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Old 15th August 2020, 06:29 PM   #201
Ziggurat
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Originally Posted by SDG View Post
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.
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Old 15th August 2020, 08:29 PM   #202
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Originally Posted by Ziggurat View Post
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?
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Old 15th August 2020, 09:03 PM   #203
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Originally Posted by SDG View Post
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.
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Old 16th August 2020, 12:56 AM   #204
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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.
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Old 16th August 2020, 04:50 AM   #205
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Originally Posted by SDG View Post
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'=?
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Old 16th August 2020, 05:16 AM   #206
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Originally Posted by Ziggurat View Post
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?




Last edited by SDG; 16th August 2020 at 05:19 AM.
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Old 16th August 2020, 05:30 AM   #207
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Originally Posted by Robin View Post
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?

Last edited by SDG; 16th August 2020 at 05:31 AM.
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Old 16th August 2020, 05:53 AM   #208
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Originally Posted by SDG View Post
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.
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Old 16th August 2020, 05:56 AM   #209
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Originally Posted by SDG View Post
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.
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Old 16th August 2020, 06:38 AM   #210
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Originally Posted by Robin View Post
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?
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Old 16th August 2020, 06:41 AM   #211
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Originally Posted by Robin View Post
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.
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Old 16th August 2020, 06:41 AM   #212
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Originally Posted by SDG View Post
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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Old 16th August 2020, 07:27 AM   #213
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Originally Posted by SDG View Post
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.
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Old 16th August 2020, 07:42 AM   #214
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Originally Posted by Ziggurat View Post
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?
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Old 16th August 2020, 08:22 AM   #215
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Originally Posted by SDG View Post
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.
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Old 16th August 2020, 09:46 AM   #216
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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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Old 16th August 2020, 02:19 PM   #217
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Originally Posted by SDG View Post
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?
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Old 16th August 2020, 02:31 PM   #218
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Originally Posted by SDG View Post
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?
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Old 16th August 2020, 03:19 PM   #219
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Old 16th August 2020, 04:23 PM   #220
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Originally Posted by SDG View Post
It is a problem for the relativity.
How is it a problem?
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Old 16th August 2020, 08:49 PM   #221
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These are your couple of last posts.


Originally Posted by Ziggurat View Post
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 View Post
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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Old 16th August 2020, 09:54 PM   #222
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Originally Posted by SDG View Post
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 isnít a problem. Rotating frames are not inertial.

What you donít understand is that Lorenz boosts (accelerating frames) are just 4D Minkowski space rotations. They arenít fundamentally any different than you facing the sun and then turning around. You think itís 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: itís where your horizontal line above the xí axis intersects the tí 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?
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Old 16th August 2020, 11:54 PM   #223
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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)
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Old 17th August 2020, 10:17 AM   #224
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The discussion is getting better, thank you, Ziggurat!


Originally Posted by Ziggurat View Post
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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Old 17th August 2020, 10:54 AM   #225
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Originally Posted by SDG View Post
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.
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Old 17th August 2020, 12:50 PM   #226
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Originally Posted by Ziggurat View Post
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 isnít a problem. Rotating frames are not inertial.

What you donít understand is that Lorenz boosts (accelerating frames) are just 4D Minkowski space rotations. They arenít fundamentally any different than you facing the sun and then turning around. You think itís 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: itís where your horizontal line above the xí axis intersects the tí 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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Old 17th August 2020, 02:15 PM   #227
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Originally Posted by SDG View Post
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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Old 17th August 2020, 02:27 PM   #228
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Originally Posted by SDG View Post
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 View Post
If i turn around...
This is irreverent.

Originally Posted by SDG View Post
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 View Post
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.

Last edited by Reality Check; 17th August 2020 at 02:38 PM.
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Old 17th August 2020, 02:52 PM   #229
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Originally Posted by Robin View Post
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?


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Old 17th August 2020, 03:12 PM   #230
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Originally Posted by SDG View 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 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.
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Old 17th August 2020, 03:22 PM   #231
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Originally Posted by Ziggurat View Post
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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Old 17th August 2020, 03:24 PM   #232
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Originally Posted by Reality Check View Post
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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Old 17th August 2020, 03:26 PM   #233
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Originally Posted by SDG View Post
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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Old 17th August 2020, 03:36 PM   #234
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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.
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Old 17th August 2020, 04:16 PM   #235
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Originally Posted by Myriad View Post
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: itís where your horizontal line above the xí axis intersects the tí axis.
Please, tell me where is this, I am confused.

Here is Robin's C


Last edited by SDG; 17th August 2020 at 04:30 PM.
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Old 17th August 2020, 05:22 PM   #236
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Confused enough?
How about this?






The axes ct and ct' are out of sync.
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Old 17th August 2020, 05:54 PM   #237
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Originally Posted by SDG View Post
How does your response addresses the t'=8s'?
I did: This is nonsense, SDG.
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Old 17th August 2020, 06:57 PM   #238
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Originally Posted by SDG View Post
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???
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Old 17th August 2020, 06:58 PM   #239
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Originally Posted by SDG View Post
That is where C is. Just do the calculations.
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Old 17th August 2020, 07:27 PM   #240
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Originally Posted by Robin View Post
What exactly is this a diagram of???
The platform and train frame before and after the acceleration.
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