, we are going to go through all of them.

Now, if you want to explain all the light sphere center positions in the context of the non-inertial twin 4 acceleration phases, feel free since you claim to understand everything. If you cannot, don't worry, I completely understand this problem and will explain it.

No observer will ever see the center of the light-sphere, all that can ever be observed is the wave front. In the context of this experiment this is therefore relevant.

1) Simultaneity is irrelevant.
It is not irrelevant, because of this part in your experimental setup ( highlights are mine ) :

"(...) when O and O' are at the same place and at the instant O' accelerates (...)"

2) The center of the light sphere is frame dependent.
Yes, that was my point - the two observers, inertial and non-inertial cannot agree on where the center is.

This makes no difference to the problem. And, your reasoning applies to a solid sphere. But, at any co-moving inertial frame, the accelerated observer will see a constant speed of light in all directions. So, this does not follow from SR.
There is no such thing as a "co-moving inertial frame" for the accelerated observer. This was already pointed out to you quite clearly on the other forum.

5) You need to make a decision as to whether the accelerated observer adopts the truths of an inertial frame in which it enters. If not, then it will disagree with the truths of the inertial frame and hence SR is in a contradiction. So, it is time you understood SR.I thought you said no further detail was needed?
I believe it is the center point of a sphere which you are referring to. The agreement in that case is only possible if the accelerated observer stops accelerating and moves uniformely at the same speed as the inertial frame's observer, i.e. if the accelerated observer's frame becomes inertial. If he continues to be accelerated, then he hasn't joined the inertial frame.

Now, if you want to explain all the light sphere center positions in the context of the non-inertial twin 4 acceleration phases, feel free since you claim to understand everything. If you cannot, don't worry, I completely understand this problem and will explain it.
Go ahead then, don't let us stop you.
Otherwise it is clear that, if the acceleration goes through all the usual four phases, and the observer ends up at rest at his point of origin, we have performed a unity transformation in space.
Just out of interest though - where did I claim to "understand everything" ?

23. You need to make a decision as to whether the accelerated observer adopts the truths of an inertial frame in which it enters. If not, then it will disagree with the truths of the inertial frame and hence SR is in a contradiction. So, it is time you understood SR.I thought you said no further detail was needed?
Ok, so this is really the heart of the issue. Is there a contradiction in Special Relativity or not ?
In order to treat this correctly under SR the accelerated observer must start off at rest, at which time he agrees with the stationary observer, because they are in the same frame. At the end of his loop ( 4 phases of acceleration, as per yourself ) he once again is at rest in the same frame as the stationary observer, hence they will once again agree on the center point. However, during his acceleration phases he is not in agreement with the stationary observer - but then again, the frame is no longer inertial and thus no longer symmetric to the stationary observer, and SR is never violated.
It all comes down to the fact that there is not necessarily an agreement between inertial and non-inertial frames under the the rules of SR. However, if both frames are inertial they will always agree.

24. Originally Posted by Markus Hanke
You need to make a decision as to whether the accelerated observer adopts the truths of an inertial frame in which it enters. If not, then it will disagree with the truths of the inertial frame and hence SR is in a contradiction. So, it is time you understood SR.I thought you said no further detail was needed?
Ok, so this is really the heart of the issue. Is there a contradiction in Special Relativity or not ?
In order to treat this correctly under SR the accelerated observer must start off at rest, at which time he agrees with the stationary observer, because they are in the same frame. At the end of his loop ( 4 phases of acceleration, as per yourself ) he once again is at rest in the same frame as the stationary observer, hence they will once again agree on the center point. However, during his acceleration phases he is not in agreement with the stationary observer - but then again, the frame is no longer inertial and thus no longer symmetric to the stationary observer, and SR is never violated.
It all comes down to the fact that there is not necessarily an agreement between inertial and non-inertial frames under the the rules of SR. However, if both frames are inertial they will always agree.
First let's go through the 6 phases. But, first, we acknowledge the mainstream article. For each phase of acceleration, the O frame elapses t = c/a sinh(a (Ta)/c).

1) The center of the light sphere moves out in front of the accelerating observer. When the acceleration is complete, the center of the light sphere is not behind the accelerating observer, but ahead of it. In addition, the time on the O clock elapses t = c/a sinh(a (Ta)/c) while the non inertial twin elapses Ta.

2) There is a period of relative motion in which the O' observer elapses tr = Tr / γ.

3) The O' twin decelerates back to the frame of O. The O frame elapses t = c/a sinh(a (Ta)/c) while the O' frame elapses Ta. During this phase, the center of the light sphere starts from O2, runs past O' and heads back to O.

4) The O' frame immediately accelerates back toward O. The O frame elapses t = c/a sinh(a (Ta)/c) while the O' frame elapses Ta. During this phase, the center of the light sphere starts at O and moves further away from O'.

5) There is a period of relative motion in which the O' observer elapses tr = Tr / γ.

6) The O' twin decelerates back to the frame of O and the location of O. The O frame elapses t = c/a sinh(a (Ta)/c) while the O' frame elapses Ta. During this phase, the center of the light sphere starts from the negative direction of O and proceeds toward O' until O is at rest with I' and then the center of the light sphere is at their common location.

So far so good?

25. Originally Posted by chinglu
First let's go through the 6 phases.
There are four phases, not six - two accelerations, two decelerations. Refer to your OP :

The non-inertial twin will accelerate at "a" for time Ta through the 4 acceleration phases.

26. For each phase of acceleration, the O frame elapses t = c/a sinh(a (Ta)/c).
The relation is correct, but it is far too troublesome to do this via elapsed time in the rest frame. All we need to consider is two simple premises :

(1) Two observers at rest in the same inertial frame, and not spatially separated, will agree on where the center of the light sphere is
(2) There is no requirement under the laws of SR for an inertial and a non-inertial frame to agree on any measurement

To compare distances between the observers before and after the acceleration event consider firstly the relation between coordinate time and proper time :

and then the relativistic velocity relation

Now differentiate the velocity to get proper acceleration

which has a magnitude of

Now, since the initial relative velocity between both observers is 0, and the acceleration is constant, the magnitude simply becomes

which can now simply be integrated to get the relationship for distance

The integration constant vanishes because both observers are initially at rest. As the accelerated observer goes through his four phases, the acceleration in terms of magnitude will be +a, -a, +a, -a; the term is squared under the root, so no change, but only as a single factor in the fraction, so the total change in position after all acceleration phases is

By premise (1) above it is immediately clear that now both observers will once again agree on the position of the centre point, and by (2) it is clear that while acceleration takes place no agreement is possible. Therefore, there is never any contradiction.

27. Originally Posted by Markus Hanke
No observer will ever see the center of the light-sphere, all that can ever be observed is the wave front. In the context of this experiment this is therefore relevant.

1) Simultaneity is irrelevant.
It is not irrelevant, because of this part in your experimental setup ( highlights are mine ) :

"(...) when O and O' are at the same place and at the instant O' accelerates (...)"

2) The center of the light sphere is frame dependent.
Yes, that was my point - the two observers, inertial and non-inertial cannot agree on where the center is.

This makes no difference to the problem. And, your reasoning applies to a solid sphere. But, at any co-moving inertial frame, the accelerated observer will see a constant speed of light in all directions. So, this does not follow from SR.
There is no such thing as a "co-moving inertial frame" for the accelerated observer. This was already pointed out to you quite clearly on the other forum.

5) You need to make a decision as to whether the accelerated observer adopts the truths of an inertial frame in which it enters. If not, then it will disagree with the truths of the inertial frame and hence SR is in a contradiction. So, it is time you understood SR.I thought you said no further detail was needed?
I believe it is the center point of a sphere which you are referring to. The agreement in that case is only possible if the accelerated observer stops accelerating and moves uniformely at the same speed as the inertial frame's observer, i.e. if the accelerated observer's frame becomes inertial. If he continues to be accelerated, then he hasn't joined the inertial frame.

Now, if you want to explain all the light sphere center positions in the context of the non-inertial twin 4 acceleration phases, feel free since you claim to understand everything. If you cannot, don't worry, I completely understand this problem and will explain it.
Go ahead then, don't let us stop you.
Otherwise it is clear that, if the acceleration goes through all the usual four phases, and the observer ends up at rest at his point of origin, we have performed a unity transformation in space.
Just out of interest though - where did I claim to "understand everything" ?

1) If you want to claim the accelerated observer does not see the center of the light sphere, then you are wrong. The mathematics suggest that the center moves through O' in phase 2. Please prove this is false.

2) You obviously do not understand calculus. The idiot that you support does not understand how the integral is performed.

You integrate with a co-moving inertial frame at any infinitesimal time slice. What exactly do you think you are integrating with?

Anyway, here is a simple example to teach you how the integral works. You will note the link specifies integrating with a co-moving inertial frame.

SR treatment of arbitrarily accelerated motion

28. Originally Posted by Markus Hanke
Originally Posted by chinglu
First let's go through the 6 phases.
There are four phases, not six - two accelerations, two decelerations. Refer to your OP :

The non-inertial twin will accelerate at "a" for time Ta through the 4 acceleration phases.
I only specified the the accelerations phases in the OP. Everyone knows there are 2 relative motions phases.

Do you disagree?

29. Originally Posted by Markus Hanke
For each phase of acceleration, the O frame elapses t = c/a sinh(a (Ta)/c).
The relation is correct, but it is far too troublesome to do this via elapsed time in the rest frame. All we need to consider is two simple premises :

(1) Two observers at rest in the same inertial frame, and not spatially separated, will agree on where the center of the light sphere is
(2) There is no requirement under the laws of SR for an inertial and a non-inertial frame to agree on any measurement

To compare distances between the observers before and after the acceleration event consider firstly the relation between coordinate time and proper time :

and then the relativistic velocity relation

Now differentiate the velocity to get proper acceleration

which has a magnitude of

Now, since the initial relative velocity between both observers is 0, and the acceleration is constant, the magnitude simply becomes

which can now simply be integrated to get the relationship for distance

The integration constant vanishes because both observers are initially at rest. As the accelerated observer goes through his four phases, the acceleration in terms of magnitude will be +a, -a, +a, -a; the term is squared under the root, so no change, but only as a single factor in the fraction, so the total change in position after all acceleration phases is

By premise (1) above it is immediately clear that now both observers will once again agree on the position of the centre point, and by (2) it is clear that while acceleration takes place no agreement is possible. Therefore, there is never any contradiction.
I am not checking your math here since some of the symbols I do not know.

However, are you able to write an equation based on the 6 phases for the stay at home twin O?

I am able to.

Acceleration phases

ta = 4c/a sinh(a (Ta)/c).

Relative motion phases

tr = 2Tr / γ.

So, total time for the stay at home twin is ta + tr

Now, do you agree or disagree?

30. Originally Posted by chinglu
I am not checking your math here since some of the symbols I do not know.
I must say I am very surprised; this is simple vector algebra. If you do not even understand this, then you are not in a position from which you can judge the internal consistency of the theory of relativity.

However, are you able to write an equation based on the 6 phases for the stay at home twin O?
I have already explained that all you need to do is check if the two end up in the same frame of reference after all your phases. If you include two periods of uniform movement in opposite directions, the only thing that changes is

to account for two equal periods of uniform movement in opposite directions. The result is still the same, obviously.

ta = 4c/a sinh(a (Ta)/c).

Relative motion phases

tr = 2Tr / γ.

So, total time for the stay at home twin is ta + tr
ta would thus be the clock reading of the stay-at-home twin as measured by an observer accelerating at a.
Why do you take the reverse of the gamma factor, thereby accelerating a clock reading tr ? Whose clock is that, and what is measured ?

31. Originally Posted by Markus Hanke
Originally Posted by chinglu
I am not checking your math here since some of the symbols I do not know.
I must say I am very surprised; this is simple vector algebra. If you do not even understand this, then you are not in a position from which you can judge the internal consistency of the theory of relativity.

However, are you able to write an equation based on the 6 phases for the stay at home twin O?
I have already explained that all you need to do is check if the two end up in the same frame of reference after all your phases. If you include two periods of uniform movement in opposite directions, the only thing that changes is

to account for two equal periods of uniform movement in opposite directions. The result is still the same, obviously.

ta = 4c/a sinh(a (Ta)/c).

Relative motion phases

tr = 2Tr / γ.

So, total time for the stay at home twin is ta + tr
ta would thus be the clock reading of the stay-at-home twin as measured by an observer accelerating at a.
Why do you take the reverse of the gamma factor, thereby accelerating a clock reading tr ? Whose clock is that, and what is measured ?
1) I did not see the SR velocity add equation for your integral using your vector algebra. So, I concluded as I did.

2) No, we need to know mathematically the time elapsed on the clock of O based on the O' non-inertial observer. The fact that the light sphere will be centered at O and O' at the end s not in debate at this point.

3) Why do you take the reverse of the gamma factor, thereby accelerating a clock reading tr ? Whose clock is that, and what is measured
I did not, γ > 1. This is time dilation from the view of the O frame for the O' frame for the relative motion phases. Check your math.

32. Originally Posted by chinglu
No, we need to know mathematically the time elapsed on the clock of O based on the O' non-inertial observer.
I thought we are debating this ( from your OP ), and that is what my calculation is based on :

How will the center of the light sphere travel in the view of the accelerating twin from the frame of O to the frame of O2?
For this we do not need any time readings, we only need to know the displacement of the centre point before and after the acceleration event.

The fact that the light sphere will be centered at O and O' at the end s not in debate at this point.
I take it then that you agree that after the acceleration event all observers agree on where the centre point of the sphere is located, as shown in post 25.

I did not, γ > 1. This is time dilation from the view of the O frame for the O' frame for the relative motion phases. Check your math.
I think this was your maths, not mine.
But anyway, I see what you are doing now; had you written all this in terms of proper time for each observer, it would have been much clearer. Please go on and bring your "presentation" to a conclusion - I still don't know where you are going with this. I hope it is not simply about the fact that the clock readings will differ in the end, because we already understand that, due to one of the frames being accelerated.

33. Originally Posted by chinglu
No, we need to know mathematically the time elapsed on the clock of O based on the O' non-inertial observer.
I can't really agree with you on the maths until I do the calculations myself, so here it goes.
This is a straightforward exercise if you base all clock readings on proper time for each reference frame. All you do then is allow v to depend on time, and then evaluate the sum of proper time integrals for each acceleration phase and inertial motion phase :

where n = 1...6 denotes the respective phases of acceleration or motion. The integration constant can always be made zero by choosing an appropriate coordinate system, therefore I will henceforth ignore it. The coordinate velocity as a function of proper acceleration is

Acceleration phases ( tricky integral !!! ) :

Inertial motion phases :

The total elapsed proper time is thus

wherein a is the proper acceleration, and t is the coordinate time elapsed for the stay-at-home twin.

Now, unless my integral calculus fails me completely ( which I don't think because MAPLE agrees with my maths ), or I am missing something crucial in your posts, the proper acceleration time integral actually evaluates to a arcsinh(...) function, not a hyperbolic sinus as in your calculation. Can you explain where you got the sinh(...) from ? I really need to see your calculation, not just the end result. And spare us the reference, I want to see your own calculations.

And I still maintain that it is much easier to just consider the distances as done in post 25 !

34. In order to get some context, I have taken the liberty to read both of these threads ( as they took place before I joined this forum ), as well as some threads you have participated in on other forums, and I must say I am dismayed by what I saw. I want to give you my honest opinion - your attempts to 'disprove' a well established, empirically tested and verified theory such as relativity are, quite frankly, an embarrassment. All your arguments are based on one or several of these :

1. Failure to understand basic principles of relativity
2. Wrong application of relativistic laws and transformations, such as mixing reference frames
3. Refusal to acknowledge explanations given to you, or mistakes pointed out to you
4. Deliberate obfuscation of normally very simple relationships
5. Given the overall context it is clear that you are an anti-relativity adherer, and quite clearly have ulterior motives

Based on what I have seen on all of the other threads I have no reason to believe that this thread is going to end any different. I therefore wish to point out to you the following :

1. You will not refute or disprove relativity by any thought experiment, proposition or hypothesis you can raise here. The entire theory is empirically very well tested and experimentally verified, and completely self-consistent. I would urge you to refer to the sticky I have created with modern tests of relativity : Modern Tests of Relativity. None of these experiments would work if relativity had the flaws you say it has. The evidence is overwhelmingly against you.
2. Show me just one empirical, peer reviewed, independently verified and well documented real world experiment which shows that relativity is violated
3. This is a forum of amateur people interested in science. I am not a scientist, and neither are most other participants here. Even if you could present something that none of us can refute ( which is unlikely ), it would have no value in the bigger scheme of things, because we are not scientists. If you are so convinced that you have a genuine case against relativity you will need to follow the established route and publish a proper scientific paper, and have it peer-reviewed. By the very fact that you are wasting your time on forums such as this it is clear to me that you do not actually have any case - you are really just here to fight a personal battle ( "I am right, you are wrong !" ), thereby wasting everyone's time. I do not particularly appreciate people like that, and I am sure I am not alone with this opinion amongst our forum members.
4. Consider that you are not the first anti-relativity spokesperson here, and you will not be the last one. All of them thus far have either been shown to be wrong, or were banned for violating forum protocol. I see no reason to believe that your case is any different.

Anyway, that is my two cents' worth after having read your other threads. As you can see, you come across as neither likeable nor particularly knowledgeable in the subject matter.

As for this scenario we are discussing :

1. The centre of the light sphere will be in the same place after the acceleration then it is before
2. The world line of the two twins is of different lengths, i.e. the accelerating twin will have aged less then the stationary one
3. The observers will only agree on where the centre point is if they are at rest in the same frame - otherwise no agreement is required
4. The observers will also not necessarily agree on the shape of the light sphere - it is always a sphere only within the same, given inertial frame. There is no requirement for it to be spherical or for the centre points to agree if the observers are in different frames. The only requirement is that Maxwell's laws are the same in all frames. I have shown this to be the case over in your spherical light wave thread - go and take a look, you will find the maths there. I can of course give full references for those calculations, if required, as I can for all my maths.

Nothing in there can be refuted.
You are quite wrong. In fact you had been proven wrong on both threads by several people, the only thing is that you refused to acknowledge that, and kept deliberately obfuscating things to make it look like you still had a case. You didn't. This is an observation by an independent reader who was not involved in those threads. However, just for good measure I have posted my own opinions over there too. You may wish to check.

35. Originally Posted by Markus Hanke
Originally Posted by chinglu
No, we need to know mathematically the time elapsed on the clock of O based on the O' non-inertial observer.
I can't really agree with you on the maths until I do the calculations myself, so here it goes.
This is a straightforward exercise if you base all clock readings on proper time for each reference frame. All you do then is allow v to depend on time, and then evaluate the sum of proper time integrals for each acceleration phase and inertial motion phase :

where n = 1...6 denotes the respective phases of acceleration or motion. The integration constant can always be made zero by choosing an appropriate coordinate system, therefore I will henceforth ignore it. The coordinate velocity as a function of proper acceleration is

Acceleration phases ( tricky integral !!! ) :

Inertial motion phases :

The total elapsed proper time is thus

wherein a is the proper acceleration, and t is the coordinate time elapsed for the stay-at-home twin.

Now, unless my integral calculus fails me completely ( which I don't think because MAPLE agrees with my maths ), or I am missing something crucial in your posts, the proper acceleration time integral actually evaluates to a arcsinh(...) function, not a hyperbolic sinus as in your calculation. Can you explain where you got the sinh(...) from ? I really need to see your calculation, not just the end result. And spare us the reference, I want to see your own calculations.

And I still maintain that it is much easier to just consider the distances as done in post 25 !
Well, your answer fails because we are letting Ta be the timing which is in the context of the accelerating frame.

But, I can use your data. It means the same thing.

OK, now assume we have no relative motion phases.

Then, ta > Ta and we can calculate an adjustment of tadjustment = ta - Ta to add to the accelerating clock in order to make the times the same when the experiment is finished.

What we know is that if relative motion is eliminated, the two frames disagree on time and hence, they disagree on the speed of light when they are again at the same place. That is because they agree on the distance the light pulse traveled.

Now, this is all good with SR and GR that acceleration can cause frames to disagree on the speed of light.

So, now perform the entire twins experiment.

We know the total time on the stay at home twin is ta + tr. The O' twins is Tr + Ta.

We add ta - Ta to O's total time to adjust for the acceleration phases.

Are you following me?

OK, once that adjustment is performed, the frames again will disagree on the speed of light because of time dilation.

Yet, under the relative motion phases, frames cannot disagree on the speed of light because under relative motion all frames must measure c for common light spheres.

However, because of time dilation, the frames agree on distance, so they disagree on the speed of light for the relative motion phases, which is a contradiction.

36. Well, your answer fails because we are letting Ta be the timing which is in the context of the accelerating frame.
My answer fails - why ? Because you say so ? The relation I gave you is the proper time of the accelerating observer, just as you demanded. If you are not happy with the maths then please point out exactly where you think it is wrong.
You on the other hand have failed to show me your calculation, and have failed to explain where you get the sinh(...) function from when the integral clearly evaluates to to an arcsinh(...). I am still awaiting your maths on that.

But, I can use your data. It means the same thing.
So now you accept my maths...?

OK, now assume we have no relative motion phases.
And then you are changing the setup of your very own thought experiment...for the second time during the course of this thread.

What we know is that if relative motion is eliminated, the two frames disagree on time and hence, they disagree on the speed of light when they are again at the same place.
So let me get this straight - you want to measure the speed of light by emitting a pulse from within an inertial frame. Then you leave that inertial frame, accelerating away; after your return you use your proper time to compare measurements ? And then you think because the times don't agree you have magically discovered a contradiction in SR ?

It is quite obvious that you do not understand the concepts of proper speed, proper distance and proper time as compared to coordinate speed, coordinate distance and coordinate time, all of which are related via the concept of rapidity.
The coordinate speed of light (both instantaneous and average) is slowed in the presence of acceleration and/or gravitational fields. The local instantaneous proper speed of light is always c. However, the average proper speed of light will differ from c in the presence of acceleration or a gravitational potential over the length of the worldline - this is a well understood relativistic phenomenon, and does not constitute a contradiction at all. Refer here for an in-depth treatment :

[gr-qc/9909081v7] Propagation of light in non-inertial reference frames

As you can clearly see, there are no contradictions anywhere.

so they disagree on the speed of light for the relative motion phases
According to yourself you had just eliminated the relative motion phases ?? This is now the third change of your setup. Seems as if you are so desparate to discover some imaginary contradiction that you are beginning to mix up your very own thought experiment.
Anyway, to check how the two observers relate to one another you need to use the relations for proper velocity and rapidity, like so :

wherein is the rapidity. This relation is true only if both observers see the same speed of light. So is it true or not in this scenario ? Let's check for one phase of acceleration - if it's true for one of the phases we know it is true for each of them, because the acceleration is constant :

Obviously the same relation hold for both observers, so long as the proper relations are used ! So once again, there are no contradictions in this scenario, and both observers see the same speed of light for the rest frame. For the ( global ) world line of the accelerating observer the speed of light does vary because there is acceleration involved, but when he gets back into the frame of the resting twin he once again sees the same speed of light, despite his acceleration, because his proper time measurement relates to rapidity, and not to "normal" velocity - that was the mistake you made. In essence, you were mixing reference frames - once again.

37. Before we are wasting any more time on this - it can be shown mathematically that Special Relativity is an internally self-consistent system, i.e. that it is in fact mathematically impossible to derive any form of contradiction from its axioms. This proof was first given by Hermann Minkowski back in 1908, but I have decided to present it again in its own thread with slightly more modern maths using tensor analysis. See here :

General Proof that Special Relativity is Self-Consistent

It is very straightforward. You will find appropriate references as well as empirical evidence to back up the calculation in there as well.
In essence, what this means is that anyone who claims to have discovered a contradiction in SR has merely succeeded in demonstrating his/her lack of understanding of the basic axioms and formalisms of Special Relativity, because there simply are no logical contradictions. The only way to actually disprove SR would be by showing empirical evidence that Lorentz invariance is indeed violated. This might well be the case in high energy domains in the order of the Planck length, but certainly not in the low energy realm, which is what we are talking about here with your twin scenario.

38. There are far too many posts, I won't bother reading them. Is this about Uniform motion or acceleration? If it's about acceleration then I'm out because I know nothing about General Relativity.

39. chinglu, just because I am a nice guy I will also prove to you that the speed of light is indeed always the same locally, even for the accelerating observer. Consider the classical Maxwell wave equations ( these are the equations for light waves ) :

and

Now, to find the covariant formulation we introduce the electromagnetic tensor F :

and because

the Maxwell wave equations now become

with general solutions of the form

for the E and B fields, wherein f is some arbitrary but analytically well-behaved function. Since the wave vector is of the form

it is clear that the wave propagates at the speed of light. Note that the wave vector k is a constant.

Since both the d'Alembert operator and the field tensor are evidently covariant, these wave equations are locally valid for both stationary and accelerated frames; thus is has been proven that the speed of light is locally always c, even for the accelerated twin, and even after going back to rest after acceleration.
You could do a similar prove for full general relativity - in that context the metric would be allowed to vary, and thus over the total world line of the twin the speed of light would also vary, however, since he ends up at rest after all acceleration phases the speed of light is once again the same as the stationary twin.

40. Markus Hanke are you a professor?

41. Originally Posted by Wise Man
Markus Hanke are you a professor?
Ha ha, no I am not
I'm someone who has spent most of my life trying to understand these things, merely out of personal interest...I'm just your average bookworm

42. Originally Posted by Markus Hanke
Before we are wasting any more time on this - it can be shown mathematically that Special Relativity is an internally self-consistent system, i.e. that it is in fact mathematically impossible to derive any form of contradiction from its axioms. This proof was first given by Hermann Minkowski back in 1908, but I have decided to present it again in its own thread with slightly more modern maths using tensor analysis. See here :

General Proof that Special Relativity is Self-Consistent

It is very straightforward. You will find appropriate references as well as empirical evidence to back up the calculation in there as well.
In essence, what this means is that anyone who claims to have discovered a contradiction in SR has merely succeeded in demonstrating his/her lack of understanding of the basic axioms and formalisms of Special Relativity, because there simply are no logical contradictions. The only way to actually disprove SR would be by showing empirical evidence that Lorentz invariance is indeed violated. This might well be the case in high energy domains in the order of the Planck length, but certainly not in the low energy realm, which is what we are talking about here with your twin scenario.

I will show you that later if you like. You need to apply SR with a partial derivative from all possible positions given any light pulse and 2 frames

What you have now to contend with is the fact that the twins experiment I presented reveals if any time dilation occurs to the moving twin during the relative motion phase, then the twins will disagree on the speed of light during the relative motion phases. That is a contradiction in SR.

That is now the subject of this thread.

43. [QUOTE=chinglu;346375]
Originally Posted by Markus Hanke
That is a contradiction in SR.

That is now the subject of this thread.
And that is where this thread gets bumped to pseudoscience.

44. Originally Posted by chinglu
So you keep saying about all my posts, without ever giving any proper refutations.
It's not my proof, it is Hermann Minkowski's, as clearly stated. I have merely reformulated this in tensor analysis, and presented it here.
But if you think you are smart enough, then go ahead, show the world mathematically why the proof is wrong. You are not up against me with this, but rather Prof Minkowski himself. Good luck with that.

You need to apply SR with a partial derivative from all possible positions given any light pulse and 2 frames
That's precisely what I have done in post 38. Or do you understand the concept of covariance ? It seems not.
Just show me where the maths in post 38 is wrong, that's all you need to do.

What you have now to contend with is the fact that the twins experiment I presented reveals if any time dilation occurs to the moving twin during the relative motion phase, then the twins will disagree on the speed of light during the relative motion phases. That is a contradiction in SR.
And what you now have to contend with is the fact that your thread got relegated to pseudoscience. Which is where it belongs, together with all the other anti-relativity cranks - I should mention also that Janus is a lot more knowledgeable about these matters as I am, so his judgement about your material is rather clear evidence of its nonsense status.

Btw, I must remind you that, in post 34, you had eliminated the relative motion phases.

That is now the subject of this thread.
You have already been proven wrong, there is no contradiction. Your refusal to acknowledge that fact does not make it any less right. I have shown this both in your specific scenario, as well as in the general case.

The subject of this thread will now be your full mathematical refutation of Minkowski's proof as I have presented it, as well as your full proof that Maxwell's equations are also wrong ( see post 38 ). In effect, you are now trying to refute about 80% of currently accepted physics, as well as afterwards trying to explain why all empirical evidence agrees with those physics which you are refuting. If you can't do that then you have exactly nothing.
Good luck - this should be fun

45. So, to summarize the current state of this thread, we are currently awaiting the following from you chinglu :

1. Full mathematical refutation of Minkowski's self-consistency proof
2. Full mathematical refutation of Maxwell's equations as presented in post 38
3. A mathematical explanation as to how you arrive at a sinh(...) function, as asked twice before
4. Empirical evidence, peer reviewed, repeatable and independently verified, of a real-world experiment which shows a violation of Special Relativity
5. A full explanation as to why all the experiments referenced in my thread "Modern Tests of Relativity" fully confirm all predictions of relativity since, as you claim, the theory is wrong

I shall be very much looking forward to your calculations. Needless to say that your maths will be scrutinized by us in extreme detail...

46. Originally Posted by Markus Hanke
So, to summarize the current state of this thread, we are currently awaiting the following from you chinglu :

1. Full mathematical refutation of Minkowski's self-consistency proof
2. Full mathematical refutation of Maxwell's equations as presented in post 38
3. A mathematical explanation as to how you arrive at a sinh(...) function, as asked twice before
4. Empirical evidence, peer reviewed, repeatable and independently verified, of a real-world experiment which shows a violation of Special Relativity
5. A full explanation as to why all the experiments referenced in my thread "Modern Tests of Relativity" fully confirm all predictions of relativity since, as you claim, the theory is wrong

I shall be very much looking forward to your calculations. Needless to say that your maths will be scrutinized by us in extreme detail...
Let's see where we are.

We can remove the effects of the acceleration from the total time for the twin's paradox.

Then we have the relative motion phase in which you agree the frames disagree on time during the relative motions phases.

Now exactly how does the mainstream claim the speed of light is the same for inertial frames given the same light sphere for the twins experiment and yet support time dilation?

Can you explain that?

47. Originally Posted by Markus Hanke
So, to summarize the current state of this thread, we are currently awaiting the following from you chinglu :

1. Full mathematical refutation of Minkowski's self-consistency proof
2. Full mathematical refutation of Maxwell's equations as presented in post 38
3. A mathematical explanation as to how you arrive at a sinh(...) function, as asked twice before
4. Empirical evidence, peer reviewed, repeatable and independently verified, of a real-world experiment which shows a violation of Special Relativity
5. A full explanation as to why all the experiments referenced in my thread "Modern Tests of Relativity" fully confirm all predictions of relativity since, as you claim, the theory is wrong

I shall be very much looking forward to your calculations. Needless to say that your maths will be scrutinized by us in extreme detail...

4. Empirical evidence, peer reviewed, repeatable and independently verified, of a real-world experiment which shows a violation of Special Relativity
Can you and Janus show exactly where the experimental evidence of SR proves the center of the light sphere moves during acceleration?

I want to see that. If that is not in the evidence, then I think any intelligent person would agree the center of the light sphere cannot move around in the coordinates of an accelerating frame. That is absolutely absurd and that is pure crackpottery.

48. Originally Posted by chinglu
Let's see where we are.

We can remove the effects of the acceleration from the total time for the twin's paradox.

Then we have the relative motion phase in which you agree the frames disagree on time during the relative motions phases.

Now exactly how does the mainstream claim the speed of light is the same for inertial frames given the same light sphere for the twins experiment and yet support time dilation?

Can you explain that?
So it sounds like you are saying that you cannot answer any of Markus's challenges and therefore you concede that your claims are completely false? Is that right?

But rather than admit that, you just keep repeating the same question that has already been answered.

49. Originally Posted by Strange
Originally Posted by chinglu
Let's see where we are.

We can remove the effects of the acceleration from the total time for the twin's paradox.

Then we have the relative motion phase in which you agree the frames disagree on time during the relative motions phases.

Now exactly how does the mainstream claim the speed of light is the same for inertial frames given the same light sphere for the twins experiment and yet support time dilation?

Can you explain that?
So it sounds like you are saying that you cannot answer any of Markus's challenges and therefore you concede that your claims are completely false? Is that right?

But rather than admit that, you just keep repeating the same question that has already been answered.

I can answer any of his challenges. However, this thread proves a simple case where SR is refuted.

So you people need to prove that SR is valid with this experiment. I have proven it is invalid.

More specifically, under the relative motion phases, exactly how can time dilation be true and also the twins measure exactly the same light sphere and speed of light when they are again common.

Can any of you people do this?

50. Originally Posted by chinglu
I can answer any of his challenges.
Good. Go on then.

51. Originally Posted by Strange
Originally Posted by chinglu
I can answer any of his challenges.
Good. Go on then.
I will prove the inconsistency of the Minkowski space-time.

But, this is a different issue from this thread.

First, the mainstream must explain this twin's paradox under the context of SR and make it work.

52. Originally Posted by chinglu
Now exactly how does the mainstream claim the speed of light is the same for inertial frames given the same light sphere for the twins experiment and yet support time dilation?
Can you explain that?
Of I course I can explain that - simply by using the correct relations and laws. What you are doing in your scenario is mixing an accelerating clock with a stationary ruler. You cannot do that - if you determine the speed of light in inertial frames, you must use inertial clocks and rulers. You then get time dilation between frames, but also length contraction, and the factor by which both measurements change is the same. It's that simple.

It is ridiculous that you are even questioning that, because it is very easy to test experimentally, and has been done hundreds of time. The outcome of all these experiments clearly show that the speed of is constant, just as SR postulates - it is a petty that you are totally ignoring the evidence. Also, a much more practical reason is that all observers see the same vacuum, regardless of their state of relative motion; and since light speed is directly determined by vacuum permittivity and permeability

it cannot change just because two observers are in relative motion. You do not even need SR to understand that, it is really just elementary electrodynamics !

I want to see that. If that is not in the evidence, then I think any intelligent person would agree the center of the light sphere cannot move around in the coordinates of an accelerating frame. That is absolutely absurd and that is pure crackpottery.
How is that absurd, considering that the light pulse was emitted while at rest ? How is that even an issue ? You emit the light pulse while at rest, you measure the speed while at rest, that's all there is to it. Your mistake is that you use a clock which does not remain at rest, but undergoes acceleration, and then compare this to a ruler measurement taken at rest. Now that is what I call absurd, because it means you are mixing reference frames. If you wish to do this then you must consider that any point on the expanding light sphere, as seen from the accelerating frame, behaves as if it is within a gravitational field. The result is that its trajectory becomes curved, like here :

Tests of general relativity - Wikipedia, the free encyclopedia
Gravitational lens - Wikipedia, the free encyclopedia

You have not considered this at all in your reasoning, you only talk about time.

So you people need to prove that SR is valid with this experiment.
Already done - see post 38, 36, 35. The proof stands as it is, and it does not need your agreement to do so. You on the other hand have not answered any of my proofs.
Besides, you are the one rejecting established science, so the onus is on you to provide proof, not the other way around.

More specifically, under the relative motion phases, exactly how can time dilation be true and also the twins measure exactly the same light sphere and speed of light when they are again common.
Relative motion phases means inertial frames. In inertial frames you have both time dilation and length contraction. Both carry the same gamma factor, so the relation between the two is always c.
Besides - "when they are again common" means they are again in the same frame of reference. If they are again in the same frame, they will also once again measure the same speed of light, because they use the same rulers and clocks.

But, this is a different issue from this thread.
Not at all, it is exactly what this thread is about. So long as Minkowski's general proof stands you are wrong by default, and each "maths" you present only shows off your ignorance of how to correctly apply relativistic laws.
So, what you need to do now is show the world mathematically that Minkowski's proof as I have presented it is wrong. If you can do that in a mathematically consistent way, then you have a good case which is worth considering.
If not, then all you have is a serious delusion.

53. Originally Posted by chinglu
I will prove the inconsistency of the Minkowski space-time.
I'm waiting for that. In the meantime I have two more challenges for you :

1. Consider heavy muons moving at speeds close to the speed of light as they are generated by cosmic radiation in the high atmosphere. They are very heavy, and thus have a very short lifespan, far too short in fact to cross the width of the atmosphere before they decay. Yet we do indeed detect them here at the surface. Explain this without recourse to any laws of relativity.
2. Consider two spherical ( at rest ) heavy gold ions accelerated in opposite directions close to the speed of light within particle accelerators. When they collide they behave not as spherical objects, but like flattened discs, as is evident in the resulting quark plasma. Explain this without recourse to relativity.

Now, I suggest you stop stalling and bring on the following, or else the list will grow too long :

1. Full mathematical refutation of Minkowski's self-consistency proof
2. Full mathematical refutation of Maxwell's equations as presented in post 38
3. A mathematical explanation as to how you arrive at a sinh(...) function, as asked twice before
4. Empirical evidence, peer reviewed, repeatable and independently verified, of a real-world experiment which shows a violation of Special Relativity
5. A full explanation as to why all the experiments referenced in my thread "Modern Tests of Relativity" fully confirm all predictions of relativity since, as you claim, the theory is wrong
6. Explain the muon observation
7. Explain the heavy ion collider experiment

I shall be very much looking forward to your calculations. Needless to say that your maths will be scrutinized by us in extreme detail...

54. Originally Posted by chinglu
I will prove the inconsistency of the Minkowski space-time.

Come on then. It can't be difficult. There are only 5 questions. It won't take someone like you long. Here, I'll get you started:

1.

2.

3.

4.

5.

All you have to do is fill in the blanks. You said you can do that. And I'm sure you weren't lying.

55. Originally Posted by Strange
All you have to do is fill in the blanks. You said you can do that. And I'm sure you weren't lying.
Ha ha
You know, number (1) would mean he has to prove that the metric tensor on a flat manifold is not constant.
Now that would be an achievement !

56. We can remove the effects of the acceleration from the total time for the twin's paradox.
No, we will not and we need not, because I just remembered the concept of Rindler coordinates, which I had read about in a textbook a while back. Using that coordinate system for accelerated frames in Minkowski space, it is now very easy to show that the observers agree on the speed of light when unaccelerated, and disagree if accelerated, just like GR says.

So, the Rindler coordinates are defined as ( in 2D, spacelike separation, x and y directions not considered ) :

and

which means that a ray of light appears to move in a hyperbolic trajectory, which is exactly what SR says about an accelerated observer. Now calculate the speed - remember that the time coordinate cT is now in the Rindler system, i.e. accelerated with constant acceleration a. The speed function when looking back at the stationary frame then becomes simply the first derivate with respect to (cT), which is

For the beginning and the end of the acceleration, i.e. when both twins are together and thus X=0, this speed becomes

Quod erad demonstrandum. Now, before you reply remember that these are Rindler coordinates, not Cartesian coordinates. The time coordinate (cT) is already dilated as compared to the rest frame, as you can see above. This coordinate system greatly simplifies the whole matter, and shows us immediately that, when the twins are together, the measured speed is always the speed of light c, independent if the time coordinates of either twins.

Yet another proof that shows the relativity works just fine, without contradiction !
I only wish I had remembered the Rindler coordinates earlier, this calculation is so much more straightforward !

So you people need to prove that SR is valid with this experiment.
Done and dusted, and with minimum effort.
You on the other hand...

57. Originally Posted by Markus Hanke
Originally Posted by chinglu
Now exactly how does the mainstream claim the speed of light is the same for inertial frames given the same light sphere for the twins experiment and yet support time dilation?
Can you explain that?
Of I course I can explain that - simply by using the correct relations and laws. What you are doing in your scenario is mixing an accelerating clock with a stationary ruler. You cannot do that - if you determine the speed of light in inertial frames, you must use inertial clocks and rulers. You then get time dilation between frames, but also length contraction, and the factor by which both measurements change is the same. It's that simple.

It is ridiculous that you are even questioning that, because it is very easy to test experimentally, and has been done hundreds of time. The outcome of all these experiments clearly show that the speed of is constant, just as SR postulates - it is a petty that you are totally ignoring the evidence. Also, a much more practical reason is that all observers see the same vacuum, regardless of their state of relative motion; and since light speed is directly determined by vacuum permittivity and permeability

it cannot change just because two observers are in relative motion. You do not even need SR to understand that, it is really just elementary electrodynamics !

I want to see that. If that is not in the evidence, then I think any intelligent person would agree the center of the light sphere cannot move around in the coordinates of an accelerating frame. That is absolutely absurd and that is pure crackpottery.
How is that absurd, considering that the light pulse was emitted while at rest ? How is that even an issue ? You emit the light pulse while at rest, you measure the speed while at rest, that's all there is to it. Your mistake is that you use a clock which does not remain at rest, but undergoes acceleration, and then compare this to a ruler measurement taken at rest. Now that is what I call absurd, because it means you are mixing reference frames. If you wish to do this then you must consider that any point on the expanding light sphere, as seen from the accelerating frame, behaves as if it is within a gravitational field. The result is that its trajectory becomes curved, like here :

Tests of general relativity - Wikipedia, the free encyclopedia
Gravitational lens - Wikipedia, the free encyclopedia

You have not considered this at all in your reasoning, you only talk about time.

So you people need to prove that SR is valid with this experiment.
Already done - see post 38, 36, 35. The proof stands as it is, and it does not need your agreement to do so. You on the other hand have not answered any of my proofs.
Besides, you are the one rejecting established science, so the onus is on you to provide proof, not the other way around.

More specifically, under the relative motion phases, exactly how can time dilation be true and also the twins measure exactly the same light sphere and speed of light when they are again common.
Relative motion phases means inertial frames. In inertial frames you have both time dilation and length contraction. Both carry the same gamma factor, so the relation between the two is always c.
Besides - "when they are again common" means they are again in the same frame of reference. If they are again in the same frame, they will also once again measure the same speed of light, because they use the same rulers and clocks.

But, this is a different issue from this thread.
Not at all, it is exactly what this thread is about. So long as Minkowski's general proof stands you are wrong by default, and each "maths" you present only shows off your ignorance of how to correctly apply relativistic laws.
So, what you need to do now is show the world mathematically that Minkowski's proof as I have presented it is wrong. If you can do that in a mathematically consistent way, then you have a good case which is worth considering.
If not, then all you have is a serious delusion.
Of I course I can explain that - simply by using the correct relations and laws. What you are doing in your scenario is mixing an accelerating clock with a stationary ruler. You cannot do that - if you determine the speed of light in inertial frames, you must use inertial clocks and rulers. You then get time dilation between frames, but also length contraction, and the factor by which both measurements change is the same. It's that simple.

I excluded the acceleration component as done with GPS. I then isolated the time dilation component.

Why does this math fail since GPS uses it?

58. Originally Posted by Markus Hanke
We can remove the effects of the acceleration from the total time for the twin's paradox.
No, we will not and we need not, because I just remembered the concept of Rindler coordinates, which I had read about in a textbook a while back. Using that coordinate system for accelerated frames in Minkowski space, it is now very easy to show that the observers agree on the speed of light when unaccelerated, and disagree if accelerated, just like GR says.

So, the Rindler coordinates are defined as ( in 2D, spacelike separation, x and y directions not considered ) :

and

which means that a ray of light appears to move in a hyperbolic trajectory, which is exactly what SR says about an accelerated observer. Now calculate the speed - remember that the time coordinate cT is now in the Rindler system, i.e. accelerated with constant acceleration a. The speed function when looking back at the stationary frame then becomes simply the first derivate with respect to (cT), which is

For the beginning and the end of the acceleration, i.e. when both twins are together and thus X=0, this speed becomes

Quod erad demonstrandum. Now, before you reply remember that these are Rindler coordinates, not Cartesian coordinates. The time coordinate (cT) is already dilated as compared to the rest frame, as you can see above. This coordinate system greatly simplifies the whole matter, and shows us immediately that, when the twins are together, the measured speed is always the speed of light c, independent if the time coordinates of either twins.

Yet another proof that shows the relativity works just fine, without contradiction !
I only wish I had remembered the Rindler coordinates earlier, this calculation is so much more straightforward !

So you people need to prove that SR is valid with this experiment.
Done and dusted, and with minimum effort.
You on the other hand...
The problem with your argument is that you are claiming for whatever reasons that when the frames are in relative motion the will disagree on the speed of light for the same light pulse.

How exactly are you able to prove under SR, that 2 frames in relative motion agree on a speed of light c and yet also do not agree on a speed of light c.

Can you explain that?

59. Originally Posted by chinglu
The problem with your argument is that you are claiming for whatever reasons that when the frames are in relative motion the will disagree on the speed of light for the same light pulse.
Do you realize that these are Rindler coordinates ?
I followed your setup in which one twin stays at home, whereas the other twin accelerates away, then comes back to join the stay-at-home twin. All I have shown is that while acceleration takes place the moving twin may not necessarily agree with the stationary twin on the speed of light measurement ( equivalence principle ); however, when they are back in the same frame without any relative motion they will once again agree.

I excluded the acceleration component as done with GPS.
The GPS time dilation calculation does not exclude acceleration at all - it accounts for both SR time dilation and gravitational time dilation under GR :

60. Originally Posted by chinglu
I will prove the inconsistency of the Minkowski space-time.
Still waiting ...

61. You have not replied to any of the points in post 52. I suggest you start addressing those instead of going over the same old scenario ( which has been sufficiently answered ) over and over again.

EDIT : I guess we aren't going to see his 'proof' after all - petty, I was actually looking forward to that !!

62. Originally Posted by Markus Hanke
No, we will not and we need not, because I just remembered the concept of Rindler coordinates, which I had read about in a textbook a while back
I mentioned them in post #13 of this thread, too.

63. Originally Posted by SpeedFreek
Originally Posted by Markus Hanke
No, we will not and we need not, because I just remembered the concept of Rindler coordinates, which I had read about in a textbook a while back
I mentioned them in post #13 of this thread, too.

You're right, I completely missed that !

64. Even my one line answers, like the one in post #11, are designed to lead the thread in the relevant direction, by pointing out the key issues involved.

65. Originally Posted by SpeedFreek
Even my one line answers, like the one in post #11, are designed to lead the thread in the relevant direction, by pointing out the key issues involved.
I don't think any of chinglu's threads ever went into a relevant direction...he just kept repeating his own delusions over and over again while constantly changing his own thought experiment setups.

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