Crash in Special Relativty
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dseppala
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Joined: Fri Apr 15, 2011 1:10 pm Posts: 359

Crash in Special Relativty
In an inertial reference frame which I'll call F0, let's say scientists build a device that can accelerate particles at a uniform acceleration rate. Let's say this device accelerates particles along the x axis. No matter where along the xaxis the particle starts it acceleration, the particle accelerates at a constant rate (as long as its speed is less than c). No matter what the initial speed of the particle along the xaxis is, this device accelerates the particle at the same constant acceleration rate.
Now there is another inertial reference frame moving with velocity V parallel to the xaxis of F0. I'll call this frame F1. Observers in F1 measure the motion of particles in this uniform acceleration device of F0. These F1 observers also agree that the acceleration pattern is the same pattern independent of the particle's starting point along the x axis of F0. Per Einstein, F1 doesn't necessarily agree that the acceleration and velocity of the particle is the same as measured in the two inertial reference frames, but F1 agrees that the scientists in F0 have built an ideal uniform acceleration device.
With this device a problem with Einstein's concept of simultaneous events occurs. Let's say there are two particles separated by some distance L along the xaxis, moving with some constant relative velocity V1 along the xaxis. Let observers in F1 start the acceleration of each particle in the F0 device simultaneously as measured in their F1 frame. Per Einstein's concept of simultaneous events, observers in F0 will measure that the two particles in the device did not start their accelerations simultaneously.
Per F1 once the identical acceleration of each particle starts simultaneously, the separation of the two particles is always measured to be the constant distance L. However in frame F0, since one particle started accelerating before the other particle, the distance between the two particles is constantly changing. We can set up the experiment so that the two particles are moving closer and closer to each other. In fact, within the bounds of the particles having a velocity less than c as measured in F0, we can set up this problem so that the two particles crash into each other some time after the accelerations of each particle have started.
If the two particles can crash into each other, how does frame F1 resolve the situation that both particles had the same initial velocity, both particles started their acceleration simultaneously, and the acceleration pattern of any particle is identical at any and every point along the xaxis, yet somehow one particle must have traveled faster than the other particle for the crash to occur? How do observers in F1 explain this crash?
David Seppala Bastrop TX

Sat Mar 18, 2017 8:36 pm 


Darwin123
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Joined: Sat Dec 13, 2014 4:00 pm Posts: 656

Re: Crash in Special Relativty
dseppala wrote: In an inertial reference frame which I'll call F0, let's say scientists build a device that can accelerate particles at a uniform acceleration rate. Let's say this device accelerates particles along the x axis. No matter where along the xaxis the particle starts it acceleration, the particle accelerates at a constant rate (as long as its speed is less than c). No matter what the initial speed of the particle along the xaxis is, this device accelerates the particle at the same constant acceleration rate.
Now there is another inertial reference frame moving with velocity V parallel to the xaxis of F0. I'll call this frame F1. Observers in F1 measure the motion of particles in this uniform acceleration device of F0. These F1 observers also agree that the acceleration pattern is the same pattern independent of the particle's starting point along the x axis of F0. Per Einstein, F1 doesn't necessarily agree that the acceleration and velocity of the particle is the same as measured in the two inertial reference frames, but F1 agrees that the scientists in F0 have built an ideal uniform acceleration device. This situation could not happen. A body having uniform acceleration in one inertial frame can not have uniform acceleration in another inertial frame. You said that the 'F1 observers agree' and that the 'F1 observer also agree'. I do not believe that. The speed of light time delays make such a statement impossible. You have to support your assertion. Otherwise, the rest of you problem makes no sense.

Sat Mar 18, 2017 9:14 pm 


OZLOFT
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Joined: Fri Nov 07, 2008 8:21 am Posts: 3759

Re: Crash in Special Relativty
That an object can accelerate at a constant rate from a reference frame (F0) is certainly possible in physics, observers attached to the object... ...feeling this constant force during the prolonged acceleration process. If another reference frame (spaceship F1) is passing by at velocity v relative to F0 this will NOT cause the acceleration undergone by said object to become NONuniform. So what does Tricky Dicky123 actually mean? Darwin123 wrote: You (dseppala) said that the 'F1 observers agree' and that the 'F1 observer also agree'. I do not believe that. The speed of light time delays make such a statement impossible. The speed of light and resulting delays in its transit does not alter what is true. The speed of light has nothing to do with whether an object accelerating from F0 has a uniform acceleration or not  since it is NOT a question of what time signals from the object and the various reference frames are received! So to make his bizarre claim TD123 has to support his assertion... Darwin123 wrote: You (dseppala) have to support your assertion. Otherwise, the rest of your problem makes no sense. ...with SR of course , otherwise his objections make no sense! But rather than just make snide assertions at what RTD123 really means, I can help you with the answer dseppala. TD123's criticisms are perfectly valid if and only if he accepts that the object undergoing uniform acceleration relative to F0 but not relative to F1 means that F0 is in one parallel universe where the object undergoes uniform acceleration whereas F1 is in another parallel universe where the object is undergoing NONuniform acceleration. In that way I can accept that TD123's criticism is 100% valid. Otherwise, if he trying to tell you that the object is undergoing nonuniform AND uniform acceleration in the very same situation, then his criticism is 100% BS. So you have a choice dseppala  think for yourself and reject TD123's criticism as selfserving Einsteinian BS ( ) or bow down and worship the mighty Big Al Einstein of the LightProhibition Era and his latterday prophet RTD123 ( )! Yours faithfully OZLOFT

Sun Mar 19, 2017 2:37 am 


dseppala
Board Warrior
Joined: Fri Apr 15, 2011 1:10 pm Posts: 359

Re: Crash in Special Relativty
Darwin, I should have clarified things for you. I said in frame F0 they built a device that has uniform acceleration as measured in F0. I said this so that its easy to do a calculation of velocity and distance as a function of time in F0. The main point is that at any position along the x axis, a particle that has some initial velocity (or zero) and starts accelerating will follow the exact same acceleration pattern as any other identical particle with the same initial velocity that starts at some other x coordinate of F0. The acceleration pattern is independent of the x position in the device where the acceleration starts. . Frame F1 will also measure that the acceleration pattern is independent of the x position of the device where the acceleration of the particle starts. If this is true as measured by observers in F1, then observers in F1 will always measure that distance between two particles that start accelerating simultaneously (as measured by observers in F1) to be constant.
David Seppala Bastrop TX

Sun Mar 19, 2017 8:08 am 


Darwin123
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Joined: Sat Dec 13, 2014 4:00 pm Posts: 656

Re: Crash in Special Relativty
dseppala wrote: Darwin, I should have clarified things for you. I said in frame F0 they built a device that has uniform acceleration as measured in F0. I said this so that its easy to do a calculation of velocity and distance as a function of time in F0. The main point is that at any position along the x axis, a particle that has some initial velocity (or zero) and starts accelerating will follow the exact same acceleration pattern as any other identical particle with the same initial velocity that starts at some other x coordinate of F0. The acceleration pattern is independent of the x position in the device where the acceleration starts. That is not true. Okay, ltas do this. Suppose that in inertial frame F0, the particle satisfies a Newtonian type acceleration: x=X_0+V_0 t+0.5At^2. Of course: x is position (m), and t is time (s) in frame F0 only. I will use W to designate the velocity of frame F1 relative to frame F0. Please note that W is not the same as v, the velocity of the particle in frame F0. Mechanical forces are being applied in such a way that the above equation is satisfied in frame F_0 only. I am using small case letters for parameters that are measured in frame F0 and capital letters for parameters that won't change with frame. So the parameters X_0, V_0 and A are constants that won't change in the problem. Now, let us use primes to designate the values of parameters that will be measured in frame F1. According to the Lorentz transformation: x=γ[(x'Wt'] t=γ[t'+Wx'/c^2] γ=1/sqrt[1W/c^2] I purposely used the inverse of the Lorentz transform that is usually presented. However, it is a matter of simple algebra to show that this is equivalent to the Lorentz transform that I have seen you use in the past. Now do a direct substitution of the Lorentz transform into the equation of x. γ[(x'Wt']=X_0+V_0 γ[t'+Wx'/c^2] + 0.5 A γ[t'+Wx'/c^2]^2 Note that I have not changed X_0, V_0 or A. These parameters are independent of measured time and measured position, and so are not changed by the LT. x'=W t'+X_0/γ+V_0 [t'+Wx'/c^2] + 0.5 A γ[t'+Wx'/c^2]^2 Okay, I still haven't changed X_0, V_0 or A. I have merely applied the LT to the expression for particle position. Let me collect terms. I am NOT even going to use calculus. x'=X_0/γ+ t' {V_0+ 0.5 A γ[t'+2Wx'/c^2]} + 0.5 x' A{W/c^2 + γ[Wx'/c^2]^2} The acceleration as measured in frame F1 is very obviously NOT independent of position and time as measured in frame F1. It should be evident. However, I complete the problem mathematically to make it clear. So now let us define new parameters X_0', V_0', and A' such that: x'=X_0'+V_0' t+0.5A' t'^2. V_0'=V_0+ 0.5 A γ[t'+2Wx'/c^2] So, X_0'=X_0/γ V_0'={V_0+ 0.5 A γ[t'+2Wx'/c^2]} A'=x' A{W/c^2 + γ[Wx'/c^2]^2}/t'^2 You should check my results. I did this rather quickly in my head, so I made have made a small mistake. If you find a mistake, then it most likely won't affect this conclusion. So the above 3 expressions are the LT for parameters X_0, V_0 and A. You implied that the parameters were uniform in frame F0, which is possible. We see that the corresponding parameters after the LT are not independent of position and time in frame F1. Obviously, V_0' and A' are not independent of position and time as measured in F1. So your initial hypothesis already contradicts SR. So the logical contradiction was in YOUR initial hypothesis, not relativity. Here is the reason. The measuring instruments of F0 and F1 are different. The Doppler lidar used in frame F0 is completely different from the Doppler lidar used in frame. I just used the kinematics for the above demonstration. I stuck with 'time and space' rather than look at the mechanical forces. In setting up your problem, you ignored the mechanical forces. This was your mistake. You implicitly assumed that the mechanical forces that were causing the acceleration were uniform in both reference frames. However, this is impossible according to the dynamic part of SR. To set up your initial system in frame F0, you had to make implicit assumptions on the mechanical forces acting on the particle. There would be no acceleration on the particle if there were no mechanical forces. However, the mechanical forces are subject to the LT. You assumed that the mechanical forces were not affected by the LT. You made the common mistake of applying the kinematics of SR while ignoring the kinetics of SR. Anyway, that is how a relativist answers the question. Once all the measured conditions in F0, the corresponding conditions in F1 are fixed by the LT. The change from F0 to F1 merely corresponds to a change in measuring instruments, not a change in the particles being examined by the measuring instruments. Once you have chosen the arbitrary parameters of motion for a particle in inertial frame F0, You have no freedom to choose the corresponding parameters in F1. Therefore, you can not assume that the acceleration of a particle under examination is the uniform in both F0 and F1. Pay attention to what I just said. I did not say that the accelerations can be uniform but have different values in F0 and F1. I said that the accelerations could not both be uniform in F0 and F1, even if one accepts the fact that the accelerations are different. Uniform acceleration in one frame is uniform acceleration in another frame only in Newtonian physics (Galilean transformation). Uniform acceleration in one inertial frame IS NOT uniform acceleration in another frame according to SR (Lorentz transformation). The acceleration of a particle can not be uniform and constant in both F0 and F1 that are moving relative to each other. If you have a particle as measured in F0 that is accelerating at a constant rate, then the same particle as measured in F1 is not accelerating at a constant rate. Hence, your hypotheses contradicted SR from the very beginning. You assumed that SR was inconsistent before you showed that SR was inconsistent.

Sun Mar 19, 2017 11:00 am 


dseppala
Board Warrior
Joined: Fri Apr 15, 2011 1:10 pm Posts: 359

Re: Crash in Special Relativty
Again Darwin, I should have been clearer for you.
Let's just look at frame F0. I say that scientists in that frame built a device that produces uniform acceleration of a particle along the xaxis. IN FRAME F0, if a particle with velocity V starts accelerating at x=0, and another identical particle with velocity V starts accelerating at x = x1, they have the identical acceleration pattern. If an identical particle with velocity V starts accelerating at x = x2, it has the identical acceleration pattern as the first two particles.
I am stating that as measured in F0, if a particle with velocity V starts accelerating at any point along the xaxis, the acceleration pattern as measured in frame F0 is always identical, independent of the particle's starting location. The starting position as measured in frame F0 has zero effect on the acceleration pattern as measured in frame F0.
Without going into anything as measured in frame F1 and only looking at the acceleration pattern as measured in frame F0, do you agree that the starting position of the particle along the xaxis has zero effect on the acceleration pattern as measured in frame F0?
If so, then we can proceed to the next step in the problem. If not, tell me why you don't agree?
David Seppala Bastrop TX

Sun Mar 19, 2017 1:43 pm 


cryptic
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Joined: Sat Jun 06, 2009 10:12 am Posts: 1054

Re: Crash in Special Relativty
Darwin123 wrote: ... In setting up your problem, you ignored the mechanical forces. This was your mistake. You implicitly assumed that the mechanical forces that were causing the acceleration were uniform in both reference frames. However, this is impossible according to the dynamic part of SR. ... Suppose that the particle is accelerated in a homogeneous gravitational field!

Sun Mar 19, 2017 2:01 pm 


Darwin123
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Joined: Sat Dec 13, 2014 4:00 pm Posts: 656

Re: Crash in Special Relativty
dseppala wrote: Again Darwin, I should have been clearer for you. ... I am stating that as measured in F0, if a particle with velocity V starts accelerating at any point along the xaxis, the acceleration pattern as measured in frame F0 is always identical, independent of the particle's starting location. And I am stating that the acceleration pattern in F0 and F1 IS NOT identical. The acceleration of the particle in inertial frame F1 IS NOT independent of the particles starting location, even if the acceleration in intertial frame F0 IS independent of particle position. I understand very well that with a lot of work, you can make a particle with constant acceleration as measured in F0. I really, really understand that. However, the acceleration of the same particle will not be constant as measured in F1. The same particle exists in both inertial frames. However, different inertial frames have different measuring instruments. I understand exactly what you are claiming. I hypothesize that you don't have any idea in your own head what you mean by 'acceleration pattern'. I hypothesized that what you meant by 'acceleration pattern' is uniform acceleration within a frame. If 'uniform' is not what you meant by 'pattern', then maybe I don't know what you are talking about. However, it is quite possible that you don't know what you mean, either. There is no quantitive definition of 'acceleration pattern' anywhere. If you don't believe me, Google it. I have agreed with you that it is possible to force a particle to travel with uniform acceleration in frame F0. If one chooses a complex and unique 'pattern' of mechanical forces acting on the particle, the the acceleration of particle AS MEASURED IN inertial frame F0 could be precisely uniform over a finite amount of time. So I gladly made that assumption. The formula for distance, x, of a uniformly accelerating particle is determined solely by the initial velocity and the initial acceleration. It doesn't matter whether the forces are Newtonian or relativistic as long as the acceleration is uniform and constant. So I wrote an expression giving the position of the particle as measured in frame F0. The only three degrees of freedom in this expression are X_0, V_0 and A. These three parameters are arbitrary and constant. These three parameters can't vary with x for a uniformly accelerating particle. The only other degree of freedom that is arbitrary is the relative velocity between F0 and F1. Again, W can't vary with position according to your own assumptions. So the only arbitrary constants in your example is X_0, V_0, A and W. Once you choose these four parameters, you have no freedom to choose any others according to your assumptions. After you choose these four parameters, you have no more choice. The values for position and time for that particle in F1 have to be determined by the Lorentz transform, and only the Lorentz transform (LT). Therefore, you can not decide that the acceleration of the particle in F1 is uniform. I showed that if the acceleration is uniform in frame F0, then the acceleration in frame F1 can not be uniform. could not be uniform. The acceleration of the particle in frame F1 will vary explicitly with position and time. Quite definitely, the particles acceleration will vary implicitly with position. Again, I accuse you of not knowing what your own word means. If 'acceleration pattern' means uniformity, then you are WRONG. If it means something else, then the burden is on you to tell us what it is. According to your latest response, the word 'pattern' does not mean 'uniformity'. I showed you that the acceleration in frame F1 is not uniform. You said that I didn't understand you. So maybe you are right. However, the burden is on you to tell the rest of us what the word 'pattern' means. Please tell us in quantitative terms what the phrase 'acceleration pattern' means. If you can't provide us a formal definition of 'acceleration pattern'. then I will not respond to any more of your posts. I don't think anyone else knows what you mean by 'acceleration pattern'. I invite other members of this forum to explain to us what DSEPPELA means by an 'acceleration pattern'. Here are some helpful links that I was able to look up for you. Choose any ‘acceleration pattern’ that you like. Just tell me how it is relevant. ‘Acceleration pattern’ as graph of position versus time. Notet hat you did not include a graph in your problem. http://www.chegg.com/homeworkhelp/ques ... tq4891494Acceleration pattern as ‘athletic style’ http://speedendurance.com/2009/02/17/vi ... velocity/‘Pattern’ as bone oscillation. https://www.ncbi.nlm.nih.gov/pubmed/12855299Heart rate patterns https://www.ncbi.nlm.nih.gov/pubmed/22008491Nonuniform rates of acceleration in people listening to Smartphones. http://journals.plos.org/plosone/articl ... ne.0094811Acceleration pattern in software. https://www.ibm.com/developerworks/comm ... _0?lang=en

Sun Mar 19, 2017 3:31 pm 


OZLOFT
Consumed by Physics
Joined: Fri Nov 07, 2008 8:21 am Posts: 3759

Re: Crash in Special Relativty
We imagine numerous spaceships starting off from different positions in F0 frame and undergoing equal & uniform accelerations. The observers inside the spaceships will note the constancy of the accelerating forces experienced and measure them accordingly with accurate instruments. They then broadcast these results to observers stationary in the F0 frame and others moving at W, i.e. stationary in the F1 frame! Will the observers stationary in the F0 frame and the F1 frame hear the same words & results from the observers in the accelerating spaceships?Let's work this out, using Einstein's "principles" which Darwin123 very faithfully reproduces. RTD123 wrote: And I am stating that the acceleration pattern in F0 and F1 IS NOT identical. The acceleration of the particle [or spaceship] in inertial frame F1 IS NOT independent of the particles [or spaceships'] starting location, even if the acceleration in inertial frame F0 IS independent of particle position.
I understand very well that with a lot of work, you can make a particle with constant acceleration as measured in F0. I really, really understand that. However, the acceleration of the same particle will not be constant as measured in F1. And I really really think that D123 has not considered what the F1 acceleratingspaceship observers and their instruments will broadcast about the constancy of the accelerating forces (i.e. the acceleration pattern). RTD123 wrote: The same particle [or spaceship] exists in both inertial frames. However, different inertial frames have different measuring instruments.
I understand exactly what you are claiming. I hypothesize that you don't have any idea in your own head what you mean by 'acceleration pattern'. Why abuse the poor guy gratuitously? He isn't OZLOFT who is the one who really deserves such dismissive treatment! RTD123 wrote: I hypothesized that what you meant by 'acceleration pattern' is uniform acceleration within a frame. If 'uniform' is not what you meant by 'pattern', then maybe I don't know what you are talking about. However, it is quite possible that you don't know what you mean, either.
There is no quantitive definition of 'acceleration pattern' anywhere. If you don't believe me, Google it. Quite right there, D123  because Einsteinians don't want this issue exposed, for the simple reason that it reveals how absurd SR really is since they cannot resolve the issue satisfactorily! RTD123 wrote: I have agreed with you that it is possible to force a particle to travel with uniform acceleration in frame F0. If one chooses a complex and unique 'pattern' of mechanical forces acting on the particle, the the acceleration of particle AS MEASURED IN inertial frame F0 could be precisely uniform over a finite amount of time.
So I gladly made that assumption. The formula for distance, x, of a uniformly accelerating particle is determined solely by the initial velocity and the initial acceleration. It doesn't matter whether the forces are Newtonian or relativistic as long as the acceleration is uniform and constant. So I wrote an expression giving the position of the particle as measured in frame F0.
The only three degrees of freedom in this expression are X_0, V_0 and A. These three parameters are arbitrary and constant. These three parameters can't vary with x for a uniformly accelerating particle. The only other degree of freedom that is arbitrary is the relative velocity between F0 and F1. Again, W [i.e. the velocity of F1 compared to F0] can't vary with position according to your own assumptions.
So the only arbitrary constants in your example is X_0, V_0, A and W. Once you choose these four parameters, you have no freedom to choose any others according to your assumptions. After you choose these four parameters, you have no more choice. The values for position and time for that particle in F1 have to be determined by the Lorentz transform, and only the Lorentz transform (LT). Therefore, you can not decide that the acceleration of the particle in F1 is uniform. Well why don't we ask the observers in the acceleratingspaceships whether they find uniform acceleration or not? RTD123 wrote: I showed that if the acceleration is uniform in frame F0, then the acceleration in frame F1 can not be uniform. could not be uniform. The acceleration of the particle in frame F1 will vary explicitly with position and time. Quite definitely, the particles' [and accelerating spaceships'] acceleration will vary implicitly with position. Which means that in the F1 universe, the acceleratingspaceships observers will broadcast that their accelerational forces varied (from each other as well as over time), whereas in the F0 universe, the same observers broadcast that their accelerational forces DID NOT vary. Well Hugh Everett! Seems Darwin123 owes you a big apology! RTD123 wrote: Again, I accuse you of not knowing what your own word means. If 'acceleration pattern' means uniformity, then you are WRONG. If it means something else, then the burden is on you to tell us what it is. Again, why the gratuitous abuse of dseppala (instead of abusing OZ, who deserves it)? RTD123 wrote: According to your latest response, the word 'pattern' does not mean 'uniformity'. I showed you that the acceleration in frame F1 is not uniform. You said that I didn't understand you. So maybe you are right. However, the burden is on you to tell the rest of us what the word 'pattern' means. Please tell us in quantitative terms what the phrase 'acceleration pattern' means. If you can't provide us a formal definition of 'acceleration pattern'. then I will not respond to any more of your posts. I don't think anyone else knows what you mean by 'acceleration pattern'. Well OZLOFT does! If I remember my calculus (Leibniz notation) correctly, a constant acceleration pattern (where K means a 'constant') will be: d²x/dx² = K; d³x/dx³ = 0: the latter term means "rate of change of acceleration[al force]." So what dseppala means is that during the timeperiod of the experiment, the acceleration undergone by the accelerating spaceships (or particles) was uniform, meaning that instruments and observers in these spaceships will record this FACT! In the Einsteinworld however you seem either to expect them to lie about their findings or you (and they) have to resort covertly to Hugh Everett's parallel universes theory! RTD123, all sweetness and light, wrote: I invite other members of this forum to explain to us what DSEPPELA means by an 'acceleration pattern'. Thanks, D123. So I know you'll appreciate my mathematical elucidations above! But for those who still think I'm a jerk for taunting D123 by posting the equations above, I have a suitable reference. https://en.wikipedia.org/wiki/Jerk_(physics) Yours faithfully OZLOFT
Last edited by OZLOFT on Tue Mar 21, 2017 12:12 am, edited 6 times in total.

Sun Mar 19, 2017 10:07 pm 


dseppala
Board Warrior
Joined: Fri Apr 15, 2011 1:10 pm Posts: 359

Re: Crash in Special Relativty
Darwin wrote: Quote: And I am stating that the acceleration pattern in F0 and F1 IS NOT identical. The acceleration of the particle in inertial frame F1 IS NOT independent of the particles starting location, even if the acceleration in intertial frame F0 IS independent of particle position. I do not disagree with that statement, but it is not relevant to the problem. In F0 the acceleration of a particle is independent of the x position where the particle starts its acceleration. The question for you, as measured in F1 does the x' coordinate where the acceleration of the particle starts affect the acceleration as measured in F1? If you agree that the particle, AS MEASURED IN F1 can start at x' = 10, or x' = 10000 or x' = 380 or x' = whatever and the acceleration pattern will be identical AS MEASURED IN F1 at each of those starting points, then we can go to the next step. What I am saying is that observers in F1 measure that the starting x' coordinate where the acceleration starts has zero effect on the acceleration of the particle. If you agree, then we will let observers in F1 simultaneously, as measured in F1, start the acceleration of two particles at two different x' coordinates separated as distance D. Now do the observers in F1 measure that the distance between these two particles always remains equal to D? David Seppala Bastrop TX

Mon Mar 20, 2017 7:17 am 


Darwin123
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Joined: Sat Dec 13, 2014 4:00 pm Posts: 656

Re: Crash in Special Relativty
OZLOFT wrote: Well why don't we ask the observers in the acceleratingspaceships whether they find uniform acceleration or not?
Because that would violate both logic and the English language. 1. DSEPPELA said very specifically that both F0 and F1 are INERTIAL frames. 2. An accelerating spaceship can not be an inertial frame. 3. QED: Neither accelerating space ship is either frame F0 or frame F1. By acceleration, you obviously mean proper acceleration. Let us look at this in terms of proper acceleration. A space ship exists according to both frame F0 and frame F1. If the space ship is accelerating in frame F0, then it is also accelerating in frame F1. Thus, there is a mechanical force with a source acting on the space ship. Therefore, the observer sitting in the space ship has a proper acceleration that is a nonzero vector. Therefore, the observer accelerating space ship can not be an inertial observer. The rules of physics as construed by the accelerating observers are significantly different from the rules of physics in an inertial frame as in a noninertial frame, in both Galilean and Lorentz invariant physics. For instance, the accelerating space ships can't even prove that momentum in both Lorentzian and Galilean physics. However, we have been discussing the precise same issues before. Therefore, I don't wish to debate you further in this thread. I will not respond to you any more in this thread. I am waiting for DSEPPELAs response. You still didn't tell me what DSEPELLA meant by 'acceleration pattern'. I did not ask for a new problem with more reference frames. I asked the reader to explain what a 'proper acceleration' is. So your attempted to derail the discussion. You made a non sequitur. I will not respond any more to non sequiturs. Since F0 and F1 are both inertial frames, the force laws are the same in frame F0 and frame F1. The force laws in accelerating space ships are different from the force laws in frame F0 and Frame F1. So the laws of physics that he is using don't have to apply for observers in the accelerating space ships. The observers in the space ships are not part of the problem presented by the original post. If anyone wants to bring in reference frames other than F0 and F1, then I suggest that he start another thread. However, DSEPELLA's example include only inertial frame F0 and inertial frame F1. Therefore, there are no mechanical forces acting on any of the observers in this problem. The only forces that are acting in this problem are acting on the particles being observed. The ONLY observers in this problem are F0 and F1. I will only respond to DSEPPELA in this thread from now on.

Mon Mar 20, 2017 12:16 pm 


OZLOFT
Consumed by Physics
Joined: Fri Nov 07, 2008 8:21 am Posts: 3759

Re: Crash in Special Relativty
So now I see! Dseppala is guilty of improper acceleration! Darwin123 wrote: OZLOFT wrote: Well why don't we ask the observers in the acceleratingspaceships whether they find uniform acceleration or not? Because that would violate both logic and the English language. That's meaningless since we are allowed to have spaceships with observers in them  i.e. the question is violating nature, not your abstract logic which you now demonstrate in spades! Darwin123 wrote: #1. DSEPPELA said very specifically that both F0 and F1 are INERTIAL frames. #2. An accelerating spaceship can not be an inertial frame. #3. QED: Neither accelerating space ship is either frame F0 or frame F1. But dseppala stipulated  and you had to admit its possibility  that uniform acceleration from many points at F0 can take place! The uniform acceleration is a force occurring to the spaceships, not frames F0 or F1 as you backhandedly admit in #3. But the uniform acceleration being a fact concerning the spaceships, it will be experienced by observers there and become a universally true fact  i.e. true for all other observers no matter what their velocity i.e. whether they are at rest relative to F0, F1 or even Donald Trump! Darwin123 wrote: By acceleration, you obviously mean proper acceleration. Let us look at this in terms of proper acceleration.
A space ship exists according to both frame F0 and frame F1. If the space ship is accelerating in frame F0, then it is also accelerating in frame F1. Thus, there is a mechanical force with a source acting on the space ship. Therefore, the observer sitting in the space ship has a proper acceleration that is a nonzero vector. Therefore, the observer accelerating space ship can not be an inertial observer.
The rules of physics as construed by the accelerating observers are significantly different from the rules of physics in an inertial frame as in a noninertial frame, in both Galilean and Lorentz invariant physics. For instance, the accelerating space ships can't even prove that momentum in both Lorentzian and Galilean physics. But they can measure the force of acceleration (it feels like gravity!) and work out whether it is uniform i.e. constant in its application. They can also measure spectral lines from stars in front of and behind them (e.g. via the Fehrenbach Method) and from that work out their change in velocity relative to said stars and relative to the beginning of the acceleration when at rest relative to F0. But you will be happy to know in reply to your words below that I do NOT want you to reply to me in this thread, since dseppala is doing an excellent job on his own. When the good cop is raking the suspect over, why is there any need for a bad cop like OZ? Darwin123 wrote: However, we have been discussing the precise same issues before. Therefore, I don't wish to debate you further in this thread. I will not respond to you any more in this thread. I am waiting for DSEPPELAs response.
You still didn't tell me what DSEPELLA meant by 'acceleration pattern'. Yes I did, you jerk, I told you that 'acceleration pattern' is referring to the third differential of Newton's laws of motion i.e. the rate of change of acceleration. Darwin123 wrote: I did not ask for a new problem with more reference frames. I asked the reader to explain what a 'proper acceleration' is. So your attempted to derail the discussion. You made a non sequitur. I will not respond any more to non sequiturs.
Since F0 and F1 are both inertial frames, the force laws are the same in frame F0 and frame F1. The force laws in accelerating space ships are different from the force laws in frame F0 and Frame F1. So the laws of physics that he is using don't have to apply for observers in the accelerating space ships. The observers in the space ships are not part of the problem presented by the original post. Their relevance is in demonstrating the physical impossibility of SR when applied to actual physical situations  such as what dseppala outlines. Darwin123 wrote: If anyone wants to bring in reference frames other than F0 and F1, then I suggest that he start another thread. However, DSEPELLA's example include only inertial frame F0 and inertial frame F1. Therefore, there are no mechanical forces acting on any of the observers in this problem. The only forces that are acting in this problem are acting on the particles being observed. The ONLY observers in this problem are F0 and F1. These words imply that what happens to particles wrt* F0 (e.g. a uniform acceleration pattern i.e. without jerks) are not what happens to the very same particles wrt F1. But I'm sure that dseppala is already wise to that Einsteinian escape hatch! Darwin123 wrote: I will only respond to DSEPPELA in this thread from now on. Glad to hear it, D123, since I am now fully confident that antirelativists on this website can handle pretty well all the Einsteinian chicanery  including " improper acceleration." But as dseppala has already replied to YOU, D123 in the posting above your reply to me immediately above, it is now YOUR turn to reply to dseppala's query there! Yours faithfully OZLOFT *wrt  "with respect to".

Mon Mar 20, 2017 12:44 pm 


Darwin123
Obsessed With the Question
Joined: Sat Dec 13, 2014 4:00 pm Posts: 656

Re: Crash in Special Relativty
dseppala wrote: If you agree that the particle, AS MEASURED IN F1 can start at x' = 10, or x' = 10000 or x' = 380 or x' = whatever and the acceleration pattern will be identical AS MEASURED IN F1 at each of those starting points, then we can go to the next step. What I am saying is that observers in F1 measure that the starting x' coordinate where the acceleration starts has zero effect on the acceleration of the particle. I have repeatedly disagreed that 'the acceleration pattern will be identical as measured in F1'. I have disagreed several times. So according to this new claim, 'we can't go on to the nent step' and never could. You should have said something about that before. I don't know what you mean by the last sentence. The last sentence is badly constructed, for one thing. I don't know 'how one can measure' that 'where the accelerations starts has zero effect on the acceleration of the particle'. Since this sentence is completely incoherent, I have to make an educated guess as to what you mean. You seem to be saying that there are two events characterized by mechanical forces that occur at the exact same time (t') in F1, even though they occur in different places (xA'>xB') in F1. I will agree that the mechanical forces at different points x' are not causally connected at any one time, t'. In order for two events to be causally connected, an electromagnetic pulse has to pass from one point to another. If you don't provide the light enough time to travel, then two events in different places can't be causally connected. Two events occurring at the exact same time can be causally connected to a third event, but not to each other. I don't see what the relevance. You seem to be saying that if one were to graph the acceleration of the particle against time, the shape of the graph in frames F0 and F1 would be the same except for a time delay. The Lorentz transformation, according to you, would only create a time delay in the acceleration. Presumably, this is because the mechanical forces as measured in frame F0 and F1 are the same. In other words, you claim that the mechanical forces that cause the acceleration of the particle are not changed by the Lorentz transformation.This is not true! I showed mathematically that this is not true! Frame F0 and F1 contains the devices necessary to make a measurement. The length and rate of each device is affected by its motion, caused by delays in their internal forces. There is a time delay between the forces that communicate between the forces in F0 and F1. The LT characterizes the differences between the devices in F0 and F1. The observation of the particle does not affect the devices. There are articles online that give generalized expressions for how acceleration will differ in inertial frame F0 and inertial F1. I already cited such articles. You apparently don't remember our previous conversations. So I have given you a rather rough demonstration of how acceleration (proper OR improper) changes between F0 and F1. So you are free to find a mathematical error in either my derivation or someone else's derivation. I suspect there may be an error with mine, as I did it very quickly in my head. However, I can cite the same articles that I cited before. The authors probably spent more time on the derivation. However, no author found an 'acceleration pattern that was the same' in both inertial frames. The hypothesis that there is 'the same' acceleration in frames F0 and F1 is yours and yours alone. This stated assumption with no justification can not be part of a compelling argument. If you want further response, stop saying 'if we agree'. I don't agree with you. You can try to tell me why I am wrong. However, I have no urge 'to go on'. If I am unconvinced of your hypothesis, then I refuse to go on. The statement 'If you agree' does not obligate me 'to agree'. Your logic crashed, not special relativity.

Mon Mar 20, 2017 1:30 pm 


dseppala
Board Warrior
Joined: Fri Apr 15, 2011 1:10 pm Posts: 359

Re: Crash in Special Relativty
I wrote: Quote: If you agree that the particle, AS MEASURED IN F1 can start at x' = 10, or x' = 10000 or x' = 380 or x' = whatever and the acceleration pattern will be identical AS MEASURED IN F1 at each of those starting points, then we can go to the next step. What I am saying is that observers in F1 measure that the starting x' coordinate where the acceleration starts has zero effect on the acceleration of the particle. Darwin replied: Quote: I have repeatedly disagreed that 'the acceleration pattern will be identical as measured in F1'. I have disagreed several times. So according to this new claim, 'we can't go on to the nent step' and never could. You should have said something about that before. Darwin, read my English very carefully  you seem to be adding things that I did not state. I did not say that the acceleration pattern as measured in F1 is identical to the acceleration pattern measured in F0. What I'm saying is that the acceleration pattern of a particle is independent of the starting position of the acceleration along the x axis. If you disagree with that, please explain the difference as measured by observers in F1 when a particle with velocity Vp starts accelerating at x' = 10 compared to an identical particle with velocity Vp starts accelerating at x' = 10000 as measured in F1. David Seppala Bastrop TX

Mon Mar 20, 2017 5:26 pm 


Darwin123
Obsessed With the Question
Joined: Sat Dec 13, 2014 4:00 pm Posts: 656

Re: Crash in Special Relativty
dseppala wrote: Darwin, read my English very carefully  you seem to be adding things that I did not state. I did not say that the acceleration pattern as measured in F1 is identical to the acceleration pattern measured in F0. What I'm saying is that the acceleration pattern of a particle is independent of the starting position of the acceleration along the x axis. If you disagree with that, please explain the difference as measured by observers in F1 when a particle with velocity Vp starts accelerating at x' = 10 compared to an identical particle with velocity Vp starts accelerating at x' = 10000 as measured in F1.
1. The inertial mass of the particle as measured in F1 at x'=10000 is probably larger than the inertial mass of the particle at measured by F1 at x'=10. 2. The mechanical force on the particle as measured in F1 when starting x'=10000 is probably different than the mechanical force of the particle at measured by F1 when starting at x'=10. Consider a particle that starts of at x=0 in F0 and accelerates at a constant acceleration up to x=10000 in F0. That is what you started with, after all. The rest mass of the particle remains constant. However, the kinetic energy of the particle as measured in F0 is increasing. Special relativity has equivalence of energy and mass. So the inertial mass of the particle as measured in F0 is increasing with x. The kinetic energy of the particle increased because there was a mechanical force doing work on the particle that added kinetic energy to the particle. The total energy, and therefore the total inertial mass of the system, has to be conserved in any inertial frame including F0. So there was inertial mass being transferred from some other particles into the particle that you said was accelerating at a constant rate in inertial frame F0. The acceleration of the particle is constant and uniform. The acceleration of the particle is the mechanical force on the particle divided by its inertial mass. So the mechanical force as measured by F0 has to also be increasing with x. So in frame F0, both the mechanical force on the particle AND the inertial mass of the particle are increasing with x. The only reason that the acceleration is constant ia because the increase in both quantities are cancelling out. So you can not assume, even in frame F0, that the mechanical force is independent of position. We know that as the particles velocity approaches the speed of light, the inertial mass approaches infinity. So the particle will never exceed the speed of light. So the time over which the acceleration is constant is limited. However, the mechanical force as measured in F0 also approaches infinity until the speed of the particle reaches infinity. The mechanical force on the particle can not be independent of position. The particle will reach the speed of light at a position xA and time tA. We switch instruments. We decide to measure the progress of the same particle from the standpoint of inertial frame F1. Inertial frame F1 is moving at a velocity W relative to frame F0. The speed, W, can be very close to the speed of light. Also, W may be >0. You didn't way W was close to the speed of light or positive but less assume that to facilitate the discussion. Again the mass and the mechanical force as measured in F1 have to be increasing with distance, x'. However, they don't have to be strictly proportional to each other. They can't be proportional to each other as the particle speed, V', reaches the speed of light. In fact, the particle has to approach the speed of light at an xA' and tA' that are different from xA and tA. So the acceleration of the particle CAN'T be constant in frame F1. The mechanical forces that accelerate the particle are Lorentz invariant. So they can't move a particle faster than the speed of light. That means that at some point where the particles speed starts to approach the speed of light, the acceleration of the particle has to decrease. This point, where the acceleration is decreasing, is different in frame F0 and F1. So the acceleration pattern of the particle has to be significantly different in frames F0 and F1 in some region where the particle approaches the speed of light. So the acceleration of the particle can't be constant for all time in either F0 or F1. I agreed to assume that the acceleration is constant over some small distance in frame F0. There may be a finite region in F1 where the acceleration is constant. However, there is no reason to think that the two regions overlap. Your assumption that the acceleration could be constant forever is wrong. For particles moving close to the speed of light, the effort of choosing forces to keep the acceleration constant becomes very hard. At some point, the acceleration has to decrease in either frame. So 'where the acceleration starts' very much affects the 'acceleration pattern'. You are making the mistake of thinking that if the acceleration of the particle is independent of position, the mechanical force acting on it has to be independent of position. This would be true if the inertial mass were independent of position. The inertial mass of the particle is not independent of position and it is not independent of inertial frame. The mechanical force causing the acceleration is not independent of position and is not independent of inertial frame. The uniform acceleration of the particle in frame F0 is a fortuitous accident. Or rather: The uniform acceleration of the particle in frame F0 is the achievement of the experimenter. He had set up the mechanical forces very carefully to get that uniform acceleration. Furthermore, the acceleration CAN'Tremain constant in ANY frame. As the particle approaches the speed of light, its inertial mass becomes enormous. So there is no way an experimenter can set up a mechanical force to keep it constantly acceleration in this limit. We managed to avoid talking about forces because you specified a constant acceleration in inertial frame F0. So I merely had to apply the Lorentz transformation to position and time of the particle. No forces, no problem. If you insist on asking me to explain 'the differences in acceleration', then I have to start explaining 'the differences in inertial mass and mechanical force'.

Mon Mar 20, 2017 8:43 pm 



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