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Quote from: Malamute Lover on 06/08/2020 22:39:50...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.L_{0} - length of the spaceship in its rest frame.5L_{0} - 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?JanoEdit: 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.

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

Quote from: Malamute Lover on 09/08/2020 20:58:06The 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 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.

Quote from: Halc on 10/08/2020 13:55:03Quote from: Jaaanosik on 10/08/2020 12:19:19Quote from: Malamute Lover on 09/08/2020 20:58:06SP1 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.QuoteIf 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

Quote from: Jaaanosik on 10/08/2020 12:19:19Quote from: Malamute Lover on 09/08/2020 20:58:06SP1 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.QuoteIf 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.

Quote from: Malamute Lover on 09/08/2020 20:58:06SP1 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?

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

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

Quote from: Jaaanosik on 10/08/2020 17:01:20Halc,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.

Halc,I see your point about acceleration and stopping.

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

QuoteHaving 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.

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

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.

Quote from: Halc on 10/08/2020 18:38:50Quote from: Jaaanosik on 10/08/2020 17:01:20Halc,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.QuoteHaving 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.Quotethe 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

Quote from: Jaaanosik on 10/08/2020 17:01:20Halc,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.QuoteHaving 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.

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

Great. An attempt to obfuscate by specifying everything in proper terms instead of coordinate terms.Let’s see if the numbers hold up...Quote from: Malamute Lover on 10/08/2020 19:44:10SP1a [0,0]SP1 proper acceleration is 100gSP1 duration of proper acceleration is 4 daysSP1 final proper speed is 1.130 cSP1 proper acceleration has been continuous, so average proper speed is 0.565 cSP1 proper distance covered is 2.261 LDSP1b proper [2.261,4]Still using Newtonian mathematics I see. This is wrong. The actual proper speed is 1.387cNow 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.

SP1a [0,0]SP1 proper acceleration is 100gSP1 duration of proper acceleration is 4 daysSP1 final proper speed is 1.130 cSP1 proper acceleration has been continuous, so average proper speed is 0.565 cSP1 proper distance covered is 2.261 LDSP1b proper [2.261,4]

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.QuoteSP2a [-10,0]SP2 proper acceleration is 100gSP2 duration of proper acceleration is 4 daysSP2 final proper speed is 1.130 cSP2 proper acceleration has been continuous, so average proper speed is 0.565 cSP2 proper distance covered is 2.261 LDSP2b proper [-7.739,4]2.261 -(-7.739) = 10 LDOK, 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.

SP2a [-10,0]SP2 proper acceleration is 100gSP2 duration of proper acceleration is 4 daysSP2 final proper speed is 1.130 cSP2 proper acceleration has been continuous, so average proper speed is 0.565 cSP2 proper distance covered is 2.261 LDSP2b proper [-7.739,4]2.261 -(-7.739) = 10 LD

QuoteSame 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.

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

QuoteThe 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.

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

…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.

QuoteSP1 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.QuoteSP1 end proper acceleration is 100gSP1 end duration of proper acceleration is 4 daysSP1 end final proper speed is 1.130 cSP1 end proper acceleration has been continuous, so average proper speed is 0.565 cSP1 end proper distance covered is 2.261 LDSP1 end at SP1b proper [2.261,4]SP2 end [0,-10]SP2 end proper acceleration is 100gSP2 end duration of proper acceleration is 4 daysSP2 end final proper speed is 1.130 cSP2 end proper acceleration has been continuous, so average proper speed is 0.565 cSP2 end proper distance covered is 2.261 LDSP2 end at SP2b proper [-7.739,4]2.261 -(-7.739) = 10 LDThis 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.

SP1 end [0,0]

SP1 end proper acceleration is 100gSP1 end duration of proper acceleration is 4 daysSP1 end final proper speed is 1.130 cSP1 end proper acceleration has been continuous, so average proper speed is 0.565 cSP1 end proper distance covered is 2.261 LDSP1 end at SP1b proper [2.261,4]SP2 end [0,-10]SP2 end proper acceleration is 100gSP2 end duration of proper acceleration is 4 daysSP2 end final proper speed is 1.130 cSP2 end proper acceleration has been continuous, so average proper speed is 0.565 cSP2 end proper distance covered is 2.261 LDSP2 end at SP2b proper [-7.739,4]2.261 -(-7.739) = 10 LD

QuoteSame 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.

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.

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/

Quote from: Halc on 03/08/2020 04:36:30 QuoteLet us define SP1x as Time = 2 on the SP1 clockAn event presumably?No, a time, a clock reading. The clocks are all synchronized. Location of the clock does not matter. QuoteSP1 proper acceleration is 100gSP1 duration of proper acceleration is 2 daysSP1 proper speed is 0.565 cSP1 proper acceleration has been continuous, so average proper speed is 0.2825 cSP1 at SP1x proper distance covered is 0.565 LDSP1 at SP1x proper [0.565,2]Let us define SP2x as Time = 2 on the SP2 clockSP2 proper acceleration is 100gSP2 duration of proper acceleration is 2 daysSP2 proper speed is 0.565 cSP2 proper acceleration has been continuous, so average proper speed is 0.2825 cSP2 at SP2x proper distance covered is 0.565 LDSP2 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.

QuoteLet us define SP1x as Time = 2 on the SP1 clockAn event presumably?

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

SP1 proper acceleration is 100gSP1 duration of proper acceleration is 2 daysSP1 proper speed is 0.565 cSP1 proper acceleration has been continuous, so average proper speed is 0.2825 cSP1 at SP1x proper distance covered is 0.565 LDSP1 at SP1x proper [0.565,2]Let us define SP2x as Time = 2 on the SP2 clockSP2 proper acceleration is 100gSP2 duration of proper acceleration is 2 daysSP2 proper speed is 0.565 cSP2 proper acceleration has been continuous, so average proper speed is 0.2825 cSP2 at SP2x proper distance covered is 0.565 LDSP2 at SP2x proper [-9.435,2]

QuoteNow the rodSP1 end proper acceleration is 100gAgain 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.

Now the rodSP1 end proper acceleration is 100g

QuoteSP1 end duration of proper acceleration is 2 daysSP1 end final proper speed is 0.565 cSP1 end proper acceleration has been continuous, so average proper speed is 0.2825 cSP1 end at SP1x proper distance covered is 0.565 LDSP1 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 = 0You’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.

SP1 end duration of proper acceleration is 2 daysSP1 end final proper speed is 0.565 cSP1 end proper acceleration has been continuous, so average proper speed is 0.2825 cSP1 end at SP1x proper distance covered is 0.565 LDSP1 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

QuoteThe 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?

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

QuoteLet 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.

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]

QuoteHow long is the rod at the 2 day mark?Frame dependent question.

How long is the rod at the 2 day mark?

QuoteRh proper acceleration is 100gYou’ve not established that with any numbers in the frame of the rod. So your argument falls apart here.

Rh proper acceleration is 100g

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.

QuoteThe 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?

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

QuoteSince 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.

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.

Quote from: Jaaanosik on 10/08/2020 20:34:03If 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.

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.

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.

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.

Quote from: Halc on 11/08/2020 01:09:42Quote from: Malamute Lover on 10/08/2020 19:44:10SP1 final proper speed is 1.130 cStill using Newtonian mathematics I see. This is wrong. The actual proper speed is 1.387cNow 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.

Quote from: Malamute Lover on 10/08/2020 19:44:10SP1 final proper speed is 1.130 cStill using Newtonian mathematics I see. This is wrong. The actual proper speed is 1.387cNow 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.

SP1 final proper speed is 1.130 c

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.

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

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

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.

Quote from: HalcQuoteThe rod has been defined as having constant acceleration along its length.It most certainly has not. The rod has been defined to be stationary in the frame of SP2 at all times.In post #8 (new thread numbering) you agreed to add thrusters to the rod to avoid the speed of sound problem

QuoteThe 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.

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.

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.

Quote from: Jaaanosik on 07/08/2020 15:34:25Edit: This is the Special Relativity acceleration 101 (One-o-One) in transition towards general frames. What do you know, this is the post #101 in this thread. Sorry Jano. I split the topic and it isn't post 101 anymore Quote from: Malamute Lover on 12/08/2020 22:37:27The observers at the end of the ruler never see any change in the length of the ruler. SP1 and SP2 never see any change in their separation and the position of the rod.Proof by bold I see.The former statement is true because the front of the ruler experiences less acceleration than the rear, as can be plainly seen in the diagram you posted.SP1 on the other hand accelerates at the same g-force as SP2 (the rear of the ruler), so it pulls away from the end of the ruler that is accelerating at a lower pace.Your numbers posted reflect this, and thus contradict what you’re asserting in bold here. SP1 and SP2 cover the same proper distance in their respective 4 subjective seconds.Look at the other picture with the number to see how far the end of the ruler gets in 4 subjective time units, which is even less. The left end of the ruler (red line) after 1.5 time units has moved from proper location 0 (you seem to only be able to think in such terms) to location 1.4, but the right end of the ruler (orange line) has moved from 1.5 to just short of 2, or about a third as far after two seconds on its clock. The clocks are not staying in sync with each other, not even in the inertial frame. But they do stay in sync (in the original frame) with SP1 and SP2.Quote from: Malamute Lover on 13/08/2020 00:40:54The 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.Quote from: Malamute Lover on 13/08/2020 03:52:20Quote from: Halc on 11/08/2020 01:09:42Quote from: Malamute Lover on 10/08/2020 19:44:10SP1 final proper speed is 1.130 cStill using Newtonian mathematics I see. This is wrong. The actual proper speed is 1.387cNow me just saying that is words, so let me let the numbers speak.You apparently got the proper speed the Newtonian way by multiplying 980m/sec2 by 345600 seconds which indeed yields a proper speed (dx/dt’) of 1.13c. This corresponds to a coordinate speed (dx/dt) of about .749c, not .8112c. Doing it this way leads to contradictions:Suppose we accelerated for 8 days instead of 4. Using your linear method, that would double the proper speed to 2.26c, but that contradicts the relativistic computation of proper speed:.749c coordinate speed doubled using relativistic addition is .9596c, which has a λ of 3.5558 yielding a proper speed of 3.4123c which contradicts the 2.26c using your method. Proper speeds don’t add in a linear way like you’re doing. You can’t use Newtonian method of just multiplying proper time by proper acceleration.For proper separation and proper speed that is exactly what you do.I notice that you did not point out any error in the mathematics used to demonstrate your proper speed calculation to be faulty. Therefore I presume that you agree that your numbers are self-inconsistent. Math doesn’t lie.QuoteThe 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.QuoteI 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.QuoteThere 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. Coordinate speed is dx’/dt’, and proper speed is (by definion) dx/dt’, so all you need to do to get the relativistic proper speed is multiply coordinate speed by dx/dx’ which is λ.QuoteSaying 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.QuoteConstant 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.QuoteYou 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, thanks. That was unclear because of course I don’t consider SP1 to be at the end of the rod.I will have at your numbers in that light.QuoteSP1 end proper acceleration is 100gSP1 end duration of proper acceleration is 4 daysSP1 end final proper speed is 1.130 cSP1 end proper acceleration has been continuous, so average proper speed is 0.565 cSP1 end proper distance covered is 2.261 LDSP1 end at SP1b proper [2.261,4]SP2 end [0,-10]SP2 end proper acceleration is 100gSP2 end duration of proper acceleration is 4 daysSP2 end final proper speed is 1.130 cSP2 end proper acceleration has been continuous, so average proper speed is 0.565 cSP2 end proper distance covered is 2.261 LDSP2 end at SP2b proper [-7.739,4]2.261 -(-7.739) = 10 LDYou don’t give any indication of duration in frame M or actual speed in frame M, so I’ll use the numbers you gave before, which is time 4.9082 days and .8112c. The exact time and speed doesn’t really matter so long as its the same for both SP1 and SP2 in frame M.I cannot convert your proper speed to coordinate speed because you’re using Newtonian speed, which of course is the same as coordinate speed in Newtonian mechanics (things can move faster than light given those physics). But I’ll use your proper distance of 2.261 LD.SP2 end is attached to SP2, so of course those numbers will be the same.So I’m going to put some stationary observer here and there in frame M with clocks synced in that frame. Each never moves. You refuse to tell me how far out of sync the ship clocks are in their proper frame, so I cannot talk about what time each pilot sees on the other clock, so I’m forced to use stationary observers. Your total inability to produce any figures in frame N is a serious indication that you don’t know what’s going on.There are 1000 grid markers every light day, all stationary in the original frame M. You report each ship’s proper speed, meaning you are having them count these markers going by. So each has counted 2261 markers after a subjective 4 seconds.My first observers are at markers -7739 and 2261. They are present at the point where each ship stops accelerating as it flies by at .8112c. The time on their clocks is 4.908, and they can see the time of 4 on the ship/rod clock as it passes by. The guy at the 2261 mark does nothing except note the location of the event where both 10-LD rod and SP1 pass by together per your assertion.The observer 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 almost 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.