Is the frame dragging part of the Special Relativity?

[SDG]

Originally Posted by Ziggurat

I understand the diagram. I don't understand you. I cannot correct that. Either you can make yourself comprehensible, or you cannot.

Your quote:

Quote:

And I have no idea what you're trying to convey by the distinction between "deceleration to" and "acceleration from". There is no real distinction between deceleration and acceleration. And since your P point is in the middle of continuous acceleration, it involves both acceleration to and acceleration from that point.

The textbook:

Quote:

The event C1 is the sudden
change of direction of acceleration by 180ı (thrust reversing). O0 is subsequently slowed down
until P and then sped up towards O, until C2. At this point, a new thrust reversing occurs, so that
O0 is slowed down until it reaches B.

There is the acceleration from A to C1 and deceleration from C1 to P.
The traveling twin/triplet accelerates from stay home twin/triplet frame and then he decelerates to stay home twin/triplet frame at P.
The traveling twin/triplet world line tangents are the same in direction and magnitude at A and P.

[SDG]

Originally Posted by Ziggurat

...
The front of the train doesn't have to trace out the same worldline as the back of the train. It doesn't have to have the same proper time. And proper time isn't coordinate time.

Is this a new theory?
The front clock has to have the same proper time ticks as the back clock.
If they do not they are not in the same inertial frame.
The worldlines have to be parallel.
If they are not they are not in the same inertial frame.

[SDG]

Originally Posted by Ziggurat

I didn't say space-time coordinates. I said spatial coordinates. In other words, in that frame, those two events happen at the same place. Not the same time.

We are talking about events in the space-time diagram hence I make the point of bi-located in the space-time.
So we agree here.

Quote:

The front of the train doesn't stay at A. Not even if it crashes. Are you talking about a scenario where the train crashes? Is that what you're trying to get at? What happens to the back of the train if the front of the train stops? Because it you are, well, there's no paradox there, but there is something quite interesting which we could talk about. If not, then I have no idea what you're going on about.

I am not talking about any crash.
The event A is just that. Coordinates and no change to these coordinates.
Otherwise it is not event A anymore.

[Ziggurat]

Originally Posted by SDG

There is the acceleration from A to C1 and deceleration from C1 to P.

Acceleration and deceleration aren't fundamentally any different, that's my point. Yes, there is a distinction, but it's a reference frame dependent one. Deceleration in one frame will be acceleration in another.

[Ziggurat]

Originally Posted by SDG

Is this a new theory?
The front clock has to have the same proper time ticks as the back clock.

Only if they stay in the same inertial reference frame.

Quote:

If they do not they are not in the same inertial frame.
The worldlines have to be parallel.
If they are not they are not in the same inertial frame.

The world lines are parallel if they're both stationary in that inertial frame. If the train accelerates, though, they don't have to remain parallel. They don't actually have to experience the same acceleration.

[SDG]

Originally Posted by Ziggurat

Only if they stay in the same inertial reference frame.



The world lines are parallel if they're both stationary in that inertial frame. If the train accelerates, though, they don't have to remain parallel. They don't actually have to experience the same acceleration.

This thread is moving fast and it has parallel sub-discussions.
There is no acceleration in this one, just the events A, B, C.



Edit: I simplified the diagram.

[Myriad]

Please note that the train is 645,300 miles long. Assuming it's a passenger train, that means it has about 40 million coaches, weighing 2.5 billion tons in Earth standard gravity and seating 3.36 billion people. It takes the conductor 532 years to punch all the tickets for one trip. At lunch time, the line for the cafe car is 318,000 miles long* (queued up through about half of the 40 million cars) and the average wait time for a coffee and sandwich is 3,200 years.

This railroad has much bigger problems to worry about than whether some guy's watch is three seconds off from somebody else's.


*All figures assume social distancing measures are not in effect.

[SDG]

Originally Posted by Ziggurat

Acceleration and deceleration aren't fundamentally any different, that's my point. Yes, there is a distinction, but it's a reference frame dependent one. Deceleration in one frame will be acceleration in another.

Right, and my point is to call it proper way based on the original stay home frame.
Therefore the traveler is in the same frame as the stay home twin/triplet at the event P.
Do we have agreement about the event P?

[Ziggurat]

Originally Posted by SDG

This thread is moving fast and it has parallel sub-discussions.
There is no acceleration in this one, just the events A, B, C.

OK, so what's your problem here?

[SDG]

Originally Posted by Ziggurat

OK, so what's your problem here?

The SR is reciprocal and therefore the observers do not agree on the other's proper time.
The conclusion is derived based on who is looking.
Translating the problem to triplets and two observers do not agree on the proper time of the third one but the world line can have only one proper time.

[Ziggurat]

Originally Posted by SDG

The SR is reciprocal and therefore the observers do not agree on the other's proper time.

Show your work.

Quote:

Translating the problem to triplets and two observers do not agree on the proper time of the third one but the world line can have only one proper time.

That is false. We've been over this already. You did the problem wrong. You mixed and matched different reference frames without accounting for the changes.

[SDG]

Originally Posted by Ziggurat

Show your work.

This is from my post #9: "Any light round trip will appear slower (time dilated) compared to any other moving frame?"
It was a rhetorical question, it is easy to prove.

Quote:

That is false. We've been over this already. You did the problem wrong. You mixed and matched different reference frames without accounting for the changes.

I did not mixed reference frames. It goes back to the acceleration sub-discussion and the event P, the post #112.
When we settle that you'll see the answer.

[Ziggurat]

Originally Posted by SDG

This is from my post #9: "Any light round trip will appear slower (time dilated) compared to any other moving frame?"
It was a rhetorical question, it is easy to prove.




I did not mixed reference frames. It goes back to the acceleration sub-discussion and the event P, the post #112.
When we settle that you'll see the answer.

We aren't going to settle it because your questions are never comprehensible.

Pick one problem. Do the calculations for that problem, showing your work. Demonstrate what you think the error is. Then I can show you how to do the problem correctly.

[theprestige]

Originally Posted by Ziggurat

Show your work.



That is false. We've been over this already. You did the problem wrong. You mixed and matched different reference frames without accounting for the changes.

I'm not expert, but it seems like every misunderstanding of SR sooner or later turns out to be improper transitions between frames.

[SDG]

Originally Posted by Ziggurat

We aren't going to settle it because your questions are never comprehensible.

Pick one problem. Do the calculations for that problem, showing your work. Demonstrate what you think the error is. Then I can show you how to do the problem correctly.

Which one would you like to do first. The roundtrip or the acceleration/the event P?

[Robin]

Decided to remove this until SDG has chosen a problem and defined it.

[theprestige]

Originally Posted by SDG

Which one would you like to do first. The roundtrip or the acceleration/the event P?

LOL.

"Pick one problem."

"Which one?"

SDG is giving us a masterclass in how to avoid substantive discussion.

[cjameshuff]

Originally Posted by theprestige

I'm not expert, but it seems like every misunderstanding of SR sooner or later turns out to be improper transitions between frames.

SDG isn't the first anti-relativist I've seen who will simply refuse to understand this detail. Typically they ignore the acceleration in the standard twin's paradox in the name of "simplifying" the problem (thus simplifying the solution out of existence), or complain about how it's accounted for. If you point out that you can remove the acceleration by adding another "inbound" party that communicates with the "outbound" party as they pass each other, and the results are the same, they ignore this and/or start thrashing around aimlessly in an attempt to confuse things...just like SDG is doing now.

[Ziggurat]

Originally Posted by SDG

Which one would you like to do first. The roundtrip or the acceleration/the event P?

That's up to you. You've got the burden of proof here, so pick whichever one you want. But do the math. Show your numbers. And be clear, much more clear than you have been, about exactly what you're referring to.

[Myriad]

In the previous thread, SDG's main argument boiled down to claiming a "contradiction" caused by measurements made in one reference frame failing to agree with what the observer could calculate (via the Lorentz transformation) would be measured by another observer in a different frame. I think that's what the train example in this thread comes down to too, but it's too muddled to tell.

In this thread, SDG also seems to be suggesting that "entering the inertial reference frame" of some other observer (that is, matching velocities with that observer, however momentarily) means that clocks moving with you must suddenly disregard their previous elapsed times and spontaneously synch up (not just in pace but in measured proper time) with all other stationary clocks in that frame. Or something like that. As if different inertial frames were like the "alternate timelines" that characters in the Marvel Universe travel between.

Describing two events or observers as being "in the same frame" when one actually means they have the same velocity, while it might seem like everyone understands that the latter is what is meant, is probably contributing to the misdirection.

[Robin]

Just a note, on seeing some of the earlier scenarios, if the scenario does not involve acceleration then all world lines must be single straight lines. Any curves, bends or links in world lines implies acceleration.

[Robin]

Also note that when a textbook divides a journey into straight line segments with instantaneous change in velocity between each segment, they are not implying that such a thing can happen in real.life, merely developing the argument to construct the integral which can provide the actual calculation.

[Robin]

Originally Posted by Myriad

Describing two events or observers as being "in the same frame" when one actually means they have the same velocity, while it might seem like everyone understands that the latter is what is meant, is probably contributing to the misdirection.

In particular the claim, that when a comoving frame instantaneously has the same simultaneity line as some inertial frame then it's clocks must be in sync with that frame, is just plain wrong.

[Robin]

Also, in the scenario definition, each event must refer to a single spacetime location in each frame. So you can't have an event refer to something observed at the front and also.at the back of a train, these have to be separate events.

[Robin]

Originally Posted by SDG

This is from my post #9: "Any light round trip will appear slower (time dilated) compared to any other moving frame?"
It was a rhetorical question, it is easy to prove.

And let's get this out of the way.

A light round trip won't be a round trip in any other frame.

And it won't appear slower, it will appear to have the same speed, (distance travelled divided by time elapsed)

It will take longer or shorter in another frame, depending on the path.

[Robin]

Originally Posted by SDG

This thread is moving fast and it has parallel sub-discussions.
There is no acceleration in this one, just the events A, B, C.

https://i.imgur.com/y6QxAd6.png

Edit: I simplified the diagram.

Note that the object here is comoving with the t' axis but parallel to the x axis. Shouldn't it be parallel to the x' axis?

[JeanTate]

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.

Late, and rather OT: for the Sun and Mercury, yes. For ~solar-mass two neutron stars in close co-orbits, very noticeable. And carefully measured in the Double Pulsar, and maybe more now. Fits GR perfectly, provides great insights into NS masses, moments of inertia, and much more ...

[SDG]

The round trip of the light.
This is the best to ideal clock for a world line that we can have.








I hope everybody understands this.
This is a round trip on the train frame.
Agreed?

[SDG]

Now we are going to do a round trip of the light on the platform.










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

[Myriad]

Originally Posted by SDG

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

1. The observer on the platform measures the light traveling 4cs in 4s. (That's the blue arrows in the first of your two new diagrams.) The observer on the train measures the light traveling 2cs in 2s. (That's the red arrows in that same diagram.) What observer in what reference frame measures the light making the round trip in 8 seconds? Not any observer on the platform. Not any observer on the train. So, who?

2. How loud does the whistle need to be, for a train that's over 372,500 miles wide?

[Robin]

Originally Posted by SDG

Now we are going to do a round trip of the light on the platform.







https://i.imgur.com/Vx4RDcN.png





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







The 4s on the platform takes 8s' on the train.

Any questions?

It would help if you could be more explicit about what is happening here.

[Robin]

Also, you have two spatial dimensions on the original diagram so you would need two spatial dimensions on the spacetime diagram, or else you can't describe what is happening.

[Robin]

And the train frame spacetime diagram has the platform moving to the right which does not match the original diagram.

[SDG]

Originally Posted by Robin

It would help if you could be more explicit about what is happening here.

The platform observer sends the light in the direction as in the first example.
The light beam returns back to the platform observer.
2cs out + 2cs back = 4cs .... 4s

The return trip of the platform light beam takes 7s' for the train frame.
Just check the invariance of the space-time interval. It checks out.

[Robin]

Originally Posted by SDG

The platform observer sends the light in the direction as in the first example.
The light beam returns back to the platform observer.
2cs out + 2cs back = 4cs .... 4s

The return trip of the platform light beam takes 7s' for the train frame.
Just check the invariance of the space-time interval. It checks out.

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.

[HansMustermann]

Originally Posted by Myriad

In this thread, SDG also seems to be suggesting that "entering the inertial reference frame" of some other observer (that is, matching velocities with that observer, however momentarily) means that clocks moving with you must suddenly disregard their previous elapsed times and spontaneously synch up (not just in pace but in measured proper time) with all other stationary clocks in that frame. Or something like that. As if different inertial frames were like the "alternate timelines" that characters in the Marvel Universe travel between.

Which really boils down to confusing the measured interval with the coordinate. I.e., the same confusion as if I were to confuse my position, as shown by the GPS on my phone, with the distance travelled, as shown by my car's mileage counter. The distance calculated by the two of them will be very different for the same trip from Berlin to Paris, if I went in a straight line or if I went through Rome first.

[Robin]

Originally Posted by HansMustermann

Which really boils down to confusing the measured interval with the coordinate. I.e., the same confusion as if I were to confuse my position, as shown by the GPS on my phone, with the distance travelled, as shown by my car's mileage counter. The distance calculated by the two of them will be very different for the same trip from Berlin to Paris, if I went in a straight line or if I went through Rome first.

If it is the same as the last thread, then SDG will take two events in the same spatial location and show that the duration between them dilates into another frame according to the Lorentz factor alone.

Then he takes two events in the second frame with the same time difference between them but different locations and says that this time difference ought to dilate to the first frame in the same way.

When it doesn't he says this is a contradiction in SR.

[HansMustermann]

Yeah, I was getting the distinct impression that his train example just points at not understanding that t' = γ(t - vx/c2), as opposed to just γt, but honestly, it's so poorly explained that it's hard to be sure.

[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.

[SDG]

An interesting point.
The velocity v_y is c/7 for the train observer on the second leg.
The light is crossing the y+ direction at c and the y- direction at c/7.
Please, have a look at v_y in my first post diagram.