0 Members and 1 Guest are viewing this topic.
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 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 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.