Is the frame dragging part of the Special Relativity?

[SDG]

Hi all,
is the frame dragging embedded in the Special Relativity?




The platform observer at x=0cs, t=0s sends a light across, the blue arrow up, 90 degree up at the top part of the figure.
The train observer at x'=0cs', t'=0s' sends a light across, the red arrow up, 90 degree up at the top part of the figure.
This happens when they are aligned, x=x'=0 and t=t'=0.
Now frame dragging happens.
The red arrow up is frame dragged across the platform frame.
The red arrow up appears straight up for the train observer but as the blue arrow to the right from the platform x=0cs origin position.
The blue arrow up is frame dragged across the train frame in the reciprocal way.
The blue arrow up appears straight up for the platform observer but as the red arrow to the left from the train x'=0cs' origin position.

Your thoughts,
SDG

[phunk]

Frame dragging is predicted by GR not SR and isn't what you just described.

[SDG]

Would you have a better name for what I described?
SDG

[theprestige]

"special relativity"?

[HansMustermann]

Yeah, it's not frame dragging. It's two different frames of reference, moving relative to each other. It's the bog standard premise and whole point of SR.

[HansMustermann]

But to answer what it is, basically in Newtonian mechanics, gravity only depends on the distance between two bodies. If you're talking about how a moon will move around a planet, only their masses and distance matter. How fast the planet spins, or whether it even spins at all, it plays no role whatsoever.

In GR, it turns out that it matters. It's pretty much lost in the decimals, but it matters. Basically if the Earth were to spin ten times faster, the moon's orbit would be a little different. Again, infinitesimally, but still...

And basically that's all there is to it. It matters if something creating a gravity well spins or not. That's what we call frame dragging.

[theprestige]

What I think is neat is that the Sun-Mercury system is massy enough, and energetic enough, that we could actually find the frame-dragging phenomenon "in the decimals", as far back as 1859. It took GR to explain this measurable deviation from the Newtonian prediction.

[Ziggurat]

Originally Posted by theprestige

What I think is neat is that the Sun-Mercury system is massy enough, and energetic enough, that we could actually find the frame-dragging phenomenon "in the decimals", as far back as 1859. It took GR to explain this measurable deviation from the Newtonian prediction.

Nope. The perihelion precession of Mercury's orbit is indeed observably affected by general relativity, but this is mostly not frame dragging. The frame dragging component is 0.002 arcseconds per century, whereas the non-rotating GR contribution (the component which would exist if the sun didn't rotate at all) is about 43 arcseconds per century (source). The overall GR contribution was observable back in 1859, but the frame dragging component was not. Even today the error margin on Mercury's perihelion precession is much larger (0.65 arcsec/century) than the frame dragging contribution.

Frame dragging is very difficult to observe, and wasn't experimentally confirmed until Gravity Probe B.

[SDG]

Originally Posted by HansMustermann

Yeah, it's not frame dragging. It's two different frames of reference, moving relative to each other. It's the bog standard premise and whole point of SR.

What are the consequences of this premise?
There is no hierarchy of frames in the SR?
Any light round trip will appear slower (time dilated) compared to any other moving frame?
The twin paradox is not resolved?
What is the SR good for?
Jano

[theprestige]

Originally Posted by Ziggurat

Nope. The perihelion precession of Mercury's orbit is indeed observably affected by general relativity, but this is mostly not frame dragging. The frame dragging component is 0.002 arcseconds per century, whereas the non-rotating GR contribution (the component which would exist if the sun didn't rotate at all) is about 43 arcseconds per century (source). The overall GR contribution was observable back in 1859, but the frame dragging component was not. Even today the error margin on Mercury's perihelion precession is much larger (0.65 arcsec/century) than the frame dragging contribution.

Frame dragging is very difficult to observe, and wasn't experimentally confirmed until Gravity Probe B.

Thank you for the correction! I will adjust my celebrations to match the actual discoveries.

[Ziggurat]

Originally Posted by SDG

What are the consequences of this premise?

Which premise? You have described a scenario, not a premise. "The speed of light is constant in all inertial frames" is a premise.

Quote:

There is no hierarchy of frames in the SR?

No hierarchy of inertial frames. Non-inertial frames... things get messy. It's possible to do, but the math can get ugly, fast.

Quote:

Any light round trip will appear slower (time dilated) compared to any other moving frame?

Yes.

Quote:

The twin paradox is not resolved?

It's resolved.

Quote:

What is the SR good for?
Jano

SR is good for relativistic motion (ie, a significant fraction of the speed of light) in the weak gravitational field limit. You only need general relativity if you want to handle strong gravity (black holes being the obvious extreme), although "strong" here is context dependent. The sun's gravity is sufficiently strong for GR effects to show up in accurate measurements of the precession of Mercury's perihelion, but (so far) not in Jupiter's.

[SDG]

Originally Posted by Ziggurat

Which premise? You have described a scenario, not a premise. "The speed of light is constant in all inertial frames" is a premise.

The scenario is based on that premise and it is easy to show that it is the case.
Having said that, what the train observer considers as 'c' the platform observer sees 'c/2'.
This is reciprocal when we switch frames.

Quote:

No hierarchy of inertial frames. Non-inertial frames... things get messy. It's possible to do, but the math can get ugly, fast.

We can leave the acceleration for later.

Quote:

Quote:

Any light round trip will appear slower (time dilated) compared to any other moving frame?

Yes.

Quote:

The twin paradox is not resolved?

It's resolved.

...

There is a contradiction between 'Yes' answer and 'It's resolved'.
If any light round trip is time dilated then the twin paradox is not resolved.
Each twin has own light round trips as own proper time clocks and they are time dilated from the other moving frame.
That's the contradiction in your replies.
Jano

[Myriad]

Originally Posted by SDG

We can leave the acceleration for later.

No we can't. If you're claiming a twin paradox, then the formerly synchronized clocks that get separated at relativistic speeds have to come back together. That can't happen without one or both of them accelerating.

[Ziggurat]

Originally Posted by SDG

There is a contradiction between 'Yes' answer and 'It's resolved'.
If any light round trip is time dilated then the twin paradox is not resolved.
Each twin has own light round trips as own proper time clocks and they are time dilated from the other moving frame.
That's the contradiction in your replies.
Jano

The traveling twin is not in an inertial reference frame. That breaks the symmetry. There is no actual paradox.

[SDG]

Originally Posted by Ziggurat

The traveling twin is not in an inertial reference frame. That breaks the symmetry. There is no actual paradox.

Based on comoving inertial frames we can extrapolate:



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

[HansMustermann]

Originally Posted by SDG

What are the consequences of this premise?
There is no hierarchy of frames in the SR?
Any light round trip will appear slower (time dilated) compared to any other moving frame?
The twin paradox is not resolved?
What is the SR good for?
Jano

If we're talking SR, we're talking inertial frames, i.e., they don't accelerate at all in respect to each other. And yes, the whole idea of SR is that there is no preferred frame. Any frame behaves just like any other frame, and is just as good as any other frame.

The twin paradox though is actually fairly easy to solve, even while sticking strictly to SR. The key is simply that there are not two frames of reference, but THREE. One for the guy sitting at home, one for the ship as it's flying away, and one for the ship as it's coming back. The solution is basically just to notice that one of the guys is switching reference frames.

As for what it's good for, well, just like any other physics it's good when you need to calculate anything where SR applies, really.

[HansMustermann]

Originally Posted by SDG

The scenario is based on that premise and it is easy to show that it is the case.
Having said that, what the train observer considers as 'c' the platform observer sees 'c/2'.
This is reciprocal when we switch frames.

No. Both see c as exactly c. Speeds don't add or subtract linearly like that in SR.

[SDG]

Originally Posted by HansMustermann

If we're talking SR, we're talking inertial frames, i.e., they don't accelerate at all in respect to each other. And yes, the whole idea of SR is that there is no preferred frame. Any frame behaves just like any other frame, and is just as good as any other frame.

The twin paradox though is actually fairly easy to solve, even while sticking strictly to SR. The key is simply that there are not two frames of reference, but THREE. One for the guy sitting at home, one for the ship as it's flying away, and one for the ship as it's coming back. The solution is basically just to notice that one of the guys is switching reference frames.

As for what it's good for, well, just like any other physics it's good when you need to calculate anything where SR applies, really.

The twins do not meet again. This is not a solution.
SDG

[SDG]

Originally Posted by HansMustermann

No. Both see c as exactly c. Speeds don't add or subtract linearly like that in SR.

You are mistaken.
The platform observers sees: c^2 = (c/2)^2 + v^2
If we deny the frame dragging then we have to take away v^2.
The result is c!=c.
Is this what we want?
SDG

[HansMustermann]

To understand why both see it as c, you need to also understand that clocks are not synchronized in the two reference frames. When you calculate how fast something moves, basically you measure when it passes by some kind of points, then basically do a v=(x₂-x₁)/(t₂-t₁). And in the other frame of reference that would be v'=(x'₂-x'₁)/(t'₂-t'₁).

When you do the Lorentz transformations for BOTH space and time, what one sees as c, the other also sees as exactly c. Not c/2, not anything else than exactly c.


But basically here's an illustration I've used before about how not only intervals are different in different frames, but even two events being synchronous doesn't carry over to other frames.

Let's say we have a 10m long hangar which is basically a straight tunnel with force fields at both ends that can be opened or closed. They're currently open. Let's say you're sitting at the controls of those doors, while I'm flying a 20m long spaceship at 0.866c through that hangar.

Well, from your point of view, my ship is only 10m long, because I chose the speed so the transformation works that way.

So there is one moment when, as seen from your frame, my ship fits snugly inside your hangar. I mean, both are 10m long in your frame of reference. So for an instant you could close both doors simultaneously, with my 20m ship fitting perfectly inside your 10m hangar.

Well, now from my perspective as I'm sitting in the captain's chair on that ship, my ship is still 20m long, and it's your hangar that is half the length. So it's only 5m long. There's no way in hell my 20m ship will ever fit neatly inside your 5m hangar, as seen from my frame.

So what happens from my point of view when you close the doors?

Well, from my point of view the two force fields don't close simultaneously. First the one in front closes for an instant in front of the tip of my ship, then it opens, my ship keeps passing through, and then the one behind closes right behind my ship for an instant.

Basically, when they said that worse things happen at sea... yeah, even worse happen in relativity

But yeah, that's the kind of effects you have to contend with, in order to understand why what looks like c to one of them, also looks like c to the other.

[HansMustermann]

Originally Posted by SDG

You are mistaken.
The platform observers sees: c^2 = (c/2)^2 + v^2
If we deny the frame dragging then we have to take away v^2.

To use a cliched quote, "You keep using that word. I do not think it means what you think it means."

As a couple of us have been trying to tell you, the Lorentz transformations are NOT the same as frame dragging. They don't have ANYTHING to do with what is actually called frame dragging. Just because one frame moves in relation to the other (by being dragged or otherwise), doesn't mean you can call it frame dragging. Jargon terms can mean something very different than just the sum of the words.

So anyway, it's not frame dragging unless you're calculating gravity around a massive rotating body (and I mean, as in, you'd have to be pretty damn close to a rotating black hole or at least neutron star to even need to apply that correction,) but the normal Lorentz transformations still apply anyway.

Originally Posted by SDG

The result is c!=c.

Well, if you get the obviously wrong result, that should be your clue that you just disproved your premises via an ad absurdum disproof.

Originally Posted by SDG

Is this what we want?
SDG

If you mean if we want a wrong calculation that is based on not even starting to understand SR, yeah, that's not what we want

[Puppycow]

Originally Posted by theprestige

What I think is neat is that the Sun-Mercury system is massy enough, and energetic enough, that we could actually find the frame-dragging phenomenon "in the decimals", as far back as 1859. It took GR to explain this measurable deviation from the Newtonian prediction.

Maybe you already know this, but it's a good for anyone.

The Planet That Wasn't By Isaac Asimov

Reading this taught me that the planet Vulcan wasn't originally something invented by the writers of Star Trek. It was something that astronomers hypothesized to explain why Mercury's orbit was so weird and didn't conform to the Newtonian prediction. So they spent a lot of time looking for this planet they called Vulcan that didn't actually exist.

[SDG]

Originally Posted by HansMustermann

To use a cliched quote, "You keep using that word. I do not think it means what you think it means."

As a couple of us have been trying to tell you, the Lorentz transformations are NOT the same as frame dragging. They don't have ANYTHING to do with what is actually called frame dragging. Just because one frame moves in relation to the other (by being dragged or otherwise), doesn't mean you can call it frame dragging. Jargon terms can mean something very different than just the sum of the words.

So anyway, it's not frame dragging unless you're calculating gravity around a massive rotating body (and I mean, as in, you'd have to be pretty damn close to a rotating black hole or at least neutron star to even need to apply that correction,) but the normal Lorentz transformations still apply anyway.



Well, if you get the obviously wrong result, that should be your clue that you just disproved your premises via an ad absurdum disproof.



If you mean if we want a wrong calculation that is based on not even starting to understand SR, yeah, that's not what we want

OK, it is not frame dragging.
Can we call it - photons comoving with the moving frame?


c^2 = (c/2)^2 + v^2

This equation is still valid.
Is SR built on classical velocity addition?
SDG

[SDG]

Originally Posted by HansMustermann

To understand why both see it as c, you need to also understand that clocks are not synchronized in the two reference frames. When you calculate how fast something moves, basically you measure when it passes by some kind of points, then basically do a v=(x₂-x₁)/(t₂-t₁). And in the other frame of reference that would be v'=(x'₂-x'₁)/(t'₂-t'₁).

When you do the Lorentz transformations for BOTH space and time, what one sees as c, the other also sees as exactly c. Not c/2, not anything else than exactly c.


But basically here's an illustration I've used before about how not only intervals are different in different frames, but even two events being synchronous doesn't carry over to other frames.

Let's say we have a 10m long hangar which is basically a straight tunnel with force fields at both ends that can be opened or closed. They're currently open. Let's say you're sitting at the controls of those doors, while I'm flying a 20m long spaceship at 0.866c through that hangar.

Well, from your point of view, my ship is only 10m long, because I chose the speed so the transformation works that way.

So there is one moment when, as seen from your frame, my ship fits snugly inside your hangar. I mean, both are 10m long in your frame of reference. So for an instant you could close both doors simultaneously, with my 20m ship fitting perfectly inside your 10m hangar.

Well, now from my perspective as I'm sitting in the captain's chair on that ship, my ship is still 20m long, and it's your hangar that is half the length. So it's only 5m long. There's no way in hell my 20m ship will ever fit neatly inside your 5m hangar, as seen from my frame.

So what happens from my point of view when you close the doors?

Well, from my point of view the two force fields don't close simultaneously. First the one in front closes for an instant in front of the tip of my ship, then it opens, my ship keeps passing through, and then the one behind closes right behind my ship for an instant.

Basically, when they said that worse things happen at sea... yeah, even worse happen in relativity

But yeah, that's the kind of effects you have to contend with, in order to understand why what looks like c to one of them, also looks like c to the other.

There is no length contraction along the y axis.
v_y=c/2 is correct for the light crossing the y axis in the platform frame.
If you do not understand this then you do not understand the SR.
Please, have a careful look:

[Roger Ramjets]

Originally Posted by SDG

The blue arrow up appears straight up for the platform observer but as the red arrow to the left from the train x'=0cs' origin position.

OK, so let's say we angle the beam to go 'straight up' (laterally) from the platform observer's point of view - what direction does he need to look in to see it?

[SDG]

Originally Posted by Roger Ramjets

OK, so let's say we angle the beam to go 'straight up' (laterally) from the platform observer's point of view - what direction does he need to look in to see it?

The blue arrow up from the platform observer is 90 degrees to the velocity vector of the train.
The train moves along x axis and the up direction is in y axis.
The x, y are platform axes.
The red arrow up moves along y' axis that is 90 degree to x' axis.
The blue arrow up in platform frame is observed as the red arrow to the left in the train frame.

[Myriad]

Originally Posted by SDG

There is no length contraction along the y axis.
v_y=c/2 is correct for the light crossing the y axis in the platform frame.
If you do not understand this then you do not understand the SR.
Please, have a careful look:

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

An observer who measures the y component of the light's velocity as c/2 will also observe the light moving at an angle instead of along the y axis, and will still measure the light's overall velocity as c.

[smartcooky]

Originally Posted by Myriad

No we can't. If you're claiming a twin paradox, then the formerly synchronized clocks that get separated at relativistic speeds have to come back together. That can't happen without one or both of them accelerating.

Wait! We have two clocks;

Clock A: at rest
Clock B: moving at, say, 20%C

As Clock A passes Clock B, we observe that both clocks are exactly sycnhronized. Clock A then proceeds on at 20%C and completes a large circle, traveling at 20%C all the way, never accelerating or decelerating. After a period of time, it passes Clock B again.

Are the clocks still synchronized?

[SDG]

Originally Posted by Myriad

An observer who measures the y component of the light's velocity as c/2 will also observe the light moving at an angle instead of along the y axis, and will still measure the light's overall velocity as c.

Agreed, but the photons are comoving with the train frame.
Yes, moving at an angle, this is aberration of the light.
To make things easier, I will not call it frame dragging but comoving.
The effect is obvious the final light speed is c, the y component is c/2 and the x component is v, based on the comoving frame velocity.
c^2 = (c/2)^2 + v^2

[SDG]

Originally Posted by smartcooky

Wait! We have two clocks;

Clock A: at rest
Clock B: moving at, say, 20%C

As Clock A passes Clock B, we observe that both clocks are exactly sycnhronized. Clock A then proceeds on at 20%C and completes a large circle, traveling at 20%C all the way, never accelerating or decelerating. After a period of time, it passes Clock B again.

Are the clocks still synchronized?

No, they will not be synchronized.
The large circle is a constant acceleration for the the clock A.

[cjameshuff]

Originally Posted by smartcooky

Wait! We have two clocks;

Clock A: at rest
Clock B: moving at, say, 20%C

As Clock A passes Clock B, we observe that both clocks are exactly sycnhronized. Clock A then proceeds on at 20%C and completes a large circle, traveling at 20%C all the way, never accelerating or decelerating. After a period of time, it passes Clock B again.

Are the clocks still synchronized?

How do you propose to travel in a circle without accelerating?

[Roger Ramjets]

Originally Posted by SDG

Agreed, but the photons are comoving with the train frame.
Yes, moving at an angle, this is aberration of the light.
To make things easier, I will not call it frame dragging but comoving.

So not frame dragging then. Does this mean answer to your question is 'no'?

[smartcooky]

Originally Posted by SDG

No, they will not be synchronized.
The large circle is a constant acceleration for the the clock A.

Originally Posted by cjameshuff

How do you propose to travel in a circle without accelerating?

Ah, yes... of course

What was I thinking!?

[HansMustermann]

Originally Posted by SDG

OK, it is not frame dragging.
Can we call it - photons comoving with the moving frame?

What meaning does "comoving" even convey there?

What you actually have, as the very minimum to measure a speed, is that in a frame of reference something passed through point {x₁, y₁, z₁} at moment t₁, and then through point {x₂, y₂, z₂} at moment t₂. You take the distance, divide by t₂-t₁, get the speed.

Meanwhile in another frame, let's call it prime, it passed through point {x'₁, y'₁, z'₁} at moment t'₁, and then through point {x'₂, y'₂, z'₂} at moment t'₂. You take the distance, divide by t'₂-t'₁, get the speed.

Whether you're talking SR or plain old Newton, the body whose motion you're measuring is simply passing through those points at those times. Thinking about it as "comoving" with anything isn't necessary or bringing you any bit of extra information at all. Because it doesn't matter if the thing you're measuring is walking inside the train, or on the ground, or in a second train. It changes nothing. It still passes through those coordinates at that time. You still just need to transform the coordinates and times, do the division, and that's it.

Originally Posted by SDG

c^2 = (c/2)^2 + v^2

This equation is still valid.

As long as you keep it inside one frame, and it's still SR, sure, Pythagoras still applies when you project any vector into orthogonal components.

Not sure what it's supposed to illustrate, though. The reason both still measure exactly c for that scenario, even though one sees the light travelling diagonally over a larger distance, is that their clocks aren't synchronized between the two frames. The guy in the train sees the light moving only half the distance that the guy on the ground sees, but also in only half the time that the guy on the ground measures. The guy on the ground sees the light moving diagonally twice the distance that the guy on the train sees, but in twice the time that the guy in the train measured.

So while it's mathematically correct, I'm not seeing anything particularly enlightening there. I'm not seeing what information is calling it "comoving" supposed to convey.

Originally Posted by SDG

Is SR built on classical velocity addition?
SDG

Not really, no. In fact, that's kinda the whole point.

[HansMustermann]

As for the 'twins paradox', instead of thinking it as one ship that goes from 0 to a point at distance r, and then back, which involves acceleration somewhere in between, you can think of it as two ships which move at constant speed and never accelerate. One starts from the origin and keeps going in the positive direction of the x axis at constant speed. One starts from 2r and goes towards the origin. When they pass each other by at point r, they synchronize their clocks. When the second one passes the origin by, it compares its clock to the one of the guy who never moved from the origin.

That version is perfectly compatible with SR, since what we have is three perfectly inertial frames.

And then it turns out it's no paradox at all.

[SDG]

Originally Posted by HansMustermann

...
Not sure what it's supposed to illustrate, though. The reason both still measure exactly c for that scenario, even though one sees the light travelling diagonally over a larger distance, is that their clocks aren't synchronized between the two frames. The guy in the train sees the light moving only half the distance that the guy on the ground sees, but also in only half the time that the guy on the ground measures. The guy on the ground sees the light moving diagonally twice the distance that the guy on the train sees, but in twice the time that the guy in the train measured.

So while it's mathematically correct, I'm not seeing anything particularly enlightening there. I'm not seeing what information is calling it "comoving" supposed to convey.
...

The enlightening part is that the time dilation determines the v_y.
This example is generic, it applies to any two moving inertial frames.
The illustration is the blue arrow up traveling 1s in the platform frame and it takes 2s' in the train frame.
This is reciprocal to the first solution, this is the paradox and this is the foundation for the twin paradox.
The SR is reciprocal, paradoxical and it is a problem.

[SDG]

Originally Posted by HansMustermann

As for the 'twins paradox', instead of thinking it as one ship that goes from 0 to a point at distance r, and then back, which involves acceleration somewhere in between, you can think of it as two ships which move at constant speed and never accelerate. One starts from the origin and keeps going in the positive direction of the x axis at constant speed. One starts from 2r and goes towards the origin. When they pass each other by at point r, they synchronize their clocks. When the second one passes the origin by, it compares its clock to the one of the guy who never moved from the origin.

That version is perfectly compatible with SR, since what we have is three perfectly inertial frames.

And then it turns out it's no paradox at all.

No it is not because the twins do not meet again.






The reversal of the frames at r that you described is generating time discontinuation at the stay at home twin.
It is seen as the jump from A to B.
That's not realistic. The time does not jump like that. Is this some kind of magic?

[Myriad]

Originally Posted by SDG

Agreed, but the photons are comoving with the train frame.
Yes, moving at an angle, this is aberration of the light.
To make things easier, I will not call it frame dragging but comoving.
The effect is obvious the final light speed is c, the y component is c/2 and the x component is v, based on the comoving frame velocity.
c^2 = (c/2)^2 + v^2

The equation c^2 = (c/2)^2 + v^2 is valid if and only if v = .866c (or v = -.866c). That is to say, it can be solved for v, and doing so merely recapitulates your initially assumed conditions.

What is the "aberration?" Are you talking about the direction of travel of the light beams (assuming they're narrow focused beams, not just omnidirectional light flashes)?

If the train observer aligns his beam emitter (stationary in his reference frame) perpendicular to the train in his reference frame, and the platform observer aligns his beam emitter (stationary in his reference frame) perpendicular to the platform in his reference frame, both observers will observe the beams diverging once they're emitted. If there are detectors arranged along the center line of the train and stationary with respect to the train, the beam from the train's emitter will be detected at a detector positioned perpendicular to the train's emitter, while the beam from the platform's emitter will be detected at a detector farther to the rear of the train.

Both observers will agree which detector detects which beam. Both observers will agree that the beams diverge. What they will disagree on is how the emitters are aimed. The platform observer won't agree that the train's emitter is perpendicular to the train (it'll seem to be aimed at a "forward" angle). The train observer won't agree that the platform's emitter is perpendicular to the platform (it'll seem to be aimed at a "backward" angle). That's because to be aimed in a direction other than parallel to the x axis, the emitters have to have a length along the y axis. The Lorentz transformation thus induces a rotation of the direction of the emitters.

(But... but... but... the Lorentz transformation doesn't affect the y coordinate! It says right there, y' = y! But since the emitter has length, light takes time to travel from the back of the emitter to the front, and in that time, as observed from a relatively moving frame, the x position of the emitter changes and is affected by the Lorentz transformation. That's observed as a change in the "aim" direction. Such complications are why typical elementary SR thought experiments usually involve long thin moving components like rockets and trains, and flashes of light rather than aimed beams.)

[Myriad]

Originally Posted by SDG

The reversal of the frames at r that you described is generating time discontinuation at the stay at home twin.

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.

Quote:

It is seen as the jump from A to B.
That's not realistic. The time does not jump like that. Is this some kind of magic?

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.)

[SDG]

Originally Posted by Myriad

The equation c^2 = (c/2)^2 + v^2 is valid if and only if v = .866c (or v = -.866c). That is to say, it can be solved for v, and doing so merely recapitulates your initially assumed conditions.

What is the "aberration?" Are you talking about the direction of travel of the light beams (assuming they're narrow focused beams, not just omnidirectional light flashes)?

If the train observer aligns his beam emitter (stationary in his reference frame) perpendicular to the train in his reference frame, and the platform observer aligns his beam emitter (stationary in his reference frame) perpendicular to the platform in his reference frame, both observers will observe the beams diverging once they're emitted. If there are detectors arranged along the center line of the train and stationary with respect to the train, the beam from the train's emitter will be detected at a detector positioned perpendicular to the train's emitter, while the beam from the platform's emitter will be detected at a detector farther to the rear of the train.

Both observers will agree which detector detects which beam. Both observers will agree that the beams diverge. What they will disagree on is how the emitters are aimed. The platform observer won't agree that the train's emitter is perpendicular to the train (it'll seem to be aimed at a "forward" angle). The train observer won't agree that the platform's emitter is perpendicular to the platform (it'll seem to be aimed at a "backward" angle). That's because to be aimed in a direction other than parallel to the x axis, the emitters have to have a length along the y axis. The Lorentz transformation thus induces a rotation of the direction of the emitters.

(But... but... but... the Lorentz transformation doesn't affect the y coordinate! It says right there, y' = y! But since the emitter has length, light takes time to travel from the back of the emitter to the front, and in that time, as observed from a relatively moving frame, the x position of the emitter changes and is affected by the Lorentz transformation. That's observed as a change in the "aim" direction. Such complications are why typical elementary SR thought experiments usually involve long thin moving components like rockets and trains, and flashes of light rather than aimed beams.)

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?