[Jaaanosik]

...
The graph shows an accelerating observer and a non-accelerating observer. What does this have to do with the subject, which is two observers in a common inertial reference frame accelerating at the same rate in the same direction beginning at an agreed on moment on synchronized clocks?

In that scenario, the entire ensemble will not be aware of any contraction. An external observer in an inertial frame would observe increasing Lorentz contraction on the entire ensemble. Are you still imagining that there is an Absolute Space in which the shrinkage is happening? There isn't. Contraction is relative, not objective.

Malamute Lover,
let us talk some concrete analysis.
L₀ - length of the spaceship in its rest frame.
5L₀  - distance between the spaceships.
Let us choose the left frame as our origin.




The right spaceships has to get much further away into the future, in the original rest frame, in order to maintain the simultaneity line of the left spaceship frame.



Do you trust the Minkowski space-time diagrams?
Does this make sense now?
Jano


Edit:
This is the Special Relativity acceleration 101 (One-o-One) in transition towards general frames.
What do you know, this is the post #101 in this thread.

[Halc]

The traveling twin thinks he is continuing to accelerate at the same rate which he can feel - and therefore exceeds c

The driver thinks he is continuing to accelerate at the same rate which he can feel - and therefore exceeds the speed limit.

Where is the problem?

No problem with that. Proper velocity has no upper limit and can exceed c, as I explained above.

What is holding them in a circular orbit rather than flying out in a straight line and crashing into the track?

Wheels

The track is holding them. If the track disappeared, they would fly off in straight lines despite having wheels.

That they would.  Same thing if the spokes disappeared in the spoke example.

The twins paradox has no answer in SR because acceleration is involved. SR cannot deal with acceleration. There is centripetal acceleration even with a constant speed. SR cannot handle this problem. You need GR and that is what GR says will happen. If you insist on staying in SR, then you have just created another fake twins paradox with no way to answer it.

You're welcome to invoke GR if that's the only way you can figure it out.  SR contains Lorentz transformations between frames, and that's all I used in my example.

Since the situations are symmetric at the beginning and in the acceleration applied according to synchronized clocks points directly at the end results having the same appearance to the participants. Clear concepts first. No need for math in this.

Until you actually do the math as I did and it shows otherwise.  Do the trivial math then. I gave coordinates of the four important events (SP1a/b and SP2a/b).  If I gave incorrect coordinates, then correct them. Because with my numbers, a 10-LD rod length-contracted down to size 5.8477 isn't going to bridge the gap between SP1b and SP2b which are still 10 LD apart. My string has broken. You have to have some kind of different numbers to have the string not break, but you've shown nothing but Newtonian math, examples with near stationary motion,  plus a lot of assertions that don't add up.

like the fact that the rod attached to SP2 will not yet move at time SP1a.

Sp1a is an event, not a time. The 10 LD mark on the rod happens to be present at that event, and it indeed begins acceleration at that event, but since it has zero velocity at the moment it begins to accelerate, I'll have to agree that it does not yet move at that event.

I am reminded of Poul Andersen's novel Tau Zero from years back. A ship with a gigantic fuel supply and a hyper-efficient engine suffers a problem. The engine won't turn off.

And nobody thinks to just put the ship into a slow turn so the speed never gets stupid.

Not objective. Another observer in a different reference frame will observe a different a time dilation factor between you and the GPS satellite.

Oopsie.  Math shows otherwise.

I have shown the inconsistency with your concepts.

You've shown my numbers to be inconsistent with your assertions is all. Without showing light speed to be not c somewhere, or to show that events seen in one frame differ from the observations made from the same events in the other frame, or that 1+2 for one observer is different than 1+2 for the other, you've not demonstrated any inconsistency at all. You've just make empty assertions with no numbers to back them. And that's all you'll continue to do, because if you post 'corrected' numbers for my scenario, I'll show exactly those inconsistencies.

Use my scenario: 100g for 4 days.  You say there's no need for mathematics, so it should be trivial for you.  Show the numbers in both frames, as I did.  But you can't.

Sloppy.

Ooh, I hit a nerve.

You are comparing the 4.9082 elapsed time that an unaccelerated observer would see (unaccelerated with respect to the original frame of SP1 and SP2) with the 4 elapsed time that both SP1 and SP2 see.

The coordinate system of M is what your stationary observer in M would compute, yes. That coordinate system is unaffected by what one pair of ships happen to be doing.

The unaccelerated observer sees 4.9082 for both.

He can work out that time for both. He?s probably not present at either event, so what he sees of those events comes some time later. I expended no effort computing what anybody sees except things happening in their presence. But yes, if there was a series of unaccelerated observers with clocks synced in M, that just happen to be present at SP1b and SP2b, their respective clocks would read 4.9082 at those events.

The accelerated observers (SP1 and SP2) both see 4. The rod is stopped relative to both SP1 and SP2 in all frames of reference.

Sloppy.

I said no such thing, and if you look at my numbers in M, the rod is very much still moving relative to SP1 at event SP1b.
The rod is stopped by definition in frame N when SP2 stops in that frame. Outside SP1?s window, event eW thus happens immediately in N, but takes over 3 weeks to happen in M.  Again, your statement suggests a complete lack of awareness of relativity of simultaneity, which for somebody claiming to know even the most basic fundamentals of relativity, is not so much sloppy as plain uneducated.  If two separated events (SP2b and eW) happen simultaneously in one frame,  they?re not going to be simultaneous in the other. There?s exceptions to this, but not in a 1D example.

In the post-acceleration frame:
Both ships are initially moving at -.8112c at come to a halt after 4 subjective days of acceleration.
The easy part. By symmetry, SP1 begins acceleration SP1a at (0, 0) (our chosen common event) and SP1b coordinates are (-2.5123, 4.9082).  If you don?t use proper acceleration, the symmetry is lost and it takes much more work to compute that.

If you use proper acceleration the elapsed time for both is 4.

The elapsed ship proper time is 4.  Elapsed time time in frame N is 4.9082. I explicitly said these are N coordinates (or rather the post-acceleration frame which I?m now calling N).

If you do not use proper acceleration, that is, if you use an unaccelerated frame then the elapsed time for both is 4.9082.

Frame N is an inertial reference frame, as is M. Neither are accelerated frames. The use of proper acceleration or not affects the flight plan and the symmetry of the mathematics, but has no effect on what has been defined as an inertial reference frame.

[Malamute Lover (OP)]

Sloppy.

Ooh, I hit a nerve.

Merely an accurate observation as will be shown in detail later.

In the original frame [M]:
SP1a is at x=0,t=0 at the start (or 0,0).  SP2a is at (-10, 0)    Units in (light days, days)
After end of acceleration,
SP1b coordinates are (2.5123, 4.9082) and
SP2b coordinates are (-7.4877, 4.9082). Both ships are moving at .8112c which has a Lorentz factor λ of  1.71008.  The SP1 clock reads 4 at SP1b and the SP2 clock reads 4 at SP2b.
At time 4.9082, The rod is still accelerating at various rates.  It is stationary relative to SP2 only in SP2’s frame, so SP1 observes a rod moving outside his window at event SP1b.

For brevity, I will call the two frames M and N, conveniently M (maroon) for the original frame, and N (navy) for the frame of either ship after the end of acceleration. This corresponds to the colors I used for the text.

If both clocks read 4 and they are initially synchronized and undergo identical acceleration histories, then both engines have stopped. Why should the rod still be accelerating?

Your question again displays a lack of awareness of relativity of simultaneity. The event of the halting of the 17.1 LD mark on the rod (which I shall call event eW for where SP1 sees the rod stop outside his Window) is simultaneous with distant event SP2b in N, so it cannot be simultaneous with event SP2b in M. In fact, the coordinates of eW in M is (21.7559, 28.6306), so it actually takes 23.7224 days to stop in frame M after SP2 shuts down. If the rod is longer than this, extending well past SP1, then those parts have not yet stopped at that time in M.



I chose my numbers (high acceleration, separation greater than distance to Rindler horizon) to highlight such effects.

You are comparing the 4.9082 elapsed time that an unaccelerated observer would see (unaccelerated with respect to the original frame of SP1 and SP2) with the 4 elapsed time that both SP1 and SP2 see.

The coordinate system of M is what your stationary observer in M would see, yes. That coordinate system is unaffected by what one pair of ships happen to be doing.[

The unaccelerated observer sees 4.9082 for both.

He can work out that time for both. He’s probably not present at either event, so what he sees of those events comes some time later. I expended no effort computing what anybody sees except things happening in their presence. But yes, if there was a series of unaccelerated observers with clocks synced in M, that just happen to be present at SP1b and SP2b, their respective clocks would read 4.9082 at those events.

The accelerated observers (SP1 and SP2) both see 4. The rod is stopped relative to both SP1 and SP2 in all frames of reference.

Sloppy.

I said no such thing, and if you look at my numbers in M, the rod is very much still moving relative to SP1 at event SP1b.
The rod is stopped by definition in frame N when SP2 stops in that frame. Outside SP1’s window, event eW thus happens immediately in N, but takes over 3 weeks to happen in M.  Again, your statement suggests a complete lack of awareness of relativity of simultaneity, which for somebody claiming to know even the most basic fundamentals of relativity, is not so much sloppy as plain uneducated.  If two separated events (SP2b and eW) happen simultaneously in one frame,  they’re not going to be simultaneous in the other. There’s exceptions to this, but not in a 1D example.

In the post-acceleration frame:
Both ships are initially moving at -.8112c at come to a halt after 4 subjective days of acceleration.
The easy part. By symmetry, SP1 begins acceleration SP1a at (0, 0) (our chosen common event) and SP1b coordinates are (-2.5123, 4.9082).  If you don’t use proper acceleration, the symmetry is lost and it takes much more work to compute that.

If you use proper acceleration the elapsed time for both is 4.

The elapsed ship proper time is 4.  Elapsed time time in frame N is 4.9082. I explicitly said these are N coordinates (or rather the post-acceleration frame which I’m now calling N).

If you do not use proper acceleration, that is, if you use an unaccelerated frame then the elapsed time for both is 4.9082.

Frame N is an inertial reference frame, as is M. Neither are accelerated frames. The use of proper acceleration or not affects the flight plan and the symmetry of the mathematics, but has no effect on what has been defined as an inertial reference frame.

If the flight plan is constant acceleration relative to M, then proper acceleration increases over time resulting in a lower average speed in the M frame than the same phase in the N frame where most of the acceleration (reduction of speed) is at the end, resulting in a higher average speed. The symmetry is lost and the elapsed time of accelration in frame M would be different than in frame N. I did not care to  attempt to compute that.
Also, 100g of acceleration for 4 days from a stop relative to some frame is impossible. It involves infinite proper acceleration before 4 days.

For SP2 we need to do a Relativity of Simultaneity calculation to compute how far out of sync their clocks will be in this new frame.

The clocks are not out of sync at all. You are mixing two different reference frames. That does not work.

Your baseless assertion, not mine. You’ve shown no self-inconsistency with my numbers, only asserting inconsistency with your fictional ones where there is no RoS, which is of consequence to me only if I can see your alternate version of all the numbers.

The numbers done the right way.

SP1a is at x=0,t=0 at the start (or 0,0).  SP2a is at (-10, 0)    Units in (light days, days)

After end of acceleration,
Because the inertial observer mentioned below does not see proper acceleration, the spatial (light days) coordinates are wrong but I have no intention of calculating that since it is not relevant.
SP1b coordinates are (2.5123, 4.9082) relative to an unaccelerated observer in the original inertial frame of SP1 and SP2

SP2b coordinates are (-7.4877, 4.9082) relative to an unaccelerated observer in the original inertial frame of SP1 and SP2

Both ships are moving at .8112c which has a Lorentz factor λ of 1.71008 relative to an unaccelerated observer in the original inertial frame of SP1 and SP2

But since both ships have been accelerating at the same rate for the same time in the same direction beginning at an agreed time on synchronized clocks, they are in the same inertial frame and there is no difference in their λ. Neither is there any difference in their proper separation from the start, although that unaccelerated observer sees the whole ensemble contracted.

The SP1 clock reads 4 at SP1b and the SP2 clock reads 4 at SP2b.

At time 4.9082 relative to an unaccelerated observer in the original inertial frame of SP1 and SP2, which is time 4 according to synchronized clocks on the two ships, both ships and the rod are stationary relative to each other. Since both ships have stopped accelerating at the same time, they are in a common inertial frame.

Relativity of Simultaneity refers to the difference in apparent time separation between two events as viewed by two observers not in the same inertial frame. A light is turned on at point A and at Point B at Midnight UTC (synchronized clocks). Observer Alice is on a point midway on a straight line between A and B. Alice sees the light from both A and B at the same time. Bob is on a fast motorcycle on that straight line heading toward A and passes Alice at Midnight UTC on Alice’s watch. Bob will see the light from A earlier than the light from B.

A and B send each other information about the time they turned the lights on and when they saw the light from the other. Since they have synchronized clocks, A and B will then agree that they turned their lights on at the same time, even though each saw the other’s light go on later than his own. If they agree on the time delay value, they know they are in a common inertial frame and can even calculate the proper separation.

The Lorentz formula for Relativity of Simultaneity allows quantifying the degree of disagreement between two observers about the relative timing of two events, generalized to different locations in space for the two observers.  It does not represent anything real, as can be seen in the above example. SP1 and SP2 observations of each other disagree about who started first but at the end when they send each other copies of their logs they agree everything was really symmetrical and simultaneous according to their synchronized clocks.

It appears that you do not understand Relativity of Simultaneity either. It is not applicable here.

I see that you have given up on the first picture you presented that attempted to show changing real lengths after I explained how it was wrong. And I see that you have again presented the second picture but this time without the explanatory text that, if you bothered to read and understand it, would have shown you that you were wrong again. Here it is, with the text.



The horizontal lines show the length of the ruler as it passes by the inertial observer at the origin, with the changes due to changing angles of observation. The horizontal distance (call it W) between where the dashed lines intersect the edges of the colored area correspond to the length of the ruler as seen by the observers at the ends of the ruler. That is the rectangles mentioned in the text formed by the points of intersection, those points being opposite corners of the rectangle.  Note that the dashed lines change angles, being steeper where the colored area becomes narrower as seen by the inertial observer. Note especially that because of the changing angles, W is constant. The observers at the end of the ruler never see any change in the length of the ruler. SP1 and SP2 never see any change in their separation and the position of the rod.


What you have is a handful of buzzwords that you do not understand, some online calculating tools that you do not know how to properly apply, and some pretty pictures that do not show what you think they do. The truth is that you do not understand the basic concepts of Relativity Theory.

Give it up. You are wrong.

[Malamute Lover (OP)]

The traveling twin thinks he is continuing to accelerate at the same rate – which he can feel - and therefore exceeds c

The driver thinks he is continuing to accelerate at the same rate – which he can feel - and therefore exceeds the speed limit.

Where is the problem?

No problem with that. Proper velocity has no upper limit and can exceed c, as I explained above.

Here is what you said.

The traveling twin thinks he is continuing to accelerate at the same rate – which he can feel - and therefore exceeds c

c is a speed. Acceleration is not. Different units. You’re equivocating the two here. It is meaningless to say that acceleration exceeds some speed.

As I showed, I was not equating speed and acceleration. You were wrong about what you said but rather than admit such a thing, you switch to another subject.

What is holding them in a circular orbit rather than flying out in a straight line and crashing into the track?

Wheels

The track is holding them. If the track disappeared, they would fly off in straight lines despite having wheels.

That they would.  Same thing if the spokes disappeared in the spoke example.

The subject was that the track would necessarily move in a direction opposite to the cars courtesy of Newton #3. But you skip that part because you cannot deal with the consequences.

The twins paradox has no answer in SR because acceleration is involved. SR cannot deal with acceleration. There is centripetal acceleration even with a constant speed. SR cannot handle this problem. You need GR and that is what GR says will happen. If you insist on staying in SR, then you have just created another fake twins paradox with no way to answer it.

You’re welcome to invoke GR if that’s the only way you can figure it out.  SR contains Lorentz transformations between frames, and that’s all I used in my example.

I did invoke GR and demonstrated that everything, cars and track, are going to contract. No gaps. SR cannot handle ongoing acceleration and that is what you have when circular motion.

Since the situations are symmetric at the beginning and in the acceleration applied according to synchronized clocks points directly at the end results having the same appearance to the participants. Clear concepts first. No need for math in this.

Until you actually do the math as I did and it shows otherwise.  Do the trivial math then. I gave coordinates of the four important events (SP1a/b and SP2a/b).  If I gave incorrect coordinates, then correct them. Because with my numbers, a 10-LD rod length-contracted down to size 5.8477 isn’t going to bridge the gap between SP1b and SP2b which are still 10 LD apart. My string has broken. You have to have some kind of different numbers to have the string not break, but you’ve shown nothing but Newtonian math, examples with near stationary motion,  plus a lot of assertions that don’t add up.

I have addressed your math earlier and shown just how wrong it is. The contraction of the entire ensemble, ships and rod, is only as seen by the inertial observer. To the ships traveling at the same speeds because of identical acceleration histories, contraction does not exist because there is no relative motion.

like the fact that the rod attached to SP2 will not yet move at time SP1a.

Sp1a is an event, not a time. The 10 LD mark on the rod happens to be present at that event, and it indeed begins acceleration at that event, but since it has zero velocity at the moment it begins to accelerate, I’ll have to agree that it does not yet move at that event.

SP1a is the time on SP1’s clock when SP1 acceleration begins. SP2a is the time on SP2’s clock when SP2 acceleration begins. Observing each other’s clocks will not give consistent results because of light speed delays. This is the case even when in common inertial frames. In inertial frames. each will see the other’s clock as behind but ticking at the same rate.  Acceleration makes this more complicated leading to strange and contradictory illusions as previously described. Only when the acceleration has ended can they see that the elapsed time on each clock is the same, even though each will still thinks the other’s clock is behind but running at the right rate. Comparing clocks must be done with a shovel of salt.

Bringing two clocks into a common inertial frame will allow comparison, with the slower clock the one that experienced more acceleration.

Not necessarily true. I can sync a pair of clocks, separate them and have clock A undergo 10x the acceleration of clock B, and yet clock B be the slower of the two when they’re compared in each other’s presence. It’s the temporal length of their respective worldlines that counts, not the amount of acceleration.

… with the one having experienced the most acceleration having less elapsed time. If you think you can set up a situation where that is not the case, show me. Don’t need to do complicated calculations. Keep it simple.

Clock A accelerates at 1g for however long it takes it to get to the nearest star and back.  Clock B meanwhile gets put into a cyclotron on Earth and is subjected to 10g of continuous acceleration the whole time Clock A is gone.  Clock A will have logged less time when they reunite, despite having accelerated far less. I didn’t do the numbers, so I can’t say how much.  Can do if you think it matters.  Clock B will age negligibly slower than one not in the cyclotron, depending on the radius of the thing.

Wrong. Clock B will have gone much faster and been time dilated much more than Clock A and will be way behind Clock A. It will be obvious which one had experienced more acceleration. Where they went or did not go is irrelevant as long as they end up side by side in a common inertial frame. What makes you think otherwise?

I am reminded of Poul Andersen’s novel Tau Zero from years back. A ship with a gigantic fuel supply and a hyper-efficient engine suffers a problem. The engine won’t turn off.

And nobody thinks to just put the ship into a slow turn so the speed never gets stupid.

And why will that matter? They will still be accelerating until all that fuel runs out. Comparing this and your response above, I get the impression that you think that straight line distance covered has something to do with time dilation. The ultra-high speed particles in accelerators are seriously time dilated as can be seen by the decay times of the unstable byproducts being much longer than usual.

Not objective. Another observer in a different reference frame will observe a different time dilation factor between you and the GPS satellite.

Oopsie.  Math shows otherwise.

You said that it was knowledge of the clock rate adjustment on a GPS satellite that allowed you to know the time dilation, which does not really prove anything. You could have been told wrong.

 If you were able to see an unadjusted clock riding on the GPS satellite you could then compare the tick rate of that clock with your own and know the time dilation difference per day, although dividing that up into the parts due to relative speed and to gravitational differences would require other information.

An observer in geosynchronous orbit (no motion relative to you to keep it simple) with an even faster clock than the unadjusted GPS clock and faster still than your clock would not see the same difference between you and GPS because its clock is ticking faster than either of the other two. The scale has changed and a different value will result. Moving that other satellite clock up and down will change the tick rate faster and slower. The values of the daily time dilation difference will vary because the scale is changing.

If you think otherwise, show the math.

Even if you exclude other observers from consideration, the GPS satellite will consider you time dilated since you are moving relative to it and are deeper in a gravity well and your clock is running slower.

Relative to the ISS, my clock runs objectively faster, despite my continuous speed relative to it, and despite my greater gravity well depth.  I’m only slower relative to GPS because the gravity difference is enough to outweigh the motion difference.  GPS satellites don’t move all that fast.

True. ISS has an average altitude of 322 km and a speed of about 28,000 km/h. GPS satellites are way up at 20,200 km and 14,000 km/h. That works out to a very convenient two orbits a day, simplifying ongoing recalibration of position.

Acceleration is in the domain of General Relativity.

Gravity actually

Gravity and acceleration are dealt with by the same mathematical framework. Mathematically gravity is acceleration and vice versa. SR cannot handle acceleration. GR can.

I have shown the inconsistency with your concepts.

You’ve shown my numbers to be inconsistent with your assertions is all. Without showing light speed to be not c somewhere, or to show that events seen in one frame differ from the observations made from the same events in the other frame, or that 1+2 for one observer is different than 1+2 for the other, you’ve not demonstrated any inconsistency at all. You’ve just make empty assertions with no numbers to back them. And that’s all you’ll continue to do, because if you post ‘corrected’ numbers for my scenario, I’ll show exactly those inconsistencies.

Use my scenario: 100g for 4 days.  You say there’s no need for mathematics, so it should be trivial for you.  Show the numbers in both frames, as I did.  But you can’t.

No, you showed mixed numbers, partly from the accelerated frame and partly from an inertial frame and assigned different numbers to SP1 and SP2 despite identical acceleration histories. You do not understand Relativity Theory at all.

I have posted corrected numbers. Let’s see what mistakes you make this time.

Be back when I can,

[Halc]

The numbers done the right way.

All the numbers are the same as mine, if only the frame M coordinates. You offer no frame N coordinates at all.

If you are correcting something, I expected different numbers. It appears you just copied my stuff, which contradicts your assertion that I had not done them correctly.

We each say the other is wrong, so it is a he-said/she-said sort of situation. But numbers don’t lie. If my numbers are wrong, you need to produce different ones. Producing the same ones means you are in agreement with my assessment of the situation. My numbers show the string breaking as there is over 17 light-days of proper separation between the ships after acceleration. You’ve not shown otherwise since you don’t compute the proper distance between the ships at all.

SP1a is at x=0,t=0 at the start (or 0,0).  SP2a is at (-10, 0)    Units in (light days, days)
After end of acceleration,

SP1b coordinates are (2.5123, 4.9082) relative to an unaccelerated observer in the original inertial frame pf SP1 and SP2

SP2b coordinates are (-7.4877, 4.9082) relative to an unaccelerated observer in the original inertial frame of SP1 and SP2

Both ships are moving at .8112c which has a Lorentz factor λ of 1.71008 relative to an unaccelerated observer in the original inertial frame of SP1 and SP2

All in agreement, and seemingly copied from my post, except for some of the rearangement of the wording.

But since both ships have been accelerating at the same rate for the same time in the same direction beginning at an agreed time on synchronized clocks, they are in the same inertial frame and there is no difference in their λ.

Agree. They both are moving at the same speed, and thus are stationary relative to each other after both have finished accelerating.

Neither is there any difference in their proper separation from the start, although that unaccelerated observer sees the whole ensemble contracted.

Words, which lie. Show numbers, which don’t.  Without that, all I can do is assert that you’re wrong, and my numbers show it, and your lack of different numbers lends zero support to your assertion. I see no computation of their proper separation after acceleration.

The SP1 clock reads 4 at SP1b and the SP2 clock reads 4 at SP2b.

Yes, kind of by definition, since that’s exactly the flight plan.

At time 4.9082 relative to an unaccelerated observer in the original inertial frame of SP1 and SP2,

That took a whole line to say “At 4.9082 in M”. That’s why I gave the frames names, so I didn’t have to type all that each time.

which is time 4 according to synchronized clocks on the two ships, both ships and the rod are stationary relative to each other.

They’re both moving at the same speed in M at that time in M. That’s different than saying they’re stationary relative to each other, because you have to use the ship frame to state that they’re stationary at that time. I go so far as to agree that relative to the ships at event SP2b, both ships and the rod are stationary relative to each other, but I cannot agree with your statement above. Again, without numbers in N, your assertion is just words. I have no coordinates in N to work with, so I cannot show you the contradictions that result from your assertions.

Since both ships have stopped accelerating at the same time, they are in a common inertial frame

They’re ‘in’ all the frames, but each is stationary in only one. I think you meant that, but being clear is nice.  It’s because they both accelerate to the same speed that makes them stationary in the same frame after acceleration. When they do it or how long they take to do it makes no difference, so long as net acceleration from M is .8112c.  We’re more controlled than that of course, but them being stationary relative to each other has nothing to do with when they stopped accelerating, especially since the ‘when’ part is completely frame dependent.

Relativity of Simultaneity refers to the difference in apparent time separation between two events as viewed by two observers not in the same inertial frame.

It has to do with the frame, and not the observers, and again, all observers are in all frames. But yes, something like how you word it.

A light is turned on at point A and at Point B at Midnight UTC (synchronized clocks). Observer Alice is on a point midway on a straight line between A and B. Alice sees the light from both A and B at the same time.

No frame reference. I presume the clocks are stationary in the frame in which they are in sync, and Alice is also stationary in this same frame. Just making sure.

Bob is on a fast motorcycle on that straight line heading toward A and passes Alice at Midnight UTC on Alice’s watch. Bob will see the light from A earlier than the light from B. A and B send each other information about the time they turned the lights on and when they saw the light from the other. Since they have synchronized clocks, A and B will then agree that they turned their lights on at the same time, even though each saw the other’s light go on later than his own.

If they agree on the time delay value, they know they are in a common inertial frame and can even calculate the proper separation.

They already know they were stationary in the one frame, so the subsequent timings just confirm again that those clocks are indeed synced in that frame, yes.

You didn’t mention what Bob concludes since he doesn’t know where in his frame the light pulses originated, and therefore seems unable to assess which one flashed first. The one might have been before the other, or just closer. So what’s the purpose of Bob here?  A stationary observer off to one side of Alice would similarly observe one pulse before the other. RoS seems not to have been well demonstrated with this example.
You didn’t label any events, but there are the two events where the light pulses were emitted, and various observation events.  RoS says that the two emission events, if simultaneous in Alice’s frame are not simultaneous in Bob’s frame, even if that wasn’t demonstrated by your example.

I see that you have given up on the first picture you presented that attempted to show changing real lengths after I explained how it was wrong.

I don’t even remember posting a picture. What post is this?  Have you now switched to replies to Jaaanosik without indication?

His picture shows a continuously accelerating rigid object (a ruler apparently), much like SP2 at 100g and the first  ~5 light days of rod in front of it. It shows some lines of proper simultaneity for the object, but doesn’t bother with putting units on anything. The picture seems accurate enough.
I haven’t much been paying attention to what Jaaanosik has been trying to convey with this picture, but he used it in post 85 to illustrate length contraction, which it does.
Notice that SP2 would be the left end of the ruler (@) accelerating at 100g, and the right end of the ruler (@') corresponds approximately to the 5 LD mark on SP2’s rod.  It is not accelerating at 100g as evidenced by the different hyperbolic curves of the two lines.  So the two worldlines, accelerating differently, converge in the original frame.  SP1 is not like that since it follows the same curve as the left side of the ruler, so there is no contraction of the distance between them, but the rod in front of SP2 does contract, so 10 LD is insufficient to reach SP1 once they’re moving.

… and therefore exceeds c

As I showed, I was not equating speed and acceleration. You were wrong about what you said but rather than admit such a thing, you switch to another subject.

I was commenting on the bit I quote just above. You didn’t say what exceeded c. Turns out you were talking about proper velocity, but apparently without knowing the term, and once that was clarified, I agreed that yes, that can exceed c.

The subject was that the track would necessarily move in a direction opposite to the cars courtesy of Newton #3. But you skip that part because you cannot deal with the consequences.

It would not since there is not just the one car, but a series of them equally spaced. The vector sum of the forces on the track is zero and it stays put. Surely you know this, and are just trying to nitpick. I was started to back off the troll assessment when I thought you were putting out numbers, but no, you just copied mine and put out comments like this one above.

I have addressed your math earlier and shown just how wrong it is.

You quoted a subset of the exact same numbers, either showing how much you agree with what I’ve said, or you are unable to do the math so you just copied them. Numbers don’t lie.
Then you added a bunch of words unbacked by different numbers.
 

The contraction of the entire ensemble, ships and rod, is only as seen by the inertial observer.

Wrong. It is contracted only relative to a frame in which it is moving.  SP2 is inertial after it shuts down, and yet it is not then contracted in the SP2 frame (N), so that contradicts your statement above. The statement also implies that any accelerating observer will not see SP2 contracted, which is wrong as well. Some observer could be accelerating the other way for instance. SP2 will be contracted relative to him.

To the ships traveling at the same speeds because of identical acceleration histories, contraction does not exist because there is no relative motion.

Agree if you add ‘Relative’ in front of that statement. For instance, I claim no length contraction of the rod in frame N once SP2 has stopped accelerating. Look at the numbers. Don’t take my word for it.  If the rod were contracted, a length 17.1 rod would not span from N coordinates -19.6131 to -2.5123 where SP1 is parked. It needs its full uncontracted length to reach between them. Numbers don’t lie.

Sp1a is an event, not a time.

SP1a is the time on SP1’s clock when SP1 acceleration begins.

OK, so you don’t know what an event is.  An event is an objective point in spacetime, a frame-invariant fact. The location and time of an event is relation with a coordinate system, typically an inertial frame.

Anyway, the coordinates of SP1a is (0, 0) in both frames since we’ve chosen that event as our common origin. That’s two values.  A time is one value.  SP1a occurs at time 0 in both frame, but time zero does not define an event since SP2a also occurs at time 0, at least in frame M, and that is a different event.

Clock A accelerates at 1g for however long it takes it to get to the nearest star and back.  Clock B meanwhile gets put into a cyclotron on Earth and is subjected to 10g of continuous acceleration the whole time Clock A is gone.  Clock A will have logged less time when they reunite, despite having accelerated far less. I didn’t do the numbers, so I can’t say how much.  Can do if you think it matters.  Clock B will age negligibly slower than one not in the cyclotron, depending on the radius of the thing.

Wrong.

Thank you. I’d question my statement if you agreed with me. You’re getting predictable.

Clock B will have gone much faster and been time dilated much more than Clock A and will be way behind Clock A. It will be obvious which one had experienced more acceleration. Where they went or did not go is irrelevant as long as they end up side by side in a common inertial frame. What makes you think otherwise?

Want numbers?  You first this time.  Pick a real distance to a star and say a cyclotron with a 20m radius, which seems reasonable. Otherwise, just more words from you. You seem to have the mistaken impression that the magnitude of time dilation is a function of the magnitude of acceleration, so I picked a counterexample as requested.

No, you showed mixed numbers, partly from the accelerated frame and partly from an inertial frame and assigned different numbers to SP1 and SP2 despite identical acceleration histories.

They’re not mixed. They’re carefully separated in discussions relative to each frame, just like Einstein talks about the frame of the train and the frame of the platform, but with no statement mixing the two frames.
All events are consistent between the two frames. That shows I did it correctly.
If thinking in terms of observers helps you, just consider the N stuff to be relative to some inertial observer stationary the whole time in frame N, watching the ships go from -.8112c to a halt over the course of 4.9082 days each.  All the N coordinates are relative to that observer. So if my numbers are wrong relative to that unaccelerated observer (who is at location 0 the whole time), then give your own numbers.  I never once referenced an accelerated reference frame, so I used nothing but inertial SR mathematics to arrive at my numbers. Feel free to do it your own way.

I have posted corrected numbers.

You mean correct numbers, not corrected, since they were my numbers.

[Jaaanosik]

The observers at the end of the ruler never see any change in the length of the ruler. SP1 and SP2 never see any change in their separation and the position of the rod.

This is not so straight forward.
Are you suggesting a rod between the two spaceships?
If yes, then you need to understand/explain the post #101.
Do you see where is the simultaneity line?

[Halc]

SP1 and SP2 never see any change in their separation and the position of the rod.

This is not so straight forward.
Are you suggesting a rod between the two spaceships?

Yes, the rod was first suggested in post 3 and defined exactly in post 8 where I affixed it to the front of the rear ship rather than having it trail behind the lead ship, which leads to it breaking even if not attached to anything.
So our scenario since post 14 has been a pair of negligible length ships (shorter than about 10 light minutes, the precision of the calculations) initially separated by 10 light days, which accelerate at 100g for 4 days ship time.  The rod in front of the rear ship is 20 light days long, so it extends past the front ship, who therefore at any time can determine the separation between the two by just reading the markings on it.  No need to send time-consuming light signals between the ships.

If yes, then you need to understand/explain the post #101.
Do you see where is the simultaneity line?

Why did you post the same picture twice in that post?

It is sort of appropriate, except it depicts continuous acceleration, whereas our scenario has them stop accelerating after 4 subjective days. M-L claims that the proper distance between the ships will somehow revert to the original 10 light-day separation after they stop accelerating.  Your picture shows continuous acceleration, and thus perhaps doesn't clearly show it.  It also lacks numbers on the ruler, making it difficult to assess the separation in a frame not depicted.
The picture BTW seems entirely accurate to me, even if it doesn't depict our exact scenario.

Based on the most recent post, M-L seems in total denial of the existence of the frame in which both ships are subsequently stopped, as he refuses to compute any numbers for it.  It would of course show that length contraction is real, so I suppose that's why he feels the need to take that stance.

I can think of three possibilities why he would do that:
1) He actually has as little clue about relativity as he presents himself to have. The only relativistic math I've seen in any post is a chart of speed-to-Lorentz-factors.  Never any math applied to a specific scenario except very slow speed examples using Newtonian mathematics.
2) He has a pathological fear of being shown wrong, and refuses to show math that would illustrate it.
I personally love being shown to be wrong, because I'm here to learn, not to be correct.
My numbers could very well have mistakes in them. I did them quickly without double-check. They need interval checks for instance, the first thing I'd do with somebody else's numbers.
3) He knows his relativity quite well, but is a troll getting his jollies from yanking our collective chains.

[Jaaanosik]

SP1 and SP2 never see any change in their separation and the position of the rod.

This is not so straight forward.
Are you suggesting a rod between the two spaceships?

Yes, the rod was first suggested in post 3 and defined exactly in post 8 where I affixed it to the front of the rear ship rather than having it trail behind the lead ship, which leads to it breaking even if not attached to anything.
So our scenario since post 14 has been a pair of negligible length ships (shorter than about 10 light minutes, the precision of the calculations) initially separated by 10 light days, which accelerate at 100g for 4 days ship time.  The rod in front of the rear ship is 20 light days long, so it extends past the front ship, who therefore at any time can determine the separation between the two by just reading the markings on it.  No need to send time-consuming light signals between the ships.

If yes, then you need to understand/explain the post #101.
Do you see where is the simultaneity line?

Why did you post the same picture twice in that post?

It is sort of appropriate, except it depicts continuous acceleration, whereas our scenario has them stop accelerating after 4 subjective days. M-L claims that the proper distance between the ships will somehow revert to the original 10 light-day separation after they stop accelerating.  Your picture shows continuous acceleration, and thus perhaps doesn't clearly show it.  It also lacks numbers on the ruler, making it difficult to assess the separation in a frame not depicted.
The picture BTW seems entirely accurate to me, even if it doesn't depict our exact scenario.

Based on the most recent post, M-L seems in total denial of the existence of the frame in which both ships are subsequently stopped, as he refuses to compute any numbers for it.  It would of course show that length contraction is real, so I suppose that's why he feels the need to take that stance.

I can think of three possibilities why he would do that:
1) He actually has as little clue about relativity as he presents himself to have. The only relativistic math I've seen in any post is a chart of speed-to-Lorentz-factors.  Never any math applied to a specific scenario except very slow speed examples using Newtonian mathematics.
2) He has a pathological fear of being shown wrong, and refuses to show math that would illustrate it.
I personally love being shown to be wrong, because I'm here to learn, not to be correct.
My numbers could very well have mistakes in them. I did them quickly without double-check. They need interval checks for instance, the first thing I'd do with somebody else's numbers.
3) He knows his relativity quite well, but is a troll getting his jollies from yanking our collective chains.

Halc,
I see your point about acceleration and stopping.
This might be one way how to look at it.
Having said that, this is not a uniform acceleration between the two spaceships as per the textbook definition.
It is a correct uniform acceleration/deceleration per individual spaceships though.
Jano

[Halc]

Halc,
I see your point about acceleration and stopping.

No, not stop moving, just cease accelerating. The ships continue at the same inertial speed off to the right, not slowing back to the original frame.

You picture also shows one rigid object accelerating up and down to a stop, not two separate objects accelerating identically.
There's seemingly strain being put on the object in your picture. If it was brittle (not bendy), then it would perhaps break given what being done to it. Hard to tell since the speed never gets high enough to make it obvious.

Having said that, this is not a uniform acceleration between the two spaceships as per the textbook definition.

Not sure what you consider uniform. Our ships accelerate at a constant proper acceleration of 100g for 4 days ship time. That's identical acceleration, not necessarily uniform, depending on your definition of uniform. If it means the object is always stationary along its length in its own frame, then yes, it illustrates that sort of motion.

Your prior picture I think showed uniform acceleration, where the left side of the ruler accelerates a higher g force than the right side.  Those two endpoints are not representative of the motion of our two ships.

[Malamute Lover (OP)]

SP1a [0,0]
SP1 proper acceleration is 100g
SP1 duration of proper acceleration is 4 days
SP1 final proper speed is 1.130 c
SP1 proper acceleration has been continuous, so average proper speed is 0.565 c
SP1 proper distance covered is 2.261 LD
SP1b proper [2.261,4]

SP2a [-10,0]
SP2 proper acceleration is 100g
SP2 duration of proper acceleration is 4 days
SP2 final proper speed is 1.130 c
SP2 proper acceleration has been continuous, so average proper speed is 0.565 c
SP2 proper distance covered is 2.261 LD
SP2b proper [-7.739,4]

2.261 -(-7.739) = 10 LD

Same proper separation between SP1 and SP2 at end of acceleration as at start of acceleration.

The rod has been defined as having constant acceleration along its length. Attached thrusters or whatever.

SP1 end [0,0]
SP1 end proper acceleration is 100g
SP1 end duration of proper acceleration is 4 days
SP1 end final proper speed is 1.130 c
SP1 end proper acceleration has been continuous, so average proper speed is 0.565 c
SP1 end proper distance covered is 2.261 LD
SP1 end at SP1b proper [2.261,4]

SP2 end [0,-10]
SP2 end proper acceleration is 100g
SP2 end duration of proper acceleration is 4 days
SP2 end final proper speed is 1.130 c
SP2 end proper acceleration has been continuous, so average proper speed is 0.565 c
SP2 end proper distance covered is 2.261 LD
SP2 end at SP2b proper [-7.739,4]

2.261 -(-7.739) = 10 LD

Same proper separation between SP1 end of rod and SP2 end of rod at end of acceleration as at start of acceleration. The rod is the same length at the beginning and at the end.

Let us define SP1x as Time = 2 on the SP1 clock

SP1 proper acceleration is 100g
SP1 duration of proper acceleration is 2 days
SP1 proper speed is 0.565 c
SP1 proper acceleration has been continuous, so average proper speed is 0.2825 c
SP1 at SP1x proper distance covered is 0.565 LD
SP1 at SP1x proper [0.565,2]

Let us define SP2x as Time = 2 on the SP2 clock

SP2 proper acceleration is 100g
SP2 duration of proper acceleration is 2 days
SP2 proper speed is 0.565 c
SP2 proper acceleration has been continuous, so average proper speed is 0.2825 c
SP2 at SP2x proper distance covered is 0.565 LD
SP2 at SP2x proper [-9.435,2]


Now the rod

SP1 end proper acceleration is 100g
SP1 end duration of proper acceleration is 2 days
SP1 end final proper speed is 0.565 c
SP1 end proper acceleration has been continuous, so average proper speed is 0.2825 c
SP1 end at SP1x proper distance covered is 0.565 LD
SP1 end at SP1x proper [0.565,2]

The difference between SP1’s proper distance covered and the SP1 end of the rod say SP1x is 0.565 - 0.565 = 0

The SP1 end of the rod is motionless outside SP1’s window for any SP1 time you choose.

Since we have already seen that the final proper separation is the same as the starting separation, the rod cannot be moving after SP1b.

Does the proper length of the rod ever change?

Let Rh be the point on the rod 5 LD forward of SP2 and 5 LD aft of SP1 at t=0 before any acceleration begins.  That is, at the mid-point of the rod. Coordinates are [-5,0]

How long is the rod at the 2 day mark?

Rh proper acceleration is 100g
Rh duration of proper acceleration is 2 days
Rh proper speed is 0.565 c
Rh proper acceleration has been continuous, so average proper speed is 0.2825 c
Rh at 2 days proper distance covered is 0.565 LD
Rh at 2 days proper [-5.435,2]



Distance to SP1 at 2 day mark is 0.565 - (-4.435) = 5 LD

Distance to SP2 at 2 day mark is -4.435 - (-9.435) = 5 LD

The proper length of the rod does not change at any point in time.


As I have been saying all along:

Since both ships have been accelerating at the same rate for the same time in the same direction beginning at an agreed time on synchronized clocks, they are in the same inertial frame and there will be no difference in proper lengths.

If you had your head on straight about basic Relativity concepts this would have been obvious from the start. But you don’t. You blithely mix up proper values and external observer values and think you are saying something meaningful. You’re not.

[Jaaanosik]

Halc,
I see your point about acceleration and stopping.

No, not stop moving, just cease accelerating. The ships continue at the same inertial speed off to the right, not slowing back to the original frame.

You picture also shows one rigid object accelerating up and down to a stop, not two separate objects accelerating identically.
There's seemingly strain being put on the object in your picture. If it was brittle (not bendy), then it would perhaps break given what being done to it. Hard to tell since the speed never gets high enough to make it obvious.

Having said that, this is not a uniform acceleration between the two spaceships as per the textbook definition.

Not sure what you consider uniform. Our ships accelerate at a constant proper acceleration of 100g for 4 days ship time. That's identical acceleration, not necessarily uniform, depending on your definition of uniform. If it means the object is always stationary along its length in its own frame, then yes, it illustrates that sort of motion.

Your prior picture I think showed uniform acceleration, where the left side of the ruler accelerates a higher g force than the right side.  Those two endpoints are not representative of the motion of our two ships.

Then this is the better diagram.

the proper distance between the ships will somehow revert to the original 10 light-day separation after they stop accelerating.

If both spaceship pause acceleration then the distance between the spaceships will not revert to the original 10 light-day interval according to the SR.
Jano

[Halc]

Great. An attempt to obfuscate by specifying everything in proper terms instead of coordinate terms.
Let’s see if the numbers hold up...

SP1a [0,0]
SP1 proper acceleration is 100g
SP1 duration of proper acceleration is 4 days
SP1 final proper speed is 1.130 c
SP1 proper acceleration has been continuous, so average proper speed is 0.565 c
SP1 proper distance covered is 2.261 LD
SP1b proper [2.261,4]

Still using Newtonian mathematics I see. This is wrong. The actual proper speed is 1.387c
Now me just saying that is words, so let me let the numbers speak.
You apparently got the proper speed the Newtonian way by multiplying 980m/sec2 by 345600 seconds which indeed yields a proper speed (dx/dt’) of 1.13c. This corresponds to a coordinate speed (dx/dt) of about .749c, not .8112c.  Doing it this way leads to contradictions:
Suppose we accelerated for 8 days instead of 4. Using your linear method, that would double the proper speed to 2.26c, but that contradicts the relativistic computation of proper speed:
.749c coordinate speed doubled using relativistic addition is .9596c, which has a λ of 3.5558 yielding a proper speed of 3.4123c which contradicts the 2.26c using your method.  Proper speeds don’t add in a linear way like you’re doing. You can’t use Newtonian method of just multiplying proper time by proper acceleration.

The ‘proper distance’ covered (as you’re using the term) is 2.5123 as you had in your prior post, but I see you’ve changed it to 2.261 now.

SP2a [-10,0]
SP2 proper acceleration is 100g
SP2 duration of proper acceleration is 4 days
SP2 final proper speed is 1.130 c
SP2 proper acceleration has been continuous, so average proper speed is 0.565 c
SP2 proper distance covered is 2.261 LD
SP2b proper [-7.739,4]

2.261 -(-7.739) = 10 LD

OK, that much is the same I computed.  In frame M (your figures are all relative to frame M), the ships are always 10 LD apart.
You’re suggesting that a moving 10 LD rod can bridge the 10 LD gap between (your) coordinates -7.739 and 2.261 in frame M. That cannot be unless the rod is not length contracted, so you seem to be in denial of relativistic length contraction. That’s at least one contradiction I can point out without any numbers in N.

Same proper separation between SP1 and SP2 at end of acceleration as at start of acceleration.

You’re using ‘proper separation’ incorrectly.  Yes, the proper separation between the grid markers stationary in frame M is 10 in frame M, but proper separation only applies to mutually stationary things, so that’s only the proper distance between the M-coordinate grid markers, not the proper separation between the ships because they’re now moving in the M frame which is the only frame you’ve referenced in this entire post. The proper separation of the ships hasn’t been computed. Proper separation of the ships is by definition the distance between the ships in the frame in which they’re both stationary, which is frame N. You’ve performed no computations in frame N. so you’ve not computed their proper separation.

The rod has been defined as having constant acceleration along its length.

It most certainly has not. The rod has been defined to be stationary in the frame of SP2 at all times.

Asserting it to have constant acceleration along its length is just words. Back it with mathematics in the frame of the rod, and you’ll see the contradiction that results from this assertion.  Jaaasonik’s picture he keeps posting over and over illustrates the situation exactly, showing a rigid rod accelerating with constant proper length.  It accelerates at a different rate along its length. The picture illustrates that nicely, but you seem to be in denial of any illustrations taken from accepted texts, preferring instead to make up your own facts.

SP1 end [0,0]

What does this mean?  That SP1 has gone nowhere in 4 days?  This whole section lacks a frame reference, so it’s not immediately clear what you’re trying to convey here.

SP1 end proper acceleration is 100g
SP1 end duration of proper acceleration is 4 days
SP1 end final proper speed is 1.130 c
SP1 end proper acceleration has been continuous, so average proper speed is 0.565 c
SP1 end proper distance covered is 2.261 LD
SP1 end at SP1b proper [2.261,4]

SP2 end [0,-10]
SP2 end proper acceleration is 100g
SP2 end duration of proper acceleration is 4 days
SP2 end final proper speed is 1.130 c
SP2 end proper acceleration has been continuous, so average proper speed is 0.565 c
SP2 end proper distance covered is 2.261 LD
SP2 end at SP2b proper [-7.739,4]

2.261 -(-7.739) = 10 LD

This just seems to be a repeat of the numbers above, but with the word ‘end’ stuck in at various places.  It’s all in frame M, so indeed, you seem to be asserting that between start and end, neither ship has changed coordinates, which means it hasn’t moved. I don’t think you mean that, but it is totally unclear what you’re trying to convey with this repeat of the same numbers. They’re all still mostly proper numbers relative to frame M, same as the first time.

Same proper separation between SP1 end of rod and SP2 end of rod at end of acceleration as at start of acceleration. The rod is the same length at the beginning and at the end.

Repeating the numbers and repeating a wrong conclusion doesn’t make it right the second time. You’re still not computing proper distance correctly. Look up the definition of it.

Let us define SP1x as Time = 2 on the SP1 clock

An event presumably?

SP1 proper acceleration is 100g
SP1 duration of proper acceleration is 2 days
SP1 proper speed is 0.565 c
SP1 proper acceleration has been continuous, so average proper speed is 0.2825 c
SP1 at SP1x proper distance covered is 0.565 LD
SP1 at SP1x proper [0.565,2]
Let us define SP2x as Time = 2 on the SP2 clock

SP2 proper acceleration is 100g
SP2 duration of proper acceleration is 2 days
SP2 proper speed is 0.565 c
SP2 proper acceleration has been continuous, so average proper speed is 0.2825 c
SP2 at SP2x proper distance covered is 0.565 LD
SP2 at SP2x proper [-9.435,2]

This is all Newtonian math, which has been shown above to lead to contradictions.  The separation between them in M is still 10 of course, even with these wrong numbers. But that’s the proper distance of the grid markers in M, not the proper distance between the ships since A) the ships are not stationary, and B) no frame has been demonstrated to exist where both ships are simultaneously moving at this half-way speed.

Now the rod

SP1 end proper acceleration is 100g

Again I have no idea how this statement is distinct from “SP1 proper acceleration is 100g”. I don’t know what ‘end’ means inserted in like that in all these statements.

SP1 end duration of proper acceleration is 2 days
SP1 end final proper speed is 0.565 c
SP1 end proper acceleration has been continuous, so average proper speed is 0.2825 c
SP1 end at SP1x proper distance covered is 0.565 LD
SP1 end at SP1x proper [0.565,2]

The difference between SP1’s proper distance covered and the SP1 end of the rod say SP1x is 0.565 - 0.565 = 0

You’re assuming the end of the rod is accelerating at 100g, which it cannot be. Compute the location of the rod in frame N and this assumption will lead to contradictions. You perhaps know this, because you avoid doing such computations like the plague. If your assertions were correct, the numbers would show them to be since they’d be self consistent. Actually, I just think you’re not up to the math since all I’ve ever seen you use is Newtonian physics.

The SP1 end of the rod is motionless outside SP1’s window for any SP1 time you choose.

My numbers in N (with which no inconsistency has been demonstrated) show otherwise. Your complete lack of numbers relative to N lend zero support to these mere words. This seems to the main point of contention, so you really need to back this one up. How far out of sync are the ship clocks in frame in which both ships are now stationary?

Let Rh be the point on the rod 5 LD forward of SP2 and 5 LD aft of SP1 at t=0 before any acceleration begins.  That is, at the mid-point of the rod. Coordinates are [-5,0]

Ooh, actual coordinates of an event. Things are improving.

How long is the rod at the 2 day mark?

Frame dependent question.

Rh proper acceleration is 100g

You’ve not established that with any numbers in the frame of the rod. So your argument falls apart here.
The ‘ruler’ in Jaaasonik’s picture shows a rod about exactly that length of about 5 LD at 100g. That means the left edge of the yellow region is SP2 and the right edge is Rh, which according to the textbook from which that illustration was taken, has significantly lower acceleration. You seem to have been denying the textbooks when it comes to SR, so that leaves your numbers to back your assertions. Still waiting for the N numbers.

The proper length of the rod does not change at any point in time.

By definition, yes. An actual statement with which I agree. That sort of wrecks your batting average, no?

Since both ships have been accelerating at the same rate for the same time in the same direction beginning at an agreed time on synchronized clocks, they are in the same inertial frame and there will be no difference in proper lengths.

But there will be a change in their proper separation, and the rod which did not change its proper length will not be able to span this new proper separation. The proper length of anything cannot change unless the object is deformed. It would then not be rigid, and we’ve defined all our objects to be rigid.

If both spaceship pause acceleration then the distance between the spaceships will not revert to the original 10 light-day interval according to the SR.

The distance between them is frame dependent.

[Malamute Lover (OP)]

The numbers done the right way.

All the numbers are the same as mine, if only the frame M coordinates. You offer no frame N coordinates at all.

If you are correcting something, I expected different numbers. It appears you just copied my stuff, which contradicts your assertion that I had not done them correctly.

We each say the other is wrong, so it is a he-said/she-said sort of situation. But numbers don’t lie. If my numbers are wrong, you need to produce different ones. Producing the same ones means you are in agreement with my assessment of the situation. My numbers show the string breaking as there is over 17 light-days of proper separation between the ships after acceleration. You’ve not shown otherwise since you don’t compute the proper distance between the ships at all.

You asked for proper numbers, as seen by SP1 and SP2. As can be clearly seen from those numbers, at all times the proper lengths and location of SP1, the rod and SP2 never change. The rod stops when SP1 stops. The proper separation never changes. Your numbers are wrong because you mix proper numbers and external observer numbers. It is clear that you have no knowledge of how Relativity Theory works.

SP1a is at x=0,t=0 at the start (or 0,0).  SP2a is at (-10, 0)    Units in (light days, days)
After end of acceleration,

SP1b coordinates are (2.5123, 4.9082) relative to an unaccelerated observer in the original inertial frame pf SP1 and SP2

SP2b coordinates are (-7.4877, 4.9082) relative to an unaccelerated observer in the original inertial frame of SP1 and SP2

Both ships are moving at .8112c which has a Lorentz factor λ of 1.71008 relative to an unaccelerated observer in the original inertial frame of SP1 and SP2

All in agreement, and seemingly copied from my post, except for some of the rearangement of the wording.

You are still missing the point. The λ of SP1, the rod and SP2 are identical as seen by an unaccelerated observer. From their own viewpoints, all lengths are the same. And their relative λ values are zero. You apply the λ only to SP2. That is wrong.  You cannot mix, you can only match.

But since both ships have been accelerating at the same rate for the same time in the same direction beginning at an agreed time on synchronized clocks, they are in the same inertial frame and there is no difference in their λ.

Agree. They both are moving at the same speed, and thus are stationary relative to each other after both have finished accelerating.

And as I have shown in a recent post, within their own common reference frame, no separations have changed. Relative to an unaccelerated observer, all separations have changed to the same degree. The rod fits between SP1 and SP2 exactly as before.

Neither is there any difference in their proper separation from the start, although that unaccelerated observer sees the whole ensemble contracted.

Words, which lie. Show numbers, which don’t.  Without that, all I can do is assert that you’re wrong, and my numbers show it, and your lack of different numbers lends zero support to your assertion. I see no computation of their proper separation after acceleration.[/quote]

I have provided lots of numbers about proper separation. It is identical at all times.

The SP1 clock reads 4 at SP1b and the SP2 clock reads 4 at SP2b.

Yes, kind of by definition, since that’s exactly the flight plan.

At time 4.9082 relative to an unaccelerated observer in the original inertial frame of SP1 and SP2,

That took a whole line to say “At 4.9082 in M”. That’s why I gave the frames names, so I didn’t have to type all that each time.

I was emphasizing that SP1, rod and SP2 are all in one reference frame and the observer is in another, a point you are having a problem with. BTW did you know that SP1, rod, SP2 and M were all originally accelerated from frame Z before t=0? What separation change between SP1 and SP2 does observer Z see? After all he sees them as going even faster with an even bigger λ. You are insisting on a change in proper separation as seen by SP1 and SP2. So who is right, M or Z?

which is time 4 according to synchronized clocks on the two ships, both ships and the rod are stationary relative to each other.

They’re both moving at the same speed in M at that time in M. That’s different than saying they’re stationary relative to each other, because you have to use the ship frame to state that they’re stationary at that time. I go so far as to agree that relative to the ships at event SP2b, both ships and the rod are stationary relative to each other, but I cannot agree with your statement above. Again, without numbers in N, your assertion is just words. I have no coordinates in N to work with, so I cannot show you the contradictions that result from your assertions.

I have provided numbers. The fact that you needed numbers to understand the result shows that you are seriously lacking in comprehension of the basic concepts of Relativity Theory.

Since both ships have stopped accelerating at the same time, they are in a common inertial frame

They’re ‘in’ all the frames, but each is stationary in only one. I think you meant that, but being clear is nice.  It’s because they both accelerate to the same speed that makes them stationary in the same frame after acceleration. When they do it or how long they take to do it makes no difference, so long as net acceleration from M is .8112c.  We’re more controlled than that of course, but them being stationary relative to each other has nothing to do with when they stopped accelerating, especially since the ‘when’ part is completely frame dependent.

They are stationary in the common inertial frame that they (SP1, rod, SP2) are all in, as I have demonstrated in the recent proper numbers post. They are not moving relative to each other at any time. Proper separation and proper length of the rod never change.

Relativity of Simultaneity refers to the difference in apparent time separation between two events as viewed by two observers not in the same inertial frame.

It has to do with the frame, and not the observers, and again, all observers are in all frames. But yes, something like how you word it.

Wrong. An observer is in the frame he is in. He is not in any other frame. If he were in all frames, then he would have all length contractions, and all time dilations and all mass-energy values simultaneously.  Not to mention that he would be traveling in all directions at all speeds.

The frame an observer is in is simply the one and only proper frame he is in where he will measure the speed of light to be the textbook value. He might be observed from all sorts of other frames but those are relative, not objective. This is why it is called Relativity Theory.

You don’t even talk the talk, but you want to claim you walk the walk.

A light is turned on at point A and at Point B at Midnight UTC (synchronized clocks). Observer Alice is on a point midway on a straight line between A and B. Alice sees the light from both A and B at the same time.

No frame reference. I presume the clocks are stationary in the frame in which they are in sync, and Alice is also stationary in this same frame. Just making sure.

That Alice is in the midpoint and see the lights at the same time tells us that they are all in a common inertial frame. No need to specify. Seems that you do not understand that point.  Oh wait! You don’t! You think that the proper length of the rod will contract. Or stretch. Which one was it again?

Bob is on a fast motorcycle on that straight line heading toward A and passes Alice at Midnight UTC on Alice’s watch. Bob will see the light from A earlier than the light from B. A and B send each other information about the time they turned the lights on and when they saw the light from the other. Since they have synchronized clocks, A and B will then agree that they turned their lights on at the same time, even though each saw the other’s light go on later than his own.

If they agree on the time delay value, they know they are in a common inertial frame and can even calculate the proper separation.

They already know they were stationary in the one frame, so the subsequent timings just confirm again that those clocks are indeed synced in that frame, yes.

No, A and B did not know for sure that they were in the same inertial frame. Alice knew because she was in the midpoint because repeated readings in each direction with her laser range finder always had the same value.  The lightkeepers at A and B did not know until they compared notes.

You didn’t mention what Bob concludes since he doesn’t know where in his frame the light pulses originated, and therefore seems unable to assess which one flashed first. The one might have been before the other, or just closer. So what’s the purpose of Bob here?  A stationary observer off to one side of Alice would similarly observe one pulse before the other. RoS seems not to have been well demonstrated with this example.

Bob sees the light flashes at different times. He has no idea of how far off they are or if they are synchronized or moving (only Alice knows before the fact). He only knows that he saw them at different times. His observation is that they were not simultaneous.

You didn’t label any events, but there are the two events where the light pulses were emitted, and various observation events.  RoS says that the two emission events, if simultaneous in Alice’s frame are not simultaneous in Bob’s frame, even if that wasn’t demonstrated by your example.

A flashed at Midnight UTC.  B flashed at Midnight UTC. Alice observed both flashes simultaneously at some time after Midnight UTC, the exact time difference being irrelevant. Bob passed Alice at Midnight UTC. His exact speed is irrelevant as long as it is non-zero with respect to Alice and that it is in the direction of A.

If you need exact numbers and math to understand this, then you do not understand the principle behind Relativity of Simultaneity and do not know how to apply it properly.

[Malamute Lover (OP)]

The numbers done the right way.

All the numbers are the same as mine, if only the frame M coordinates. You offer no frame N coordinates at all.

If you are correcting something, I expected different numbers. It appears you just copied my stuff, which contradicts your assertion that I had not done them correctly.

We each say the other is wrong, so it is a he-said/she-said sort of situation. But numbers don’t lie. If my numbers are wrong, you need to produce different ones. Producing the same ones means you are in agreement with my assessment of the situation. My numbers show the string breaking as there is over 17 light-days of proper separation between the ships after acceleration. You’ve not shown otherwise since you don’t compute the proper distance between the ships at all.

You asked for proper numbers, as seen by SP1 and SP2. As can be clearly seen from those numbers, at all times the proper lengths and location of SP1, the rod and SP2 never change. The rod stops when SP1 stops. The proper separation never changes. Your numbers are wrong because you mix proper numbers and external observer numbers. It is clear that you have no knowledge of how Relativity Theory works.

SP1a is at x=0,t=0 at the start (or 0,0).  SP2a is at (-10, 0)    Units in (light days, days)
After end of acceleration,

SP1b coordinates are (2.5123, 4.9082) relative to an unaccelerated observer in the original inertial frame pf SP1 and SP2

SP2b coordinates are (-7.4877, 4.9082) relative to an unaccelerated observer in the original inertial frame of SP1 and SP2

Both ships are moving at .8112c which has a Lorentz factor λ of 1.71008 relative to an unaccelerated observer in the original inertial frame of SP1 and SP2

All in agreement, and seemingly copied from my post, except for some of the rearangement of the wording.

You are still missing the point. The λ of SP1, the rod and SP2 are identical as seen by an unaccelerated observer. From their own viewpoints, all lengths are the same. And their relative λ values are zero. You apply the λ only to SP2. That is wrong.  You cannot mix, you can only match.

But since both ships have been accelerating at the same rate for the same time in the same direction beginning at an agreed time on synchronized clocks, they are in the same inertial frame and there is no difference in their λ.

Agree. They both are moving at the same speed, and thus are stationary relative to each other after both have finished accelerating.

And as I have shown in a recent post, within their own common reference frame, no separations have changed. Relative to an unaccelerated observer, all separations have changed to the same degree. The rod fits between SP1 and SP2 exactly as before.

Neither is there any difference in their proper separation from the start, although that unaccelerated observer sees the whole ensemble contracted.

Words, which lie. Show numbers, which don’t.  Without that, all I can do is assert that you’re wrong, and my numbers show it, and your lack of different numbers lends zero support to your assertion. I see no computation of their proper separation after acceleration.

I have provided lots of numbers about proper separation. It is identical at all times.

The SP1 clock reads 4 at SP1b and the SP2 clock reads 4 at SP2b.

Yes, kind of by definition, since that’s exactly the flight plan.

At time 4.9082 relative to an unaccelerated observer in the original inertial frame of SP1 and SP2,

That took a whole line to say “At 4.9082 in M”. That’s why I gave the frames names, so I didn’t have to type all that each time.

I was emphasizing that SP1, rod and SP2 are all in one reference frame and the observer is in another, a point you are having a problem with. BTW did you know that SP1, rod, SP2 and M were all originally accelerated from frame Z before t=0? What separation change between SP1 and SP2 does observer Z see? After all he sees them as going even faster with an even bigger λ. You are insisting on a change in proper separation as seen by SP1 and SP2. So who is right, M or Z?

which is time 4 according to synchronized clocks on the two ships, both ships and the rod are stationary relative to each other.

They’re both moving at the same speed in M at that time in M. That’s different than saying they’re stationary relative to each other, because you have to use the ship frame to state that they’re stationary at that time. I go so far as to agree that relative to the ships at event SP2b, both ships and the rod are stationary relative to each other, but I cannot agree with your statement above. Again, without numbers in N, your assertion is just words. I have no coordinates in N to work with, so I cannot show you the contradictions that result from your assertions.

I have provided numbers. The fact that you needed numbers to understand the result shows that you are seriously lacking in comprehension of the basic concepts of Relativity Theory.

Since both ships have stopped accelerating at the same time, they are in a common inertial frame

They’re ‘in’ all the frames, but each is stationary in only one. I think you meant that, but being clear is nice.  It’s because they both accelerate to the same speed that makes them stationary in the same frame after acceleration. When they do it or how long they take to do it makes no difference, so long as net acceleration from M is .8112c.  We’re more controlled than that of course, but them being stationary relative to each other has nothing to do with when they stopped accelerating, especially since the ‘when’ part is completely frame dependent.

They are stationary in the common inertial frame that they (SP1, rod, SP2) are all in, as I have demonstrated in the recent proper numbers post. They are not moving relative to each other at any time. Proper separation and proper length of the rod never change.

Relativity of Simultaneity refers to the difference in apparent time separation between two events as viewed by two observers not in the same inertial frame.

It has to do with the frame, and not the observers, and again, all observers are in all frames. But yes, something like how you word it.

Wrong. An observer is in the frame he is in. He is not in any other frame. If he were in all frames, then he would have all length contractions, and all time dilations and all mass-energy values simultaneously.  Not to mention that he would be traveling in all directions at all speeds.

The frame an observer is in is simply the one and only proper frame he is in where he will measure the speed of light to be the textbook value. He might be observed from all sorts of other frames but those are relative, not objective. This is why it is called Relativity Theory.

You don’t even talk the talk, but you want to claim you walk the walk.

A light is turned on at point A and at Point B at Midnight UTC (synchronized clocks). Observer Alice is on a point midway on a straight line between A and B. Alice sees the light from both A and B at the same time.

No frame reference. I presume the clocks are stationary in the frame in which they are in sync, and Alice is also stationary in this same frame. Just making sure.

That Alice is in the midpoint and see the lights at the same time tells us that they are all in a common inertial frame. No need to specify. Seems that you do not understand that point.  Oh wait! You don’t! You think that the proper length of the rod will contract. Or stretch. Which one was it again?

Bob is on a fast motorcycle on that straight line heading toward A and passes Alice at Midnight UTC on Alice’s watch. Bob will see the light from A earlier than the light from B. A and B send each other information about the time they turned the lights on and when they saw the light from the other. Since they have synchronized clocks, A and B will then agree that they turned their lights on at the same time, even though each saw the other’s light go on later than his own.

If they agree on the time delay value, they know they are in a common inertial frame and can even calculate the proper separation.

They already know they were stationary in the one frame, so the subsequent timings just confirm again that those clocks are indeed synced in that frame, yes.

No, A and B did not know for sure that they were in the same inertial frame. Alice knew because she was in the midpoint because repeated readings in each direction with her laser range finder always had the same value.  The lightkeepers at A and B did not know until they compared notes.

You didn’t mention what Bob concludes since he doesn’t know where in his frame the light pulses originated, and therefore seems unable to assess which one flashed first. The one might have been before the other, or just closer. So what’s the purpose of Bob here?  A stationary observer off to one side of Alice would similarly observe one pulse before the other. RoS seems not to have been well demonstrated with this example.

Bob sees the light flashes at different times. He has no idea of how far off they are or if they are synchronized or moving (only Alice knows before the fact). He only knows that he saw them at different times. His observation is that they were not simultaneous.

You didn’t label any events, but there are the two events where the light pulses were emitted, and various observation events.  RoS says that the two emission events, if simultaneous in Alice’s frame are not simultaneous in Bob’s frame, even if that wasn’t demonstrated by your example.

A flashed at Midnight UTC.  B flashed at Midnight UTC. Alice observed both flashes simultaneously at some time after Midnight UTC, the exact time difference being irrelevant. Bob passed Alice at Midnight UTC. His exact speed is irrelevant as long as it is non-zero with respect to Alice and that it is in the direction of A.

If you need exact numbers and math to understand this, then you do not understand the principle behind Relativity of Simultaneity and do not know how to apply it properly.

[Malamute Lover (OP)]

I see that you have given up on the first picture you presented that attempted to show changing real lengths after I explained how it was wrong.

I don’t even remember posting a picture. What post is this?  Have you now switched to replies to Jaaanosik without indication?

OOPS! You are right. I was preparing on reply and got distracted by some RLBS and when I went back, I started a different reply but got the two mixed up. It was Jano who posted the pictures. Apoplectic Apologies.

If you want to see the other picture Jano posted here it is.
https://i.imgur.com/XSs3QXa.png
It relates a fixed observer and a moving observer.

His picture shows a continuously accelerating rigid object (a ruler apparently), much like SP2 at 100g and the first  ~5 light days of rod in front of it. It shows some lines of proper simultaneity for the object, but doesn’t bother with putting units on anything. The picture seems accurate enough.
I haven’t much been paying attention to what Jaaanosik has been trying to convey with this picture, but he used it in post 85 to illustrate length contraction, which it does.
Notice that SP2 would be the left end of the ruler (@) accelerating at 100g, and the right end of the ruler (@') corresponds approximately to the 5 LD mark on SP2’s rod.  It is not accelerating at 100g as evidenced by the different hyperbolic curves of the two lines.  So the two worldlines, accelerating differently, converge in the original frame.  SP1 is not like that since it follows the same curve as the left side of the ruler, so there is no contraction of the distance between them, but the rod in front of SP2 does contract, so 10 LD is insufficient to reach SP1 once they’re moving.

Consider the ruler to be the rod with observers at each end (SP1 and SP2). The length of the ruler relative to the observer at the origin (0) changes as the ruler accelerates uniformly (constant rate).  But the length of the ruler as determined by the observers at each of the ruler ends is determined by the width of a rectangle whose opposite corners are where the dashed lines intercept the colored area. That is constant. The observers on SP1 and SP2 never see any change in the length despite ongoing acceleration. It is always 10 LD. It is the observer at the origin who sees changing length contraction as the relative speeds differ. 

BTW can anyone explain why the shape of the colored area first grows then shrinks as time goes on? Time is the vertical axis, increasing from bottom to top. The answer is somewhat subtle.

… and therefore exceeds c

As I showed, I was not equating speed and acceleration. You were wrong about what you said but rather than admit such a thing, you switch to another subject.

I was commenting on the bit I quote just above. You didn’t say what exceeded c. Turns out you were talking about proper velocity, but apparently without knowing the term, and once that was clarified, I agreed that yes, that can exceed c.

Here is what I said.
The traveling twin thinks he is continuing to accelerate at the same rate – which he can feel - and therefore exceeds c.

The traveling twin feels the continued acceleration and thinks he exceeds c. The reference frame is obvious – the traveling twin. And it was obviously proper acceleration and speed since no other frame was mentioned. This was all obvious but you nitpick because you are having trouble with your arguments and need to distract from that.

The subject was that the track would necessarily move in a direction opposite to the cars courtesy of Newton #3. But you skip that part because you cannot deal with the consequences.

It would not since there is not just the one car, but a series of them equally spaced. The vector sum of the forces on the track is zero and it stays put. Surely you know this, and are just trying to nitpick. I was started to back off the troll assessment when I thought you were putting out numbers, but no, you just copied mine and put out comments like this one above.

Completely and totally wrong. Each car is pushing the track in the opposite direction from its motion. Adding more cars does not change that.

Imagine a car on a gravel road that is not getting much traction. It is throwing gravel opposite to its direction of motion. Put that gravel road on the surface of a flat circular track. The track is supported on ball bearings. Put a bunch of cars on the track. Do they always throw gravel opposite to the direction of their individual motions? Yes. Are they all imparting energy to the gravel opposite to the direction they are moving?  Yes. If they are all traveling counterclockwise are they not all throwing gravel clockwise? Yes.

Now instead of gravel, substitute macadam. Are the cars all imparting energy into the macadam opposite to their individual directions of motion? Yes. The track is supported on ball bearings, so it is going to move in the opposite direction from the cars. If the cars are going counterclockwise, the track is going to move clockwise.  How fast it is going to go will depend on the relative mass of the track and the cars because angular momentum is conserved. And Newton #3 is validated.

Trying to apply math without first understanding the concepts behind it is going to give wrong results because the math will be misapplied.

I have addressed your math earlier and shown just how wrong it is.

You quoted a subset of the exact same numbers, either showing how much you agree with what I’ve said, or you are unable to do the math so you just copied them. Numbers don’t lie.
Then you added a bunch of words unbacked by different numbers.

I have provided lots of numbers about proper acceleration, length etc. Deal with it.
 

The contraction of the entire ensemble, ships and rod, is only as seen by the inertial observer.

Wrong. It is contracted only relative to a frame in which it is moving.  SP2 is inertial after it shuts down, and yet it is not then contracted in the SP2 frame (N), so that contradicts your statement above. The statement also implies that any accelerating observer will not see SP2 contracted, which is wrong as well. Some observer could be accelerating the other way for instance. SP2 will be contracted relative to him.

WRONG. Contraction is relative to observers in other frames with the degree of contraction related to speed differences in accordance with the Lorentz formula. An observer never sees contraction in his own environment because the speed difference is zero. The contraction of the entire ensemble will only be seen by the observer who never accelerated with respect to SP1,rod,SP2. Since is said the[/i[ inertial observer, there is no doubt about who was meant, the same one we have been discussing all along. Different observers at different relative speeds will see different contraction factors. SP1,rod,SP2 will not be contracted with respect to themselves because there is never any speed difference among them. Nitpicking again to avoid dealing with the actual arguments.

To the ships traveling at the same speeds because of identical acceleration histories, contraction does not exist because there is no relative motion.

Agree if you add ‘Relative’ in front of that statement. For instance, I claim no length contraction of the rod in frame N once SP2 has stopped accelerating. Look at the numbers. Don’t take my word for it.  If the rod were contracted, a length 17.1 rod would not span from N coordinates -19.6131 to -2.5123 where SP1 is parked. It needs its full uncontracted length to reach between them. Numbers don’t lie.

Sp1a is an event, not a time.

SP1a is the time on SP1’s clock when SP1 acceleration begins.

OK, so you don’t know what an event is.  An event is an objective point in spacetime, a frame-invariant fact. The location and time of an event is relation with a coordinate system, typically an inertial frame.

SP1a time is obviously the time at event SP1a. But since you want to be ultraprecise…

There are no objective points in spacetime. SP1 and SP2 can be inertial with respect to each other but could be moving with respect to another inertial observer. There is no Newtonian Absolute Space and no Newtonian Absolute Time. There are only coordinates of convenience relative to some observer. An absolute distance in spacetime between events can be calculated in SR, although the general opinion among relativity types is that even this is simply a handy tool for calculation purposes and has no physical significance. In GR it is not as simple as the complex Pythagorean Theorem used in SR (where time is in i units.  The possibility of ongoing acceleration screws the pooch. And in the real world there is always a gravitational gradient.

Clock A accelerates at 1g for however long it takes it to get to the nearest star and back.  Clock B meanwhile gets put into a cyclotron on Earth and is subjected to 10g of continuous acceleration the whole time Clock A is gone.  Clock A will have logged less time when they reunite, despite having accelerated far less. I didn’t do the numbers, so I can’t say how much.  Can do if you think it matters.  Clock B will age negligibly slower than one not in the cyclotron, depending on the radius of the thing.

Wrong.

Thank you. I’d question my statement if you agreed with me. You’re getting predictable.

Clock B will have gone much faster and been time dilated much more than Clock A and will be way behind Clock A. It will be obvious which one had experienced more acceleration. Where they went or did not go is irrelevant as long as they end up side by side in a common inertial frame. What makes you think otherwise?

Want numbers?  You first this time.  Pick a real distance to a star and say a cyclotron with a 20m radius, which seems reasonable. Otherwise, just more words from you. You seem to have the mistaken impression that the magnitude of time dilation is a function of the magnitude of acceleration, so I picked a counterexample as requested.

Clocks can only be compared when they are nearby in a common inertial frame. To get to a given point and back at a constant g factor means accelerating to the midpoint then decelerating, then accelerating again in the other direction to the midpoint then decelerating again ending up in the place and inertial frame where the trip started. Constant acceleration for half the trip and constant deceleration for half the trip.

The acceleration will slow the onboard clock relative to an inertial observer (in a common inertial frame with Earth) as the relative speed between them increases. The deceleration will speed up the clock relative to that inertial observer as the relative speed between them decreases eventually bringing the relative clock rate back to that of the original inertial frame.

 In the following scenarios, all accelerations are proper, as felt by the accelerating entity, and unless specified otherwise all times are as per an unaccelerated clock on Earth.

Let us have a clock accelerate at 1 g for 1 year then decelerate at 1 g  for 1 year . The clock will then reverse direction, accelerate at 1 g  for 1 year then decelerate at 1 g  for 1 year.

To make the situation symmetrical, let us have a clock in a cyclotron accelerate at 10 g for 1 year, decelerate at 10 g for 1 year, accelerate at 10 g for 1 year and decelerate at 10 g for 1 year.

What elapsed time do each of the traveling clocks show as compared to the Earth clock?

The formula for this is:

t = c/a*ASINH(a*T/c)

Where:

t is proper time of the accelerated clock
c is lightspeed
a is proper acceleration
T is unaccelerated clock time (Earth time)

The Earth clock elapsed time is 4 years.

The 1 g acceleration clock elapsed time is 3.50 years

The 10 g acceleration clock elapsed time is 1.17 years

The 10 g acceleration clock shows the least elapsed time indicating that it experienced the most proper acceleration.
 

No, you showed mixed numbers, partly from the accelerated frame and partly from an inertial frame and assigned different numbers to SP1 and SP2 despite identical acceleration histories.

They’re not mixed. They’re carefully separated in discussions relative to each frame, just like Einstein talks about the frame of the train and the frame of the platform, but with no statement mixing the two frames.
All events are consistent between the two frames. That shows I did it correctly.
If thinking in terms of observers helps you, just consider the N stuff to be relative to some inertial observer stationary the whole time in frame N, watching the ships go from -.8112c to a halt over the course of 4.9082 days each.  All the N coordinates are relative to that observer. So if my numbers are wrong relative to that unaccelerated observer (who is at location 0 the whole time), then give your own numbers.  I never once referenced an accelerated reference frame, so I used nothing but inertial SR mathematics to arrive at my numbers. Feel free to do it your own way.

Nope. You applied Lorentz time dilation to SP2 but not to SP1. Mixed numbers.

I have posted corrected numbers.

You mean correct numbers, not corrected, since they were my numbers.

No. I corrected your erroneous mismatch in time values for SP1 and SP2. You were mixing numbers from different reference frames. And once more, look at the proper distance numbers I posted showing that all distances are the same not just at the beginning and the end but during acceleration.

[Jaaanosik]

Malamute Lover,
The acceleration at the front of the spaceship or rod is different from the acceleration at the back.
As per the textbook, FYI,
Jano

[Halc]

These post are getting too long, so I will in general not respond to comments not backed by the language of science, which is mathematics.  If you assert that my numbers are wrong, either show a self-inconsistency with them (as I have done for yours), or show corrected number, which you have declined. Showing an inconsistency with your assertions carries no weight with me since I find your assertions to be consistently wrong.

You asked for proper numbers, as seen by SP1 and SP2.

I asked for corrected numbers, since you asserted mine to be wrong. You corrected not a one of them, so by the language that doesn’t lie, I assume you are in full agreement with them.

As can be clearly seen from those numbers, at all times the proper lengths and location of SP1, the rod and SP2 never change.

SP1 accelerated at 100g for 4 days and its proper location never changed?  I did see that at one point and pointed out the absurdity of that.
Anyway, of the new numbers you put out, I showed them to be self-inconsistent.

You are still missing the point. The λ of SP1, the rod and SP2 are identical as seen by an unaccelerated observer.

My numbers show otherwise. You lack of corrected numbers (or demonstration of self-inconsistency of my numbers) shows that you are in fact in agreement with this. Numbers don’t lie.

I have provided lots of numbers about proper separation. It is identical at all times.

You provided a small paragraph of what are claimed to be proper numbers and copied them (except for x being offset by -10 for SP2) seven times. The numbers have been shown to be self-inconsistent (same mistake in each).  You used Newtonian math to generate all those numbers.  This seems to be the limit of your ability to produce numbers, since that’s all I’ve ever seen from you if you don’t count the numbers copied unchanged from my post. You’re full of Einstein slogans, but you seem totally incapable of putting any of it into practice, always reaching for Newton’s mathematics instead.

I was emphasizing that SP1, rod and SP2 are all in one reference frame and the observer is in another

They are all in every reference frame. One cannot exit a reference frame.

You are insisting on a change in proper separation as seen by SP1 and SP2. So who is right, M or Z?

Proper separation isn’t something ‘as seen by’ anybody. It is objective. My numbers show the proper separation between the ships to be 17.1 after they both cease accelerating. Your lack of computing their proper separation shows you are in full agreement. The only thing you computed was the proper distance between the frame-M grid marker at SP1 and the frame-M grid marker at SP2, which is indeed always distance 10 at any time in M (and only in M). Since the ships are not stationary in that frame, you’ve not computed their proper separation. I have.

I have provided numbers.

You did. I showed them to be self-inconsistent.

They are stationary in the common inertial frame that they (SP1, rod, SP2) are all in

The ships are under acceleration at times, so they are not always stationary in that common frame N. You’ve not provided numbers as to when those ships become stationary in that frame, so I assume you agree with mine. My numbers show their proper separation to be 17, so you agree with that as well.

An observer is in the frame he is in. He is not in any other frame.

Then it would be wrong to say Fred is moving at 20 m/sec relative to me since he’s not in my frame at all. He must have blinked out of existence. This seems to be easily falsified by the simple observation of watching cars (with observers) go by me as I stand by the street.

If he were in all frames, then he would have all length contractions

His length contraction relative to a given frame F is a function of his velocity in F, not his existence in F.  Relative to a different frame G, yes, he’d have a different length contraction in that frame, but he’d be in both those frames or he’d not have a length at all.

Not to mention that he would be traveling in all directions at all speeds.

He is indeed traveling at one velocity in frame F and another in frame G. Not at all speeds. He can’t be moving faster than light in any inertial frame.

If you want to see the other picture Jano posted here it is.

It relates a fixed observer and a moving observer.

I like that one since it includes some inertial coordinates for the grid (‘frame M’), and clock readings for the two moving observers. It doesn’t entirely mirror our scenario since the red observer is accelerating at a far higher rate than is the orange observer. They are defined to be fixed with respect to each other, which means they maintain a constant proper separation (of 1.5 in this case).  Our two ships accelerate at the same rate, and so do not maintain a constant proper separation. They are not comoving with each other like the observers in the picture, despite your insistence otherwise.

Consider the ruler to be the rod with observers at each end (SP1 and SP2).

No. Our ships are separate accelerating objects. If they stay fixed relative to each other (as the picture above depicts), then SP1 would be accelerating far less than SP2.
If you want to assert otherwise, then pony up the number so I can tear them apart.

BTW can anyone explain why the shape of the colored area first grows then shrinks as time goes on?

Yes. The graph is depicted in some inertial frame, and the accelerating ruler becomes stationary at only one time in that frame, and is moving in all others. So only when it is stationary in that frame does the ruler project its full proper length onto that frame.

There is no Newtonian Absolute Space and no Newtonian Absolute Time. There are only coordinates of convenience relative to some observer.

And yet you refuse to compute the coordinates of convenience relative to the ships stationary in frame N.

About the cyclotron:

Clock A accelerates at 1g for however long it takes it to get to the nearest star and back.  Clock B meanwhile gets put into a cyclotron on Earth and is subjected to 10g of continuous acceleration the whole time Clock A is gone.  Clock A will have logged less time when they reunite, despite having accelerated far less.

Wrong.  Clock B will have gone much faster and been time dilated much more than Clock A and will be way behind Clock A.

I agree that if B goes much faster, then it will be more dilated. You’ve not computed how fast it’s going, so your words are totally empty. You asserted earlier it was about acceleration, not about speed:

with the slower clock the one that experienced more acceleration.

My example demonstrates that it is not in fact a function of acceleration.

It will be obvious which one had experienced more acceleration.

Yes. I set it up so B experiences 10x the magnitude of acceleration as A.

Where they went or did not go is irrelevant as long as they end up side by side in a common inertial frame.

Again correct. It is not a function of where they go.

Want numbers?  You first this time.  Pick a real distance to a star and say a cyclotron with a 20m radius, which seems reasonable. Otherwise, just more words from you. You seem to have the mistaken impression that the magnitude of time dilation is a function of the magnitude of acceleration, so I picked a counterexample as requested.

Clocks can only be compared when they are nearby in a common inertial frame.

They all come back and get compared before and after the exercise.

To get to a given point and back at a constant g factor means accelerating to the midpoint then decelerating, then accelerating again in the other direction to the midpoint then decelerating again ending up in the place and inertial frame where the trip started. Constant acceleration for half the trip and constant deceleration for half the trip.

Right. I can just hear the actual math not happening.
Deceleration is the wrong term. If it accelerates at 1 m/sec for a year in the direction of x, then for the next 2 years it accelerates at -1 m/sec in the direction of x.  The magnitude is always 1 and only the direction changes.  I’m just saying that deceleration is the same as acceleration, but with the opposite sign.

The acceleration will slow the onboard clock relative to an inertial observer (in a common inertial frame with Earth) as the relative speed between them increases.

We can put a third clock C (control) for this. It sits right next to the cyclotron but isn’t in there like the B clock is. We can compare all three when we’re done.

The deceleration will speed up the clock relative to that inertial observer as the relative speed between them decreases eventually bringing the relative clock rate back to that of the original inertial frame.
 In the following scenarios, all accelerations are proper, as felt by the accelerating entity, and unless specified otherwise all times are as per an unaccelerated clock on Earth.
Let us have a clock accelerate at 1 g for 1 year then decelerate at 1 g  for 1 year . The clock will then reverse direction, accelerate at 1 g  for 1 year then decelerate at 1 g  for 1 year.
To make the situation symmetrical, let us have a clock in a cyclotron accelerate at 10 g for 1 year

The cyclotron clock maintains continuous magnitude of acceleration until the clock from the star gets back. If the clocks measure different times, then one is going to stop before the other. If that’s 4 years as measured by clock A or C (you didn’t specify, but I kind of assumed A), then it’s something else for B since we both agree that B will read a different time than A. I say B logs more time, and you say less (due to its greater continuous acceleration).

Let us have a clock in a cyclotron accelerate at 10 g for 1 year and decelerate at 10 g for 1 year.

There is no ‘deceleration’ in a cyclotron.  The g force is continuous the whole time, but with the vector direction changing as it spins.  The speed is a function of the radius, not a function of time.  The ISS is accelerating continuously at nearly 1g, and yet it never gains speed.

What elapsed time do each of the traveling clocks show as compared to the Earth clock?

The Earth clock elapsed time is 4 years.

That answers the question of what clock measures the 4 years. I guessed wrong.

The 1 g acceleration clock elapsed time is 3.50 years

Impressive. I got that as well.

The 10 g acceleration clock elapsed time is 1.17 years

Sorry, but this assumes a straight out-and back path, not a cyclotron.  Speed at a constant acceleration is a function of radius v, not of time. V = √(ar). If the speed in the cyclotron kept going up and up like you’re computing, the g force would also keep going up, and I said it was a constant 10g. Remember, acceleration is a vector quantity and is thus a change in velocity (dv/dt), not a change in speed.

Nope. You applied Lorentz time dilation to SP2 but not to SP1. Mixed numbers.

Show me where I did that. I asked you to point out a self-inconsistency with my numbers.

I corrected your erroneous mismatch in time values for SP1 and SP2. You were mixing numbers from different reference frames.

They were never mixed. I carefully separated all the numbers with colors, each color headed with what frame was used.

[Malamute Lover (OP)]

...
The graph shows an accelerating observer and a non-accelerating observer. What does this have to do with the subject, which is two observers in a common inertial reference frame accelerating at the same rate in the same direction beginning at an agreed on moment on synchronized clocks?

In that scenario, the entire ensemble will not be aware of any contraction. An external observer in an inertial frame would observe increasing Lorentz contraction on the entire ensemble. Are you still imagining that there is an Absolute Space in which the shrinkage is happening? There isn't. Contraction is relative, not objective.

Malamute Lover,
let us talk some concrete analysis.
L₀ - length of the spaceship in its rest frame.
5L₀  - distance between the spaceships.
Let us choose the left frame as our origin.




The right spaceships has to get much further away into the future, in the original rest frame, in order to maintain the simultaneity line of the left spaceship frame.



Do you trust the Minkowski space-time diagrams?
Does this make sense now?
Jano


Edit:
This is the Special Relativity acceleration 101 (One-o-One) in transition towards general frames.
What do you know, this is the post #101 in this thread.

Malamute Lover,
The acceleration at the front of the spaceship or rod is different from the acceleration at the back.
As per the textbook, FYI,
Jano

Read the text that goes with the picture



The horizontal lines show the perceived length of the ruler as it accelerates relative to the inertial observer located on the Y axis, (the vertical time axis) as observed by that inertial observer. The changes in the colored areas, as observed by that inertial observer, are due to changing relative speeds. Notice that the length of the ruler and the width of the colored area match when the ruler matches speed with the inertial observer on the X axis.

The horizontal distance (call it W) between where the dashed lines intersect the edges of the colored area correspond to the length of the ruler as seen by the observers at the ends of the ruler. That is, the rectangles mentioned in the text formed by the points of intersection, those points being opposite corners of the rectangle.  Note that the dashed lines change angles, being steeper where the colored area becomes narrower as seen by the inertial observer. Note especially that because of the changing angles, W is constant. Again, W is the width of the rectangle whose opposite corners are where the slanted dashed lines intersect the colored areas. The observers at the end of the ruler never see any change in the length of the ruler. SP1 and SP2 never see any change in their separation and the position of the rod.

A point of interest about this graph is that the ruler is accelerating from a speed lower than the inertial observer relative to the ruler’s starting point (not on the graph). The change in the width of the colored area from smaller to larger to smaller again is because the relative speed of the inertial observer and the ruler grows lesser until they match speeds on the X axis, and then the relative speed increases as the ruler continues to accelerate.

The observers on the ruler (the observers on SP1 and SP2) would see the length of the inertial observer initially contracted because of the speed difference, then the inertial observer would be seen to be less and less contracted as the speed difference reduced, with no contraction as they matched speeds. Then as the ruler continued to accelerate and the speed difference grew, the observers on the ruler would again see the inertial observer contract.

You cannot pull a picture out of a textbook on Relativity Theory and expect to understand what it means just by looking at it.

[Malamute Lover (OP)]

The observers at the end of the ruler never see any change in the length of the ruler. SP1 and SP2 never see any change in their separation and the position of the rod.

This is not so straight forward.
Are you suggesting a rod between the two spaceships?
If yes, then you need to understand/explain the post #101.
Do you see where is the simultaneity line?

The simultaneity line concept does not apply here. It relates to disagreements about the simultaneity of different events to observers in different frames, which includes different distances to the two events even if there is no relative speed difference.  It has nothing to do with whether events are simultaneous according to observers in the same reference frame with synchronized clocks. Events that take place at the same time on synchronized clocks that have not been affected in different ways can be said to be simultaneous regardless of what different observers may see.

As I described long ago, two observers at a significant distance from each who start in the same direction at the same proper acceleration at an agreed time on synchronized clocks and who stop accelerating at an agreed time on those synchronized clocks will see odd and contradictory things along the way. But after they stop accelerating and wait for the light images from each other to stop changing, they will find themselves at the same proper distance from each other.

Relativity of Simultaneity is irrelevant.

[Malamute Lover (OP)]

SP1 and SP2 never see any change in their separation and the position of the rod.

This is not so straight forward.
Are you suggesting a rod between the two spaceships?

Yes, the rod was first suggested in post 3 and defined exactly in post 8 where I affixed it to the front of the rear ship rather than having it trail behind the lead ship, which leads to it breaking even if not attached to anything.
So our scenario since post 14 has been a pair of negligible length ships (shorter than about 10 light minutes, the precision of the calculations) initially separated by 10 light days, which accelerate at 100g for 4 days ship time.  The rod in front of the rear ship is 20 light days long, so it extends past the front ship, who therefore at any time can determine the separation between the two by just reading the markings on it.  No need to send time-consuming light signals between the ships.

If yes, then you need to understand/explain the post #101.
Do you see where is the simultaneity line?

Why did you post the same picture twice in that post?

It is sort of appropriate, except it depicts continuous acceleration, whereas our scenario has them stop accelerating after 4 subjective days. M-L claims that the proper distance between the ships will somehow revert to the original 10 light-day separation after they stop accelerating.  Your picture shows continuous acceleration, and thus perhaps doesn't clearly show it.  It also lacks numbers on the ruler, making it difficult to assess the separation in a frame not depicted.
The picture BTW seems entirely accurate to me, even if it doesn't depict our exact scenario.

Based on the most recent post, M-L seems in total denial of the existence of the frame in which both ships are subsequently stopped, as he refuses to compute any numbers for it.  It would of course show that length contraction is real, so I suppose that's why he feels the need to take that stance.

I can think of three possibilities why he would do that:
1) He actually has as little clue about relativity as he presents himself to have. The only relativistic math I've seen in any post is a chart of speed-to-Lorentz-factors.  Never any math applied to a specific scenario except very slow speed examples using Newtonian mathematics.
2) He has a pathological fear of being shown wrong, and refuses to show math that would illustrate it.
I personally love being shown to be wrong, because I'm here to learn, not to be correct.
My numbers could very well have mistakes in them. I did them quickly without double-check. They need interval checks for instance, the first thing I'd do with somebody else's numbers.
3) He knows his relativity quite well, but is a troll getting his jollies from yanking our collective chains.

Halc,
I see your point about acceleration and stopping.
This might be one way how to look at it.
Having said that, this is not a uniform acceleration between the two spaceships as per the textbook definition.
It is a correct uniform acceleration/deceleration per individual spaceships though.
Jano

I have explained what the pictures mean in an earlier post. The width of the colored area is the width as perceived by an inertial observer, not as perceived by the observers on the ruler. Since the ruler is accelerating, the perceived width will change over time. The observers on the ruler will see the same thing happen to the inertial observer since the relativity of length is related to speed difference, there being no such thing as absolute speed.