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

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**Today**at 10:20:46 »

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

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.

Proof by bold I see.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.

The former statement is true because the front of the ruler experiences less acceleration than the rear, as can be plainly seen in the diagram you posted.

SP1 on the other hand accelerates at the same g-force as SP2 (the rear of the ruler), so it pulls away from the end of the ruler that is accelerating at a lower pace.

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

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

The standard definition of uniform acceleration in Relativity Theory is proper acceleration at a constant felt rate such as what an onboard recording accelerometer would show.I will accept that. Then the ruler (or the bar in front of SP2) everywhere experiences uniform acceleration, but not identical acceleration since it is different everywhere along its length. SP1 cannot be comoving with the 10-LD mark on the rod then since SP1’s acceleration (100g) does not match the ~35g acceleration of the 10-LD mark on the rod.

I notice that you did not point out any error in the mathematics used to demonstrate your proper speed calculation to be faulty. Therefore I presume that you agree that your numbers are self-inconsistent. Math doesn’t lie.For proper separation and proper speed that isSP1 final proper speed is 1.130 cStill 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.exactlywhat you do.

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The Principle of Relativity states that the laws of physics have the same form in all arbitrary frames of reference. Uniform proper acceleration for a specified proper time will yield the expected proper speed and the expected proper distance covered.This contradicts the pictures you and Jano keep posting that shows the front of the rod not covering as much proper distance as does SP2 to which it is attached. And yet you assert that the front of the rod somehow will be where SP1 is when it stops accelerating. This is you contradicting yourself.

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

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I am using proper separation correctly. The term proper refers to what is experienced, measured etc. by the participants.Indeed, and the proper separation thus means the separation as seen by comoving observers, which means the separation in their frame, and not in another. But you’re still measuring the separation by the frame of those grid markers, not by the frame of the observers.

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There is no such thing as ‘relativistic computation of proper speed’.I actually performed such a computation above where I worked out the contradiction in your numbers. Actual speed is dx’/dt’, and proper speed is dx/dt’, so all you need to do to get the relativistic proper speed is multiply actual speed by dx/dx’ which is λ.

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Saying that shows an utter ignorance of Relativity Theory. Proper speed is what an observer will calculate from proper acceleration history as confirmed by passage of landmarks.Yes, it is, but not by shoving that history through a Newtonian linear formula. That formula assumes no time dilation as speed increases.

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Constant proper acceleration does not require any relativistic adjustment in proper speed.Then show the error in my numbers instead of just asserting that you must be correct because you assume it must be that simple.

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Which is how we keep it absolutely stationary relative to SP2 at all times in the frame of SP2.Quote from: HalcIn post #8 (new thread numbering) you agreed to add thrusters to the rod to avoid the speed of sound problemQuoteThe 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.

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If there are no thrusters on the rod and it is attached to SP2, then the force waves from the acceleration will proceed up the rod as the speed of sound in the material.Which is why we didn’t posit a lack of thrusters. It allows us to use ideal math. The ruler in the pictures being posted are considered ideal rigid objects for simplicity. We’re sticking with that.

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You omitted the ‘Now the rod’ prefix to the section. If you had included that part it would have been obvious that SP1 end meant. SP1 end of the rod. But this way you can ignore the math.Got it. That was unclear because of course I don’t consider SP1 to be at the end of the rod.

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SP1 end proper acceleration is 100gYou don’t give any indication of duration in frame M or actual speed in frame M, so I’ll use the numbers you gave before, which is time 4.9082 days and .8112c. The exact time and speed doesn’t really matter so long as its the same for both SP1 and SP2 in frame M.

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

I cannot convert your proper speed to coordinate speed because you’re using Newtonian speed, which of course is the same as coordinate speed in Newtonian mechanics (things can move faster than light given those physics). But I’ll use your proper distance of 2.261 LD.

SP2 end is attached to SP2, so of course those numbers will be the same.

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

There are 1000 grid markers every light day, all stationary in the original frame M. You report each ship’s proper speed, meaning you are having them count these markers going by. So each has counted 2261 markers after a subjective 4 seconds.

My first observers are at markers -7739 and 2261. They are present at the point where each ship stops accelerating as it flies by at .8112c. The time on their clocks is 4.908, and they can see the time of 4 on the ship/rod clock as it passes by. The guy at the 2261 mark does nothing except note the location of the event where both rod and SP1 pass by together.

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

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

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

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

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

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