Is the frame dragging part of the Special Relativity?

[Robin]

Originally Posted by SDG

What is your take on the question if the traveling twin is going to cross 3.4641cs' in 2s' or 4s'? Why faster than light?

I am not sure why you don't like the straightforward answer that many have already given you.

[SDG]

Originally Posted by Robin

I am not sure why you don't like the straightforward answer that many have already given you.

If we take the straightforward 2s' answer then there is a problem.
The inertial frame moving at 0.866c that represents an average for the accelerated frame crosses the distance in 4s'.
The 2s' answer is in disagreement with this:

[Robin]

Originally Posted by SDG

If we take the straightforward 2s' answer then there is a problem.

The inertial frame moving at 0.866c that represents an average for the accelerated frame crosses the distance in 4s'.

The 2s' answer is in disagreement with this:



https://i.imgur.com/CYocBbl.png

And you have been told many times that you can't treat an accelerated traveller as though it was an inertial traveller and here you are doing it again.

And you ask why people think you aren't trying to understand the answer.

[Robin]

Pick one scenario. Describe it clearly. Use events and give a clear description for each event. Show the maths. Stick to maths that is allowed in the theory.

Then see if there is a problem.

[Myriad]

Originally Posted by Robin

Pick one scenario. Describe it clearly. Use events and give a clear description for each event. Show the maths. Stick to maths that is allowed in the theory.

Then see if there is a problem.

He knows by now that if he does all that, the problem will go away. That's the problem.

[Robin]

Originally Posted by Myriad

He knows by now that if he does all that, the problem will go away. That's the problem.

Yes, I think that is the reason he is switching randomly between vaguely defined scenarios.

[SDG]

Originally Posted by Robin

And you have been told many times that you can't treat an accelerated traveller as though it was an inertial traveller and here you are doing it again.

And you ask why people think you aren't trying to understand the answer.

I did that in the post #317

Here it is again. The inertial traveler, the light blue axis.



It takes 4s' to cross 3.4641cs' for this inertial traveler.
This is the analysis from the point of view of this inertial traveler.

[Robin]

Originally Posted by SDG

I did that in the post #317



Here it is again. The inertial traveler, the light blue axis.



https://i.imgur.com/xIVUcQT.png



It takes 4s' to cross 3.4641cs' for this inertial traveler.

This is the analysis from the point of view of this inertial traveler.



https://i.imgur.com/CYocBbl.png

Are you saying that your second diagram there is the first diagram from the point of view of that inertial traveller shown by the light blue line?

I think not.

But again, I have pointed out again that non-inertial paths can't be treated like inertial paths and you respond yet again by trying to transform a non inertial path like an inertial path.

[steenkh]

Originally Posted by SDG

For starters, is the Lorentz contraction real or not? What do you think?

What is your take on the question if the traveling twin is going to cross 3.4641cs' in 2s' or 4s'? Why faster than light?

I am not qualified to have an opinion on these matters, but as far as I can gather, plenty of posters here are qualified, and I even believe you have got their qualified answers.

To me, it certainly looks as if you studiously ignore the answers, and just repeat your questions with new twists.

[Darat]

Originally Posted by SDG

I asked a specific question and I presented reasons why I asked the question.
Do you have an answer to the question why a traveling twin would cross 3.4641cs' distance in 2s'?
Why the traveling twin would move faster than light?

To paraphrase someone: I asked 2 specific questions....

Why do you think you have spotted a mistake that has gone unnoticed by generations of physicists and mathematicians? Why do you think you are not able to convince anyone of your correction to established physics?

[SDG]

Originally Posted by Robin

Are you saying that your second diagram there is the first diagram from the point of view of that inertial traveller shown by the light blue line?

I think not.

Yes, it is. Why it would not be?
Is the distance crossed 3.4641cs' or not?

Quote:

But again, I have pointed out again that non-inertial paths can't be treated like inertial paths and you respond yet again by trying to transform a non inertial path like an inertial path.

The red line from A to P is the accelerated frame, using X/R approach.



Is the light blue line following the same world line or not?

[Darat]

Originally Posted by SDG

...snip...
Is the light blue line following the same world line or not?

It will definitely be yes or not.

Now I've answered that how about you answer the following questions:

Why do you think you have spotted a mistake that has gone unnoticed by generations of physicists and mathematicians? Why do you think you are not able to convince anyone of your correction to established physics?

[Myriad]

Originally Posted by SDG

https://i.imgur.com/Ec2Ot1k.png

Is the light blue line following the same world line or not?

What light blue line? The one curving downward from above the t = T line, with an arrow head pointing vaguely toward event P, and with the three enormous red exclamation points next to it?

No, that's not following the same world line. (I'd ask, "same world line as what?," but it's not following any valid world line so the answer is no regardless.)

That's the only blue line in this diagram. Do you even look at the diagrams you repeatedly post?

[SDG]

Originally Posted by Darat

It will definitely be yes or not.

Now I've answered that how about you answer the following questions:

Why do you think you have spotted a mistake that has gone unnoticed by generations of physicists and mathematicians? Why do you think you are not able to convince anyone of your correction to established physics?

Is it a mistake?
I have not found a textbook that treats the twin paradox with a real stop at the mid point P.

Is it going to be a mistake when we do the analysis?

[SDG]

Originally Posted by Myriad

What light blue line? The one curving downward from above the t = T line, with an arrow head pointing vaguely toward event P, and with the three enormous red exclamation points next to it?

No, that's not following the same world line. (I'd ask, "same world line as what?," but it's not following any valid world line so the answer is no regardless.)

That's the only blue line in this diagram. Do you even look at the diagrams you repeatedly post?

This one:





The light blue line follows the same worldline as the dark red one. Agreed?

[Robin]

Originally Posted by SDG

Yes, it is. Why it would not be?

How are they related? Where is P on your second diagram?

What are the values in the first diagram?

Quote:

Is the distance crossed 3.4641cs' or not?

As I have said before, be specific. If you are going to ask a question like that specify the world line you are talking about, specify the event at the beginning and end and then we can work it out.

[Reality Check]

Originally Posted by SDG

The light blue line follows the same worldline as the dark red one. Agreed?

SDG continues to show ignorance about the textbook images he spamming the thread with.
No one with eyes will agree. The light blue line is a straight line. The dark red one is a curved line. They are different worldlines connecting the same events in spacetime.

1. 19 August 2020: SDG is wasting our time with a scenario that is explained in his textbook.
2. 19 August 2020: "The trouble with the simplified traveling twin point of view diagram." ignorance when there was no trouble.
3. 19 August 2020: Deep "they have the same size ruler" ignorance
4. 20 August 2020: Abysmal "This table is wrong. ... Because this table breaks the Equivalence Principle." ignorance.
5. 20 August 2020: "This table is wrong. ... Because this table breaks the Equivalence Principle." lie. A "1g acceleration" means an increasing Lorentz factor. This is not a "circle trajectory". A lie by quote mining (9 missing pages)

Followed by a weekend of more ignorance.

1. Fig 2.9's caption does not have an extra "comoving observer". It has Langevin's traveler O' (the travelling twin) with an infinite accelerations at A P and B. There is also a "tri-hypobolic" worldline with finite acceleration.
   These look like a simplified and more realistic way of looking at the twin's trajectory.
2. A "Do you have some answers?" when SDG has been given answers in this thread and other threads. SDG has presented no issues with relativity other than his constant refusal to learn SR even though he is citing from a relativity textbook!
3. SDG posts in a different, not sourced, uncaptioned image with a "traveling comoving observer crosses 3.4641cs'" claim.
4. The topic of this thread (see the title) was abandoned early on. This is now the "SDG in denial of relativity while citing a relativity textbook" thread.
   Darat's SDG - do you really think you have spotted something that literally generations of physicists and mathematicians have got wrong? is on topic.
5. Continues with his ignorant "Is it 4s' or 2s'?" question.
6. Derails into GR with "What local is good enough local" ignorance.
   Anyone with a relativity textbook which he has or has read about the equivalence principle knows the answer.
7. "momentarily comoving inertial reference frames?" derail.
8. An irrelevant "Where we can find parallel gravitational acceleration lines of a meaningful magnitude in the universe?" question.
9. More 2s/4s dertailing - time dilation exists! A travelling twin's clock will tick slower than the stay at home twin's clock.
10. etc.

[Reality Check]

Originally Posted by SDG

...They are very much alike and it is Einstein who came with the EP, assuming an unrealistic uniform gravitational acceleration.

24 August 2020: A "unrealistic uniform gravitational acceleration" in the equivalence principle lie.
For a start Einstein did not come up with the equivalence principle out of thin air: "Something like the equivalence principle emerged in the early 17th century, when Galileo expressed experimentally that the acceleration of a test mass due to gravitation is independent of the amount of mass being accelerated.". Einstein stated the equivalence of inertial and gravitational mass as the equivalence of gravitational fields and acceleration locally.
The lie is that uniform gravitational fields are unrealistic. The equivalence principle explicably states that the equivalence applies locally which is a scale where divergence of gravitational field lines can be ignored, i.e. the gravitational field is uniform. For example, realistically we can ignore tidal effects between our feet and head and treat Earth's surface gravity as uniform.
On the same Wikipedia page is Tests of the Einstein equivalence principle (and th weak and strong principles). Of course the indirect evidence that the equivalence principle is that fact that general relativity works extremely well !

[Robin]

SDG,

So let's take your top diagram and put some description to it, correct me if I have misunderstood you:

A (t=0 x=0) Traveller 1 (red) has v=0 and begins moving
P (t=4,x=3.464) Traveller 1 (red) has reached v=0 and continues moving in the opposite direction
B (t=8,x=0) Traveller 1 (red) has again reached a velocity of zero
Traveller 2 (blue) is moving with a constant velocity of 0.866c and is at the same position as traveller 1 at A and P.

And then plot that and put it from the perspective of the initial and final velocity of traveller 1 (left) and of traveller 2 (right):



A=(0,0,0,0) A'=(0,0,0,0)
P=(4,3.464,0,0) P'=(2,0,0,0)
B=(8,0,0,0) B'=(16,-13.856,0,0)

Note that the accelerating traveller must come very close to the speed of light at mid journey in order to arrive at the same place and time as the inertial traveller.

[SDG]

Originally Posted by Robin

SDG,

So let's take your top diagram and put some description to it, correct me if I have misunderstood you:

A (t=0 x=0) Traveller 1 (red) has v=0 and begins moving
P (t=4,x=3.464) Traveller 1 (red) has reached v=0 and continues moving in the opposite direction
B (t=8,x=0) Traveller 1 (red) has again reached a velocity of zero
Traveller 2 (blue) is moving with a constant velocity of 0.866c and is at the same position as traveller 1 at A and P.

And then plot that and put it from the perspective of the initial and final velocity of traveller 1 (left) and of traveller 2 (right):

https://robinsrevision.files.wordpre...-20.png?w=1024

A=(0,0,0,0) A'=(0,0,0,0)
P=(4,3.464,0,0) P'=(2,0,0,0)
B=(8,0,0,0) B'=(16,-13.856,0,0)

Note that the accelerating traveller must come very close to the speed of light at mid journey in order to arrive at the same place and time as the inertial traveller.

Here is the misunderstanding.



The green horizontal line is platform ruler, the length is 3.4641cs.
The train has to cross this distance in 2s'.
The ct axis has to move to the left 3.4641cs' in 2s'.
Nothing can move faster than light in the blue inertial frame.

[Robin]

Originally Posted by SDG

Here is the misunderstanding.

Now you are drawing a completely different diagram with events not shown on the original and not following any of the conventions of relativity diagrams.

This is some new theory of your own??

And where did you mention a platform in the first diagram you provided?
Where is P on the new diagram you show?? Did B move??

Again, state your scenario completely and clearly defined events and world lines and properly drawn diagrams with events consistently drawn on each and no extraneous letters or symbols.

[Myriad]

Originally Posted by SDG

The green horizontal line is platform ruler, the length is 3.4641cs.
The train has to cross this distance in 2s'.
The ct axis has to move to the left 3.4641cs' in 2s'.
Nothing can move faster than light in the blue inertial frame.

In the blue inertial frame, the platform length is not 3.4641cs.

[Robin]

SDG,

You provided this (a text book diagram with non-specific times and distances) and some extra stuff drawn on it:



I suggested that this is what you mean by it:


A=(0,0), P=(4,3.464), B=(8,0)

Now what have I missed in my diagram, that is in your diagram above it and relevant to the point you are making?

Where is the platform in this diagram?

[Robin]

Do I understand this platform to be 3.434 cs long and start at x=0?

If so then we can represent the front of the platform with a world line, shown here in green. Platform frame on the left, blue inertial traveller frame on the right:



A (t=0 x=0) Traveller 1 (red) has v=0 and begins moving
P (t=4,x=3.464) Traveller 1 (red) has reached v=0 and continues moving in the opposite direction
B (t=8,x=0) Traveller 1 (red) has again reached a velocity of zero
Traveller 2 (blue) is moving with a constant velocity of 0,866c and is at the same position as traveller 1 at A and P.
Platform is 3.464cs long and begins at x=0.
Front of platform world line shown in green

A=(0,0,0,0) A'=(0,0,0,0)
P=(4,3.464,0,0) P'=(2,0,0,0)
B=(8,0,0,0) B'=(16,-13.856,0,0)

[Robin]

So, from each point of view,

The inertial traveller has covered 3.464 cs in 4 s or;

The inertial traveller has remained in place and the front of the platform has covered 1.732 cs in 2 s.

[Robin]

SDG

The main problem with your diagram below (apart from the fact that it does not follow any of the conventions of spacetime diagrams) is that your light lines are wrong.

In the non-inertial frame of the accelerated traveller, if the time axis is straight then he must be coming under three separate gravitational fields and so at that light line won't be in a straight line on your diagram. Calculate some light paths for those areas and you will find that nothing has exceeded the speed of light:

Originally Posted by SDG

...

[SDG]

Originally Posted by Robin

Now you are drawing a completely different diagram with events not shown on the original and not following any of the conventions of relativity diagrams.

This is some new theory of your own??

And where did you mention a platform in the first diagram you provided?
Where is P on the new diagram you show?? Did B move??

Again, state your scenario completely and clearly defined events and world lines and properly drawn diagrams with events consistently drawn on each and no extraneous letters or symbols.

Robin,
It is buried in the first pages of this thread, B and P are the same events.





Here is the platform, the green horizontal lines.
If you compare it to the left side diagram below and B=P.
The traveler twin stops at P/B event. The ct' stays at x=3.4641cs in the platform frame.
The ct axis is at x'=-3.4641cs'.
How long for the ct to cross 3.4641cs' distance to the left in the blue inertial frame that goes through A, P events?
Your other diagrams do not capture stopping at P.
When the stop happens than x, x' rulers are equal and ct, ct' axes are parallel.

[SDG]

Originally Posted by Robin

Do I understand this platform to be 3.434 cs long and start at x=0?

If so then we can represent the front of the platform with a world line, shown here in green. Platform frame on the left, blue inertial traveller frame on the right:



A (t=0 x=0) Traveller 1 (red) has v=0 and begins moving
P (t=4,x=3.464) Traveller 1 (red) has reached v=0 and continues moving in the opposite direction
B (t=8,x=0) Traveller 1 (red) has again reached a velocity of zero
Traveller 2 (blue) is moving with a constant velocity of 0,866c and is at the same position as traveller 1 at A and P.
Platform is 3.464cs long and begins at x=0.
Front of platform world line shown in green

A=(0,0,0,0) A'=(0,0,0,0)
P=(4,3.464,0,0) P'=(2,0,0,0)
B=(8,0,0,0) B'=(16,-13.856,0,0)

OK, so A'->P' 2s' and P'->B' 14s'.
The traveling twin is 2x older.

The left diagram, the red line goes straight up after P because we wan to know what happens when the traveling twin stops at P.
The axis t is supposed to be parallel to t' after 2s' or 4s' and it is x'=-3.4641cs' to the left.

[Robin]

Originally Posted by SDG

Robin,
It is buried in the first pages of this thread, B and P are the same events.

Here is the diagram that you supplied that I was trying to interpret and it has B and P in different positions, so they can't be the same event. You are still mixing and matching scenarios so no wonder you are getting inconsistent answers.

[Robin]

Originally Posted by SDG

The left diagram, the red line goes straight up after P because we wan to know what happens when the traveling twin stops at P.
The axis t is supposed to be parallel to t' after 2s' or 4s' and it is x'=-3.4641cs' to the left.

Again you are mixing and matching.

You supplied a diagram which you claimed was from the point of view of the inertial traveller and that is the scenario I am showing.

[SDG]

Originally Posted by Robin

Here is the diagram that you supplied that I was trying to interpret and it has B and P in different positions, so they can't be the same event. You are still mixing and matching scenarios so no wonder you are getting inconsistent answers.

Please, forget B in this diagram.
What happens when the traveling twin stops at P?
P here is B in my diagrams.

[Robin]

Originally Posted by SDG

Please, forget B in this diagram.
What happens when the traveling twin stops at P?
P here is B in my diagrams.

It would be nice if you supplied diagrams which accurately represent the situation you are trying to describe, with consistent events.

[Robin]

So, to be clear, we have:



Where

A (0,0) Twin 2 (blue) accelerates away to the right
B (4,3.464) Twin 2 has decelerated to 0 and remains at v=0
C (4,0) Twin 1 notes 4 seconds passing on clock

Twin 1 Green world lines, front and back of platform
Twin 2 Blue world lines, front and back of train

[Robin]

So from the point of view of the inertial traveller (shown in orange) our million kilometre long train is doing something like this:

[Robin]

And if we absolutely must try to shoehorn the proper time path of the accelerated one million kilometre train onto one of these diagrams, it would look something like this, I believe:


Note that there is no A', B' or C' since at these events everything is travelling at the same velocity.

However the platform clock will show 4s in either case and the train time will show approximately 2s at B in either case. A stationary clock that started at the front of the platform at t=0, x=3.464 would show 4s at B

[Robin]

Interesting to know if the million kilometre long train is being pulled or shunted.

Forces can only propogate through any material at the speed of sound for that material.

So the scenario is kind of silly in practical terms.

[Pixel42]

Originally Posted by Robin

Interesting to know if the million kilometre long train is being pulled or shunted.

Forces can only propogate through any material at the speed of sound for that material.

So the scenario is kind of silly in practical terms.

The silliness is essential. Silly and unnecessarily complex scenarios are needed to enable the misinterpretations that physics crackpots require, in order to pretend that there are inconsistences in relativity.

[Robin]

Originally Posted by Robin

And if we absolutely must try to shoehorn the proper time path of the accelerated one million kilometre train onto one of these diagrams, it would look something like this, I believe:

https://robinsrevision.files.wordpre...-27.png?w=1024
Note that there is no A', B' or C' since at these events everything is travelling at the same velocity.

However the platform clock will show 4s in either case and the train time will show approximately 2s at B in either case. A stationary clock that started at the front of the platform at t=0, x=3.464 would show 4s at B

Obvious careless mistake on the second diagram here, I will fix up if I have time.

[HansMustermann]

Originally Posted by SDG

It is not dumb what I wrote.
Look, the simultaneity is a problem for the measurement.
The rest frame length L₀ is what all observers are supposed to agree upon.

Actually that just illustrates what is dumb about it: That presupposes that there is such a preferred frame. The whole POINT of relativity is that there isn't. ESPECIALLY in SPECIAL relativity.

In GR you can differentiate between inertial and non-inertial, but even then it doesn't automatically mean that one is more "right" than another. It may be easier to do the maths in one than in the other, but that's where any claims of being the right one stops. But in SR, you don't even have that: all inertial frames are just as good.

There is no such thing as a rest frame, nor any such thing of there being any frame that everyone must agree that it has the right measurements. Two frames can be at rest relative to each other, or not, but that's about it. Saying that the train is moving with speed v relative to the station, or that the station is moving with speed -v relative to the train is equally true. And not just saying it. Fundamentally, none of the two frames has any claim to being the one true frame that everyone has to agree on. Both are exactly as good as each other.

That's the whole POINT of relativity. That's why it has "relative" in the name: it's all relative. As in, relative to whatever frame you're using.

I.e., it's just as valid to say that Ross 128 is 11 ly away in the frame of Earth as it is to say it's 5.5 ly away in the frame of someone moving at v=0.866c. There is no one true distance.

[Robin]

Fix up of previous graph: