Is the frame dragging part of the Special Relativity? - Page 5

SDG · 14th August 2020, 06:14 AM

Originally Posted by Robin

Maybe you could start again and state the entire scenario explicitly.

And state what the events A, B and C are.

Also, your spacetime diagrams are wrong, they imply that the train is travelling at 0.866c relative to the platform and the platform is travelling at 0.866c relative to the train.

The t' and x' axes in the train spacetime diagram should form an obtuse angle.

You can check it yourself here: http://www.trell.org/div/minkowski.html

There is no problem with ct' and x' axes.

Ziggurat · 14th August 2020, 06:35 AM

Originally Posted by SDG

The 4s on the platform takes 8s' on the train.
Any questions?

Yes, I have a question. What's with this diagram?

What's the dashed line I circled for? Is it supposed to be the return path of the light?

One of the difficulties of the way you've set up your problem is that you have two spatial dimensions you're working with, but you can only display one of those dimensions in your space-time diagram. This means you can't plot your y coordinates at the same time as your x and t coordinates, at least not with whatever you're currently using to construct your diagrams. Any light path with a y component will therefore not trace out a 45 degree angle on your graph.

Another complication is that you've sort of mixed and matched things. You take a light path in the train frame that travels only in the y' direction and then transform it to the platform where it travels in both y and x. That's a pretty easy transformation. But then you take a light path in the platform frame which travels in both y and x and then transform it into the train frame. This is a more complex transformation. You can do that, there's nothing inherently wrong with it, but it's an added complication, I don't see what it's supposed to demonstrate, and because it's not equivalent starting scenario you're not going to produce an equivalent end scenario.

I don't see any errors yet. But you haven't gotten to what you think the contradiction is. So far, you've only shown that 2 second proper time in the train frame has 4 second coordinate time in the platform frame, while 4 second proper time in the platform frame has 8 second coordinate time in the train frame. In other words, Lorenz time dilation. So... where's the contradiction?

SDG · 14th August 2020, 06:46 AM

Originally Posted by Ziggurat

Yes, I have a question. What's with this diagram?
http://www.internationalskeptics.com...6907391b5c.png

What's the dashed line I circled for? Is it supposed to be the return path of the light?

One of the difficulties of the way you've set up your problem is that you have two spatial dimensions you're working with, but you can only display one of those dimensions in your space-time diagram. This means you can't plot your y coordinates at the same time as your x and t coordinates, at least not with whatever you're currently using to construct your diagrams. Any light path with a y component will therefore not trace out a 45 degree angle on your graph.

Another complication is that you've sort of mixed and matched things. You take a light path in the train frame that travels only in the y' direction and then transform it to the platform where it travels in both y and x. That's a pretty easy transformation. But then you take a light path in the platform frame which travels in both y and x and then transform it into the train frame. This is a more complex transformation. You can do that, there's nothing inherently wrong with it, but it's an added complication, I don't see what it's supposed to demonstrate, and because it's not equivalent starting scenario you're not going to produce an equivalent end scenario.

I don't see any errors yet. But you haven't gotten to what you think the contradiction is. So far, you've only shown that 2 second proper time in the train frame has 4 second coordinate time in the platform frame, while 4 second proper time in the platform frame has 8 second coordinate time in the train frame. In other words, Lorenz time dilation. So... where's the contradiction?

Is this: "Any light round trip will appear slower (time dilated) compared to any other moving frame?"
a true statement?

Myriad · 14th August 2020, 07:03 AM

Originally Posted by SDG

The event coordinates are in (x, y, t) and (x',y',t') formats.

A: (0, 0, 0) ------- (0', 0', 0')
D: (1.732, 1, 2) -- (0', 1', 1')
E: (0, 0, 4) ------- (-6.9282', 0', 8')

I hope this helps.

Apparently event E is the arrival of the light pulse from x = x' = 0 and t = t' = 0, after having been retro-reflected from a particular point on the train at event D (x' = 1, y = y' = 1), back at x = y = 0. That happens at t = 4.

Measured from the train frame, that same event happens at t' = 8, and about seven light-seconds behind the front of the train (x' = -6.93).

This is not a contradiction of anything.

Ziggurat · 14th August 2020, 07:04 AM

Originally Posted by SDG

Is this: "Any light round trip will appear slower (time dilated) compared to any other moving frame?"
a true statement?

It's not always true if you include non-inertial frames, and even for inertial frames it may depends on how you want to do the comparison. I'll give you a more precise way of saying it that's always true:

For inertial coordinate systems, coordinate time between two events is always longest in the reference frame where the two events occur in the same location.

Does this satisfy you? The start and end of your round trip can be the two events, and they are in the same location if it's a "round trip" and not just a trip. A "moving frame" would be a frame in which the start and end events are not in the same location. And the basis of comparison is now explicit: we're comparing coordinate time between the events in our different frames. And finally, note that proper time along a non-accelerating world line is equal to coordinate time in a reference frame where that world line is stationary.

Note also that I've specified inertial reference frames. For this thread, it's not worth the effort of trying to work from non-inertial reference frames, and the statement isn't always true if we include non-inertial frames either.

Myriad · 14th August 2020, 07:23 AM

Originally Posted by Ziggurat

It's not always true if you include non-inertial frames, and even for inertial frames it may depends on how you want to do the comparison. I'll give you a more precise way of saying it that's always true:

For inertial coordinate systems, coordinate time between two events is always longest in the reference frame where the two events occur in the same location.

Does this satisfy you? The start and end of your round trip can be the two events, and they are in the same location if it's a "round trip" and not just a trip. A "moving frame" would be a frame in which the start and end events are not in the same location. And the basis of comparison is now explicit: we're comparing coordinate time between the events in our different frames. And finally, note that proper time along a non-accelerating world line is equal to coordinate time in a reference frame where that world line is stationary.

Note also that I've specified inertial reference frames. For this thread, it's not worth the effort of trying to work from non-inertial reference frames, and the statement isn't always true if we include non-inertial frames either.

But even the simply worded statement holds true in SDG's train example. There are two different round trips described: a round trip in the platform frame, starting and ending at x = y = 0, and a different round trip in the train frame, starting and ending at x' = y' = 0. (The start points of both round trips coincide at a single event, but the end points don't, so they're different round trips as long as no one is trying to confuse things on purpose.)

The platform round trip measures 4 seconds in the platform frame, and 8 seconds in the train frame. In other words, longer in the relatively moving frame.

The train round trip measures 2 seconds in the train frame, and 4 seconds in the platform frame. In other words, longer in the relatively moving frame.

SDG · 14th August 2020, 07:23 AM

Originally Posted by Ziggurat

It's not always true if you include non-inertial frames, and even for inertial frames it may depends on how you want to do the comparison. I'll give you a more precise way of saying it that's always true:

For inertial coordinate systems, coordinate time between two events is always longest in the reference frame where the two events occur in the same location.

Does this satisfy you? The start and end of your round trip can be the two events, and they are in the same location if it's a "round trip" and not just a trip. A "moving frame" would be a frame in which the start and end events are not in the same location. And the basis of comparison is now explicit: we're comparing coordinate time between the events in our different frames. And finally, note that proper time along a non-accelerating world line is equal to coordinate time in a reference frame where that world line is stationary.

Note also that I've specified inertial reference frames. For this thread, it's not worth the effort of trying to work from non-inertial reference frames, and the statement isn't always true if we include non-inertial frames either.

Agreed, the statement is for inertial frames only.
This is what makes the SR reciprocal.

Two resolve the twin/triplet paradox we have to go to general/accelerated frames.
Those can be broken down to a sequence of comoving inertial frames.
Do you agree with that?

Edit: Just to clarify, did you mean shortest?

For inertial coordinate systems, coordinate time between two events is always shortest (not longest) in the reference frame where the two events occur in the same location.

Example:
4s in one location of the platform frame
8s in two different locations of the train frame

Ziggurat · 14th August 2020, 11:14 AM

Originally Posted by SDG

Two resolve the twin/triplet paradox we have to go to general/accelerated frames.

No, we do not. All relevant analyses can be done in non-accelerated frames. Pick multiple inertial reference frames if you want, analyze it from any of them, and the results for the proper time for each traveler will be the same.

Quote:

Those can be broken down to a sequence of comoving inertial frames.
Do you agree with that?

No. Accelerated frames in special relativity are not simply a series of comoving frames. It's far, far messier than that. You need to account for the changing lines of simultaneity, and that does... strange things. Including making coordinate time run backwards.

And it's completely unnecessary.

You CAN analyze this from the viewpoint of a series of inertial reference frame, but it's also unnecessary easier to screw it up than if you just do the whole thing from a single reference frame (meaning analyze all of it from one frame, you can still then pick another frame to analyze it from and repeat). And I suspect that's what you did to make you think that there was any contradiction.

Quote:

Edit: Just to clarify, did you mean shortest?

Yes, sorry.

The separation between two (time-like separated) events has the shortest coordinate time when they are in the same position, and increases coordinate time when you change to any other coordinates. Coordinate time between events does not depend on any path between events.

Conversely, the proper time between events is path-dependent, and is longest when the path is straight, but it does not change when you change coordinates.

SDG · 14th August 2020, 11:42 AM

Originally Posted by Ziggurat

No, we do not. All relevant analyses can be done in non-accelerated frames. Pick multiple inertial reference frames if you want, analyze it from any of them, and the results for the proper time for each traveler will be the same.

No. Accelerated frames in special relativity are not simply a series of comoving frames. It's far, far messier than that. You need to account for the changing lines of simultaneity, and that does... strange things. Including making coordinate time run backwards.
...

How do you propose to solve it in SR only?
The twins will not meet again.
The sudden instantaneous turn around is not realistic.
The stopping and coming back cannot happen without acceleration/deceleration.
The diagram below is not right.
Do you remember A, B, C events?
How can a observer move along a world line when there is an observer not moving at all?
That's the magic that SR needs?
Something similar is happening in the 'jump' below. Magic!

I used many figures from this text book:

The acceleration/deceleration, the SR in General Frames is way to go.
The twin diagram is not from this book. It is a different book.

Ziggurat · 14th August 2020, 12:26 PM

Originally Posted by SDG

How do you propose to solve it in SR only?
The twins will not meet again.

This is wrong. Very, very, very wrong.

First off, accelerated motion is not the same as an accelerated reference frame. It sounds like this is a point of confusion for you, and if it is, you need to solve it, because nothing will make sense if you are operating under the delusion that they are the same. They are not. They are very, very different.

Second, special relativity can handle accelerated. Accelerated motion in special relativity can get a bit messy, but it's perfectly doable, and doesn't produce any bizarre effects that differ fundamentally from non-accelerated motion.

Third special relativity can in fact handle accelerated reference frames. That doesn't get a bit messy, that actually gets VERY messy. And it does produce bizarre effects (like event horizons). But these bizarre effects are coordinate effects. They are analogous to the coordinate singularities you can produce in ordinary Euclidean geometry if you choose non-Cartesian coordinates. They are not physical effects.

Quote:

The sudden instantaneous turn around is not realistic.

True. But here's the thing: it doesn't matter.

You can complete your acceleration in some finite time, yes? Let's can the time it takes to go from stationary to whatever fraction of c you're going to use R (in the original rest frame). The minimum time for the journey in that frame is thus 4R. You spend the first R accelerating outwards, the second R decelerating back to zero, the third R accelerating back, and the fourth R decelerating. In this scenario, we need to account for the finite acceleration when calculating the proper time for the traveling twin.

But you could extend the journey. Spend the first R accelerating outwards. Then spend a bunch of time X at constant velocity outbound. Then turn around, and spend a bunch of time X coming back. And then spend the last R decelerating. So the total time would be 4R+2X.

Now suppose X = R. The fact that the acceleration isn't instant still matters, right? Sure.

But what if X = 100R? What if X = 1,000,000R? What if X = 10¹⁰⁰ R? At some point, X is sufficiently large compared to R that the total proper time is dominated by what happens during X, and what happens during R becomes an irrelevant perturbation that we don't care about anymore. And that's true no matter what value of R you started with. You can always construct the problem such that you just don't care.

And that's why it's OK to simplify the problem by assuming instantaneous acceleration. It doesn't matter that instantaneous acceleration is impossible. You can construct the problem such that it's an arbitrarily small part of the problem, and therefore you can ignore it.

Quote:

The diagram below is not right.

It's simplified (see above), but it's right.

Quote:

Do you remember A, B, C events?
How can a observer move along a world line when there is an observer not moving at all?

That question doesn't make sense. All observers move along worldlines. You can't NOT move along a worldline, because you can't stop time. The only sense in which you could stop moving along a world line is to stop existing, but that's not part of our scenario.

Quote:

That's the magic that SR needs?
Something similar is happening in the 'jump' below. Magic!

It's not magic. It's a coordinate effect.

One of the complications which your diagram doesn't address but may be tripping you up is the distinction between what you see and what you observe. You do not see your line of simultaneity. You can't, it's not physical. That also means that you can't see your line of simultaneity change when you reverse course. But you can observe it.

OK, so what is an observation in special relativity? It has a rather specific and non-obvious meaning in special relativity, and it causes a lot of confusion to learners who aren't careful. It's a sort of idealized measurement. It's what you would get if you could do measurements which you can't actually do, but it's also what you get if you do measurements you can do AND you account for how things like the finite speed of light affect your actual measurements.

I'll give you a simple example of the distinction. Have you ever heard a jet plane fly overhead, looked up, and noticed that it sounds like the noise is coming from behind the plane? That's what you hear: the sound is coming from a direction other than the current direction to the plane. Someone unaware of the finite speed of sound might incorrectly conclude that the sound didn't come from the plane itself but from behind the plane. But if you know the speed of sound, and you know how far away the plane is and how fast it's going, you can calculate where the plane was when the sound was emitted. You can observe that the plane did in fact emit the sound.

OK, so now let's look at what the twins would see, not what they observe. On the outward journey, both twins see the other's clock slowed down. But they don't see it slowed down by the Lorenz time dilation factor. No, it's slowed down even more. Why? Because each tick takes longer and longer to get to the other twin. What they see is a combination of both the Lorenz time dilation AND a Doppler shift (the same thing which makes the sound of an approaching car higher pitched than a receding car). Both twins see the same combined slowdown for the outbound journey.

When the twins are approaching each other, the Lorenz factor still applies, but the Doppler shift works in reverse. Instead of slowing it down more, it speed the clock back up. Both twins see the Doppler effect speed up the clocks by the same amount.

But the symmetry between them does break, as it must in order to resolve the paradox. Let's look first at the traveling twin. On the outbound part of his journey, he sees the Lorenz slowdown plus a Doppler slowdown (I'll call this L*D_r). On the return part of his journey, he sees the Lorenz slowdown plus a Doppler increase (L*D_b). He never sees his twin's clock jump in time, it only jumps in rate when he reverses (as the Doppler contribution switches from redshift to blueshift with the turnaround). For him, this change happens at the halfway mark of his journey. He sees the earthbound twin's clock at L*D_r for half the journey, and L*D_b for half the journey.

Now the FIRST symmetry to break here is that the effect from D_b and D_r do not cancel by the time the traveling twin returns. The blueshift of the return journey will produce more additional lapsed time than the redshift of the journey took away. In fact, the blue shift will be such that the total time elapsed on the earthbound twin's clock will EXCEED the total time elapsed on the traveling twin's clock, even though the Lorenz factor was applied the entire time.

OK, now on to the SECOND broken symmetry. I said that for the traveling twin, the switch in what he sees from being red shifted to being blue shifted happens at the halfway mark for him. But it DOESN'T happen at the halfway mark for the earthbound twin. He might OBSERVE that it does, but he doesn't SEE that it does. He doesn't see the traveling twin's clock become blue shifted until light from the turnaround point get back to him, and that happens well after the halfway point for him. So the earthbound twin sees the traveling twin's clock red shifted for more than half the time, and blue shifted for less than half the time. And those up, and by the time the traveling twin returns, he will have seen the traveling twin's clock elapsed less time than his own. At no point does either twin see any discontinuity. But they absolutely do not see the same thing. The symmetry is broken. There is no paradox.

If this doesn't get through to you, there may be no hope.

SDG · 14th August 2020, 01:50 PM

Originally Posted by Ziggurat

...

True. But here's the thing: it doesn't matter.

You can complete your acceleration in some finite time, yes? Let's can the time it takes to go from stationary to whatever fraction of c you're going to use R (in the original rest frame). The minimum time for the journey in that frame is thus 4R. You spend the first R accelerating outwards, the second R decelerating back to zero, the third R accelerating back, and the fourth R decelerating. In this scenario, we need to account for the finite acceleration when calculating the proper time for the traveling twin.

But you could extend the journey. Spend the first R accelerating outwards. Then spend a bunch of time X at constant velocity outbound. Then turn around, and spend a bunch of time X coming back. And then spend the last R decelerating. So the total time would be 4R+2X.

Now suppose X = R. The fact that the acceleration isn't instant still matters, right? Sure.

But what if X = 100R? What if X = 1,000,000R? What if X = 10¹⁰⁰ R? At some point, X is sufficiently large compared to R that the total proper time is dominated by what happens during X, and what happens during R becomes an irrelevant perturbation that we don't care about anymore. And that's true no matter what value of R you started with. You can always construct the problem such that you just don't care.

And that's why it's OK to simplify the problem by assuming instantaneous acceleration. It doesn't matter that instantaneous acceleration is impossible. You can construct the problem such that it's an arbitrarily small part of the problem, and therefore you can ignore it.

It's simplified (see above), but it's right.
...

Thanks, it is very good post and a lot of points to address.
I'll start with this.
I do understand what you are saying here, really, I do.
Therefore I am going to say this:

No matter of the X/R ratio, it follows from your equation that there is going to be R at the mid point P!
The simultaneity line is horizontal line as seen from the stay home twin/triplet frame.
The twin/triplets are in the same frame for a split of a second.
Agreed?

Ziggurat · 14th August 2020, 01:57 PM

Originally Posted by SDG

No matter of the X/R ratio, it follows from your equation that there is going to be R at the mid point P!
The simultaneity line is horizontal line as seen from the stay home twin/triplet frame.
The twin/triplets are in the same frame for a split of a second.
Agreed?

Sure.

SDG · 14th August 2020, 03:07 PM

Originally Posted by Ziggurat

Sure.

So what times are on the triplets clocks at the mid point when the simultaneity is the horizontal line?
Time on the clocks represents the proper time.
ct = 7s
ct' = ?s'
ct'' = ?s''

Robin · 14th August 2020, 04:04 PM

Originally Posted by SDG

You can check it yourself here: http://www.trell.org/div/minkowski.html

https://i.imgur.com/tNj7h4k.png

There is no problem with ct' and x' axes.

The relative velocity can't be the same in the platform frame as in the train frame.

If the relative velocity in the train frame is 0.866 then the relative velocity in the platform frame would be -0.866.

Try that.

Robin · 14th August 2020, 04:12 PM

Originally Posted by SDG

The event coordinates are in (x, y, t) and (x',y',t') formats.

A: (0, 0, 0) ------- (0', 0', 0')
D: (1.732, 1, 2) -- (0', 1', 1')
E: (0, 0, 4) ------- (-6.9282', 0', 8')

I hope this helps.

I was looking for a complete statement of what is happening. Is the observer on the platform or train? Which direction? etc

Something like

A: The observer on the platform switches on a light beam aimed at 90 degrees to the direction of the train. (0,0,0) (0',0',0')
D: The light beam .... (1.732,2,2) (0',1',1')

and so on .

Without that I can't get a clear idea of what is supposed to be happening.

SDG · 14th August 2020, 04:25 PM

Originally Posted by Robin

The relative velocity can't be the same in the platform frame as in the train frame.

If the relative velocity in the train frame is 0.866 then the relative velocity in the platform frame would be -0.866.

Try that.

What is the reason to switch frames?
Diagram is from platform frame point of view.

Robin · 14th August 2020, 04:43 PM

Originally Posted by SDG

What is the reason to switch frames?

Diagram is from platform frame point of view.

You have two diagrams. Under it you write: "This is a round trip on the train frame".

So they are both in the platform frame?

Robin · 14th August 2020, 05:10 PM

Originally Posted by SDG

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

So what times are on the triplets clocks at the mid point when the simultaneity is the horizontal line?
Time on the clocks represents the proper time.
ct = 7s
ct' = ?s'
ct'' = ?s''

As I said before (and you didn't want to listen) it depends upon the path it takes at the turn around.

You are taking a text book simplification and asking what the real life results would be.

If the change in velocity is instantaneous then it will depend on whether the paths break up t=[0,7](7,14] or t=[0,7)[7,14].

On the other hand if there is a very short continuous curve then the times for each twin will be approximately the same as the first leg t=[0,7].

Robin · 14th August 2020, 05:15 PM

Originally Posted by Robin

As I said before (and you didn't want to listen) it depends upon the path it takes at the turn around.

You are taking a text book simplification and asking what the real life results would be.

If the change in velocity is instantaneous then it will depend on whether the paths break up t=[0,7](7,14] or t=[0,7)[7,14].

On the other hand if there is a very short continuous curve then the times for each twin will be approximately the same as the first leg t=[0,7].

Certainly in each case the times on the clocks of each triplet will be different at t=7, just as you would expect.

When will you get around to saying what the contradiction is?

Robin · 14th August 2020, 08:21 PM

For the light round-trip, I assume that the situation is something like this, from the platform frame:

(Gridlines in 0.5s)

Ziggurat · 14th August 2020, 08:28 PM

Originally Posted by SDG

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

So what times are on the triplets clocks at the mid point when the simultaneity is the horizontal line?
Time on the clocks represents the proper time.
ct = 7s
ct' = ?s'
ct'' = ?s''

t' = 3s
t" = 1s

So where's your contradiction?

Not that it really matters, so far it doesn't seem to have confused anyone, but if you are looking at ct instead of c, then the answer would be in meters, not seconds.

SDG · 14th August 2020, 08:35 PM

Originally Posted by Robin

Certainly in each case the times on the clocks of each triplet will be different at t=7, just as you would expect.

When will you get around to saying what the contradiction is?

It was already discussed in this thread.
If ct=7s and ct'=3.5s' then there are two options for ct''.
The first 1s'' and the second 1.75s''.
The problem is the clock can show only one.

SDG · 14th August 2020, 08:42 PM

Originally Posted by Ziggurat

t' = 3s
t" = 1s

So where's your contradiction?

Not that it really matters, so far it doesn't seem to have confused anyone, but if you are looking at ct instead of c, then the answer would be in meters, not seconds.

How did you do the calculation?
Right about the meters.

Ziggurat · 14th August 2020, 09:37 PM

Originally Posted by SDG

It was already discussed in this thread.
If ct=7s and ct'=3.5s' then there are two options for ct''.
The first 1s'' and the second 1.75s''.
The problem is the clock can show only one.

No. 1.75 seconds is wrong.

You might incorrectly conclude that the answer is 1.75, because in the reference frame where the outbound single prime triplet is stationary, the double primed triplet has a gamma of 2. This might lead you to conclude that the double primed twin should reach the turnaround point at half the time that the single primed twin reaches the turnaround point.

This is wrong. Wrongety-wrong-wrong.

It's wrong because in this reference frame where the double primed triplet has a gamma of 2, the double primed twin turns around BEFORE the single primed twin. It's only simultaneous in the frame where the unprimed twin is stationary. The fact that the single primed twin changes to this frame momentarily doesn't matter, because in this frame, the double primed twin's outbound journey didn't have a gamma of 2. You can't take properties from one frame and apply them in another frame without accounting for that change in frames. But that's the only way to conclude that the answer should be 1.75 seconds. It isn't. It's 1 second. It's 1 second as measured in ANY inertial reference frame.

Robin · 14th August 2020, 09:38 PM

Originally Posted by SDG

Is this: "Any light round trip will appear slower (time dilated) compared to any other moving frame?"
a true statement?

If you mean "will appear to take longer" then it is true for inertial frames.

It holds true in your example.

So where is the problem?

Ziggurat · 14th August 2020, 09:39 PM

Originally Posted by SDG

How did you do the calculation?
Right about the meters.

There are lots of ways. You could use the time dilation equation. You could transform your coordinates to primed coordinates using the full Lorentz transformations and just read off the time. You could use the metric I gave you earlier. These will all produce the same answer, if done correctly. If you do it wrong, all bets are off.

1.75 seconds is doing it wrong.

Robin · 14th August 2020, 09:40 PM

Originally Posted by SDG

It was already discussed in this thread.
If ct=7s and ct'=3.5s' then there are two options for ct''.
The first 1s'' and the second 1.75s''.
The problem is the clock can show only one.

There is only one option for ct'' for any given path.

Robin · 14th August 2020, 09:44 PM

Originally Posted by SDG

It was already discussed in this thread.
If ct=7s and ct'=3.5s' then there are two options for ct''.
The first 1s'' and the second 1.75s''.
The problem is the clock can show only one.

I already pointed out the reason for the two values, ie:

Originally Posted by robin

If the change in velocity is instantaneous then it will depend on whether the paths break up t=[0,7](7,14] or t=[0,7)[7,14].

On the other hand if there is a very short continuous curve then the times for each twin will be approximately the same as the first leg t=[0,7]."

If the path is continuous then the value for ct'' is close to 1 s

SDG · 15th August 2020, 11:36 AM

You said this.

Originally Posted by Ziggurat

Originally Posted by SDG

...

No matter of the X/R ratio, it follows from your equation that there is going to be R at the mid point P!
The simultaneity line is horizontal line as seen from the stay home twin/triplet frame.
The twin/triplets are in the same frame for a split of a second.
Agreed?

Sure.

My understanding is that you agreed that all three triplets are in the same original stay at home frame at the turning point P.
All three triplets have the same 4-velocity vector in direction and magnitude at the turning point P.
So this:

Originally Posted by Ziggurat

No. 1.75 seconds is wrong.

You might incorrectly conclude that the answer is 1.75, because in the reference frame where the outbound single prime triplet is stationary, the double primed triplet has a gamma of 2. This might lead you to conclude that the double primed twin should reach the turnaround point at half the time that the single primed twin reaches the turnaround point.

This is wrong. Wrongety-wrong-wrong.

It's wrong because in this reference frame where the double primed triplet has a gamma of 2, the double primed twin turns around BEFORE the single primed twin. It's only simultaneous in the frame where the unprimed twin is stationary. The fact that the single primed twin changes to this frame momentarily doesn't matter, because in this frame, the double primed twin's outbound journey didn't have a gamma of 2. You can't take properties from one frame and apply them in another frame without accounting for that change in frames. But that's the only way to conclude that the answer should be 1.75 seconds. It isn't. It's 1 second. It's 1 second as measured in ANY inertial reference frame.

This appears to me as a contradiction.
Please, can you clear this one out?

SDG · 15th August 2020, 11:49 AM

Originally Posted by Robin

There is only one option for ct'' for any given path.

OK, differently:

The triplets go out and they just stop at mid point P.
They do not go back.
What are the distances between them?
When they stop they are in the same inertial frame.

Ziggurat · 15th August 2020, 11:56 AM

Originally Posted by SDG

This appears to me as a contradiction.
Please, can you clear this one out?

It's not a contradiction.

Look, it's not actually that complicated. Pick an inertial frame, any inertial frame. Do the calculations in that frame. Show me YOUR calculations.

SDG · 15th August 2020, 12:18 PM

Even simpler scenario.

I assume it is OK to use X/R calculation, right?
The traveling twin accelerates from A and decelerates to B.
Using X/R we are going to ignore the acceleration and deceleration.
But we know it happened and the traveler stopped at B.
They are again in the same reference frame.
What is the distance between them when the traveler stops at B?

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

Ziggurat · 15th August 2020, 01:02 PM

Originally Posted by SDG

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

Velocity is measured by distance over time using coordinate distance and time. If you use distance from one frame and time from another, it's not velocity. No one's velocity exceeds c.

caveman1917 · 15th August 2020, 02:00 PM

Originally Posted by Ziggurat

Third special relativity can in fact handle accelerated reference frames. That doesn't get a bit messy, that actually gets VERY messy. And it does produce bizarre effects (like event horizons). But these bizarre effects are coordinate effects. They are analogous to the coordinate singularities you can produce in ordinary Euclidean geometry if you choose non-Cartesian coordinates. They are not physical effects.

Just because something is a coordinate effect doesn't mean it isn't physical. For example Unruh radiation will be physically detectable just like any other radiation, independent of whether it's a coordinate effect.

SDG · 15th August 2020, 02:16 PM

Originally Posted by Ziggurat

Velocity is measured by distance over time using coordinate distance and time. If you use distance from one frame and time from another, it's not velocity. No one's velocity exceeds c.

The event A is in the original frame.
The event B is in the original frame.
What is physical space distance between A and B?
The traveler is in the original frame.
The traveler has something on his clock, what is it?
Are you saying that the traveler has different measuring rod when he is stationary in the original frame?
How did I mixed frames when both twins are stationary and they do not have any relative motion?

Ziggurat · 15th August 2020, 02:21 PM

Originally Posted by SDG

The event A is in the original frame.
The event B is in the original frame.
What is physical space distance between A and B?
The traveler is in the original frame.
The traveler has something on his clock, what is it?

I told you, it's time you start doing the calculations yourself. You tell me what you think the answer is, AND how you got it.

Quote:

Are you saying that the traveler has different measuring rod when he is stationary in the original frame?

Badly phrased question. Different than what?

Quote:

How did I mixed frames when both twins are stationary and they do not have any relative motion?

By thinking that a gamma of 2 applied to the double primed triplet's motion, from the point of view of the single primed triplet when he came to rest relative to the unprimed triplet. That's how you got 1.75 seconds, isn't it? By trying to apply a gamma of 2 to his own 3.5 seconds.

cjameshuff · 15th August 2020, 03:18 PM

I suppose by now someone's probably pointed the Hafele-Keating experiment and its various reproductions out to SDG to no productive result...

Myriad · 15th August 2020, 03:39 PM

Originally Posted by cjameshuff

I suppose by now someone's probably pointed the Hafele-Keating experiment and its various reproductions out to SDG to no productive result...

In this case the more relevant experiment might be the detection of muons created by cosmic ray collisions in the upper atmosphere (~10 km altitude) on the Earth's surface, even though muons decay within 2000 nanoseconds. How do these muons cross a 33,300 light-nanosecond distance without decaying? They're moving at velocities about .9999c, so they "age" about 70 times slower as observed from our frame.

So they cross 33,000 light-nanoseconds (measured from our frame) in about 470 nanoseconds (measured from their frame). Does that mean they travel faster than light? No, because of Ziggurat's point:

Originally Posted by Ziggurat

Velocity is measured by distance over time using coordinate distance and time. If you use distance from one frame and time from another, it's not velocity. No one's velocity exceeds c.

Robin · 15th August 2020, 04:39 PM

Originally Posted by SDG

OK, differently:

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

The triplets go out and they just stop at mid point P.
They do not go back.
What are the distances between them?
When they stop they are in the same inertial frame.

If P is an event then they can't stop at P, each triplet stopping will be a different event.

But say the stop at t=7 in the original frame then there is, (I .doing the calculation in a hurry) 0.866 light seconds distance between them.

So what is the point of the question?

SDG · 15th August 2020, 05:48 PM

Originally Posted by Ziggurat

I told you, it's time you start doing the calculations yourself. You tell me what you think the answer is, AND how you got it.

Badly phrased question. Different than what?

By thinking that a gamma of 2 applied to the double primed triplet's motion, from the point of view of the single primed triplet when he came to rest relative to the unprimed triplet. That's how you got 1.75 seconds, isn't it? By trying to apply a gamma of 2 to his own 3.5 seconds.

You are mixing sub-discussions.
I showed this diagram and I asked this:

I assume it is OK to use X/R calculation, right?
The traveling twin accelerates from A and decelerates to B.
Using X/R we are going to ignore the acceleration and deceleration.
But we know it happened and the traveler stopped at B.
They are again in the same reference frame.
What is the distance between them when the traveler stops at B?

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

Then you followed with this:
Velocity is measured by distance over time using coordinate distance and time. If you use distance from one frame and time from another, it's not velocity. No one's velocity exceeds c.

My response:
The event A is in the original frame.
The event B is in the original frame.
What is physical space distance between A and B?
The traveler is in the original frame.
The traveler has something on his clock, what is it?
Are you saying that the traveler has different measuring rod when he is stationary in the original frame?
How did I mixed frames when both twins are stationary and they do not have any relative motion?

So you recommend to do calculations myself. I did. It is very simple.
This is how acceleration and deceleration work:

Please, ignore the extra stuff around, the important part is the length contraction.
The X/R is not important, correct? The length will be contracted and back to normal.
The stay home and the traveling twin have the same measuring ruler when they start.
You can see they have the same measuring ruler when they finish.
The Lorentz time dilation for the traveling twin says he has 2s on his clock.
The rulers are back to the same length.
The calculation is done.

The question stands: did the traveler crossed 3.4641cs in 2 seconds?

14th August 2020, 06:14 AM	#161
SDG Muse Join Date: Jul 2018 Posts: 895	Originally Posted by Robin Maybe you could start again and state the entire scenario explicitly. And state what the events A, B and C are. Also, your spacetime diagrams are wrong, they imply that the train is travelling at 0.866c relative to the platform and the platform is travelling at 0.866c relative to the train. The t' and x' axes in the train spacetime diagram should form an obtuse angle. You can check it yourself here: http://www.trell.org/div/minkowski.html There is no problem with ct' and x' axes.
SDG Muse Join Date: Jul 2018 Posts: 895

14th August 2020, 06:35 AM	#162
Ziggurat Penultimate Amazing Join Date: Jun 2003 Posts: 47,300	Originally Posted by SDG The 4s on the platform takes 8s' on the train. Any questions? Yes, I have a question. What's with this diagram? What's the dashed line I circled for? Is it supposed to be the return path of the light? One of the difficulties of the way you've set up your problem is that you have two spatial dimensions you're working with, but you can only display one of those dimensions in your space-time diagram. This means you can't plot your y coordinates at the same time as your x and t coordinates, at least not with whatever you're currently using to construct your diagrams. Any light path with a y component will therefore not trace out a 45 degree angle on your graph. Another complication is that you've sort of mixed and matched things. You take a light path in the train frame that travels only in the y' direction and then transform it to the platform where it travels in both y and x. That's a pretty easy transformation. But then you take a light path in the platform frame which travels in both y and x and then transform it into the train frame. This is a more complex transformation. You can do that, there's nothing inherently wrong with it, but it's an added complication, I don't see what it's supposed to demonstrate, and because it's not equivalent starting scenario you're not going to produce an equivalent end scenario. I don't see any errors yet. But you haven't gotten to what you think the contradiction is. So far, you've only shown that 2 second proper time in the train frame has 4 second coordinate time in the platform frame, while 4 second proper time in the platform frame has 8 second coordinate time in the train frame. In other words, Lorenz time dilation. So... where's the contradiction?
	__________________ "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

14th August 2020, 06:46 AM	#163
SDG Muse Join Date: Jul 2018 Posts: 895	Originally Posted by Ziggurat Yes, I have a question. What's with this diagram? http://www.internationalskeptics.com...6907391b5c.png What's the dashed line I circled for? Is it supposed to be the return path of the light? One of the difficulties of the way you've set up your problem is that you have two spatial dimensions you're working with, but you can only display one of those dimensions in your space-time diagram. This means you can't plot your y coordinates at the same time as your x and t coordinates, at least not with whatever you're currently using to construct your diagrams. Any light path with a y component will therefore not trace out a 45 degree angle on your graph. Another complication is that you've sort of mixed and matched things. You take a light path in the train frame that travels only in the y' direction and then transform it to the platform where it travels in both y and x. That's a pretty easy transformation. But then you take a light path in the platform frame which travels in both y and x and then transform it into the train frame. This is a more complex transformation. You can do that, there's nothing inherently wrong with it, but it's an added complication, I don't see what it's supposed to demonstrate, and because it's not equivalent starting scenario you're not going to produce an equivalent end scenario. I don't see any errors yet. But you haven't gotten to what you think the contradiction is. So far, you've only shown that 2 second proper time in the train frame has 4 second coordinate time in the platform frame, while 4 second proper time in the platform frame has 8 second coordinate time in the train frame. In other words, Lorenz time dilation. So... where's the contradiction? Is this: "Any light round trip will appear slower (time dilated) compared to any other moving frame?" a true statement?
SDG Muse Join Date: Jul 2018 Posts: 895

14th August 2020, 07:03 AM	#164
Myriad Hyperthetical Join Date: Nov 2006 Location: Betwixt Posts: 16,921	Originally Posted by SDG The event coordinates are in (x, y, t) and (x',y',t') formats. A: (0, 0, 0) ------- (0', 0', 0') D: (1.732, 1, 2) -- (0', 1', 1') E: (0, 0, 4) ------- (-6.9282', 0', 8') I hope this helps. Apparently event E is the arrival of the light pulse from x = x' = 0 and t = t' = 0, after having been retro-reflected from a particular point on the train at event D (x' = 1, y = y' = 1), back at x = y = 0. That happens at t = 4. Measured from the train frame, that same event happens at t' = 8, and about seven light-seconds behind the front of the train (x' = -6.93). This is not a contradiction of anything.
	__________________ A zømbie once bit my sister... Last edited by Myriad; 14th August 2020 at 07:04 AM.

14th August 2020, 07:04 AM	#165
Ziggurat Penultimate Amazing Join Date: Jun 2003 Posts: 47,300	Originally Posted by SDG Is this: "Any light round trip will appear slower (time dilated) compared to any other moving frame?" a true statement? It's not always true if you include non-inertial frames, and even for inertial frames it may depends on how you want to do the comparison. I'll give you a more precise way of saying it that's always true: For inertial coordinate systems, coordinate time between two events is always longest in the reference frame where the two events occur in the same location. Does this satisfy you? The start and end of your round trip can be the two events, and they are in the same location if it's a "round trip" and not just a trip. A "moving frame" would be a frame in which the start and end events are not in the same location. And the basis of comparison is now explicit: we're comparing coordinate time between the events in our different frames. And finally, note that proper time along a non-accelerating world line is equal to coordinate time in a reference frame where that world line is stationary. Note also that I've specified inertial reference frames. For this thread, it's not worth the effort of trying to work from non-inertial reference frames, and the statement isn't always true if we include non-inertial frames either.
	__________________ "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

14th August 2020, 07:23 AM	#166
Myriad Hyperthetical Join Date: Nov 2006 Location: Betwixt Posts: 16,921	Originally Posted by Ziggurat It's not always true if you include non-inertial frames, and even for inertial frames it may depends on how you want to do the comparison. I'll give you a more precise way of saying it that's always true: For inertial coordinate systems, coordinate time between two events is always longest in the reference frame where the two events occur in the same location. Does this satisfy you? The start and end of your round trip can be the two events, and they are in the same location if it's a "round trip" and not just a trip. A "moving frame" would be a frame in which the start and end events are not in the same location. And the basis of comparison is now explicit: we're comparing coordinate time between the events in our different frames. And finally, note that proper time along a non-accelerating world line is equal to coordinate time in a reference frame where that world line is stationary. Note also that I've specified inertial reference frames. For this thread, it's not worth the effort of trying to work from non-inertial reference frames, and the statement isn't always true if we include non-inertial frames either. But even the simply worded statement holds true in SDG's train example. There are two different round trips described: a round trip in the platform frame, starting and ending at x = y = 0, and a different round trip in the train frame, starting and ending at x' = y' = 0. (The start points of both round trips coincide at a single event, but the end points don't, so they're different round trips as long as no one is trying to confuse things on purpose.) The platform round trip measures 4 seconds in the platform frame, and 8 seconds in the train frame. In other words, longer in the relatively moving frame. The train round trip measures 2 seconds in the train frame, and 4 seconds in the platform frame. In other words, longer in the relatively moving frame.
	__________________ A zømbie once bit my sister...

14th August 2020, 07:23 AM	#167
SDG Muse Join Date: Jul 2018 Posts: 895	Originally Posted by Ziggurat It's not always true if you include non-inertial frames, and even for inertial frames it may depends on how you want to do the comparison. I'll give you a more precise way of saying it that's always true: For inertial coordinate systems, coordinate time between two events is always longest in the reference frame where the two events occur in the same location. Does this satisfy you? The start and end of your round trip can be the two events, and they are in the same location if it's a "round trip" and not just a trip. A "moving frame" would be a frame in which the start and end events are not in the same location. And the basis of comparison is now explicit: we're comparing coordinate time between the events in our different frames. And finally, note that proper time along a non-accelerating world line is equal to coordinate time in a reference frame where that world line is stationary. Note also that I've specified inertial reference frames. For this thread, it's not worth the effort of trying to work from non-inertial reference frames, and the statement isn't always true if we include non-inertial frames either. Agreed, the statement is for inertial frames only. This is what makes the SR reciprocal. Two resolve the twin/triplet paradox we have to go to general/accelerated frames. Those can be broken down to a sequence of comoving inertial frames. Do you agree with that? Edit: Just to clarify, did you mean shortest? For inertial coordinate systems, coordinate time between two events is always shortest (not longest) in the reference frame where the two events occur in the same location. Example: 4s in one location of the platform frame 8s in two different locations of the train frame
SDG Muse Join Date: Jul 2018 Posts: 895	Last edited by SDG; 14th August 2020 at 07:30 AM.

14th August 2020, 11:14 AM	#168
Ziggurat Penultimate Amazing Join Date: Jun 2003 Posts: 47,300	Originally Posted by SDG Two resolve the twin/triplet paradox we have to go to general/accelerated frames. No, we do not. All relevant analyses can be done in non-accelerated frames. Pick multiple inertial reference frames if you want, analyze it from any of them, and the results for the proper time for each traveler will be the same. Quote: Those can be broken down to a sequence of comoving inertial frames. Do you agree with that? No. Accelerated frames in special relativity are not simply a series of comoving frames. It's far, far messier than that. You need to account for the changing lines of simultaneity, and that does... strange things. Including making coordinate time run backwards. And it's completely unnecessary. You CAN analyze this from the viewpoint of a series of inertial reference frame, but it's also unnecessary easier to screw it up than if you just do the whole thing from a single reference frame (meaning analyze all of it from one frame, you can still then pick another frame to analyze it from and repeat). And I suspect that's what you did to make you think that there was any contradiction. Quote: Edit: Just to clarify, did you mean shortest? Yes, sorry. The separation between two (time-like separated) events has the shortest coordinate time when they are in the same position, and increases coordinate time when you change to any other coordinates. Coordinate time between events does not depend on any path between events. Conversely, the proper time between events is path-dependent, and is longest when the path is straight, but it does not change when you change coordinates.
	__________________ "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

14th August 2020, 11:42 AM	#169
SDG Muse Join Date: Jul 2018 Posts: 895	Originally Posted by Ziggurat No, we do not. All relevant analyses can be done in non-accelerated frames. Pick multiple inertial reference frames if you want, analyze it from any of them, and the results for the proper time for each traveler will be the same. No. Accelerated frames in special relativity are not simply a series of comoving frames. It's far, far messier than that. You need to account for the changing lines of simultaneity, and that does... strange things. Including making coordinate time run backwards. ... How do you propose to solve it in SR only? The twins will not meet again. The sudden instantaneous turn around is not realistic. The stopping and coming back cannot happen without acceleration/deceleration. The diagram below is not right. Do you remember A, B, C events? How can a observer move along a world line when there is an observer not moving at all? That's the magic that SR needs? Something similar is happening in the 'jump' below. Magic! I used many figures from this text book: The acceleration/deceleration, the SR in General Frames is way to go. The twin diagram is not from this book. It is a different book.
SDG Muse Join Date: Jul 2018 Posts: 895

14th August 2020, 12:26 PM	#170
Ziggurat Penultimate Amazing Join Date: Jun 2003 Posts: 47,300	Originally Posted by SDG How do you propose to solve it in SR only? The twins will not meet again. This is wrong. Very, very, very wrong. First off, accelerated motion is not the same as an accelerated reference frame. It sounds like this is a point of confusion for you, and if it is, you need to solve it, because nothing will make sense if you are operating under the delusion that they are the same. They are not. They are very, very different. Second, special relativity can handle accelerated. Accelerated motion in special relativity can get a bit messy, but it's perfectly doable, and doesn't produce any bizarre effects that differ fundamentally from non-accelerated motion. Third special relativity can in fact handle accelerated reference frames. That doesn't get a bit messy, that actually gets VERY messy. And it does produce bizarre effects (like event horizons). But these bizarre effects are coordinate effects. They are analogous to the coordinate singularities you can produce in ordinary Euclidean geometry if you choose non-Cartesian coordinates. They are not physical effects. Quote: The sudden instantaneous turn around is not realistic. True. But here's the thing: it doesn't matter. You can complete your acceleration in some finite time, yes? Let's can the time it takes to go from stationary to whatever fraction of c you're going to use R (in the original rest frame). The minimum time for the journey in that frame is thus 4R. You spend the first R accelerating outwards, the second R decelerating back to zero, the third R accelerating back, and the fourth R decelerating. In this scenario, we need to account for the finite acceleration when calculating the proper time for the traveling twin. But you could extend the journey. Spend the first R accelerating outwards. Then spend a bunch of time X at constant velocity outbound. Then turn around, and spend a bunch of time X coming back. And then spend the last R decelerating. So the total time would be 4R+2X. Now suppose X = R. The fact that the acceleration isn't instant still matters, right? Sure. But what if X = 100R? What if X = 1,000,000R? What if X = 10¹⁰⁰ R? At some point, X is sufficiently large compared to R that the total proper time is dominated by what happens during X, and what happens during R becomes an irrelevant perturbation that we don't care about anymore. And that's true no matter what value of R you started with. You can always construct the problem such that you just don't care. And that's why it's OK to simplify the problem by assuming instantaneous acceleration. It doesn't matter that instantaneous acceleration is impossible. You can construct the problem such that it's an arbitrarily small part of the problem, and therefore you can ignore it. Quote: The diagram below is not right. It's simplified (see above), but it's right. Quote: Do you remember A, B, C events? How can a observer move along a world line when there is an observer not moving at all? That question doesn't make sense. All observers move along worldlines. You can't NOT move along a worldline, because you can't stop time. The only sense in which you could stop moving along a world line is to stop existing, but that's not part of our scenario. Quote: That's the magic that SR needs? Something similar is happening in the 'jump' below. Magic! It's not magic. It's a coordinate effect. One of the complications which your diagram doesn't address but may be tripping you up is the distinction between what you see and what you observe. You do not see your line of simultaneity. You can't, it's not physical. That also means that you can't see your line of simultaneity change when you reverse course. But you can observe it. OK, so what is an observation in special relativity? It has a rather specific and non-obvious meaning in special relativity, and it causes a lot of confusion to learners who aren't careful. It's a sort of idealized measurement. It's what you would get if you could do measurements which you can't actually do, but it's also what you get if you do measurements you can do AND you account for how things like the finite speed of light affect your actual measurements. I'll give you a simple example of the distinction. Have you ever heard a jet plane fly overhead, looked up, and noticed that it sounds like the noise is coming from behind the plane? That's what you hear: the sound is coming from a direction other than the current direction to the plane. Someone unaware of the finite speed of sound might incorrectly conclude that the sound didn't come from the plane itself but from behind the plane. But if you know the speed of sound, and you know how far away the plane is and how fast it's going, you can calculate where the plane was when the sound was emitted. You can observe that the plane did in fact emit the sound. OK, so now let's look at what the twins would see, not what they observe. On the outward journey, both twins see the other's clock slowed down. But they don't see it slowed down by the Lorenz time dilation factor. No, it's slowed down even more. Why? Because each tick takes longer and longer to get to the other twin. What they see is a combination of both the Lorenz time dilation AND a Doppler shift (the same thing which makes the sound of an approaching car higher pitched than a receding car). Both twins see the same combined slowdown for the outbound journey. When the twins are approaching each other, the Lorenz factor still applies, but the Doppler shift works in reverse. Instead of slowing it down more, it speed the clock back up. Both twins see the Doppler effect speed up the clocks by the same amount. But the symmetry between them does break, as it must in order to resolve the paradox. Let's look first at the traveling twin. On the outbound part of his journey, he sees the Lorenz slowdown plus a Doppler slowdown (I'll call this LD_r). On the return part of his journey, he sees the Lorenz slowdown plus a Doppler increase (LD_b). He never sees his twin's clock jump in time, it only jumps in rate when he reverses (as the Doppler contribution switches from redshift to blueshift with the turnaround). For him, this change happens at the halfway mark of his journey. He sees the earthbound twin's clock at LD_r for half the journey, and LD_b for half the journey. Now the FIRST symmetry to break here is that the effect from D_b and D_r do not cancel by the time the traveling twin returns. The blueshift of the return journey will produce more additional lapsed time than the redshift of the journey took away. In fact, the blue shift will be such that the total time elapsed on the earthbound twin's clock will EXCEED the total time elapsed on the traveling twin's clock, even though the Lorenz factor was applied the entire time. OK, now on to the SECOND broken symmetry. I said that for the traveling twin, the switch in what he sees from being red shifted to being blue shifted happens at the halfway mark for him. But it DOESN'T happen at the halfway mark for the earthbound twin. He might OBSERVE that it does, but he doesn't SEE that it does. He doesn't see the traveling twin's clock become blue shifted until light from the turnaround point get back to him, and that happens well after the halfway point for him. So the earthbound twin sees the traveling twin's clock red shifted for more than half the time, and blue shifted for less than half the time. And those up, and by the time the traveling twin returns, he will have seen the traveling twin's clock elapsed less time than his own. At no point does either twin see any discontinuity. But they absolutely do not see the same thing. The symmetry is broken. There is no paradox. If this doesn't get through to you, there may be no hope.
	__________________ "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

14th August 2020, 01:50 PM	#171
SDG Muse Join Date: Jul 2018 Posts: 895	Originally Posted by Ziggurat ... True. But here's the thing: it doesn't matter. You can complete your acceleration in some finite time, yes? Let's can the time it takes to go from stationary to whatever fraction of c you're going to use R (in the original rest frame). The minimum time for the journey in that frame is thus 4R. You spend the first R accelerating outwards, the second R decelerating back to zero, the third R accelerating back, and the fourth R decelerating. In this scenario, we need to account for the finite acceleration when calculating the proper time for the traveling twin. But you could extend the journey. Spend the first R accelerating outwards. Then spend a bunch of time X at constant velocity outbound. Then turn around, and spend a bunch of time X coming back. And then spend the last R decelerating. So the total time would be 4R+2X. Now suppose X = R. The fact that the acceleration isn't instant still matters, right? Sure. But what if X = 100R? What if X = 1,000,000R? What if X = 10¹⁰⁰ R? At some point, X is sufficiently large compared to R that the total proper time is dominated by what happens during X, and what happens during R becomes an irrelevant perturbation that we don't care about anymore. And that's true no matter what value of R you started with. You can always construct the problem such that you just don't care. And that's why it's OK to simplify the problem by assuming instantaneous acceleration. It doesn't matter that instantaneous acceleration is impossible. You can construct the problem such that it's an arbitrarily small part of the problem, and therefore you can ignore it. It's simplified (see above), but it's right. ... Thanks, it is very good post and a lot of points to address. I'll start with this. I do understand what you are saying here, really, I do. Therefore I am going to say this: No matter of the X/R ratio, it follows from your equation that there is going to be R at the mid point P! The simultaneity line is horizontal line as seen from the stay home twin/triplet frame. The twin/triplets are in the same frame for a split of a second. Agreed?
SDG Muse Join Date: Jul 2018 Posts: 895

14th August 2020, 01:57 PM	#172
Ziggurat Penultimate Amazing Join Date: Jun 2003 Posts: 47,300	Originally Posted by SDG No matter of the X/R ratio, it follows from your equation that there is going to be R at the mid point P! The simultaneity line is horizontal line as seen from the stay home twin/triplet frame. The twin/triplets are in the same frame for a split of a second. Agreed? Sure.
	__________________ "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

14th August 2020, 03:07 PM	#173
SDG Muse Join Date: Jul 2018 Posts: 895	Originally Posted by Ziggurat Sure. So what times are on the triplets clocks at the mid point when the simultaneity is the horizontal line? Time on the clocks represents the proper time. ct = 7s ct' = ?s' ct'' = ?s''
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14th August 2020, 04:04 PM	#174
Robin Penultimate Amazing Join Date: Apr 2004 Posts: 12,564	Originally Posted by SDG You can check it yourself here: http://www.trell.org/div/minkowski.html https://i.imgur.com/tNj7h4k.png There is no problem with ct' and x' axes. The relative velocity can't be the same in the platform frame as in the train frame. If the relative velocity in the train frame is 0.866 then the relative velocity in the platform frame would be -0.866. Try 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"

14th August 2020, 04:12 PM	#175
Robin Penultimate Amazing Join Date: Apr 2004 Posts: 12,564	Originally Posted by SDG The event coordinates are in (x, y, t) and (x',y',t') formats. A: (0, 0, 0) ------- (0', 0', 0') D: (1.732, 1, 2) -- (0', 1', 1') E: (0, 0, 4) ------- (-6.9282', 0', 8') I hope this helps. I was looking for a complete statement of what is happening. Is the observer on the platform or train? Which direction? etc Something like A: The observer on the platform switches on a light beam aimed at 90 degrees to the direction of the train. (0,0,0) (0',0',0') D: The light beam .... (1.732,2,2) (0',1',1') and so on . Without that I can't get a clear idea of what is supposed to be happening.
	__________________ 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"

14th August 2020, 04:25 PM	#176
SDG Muse Join Date: Jul 2018 Posts: 895	Originally Posted by Robin The relative velocity can't be the same in the platform frame as in the train frame. If the relative velocity in the train frame is 0.866 then the relative velocity in the platform frame would be -0.866. Try that. What is the reason to switch frames? Diagram is from platform frame point of view.
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14th August 2020, 04:43 PM	#177
Robin Penultimate Amazing Join Date: Apr 2004 Posts: 12,564	Originally Posted by SDG What is the reason to switch frames? Diagram is from platform frame point of view. You have two diagrams. Under it you write: "This is a round trip on the train frame". So they are both in the platform frame?
	__________________ 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"

14th August 2020, 05:10 PM	#178
Robin Penultimate Amazing Join Date: Apr 2004 Posts: 12,564	Originally Posted by SDG https://i.imgur.com/QbyqCQl.png So what times are on the triplets clocks at the mid point when the simultaneity is the horizontal line? Time on the clocks represents the proper time. ct = 7s ct' = ?s' ct'' = ?s'' As I said before (and you didn't want to listen) it depends upon the path it takes at the turn around. You are taking a text book simplification and asking what the real life results would be. If the change in velocity is instantaneous then it will depend on whether the paths break up t=[0,7](7,14] or t=[0,7)[7,14]. On the other hand if there is a very short continuous curve then the times for each twin will be approximately the same as the first leg t=[0,7].
	__________________ 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"

14th August 2020, 05:15 PM	#179
Robin Penultimate Amazing Join Date: Apr 2004 Posts: 12,564	Originally Posted by Robin As I said before (and you didn't want to listen) it depends upon the path it takes at the turn around. You are taking a text book simplification and asking what the real life results would be. If the change in velocity is instantaneous then it will depend on whether the paths break up t=[0,7](7,14] or t=[0,7)[7,14]. On the other hand if there is a very short continuous curve then the times for each twin will be approximately the same as the first leg t=[0,7]. Certainly in each case the times on the clocks of each triplet will be different at t=7, just as you would expect. When will you get around to saying what the contradiction is?
	__________________ 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"

14th August 2020, 08:21 PM	#180
Robin Penultimate Amazing Join Date: Apr 2004 Posts: 12,564	For the light round-trip, I assume that the situation is something like this, from the platform frame: (Gridlines in 0.5s)
	__________________ 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; 14th August 2020 at 08:36 PM.

14th August 2020, 08:28 PM	#181
Ziggurat Penultimate Amazing Join Date: Jun 2003 Posts: 47,300	Originally Posted by SDG https://i.imgur.com/QbyqCQl.png So what times are on the triplets clocks at the mid point when the simultaneity is the horizontal line? Time on the clocks represents the proper time. ct = 7s ct' = ?s' ct'' = ?s'' t' = 3s t" = 1s So where's your contradiction? Not that it really matters, so far it doesn't seem to have confused anyone, but if you are looking at ct instead of c, then the answer would be in meters, not seconds.
	__________________ "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

14th August 2020, 08:35 PM	#182
SDG Muse Join Date: Jul 2018 Posts: 895	Originally Posted by Robin Certainly in each case the times on the clocks of each triplet will be different at t=7, just as you would expect. When will you get around to saying what the contradiction is? It was already discussed in this thread. If ct=7s and ct'=3.5s' then there are two options for ct''. The first 1s'' and the second 1.75s''. The problem is the clock can show only one.
SDG Muse Join Date: Jul 2018 Posts: 895	Last edited by SDG; 14th August 2020 at 08:44 PM.

14th August 2020, 08:42 PM	#183
SDG Muse Join Date: Jul 2018 Posts: 895	Originally Posted by Ziggurat t' = 3s t" = 1s So where's your contradiction? Not that it really matters, so far it doesn't seem to have confused anyone, but if you are looking at ct instead of c, then the answer would be in meters, not seconds. How did you do the calculation? Right about the meters.
SDG Muse Join Date: Jul 2018 Posts: 895	Last edited by SDG; 14th August 2020 at 08:49 PM.

14th August 2020, 09:37 PM	#184
Ziggurat Penultimate Amazing Join Date: Jun 2003 Posts: 47,300	Originally Posted by SDG It was already discussed in this thread. If ct=7s and ct'=3.5s' then there are two options for ct''. The first 1s'' and the second 1.75s''. The problem is the clock can show only one. No. 1.75 seconds is wrong. You might incorrectly conclude that the answer is 1.75, because in the reference frame where the outbound single prime triplet is stationary, the double primed triplet has a gamma of 2. This might lead you to conclude that the double primed twin should reach the turnaround point at half the time that the single primed twin reaches the turnaround point. This is wrong. Wrongety-wrong-wrong. It's wrong because in this reference frame where the double primed triplet has a gamma of 2, the double primed twin turns around BEFORE the single primed twin. It's only simultaneous in the frame where the unprimed twin is stationary. The fact that the single primed twin changes to this frame momentarily doesn't matter, because in this frame, the double primed twin's outbound journey didn't have a gamma of 2. You can't take properties from one frame and apply them in another frame without accounting for that change in frames. But that's the only way to conclude that the answer should be 1.75 seconds. It isn't. It's 1 second. It's 1 second as measured in ANY inertial reference frame.
	__________________ "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

14th August 2020, 09:38 PM	#185
Robin Penultimate Amazing Join Date: Apr 2004 Posts: 12,564	Originally Posted by SDG Is this: "Any light round trip will appear slower (time dilated) compared to any other moving frame?" a true statement? If you mean "will appear to take longer" then it is true for inertial frames. It holds true in your example. So where is the problem?
	__________________ 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"

14th August 2020, 09:39 PM	#186
Ziggurat Penultimate Amazing Join Date: Jun 2003 Posts: 47,300	Originally Posted by SDG How did you do the calculation? Right about the meters. There are lots of ways. You could use the time dilation equation. You could transform your coordinates to primed coordinates using the full Lorentz transformations and just read off the time. You could use the metric I gave you earlier. These will all produce the same answer, if done correctly. If you do it wrong, all bets are off. 1.75 seconds is doing it wrong.
	__________________ "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

14th August 2020, 09:40 PM	#187
Robin Penultimate Amazing Join Date: Apr 2004 Posts: 12,564	Originally Posted by SDG It was already discussed in this thread. If ct=7s and ct'=3.5s' then there are two options for ct''. The first 1s'' and the second 1.75s''. The problem is the clock can show only one. There is only one option for ct'' for any given path.
	__________________ 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"

14th August 2020, 09:44 PM	#188
Robin Penultimate Amazing Join Date: Apr 2004 Posts: 12,564	Originally Posted by SDG It was already discussed in this thread. If ct=7s and ct'=3.5s' then there are two options for ct''. The first 1s'' and the second 1.75s''. The problem is the clock can show only one. I already pointed out the reason for the two values, ie: Originally Posted by robin If the change in velocity is instantaneous then it will depend on whether the paths break up t=[0,7](7,14] or t=[0,7)[7,14]. On the other hand if there is a very short continuous curve then the times for each twin will be approximately the same as the first leg t=[0,7]." If the path is continuous then the value for ct'' is close to 1 s
	__________________ 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"

15th August 2020, 11:36 AM	#189
SDG Muse Join Date: Jul 2018 Posts: 895	You said this. Originally Posted by Ziggurat Originally Posted by SDG ... No matter of the X/R ratio, it follows from your equation that there is going to be R at the mid point P! The simultaneity line is horizontal line as seen from the stay home twin/triplet frame. The twin/triplets are in the same frame for a split of a second. Agreed? Sure. My understanding is that you agreed that all three triplets are in the same original stay at home frame at the turning point P. All three triplets have the same 4-velocity vector in direction and magnitude at the turning point P. So this: Originally Posted by Ziggurat No. 1.75 seconds is wrong. You might incorrectly conclude that the answer is 1.75, because in the reference frame where the outbound single prime triplet is stationary, the double primed triplet has a gamma of 2. This might lead you to conclude that the double primed twin should reach the turnaround point at half the time that the single primed twin reaches the turnaround point. This is wrong. Wrongety-wrong-wrong. It's wrong because in this reference frame where the double primed triplet has a gamma of 2, the double primed twin turns around BEFORE the single primed twin. It's only simultaneous in the frame where the unprimed twin is stationary. The fact that the single primed twin changes to this frame momentarily doesn't matter, because in this frame, the double primed twin's outbound journey didn't have a gamma of 2. You can't take properties from one frame and apply them in another frame without accounting for that change in frames. But that's the only way to conclude that the answer should be 1.75 seconds. It isn't. It's 1 second. It's 1 second as measured in ANY inertial reference frame. This appears to me as a contradiction. Please, can you clear this one out?
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15th August 2020, 11:49 AM	#190
SDG Muse Join Date: Jul 2018 Posts: 895	Originally Posted by Robin There is only one option for ct'' for any given path. OK, differently: The triplets go out and they just stop at mid point P. They do not go back. What are the distances between them? When they stop they are in the same inertial frame.
SDG Muse Join Date: Jul 2018 Posts: 895	Last edited by SDG; 15th August 2020 at 11:59 AM.

15th August 2020, 11:56 AM	#191
Ziggurat Penultimate Amazing Join Date: Jun 2003 Posts: 47,300	Originally Posted by SDG This appears to me as a contradiction. Please, can you clear this one out? It's not a contradiction. Look, it's not actually that complicated. Pick an inertial frame, any inertial frame. Do the calculations in that frame. Show me YOUR calculations.
	__________________ "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, 12:18 PM	#192
SDG Muse Join Date: Jul 2018 Posts: 895	Even simpler scenario. I assume it is OK to use X/R calculation, right? The traveling twin accelerates from A and decelerates to B. Using X/R we are going to ignore the acceleration and deceleration. But we know it happened and the traveler stopped at B. They are again in the same reference frame. What is the distance between them when the traveler stops at B? If gamma = 2 did the traveler crossed 3.4641cs in 2 seconds?
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15th August 2020, 01:02 PM	#193
Ziggurat Penultimate Amazing Join Date: Jun 2003 Posts: 47,300	Originally Posted by SDG If gamma = 2 did the traveler crossed 3.4641cs in 2 seconds? Velocity is measured by distance over time using coordinate distance and time. If you use distance from one frame and time from another, it's not velocity. No one's velocity exceeds c.
	__________________ "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, 02:00 PM	#194
caveman1917 Philosopher Join Date: Feb 2015 Posts: 7,591	Originally Posted by Ziggurat Third special relativity can in fact handle accelerated reference frames. That doesn't get a bit messy, that actually gets VERY messy. And it does produce bizarre effects (like event horizons). But these bizarre effects are coordinate effects. They are analogous to the coordinate singularities you can produce in ordinary Euclidean geometry if you choose non-Cartesian coordinates. They are not physical effects. Just because something is a coordinate effect doesn't mean it isn't physical. For example Unruh radiation will be physically detectable just like any other radiation, independent of whether it's a coordinate effect.
caveman1917 Philosopher Join Date: Feb 2015 Posts: 7,591	__________________ "Ideas are also weapons." - Subcomandante Marcos "We must devastate the avenues where the wealthy live." - Lucy Parsons "Let us therefore trust the eternal Spirit which destroys and annihilates only because it is the unfathomable and eternal source of all life. The passion for destruction is a creative passion, too!" - Mikhail Bakunin Last edited by caveman1917; 15th August 2020 at 02:01 PM.

15th August 2020, 02:16 PM	#195
SDG Muse Join Date: Jul 2018 Posts: 895	Originally Posted by Ziggurat Velocity is measured by distance over time using coordinate distance and time. If you use distance from one frame and time from another, it's not velocity. No one's velocity exceeds c. The event A is in the original frame. The event B is in the original frame. What is physical space distance between A and B? The traveler is in the original frame. The traveler has something on his clock, what is it? Are you saying that the traveler has different measuring rod when he is stationary in the original frame? How did I mixed frames when both twins are stationary and they do not have any relative motion?
SDG Muse Join Date: Jul 2018 Posts: 895