[Malamute Lover (OP)]

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.

Correct. Good point to bring out. There is such thing as ‘stopped’ except relative to a specified frame (as you referenced). Since Absolute Space does not exist, any observer is always in motion relative to some other observer and each will see the other through the lens of Relativity.

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

The picture shows the apparent length as perceived by an observer at a different and changing relative speed. The observers on the ruler see the same thing happen to the inertial observer. Different observers in different frames, inertial or accelerating will see different things. Who is right about who will bend or break and to what degree?  The local frame is right. In most circumstances, inertial or uniform acceleration, the local frame is indistinguishable from a Newtonian frame. (The significant exception is when acceleration differential is so great (like near a black hole) that the idea of ‘local’ is not so 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.

The standard definition of uniform acceleration in Relativity Theory is proper acceleration at a constant felt rate such as what an onboard recording accelerometer would show.

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.

The picture represents the viewpoint of an inertial observer, not what the observers in the accelerating frame see. It represents perceived length contraction. Another inertial observer who is in motion with respect to the first inertial observer will see a different contraction. Also, since several observers at different relative speeds will have their time axes angled differently w.r.t. each other, the perceived instantaneous slope of the edges of the colored area is not meaningful in any absolute sense.


Some comments about terms. The textbook uses the term ‘uniformly accelerated motion’. This refers to the motion resulting from speed that changes at a uniform rate over time. All terms proper. One might refer to ‘uniform acceleration’ which is the proper acceleration rate involved. But if you see ‘accelerated frame’, check the context. It may mean the proper acceleration is still ongoing at the time of interest, or that the proper acceleration has ended at the time of interest and the frame is being compared to some other frame. In SR it typical means the latter since SR does not deal with ongoing acceleration, only the resulting changes at specified times on someone’s clock.

[Malamute Lover (OP)]

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

As previously discussed, the picture shows the perception of an inertial observer w.r.t. an accelerating frame. It does not indicate what the participants in the acceleration will see.

[Malamute Lover (OP)]

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.

For proper separation and proper speed that is exactly what you do. The Principle of Relativity states that the laws of physics have the same form in all arbitrary frames of reference. Uniform proper acceleration for a specified proper time will yield the expected proper speed and the expected proper distance covered. The motions of SP1 and SP2 are identical. Their initial proper separation and their final proper separation will be the same. If you cannot understand that they you do not understand the foundation of Relativity Theory and if you do not understand that you know nothing at all about Relativity Theory.

You insist on applying a nonsensical relativity correction to proper values but never stop to think that this relativity correction of yours depends on which observer you use as a base. And that has no influence whatsoever on what SP1 and SP2 will see.

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.

There is no such thing as ‘relativistic computation of proper speed’. Saying that shows an utter ignorance of Relativity Theory. Proper speed is what an observer will calculate from proper acceleration history as confirmed by passage of landmarks. Constant proper acceleration does not require any relativistic adjustment in proper speed.

In the first post I copied your numbers but this time I calculated it myself, avoiding your unjustified ‘relativistic correction’ which does not exist in proper measurements.

In an inertial or uniformly accelerating frame (the definition of both including ‘proper’) local physics is Newtonian.

The entire frame in which SP1 and SP2 have accelerated identically has been contracted as viewed by an inertial observer, not as viewed by SP1 and SP2 because they are traveling at the same sapped and will not see any contraction.  Different inertial observers will see different contraction factors. SP1 and SP2 will see all of those other observers as contracted because observation of Lorentz contraction is related to relative speed. You are still thinking in terms of there being some kind of absolute space underneath everything in which things really contract. There is not. If you do not understand that then you know nothing about Relativity Theory. But I think that has been amply demonstrated already.

Buzzwords and online calculators are useless unless you know to apply them. And that requires understanding fundamental concepts, an area in which you are seriously lacking.

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.

No, I am using proper separation correctly. The term proper refers to what is experienced, measured etc. by the participants. You cannot adjust actual proper separation by what some other observer sees, What about observers O, P, Q, R, S, T, U, V, W, X, Y and Z all of whom are inertial but have different speeds relative to each other and to SP1 and SP2? They will all want different contraction factors. Proper separation is what SP1 and SP2 will both see and once the light images catch up and the proper separation will be what is was at the start. What any other observer sees is irrelevant to that because different observers will see different things. That’s why it’s called Relativity Theory.

You have got buzzwords and online calculating tools but you have no idea how to apply them correctly because you do not understand the basic concepts underneath them.

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.

In post #8 (new thread numbering) you agreed to add thrusters to the rod to avoid the speed of sound problem

…outfit the rod with its own thrust as you describe so at no time is it under stress or strain. Of course those thrusters would also have to be programmed with the plan so they start and stop at the proper time. 

Later the rod got attached to SP2 instead of SP1 but still the same problem. If there are no thrusters on the rod and it is attached to SP2, then the force waves from the acceleration will proceed up the rod as the speed of sound in the material. If we say that the speed of sound is the maximum possible, the speed of light, then it will take ten days for the force waves to reach the front end. SP1 will be long gone. When the acceleration stops and the last of the waves reaches the front end, the acceleration of the rod will cease, the rod will be its original proper length and the front end will come to a halt relative to SP1 exactly where it started, outside SP1’s window.

Jaaasonik’s picture does not mean what you and he thought it means.  It is about what an inertial observer sees. You need to do the rectangle analysis as the (omitted) explanatory text explains to discover that the observers on the ruler see. Once you do that you see that they never see any contraction. The width of the rectangle is always the same.

You cannot grab a picture from page 409 in a textbook on advanced physics, ignore the explanation, and think you understand what the picture is saying.

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.

You omitted the ‘Now  the rod’ prefix to the section. If you had included that part it would have been obvious that SP1 end meant. SP1 end of the rod. But this way you can ignore the math.

Of course it is proper numbers relative to frame M. It is proper separation that we are talking about. It does not matter what frame N, O, P, Q, R, S, T, U, V, W, X, Y or Z see. It matters what SP1 and SP2 will see when acceleration stops and the light images catch up. There is no relativistic correction to proper measurements, only differing measurements made by observers in different frames. And since different observers will see different things it is clear that only frame M counts when making proper measurements.

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.

Proper distance is the same concept as proper length.

Proper length L0 is the distance between two points measured by an observer who is at rest relative to both of the points.
https://courses.lumenlearning.com/physics/chapter/28-3-length-contraction/

Now where is your definition that proper length/distance involves relativistic corrections based on an arbitrary observer not in the same frame. Put up or shut up.

[Malamute Lover (OP)]

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

An event presumably?

No, a time, a clock reading. The clocks are all synchronized. Location of the clock does not matter.

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.

Proper measurements within a frame of reference in which there is no relative motion are Newtonian. Identical proper acceleration in the same direction for the same proper duration means no relative motion. The frame is of course the common frame of observers not in motion relative to each other. There is no such thing as absolute motion. And there is no relativistic correction of proper measurements. If you do not understand that, you do not understand Relativity. But we know that already.

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 is the end of the rod outside SP1’s window. This time you did include the explanatory prefix but ignored it anyway.

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.

Frame N is irrelevant as are frames O, P, Q, R, S, T, U, V, W, X, Y and Z all of which would provide different correction values. And all of which would be length contracted and time dilated relative to M because they have different relative speeds. There is no relativistic correction of proper distance. You got some buzzwords and online calculators but have no idea how to correctly apply them because you do not understand the concepts.

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?

Frame N is irrelevant as are frames O, P, Q, R, S, T, U, V, W, X, Y and Z all of which are moving at different relative speeds and would provide different correction values. The clocks on SP1 and SP2 all agree on the same time. Since they were all initially synchronized and experienced identical proper accelerations for identical proper times, why would they not be? Move the SP1 clock back to the midpoint between SP1 and SP2 and move the SP2 clock back to that same midpoint in the same identical way and they will agree. And as my math has showed, that midpoint will be the midpoint marker of the rod.

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.

But sadly you are not.

How long is the rod at the 2 day mark?

Frame dependent question.

Evasion. SP1 and SP2 and the entire rod with thrusters have undergone identical proper acceleration in the same direction for the same proper time beginning at the same time on synchronized clocks. It is all the same reference frame.  You want it somehow not to be but you cannot demonstrate why.

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.

It is part of the previously agreed idea of thrusters all along the length of the rod.

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.

I am not denying the textbook. I am [i[understanding[/i] what it means. Which you are not. The graph shows the length contraction of the uniformly accelerating ruler perceived by the inertial observer as the relative speed between them changes. To get the length of the ruler as perceived by the two observers on the ruler, you have to construct the rectangles using the points where the slanted dashed lines as opposite corners. Do that and you will see that the width of the rectangle never changes. The length of the ruler, as perceived by observers on each end, never changes even though the ruler is accelerating. The use of rectangles in this manner is a standard technique in graphic representations of situations in Relativity Theory. But you would not know anything about that.

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?

Then you are agreeing with my argument that the entire ensemble is in a common inertial frame and the proper distance relationships never change throughout. Kind of wrecks your argument, doesn’t it.

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.

The perceived contraction of the entire ensemble will be the same for any other given observer. There will be no change in separation. Different observers at different relative speeds will see different contraction factors but none of them will see any gaps open up.

Oh wait! The difference in contraction between the front half of SP1 and the back half of SP1 has caused the spaceship to break apart! How much are the two halves separated? Depends on which observer you ask.

Obviously nonsense. But it is really no different from your claim of a gap appearing between SP1 and the rod because somebody looked at them.  Contraction does not happen in an absolute space. There is no absolute space. Contraction is what observers in other reference frames see. Observers in frame M will see the observer in frame N contracted. Who is right? Properly speaking they are both wrong. Neither reference frame will experience proper. No gap.

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.

And in their common frame of reference which they maintained during the entire experiment it is still a proper 10 light days.

[Malamute Lover (OP)]

Due to serious personal developments I will not be participating in this forum until further notice.

[Halc]

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.

Sorry Jano.  I split the topic and it isn't post 101 anymore

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.

Proof by bold I see.
The former statement is true because the front of the ruler experiences less acceleration than the rear, as can be plainly seen in the diagram you posted.
SP1 on the other hand accelerates at the same g-force as SP2 (the rear of the ruler), so it pulls away from the end of the ruler that is accelerating at a lower pace.

Your numbers posted reflect this, and thus contradict what you’re asserting in bold here. SP1 and SP2 cover the same proper distance in their respective 4 subjective seconds.

Look at the other picture with the number to see how far the end of the ruler gets in 4 subjective time units, which is even less.  The left end of the ruler (red line) after 1.5 time units has moved from proper location 0 (you seem to only be able to think in such terms) to location 1.4, but the right end of the ruler (orange line) has moved from 1.5 to just short of 2, or about a third as far after two seconds on its clock.  The clocks are not staying in sync with each other, not even in the inertial frame.  But they do stay in sync (in the original frame) with SP1 and SP2.

The standard definition of uniform acceleration in Relativity Theory is proper acceleration at a constant felt rate such as what an onboard recording accelerometer would show.

I will accept that. Then the ruler (or the bar in front of SP2) everywhere experiences uniform acceleration, but not identical acceleration since it is different everywhere along its length.  SP1 cannot be comoving with the 10-LD mark on the rod then since SP1’s acceleration (100g) does not match the ~35g acceleration of the 10-LD mark on the rod.

SP1 final proper speed is 1.130 c

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.

For proper separation and proper speed that is exactly what you do.

I notice that you did not point out any error in the mathematics used to demonstrate your proper speed calculation to be faulty. Therefore I presume that you agree that your numbers are self-inconsistent. Math doesn’t lie.

The Principle of Relativity states that the laws of physics have the same form in all arbitrary frames of reference. Uniform proper acceleration for a specified proper time will yield the expected proper speed and the expected proper distance covered.

This contradicts the pictures you and Jano keep posting that shows the front of the rod not covering as much proper distance as does SP2 to which it is attached. And yet you assert that the front of the rod somehow will be where SP1 is when it stops accelerating.  This is you contradicting yourself.

Their initial proper separation and their final proper separation will be the same.

The proper distance covered by both will be the same. The proper separation between the two will not. You seem completely unaware of the difference in meaning between the two terms. I’ve said this before, and you’ll ignore me again because you know better.

I am using proper separation correctly. The term proper refers to what is experienced, measured etc. by the participants.

Indeed, and the proper separation thus means the separation as seen by comoving observers, which means the separation in their frame, and not in another. But you’re still measuring the separation by the frame of those grid markers, not by the frame of the observers.

There is no such thing as ‘relativistic computation of proper speed’.

I actually performed such a computation above where I worked out the contradiction in your numbers.  Actual speed is dx’/dt’, and proper speed is dx/dt’, so all you need to do to get the relativistic proper speed is multiply actual speed by dx/dx’ which is λ.

Saying that shows an utter ignorance of Relativity Theory. Proper speed is what an observer will calculate from proper acceleration history as confirmed by passage of landmarks.

Yes, it is, but not by shoving that history through a Newtonian linear formula. That formula assumes no time dilation as speed increases.

Constant proper acceleration does not require any relativistic adjustment in proper speed.

Then show the error in my numbers instead of just asserting that you must be correct because you assume it must be that simple.

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.

In post #8 (new thread numbering) you agreed to add thrusters to the rod to avoid the speed of sound problem

Which is how we keep it absolutely stationary relative to SP2 at all times in the frame of SP2.

If there are no thrusters on the rod and it is attached to SP2, then the force waves from the acceleration will proceed up the rod as the speed of sound in the material.

Which is why we didn’t posit a lack of thrusters. It allows us to use ideal math.  The ruler in the pictures being posted are considered ideal rigid objects for simplicity. We’re sticking with that.

You omitted the ‘Now  the rod’ prefix to the section. If you had included that part it would have been obvious that SP1 end meant. SP1 end of the rod. But this way you can ignore the math.

Got it.  That was unclear because of course I don’t consider SP1 to be at the end of the rod.

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

You don’t give any indication of duration in frame M or actual speed in frame M, so I’ll use the numbers you gave before, which is time 4.9082 days and .8112c. The exact time and speed doesn’t really matter so long as its the same for both SP1 and SP2 in frame M.
I cannot convert your proper speed to coordinate speed because you’re using Newtonian speed, which of course is the same as coordinate speed in Newtonian mechanics (things can move faster than light given those physics).  But I’ll use your proper distance of 2.261 LD.
SP2 end is attached to SP2, so of course those numbers will be the same.

So I’m going to put some stationary observer here and there in frame M with clocks synced in that frame.  Each never moves. You refuse to tell me how far out of sync the ship clocks are in their proper frame, so I cannot talk about what time each pilot sees on the other clock, so I’m forced to use stationary observers.  Your total inability to produce any figures in frame N is a serious indication that you don’t know what’s going on.

There are 1000 grid markers every light day, all stationary in the original frame M.  You report each ship’s proper speed, meaning you are having them count these markers going by.  So each has counted 2261 markers after a subjective 4 seconds.
My first observers are at markers -7739 and 2261. They are present at the point where each ship stops accelerating as it flies by at .8112c.  The time on their clocks is 4.908, and they can see the time of 4 on the ship/rod clock as it passes by. The guy at the 2261 mark does nothing except note the location of the event where both rod and SP1 pass by together.

The guy at -7739 (SP2b) sends a light signal towards the opposite end of the rod at time 4.9082.  The third observer is at marker 45227.  This marker is 52.966 light days away from the first observer, and it thus takes about 53 days for the light to get there at time 57.874 on that observer’s clock.  Just at that exact time the right end of the rod goes by him.  Verification of that: The far end of the rod was at marker 2261 at time 4.908.  Marker 45227 is 42.966 LD from that, and the rod end is moving at .8112c, so it covers 42.966 LD in 52.966 days, the exact same time as the light signal to get to the same spot.

The signal is reflected back.  There is a 4th observer at marker 39706, 5.521 LD distant from the 3rd guy with the mirror, so it takes 5.521 days to get there at time 63.395.  Meanwhile, SP2 moving at .8112c takes 58.487 days to get to the same observer at the same time as the refleted light signal.

Total time (as measured by the stationary observers) is 58.487 to do the round trip.  Now the ship clocks, moving inertially since the acceleration stopped, are moving at .8112c in that frame, so the time dilation factor is 1.71008, so SP2 clock will have advanced not 58.487 days, but only 34.201 days while waiting for the light to reach the other end of the rod and back.

This is a contradiction with your asserted length of 10 LD for the rod. It should have taken 20 days (ship time) for light to traverse a 10-LD rod (stationary relative to the ship taking the measurement) to  SP1 where you assert the other end of the rod is, but it took 34.2 days, indicating a proper separation of exactly 17.1008 LD between the ships.

This again shows your numbers to be self-contradictory.  Either that or I made a mistake above, in which case you’ll be happy to point it out.

If I got the coordinate speed wrong, you can tell me which one is correct. It’s easy enough to change the above figures for a different speed.  The same contradiction will result, but 17.1 is unique to the 0.8112 speed.  Talk to the numbers, not to the text.