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Neutral simultaneity for two frames.


DimaMazin

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Look I'm going to answer questions as I choose to answer them. At any point in time I can back up any statement I make with literature or the mathematics..

 

I supplied links explaining proper time and coordinate time if you choose not to read those links and attempt to understand what I'm referring to that's your choice.

 I have also supplied two articles that directly show the rigid rod analysis including taking the time to find a mathematically simplified version.

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36 minutes ago, Mordred said:

At any point in time I can back up any statement I make with literature or the mathematics..

Still waiting.....

5 hours ago, Mordred said:

Each engine clock will be in coordinate time.

 

Quote

I supplied links explaining proper time and coordinate time if you choose not to read those links and attempt to understand what I'm referring to that's your choice.

Proper time:

Quote

In relativity, proper time (from Latin, meaning own time) along a timelike world line is defined as the time as measured by a clock following that line.

That's inside the car, where the engine clock is. Traveling along its world line as it accelerates.

Coordinate time:

Quote

In the theory of relativity, it is convenient to express results in terms of a spacetime coordinate system relative to an implied observer.

That's the ground, buddy. The observer is on the ground.

"Each engine clock will be in coordinate time" is 100 percent, Grade-A baloney.

B-A-L-O-N-E-Y.

Baloney.

 

Edited by Lorentz Jr
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I see so we have from the topic at hand to you arguing the definitions given by GR that were not set by me.

Let's make this simple in GR there is no rest frame unlike SR. 

In GR all observers frames of reference is the coordinate time.  This is because the coordinate time is not invariant.. it is coordinate dependent.   The rate of time will vary at any given coordinate as well as  the observer.

Got that so far ?

The only invariant reference frame is the clock that follows the null geodesic worldline it is path dependent. This means it will depend on that worldline between different geometries , flat curved, Schwartzhild metric, Kerr Metric etc.

This is why it's the proper time it is the only Lorentz invariant frame of reference in different geometries between different observers

 If you still have questions on that free to ask. 

I don't know your math skills but here you go

http://web.mit.edu/edbert/Alexandria/notes1.pdf

Now why does the wiki link specify the timelike geodesic (null geodesic) ? Well for starters c is Lorentz invariant. All observers measuring the velocity of a photon will measure the same value. Does that help to make better sense of the clock following the null geodesic ?

 

 

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1 hour ago, Mordred said:

I see so we have from the topic at hand to you arguing the definitions given by GR that were not set by me.

Wait, so now you're disagreeing with the Wikipedia links you just posted to explain proper time and coordinate time? Is that your new position? And you're also claiming authority to define those terms yourself?

1 hour ago, Mordred said:

Let's make this simple in GR there is no rest frame unlike SR.

In general relativity, all free-falling frames are physically indistinguishable from rest frames.

Every time you make an excuse for your mistakes, you provide more evidence of your ignorance.

1 hour ago, Mordred said:

In SR proper time was the rest frame and coordinate time was the inertial observer.

More ignorant fantasies. Proper time is always the time along a world line, the time experienced by a moving object. The inertial observer's frame is the rest frame.

1 hour ago, Mordred said:

In GR all observers frames of reference is the coordinate time.  This is because the coordinate time is not invariant..

Okay, so again, you know the word "invariant", but you don't know what it means. Coordinate time is time in the observer's frame by definition. It's not true "because the coordinate time is not invariant.."

1 hour ago, Mordred said:

Got that so far ?

Yes, I got that you're still trolling, you're still making stuff up and refusing to admit that you were wrong.

1 hour ago, Mordred said:

The only reference frame is the clock that follows the null geodesic worldline 

What does "the only reference frame" mean?

1 hour ago, Mordred said:

I don't know your math skills but here you go

http://web.mit.edu/edbert/Alexandria/notes1.pdf

Why don't you summarize it for me. If you're such an expert, what does that article say, in your own words?

Edited by Lorentz Jr
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G

13 minutes ago, Lorentz Jr said:

 

In general relativity, all free-falling frames are physically indistinguishable from rest frames.

 

GR does not use rest frames all reference frames are inertial . What is the definition of an inertial frame ? A frame under constant velocity. Equal free fall it is not the equivalent to a rest frame.

Do you have a reading problem ?

1 hour ago, Mordred said:

 

The only invariant reference frame is the clock that follows the null geodesic worldline it is path dependent. This means it will depend on that worldline between different geometries , flat curved, Schwartzhild metric, Kerr Metric etc.

This is why it's the proper time it is the only Lorentz invariant frame of reference in different geometries between different observers

 

 

The above propertime that clock on the worldline AS PER the WIKI link........ shall I go and quote each instance I stated the clock along the worldline for you for proper time...

15 minutes ago, Lorentz Jr said:

 

What does "the only reference frame" mean?

 

What don't you understand about that ? Is it the term reference frame ? A reference frame is an inertial frame of reference 

17 minutes ago, Lorentz Jr said:

 

Why don't you summarize it for me. If you're such an expert, what does that article say, in your own words?

I already have summarized it's not my fault you refuse to read or understand

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20 minutes ago, Mordred said:

all reference frames are inertial .

What is the definition of an inertial frame ? A frame under constant velocity.

Equal free fall it is not the equivalent to a rest frame.

Do you have a reading problem ?

The above propertime that clock on the worldline

What don't you understand about that ? Is it the term reference frame ?

A reference frame is an inertial frame of reference

Okay, well, I and our medieval friend here are obviously not getting along. That's too bad. Not really unexpected, but disappointing. Good night, Mordy. Maybe we'll chat again some other time.

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Good night hopefully the discussion goes better another time 

14 minutes ago, Mordred said:

Good night hopefully the discussion goes better another time 

Lol maybe in conformal time lmao

2 hours ago, Mordred said:

I see so we have from the topic at hand to you arguing the definitions given by GR that were not set by me.

Let's make this simple in GR there is no rest frame unlike SR. 

In GR all observers frames of reference is the coordinate time.  This is because the coordinate time is not invariant.. it is coordinate dependent.   The rate of time will vary at any given coordinate as well as  the observer.

Got that so far ?

The only invariant reference frame is the clock that follows the null geodesic worldline it is path dependent. This means it will depend on that worldline between different geometries , flat curved, Schwartzhild metric, Kerr Metric etc.

This is why it's the proper time it is the only Lorentz invariant frame of reference in different geometries between different observers

 If you still have questions on that free to ask. 

I don't know your math skills but here you go

http://web.mit.edu/edbert/Alexandria/notes1.pdf

Now why does the wiki link specify the timelike geodesic (null geodesic) ? Well for starters c is Lorentz invariant. All observers measuring the velocity of a photon will measure the same value. Does that help to make better sense of the clock following the null geodesic ?

 

 

Correction to above timelike isn't as I described. Null is light like timelike v<c

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5 hours ago, Mordred said:

Sure I can go with that 

Here is the Born Rigidity examination

https://fnegari.github.io/files/notes/009.pdf

From this article it should be clear just how tricky any rigid rod examination can get

 

This doesn't refute the validity of Born rigidity especially in terms of kinematics. If you understood Born rigidity you'd know this thread is concerned with the "irrotational motions" class of Born rigidity, which is broken if the train turns. So bringing it up is a strawman argument. It sounds like you're trying to argue that Born rigidity can't be satisfied, by bringing up irrelevant counter-examples where it is not satisfied.

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I have yet to see a treatment where Born rigidity holds true. Care to provide one ?

Lol course we can also examine Ehrenfests treatment

https://en.m.wikipedia.org/wiki/Ehrenfest_paradox

Though that will be a bit off topic though still involving a rigid body.

Edit if you have a good example of Born rigidity that doesn't violate spacetime causality feel free to post it.

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1 hour ago, Mordred said:

I have yet to see a treatment where Born rigidity holds true. Care to provide one ?

Lol course we can also examine Ehrenfests treatment

https://en.wikipedia.org/wiki/Born_rigidity

The "Class A" section contains an entire class of applicable motions.

Quote

Born rigidity is satisfied if the orthogonal spacetime distance between infinitesimally separated curves or worldlines is constant,[7] or equivalently, if the length of the rigid body in momentary co-moving inertial frames measured by standard measuring rods (i.e. the proper length) is constant and is therefore subjected to Lorentz contraction in relatively moving frames.[8] Born rigidity is a constraint on the motion of an extended body, achieved by careful application of forces to different parts of the body. A body rigid in itself would violate special relativity, as its speed of sound would be infinite.

Instantaneous (not generally simultaneous) acceleration using the solution I proposed (different from OP's I think), does not satisfy Born rigidity, because it requires different parts of the train to accelerate at different times, in the 2 rest frames (before and after acceleration) of the train. Therefore the train's length changes in those frames, and its proper length doesn't even seem defined while accelerating. However, it should be the only solution that involves a single instantaneous acceleration at each point of the train, with which the train has the same proper length before and after the acceleration.

It would seem not even Born rigidity can be satisfied with an instantaneous acceleration. However, OP's proposal never mentioned anything about a requirement of rigidity. For that matter, they neither required that the proper length of the train is the same before and after, I assumed that.

 

15 hours ago, DimaMazin said:

Because velosity of the wave of the acceleration = (gamma +1)v/gamma

For example  gamma=2

v=31/2c/2

velosity of wave acceleration=3*31/2c/4

The way I figure it, if you have the back of the train accelerate at time 0, the rest of the train accelerates over time until the front of the train finally accelerates when the length of the train has become L/gamma in the initial frame, where L is its original proper length. By that time, the back of the train has traveled a distance of (L - L/gamma) at velocity v. Therefore the time when the front of the train accelerates would be t=d/v = (L - L/gamma)/v.

Then the velocity of the "wave" would have to be d/t = Lv/(L - L/gamma) = v/(1 - 1/gamma)

For gamma=2, it seems intuitive that the wave would have to travel at 2v, to reach the front of the train at the same time that the back of the train traveling at v gets halfway there. With v = 3^0.5 c/2, I get a velocity of the wave equal to 3^0.5 c.

Like you said this is faster than c, so it's non-causal and needs local forces to accelerate the parts of the train.

Edited by md65536
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Yeah I've read proposals on  different parts of the rigid body requiring different accelerations at each part so I do understand where your coming from on that aspect. One of the distinctions between Newton rigid bodies and relativistic rigid bodies is literally in how one defines the criteria of a rigid body. The relativistic rigid body allows for length contraction, while the criteria for the Newtonian version must maintain its length. If I recall the articles I've come across though there is still causation issues in signals propagating greater than c in particular with acceleration/rapidity which relates to the hyperbolic rotation of the Minkowskii metric. The different accelerations will result in differing Hyperbolic rotations throughout the rigid body. If I recall but would have to double check the Born rigidity article under SR article I linked also refers to that.

 

to be honest though as far as the OP is concerned I'm still trying to fathom what he means by neutral simultaneity. Might just be a translation error  

Would those involved in this discussion prefer I stick to SR treatment as opposed to GR treatment as things can get immensely complex under GR for rigid bodies ?

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On 12/10/2022 at 8:52 AM, DimaMazin said:

Local forces accelerate the train. Therefore no force travels faster than c.

Why velosity of wave of acceleration can be faster than c? Because local forces don't need to wait when light crosses already contracted part of the train.

As mentioned, this isn't a rigid body and that's fine because it's not being accelerated as a rigid body, rather all the parts of the train are accelerated independently, as if it is modelled as a soup of particles.

I also think it's not describing a practical impossibility, where every smallest part of the train needs to be a train engine, because the cause of the acceleration can be external. For example the train could be a metal object accelerated by a rail gun.

 

I think it's possible that the train might not feel any stress from being stretched or compressed, it might only feel the local proper acceleration, temporal distortion, etc. This is because the acceleration "wave" crosses the train at a rate faster than c. For any particle on the train, even though the one "behind" it has started moving first, the particle has accelerated before the one behind has caught up and before any causal effect of the particle behind it has reached it. And even though the particle is moving before the one in front moves, by the time the particle gets to the location of the one in front, the one in front has already accelerated. This is true no matter how small you make the separation of the particles. (I guess this assumes that any field effects that can be felt, like EM field, are also accelerated along with the particles. Is this impossible?) On second thought... after a particle has accelerated it is now in the new inertial frame, in which the particle in front has already accelerated first! It has accelerated to rest.

This gives an observer in the middle of the train (and who accelerates along with it) a bizarre account of what happens! According to her, the rear of the train begins to compress forward (not fast enough to see or feel it) while the front of the train remains stationary. Then she feels proper acceleration, then the rear of the train begins to stretch backward (but not faster than can be felt??? It seems this compression would have to be visible) while the front of the train again remains stationary! That seems surprising enough that I wonder what details I'm missing or getting wrong.

Edit: Due to delay of light I think she'd see the front of the train appear to be approaching and stretched out (due to aberration of light). These are images of the front of the train before those parts accelerated, seen in the post-acceleration frame.

But if it can be seen it can be felt?

Edited by md65536
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29 minutes ago, md65536 said:

I also think it's not describing a practical impossibility, where every smallest part of the train needs to be a train engine, because the cause of the acceleration can be external. For example the train could be a metal object accelerated by a rail gun.

I think it's possible that the train might not feel any stress from being stretched or compressed, it might only feel the local proper acceleration, temporal distortion, etc. This is because the acceleration "wave" crosses the train at a rate faster than c.

What kind of real wave do you think travels faster than c, and how do you think a simulated wave going faster than c could be implemented without local engines?

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14 hours ago, md65536 said:

 

 

The way I figure it, if you have the back of the train accelerate at time 0, the rest of the train accelerates over time until the front of the train finally accelerates when the length of the train has become L/gamma in the initial frame, where L is its original proper length. By that time, the back of the train has traveled a distance of (L - L/gamma) at velocity v. Therefore the time when the front of the train accelerates would be t=d/v = (L - L/gamma)/v.

Then the velocity of the "wave" would have to be d/t = Lv/(L - L/gamma) = v/(1 - 1/gamma)

For gamma=2, it seems intuitive that the wave would have to travel at 2v, to reach the front of the train at the same time that the back of the train traveling at v gets halfway there. With v = 3^0.5 c/2, I get a velocity of the wave equal to 3^0.5 c.

Like you said this is faster than c, so it's non-causal and needs local forces to accelerate the parts of the train.

Good  analysis +1

5 minutes ago, Lorentz Jr said:

What kind of real wave do you think travels faster than c, and how do you think a simulated wave going faster than c could be implemented without local engines?

Unfortunately no real wave travels faster than c. I will have time to respond further later on RL

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9 minutes ago, Mordred said:

Unfortunately no real wave travels faster than c.

I know that, Professor. I was asking md65536 what she thinks, because what she posted is inconsistent.

Please either learn how to read or stay out of other people's conversations.

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21 minutes ago, Lorentz Jr said:

What kind of real wave do you think travels faster than c, and how do you think a simulated wave going faster than c could be implemented without local engines?

Nothing physical is moving faster than c, it's just the timing of the local accelerations across the length of the train moving faster than c. It's only the phase velocity of this wave that's traveling faster than c. That's fine because the propagation of the wave isn't causal; the parts of the train are accelerating independently of each other.

I can avoid calling it a wave to be less confusing.

Yes, I've assumed "local engines" all along, that was implied by OP. I'm just saying the engines don't have to be part of the train itself. I'm trying to talk about the effects of SR without getting hung up on the difficulties of building a physical version of a thought experiment.

 

I don't want to debate whether it's possible for 2 events to be simultaneous ie. not causally connected.

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1 hour ago, Lorentz Jr said:

I know that, Professor. I was asking md65536 what she thinks, because what she posted is inconsistent.

Please either learn how to read or stay out of other people's conversations.

If I choose to reply I will do so.

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1 hour ago, md65536 said:

Yes, I've assumed "local engines" all along, that was implied by OP. I'm just saying the engines don't have to be part of the train itself. I'm trying to talk about the effects of SR without getting hung up on the difficulties of building a physical version of a thought experiment.

Okay, that's good. Although even a rail gun would have to propel each car according to the same schedule. The main differences would be the clocks in the rail gun are in the ground frame instead of in the cars, and the different points on the gun have to keep changing from one car's schedule to the next one. And it would have be a very long rail gun. 😄

 

2 hours ago, Lorentz Jr said:

What kind of real wave do you think travels faster than c, and how do you think a simulated wave going faster than c could be implemented without local engines?

2 hours ago, Mordred said:

Unfortunately no real wave travels faster than c.

2 hours ago, Lorentz Jr said:

I know that, Professor. I was asking md65536 what she thinks, because what she posted is inconsistent.

1 hour ago, Mordred said:

If I choose to reply I will do so.

And if you continue to insult my intelligence with elementary dictionary definitions and freshman-level factoids, I'll continue to tell you what you can do with yourself.

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23 hours ago, Lorentz Jr said:

I've been assuming the lengths of the cars are neither especially long nor especially short. In order for them to all accelerate simultaneously without getting pulled apart or crashing together, the information about their accelerations as a function of time has to be contained in each car from the start, and all the engines have to start accelerating at the same time.

23 hours ago, studiot said:

I think most participants is this thread know and agree that these conditions are neither necesary nor possible.

22 hours ago, Lorentz Jr said:

Really? Based on what? How else would one accelerate train cars without stress between them, and what do you think makes synchronized self-propulsion impossible?

Still waiting for an explanation of this comment, which is just as nonsensical as anything Professor Morty has said.

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1 hour ago, md65536 said:

Nothing physical is moving faster than c, it's just the timing of the local accelerations across the length of the train moving faster than c. It's only the phase velocity of this wave that's traveling faster than c. That's fine because the propagation of the wave isn't causal; the parts of the train are accelerating independently of each other.

I can avoid calling it a wave to be less confusing.

Yes, I've assumed "local engines" all along, that was implied by OP. I'm just saying the engines don't have to be part of the train itself. I'm trying to talk about the effects of SR without getting hung up on the difficulties of building a physical version of a thought experiment.

 

I don't want to debate whether it's possible for 2 events to be simultaneous ie. not causally connected.

I don't particularly see describing it as a non causal wave to be an issue here. As long as it's understood to represent individual components of the train separately powered via the engine.

 I've still been considering the hyperbolic Lorentz transforms in regards to the different acceleration rates.

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