1

New Theories / Re: Split: Bell's paradox: Does the string break?

« on: 13/08/2020 07:48:18 »

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

2

New Theories / Re: Bell's paradox: Does the string break?

« on: 13/08/2020 03:53:47 »

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.

3

New Theories / Re: Bell's paradox: Does the string break?

« on: 13/08/2020 03:52:20 »

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.

4

New Theories / Re: Split: Bell's paradox: Does the string break?

« on: 13/08/2020 00:43:26 »

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.

5

New Theories / Re: Split: Bell's paradox: Does the string break?

« on: 13/08/2020 00:40:54 »

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.

6

New Theories / Re: Split: Bell's paradox: Does the string break?

« on: 12/08/2020 23:39:14 »

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.

7

New Theories / Re: Split: Bell's paradox: Does the string break?

« on: 12/08/2020 23:30:32 »

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.

8

New Theories / Re: Split: Bell's paradox: Does the string break?

« on: 12/08/2020 22:37:27 »

...
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.

9

New Theories / Re: Split: Bell's paradox: Does the string break?

« on: 11/08/2020 04:08:53 »

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.

10

New Theories / Re: Split: Bell's paradox: Does the string break?

« on: 11/08/2020 04:02:35 »

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.

11

New Theories / Re: Split: Bell's paradox: Does the string break?

« on: 11/08/2020 03:59:59 »

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.

12

New Theories / Re: Split: Bell's paradox: Does the string break?

« on: 10/08/2020 19:44:10 »

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.

13

New Theories / Re: Split: Bell's paradox: Does the string break?

« on: 09/08/2020 23:19:07 »

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,

14

New Theories / Re: Split: Bell's paradox: Does the string break?

« on: 09/08/2020 20:58:06 »

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.

15

New Theories / Re: Theory D - The Ultimate Theory for the Universe

« on: 07/08/2020 04:19:31 »

Based on Newton formula I have found that the total mass should be 75 M sun mass.

I already pointed out to you that you are using the radius of the very large cold gas halo and assigning it the estimated speed of the inner edge of the accretion disk, which is much higher than the outer edge of the halo. This is why your calculation of the mass is way off. And this is why I am not bothering with this thread anymore. You are not listening to anyone but yourself.

16

Physics, Astronomy & Cosmology / Re: I rate 'many worlds' people together with flat-earthers. Am I wrong?

« on: 07/08/2020 04:10:54 »

Both possess beliefs that I find unconvincing.
The former harder to prove erroneous than the latter.

The latter has abundant falsification tests. The latter, like any valid quantum interpretation, has none.
I think that rates the flat Earthers on a different level of crazy than the MWI people.

FYI, Everett's interpretation does not posit parallel universes, despite that popular spin on it by the pop media.

Sean Carroll is a researcher approaching quantum physics from the 'many worlds' viewpoint.
- He happily admits that the underlying hypothesis is unprovable
- But suggests that it requires fewer assumptions than some other interpretations of quantum physics

Of all the quantum interpretations, I do believe Everett's hypothesis hold the record for fewest assumptions:

"All isolated systems evolve according to the Schrodinger equation"

That's it.  No more. No positing of parallel universes.
I'm not an MWI guy, but I respect the simplicity of that.

BTW, the many worlds in the Many Worlds hypothesis are not parallel, Technically they are perpendicular.

17

New Theories / Re: Split: Bell's paradox: Does the string break?

« on: 07/08/2020 03:07:20 »

Acceleration is absolute in that it is experienced by the accelerating observer. The magnitude of the acceleration is not absolute in that time dilation disassociates proper acceleration from that measured by an inertial frame observer.

The acceleration is experience by the  accelerating observer and observered by everyone else. The magnitude of the proper acceleration is absolute. The magnitude of the acceleration is indeed frame dependent.

Since there is disagreement among multiple observers about the magnitude of the acceleration, saying that the magnitude of the proper acceleration is absolute gives the wrong impression, implying that what the accelerating observer sees and predicts will necessarily be correct for everyone. Acceleration is real, having a real effect on spacetime, which gets bent by the energy being expended, But since it is spacetime that is bent, and acceleration is rate of change over time different observers will disagree about the rate of acceleration.

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.

Did you actually read what you wrote?  Or what I wrote?

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 ‘equivocation’ here. Or equating as the case may be.

This is why the pilot of the spaceship and the inertial frame observer disagree on how fast the ship is going.

They disagree because the pilot always says he is stationary. One cannot have a nonzero velocity relative to ones self.

The pilot feels acceleration and sees the landmarks go by faster and faster. He knows his speed is changing and that he is not stationary. Again, did you actually read what you wrote?

In the roller coaster thing, the cars are being held against centripetal force by an outside track which is not rotating with them.

As the cars accelerate, they push the track in the opposite direction. (Newton’s Third Law.) That is the acceleration will not be as much as expected, half of the energy going into the track.

No energy goes into the track. We’re assuming lack of friction here. No work is being done by or on the track. It just guides the cars. The cars are not thrusting against the track, so the track remains stationary.

If there is no friction, then the cars could not accelerate in the first place. The wheels would just slip.  No reaction, no action. If the cars have always been racing around at the same speed, then the spacing, if any, is what it has always been. No changing contraction to worry about.

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.

As it turns out, the track is going to contract with the cars. How? Lense-Thirring frame dragging. (The names are Austrian and are pronounced LENsuh TEERing).

Congrats, you managed to find yet another way to add obfuscation to this topic. This is an SR example, so you can assume negligible mass of the moving parts. Nobody considers the mass of the space ship in the twins scenario, despite the fact that that mass will affect the dilation of the ship clock. It also is assumed to be negligible.

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.

The mass of the spaceship and the mass of the Earth for the stay at home twin will in fact alter how much the clocks will disagree by small amounts. But the key factor is that the traveling twin underwent sufficient acceleration to make those factors insignificant and this is what made him end up standing next to a twin who looks like his grandfather.

Going around a track involves centripetal acceleration. The faster those cars go, the more centripetal acceleration and the more obvious GR effects will be. No obfuscation in applying the proper tools for the situation. Substantial acceleration? This is not an SR problem anymore.

If you do not know the rules, do not try to play the game.
...
I know SR cold

Says the guy who cannot compute the coordinates of some spaceships except using Newtonian physics.

I was demonstrating the principle that synchronizing clocks and separating them negates simultaneity. Assuming a non-existent simultaneity is what confused you. Clear concepts first, a realistic experimental setup taking those concepts into account, then plug and chug the math. Your belief that the SP2 rod would start moving at time SP1a showed that you did not understand this concept.

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.

Acceleration puts different observers in different reference frames. Time dilation is observer dependent.

It makes them stationary in different reference frames. It doesn’t put them in different reference frames since you can’t enter or exit a reference frame under SR.

Acceleration takes us out of SR.  Acceleration changes reference frames. You need GR if there is any asymmetry of situations. SR snapshots could be used in the two spaceships scenario since the situations are symmetric. But you have to pay attention to the actual physical circumstances, like the fact that the rod attached to SP2 will not yet move at time SP1a.

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.

Acceleration changes the length of the time component of a worldline. Two observers moving relative to each other will always see the other’s clock as running slower. But bring them into a common inertial frame and there will be a difference in their elapsed times from a synchronization event in the past according to their acceleration histories, 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.

To an observer on the surface (and therefore to the mechanical properties of the cylinder) the same number of revolutions a minute are taking place

Now you’ve contradicted yourself again.  We have a stationary marker by which a rotation can be measured.  If their clocks are not running at the same pace, they necessarily measure a different time for one revolution.

The circle has contracted because the geometry has gone non-Euclidean. To an external observer, the circle is smaller and faster, but the clock is slower. An observer on the circle counts the same revolutions per minute.

You only need SR to understand that part. What is your problem?

Nope. The reduced circumference circle is spinning faster because tangential speed is still the same but the increased time dilation makes an observer on the circumference think a revolution takes as long as if the circumference were still longer.

You’re failing simple grade school arithmetic now. Angular rate is measured in say RPM which isn’t a function of radius or linear speed of any of the parts.  If a rotation takes 12 seconds as measured by the central clock, then regardless of the radius and rim speed, that object is rotating at 5 RPM.  If the clock at the rim runs slower and measures 10 seconds, then the object rotates at 6 RPM. It is only a function of the measured time to do one revolution, so if the two measured times are different, the measured number of revolutions a minute must be different.

Your concepts are mistaken. You must look at the details to get them right.

To begin with, tangential speed and radius and RPM are all related. Your claim that they are not related is wrong. If you know the values of two of them, you know the third. That is also true in non-Euclidean geometry as long as the details of the curvature between circle and center are consistent all around the circle. You just need to use the proper formulas consistent with the curvature.

We should note that the central observer will always see the clock on the circle as running slower, and vice versa, because they are moving relative to each other. However, the circle has shrunken and the tangential speed is now greater. As a result the two clocks will each be seen as running slower than before by the opposite observer. But the observer on the circle never looks at the center so we need only deal with increased time dilation on the circle as viewed by the central observer when the circle shrank.

The clock on the circle is running slower than before as viewed from the center. Previously it took 12 seconds according to the central clock for a marked point on the circle to pass a bright red line pointing out from the center, a line that is stationary to the central observer. But the circle has shrunk and sped up.  Now the central clock says it takes only 10 seconds.

Imagine the circle has evenly spaced markings on it. Viewed from the center it previously took 1.2 seconds for these markings to pass that stationary red line. Now it takes only 1 second as viewed from the center since the circle has shrunk and is moving faster. Previously 5 RPM, now 6 RPM. Tangential speed is 20% faster.

However, the observer on the circle has had his clock slowed as compared to the central clock. The ticks do not come as often as seen from the center. How much slower are those circle ticks coming? Since the shrinking of the circle and consequent speed up was due to Lorentz contraction, the time dilation will also be in accordance with the Lorentz factor. Tangential speed is now 20% faster what it was as viewed from the center, so the perceived clock rate will be 1/1.2 = 0.83 as fast on the circle clock as compared to the central clock by the central observer.

Viewed from the center, the time for each marking to pass the stationary red line is 1 second but the clock on the circle is running at 0.83 speed. However the observer on the circle is unaware of this. He thinks the ticks on his clock are coming as fast as ever.  He sees the markings pass the red line every 1/.083 = 1.2 seconds same as before the circle shrunk. He is unaware that the circle has shrunk. He still thinks it is going at 5 RPM and the mechanical stresses experienced on the circle per unit of time are the same as before.

Accelerated observers feel acceleration but are unaware of length contraction or time dilation that other observers may notice. Relative. Not objective.

And “a pilot always experiences being stopped”? No, the pilot experiences being accelerated to 0.99 c from his original reference frame. If he decelerates and finds himself in a matching inertial frame that is actually accelerated from the starting frame, is he stopped?

A pilot is an observer observing from a position of rest in his current reference frame. That means he experiences his clock running at normal rate, and experiences his proper mass, not something 7x his prior mass.

What you seem to be referring to is proper velocity, which is indeed 7c. Proper velocity is the result of integration of past acceleration, and there’s no limit to that, which is why, with a fast enough ship, I can get to the far side of the galaxy before I die.

Stopped is a relative term. There is no absolute space so there is no absolute stopped. The pilot can say he is stopped if he does not feel acceleration and does not observe any motion in other objects that he considers stopped.

Length contraction, time dilation, and mass-energy increase are not observed in the accelerated reference frame. They are relative to other reference frames, not objective.

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.

If you were able to see the GPS clock it would be going at the same rate as your own clock because it was intentionally set to run slower.

Looking at a clock known to be designed to run slower and seeing it run the same rate as my normal clock is a direct observation of objective time dilation between those clocks.

Sorry, no. Not objective. Another observer in a different reference frame will observe a different a time dilation factor between you and the GPS satellite. 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. The same absolute value of time dilation but opposite signs. Not objective which would require identical values measured from either end. Time dilation is relative, not objective

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.

Einstein was not believed by everyone for some time. There are relativity deniers today, to use your phrase.

If you’re not a relativity denier, then why have you not shown any relativistic mathematics when computing the coordinates of the ships in the string example? That exercise is pretty trivial, and yet all I get is low speed Newtonian math from you, which indicates a fear to expose the inconsistencies in your view.  Stop it with these repetitive assertions and give workable numbers.
Choose a different scenario preferably, but not one from a website where somebody else has done it. Notice that I don’t use the same one twice. You must use a scenario with relativistic dilation. Ideally, the situation should become evident with a couple digits of precision, but I did mine with around 5 digits.  A speed of 100 m/sec would require more digits than my calculator has. It demonstrates nothing.

I’ve put up numbers. You’ve not found any inconsistency in them except disagreements with your personal vision, which I do not accept as evidence that I did it wrong. If it’s wrong, then somewhere there will be a pair of events that don’t relate properly with fixed light speed.  I can’t do that with yours because you’ve given no relativistic example to work with. You apparently don’t know your physics at all because you continue to decline to do this.

I have shown the inconsistency with your concepts. Your math is wrong because it does not reflect the real world. You assume simultaneity at a distance which is invalid even in SR. If I showed math that was based on the correct concepts you would just say it was wrong because you do not understand the concepts. Get those straight first.

18

New Theories / Re: Split: Bell's paradox: Does the string break?

« on: 06/08/2020 22:39:50 »

Malamute Lover,
here are a couple of diagrams:





This is a uniform acceleration as per the textbook.
You can see the length contraction of the ruler.
This discussion would be better in the SR reciprocal thread though,
Jano

What did you think this means?

It is showing the worldlines of a uniformly accelerated observer and an observer who is at fixed coordinates.

Did the term ‘comoving’ confuse you? It just means that the observer is moving with the expansion of the universe but that (for an observer in an inertial reference frame) the coordinates remain the same and that for another observer not in the same reference frame, the coordinate system used for the first observer remains in effect. Basically, it means to ignore the expansion of the universe. It simplifies the math by keeping things local and not in reference to any noticeably red-shifted distant object.

As expected, the two worldlines are different because the situations are different. But the worldlines for spaceships SP1 and SP2, uniformly accelerating at the same rate, the worldline shapes would be identical.

So why the string between the spaceships would not try to shrink like the ruler does?
The end-result will be a broken string.
That's the logical conclusion when the spaceships worldlines are going to be identical,
Jano

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.

19

New Theories / Re: Split: Bell's paradox: Does the string break?

« on: 06/08/2020 22:30:57 »

Although I did not explicitly mention it, imagine that each ship is displaying a clock so that the other one can see it.

Can do that. I just didn’t find it useful for either of them to look at the other.

Each will start off seeing the other as being one second behind. Since this (initially) matches the apparent increase or decrease of separation, each one can consider that as evidence that their clocks are not synchronized after all.

Their motion is identical in the initial frame and their clocks are synced in that frame. Thus their clock must remain synced in that frame.
I agree that they will not observe that one second lag since they would need to be stationary in that frame to observe that.

Spatial separation rules out giving meaning to synchronized clocks if acceleration is involved.

Oopsie.  They must stay synced in the original frame if their acceleration is identical, which it is.  Per RoS, any clocks synced in that frame are not synced in another, so indeed they’ll be out of sync in either ship frame.

My point was that the one second lag in the onset of acceleration, due to the distance between them, led to observations on each side that contradicted the other side. SP1 was convinced by all evidence that SP2 was increasingly lagging behind and that SP2’s clock was running slow. Not just one second behind, but increasingly behind. SP2 was convinced by all evidence that SP1 was going too slow and SP2 was gaining on SP1 and that SP1’s clock was increasingly behind. Not until they got back into a common inertial reference frame was the illusion broken. The distance between the ships is the same as at the start and the string is unbroken.

When acceleration is involved , different observers separated by space can see different things that are not necessarily the case. And this is not even in relativistic territory.

If you think you can provide some way whereby the two accelerating ships can each appear to the other to be at a constant speed and distance relative to each other even if they are, please provide it. What is each ship supposed to do to accomplish this?

I never suggested any such thing. I said I don’t care how they appear to each other.  I care about what they’re doing.
I had a rod with distance marks affixed to the front of SP2 so all SP1 has to do is look out his window at the rod in his presence to know the separation in SP2’s frame. No dependence on light speed. The rod, being attached to SP2, is always stationary in SP2’s frame.  I put it on SP2 because it might break if we put it on SP1.

As we will see, the rod is not always stationary in SP2’s frame.

The point of my exercise was that at the end of the story, the two ships are found to be in a common inertial frame at their original separation and with an unbroken string between them despite their interpretation of what they are seeing.

That is your goal, but you need to perform invalid mathematics to get that story. SR predicts otherwise.

My math was completely valid. The point was to show that measurements made at a distance during acceleration, even common acceleration are not necessarily reflective of the measurements made in common inertial frames.

You want math? I will use some of your numbers.

Two spaceships 300,000 km apart. Call it 1 light second.

Each should experience identical proper acceleration regardless of how long they accelerate or what speed they reach relative to their original inertial frame.

OK, this is the first you’ve stated the use of proper acceleration. Your mathematics in prior posts used constant acceleration, not constant proper acceleration. They yield different numbers. Proper is easier to use due to the frame symmetry of it all, but it works either way if you don’t mind the more complex arithmetic.

We have both been talking about proper acceleration all along as is very obvious from the context and the discussions. Nit picking again. Acceleration of the two ships would not have the same values for several observers in varying frames but for each observer it would still be the same for each of the ships independent of externally measured acceleration value.

If after coming back into a common inertial frame, they each emailed a copy of their acceleration record to each other, would either one see a difference between the other’s record and his own?

If you think there will be a difference in the acceleration records because of relativistic speeds, show me the math.

You've not invoked relativity theory at all. You're adding velocities the Newtonian way, and you're using the original rest frame to compute what each observer sees, not the frame of the observer. The inconsistencies will pile up once you move faster than a city bus.

That is right, I did not incorporate relativistic considerations.  My point was that an apparent speed difference is not a necessarily a real difference if the clocks do not correspond.

I’ve not considered appearances at all.  I want the coordinates of the two events where the ships turn off their engines relative to the ship frame. Newtonian physics does not answer that. Motion does not involve time or length dilation. It assumes an absolute frame, something at least you claim to deny, but here you are using only absolute physics, and only at low speeds to boot. The theory had been falsified 150 years ago.

No, I am demonstrating that to properly assess what is happening one must consider that the two spaceships are separated in space and that the speed of light is finite. This has ramifications as we will shortly see.

So let’s see some relativity math that does not depend on illusory difference in distance.

As suspected, you have no clue how to go about it.  Yes, let’s see some. I’ll do the brutal method since it illustrates the concepts you’re trying to avoid by keeping the speed low.

Two ship 10 light days apart each accelerate at 100g proper acceleration for exactly 4 days as measured on ship clock.  That’s today's scenario.
SP1 initiates acceleration at event SP1a and shuts the engine off at SP1b.  We’ve chosen SP1a event as the origin of both frames.
SP2  initiates acceleration at event SP2a and shuts the engine off at SP2b.

In the original frame:
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.[/color
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.

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.
ΔT = λ * separation distance (in original frame) * Δv = 1.71008 * 10 * .8112 = 13.8721
Therefore:
event SP2a coordinates are (-17.1008, 13.8721)   t = 0 on SP2 clock, v = -.8112
event SP2b coordinates are (- 19.6131, 18.7803)  t = 4 on SP2 clock, v = 0
and while we’re at it, we can compute
where SP2 is at time zero: (-5.8477, 0) t=-8.112 on SP2 clock, v = -.8112

At time 18.78 in this frame, both ships become stationary. SP2 clock reads 4, and SP1 clock reads 17.8721, and only at this time does the rod outside his window stop moving.  When it does, he reads 17.1008 on it (the difference between their x coordinates).  If there was a 10 light-day string between them, it would have broken long ago since it cannot span the 17+ light-day gap.

There’s the mathematics. You’ve shown only Newtonian mathematics, which has been falsified a century and a half ago.
If you see an error in the above numbers where they conflict with SR, point it out. If however you feel the numbers do not correspond to your naive beliefs, then keep it to yourself.

Notice that at no point did I bother to have either ship ‘observe’ the other. The only observation done was SP1 looking at the markings on the rod in his presence.

The separation as measured by the measuring rods they brought along is the same at the end as at the beginning. Why would that not be the case?

The above analysis shows it not to be the case.  Any other value would contradict light speed being a frame independent constant. It is that assumption alone from which all the rules I used to compute the above numbers were derived.

For clarity, the rod is attached to SP2 who is pushing it and it is not attached to SP as per your comments above.

In that case, you are wrong. You are assuming that the end of the rod will stay stationary outside SP1’s window until he stops accelerating. It will not. It will be left behind.

The acceleration force cannot move through the rod at greater than the speed of sound in the material. 100 G acceleration would probably deform most materials, compromising its length. So let us assume that the rod is as strong as it needs to be and that the speed of sound in the material is the speed of light. 

In that case, SP1 will be leaving the rod behind when it starts accelerating because the accelerating force from SP2 has not yet reached the front end of the rod. It will not reach the front end of the rod for ten days as counted by an observer at x=0 and in the same inertial frame as SP1 started from. Will the rod get compressed by the speed differentials? No, it is as stiff as possible and can transmit acceleration force at lightspeed. The rod will get progressively Lorentz contracted from the SP2 end forward as seen by an inertial observer.

When SP2 stops accelerating, all points on the rod will, over time, stop receiving acceleration forces and achieve a common metric everywhere, and will be the original proper length as measured by the measuring rods that SP1 and SP2 brought along. 

SP1 saw SP2 start accelerating ten days after himself. He will also see SP2 stop accelerating ten days after himself. At the end of the ten days, he will look out the window and see that the rod he left behind ten days earlier has now caught up with him and is where it started. Everything is symmetric and everything will end up in the same relative positions, although all equally Lorentz contracted as per an inertial observer.

You assumed that SP1a and SP2a were simultaneous and that the rod would start moving at SP1a. A basic lesson of Relativity Theory, which you should learn someday, is that at any non-trivial distance, the idea of even approximate simultaneity has no meaning.   

If you are going to do thought experiments, get your thoughts together before starting the math.

20

Physics, Astronomy & Cosmology / Re: Is angular momentum frame dependent?

« on: 06/08/2020 00:59:22 »

To an observer on the surface (and therefore to the mechanical properties of the cylinder) the same number of revolutions a minute are taking place

Now you’ve contradicted yourself again.  We have a stationary marker by which a rotation can be measured.  Both observers agree on what one rotation is, but if their clocks are not running at the same pace, they necessarily measure a different time for one revolution.

Nope. The reduced circumference circle is spinning faster because tangential speed is still the same but the increased time dilation makes an observer on the circumference think a revolution takes as long as if the circumference were still longer.

An observer inside with a faster clock will be surprised to not see the cylinder shatter just as he is surprised when his Twin comes back much younger than him.

The twin apparently doesn’t know his physics then, because if he did, there would be no surprise.  The cylinder doesn’t shatter because it was always spinning.

Based on his first person experience, the traveling twin is convinced Einstein is wrong and speeds greater than c are possible. Not until he gets back into a common inertial frame with his now much older twin is he given evidence that Einstein was right after all and that observations on different reference frames are relative.

At this point you go into a bend about gravity and GR, which seems a diversion from the more simple SR topic that you need to master first. 

I know SR cold which is why I know that when acceleration is in the picture you need to bring in GR and non-Euclidean spacetime. The circumference is smaller and the spokes do not reach as far out because the spacetime they are on is curved.

One comment though:

It will be crushed around the circumference. Where does the energy come from to crush it?

It doesn’t take energy to crush something. It seemingly takes force, which means a strong but brittle object can be crushed by expenditure of arbitrarily small energy. The less brittle it is, the more that energy goes into strain and not into failure, so it takes more. I’m assuming insanely brittle and strong materials for our objects else they’d not be able to withstand the centripetal stresses being put on them. Anything else would just fly apart.

(We are now talking about the material circle constructed around a neutron star being raised out of the gravity well in a differently shaped spacetime.)

There must be some force involved or there would be no crushing at all. Where does the energy come from? It comes from the same source as the energy used to raise it out of the gravity well. It would resist being forced into a differently shaped spacetime. Note that there is no spinning involved here so this is a different problem from the spinning wheel situations.

Whether and how much contraction an observer will see depends on the observer.

The cases I enumerate above are observed by anybody. They were chosen for that purpose. The measuring rod between ships is also a real consequence.

You cannot directly see curved spacetime. Our 3D Euclidean oriented brains do not work that way. You can only see the consequences. I have addressed all of the cases you enumerated and shown how your assumptions are wrong because you do not take into account GR and curved spacetime, which must be considered because acceleration is involved.

An observer on the spinning wheel sees nothing different because contraction is only visible from another reference frame.

In all three cases, the observer on the wheel very much sees differences, which are pointed out in the cases above. You seem to agree that one ring fits through the other, something to which all observers agree. You don’t seem to have any fake physics that lets you deny the bumper-car-track thing, nor do you seem to deny the non-Euclidean dimensions of the ‘cylinder’.  OK, the  non-Euclidean dimensions are frame dependent.  A stationary observer will measure normal dimensions, but the inability of the object to change its angular speed is an objective observation.

The observer on the wheel does not see length contraction which can only be seen from a different reference frame. The observer on the wheel is time dilated relative to another observer and thinks the wheel is the same circumference because it takes the same subjective time to make a revolution relative to a fixed marker. But from a different reference frame nearer to the center, the circumference, being length contracted, is smaller than would be expected from the length of the spokes, a clear indication that spacetime is curved. As with the orbit of Mercury deep in the Sun’s gravity well, the curvature cannot be seem directly but can be inferred from the precession of the elliptical orbit.

An object certainly can change its angular speed by changing the radius. When a spinning skater pulls her arms in, angular momentum is conserved which requires and increase in angular velocity.

First some comments about time dilation. The pilot of a rocket ship that has accelerated to 0.99 c will experience a time dilation factor of about 7. The clock on the ship will be running 7 times slower than before the acceleration.

No frame reference, so that statement is ambiguous.  It will be running 7 times slower relative to the frame in which the ship is moving at .99c.  Not saying you're wrong, just sloppy.  It's running at normal rate relative to the ship of course.

I did provide a reference frame. The frame in which the ship is moving at .99 c is the one before the acceleration, which I explicitly mentioned right there: “accelerated to 0.99 c”. No sloppiness. We are not going ad hom., are we?

It’s not an ad-hom.  It’s sloppy because you’re describing ‘what the pilot experiences’ and a pilot always experiences being stopped.  You’re referencing the pilot frame and also the original frame, which makes it confusing. That’s sloppy.

You first accused me of not mentioning the original frame, which I did implicitly by mentioning acceleration from it. Now you are berating me for mentioning both frames.  You are just looking for something, anything, to nitpick about even if you contradict yourself. That sure sounds like ad hom.

And “a pilot always experiences being stopped”? No, the pilot experiences being accelerated to 0.99 c from his original reference frame. If he decelerates and finds himself in a matching inertial frame that is actually accelerated from the starting frame, is he stopped? Do you have any understanding of the subject at all or was this another ad hom attempt that misfired?

Secondly, the bolded statement is wrong since no observer can experience time dilation. I can look at the GPS clocks and objectively notice I’m running slower than them, but I still don’t experience that dilation.

If you were able to see the GPS clock, it would be going at the same rate as your own clock because it was intentionally set to run slower.

To account for the increased rate of the GPS clock due to lower gravitational level at the 20,000 km orbital altitude (45 μ s a day fast) and the time dilation at 14,000 km/s orbital speed  (7 μ s a day slow), the clock on a GPS satellite is set to run 38 μ s a day slow to match mean sea level clocks. If there were another unadjusted clock on the satellite, then it would be seen to be 38 μ s a day fast.  Note however that even on the adjusted clock running at the same rate as yours, there will be a difference in time readings of .067 to .089 seconds depending which of the 24 active satellites you were comparing against. The time difference is how GPS works.

You have to bring two observers into the same reference frame to judge which one is right, that is, which one underwent acceleration.

Acceleration is absolute (at least in Minkowski spacetime). An accelerometer works inside a box. All observers will agree if something has accelerated. That’s twice you’ve made this mistake.

The mistake is on your side.

Because the accelerometer on the spaceship is time dilated relative to the unaccelerated observer, it’s history will show a steady acceleration all the way up to 7 c just like the pilot felt, while an outside observer will see the spaceship’s acceleration decrease as it approaches 0.99 c. The fact of acceleration is absolute. But because acceleration slows the clock rate as seen by an outside observer but not by anyone on the spaceship, the rate of acceleration is relative.

To prove whose subjective experience is in error requires bringing both clocks into a common inertial frame and comparing them. In this case, the pilot will realize that his clock ran slow even if it did not seem that way to him. He will then accept that Einstein was right after all.

This is basic stuff. Why are you having a problem with it?

His observation is that he is traveling faster than c. Known landmarks are whizzing by at 7 c. Why can’t he conclude that Einstein was wrong?

He is free to propose a different theory, but none has been found so far. So are you a relativity denier then? It seems to be your goal here. You resist it at every step of the way.
Such deniers are dime a dozen on sites like this, but then don't go telling me that your stories conform to an established theory and mine don't. I've pointed out several self contradictions with your assertions.

Einstein didn’t just suggest that light speed yielded the same value in any frame. That because quite apparent by all the attempts to measure the difference as was predicted by the prevailing view of the time. So he can’t conclude Einstein was wrong, he’d have to conclude that all the decades of light speed measurement were wrong. Einstein didn’t perform any of those measurements.

Acceleration is in the domain of General Relativity. A constant speed of light only leads to Special Relativity.  Because the pilot only studied SR, not GR (and how many have studied GR?) he did not know that acceleration was the explanation of the twin paradox. From his SR based viewpoint, it was perfectly valid for him to say that his twin was traveling away from him and should have been time dilated. In SR, the traveling twin will in fact see the other twin’s clock run slower and consider that proof of the other twin being the one in motion.

To prove whose subjective experience is the more credible, it is necessary to bring both clocks into a common inertial frame to determine who experienced less time.  That is the one who was most time dilated.  I say ‘most’ time dilated because the stay at home twin might have flown X-15 rocket planes a lot and the clock he carried with him at all times is going to be behind the clock on his wall at home. But even that clock is going to have less elapsed time than the one in high orbit around the earth.  Keep in mind that all these clocks will run at the same rate when brought into a common inertial frame.

The pilot disbelieving his own subjective experience because he knew Relativity Theory is just silly. Anyway he only knew the easy part. Which you seem to be having trouble with.

BTW the reason for the attempts to determine if light speed would always be measured at the same value regardless of the state of motion was that Maxwell’s Field Theory of Electromagnetism said that it would and the speed that fell out of the equations was equal to the already measured speed of light. This is why Michaelson and Morley conducted their experiments.  Michaelson continued to believe in the luminiferous ether despite the experiments, and also despite Maxwell’s explanation of why the ether was not necessary.  Einstein was not believed by everyone for some time. There are relativity deniers today, to use your phrase. But there are many more relativity misunderstanders, which includes you, I am afraid.