Is the frame dragging part of the Special Relativity? - Page 6

Ziggurat · 15th August 2020, 06:29 PM

Originally Posted by SDG

You are mixing sub-discussions.

You can't stick with a single topic, and *I'm* the one mixing sub-discussions?

Quote:

If gamma = 2 did the traveler crossed 3.4641cs in 2 seconds?

This is an ambiguous question. What time are you referring to? Proper time or coordinate time?

I'm not going to bother double-checking your math right now, but it looks plausible. In which case, the traveler went 3.46 cs distance in the unprimed reference frame, and experienced 2 seconds of proper time in doing so.

So where's the conflict? You still haven't produced one.

SDG · 15th August 2020, 08:29 PM

Originally Posted by Ziggurat

You can't stick with a single topic, and *I'm* the one mixing sub-discussions?

This is an ambiguous question. What time are you referring to? Proper time or coordinate time?

I'm not going to bother double-checking your math right now, but it looks plausible. In which case, the traveler went 3.46 cs distance in the unprimed reference frame, and experienced 2 seconds of proper time in doing so.

So where's the conflict? You still haven't produced one.

Where is 8s of traveler coordinate time to account for 4s of stay home twin proper time?

Ziggurat · 15th August 2020, 09:03 PM

Originally Posted by SDG

Where is 8s of traveler coordinate time to account for 4s of stay home twin proper time?

It's where the line of simultaneity for the outbound prime reference frame would intersect both the 4 second mark for the earthbound twin and that outbound journey, if the traveling twin hadn't turned around.

But he did. And that changed everything.

This is really, really basic special relativity. I don't know what to tell you if you still can't get it.

HansMustermann · 16th August 2020, 12:56 AM

Well, what I could say as a hint is what I've said before: don't forget that t' = γ(t - vx/c2), not just γt. That seems to account for pretty much every confusion about the twins paradox I've seen so far.

Robin · 16th August 2020, 04:50 AM

Originally Posted by SDG

Where is 8s of traveler coordinate time to account for 4s of stay home twin proper time?

What is the co-ordinate time of event C in your diagram in the traveler's frame (when at full speed)?

C: t=4, x=0, t'=?, x'=?

SDG · 16th August 2020, 05:16 AM

Originally Posted by Ziggurat

It's where the line of simultaneity for the outbound prime reference frame would intersect both the 4 second mark for the earthbound twin and that outbound journey, if the traveling twin hadn't turned around.

But he did. And that changed everything.

This is really, really basic special relativity. I don't know what to tell you if you still can't get it.

Fair enough.
The platform 1s of proper time is 2s' of train coordinate time.
How do 3s of platform proper time flow when there is no change of the traveling twin coordinate time?

SDG · 16th August 2020, 05:30 AM

Originally Posted by Robin

What is the co-ordinate time of event C in your diagram in the traveler's frame (when at full speed)?

C: t=4, x=0, t'=?, x'=?

The last thread of discussion is that the traveler stops.
Therefore t'=2s and x'=-3.4641cs.
There is no more length contraction after the stop.
There are question marks as per my post above.
It's like the platform origin crossed 3.4641cs to the left in 2s', right?

Robin · 16th August 2020, 05:53 AM

Originally Posted by SDG

The last thread of discussion is that the traveler stops.
Therefore t'=2s and x'=-3.4641cs.
There is no more length contraction after the stop.

There is no t' when the traveller stops. The twins have the same velocity after that.

Robin · 16th August 2020, 05:56 AM

Originally Posted by SDG

Fair enough.
The platform 1s of proper time is 2s' of train coordinate time.
How do 3s of platform proper time flow when there is no change of the traveling twin coordinate time?

That would only happen if the traveller's velocity changed instantaneously to zero.

If we are going to discuss a simplification we should accept that the simplification will break down at some stage. That doesn't imply a problem for relativity.

SDG · 16th August 2020, 06:38 AM

Originally Posted by Robin

There is no t' when the traveller stops. The twins have the same velocity after that.

There has to be a time dilation.
If it was a train car moving then front and the back observers have 2s on their clocks. Agreed?
How did the platform x=0 observer moved across 3.4641cs train car in 2s?

SDG · 16th August 2020, 06:41 AM

Originally Posted by Robin

That would only happen if the traveller's velocity changed instantaneously to zero.

If we are going to discuss a simplification we should accept that the simplification will break down at some stage. That doesn't imply a problem for relativity.

The same problem arises when we use accelerations without simplifications.
It is a problem for the relativity.

Myriad · 16th August 2020, 06:41 AM

Originally Posted by SDG

How did the platform x=0 observer moved across 3.4641cs train car in 2s?

Oh, I know this one. Because he wanted to see time fly.

Ziggurat · 16th August 2020, 07:27 AM

Originally Posted by SDG

Fair enough.
The platform 1s of proper time is 2s' of train coordinate time.
How do 3s of platform proper time flow when there is no change of the traveling twin coordinate time?

You did the acceleration instantly. Do it in a finite amount of time, and in that accelerating frame, the platform time passes very rapidly but it's still continuous.

I told you, accelerated frames are weird.

SDG · 16th August 2020, 07:42 AM

Originally Posted by Ziggurat

You did the acceleration instantly. Do it in a finite amount of time, and in that accelerating frame, the platform time passes very rapidly but it's still continuous.

I told you, accelerated frames are weird.

So now we cannot simplify, right.
So no simplification.

There is a train car with L₀=3.4641cs and the platform frame.
Front of the train car is x'=0 and t'=0 and it is align with the platform origin x=0 and t=0.

So the green B is platform origin and train car front (the train origin) aligned.
There is no motion at the beginning, the motion starts at t=t'=0.
The acceleration, then the deceleration as per the diagram.
It takes 4s of the platform time.
Then everything stops again.
The back of the train car is aligned with the platform origin.
There is a time dilation so the front train clock and the back train clock have both 2s on them, agreed?

How did the platform origin crossed 3.4641cs of the train frame in 2s of the train frame?

Ziggurat · 16th August 2020, 08:22 AM

Originally Posted by SDG

So now we cannot simplify, right.
So no simplification.

We can simplify, if you understand the simplification. You aren't comfortable with the results of that simplification, but so what? Your discomfort doesn't matter. So what if there's a time jump? It's a coordinate effect. As I explained previously, there is no discontinuity in what the traveling twin sees. Even if he reverses direction instantly, what he sees is still continuous. So why do you care if there's a coordinate time jump at a location outside his light cone? Why does it matter?

Quote:

How did the platform origin crossed 3.4641cs of the train frame in 2s of the train frame?

Look at the sun. It's about 8 light minutes away. Turn 180 degrees. Now it's 8 light minutes away in the other direction. How did the sun travel 16 light minutes in 1 second?

You changed reference frames.

Myriad · 16th August 2020, 09:46 AM

This effect of special relativity has been a trope in science fiction for a hundred years. A ~~train~~ spaceship travels to a distant ~~platform~~ star system, accelerating to a high fraction of the speed of light near the start and decelerating at the destination. From the stationary frame of the origin ~~platform~~ planet or the destination ~~platform~~ planet, the ~~train~~ ship always travels slower than the speed of light, but the traveler's "subjective time" (time on the ~~train's~~ ship's clock) is less than the time it takes light to make the trip in the origin's or destination's frame.

Disregarding the acceleration and deceleration times, a trip at a relative velocity greater than .707c (c/sqrt(2)) will be "faster than light" in subjective time. The traveler's subjective time can be arbitrarily short if arbitrarily high fractions of c are reached (as in e.g. Poul Anderson's Tao Zero).

Of course, from the traveler's frame, the distance traveled is shortened, so the traveler never observes the ship traveling faster than light either.

Robin · 16th August 2020, 02:19 PM

Originally Posted by SDG

There has to be a time dilation.

If it was a train car moving then front and the back observers have 2s on their clocks. Agreed?

How did the platform x=0 observer moved across 3.4641cs train car in 2s?

When the traveller stops and compares his clock to a stationary clock, do you think they will read the same?

What will the stationary be clock read?

Robin · 16th August 2020, 02:31 PM

Originally Posted by SDG

So now we cannot simplify, right.

So no simplification.

There is a train car with L₀=3.4641cs and the platform frame.

Front of the train car is x'=0 and t'=0 and it is align with the platform origin x=0 and t=0.

https://i.imgur.com/3TyzBkF.png

So the green B is platform origin and train car front (the train origin) aligned.

There is no motion at the beginning, the motion starts at t=t'=0.

The acceleration, then the deceleration as per the diagram.

It takes 4s of the platform time.

Then everything stops again.

The back of the train car is aligned with the platform origin.

There is a time dilation so the front train clock and the back train clock have both 2s on them, agreed?

How did the platform origin crossed 3.4641cs of the train frame in 2s of the train frame?

Again, if each observer stops.adjacent to a clock that has been stationary on the platform all along, what will it read?

HansMustermann · 16th August 2020, 03:19 PM

Deleted.

Roger Ramjets · 16th August 2020, 04:23 PM

Originally Posted by SDG

It is a problem for the relativity.

How is it a problem?

SDG · 16th August 2020, 08:49 PM

These are your couple of last posts.

Originally Posted by Ziggurat

We can simplify, if you understand the simplification. You aren't comfortable with the results of that simplification, but so what? Your discomfort doesn't matter. So what if there's a time jump? It's a coordinate effect. As I explained previously, there is no discontinuity in what the traveling twin sees. Even if he reverses direction instantly, what he sees is still continuous. So why do you care if there's a coordinate time jump at a location outside his light cone? Why does it matter?

Look at the sun. It's about 8 light minutes away. Turn 180 degrees. Now it's 8 light minutes away in the other direction. How did the sun travel 16 light minutes in 1 second?

You changed reference frames.

Originally Posted by Ziggurat

You did the acceleration instantly. Do it in a finite amount of time, and in that accelerating frame, the platform time passes very rapidly but it's still continuous.

I told you, accelerated frames are weird.

When I read them my understanding is that you claim the time dilation is a function of acceleration and not velocity. What do I miss?

If I look at the sun and the sun is in +x direction then I turn 180 degrees I am facing -x direction and the sun stayed in the +x direction. What are you talking about?
Even if I change frames + is + and - is -. The relative velocity is +v and -v between two moving frames, isn't it?

Does this work for the Lorentz transformation?

Ziggurat · 16th August 2020, 09:54 PM

Originally Posted by SDG

When I read them my understanding is that you claim the time dilation is a function of acceleration and not velocity. What do I miss?

Time dilation is different than the jump in time due to the change in reference frames.

Quote:

If I look at the sun and the sun is in +x direction then I turn 180 degrees I am facing -x direction and the sun stayed in the +x direction. What are you talking about?

You can define x however you want. +x can be in reference to which direction you face. The sun does indeed go from +x to -x.

But again, this isn’t a problem. Rotating frames are not inertial.

What you don’t understand is that Lorenz boosts (accelerating frames) are just 4D Minkowski space rotations. They aren’t fundamentally any different than you facing the sun and then turning around. You think it’s wrong to change frames when you rotate in Euclidean space, but you insist on changing frames when you rotate in Minkowski space. That there is your contradiction.

Quote:

Does this work for the Lorentz transformation?

In your first drawing, you have three events marked, A, B and C. In your second drawing, A is still marked, but B and C are not. We can locate C, though: it’s where your horizontal line above the x’ axis intersects the t’ axis.

But B is NOT where that line intersects the t axis. B is located at the point where the t axis intersects with the horizontal line given by t=8s.

So what are you trying to say with this graph?

Robin · 16th August 2020, 11:54 PM

Still saved from the last time we had this discussion:

A=(0,0,0,0) A'=(0,0,0,0)
B=(4,3.464,0,0) B'=(2,0,0,0)
C=(4,0,0,0) C'=(8,-6.928,0,0)

SDG · 17th August 2020, 10:17 AM

The discussion is getting better, thank you, Ziggurat!

Originally Posted by Ziggurat

Time dilation is different than the jump in time due to the change in reference frames.
...

There is so much going on in just this one sentence, here is why.

The first couple of milliseconds of the acceleration.
The back train car observer accelerates more yet stays back and the front observer moves 'ahead'.
The last couple of milliseconds of the deceleration.
The back train car observer decelerates less yet moves 'ahead' now and the front observer is already 'ahead' and waits for the back observer to catch up.

Isn't this like the equivalence principle?
The accelerated frame undergoes different acceleration at the back and the front?
Where is the time dilation coming from?
Is it really relative motion that is causing the time dilation or it is the delta in the acceleration/deceleration?

Ziggurat · 17th August 2020, 10:54 AM

Originally Posted by SDG

The first couple of milliseconds of the acceleration.
The back train car observer accelerates more yet stays back and the front observer moves 'ahead'.

Depends which reference frame you want to work in. In a frame which is accelerating with the train observer, yes, they will observe that the front of the train gets stretched out compared to the back of the frame. In the platform frame, they remain the same distance apart at all times.

I keep telling you, accelerating frames are strange.

Quote:

Where is the time dilation coming from?

Time dilation is due to velocity difference only. Time differences due to accelerated reference frames are not called "time dilation".

Quote:

Is it really relative motion that is causing the time dilation or it is the delta in the acceleration/deceleration?

It's really relative motion. Time dilation appears in scenarios which involve no acceleration at all.

SDG · 17th August 2020, 12:50 PM

Originally Posted by Ziggurat

Time dilation is different than the jump in time due to the change in reference frames.

You can define x however you want. +x can be in reference to which direction you face. The sun does indeed go from +x to -x.

But again, this isn’t a problem. Rotating frames are not inertial.

What you don’t understand is that Lorenz boosts (accelerating frames) are just 4D Minkowski space rotations. They aren’t fundamentally any different than you facing the sun and then turning around. You think it’s wrong to change frames when you rotate in Euclidean space, but you insist on changing frames when you rotate in Minkowski space. That there is your contradiction.

It all depends how we define our frames, if they are inertial or not.
If i turn around and my frame has +x where I look, what I face, then no problem.
We have a discussion about inertial frames and that's why I said the sun would stay behind me.
Changing, 'jumping' frames is not easy.
It involves 4-momentum, mass and energy of the bodies that are 'jumping' frames and there are limitations.

Quote:

In your first drawing, you have three events marked, A, B and C. In your second drawing, A is still marked, but B and C are not. We can locate C, though: it’s where your horizontal line above the x’ axis intersects the t’ axis.

But B is NOT where that line intersects the t axis. B is located at the point where the t axis intersects with the horizontal line given by t=8s.

So what are you trying to say with this graph?

B is on ct' axis when t'=2s'.
Do you agree with how the right diagram is drawn?

Reality Check · 17th August 2020, 02:15 PM

Originally Posted by SDG

How did the platform origin crossed 3.4641cs of the train frame in 2s of the train frame?.

Read the SR part of time dilation. You changed reference frames so time dilated. This is textbook SR.

Reality Check · 17th August 2020, 02:27 PM

Originally Posted by SDG

It all depends how we define our frames, if they are inertial or not.

This is wrong, SDG. Frames of reference in special relativity are inertial (but see below). It is GR that deals with non-inertial frames. We do not really define frames as inertial or non-inertial. It is the physical situation being considered that defines this. For example you have a train moving at a constant speed and a platform, thus inertial frames.

Originally Posted by SDG

If i turn around...

This is irreverent.

Originally Posted by SDG

Changing, 'jumping' frames is not easy. ...

Changing (jumping between) frames is extremely easy. You just write that you have changed frames! There are no bodies "'jumping' frames", there are observers in their respective frames. There are no imitations.

Originally Posted by SDG

Do you agree with how the right diagram is drawn?

This question is irreverent if you are getting the physics wrong. You have an impossible scenario. An theoretical train moving at a speed v cannot have cars that "accelerate more" or "accelerate less". Every car is moving at the same speed v. If the theoretical train accelerates than every car will accelerate equally and we have a general relativity scenario. If you want to make this into a hideously complex GR scenario then include the flexibility of the train (couplings) and the finite speed of the acceleration being applied to each car.

There is one way that acceleration can be included in SR - make it instantaneous. This is one resolution of the twin paradox. The travelling twin makes an instantaneous change in direction to return to Earth which puts them into a inertial frame with an opposite velocity.

SDG · 17th August 2020, 02:52 PM

Originally Posted by Robin

Still saved from the last time we had this discussion:

A=(0,0,0,0) A'=(0,0,0,0)
B=(4,3.464,0,0) B'=(2,0,0,0)
C=(4,0,0,0) C'=(8,-6.928,0,0)

As pointed out by Ziggurat,

Quote:

It's where the line of simultaneity for the outbound prime reference frame would intersect both the 4 second mark for the earthbound twin and that outbound journey, if the traveling twin hadn't turned around.

But he did. And that changed everything.

We do not have C' as you have it in your post.
The traveling twin did not turned around, just stopped motion in the outward direction, to be precise, nevertheless there is no C' with t'=8s.
This is the question, where is C/C' in the right side diagram of this figure?

Reality Check · 17th August 2020, 03:12 PM

Originally Posted by SDG

The traveling twin did not turned around, just stopped motion in the outward direction, to be precise, nevertheless there is no C' with t'=8s.

This is nonsense, SDG. The twin paradox requires that the clocks of each twin be compared. What you have is a scenario where Alice stays at home, Bob travels some distance and then stops relative to Alice. Alice and Bob end up in the same inertial frame of reference. This leads to the trivial situation where Alice and Bob's clocks tick along at the same rate but have different times from GR acceleration effects. Nothing to do with your attempt to do special relativity.

SDG · 17th August 2020, 03:22 PM

Originally Posted by Ziggurat

Depends which reference frame you want to work in. In a frame which is accelerating with the train observer, yes, they will observe that the front of the train gets stretched out compared to the back of the frame. In the platform frame, they remain the same distance apart at all times.

I keep telling you, accelerating frames are strange.
...

The platform frame sees the train car to contract and then to prolong.
I know, the accelerating frames are 'strange'.
The mass, energy centroids are not identical and they are frame dependent, go figure.

SDG · 17th August 2020, 03:24 PM

Originally Posted by Reality Check

This is nonsense, SDG. The twin paradox requires that the clocks of each twin be compared. What you have is a scenario where Alice stays at home, Bob travels some distance and then stops relative to Alice. Alice and Bob end up in the same inertial frame of reference. This leads to the trivial situation where Alice and Bob's clocks tick along at the same rate but have different times from GR acceleration effects. Nothing to do with your attempt to do special relativity.

How does your response addresses the t'=8s'?

Myriad · 17th August 2020, 03:26 PM

Originally Posted by SDG

This is the question, where is C/C' in the right side diagram of this figure?

Ziggurat and Robin have both given you the clear and correct answer to this question. What don't you like about it?

Everyone and everything exists in every reference frame all the time. Observers don't pop "into" and "out of" reference frames like the TARDIS. The measurements attributed to the primed reference frame, whether made directly by observers stationary with respect to that frame or inferred from Special Relativity via Lorentz transformations, don't become invalid when the train stops.

HansMustermann · 17th August 2020, 03:36 PM

Hear, hear. SR doesn't even have the event horizons that an accelerating frame would have, so there is no time where any point exists in one frame but not in another. If they are SR frames, that is.

SDG · 17th August 2020, 04:16 PM

Originally Posted by Myriad

Ziggurat and Robin have both given you the clear and correct answer to this question. What don't you like about it?

Everyone and everything exists in every reference frame all the time. Observers don't pop "into" and "out of" reference frames like the TARDIS. The measurements attributed to the primed reference frame, whether made directly by observers stationary with respect to that frame or inferred from Special Relativity via Lorentz transformations, don't become invalid when the train stops.

Well,

Quote:

We can locate C, though: it’s where your horizontal line above the x’ axis intersects the t’ axis.

Please, tell me where is this, I am confused.

Here is Robin's C

SDG · 17th August 2020, 05:22 PM

Confused enough?

How about this?

The axes ct and ct' are out of sync.

Reality Check · 17th August 2020, 05:54 PM

Originally Posted by SDG

How does your response addresses the t'=8s'?

I did: This is nonsense, SDG.

Robin · 17th August 2020, 06:57 PM

Originally Posted by SDG

Confused enough?

How about this?

https://i.imgur.com/X4SfqRS.png

The axes ct and ct' are out of sync.

What exactly is this a diagram of???

Robin · 17th August 2020, 06:58 PM

Originally Posted by SDG

Here is Robin's C

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

That is where C is. Just do the calculations.

SDG · 17th August 2020, 07:27 PM

Originally Posted by Robin

What exactly is this a diagram of???

The platform and train frame before and after the acceleration.

15th August 2020, 06:29 PM	#201
Ziggurat Penultimate Amazing Join Date: Jun 2003 Posts: 47,309	Originally Posted by SDG You are mixing sub-discussions. You can't stick with a single topic, and I'm the one mixing sub-discussions? Quote: If gamma = 2 did the traveler crossed 3.4641cs in 2 seconds? This is an ambiguous question. What time are you referring to? Proper time or coordinate time? I'm not going to bother double-checking your math right now, but it looks plausible. In which case, the traveler went 3.46 cs distance in the unprimed reference frame, and experienced 2 seconds of proper time in doing so. So where's the conflict? You still haven't produced one.
	__________________ "As long as it is admitted that the law may be diverted from its true purpose -- that it may violate property instead of protecting it -- then everyone will want to participate in making the law, either to protect himself against plunder or to use it for plunder. Political questions will always be prejudicial, dominant, and all-absorbing. There will be fighting at the door of the Legislative Palace, and the struggle within will be no less furious." - Bastiat, The Law

15th August 2020, 08:29 PM	#202
SDG Muse Join Date: Jul 2018 Posts: 895	Originally Posted by Ziggurat You can't stick with a single topic, and I'm the one mixing sub-discussions? This is an ambiguous question. What time are you referring to? Proper time or coordinate time? I'm not going to bother double-checking your math right now, but it looks plausible. In which case, the traveler went 3.46 cs distance in the unprimed reference frame, and experienced 2 seconds of proper time in doing so. So where's the conflict? You still haven't produced one. Where is 8s of traveler coordinate time to account for 4s of stay home twin proper time?
SDG Muse Join Date: Jul 2018 Posts: 895

15th August 2020, 09:03 PM	#203
Ziggurat Penultimate Amazing Join Date: Jun 2003 Posts: 47,309	Originally Posted by SDG Where is 8s of traveler coordinate time to account for 4s of stay home twin proper time? It's where the line of simultaneity for the outbound prime reference frame would intersect both the 4 second mark for the earthbound twin and that outbound journey, if the traveling twin hadn't turned around. But he did. And that changed everything. This is really, really basic special relativity. I don't know what to tell you if you still can't get it.
	__________________ "As long as it is admitted that the law may be diverted from its true purpose -- that it may violate property instead of protecting it -- then everyone will want to participate in making the law, either to protect himself against plunder or to use it for plunder. Political questions will always be prejudicial, dominant, and all-absorbing. There will be fighting at the door of the Legislative Palace, and the struggle within will be no less furious." - Bastiat, The Law

16th August 2020, 12:56 AM	#204
HansMustermann Penultimate Amazing Join Date: Mar 2009 Posts: 17,158	Well, what I could say as a hint is what I've said before: don't forget that t' = γ(t - vx/c2), not just γt. That seems to account for pretty much every confusion about the twins paradox I've seen so far.
	__________________ Which part of "Ph'nglui mglw'nafh Cthulhu R'lyeh wgah'nagl fhtagn" don't you understand?

16th August 2020, 04:50 AM	#205
Robin Penultimate Amazing Join Date: Apr 2004 Posts: 12,597	Originally Posted by SDG Where is 8s of traveler coordinate time to account for 4s of stay home twin proper time? What is the co-ordinate time of event C in your diagram in the traveler's frame (when at full speed)? C: t=4, x=0, t'=?, x'=?
	__________________ The non-theoretical character of metaphysics would not be in itself a defect; all arts have this non-theoretical character without thereby losing their high value for personal as well as for social life. The danger lies in the deceptive character of metaphysics; it gives the illusion of knowledge without actually giving any knowledge. This is the reason why we reject it. - Rudolf Carnap "Philosophy and Logical Syntax" Last edited by Robin; 16th August 2020 at 05:04 AM.

16th August 2020, 05:16 AM	#206
SDG Muse Join Date: Jul 2018 Posts: 895	Originally Posted by Ziggurat It's where the line of simultaneity for the outbound prime reference frame would intersect both the 4 second mark for the earthbound twin and that outbound journey, if the traveling twin hadn't turned around. But he did. And that changed everything. This is really, really basic special relativity. I don't know what to tell you if you still can't get it. Fair enough. The platform 1s of proper time is 2s' of train coordinate time. How do 3s of platform proper time flow when there is no change of the traveling twin coordinate time?
SDG Muse Join Date: Jul 2018 Posts: 895	Last edited by SDG; 16th August 2020 at 05:19 AM.

16th August 2020, 05:30 AM	#207
SDG Muse Join Date: Jul 2018 Posts: 895	Originally Posted by Robin What is the co-ordinate time of event C in your diagram in the traveler's frame (when at full speed)? C: t=4, x=0, t'=?, x'=? The last thread of discussion is that the traveler stops. Therefore t'=2s and x'=-3.4641cs. There is no more length contraction after the stop. There are question marks as per my post above. It's like the platform origin crossed 3.4641cs to the left in 2s', right?
SDG Muse Join Date: Jul 2018 Posts: 895	Last edited by SDG; 16th August 2020 at 05:31 AM.

16th August 2020, 05:53 AM	#208
Robin Penultimate Amazing Join Date: Apr 2004 Posts: 12,597	Originally Posted by SDG The last thread of discussion is that the traveler stops. Therefore t'=2s and x'=-3.4641cs. There is no more length contraction after the stop. There is no t' when the traveller stops. The twins have the same velocity after that.
	__________________ The non-theoretical character of metaphysics would not be in itself a defect; all arts have this non-theoretical character without thereby losing their high value for personal as well as for social life. The danger lies in the deceptive character of metaphysics; it gives the illusion of knowledge without actually giving any knowledge. This is the reason why we reject it. - Rudolf Carnap "Philosophy and Logical Syntax"

16th August 2020, 05:56 AM	#209
Robin Penultimate Amazing Join Date: Apr 2004 Posts: 12,597	Originally Posted by SDG Fair enough. The platform 1s of proper time is 2s' of train coordinate time. How do 3s of platform proper time flow when there is no change of the traveling twin coordinate time? That would only happen if the traveller's velocity changed instantaneously to zero. If we are going to discuss a simplification we should accept that the simplification will break down at some stage. That doesn't imply a problem for relativity.
	__________________ The non-theoretical character of metaphysics would not be in itself a defect; all arts have this non-theoretical character without thereby losing their high value for personal as well as for social life. The danger lies in the deceptive character of metaphysics; it gives the illusion of knowledge without actually giving any knowledge. This is the reason why we reject it. - Rudolf Carnap "Philosophy and Logical Syntax"

16th August 2020, 06:38 AM	#210
SDG Muse Join Date: Jul 2018 Posts: 895	Originally Posted by Robin There is no t' when the traveller stops. The twins have the same velocity after that. There has to be a time dilation. If it was a train car moving then front and the back observers have 2s on their clocks. Agreed? How did the platform x=0 observer moved across 3.4641cs train car in 2s?
SDG Muse Join Date: Jul 2018 Posts: 895

16th August 2020, 06:41 AM	#211
SDG Muse Join Date: Jul 2018 Posts: 895	Originally Posted by Robin That would only happen if the traveller's velocity changed instantaneously to zero. If we are going to discuss a simplification we should accept that the simplification will break down at some stage. That doesn't imply a problem for relativity. The same problem arises when we use accelerations without simplifications. It is a problem for the relativity.
SDG Muse Join Date: Jul 2018 Posts: 895

16th August 2020, 06:41 AM	#212
Myriad Hyperthetical Join Date: Nov 2006 Location: Betwixt Posts: 16,924	Originally Posted by SDG How did the platform x=0 observer moved across 3.4641cs train car in 2s? Oh, I know this one. Because he wanted to see time fly.
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16th August 2020, 07:27 AM	#213
Ziggurat Penultimate Amazing Join Date: Jun 2003 Posts: 47,309	Originally Posted by SDG Fair enough. The platform 1s of proper time is 2s' of train coordinate time. How do 3s of platform proper time flow when there is no change of the traveling twin coordinate time? You did the acceleration instantly. Do it in a finite amount of time, and in that accelerating frame, the platform time passes very rapidly but it's still continuous. I told you, accelerated frames are weird.
	__________________ "As long as it is admitted that the law may be diverted from its true purpose -- that it may violate property instead of protecting it -- then everyone will want to participate in making the law, either to protect himself against plunder or to use it for plunder. Political questions will always be prejudicial, dominant, and all-absorbing. There will be fighting at the door of the Legislative Palace, and the struggle within will be no less furious." - Bastiat, The Law

16th August 2020, 07:42 AM	#214
SDG Muse Join Date: Jul 2018 Posts: 895	Originally Posted by Ziggurat You did the acceleration instantly. Do it in a finite amount of time, and in that accelerating frame, the platform time passes very rapidly but it's still continuous. I told you, accelerated frames are weird. So now we cannot simplify, right. So no simplification. There is a train car with L₀=3.4641cs and the platform frame. Front of the train car is x'=0 and t'=0 and it is align with the platform origin x=0 and t=0. So the green B is platform origin and train car front (the train origin) aligned. There is no motion at the beginning, the motion starts at t=t'=0. The acceleration, then the deceleration as per the diagram. It takes 4s of the platform time. Then everything stops again. The back of the train car is aligned with the platform origin. There is a time dilation so the front train clock and the back train clock have both 2s on them, agreed? How did the platform origin crossed 3.4641cs of the train frame in 2s of the train frame?
SDG Muse Join Date: Jul 2018 Posts: 895

16th August 2020, 08:22 AM	#215
Ziggurat Penultimate Amazing Join Date: Jun 2003 Posts: 47,309	Originally Posted by SDG So now we cannot simplify, right. So no simplification. We can simplify, if you understand the simplification. You aren't comfortable with the results of that simplification, but so what? Your discomfort doesn't matter. So what if there's a time jump? It's a coordinate effect. As I explained previously, there is no discontinuity in what the traveling twin sees. Even if he reverses direction instantly, what he sees is still continuous. So why do you care if there's a coordinate time jump at a location outside his light cone? Why does it matter? Quote: How did the platform origin crossed 3.4641cs of the train frame in 2s of the train frame? Look at the sun. It's about 8 light minutes away. Turn 180 degrees. Now it's 8 light minutes away in the other direction. How did the sun travel 16 light minutes in 1 second? You changed reference frames.
	__________________ "As long as it is admitted that the law may be diverted from its true purpose -- that it may violate property instead of protecting it -- then everyone will want to participate in making the law, either to protect himself against plunder or to use it for plunder. Political questions will always be prejudicial, dominant, and all-absorbing. There will be fighting at the door of the Legislative Palace, and the struggle within will be no less furious." - Bastiat, The Law

16th August 2020, 09:46 AM	#216
Myriad Hyperthetical Join Date: Nov 2006 Location: Betwixt Posts: 16,924	This effect of special relativity has been a trope in science fiction for a hundred years. A ~~train~~ spaceship travels to a distant ~~platform~~ star system, accelerating to a high fraction of the speed of light near the start and decelerating at the destination. From the stationary frame of the origin ~~platform~~ planet or the destination ~~platform~~ planet, the ~~train~~ ship always travels slower than the speed of light, but the traveler's "subjective time" (time on the ~~train's~~ ship's clock) is less than the time it takes light to make the trip in the origin's or destination's frame. Disregarding the acceleration and deceleration times, a trip at a relative velocity greater than .707c (c/sqrt(2)) will be "faster than light" in subjective time. The traveler's subjective time can be arbitrarily short if arbitrarily high fractions of c are reached (as in e.g. Poul Anderson's Tao Zero). Of course, from the traveler's frame, the distance traveled is shortened, so the traveler never observes the ship traveling faster than light either.
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16th August 2020, 02:19 PM	#217
Robin Penultimate Amazing Join Date: Apr 2004 Posts: 12,597	Originally Posted by SDG There has to be a time dilation. If it was a train car moving then front and the back observers have 2s on their clocks. Agreed? How did the platform x=0 observer moved across 3.4641cs train car in 2s? When the traveller stops and compares his clock to a stationary clock, do you think they will read the same? What will the stationary be clock read?
	__________________ The non-theoretical character of metaphysics would not be in itself a defect; all arts have this non-theoretical character without thereby losing their high value for personal as well as for social life. The danger lies in the deceptive character of metaphysics; it gives the illusion of knowledge without actually giving any knowledge. This is the reason why we reject it. - Rudolf Carnap "Philosophy and Logical Syntax"

16th August 2020, 02:31 PM	#218
Robin Penultimate Amazing Join Date: Apr 2004 Posts: 12,597	Originally Posted by SDG So now we cannot simplify, right. So no simplification. There is a train car with L₀=3.4641cs and the platform frame. Front of the train car is x'=0 and t'=0 and it is align with the platform origin x=0 and t=0. https://i.imgur.com/3TyzBkF.png So the green B is platform origin and train car front (the train origin) aligned. There is no motion at the beginning, the motion starts at t=t'=0. The acceleration, then the deceleration as per the diagram. It takes 4s of the platform time. Then everything stops again. The back of the train car is aligned with the platform origin. There is a time dilation so the front train clock and the back train clock have both 2s on them, agreed? How did the platform origin crossed 3.4641cs of the train frame in 2s of the train frame? Again, if each observer stops.adjacent to a clock that has been stationary on the platform all along, what will it read?
	__________________ The non-theoretical character of metaphysics would not be in itself a defect; all arts have this non-theoretical character without thereby losing their high value for personal as well as for social life. The danger lies in the deceptive character of metaphysics; it gives the illusion of knowledge without actually giving any knowledge. This is the reason why we reject it. - Rudolf Carnap "Philosophy and Logical Syntax"

16th August 2020, 03:19 PM	#219
HansMustermann Penultimate Amazing Join Date: Mar 2009 Posts: 17,158	Deleted.
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16th August 2020, 04:23 PM	#220
Roger Ramjets Philosopher Join Date: Jun 2008 Posts: 5,032	Originally Posted by SDG It is a problem for the relativity. How is it a problem?
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16th August 2020, 08:49 PM	#221
SDG Muse Join Date: Jul 2018 Posts: 895	These are your couple of last posts. Originally Posted by Ziggurat We can simplify, if you understand the simplification. You aren't comfortable with the results of that simplification, but so what? Your discomfort doesn't matter. So what if there's a time jump? It's a coordinate effect. As I explained previously, there is no discontinuity in what the traveling twin sees. Even if he reverses direction instantly, what he sees is still continuous. So why do you care if there's a coordinate time jump at a location outside his light cone? Why does it matter? Look at the sun. It's about 8 light minutes away. Turn 180 degrees. Now it's 8 light minutes away in the other direction. How did the sun travel 16 light minutes in 1 second? You changed reference frames. Originally Posted by Ziggurat You did the acceleration instantly. Do it in a finite amount of time, and in that accelerating frame, the platform time passes very rapidly but it's still continuous. I told you, accelerated frames are weird. When I read them my understanding is that you claim the time dilation is a function of acceleration and not velocity. What do I miss? If I look at the sun and the sun is in +x direction then I turn 180 degrees I am facing -x direction and the sun stayed in the +x direction. What are you talking about? Even if I change frames + is + and - is -. The relative velocity is +v and -v between two moving frames, isn't it? Does this work for the Lorentz transformation?
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16th August 2020, 09:54 PM	#222
Ziggurat Penultimate Amazing Join Date: Jun 2003 Posts: 47,309	Originally Posted by SDG When I read them my understanding is that you claim the time dilation is a function of acceleration and not velocity. What do I miss? Time dilation is different than the jump in time due to the change in reference frames. Quote: If I look at the sun and the sun is in +x direction then I turn 180 degrees I am facing -x direction and the sun stayed in the +x direction. What are you talking about? You can define x however you want. +x can be in reference to which direction you face. The sun does indeed go from +x to -x. But again, this isn’t a problem. Rotating frames are not inertial. What you don’t understand is that Lorenz boosts (accelerating frames) are just 4D Minkowski space rotations. They aren’t fundamentally any different than you facing the sun and then turning around. You think it’s wrong to change frames when you rotate in Euclidean space, but you insist on changing frames when you rotate in Minkowski space. That there is your contradiction. Quote: Does this work for the Lorentz transformation? In your first drawing, you have three events marked, A, B and C. In your second drawing, A is still marked, but B and C are not. We can locate C, though: it’s where your horizontal line above the x’ axis intersects the t’ axis. But B is NOT where that line intersects the t axis. B is located at the point where the t axis intersects with the horizontal line given by t=8s. So what are you trying to say with this graph?
	__________________ "As long as it is admitted that the law may be diverted from its true purpose -- that it may violate property instead of protecting it -- then everyone will want to participate in making the law, either to protect himself against plunder or to use it for plunder. Political questions will always be prejudicial, dominant, and all-absorbing. There will be fighting at the door of the Legislative Palace, and the struggle within will be no less furious." - Bastiat, The Law

16th August 2020, 11:54 PM	#223
Robin Penultimate Amazing Join Date: Apr 2004 Posts: 12,597	Still saved from the last time we had this discussion: A=(0,0,0,0) A'=(0,0,0,0) B=(4,3.464,0,0) B'=(2,0,0,0) C=(4,0,0,0) C'=(8,-6.928,0,0)
	__________________ The non-theoretical character of metaphysics would not be in itself a defect; all arts have this non-theoretical character without thereby losing their high value for personal as well as for social life. The danger lies in the deceptive character of metaphysics; it gives the illusion of knowledge without actually giving any knowledge. This is the reason why we reject it. - Rudolf Carnap "Philosophy and Logical Syntax" Last edited by Robin; 17th August 2020 at 12:07 AM.

17th August 2020, 10:17 AM	#224
SDG Muse Join Date: Jul 2018 Posts: 895	The discussion is getting better, thank you, Ziggurat! Originally Posted by Ziggurat Time dilation is different than the jump in time due to the change in reference frames. ... There is so much going on in just this one sentence, here is why. The first couple of milliseconds of the acceleration. The back train car observer accelerates more yet stays back and the front observer moves 'ahead'. The last couple of milliseconds of the deceleration. The back train car observer decelerates less yet moves 'ahead' now and the front observer is already 'ahead' and waits for the back observer to catch up. Isn't this like the equivalence principle? The accelerated frame undergoes different acceleration at the back and the front? Where is the time dilation coming from? Is it really relative motion that is causing the time dilation or it is the delta in the acceleration/deceleration?
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17th August 2020, 10:54 AM	#225
Ziggurat Penultimate Amazing Join Date: Jun 2003 Posts: 47,309	Originally Posted by SDG The first couple of milliseconds of the acceleration. The back train car observer accelerates more yet stays back and the front observer moves 'ahead'. Depends which reference frame you want to work in. In a frame which is accelerating with the train observer, yes, they will observe that the front of the train gets stretched out compared to the back of the frame. In the platform frame, they remain the same distance apart at all times. I keep telling you, accelerating frames are strange. Quote: Where is the time dilation coming from? Time dilation is due to velocity difference only. Time differences due to accelerated reference frames are not called "time dilation". Quote: Is it really relative motion that is causing the time dilation or it is the delta in the acceleration/deceleration? It's really relative motion. Time dilation appears in scenarios which involve no acceleration at all.
	__________________ "As long as it is admitted that the law may be diverted from its true purpose -- that it may violate property instead of protecting it -- then everyone will want to participate in making the law, either to protect himself against plunder or to use it for plunder. Political questions will always be prejudicial, dominant, and all-absorbing. There will be fighting at the door of the Legislative Palace, and the struggle within will be no less furious." - Bastiat, The Law

17th August 2020, 12:50 PM	#226
SDG Muse Join Date: Jul 2018 Posts: 895	Originally Posted by Ziggurat Time dilation is different than the jump in time due to the change in reference frames. You can define x however you want. +x can be in reference to which direction you face. The sun does indeed go from +x to -x. But again, this isn’t a problem. Rotating frames are not inertial. What you don’t understand is that Lorenz boosts (accelerating frames) are just 4D Minkowski space rotations. They aren’t fundamentally any different than you facing the sun and then turning around. You think it’s wrong to change frames when you rotate in Euclidean space, but you insist on changing frames when you rotate in Minkowski space. That there is your contradiction. It all depends how we define our frames, if they are inertial or not. If i turn around and my frame has +x where I look, what I face, then no problem. We have a discussion about inertial frames and that's why I said the sun would stay behind me. Changing, 'jumping' frames is not easy. It involves 4-momentum, mass and energy of the bodies that are 'jumping' frames and there are limitations. Quote: In your first drawing, you have three events marked, A, B and C. In your second drawing, A is still marked, but B and C are not. We can locate C, though: it’s where your horizontal line above the x’ axis intersects the t’ axis. But B is NOT where that line intersects the t axis. B is located at the point where the t axis intersects with the horizontal line given by t=8s. So what are you trying to say with this graph? B is on ct' axis when t'=2s'. Do you agree with how the right diagram is drawn?
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17th August 2020, 02:15 PM	#227
Reality Check Penultimate Amazing Join Date: Mar 2008 Location: New Zealand Posts: 27,450	Originally Posted by SDG How did the platform origin crossed 3.4641cs of the train frame in 2s of the train frame?. Read the SR part of time dilation. You changed reference frames so time dilated. This is textbook SR.
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17th August 2020, 02:27 PM	#228
Reality Check Penultimate Amazing Join Date: Mar 2008 Location: New Zealand Posts: 27,450	Originally Posted by SDG It all depends how we define our frames, if they are inertial or not. This is wrong, SDG. Frames of reference in special relativity are inertial (but see below). It is GR that deals with non-inertial frames. We do not really define frames as inertial or non-inertial. It is the physical situation being considered that defines this. For example you have a train moving at a constant speed and a platform, thus inertial frames. Originally Posted by SDG If i turn around... This is irreverent. Originally Posted by SDG Changing, 'jumping' frames is not easy. ... Changing (jumping between) frames is extremely easy. You just write that you have changed frames! There are no bodies "'jumping' frames", there are observers in their respective frames. There are no imitations. Originally Posted by SDG Do you agree with how the right diagram is drawn? This question is irreverent if you are getting the physics wrong. You have an impossible scenario. An theoretical train moving at a speed v cannot have cars that "accelerate more" or "accelerate less". Every car is moving at the same speed v. If the theoretical train accelerates than every car will accelerate equally and we have a general relativity scenario. If you want to make this into a hideously complex GR scenario then include the flexibility of the train (couplings) and the finite speed of the acceleration being applied to each car. There is one way that acceleration can be included in SR - make it instantaneous. This is one resolution of the twin paradox. The travelling twin makes an instantaneous change in direction to return to Earth which puts them into a inertial frame with an opposite velocity.
	__________________ NASA Finds Direct Proof of Dark Matter (another observation) (and Abell 520) Electric comets still do not exist! Last edited by Reality Check; 17th August 2020 at 02:38 PM.

17th August 2020, 02:52 PM	#229
SDG Muse Join Date: Jul 2018 Posts: 895	Originally Posted by Robin Still saved from the last time we had this discussion: A=(0,0,0,0) A'=(0,0,0,0) B=(4,3.464,0,0) B'=(2,0,0,0) C=(4,0,0,0) C'=(8,-6.928,0,0) As pointed out by Ziggurat, Quote: It's where the line of simultaneity for the outbound prime reference frame would intersect both the 4 second mark for the earthbound twin and that outbound journey, if the traveling twin hadn't turned around. But he did. And that changed everything. We do not have C' as you have it in your post. The traveling twin did not turned around, just stopped motion in the outward direction, to be precise, nevertheless there is no C' with t'=8s. This is the question, where is C/C' in the right side diagram of this figure?
SDG Muse Join Date: Jul 2018 Posts: 895	Last edited by SDG; 17th August 2020 at 02:54 PM.

17th August 2020, 03:12 PM	#230
Reality Check Penultimate Amazing Join Date: Mar 2008 Location: New Zealand Posts: 27,450	Originally Posted by SDG The traveling twin did not turned around, just stopped motion in the outward direction, to be precise, nevertheless there is no C' with t'=8s. This is nonsense, SDG. The twin paradox requires that the clocks of each twin be compared. What you have is a scenario where Alice stays at home, Bob travels some distance and then stops relative to Alice. Alice and Bob end up in the same inertial frame of reference. This leads to the trivial situation where Alice and Bob's clocks tick along at the same rate but have different times from GR acceleration effects. Nothing to do with your attempt to do special relativity.
	__________________ NASA Finds Direct Proof of Dark Matter (another observation) (and Abell 520) Electric comets still do not exist!

17th August 2020, 03:22 PM	#231
SDG Muse Join Date: Jul 2018 Posts: 895	Originally Posted by Ziggurat Depends which reference frame you want to work in. In a frame which is accelerating with the train observer, yes, they will observe that the front of the train gets stretched out compared to the back of the frame. In the platform frame, they remain the same distance apart at all times. I keep telling you, accelerating frames are strange. ... The platform frame sees the train car to contract and then to prolong. I know, the accelerating frames are 'strange'. The mass, energy centroids are not identical and they are frame dependent, go figure.
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17th August 2020, 03:24 PM	#232
SDG Muse Join Date: Jul 2018 Posts: 895	Originally Posted by Reality Check This is nonsense, SDG. The twin paradox requires that the clocks of each twin be compared. What you have is a scenario where Alice stays at home, Bob travels some distance and then stops relative to Alice. Alice and Bob end up in the same inertial frame of reference. This leads to the trivial situation where Alice and Bob's clocks tick along at the same rate but have different times from GR acceleration effects. Nothing to do with your attempt to do special relativity. How does your response addresses the t'=8s'?
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17th August 2020, 03:26 PM	#233
Myriad Hyperthetical Join Date: Nov 2006 Location: Betwixt Posts: 16,924	Originally Posted by SDG This is the question, where is C/C' in the right side diagram of this figure? Ziggurat and Robin have both given you the clear and correct answer to this question. What don't you like about it? Everyone and everything exists in every reference frame all the time. Observers don't pop "into" and "out of" reference frames like the TARDIS. The measurements attributed to the primed reference frame, whether made directly by observers stationary with respect to that frame or inferred from Special Relativity via Lorentz transformations, don't become invalid when the train stops.
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17th August 2020, 03:36 PM	#234
HansMustermann Penultimate Amazing Join Date: Mar 2009 Posts: 17,158	Hear, hear. SR doesn't even have the event horizons that an accelerating frame would have, so there is no time where any point exists in one frame but not in another. If they are SR frames, that is.
	__________________ Which part of "Ph'nglui mglw'nafh Cthulhu R'lyeh wgah'nagl fhtagn" don't you understand?

17th August 2020, 04:16 PM	#235
SDG Muse Join Date: Jul 2018 Posts: 895	Originally Posted by Myriad Ziggurat and Robin have both given you the clear and correct answer to this question. What don't you like about it? Everyone and everything exists in every reference frame all the time. Observers don't pop "into" and "out of" reference frames like the TARDIS. The measurements attributed to the primed reference frame, whether made directly by observers stationary with respect to that frame or inferred from Special Relativity via Lorentz transformations, don't become invalid when the train stops. Well, Quote: We can locate C, though: it’s where your horizontal line above the x’ axis intersects the t’ axis. Please, tell me where is this, I am confused. Here is Robin's C
SDG Muse Join Date: Jul 2018 Posts: 895	Last edited by SDG; 17th August 2020 at 04:30 PM.