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Old 12th August 2020, 06:35 AM   #41
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Originally Posted by Myriad View Post
The stay at home twin experiences no time discontinuity. The only discontinuity is in what the traveling twin observes is happening to the stay at home twin.




It's the same kind of magic that allows the traveling twin at r to instantaneously reverse direction in the first place, requiring infinite acceleration and infinite energy.

(In the "no acceleration" version, where there are two travelers and they simply cross paths and transfer clock information at r, neither of the travelers observes any discontinuity for the stay at home twin. They merely disagree with one another about what time it is back on Earth at the moment of the exchange... which is expected, since they're observing Earth from different inertial reference frames.)

That's right, it is magic, therefore not realistic.
If you look at the event C in the diagram.
No time passing at C when the time is moving from A to B?
Is C some kind of 'black hole'?
There are no 'black holes' in the SR. Therefore the event C is unrealistic.

Last edited by SDG; 12th August 2020 at 06:42 AM.
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Old 12th August 2020, 06:38 AM   #42
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Well, they don't have to agree about what time it is on Earth. They just have to synchronize their own clocks. Like, if the one going away started the journey at 6AM, when it passes the other guy by, he transmits a signal that says, "hey, my clock says it's 12:34 right now", and the other one simply also sets his own clock to 12:34 too, and lets it tick normally from there. When he passes earth by, he and the guy who stayed home exchange similar signals and yep, they're different.

Mind you, when the guy sets the clock to match there is basically a discontinuity (his clock probably showed something else), but that's how you simulate the same clock ticking on both legs of the journey. It has to continue on the trip back from where it reached on the way out.
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Old 12th August 2020, 06:39 AM   #43
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Originally Posted by SDG View Post
So based on textbooks the time dilation is being an effect of the speed and not the acceleration.
This statement is supported by the Lorentz factor that is based on the speed and not an acceleration,
SDG
True. But you still aren't looking at one inertial frame for the traveling twin. Again, that throws off the symmetry you're assuming.

Here's another way to look at it. "Proper time", the time experienced by a traveler moving along a line, is basically just a form of distance measurement. So you're basically just comparing the distance between two points along different lines.

In Euclidean geometry, how do we measure distance? For a straight line segment, the length s can be found as
s2 = x2+y2+z2
For a curved line, we chop it up into infinitesmal bits and add up the lengths of each bit. So
s2 = integral(ds2) = integral(dx2+dy2+dz2)
You can also abreviate this as simply
ds2 = dx2 + dy2 + dz2
This is called the metric for Euclidean space. One of the consequences is that the shortest path between two points is a straight line. Deviate from that, and you increase the length.

But special relativity doesn't happen in Euclidean space. The metric is different:

ds2 = d(ct)2 - dx2 - dy2 - dz2

In this metric, for time-like separated points (ie, were ds2 is positive), a straight path is the LONGEST distance between them. Curve the path, and you make it shorter. Furthermore (and this is important), it doesn't matter which reference frame you look at it from, s will always be the same. The different components may change (just like in Euclidean space, if you rotate everything your x's, y's, and z's will all change but your distances won't), but the lengths won't.

So let's compare the twins. One twin, the "stationary" one, takes a straight path between the two points in question. One twin, the "traveling" one, takes a curved path.

The curved path is shorter than the straight path. That isn't a function of the curvature itself, exactly, but the curvature is still a necessary part of that. Same as Euclidean space, except you're going shorter instead of longer.
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Last edited by Ziggurat; 12th August 2020 at 06:40 AM.
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Old 12th August 2020, 06:44 AM   #44
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Originally Posted by SDG View Post
Agreed, all this points to the reciprocity of the SR, the paradox of the SR, each others time is dilated so how to solve the twin paradox?
I keep telling you how it's solved. And you don't even have to take MY word for it. Here's a Fermi lab physicist saying the same thing: https://www.youtube.com/watch?v=GgvajuvSpF4

One of many. There's no shortage of it getting explained if you just Google or search YouTube for "twins paradox."
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Old 12th August 2020, 06:45 AM   #45
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Originally Posted by Myriad View Post
It's the same kind of magic that allows the traveling twin at r to instantaneously reverse direction in the first place, requiring infinite acceleration and infinite energy.
Careful here. Instant reversal requires infinite power, but not infinite energy.

An alternate version I like is where the traveling twin goes in a big circle. They're constantly accelerating, but at each point along the trip their relative velocities are always the same. The acceleration makes the traveling twin non-inertial, but here's the kicker: it doesn't matter what the acceleration is, only the velocity. The traveling twin could trace out one big circle (small acceleration), or he could trace out a small circle a whole bunch of times (large acceleration), the final answer won't change. The acceleration is a necessary component to the problem, but you never need to use it in any of the calculations.
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Old 12th August 2020, 06:45 AM   #46
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Originally Posted by HansMustermann View Post
Well, they don't have to agree about what time it is on Earth. They just have to synchronize their own clocks. Like, if the one going away started the journey at 6AM, when it passes the other guy by, he transmits a signal that says, "hey, my clock says it's 12:34 right now", and the other one simply also sets his own clock to 12:34 too, and lets it tick normally from there. When he passes earth by, he and the guy who stayed home exchange similar signals and yep, they're different.

Mind you, when the guy sets the clock to match there is basically a discontinuity (his clock probably showed something else), but that's how you simulate the same clock ticking on both legs of the journey. It has to continue on the trip back from where it reached on the way out.
The A, B jump is unrealistic.
The twins do not meet again.
Check the diagram, it recognizes that the time is moving faster on the spaceship compared to the stay at home.
The paradox is not resolved.
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Old 12th August 2020, 06:48 AM   #47
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Originally Posted by HansMustermann View Post
I keep telling you how it's solved. And you don't even have to take MY word for it. Here's a Fermi lab physicist saying the same thing: https://www.youtube.com/watch?v=GgvajuvSpF4

One of many. There's no shortage of it getting explained if you just Google or search YouTube for "twins paradox."
I have seen it. The argument does not hold. I showed why it does not hold.
It requires the 'black hole' at the event C.
If you get Dr. Lincoln here we can discuss.
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Old 12th August 2020, 06:51 AM   #48
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Originally Posted by Ziggurat View Post
Careful here. Instant reversal requires infinite power, but not infinite energy.

An alternate version I like is where the traveling twin goes in a big circle. They're constantly accelerating, but at each point along the trip their relative velocities are always the same. The acceleration makes the traveling twin non-inertial, but here's the kicker: it doesn't matter what the acceleration is, only the velocity. The traveling twin could trace out one big circle (small acceleration), or he could trace out a small circle a whole bunch of times (large acceleration), the final answer won't change. The acceleration is a necessary component to the problem, but you never need to use it in any of the calculations.
Correct, that's why I posted this:




Do you see how the stay at home twin and the travelling twin are in the same reference frame at t=T/2?
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Old 12th August 2020, 06:57 AM   #49
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Originally Posted by SDG View Post
I have seen it. The argument does not hold. I showed why it does not hold.
It requires the 'black hole' at the event C.
If you get Dr. Lincoln here we can discuss.
Actually, technically so far all you've shown is that you don't know what you're talking about -- as in, not even the nomenclature -- but insist that that shows a problem with everyone ELSE's understanding. That's really not helping anyone. And I mean, not even as illustration of Dunning-Kruger, since we already had people like Pixie Of Key illustrate that far better
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Old 12th August 2020, 06:58 AM   #50
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Originally Posted by SDG View Post
The A, B jump is unrealistic.
It's unrealistic for that jump to happen instantaneously, sure. That's a simplification. But the transition can happen in a very short length of time, short enough compared to the overall trip that it's negligible.

Quote:
Check the diagram, it recognizes that the time is moving faster on the spaceship compared to the stay at home.
The paradox is not resolved.
The paradox is resolved. Let's look at it from other inertial frames:

I made this diagram for triplets. The red one stays home, the green one goes off to the left, the blue one goes off to the right. Diagram 1 shows this from the "home" reference frame. Green and blue are time dilated, red is not. Red experiences more time than green or blue, as measured in the home frame.

Now let's look at it from the right moving frame (diagram 2). Red is now time dilated. Green is time dilated for part of the journey, not time dilated for part. But it's time dilated to an even greater degree than red for the portion that is time dilated, and that portion is over half the journey in this frame. So overall green will end up with less total time than red because of that greater time dilation. Same with blue, except that it's the second part of its journey which has the increased time dilation, not the first. Diagram 3 is the other direction.

No matter which inertial frame you pick, red ends up with more time than green or blue. But you have to stick with an inertial frame, or Lorenz factors won't suffice. That means you can't pick any frame in which green or blue are straight lines.

And it doesn't matter to this whether the turnarounds are instant or not. That rounds the corners of the green and blue, but doesn't fundamentally change anything.

Now if you want to know how to handle being in an accelerating reference frame, well, that's quite interesting. The math is really nasty but you can get some of the concepts without getting too into the weeds. But you've got to be able to wrap your head around what's happening in different inertial reference frames first, or you won't have a solid structure to build off of.
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Old 12th August 2020, 06:59 AM   #51
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Originally Posted by Ziggurat View Post
True. But you still aren't looking at one inertial frame for the traveling twin. Again, that throws off the symmetry you're assuming.

Here's another way to look at it. "Proper time", the time experienced by a traveler moving along a line, is basically just a form of distance measurement. So you're basically just comparing the distance between two points along different lines.

In Euclidean geometry, how do we measure distance? For a straight line segment, the length s can be found as
s2 = x2+y2+z2
For a curved line, we chop it up into infinitesmal bits and add up the lengths of each bit. So
s2 = integral(ds2) = integral(dx2+dy2+dz2)
You can also abreviate this as simply
ds2 = dx2 + dy2 + dz2
This is called the metric for Euclidean space. One of the consequences is that the shortest path between two points is a straight line. Deviate from that, and you increase the length.

But special relativity doesn't happen in Euclidean space. The metric is different:

ds2 = d(ct)2 - dx2 - dy2 - dz2

In this metric, for time-like separated points (ie, were ds2 is positive), a straight path is the LONGEST distance between them. Curve the path, and you make it shorter. Furthermore (and this is important), it doesn't matter which reference frame you look at it from, s will always be the same. The different components may change (just like in Euclidean space, if you rotate everything your x's, y's, and z's will all change but your distances won't), but the lengths won't.

So let's compare the twins. One twin, the "stationary" one, takes a straight path between the two points in question. One twin, the "traveling" one, takes a curved path.

The curved path is shorter than the straight path. That isn't a function of the curvature itself, exactly, but the curvature is still a necessary part of that. Same as Euclidean space, except you're going shorter instead of longer.
The acceleration in the Minkowski space-time diagram leads to the same problem of the 'black hole'.
This is a problem, the 'black hole' is not real just virtual.
Is it some kind of magic?
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Old 12th August 2020, 07:00 AM   #52
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Originally Posted by SDG View Post
Correct, that's why I posted this:


https://i.imgur.com/9RSrQN7.png

Do you see how the stay at home twin and the travelling twin are in the same reference frame at t=T/2?
I'm not sure what's confusing you there. Regardless of what reference frame you choose, you can plot the {x, y, z, t} coordinates for everyone involved in it. They're ALWAYS in the same reference frame, regardless of what you do. You could put your reference frame on Jupiter, and they'll still both be in it. Those coordinates will be different, based on what frame you're using, but there's no such thing as one being in the reference frame and the other not being in it.
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Old 12th August 2020, 07:04 AM   #53
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Originally Posted by SDG View Post
Do you see how the stay at home twin and the travelling twin are in the same reference frame at t=T/2?
Doesn't matter. It's not the frame the traveling twin was in during any other part of the journey.
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Old 12th August 2020, 07:12 AM   #54
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Originally Posted by Ziggurat View Post
It's unrealistic for that jump to happen instantaneously, sure. That's a simplification. But the transition can happen in a very short length of time, short enough compared to the overall trip that it's negligible.



The paradox is resolved. Let's look at it from other inertial frames:

I made this diagram for triplets. The red one stays home, the green one goes off to the left, the blue one goes off to the right. Diagram 1 shows this from the "home" reference frame. Green and blue are time dilated, red is not. Red experiences more time than green or blue, as measured in the home frame.

Now let's look at it from the right moving frame (diagram 2). Red is now time dilated. Green is time dilated for part of the journey, not time dilated for part. But it's time dilated to an even greater degree than red for the portion that is time dilated, and that portion is over half the journey in this frame. So overall green will end up with less total time than red because of that greater time dilation. Same with blue, except that it's the second part of its journey which has the increased time dilation, not the first. Diagram 3 is the other direction.

No matter which inertial frame you pick, red ends up with more time than green or blue. But you have to stick with an inertial frame, or Lorenz factors won't suffice. That means you can't pick any frame in which green or blue are straight lines.

And it doesn't matter to this whether the turnarounds are instant or not. That rounds the corners of the green and blue, but doesn't fundamentally change anything.

Now if you want to know how to handle being in an accelerating reference frame, well, that's quite interesting. The math is really nasty but you can get some of the concepts without getting too into the weeds. But you've got to be able to wrap your head around what's happening in different inertial reference frames first, or you won't have a solid structure to build off of.
Look at your diagram 1.
All 3 triplets are in the same reference frame at the 'half time'.
They have the same simultaneity line. The horizontal line.
Please, show me that same simultaneity line of one reference system in the diagrams 2 and 3.
Please, remember the triplets are supposed be in the same frame, no relative motion in 2 and 3.

Edit: The diagrams 2 and 3 contradict 1 because they do not show acceleration for the left and right triplets.
The blue triplet does not have an acceleration on the first leg in 2 and the green triplet does not have an acceleration in 3.

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Old 12th August 2020, 07:22 AM   #55
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Originally Posted by SDG View Post
The acceleration in the Minkowski space-time diagram leads to the same problem of the 'black hole'.
This is a problem, the 'black hole' is not real just virtual.
Is it some kind of magic?
It's not a black hole, it's an event horizon. No, it's not magic, yes, you could describe it as virtual, and no, it's not a problem.

Physicists' preference for inertial reference frames predates special relativity. You encounter similar things in Newtonian mechanics. F=ma works in an inertial reference frame, it doesn't work in a non-inertial reference frame.

But you can still do Newtonian mechanics in non-inertial frames. How? Well, you just have to subtract the acceleration from the non-inertiality :

F = mareal = m(aapparent - aframe)

Rewrite this a bit and you get:

maapparent = F + maframe = Fapparent

So we can account for our non-inertiality if we know how our frame is accelerating. This leads to so-called "fictitious forces":

Fapparent = Freal + Ffictitious
Ffictitious = maframe

Thes fictitious forces are always proportional to mass, and in many cases (such as rotation), they are also position-dependent.

You can do similar stuff in special relativity to account for non-inertiality, but it gets even messier than Newtonian mechanics. When you accelerate, you introduce fictitious forces, but also fictitious Lorenz factors (if you want to work within an accelerating reference frame). And these are position-dependent, because in special relativity, acceleration is a form of 4D rotation.

This actually touches on an insight that Einstein had: a uniform gravitational field is the same as acceleration. GR gets significantly messier than special relativity (even special relativity in non-inertial reference frames) because gravity isn't uniform. So you have to make these "fictitious" effects real. But it's still the same jumping off point.
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Old 12th August 2020, 07:27 AM   #56
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Originally Posted by HansMustermann View Post
I'm not sure what's confusing you there. Regardless of what reference frame you choose, you can plot the {x, y, z, t} coordinates for everyone involved in it. They're ALWAYS in the same reference frame, regardless of what you do. You could put your reference frame on Jupiter, and they'll still both be in it. Those coordinates will be different, based on what frame you're using, but there's no such thing as one being in the reference frame and the other not being in it.
Here is the problem:




The point A in the diagram is a 'virtual black hole'.
It is an intersect with all the simultaneity lines of the deceleration and the acceleration shown in the diagram.
There is not motion in A, no time flow, nothing - it appears there like a 'black hole'.
If we do this exercise in the intergalactic space there is no real black hole at A.
The event A is a contradiction to what Minkowski space-time diagram is supposed to show.

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Old 12th August 2020, 07:29 AM   #57
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Originally Posted by SDG View Post
Look at your diagram 1.
All 3 triplets are in the same reference frame at the 'half time'.
They have the same simultaneity line. The horizontal line.
But they will disagree about when that simultaneity line is, because they didn't get to that the same way. Look at diagram 2 and 3: that "half time" isn't simultaneous if you pick either the right-moving or the left-moving frame. It's ONLY simultaneous in diagram 1. But diagram 1 is the only diagram in which any triplet is stationary for the whole duration, and only one of the triplets is stationary for it.

Quote:
Edit: The diagrams 2 and 3 contradict 1 because they do not show acceleration for the left and right triplets.
The blue triplet does not have an acceleration on the first leg in 2 and the green triplet does not have an acceleration in 3.
Diagrams 2 and 3 are simply Lorentz transformations of diagram 1. There is no contradiction.
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Old 12th August 2020, 07:34 AM   #58
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Originally Posted by SDG View Post
Here is the problem:

https://i.imgur.com/XSs3QXa.png


The point A in the diagram is a 'virtual black hole'.
It is an intersect with all the simultaneity lines of the deceleration and the acceleration shown in the diagram.
There is not motion in A, no time flow, nothing - it appears there like a 'black hole'.
If we do this exercise in the intergalactic space there is no real black hole at A.
The event A is a contradiction to what Minkowski space-time diagram is supposed to show.
No it isn't a contradiction.

Give this a try. Go outside, spin around, and watch what happens to the sun (or stars at night). It moves! It spins around you! This massive thing is moving at incredible speed, and experiencing incredible acceleration. And that's just within our solar system. The speed of other galaxies massively exceeds the speed of light, and the forces required to accelerate them in a circle around you are incomprehensibly large.

Or are they? You can probably understand that the sun didn't really move. It didn't really experience centrifugal acceleration as you spun. This is just an artifact of you spinning.

Inertial frames are different than non-inertial frames. If you take the non-inertial frame to be real, you can reach some very strange conclusions. That's all that this is.

ETA: point A is not a black hole. The line delta1 is an event horizon. There is actually a difference. A black hole is surrounded by an event horizon, but it isn't synonymous with one.
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Old 12th August 2020, 07:38 AM   #59
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Originally Posted by Ziggurat View Post
It's not a black hole, it's an event horizon. No, it's not magic, yes, you could describe it as virtual, and no, it's not a problem.

Physicists' preference for inertial reference frames predates special relativity. You encounter similar things in Newtonian mechanics. F=ma works in an inertial reference frame, it doesn't work in a non-inertial reference frame.

But you can still do Newtonian mechanics in non-inertial frames. How? Well, you just have to subtract the acceleration from the non-inertiality :

F = mareal = m(aapparent - aframe)

Rewrite this a bit and you get:

maapparent = F + maframe = Fapparent

So we can account for our non-inertiality if we know how our frame is accelerating. This leads to so-called "fictitious forces":

Fapparent = Freal + Ffictitious
Ffictitious = maframe

Thes fictitious forces are always proportional to mass, and in many cases (such as rotation), they are also position-dependent.

You can do similar stuff in special relativity to account for non-inertiality, but it gets even messier than Newtonian mechanics. When you accelerate, you introduce fictitious forces, but also fictitious Lorenz factors (if you want to work within an accelerating reference frame). And these are position-dependent, because in special relativity, acceleration is a form of 4D rotation.

This actually touches on an insight that Einstein had: a uniform gravitational field is the same as acceleration. GR gets significantly messier than special relativity (even special relativity in non-inertial reference frames) because gravity isn't uniform. So you have to make these "fictitious" effects real. But it's still the same jumping off point.
Well, I think the acceleration could be part of the discussion.
I'd like to see your response to the post #54.
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Old 12th August 2020, 07:41 AM   #60
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Originally Posted by Ziggurat View Post
But they will disagree about when that simultaneity line is, because they didn't get to that the same way. Look at diagram 2 and 3: that "half time" isn't simultaneous if you pick either the right-moving or the left-moving frame. It's ONLY simultaneous in diagram 1. But diagram 1 is the only diagram in which any triplet is stationary for the whole duration, and only one of the triplets is stationary for it.



Diagrams 2 and 3 are simply Lorentz transformations of diagram 1. There is no contradiction.
If something moves left of right it has to accelerate/decelerate.



Would you agree?

The triplets are not moving, they are stationary in diagram 1 at the half time.
Is the same condition valid in 2 and 3?

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Old 12th August 2020, 07:55 AM   #61
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Originally Posted by Ziggurat View Post
No it isn't a contradiction.

Give this a try. Go outside, spin around, and watch what happens to the sun (or stars at night). It moves! It spins around you! This massive thing is moving at incredible speed, and experiencing incredible acceleration. And that's just within our solar system. The speed of other galaxies massively exceeds the speed of light, and the forces required to accelerate them in a circle around you are incomprehensibly large.

Or are they? You can probably understand that the sun didn't really move. It didn't really experience centrifugal acceleration as you spun. This is just an artifact of you spinning.

Inertial frames are different than non-inertial frames. If you take the non-inertial frame to be real, you can reach some very strange conclusions. That's all that this is.

ETA: point A is not a black hole. The line delta1 is an event horizon. There is actually a difference. A black hole is surrounded by an event horizon, but it isn't synonymous with one.
Fair enough, I'd like to focus on your triplets example.
Diagram 1, going left, green, accelerates and decelerates in the first leg.
Diagram 3, no acceleration for the green triplet in the first leg.
It is clear the diagrams do not display the same thing.
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Old 12th August 2020, 08:07 AM   #62
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Originally Posted by SDG View Post
If something moves left of right it has to accelerate/decelerate.
Left or right relative to what?
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Old 12th August 2020, 08:24 AM   #63
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Originally Posted by SDG View Post
Fair enough, I'd like to focus on your triplets example.
Diagram 1, going left, green, accelerates and decelerates in the first leg.
Diagram 3, no acceleration for the green triplet in the first leg.
It is clear the diagrams do not display the same thing.
Acceleration is curvature. Technically speaking, all three diagrams were drawn without any curvature, using simply straight line segments.

Of course, we can posit a scenario in which there is finite acceleration but on a scale too small (ie, too short a time period) to be seen in the diagrams. This is fine, because the acceleration itself doesn't need to enter into the calculations. It can be made an arbitrarily small perturbation to the problem, so we can solve the issues of interest without reference to it.

So we actually have multiple possible scenarios for what's happening at t=0. We can have all three tripplets start at rest in diagram 1, and then the green and blue accelerate outwards. We could also have one or both of blue and green start out already in motion. Hell, we could even start with all three moving to the left, and then blue stops and red reverses direction. These are all indistinguishable from each other, if the accelerations occur quickly enough.

And all of them can be covered by all three diagrams. There is no contradiction, they all display the same thing to within the specificity possible by these simple diagrams. Any difference you allege between diagram 1 and 2, for example, is either a misinterpretation or can be covered by ambiguity of each diagram individually.
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Old 12th August 2020, 08:30 AM   #64
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You know some conversation has taken a wrong turn at Albuquerque when you have objections on par with 'but... but... but the horse is not a point.'

Same thing here. We work with a simplified model, because that's all we need.

Yes, to actually change direction in zero time, you'd need infinite acceleration, but that doesn't change the fact that it shows how the key to the whole thing is that one of the observers switches frames. You could work with a more sane acceleration in between, but as Ziggurat said, that's basically just rounding the corners on the graph and making the maths a bit more complicated. Now you get to integrate over a curve instead of a triangle, but the basic idea is the same.


Also, just like Ziggurat said, having an event horizon doesn't actually mean that a black hole has formed. The light doesn't actually go into some singularity there. And when you stop accelerating, it can reach you again just the same.

And it's not even just at relativistic speeds. If I accelerate an old steam locomotive at 1m/s, I have an event horizon behind me just the same. It's too insanely far away to be even noticed, but it exists just the same. So what?

And, again, when you stop accelerating, it disappears and the light can reach you again. No information has actually been lost.

In particular, if we're talking infinite acceleration for a duration of zero, then that's also exactly how long that event horizon exists: zero. After that, you can see the guy waiting back on Earth, or really any other point in flat space, just as if nothing happened.

In essence, that's why you can ignore it in the model.
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Old 12th August 2020, 08:34 AM   #65
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Additionally, since we're talking virtual even horizons, that exists only in the opposite direction than your acceleration. Since the ship would be accelerating towards the Earth, there is no moment when it would lose sight of Earth because of that event horizon. Because the event horizon is in the other direction. In every possible sense of the word, the twins never cease to be visible in the same frame.

And that is another reason to ignore it from the model: it just happens outside the space that we're interested in.
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Old 12th August 2020, 08:35 AM   #66
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Originally Posted by Ziggurat View Post
Acceleration is curvature. Technically speaking, all three diagrams were drawn without any curvature, using simply straight line segments.

Of course, we can posit a scenario in which there is finite acceleration but on a scale too small (ie, too short a time period) to be seen in the diagrams. This is fine, because the acceleration itself doesn't need to enter into the calculations. It can be made an arbitrarily small perturbation to the problem, so we can solve the issues of interest without reference to it.

So we actually have multiple possible scenarios for what's happening at t=0. We can have all three tripplets start at rest in diagram 1, and then the green and blue accelerate outwards. We could also have one or both of blue and green start out already in motion. Hell, we could even start with all three moving to the left, and then blue stops and red reverses direction. These are all indistinguishable from each other, if the accelerations occur quickly enough.

And all of them can be covered by all three diagrams. There is no contradiction, they all display the same thing to within the specificity possible by these simple diagrams. Any difference you allege between diagram 1 and 2, for example, is either a misinterpretation or can be covered by ambiguity of each diagram individually.
There is a contradiction.
The green triplet measures acceleration on his accelerometer in the first leg in diagram 1.
The green triplet DOES NOT measure any acceleration on his accelerometer in the first leg in diagram 3.
These are two distinct and different scenarios.
Your diagrams do not describe the same stuff.

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Old 12th August 2020, 08:36 AM   #67
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Originally Posted by theprestige View Post
Left or right relative to what?
It does not matter. The accelerometer will show some acceleration.
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Old 12th August 2020, 08:42 AM   #68
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Originally Posted by HansMustermann View Post
You know some conversation has taken a wrong turn at Albuquerque when you have objections on par with 'but... but... but the horse is not a point.'

Same thing here. We work with a simplified model, because that's all we need.

Yes, to actually change direction in zero time, you'd need infinite acceleration, but that doesn't change the fact that it shows how the key to the whole thing is that one of the observers switches frames. You could work with a more sane acceleration in between, but as Ziggurat said, that's basically just rounding the corners on the graph and making the maths a bit more complicated. Now you get to integrate over a curve instead of a triangle, but the basic idea is the same.


Also, just like Ziggurat said, having an event horizon doesn't actually mean that a black hole has formed. The light doesn't actually go into some singularity there. And when you stop accelerating, it can reach you again just the same.

And it's not even just at relativistic speeds. If I accelerate an old steam locomotive at 1m/s, I have an event horizon behind me just the same. It's too insanely far away to be even noticed, but it exists just the same. So what?

And, again, when you stop accelerating, it disappears and the light can reach you again. No information has actually been lost.

In particular, if we're talking infinite acceleration for a duration of zero, then that's also exactly how long that event horizon exists: zero. After that, you can see the guy waiting back on Earth, or really any other point in flat space, just as if nothing happened.

In essence, that's why you can ignore it in the model.
Well, no, to the bold part.
If you go without an acceleration in the solution then we are left with the paradox.
The SR does not have a hierarchy of the inertial frames.
Once we switch to accelerations there is a difference.
The tag scenario is not a solution to twin paradox, only acceleration can help.
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Old 12th August 2020, 09:08 AM   #69
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Originally Posted by HansMustermann View Post
Additionally, since we're talking virtual even horizons, that exists only in the opposite direction than your acceleration. Since the ship would be accelerating towards the Earth, there is no moment when it would lose sight of Earth because of that event horizon. Because the event horizon is in the other direction. In every possible sense of the word, the twins never cease to be visible in the same frame.

And that is another reason to ignore it from the model: it just happens outside the space that we're interested in.








Please, look at the reciprocity.
There is an inertial moving frame identical with with AP red wordline.
This is an additional triplet, let's call him the red triplet.
The red triplet is right there flying by when the black triplet accelerates from the stay home triplet.
The acceleration away, the 0 to T/4 interval, it is like deceleration for the red triplet frame.
The the traveling black triplet decelerates to the event P, T/4 to T/2 interval, this is like acceleration to the red triplet frame.
The same thing repeats on the way back.
How do we solve this time dilation?

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Old 12th August 2020, 09:56 AM   #70
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I'm still not sure what the problem is, to be honest, or rather how is it tied to the reference frames.
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Old 12th August 2020, 10:01 AM   #71
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Originally Posted by SDG View Post
Well, no, to the bold part.
If you go without an acceleration in the solution then we are left with the paradox.
No you're not. Nothing you wrote so far has indicated that the maths doesn't hold or anything. You just picked on the practicality of the scenario, but that's a whole other thing.
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Old 12th August 2020, 10:09 AM   #72
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Originally Posted by HansMustermann View Post
I'm still not sure what the problem is, to be honest, or rather how is it tied to the reference frames.
The SR does not have a solution for the twin paradox.
The acceleration is not a simple solution as well as shown in my previous post.
The analysis has to be done carefully.
The triplet diagrams show the contradictions.
If a hierarchy of the reference frames cannot be achieved we have a problem.
The experiments point towards the frame hierarchy to be real.
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Old 12th August 2020, 10:10 AM   #73
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Originally Posted by HansMustermann View Post
No you're not. Nothing you wrote so far has indicated that the maths doesn't hold or anything. You just picked on the practicality of the scenario, but that's a whole other thing.
How do the twins meet again without an acceleration?
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Old 12th August 2020, 10:21 AM   #74
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Originally Posted by SDG View Post
Please, look at the reciprocity.
There is an inertial moving frame identical with with AP red wordline.
No, there isn't. There are TWO inertial frames which together cover the red worldline. But it's two different frames, not one, and the fact that it's two is critical. The two frames WILL NOT AGREE about simultaneity for the red world line of the traveling twin and the grey world line for the earthbound twin. The frame for the outbound twin will think that at the turnaround point for the traveling twin, the earthbound twin has already experienced less time. But the frame for the inbound twin will think that at the turnaround point for the traveling twin, the earthbound twin will have already experienced MORE time.

If you cannot understand that my three diagrams all depict the exact same scenario in three different inertial reference frames, then you either don't understand the scenario, or you don't understand Lorentz transformations. It really is that simple. Special relativity has been around for over 100 years. It is a mathematically elegant and fully self-consistent theory. You will not find any actual paradoxes. If you think you found one, it's because you don't understand something. And that's perfectly understandable: special relativity is counter-intuitive. Nobody understands it without study. But you are not smarter than every physicist of the last 100 years. You haven't figured out what nobody else has figured out.
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Old 12th August 2020, 10:24 AM   #75
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Originally Posted by SDG View Post
Well, no, to the bold part.
If you go without an acceleration in the solution then we are left with the paradox.
You missed the point entirely. Yes, you need acceleration in the problem. But if you want to plug in actual numbers, you can make the acceleration such that it contributes an arbitrarily small perturbation to the final answer. Therefore we can ignore what the acceleration is for our calculations for simplicity. That a twin accelerates is vital. What the acceleration is, however, is not.
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Old 12th August 2020, 10:24 AM   #76
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Originally Posted by SDG View Post
The SR does not have a solution for the twin paradox.
Yes it does, and it has been explained to you repeatedly. Just repeating the same falsehood one more time won't make it true. We're not in The Hunting Of The Snark and you're not the Bellman. Something doesn't become true just because you say it three (more) times.

Again, nowhere have you shown that the maths doesn't work or anything else of substance. Your objection has been just, basically, that it's impossible or unrealistic to perform that experiment. But that's a completely unrelated issue than whether SR holds or not.

Basically equally I could calculate the thrust force if monkeys were to fly out of my ass. More specifically, adult capuchin monkeys, at 100m/s and a rate of one monkey per second. Is it a realistic or even possible scenario? No, it isn't. But that doesn't mean that the thrust equation is wrong.

Originally Posted by SDG View Post
The acceleration is not a simple solution as well as shown in my previous post.
Err... so? No, seriously. The maths being complicated doesn't make something wrong or anything.

Originally Posted by SDG View Post
The analysis has to be done carefully.
The triplet diagrams show the contradictions.
No it doesn't. So far it just showed that you don't understand it.

Originally Posted by SDG View Post
If a hierarchy of the reference frames cannot be achieved we have a problem.
The experiments point towards the frame hierarchy to be real.
No they don't. In fact, not even in GR. So even introducing acceleration into it still won't point in the direction you think. The whole point of why it has "relativity" in the name is that it is, surprisingly enough, relative. GR lets you calculate things in an accelerating frame (or indeed in a gravity well, same thing), but it's still not a hierarchy.

If you have one guy on the Earth and one on the moon, the curvature of the space (because of acceleration, which gravity is a part of too) will be different, but one frame (or chart, before someone objects to my misusing terms) is still not in any way better than the other. None of them is inherently the preferred frame. You might prefer one or the other because it makes your maths easier, but essentially neither of them is inherently THE one true frame to use.
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Old 12th August 2020, 10:25 AM   #77
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Originally Posted by SDG View Post
There is a contradiction.
The green triplet measures acceleration on his accelerometer in the first leg in diagram 1.
No he doesn't. It's a straight line. There is no acceleration during the leg. Any acceleration before that leg can be made an arbitrarily small perturbation.

Quote:
Your diagrams do not describe the same stuff.
Yes, they do. They really do.
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Old 12th August 2020, 11:32 AM   #78
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Originally Posted by Ziggurat View Post
You missed the point entirely. Yes, you need acceleration in the problem. But if you want to plug in actual numbers, you can make the acceleration such that it contributes an arbitrarily small perturbation to the final answer. Therefore we can ignore what the acceleration is for our calculations for simplicity. That a twin accelerates is vital. What the acceleration is, however, is not.
Let us ignore the acceleration. What happened here?






Two triplets travel to the right.
Do ct and ct' triplets agree on the ct'' time when they meet at t=14s?
The relative speed between ct' and ct'' is the same as between ct and ct'.
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Old 12th August 2020, 12:02 PM   #79
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Originally Posted by Ziggurat View Post
No, there isn't. There are TWO inertial frames which together cover the red worldline. But it's two different frames, not one, and the fact that it's two is critical. The two frames WILL NOT AGREE about simultaneity for the red world line of the traveling twin and the grey world line for the earthbound twin. The frame for the outbound twin will think that at the turnaround point for the traveling twin, the earthbound twin has already experienced less time. But the frame for the inbound twin will think that at the turnaround point for the traveling twin, the earthbound twin will have already experienced MORE time.

If you cannot understand that my three diagrams all depict the exact same scenario in three different inertial reference frames, then you either don't understand the scenario, or you don't understand Lorentz transformations. It really is that simple. Special relativity has been around for over 100 years. It is a mathematically elegant and fully self-consistent theory. You will not find any actual paradoxes. If you think you found one, it's because you don't understand something. And that's perfectly understandable: special relativity is counter-intuitive. Nobody understands it without study. But you are not smarter than every physicist of the last 100 years. You haven't figured out what nobody else has figured out.
Is there an 'event horizon/black hole' at event A here?





Why would I ask?
Because when the front of the train car aligns with the platform frame where x=x'=0 and t=t'=0
and the front train car observer is going to 'ask' his comoving observer at the back what do see outside?
Meaning the comoving observer knows to record what is outside at t'=0 and x'=-3.4641cs' ( L0=3.4641cs)
The train back observer sees -6.9282cs outside and time -6s.
The same is done by the platform observers.
The 'back' platform observer records what he sees on the train.
The 'back' platform observer located at x=-1.732cs records -3.4641cs' and t'=3s' on the train.
How does the train 'shrinks' from the event B to event C with no time flowing at the front of the train car?
Is event A a 'black hole'???
The time is moving in both frames at the back but it is not moving at the front?

This all goes back to what time dilation is and how it is reciprocal.

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Old 12th August 2020, 02:00 PM   #80
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Not sure you understand what a black hole is. A black hole has an event horizon. Not every event horizon is a black hole or around a black hole.

Basically just like a bird has wings, but not everything that has wings is a bird. I just saw four wings on a helicopter, for example.

Additionally, unless I'm reading your graph wrong, there isn't even an event horizon there.

An event horizon ONLY happens when there's acceleration involved. Gravity IS acceleration, or at least equivalent, which is why a black hole has one. An accelerating body would also 'see' one behind it, although it would have to be accelerating really hard, and it's not really there for anyone else. Some guy standing on the ground next to the accelerating train would not actually see any event horizon there.

So basically not only there is no black hole there in any case, but if it's a constant speed motion (which seems to be the case in your graph) there isn't an event horizon at all either.
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