Lorentz Jr Posted December 12, 2022 Share Posted December 12, 2022 2 minutes ago, Mordred said: You asked what real waves I directly answered this question and answered it correctly I said "What do you think", meaning md65536. You directly and correctly answered the kind of question a stupid beginner might ask, which was completely different from the question I asked md65536. This is one of your favorite trolling tactics, Mordy. You ignore the question the person actually asked, and then you insult the person's intelligence by fabricating your own little fantasy question that involves the person not knowing basic things like definitions of vocabulary terms. And then you make your own retarded comments, like these: "all reference frames are inertial" "Equal free fall it is not the equivalent to a rest frame". "A reference frame is an inertial frame of reference." Not to mention mangled gibberish like "The above propertime that clock on the worldline". And you try to intimidate people by gratuitously throwing around advanced technical terms and citing your "Resident Expert" status in the forum. Go ahead, Professor. Keep replying to my comments. Or not. However you see fit. -4 Link to comment Share on other sites More sharing options...

Mordred Posted December 12, 2022 Share Posted December 12, 2022 If your addressing a specific individual you should state that but when you leave a question on a physics forum. Anyone can answer said question. That is precisely the purpose of this forum Link to comment Share on other sites More sharing options...

Lorentz Jr Posted December 12, 2022 Share Posted December 12, 2022 (edited) 9 minutes ago, Mordred said: If your addressing a specific individual you should state that What?? Are you feeling okay, Professor? What part of "md65536 said:" do you not understand? PS: You might want to try doing that yourself for a change. 😉 9 minutes ago, Mordred said: Anyone can answer said question. About another person's thoughts? That's quite a feat, Professor. 🙂 Edited December 12, 2022 by Lorentz Jr -2 Link to comment Share on other sites More sharing options...

Mordred Posted December 12, 2022 Share Posted December 12, 2022 8 minutes ago, Lorentz Jr said: "all reference frames are inertial" "Equal free fall it is not the equivalent to a rest frame". "A reference frame is an inertial frame of referenc" I Strongly suggest you study the terminology you will find those statements are 100% accurate under both SR and GR. You do not need to take my word for it pick up ant SR or GR textbook 5 minutes ago, Lorentz Jr said: What?? Are you feeling okay, Professor? What part of "md65536 said:" do you not understand? PS: You might want to try doing that yourself for a change. 😉 About another person's thoughts? That's quite a feat, Professor. 🙂 Seriously you post something for everyone to read where does reading thoughts enter the picture ? This is a forum anyone can respond at any time whether you like it or not. That's is part of the rules on a public forum that you agreed to when you signed up. Link to comment Share on other sites More sharing options...

Lorentz Jr Posted December 12, 2022 Share Posted December 12, 2022 (edited) 37 minutes ago, Mordred said: I Strongly suggest you study the terminology you will find those statements are 100% accurate under both SR and GR. You do not need to take my word for it pick up ant SR or GR textbook I already read the textbooks, Professor. You have no idea what you're talking about. The reference frame of a spaceship with its rockets blasting isn't an inertial frame of reference. The astronauts can tell that because they feel like they're being pushed into their seats. The reference frame of Earth's surface isn't an inertial frame of reference in general relativity. You can tell that because you feel like you're being pushed down onto the ground or your chair or whatever you're resting on. The reference frame of a merry-go-round or the edge of a spinning space station isn't an inertial frame of reference. People can tell that because they feel like they're being flung away from the center. You have no business posting on a real physics forum, Professor. You're a shameless, ignorant troll, and you're not fooling anyone except the most casual or uninformed readers. 4 hours ago, Lorentz Jr said: What kind of real wave do you think travels faster than c, and how do you think a simulated wave going faster than c could be implemented without local engines? 37 minutes ago, Mordred said: where does reading thoughts enter the picture ? The part where it says "think", Professor. Speaking of kindergarten-level vocabulary terms, do you understand that thinking means having thoughts? You said you "directly answered this question and answered it correctly", and my question was about another poster's thoughts. Am I going too fast for you, Professor? Quote This is a forum anyone can respond at any time whether you like it or not. That's is part of the rules on a public forum that you agreed to when you signed up. Okay, great. Keep responding with more baloney, Professor, and I'll tell you what I think of it. 🙂 Edited December 12, 2022 by Lorentz Jr -3 Link to comment Share on other sites More sharing options...

md65536 Posted December 12, 2022 Share Posted December 12, 2022 1 hour ago, Lorentz Jr said: Okay, that's good. Although even a rail gun would have to propel each car according to the same schedule. The main differences would be the clocks in the rail gun are in the ground frame instead of in the cars, and the different points on the gun have to keep changing from one car's schedule to the next one. And it would have be a very long rail gun. 😄 Just to expand the realm of possibility, the rail gun wouldn't have to be in the ground/tracks frame. Both what I described and what OP did, here, On 12/10/2022 at 8:52 AM, DimaMazin said: Because velosity of the wave of the acceleration = (gamma +1)v/gamma describe a set of events that are simultaneous in some frame, when gamma is high like 2 in the example. It's as easy to coordinate the events as it is to synchronize a (maybe large) set of clocks in that frame. Then, the rail gun would only need to be as long as the length-contracted train in that frame (by a factor of less than 2 in this example, because the train's speed is less than v in that frame). When figuring out the rate at which the acceleration events happen along the length of the train, I saw that the proper length of the train gets simplified out of the equation. That means that the rate is the same whether the train is short or long. Whether "long" means 1 m or a billion light years, the same answer applies (with the railgun length proportional to train length, as is the timing in the ground frame). (Then yes, it would be impossible to accelerate a full-size train fast enough, so just make the train out of electrons or something lighter, etc.) Link to comment Share on other sites More sharing options...

Mordred Posted December 12, 2022 Share Posted December 12, 2022 (edited) Sigh (hopefully you don't get insulted again) lets show you how this works. start with the Lorentz transformation. \[\acute{t}=\gamma(\frac{vx}{c^2})\] \[\acute{x}=\gamma(x-vt)\] \[\acute{y}=y\] \[\acute{z}=z\] the two equations most relevant are the first two note the velocity term velocity is the vector rate of change in position and direction. As you doubt everything I state and argue everything I state here https://www.physicsclassroom.com/class/1DKin/Lesson-1/Speed-and-Velocity Quote "Frames of reference can be divided into two groups: inertial (relative motion with constant velocity) and non-inertial (accelerating, moving in curved paths, rotational motion with constant angular velocity, etc.). The term "Lorentz transformations" only refers to transformations between inertial frames, usually in the context of special relativity" Oh my it specifies inertia frames https://en.wikipedia.org/wiki/Lorentz_transformation you can read it for yourself and don't take my word for it if you choose acceleration is handled via rapidity the equations are also in that link. Quote In Newtonian physics, free fall is any motion of a body where gravity is the only force acting upon it. In the context of general relativity, where gravitation is reduced to a space-time curvature, a body in free fall has no force acting on it. https://en.wikipedia.org/wiki/Free_fall Newtons laws of inertia https://en.wikipedia.org/wiki/Newton's_laws_of_motion note that the freefall link specifies no force acting upon the body it is in freefall no net force however Newtons laws of inertia still apply. The body is in motion, one can arbitrarily choose any event as the rest frame under SR it could be the twin leaving the planet or the stay at home twin both twins are in inertial frames (not accelerating, constant velocity). the same applies to freefall Alice is the observer on Earth. Bob is the Observer in freefall. The choice of who is the rest frame is arbitrary with the above transformation. Hopefully in the freefall state your know \[m_g=m_i\] equivalence principle gravitational mass is equivalent to inertial mass. if not google Principle of equivalence. we wont get into tidal forces just yet but succinctly if you have two bodies in freefall the vector direction is towards the CoM. the tidal force between the two bodies is \[\frac{d^2x}{dt^2}-\frac{MG}{r^3}x\] 45 minutes ago, Lorentz Jr said: I already read the textbooks, Professor. You have no idea what you're talking about. The reference frame of a spaceship with its rockets blasting isn't an inertial frame of reference. The astronauts can tell that because they feel like they're being pushed into their seats. The reference frame of Earth's surface isn't an inertial frame of reference. You can tell that because you feel like you're being pushed down onto the ground or your chair or whatever you're resting on. The reference frame of a merry-go-round or the edge of a spinning space station isn't an inertial frame of reference. People can tell that because they feel like they're being flung away from the center. You have no business posting on a real physics forum, professor. You're a shameless, ignorant troll, and you're not fooling anyone except the most casual or uninformed readers. now I suggest you study again the above further applies to the addition of velocities \[U=v+\acute{U}\] where v is the velocity between observers U and \[\acute{U}\] is the two observers. Edited December 12, 2022 by Mordred Link to comment Share on other sites More sharing options...

md65536 Posted December 12, 2022 Share Posted December 12, 2022 6 minutes ago, Mordred said: acceleration is handled via rapidity the equations are also in that link. You're not still using "rapidity" as a synonym of acceleration, are you? Link to comment Share on other sites More sharing options...

Mordred Posted December 12, 2022 Share Posted December 12, 2022 44 minutes ago, md65536 said: You're not still using "rapidity" as a synonym of acceleration, are you? No it can involve rapidity but is not the only methodology. I do recall that discussion lol I actually enjoyed that discussion. Link to comment Share on other sites More sharing options...

Lorentz Jr Posted December 12, 2022 Share Posted December 12, 2022 (edited) 22 hours ago, Mordred said: all reference frames are inertial A reference frame is an inertial frame of reference 1 hour ago, Mordred said: Sigh (hopefully you don't get insulted again) lets show you how this works. Great! Show me how it works, Professor! 😄 Quote Frames of reference can be divided into two groups: inertial ... and non-inertial 1 hour ago, Mordred said: Oh my it specifies inertia frames Oh my! It says some frames are inertial! Good job, Professor! You've demonstrated the faulty generalization fallacy! Some doesn't imply all! Nobody slings baloney like you do, Professor! 😋 Tragically, that's not officially allowed on this forum, Professor! You're still spewing nonsensical hogwash! 😲 Quote The use of logical fallacies to prove a point is prohibited. The use of fallacies undermines an argument, and the constant use of them is simply irritating. Great lecture, Professor! Irritating indeed! It has nothing to do with this thread or any comments I've made, but I'm sure it'll impress lots of readers who don't realize it's all fake! 😆 Edited December 12, 2022 by Lorentz Jr -1 Link to comment Share on other sites More sharing options...

Mordred Posted December 12, 2022 Share Posted December 12, 2022 Thread reported I'm tired of your attitude Link to comment Share on other sites More sharing options...

Lorentz Jr Posted December 12, 2022 Share Posted December 12, 2022 (edited) 10 minutes ago, Mordred said: Thread reported I'm tired of your attitude Good. I've been tired of your attitude since page 2. PS: What the hell does "methodology" have to do with definitions? 1 hour ago, md65536 said: You're not still using "rapidity" as a synonym of acceleration, are you? 51 minutes ago, Mordred said: No it can involve rapidity but is not the only methodology. Edited December 12, 2022 by Lorentz Jr Link to comment Share on other sites More sharing options...

Mordred Posted December 12, 2022 Share Posted December 12, 2022 Be well Link to comment Share on other sites More sharing options...

swansont Posted December 12, 2022 Share Posted December 12, 2022 15 hours ago, Lorentz Jr said: Sure, buddy. Stick to your story no matter how retarded it is. 😉 ! Moderator Note “retarded” is something that crosses the line of civility, as does the intentional misuse of a user name, and sarcastic substitutions for user names. It needs to stop. Link to comment Share on other sites More sharing options...

md65536 Posted December 13, 2022 Share Posted December 13, 2022 On 12/10/2022 at 8:52 AM, DimaMazin said: Because velosity of the wave of the acceleration = (gamma +1)v/gamma Did you mention you already figured this was wrong? I figure that the correct velocity should be greater than c for all v<c, otherwise the accelerations won't be simultaneous in any frame. As well, when v approaches 0, this velocity should approach infinity, because for vanishing v, the frame in which the accelerations are simultaneous approaches the track's frame. Link to comment Share on other sites More sharing options...

DimaMazin Posted December 13, 2022 Author Share Posted December 13, 2022 2 hours ago, md65536 said: Did you mention you already figured this was wrong? I figure that the correct velocity should be greater than c for all v<c, otherwise the accelerations won't be simultaneous in any frame. As well, when v approaches 0, this velocity should approach infinity, because for vanishing v, the frame in which the accelerations are simultaneous approaches the track's frame. Thank you for your definition which is correct. I think the same. Any such 'velocity' is proportional non-simultaneity on proportional distance. I think the 'velocity' is the same in frame S' when local forces brake the train which is in S . Link to comment Share on other sites More sharing options...

Lorentz Jr Posted December 14, 2022 Share Posted December 14, 2022 (edited) On 12/9/2022 at 1:01 PM, DimaMazin said: If we want to accelerate very long train, instantly accelerating each railway carryage into a moving frame at v, then we should accelerate them simultaneously in neutral simultaneity. Right. And this is roughly how fast the cars have to move and accelerate when they first begin, at minimally relativistic speeds (to first order in v): [math]v = \frac{a_0 t}{1 + (a_0 L n / c^2)}[/math] [math]a = \frac{a_0}{1 + (a_0 L n / c^2)}[/math] [math]L[/math] is the length of each car; n is the number of the car, starting from the last one; and [math]a_0[/math] is the acceleration of the last car (n=0). The complete solution is much more complicated, so it might have to be computed numerically on a computer, but this approximation shows that the initial speed and acceleration taper off toward the front of the train, and they have to taper off more sharply when the train accelerates more rapidly (because of the [math]a_0 L n / c^2[/math] term in the denominator). Edited December 14, 2022 by Lorentz Jr Link to comment Share on other sites More sharing options...

Lorentz Jr Posted December 17, 2022 Share Posted December 17, 2022 (edited) On 12/14/2022 at 1:32 AM, Lorentz Jr said: And this is roughly how fast the cars have to move and accelerate when they first begin, at minimally relativistic speeds (to first order in v): Well, I don't have anything better to do right now, so let's work out these results. We're trying to quickly accelerate the train to relativistic speeds without the cars crashing into each other or the couplings between them breaking apart. [######]-[######]-[######]-[######] --> The condition is that there's little or no stress on the couplings between the cars, so we're effectively simulating Born rigidity. If the propulsion for each car is near the middle of the car, maybe compression in the front half will roughly cancel out stretching in the back half. So, defining [math]L_0[/math] as the uncontracted length of each car, the contracted length should be pretty close to the special-relativistic value [math]L = L_0/\gamma[/math]. To keep the equations as uncluttered as possible, we'll pretend that [math]L_0[/math] and c are equal to 1. In other words, we'll do the problem using [math]L_0[/math] as our unit of distance and [math]L_0/c[/math] as our unit of time. These are the "natural" units for this problem. The units in the final answer will be wrong, but that's easy enough to fix, because there's only one way to fix them using [math]L_0[/math] and c. So, with these conventions, the contracted length of each car is [math]L = 1/\gamma = \sqrt{1-v^2} [/math] And the rate at which each car's length contracts is the time derivative of this formula, which is equal to the difference between the speed of the coupling in front of the car and the speed of the one in back of it: [math]\displaystyle{\frac{dL}{dt}= \frac{1}{2} \left( \frac{-2va}{\sqrt{1-v^2}}\right) = −\gamma va = v_{n+1}−v_n }[/math] The tapering off of the acceleration along the length of the train is the difference between the speed of one car and the speed of the one in front of it. This will be pretty close to the difference between the speeds of the couplings, as long as the acceleration isn't too extreme, so we can calculate that: [math]\displaystyle{\frac{\partial v}{\partial n} = \frac{(v_{n+1}−v_n)}{1}=−\gamma va}[/math] We can also ignore the [math]\gamma[/math] term, because it only becomes important at higher speeds (the approximation will be linear in v). [math]\gamma = 1 + O(v^2) [/math] [math]\displaystyle{\frac{\partial v}{\partial n} = -v \frac{\partial v}{\partial t}}[/math] Then we'll try to separate n and v by taking a wild guess that maybe we can express the speed [math]v(n,t)[/math] as the product of two simpler functions, [math]\nu(n)[/math] and [math]\tau(t)[/math]: [math]v(n,t) \equiv \nu(n)\tau(t) [/math] And then we'll plug & chug through the math. One side of the separated equation is independent of t, and the other side is independent of n, so they must both be constant. [math]\nu'\tau = -(\nu\tau)(\tau'\nu) = -\nu^2 \tau'\tau[/math] [math]\displaystyle{- \frac{\nu'}{\nu^2} = \tau' = k_1} [/math] [math]\tau = k_1 t + k_2 [/math] [math]\tau(0) = 0[/math], so [math]k_2 = 0[/math]. [math]\displaystyle{-\frac{d\nu}{dn} = k_1 w^2}[/math] [math]\displaystyle{- \frac{d\nu}{\nu^2} = k_1 dn}[/math] Now integrate both sides of the equation, [math] \displaystyle{\frac{1}{\nu} = k_1(n + k_3)} [/math] and plug [math]\nu[/math] and [math]\tau[/math] back into the definition of v: [math] \displaystyle{v = \nu\tau = \frac{k_1 t}{k_1 (n+k_3)} = \frac{t}{n+k_3}} [/math] Now define [math]a_0[/math] as the (constant) acceleration of the last car. Then [math]\displaystyle{ v_0 \equiv a_0 t = \frac{t}{0+k_3}} [/math], so [math] k_3 = 1/a_0 [/math]. And now we have the final solution for the speed, [math]\displaystyle{v(n,t)=\frac{a_0 t}{1 + a_0 n}} [/math] except we need to fix the units. The problem is the [math] a_0 [/math], which is an acceleration, in the denominator, so the way to fix it is to divide it by [math] c^2/L_0 [/math], which is also an acceleration: [math]\displaystyle{v=\frac{a_0 t}{1+\frac{a_0 L_0 n}{c^2}}} [/math] And differentiate with respect to t to get the acceleration: [math] \displaystyle{a=\frac{a_0}{1+\frac{a_0 L_0 n}{c^2}}} [/math] Edited December 17, 2022 by Lorentz Jr Link to comment Share on other sites More sharing options...

Lorentz Jr Posted December 17, 2022 Share Posted December 17, 2022 (edited) 1 hour ago, Lorentz Jr said: [math]\displaystyle{-\frac{d\nu}{dn} = k_1 w^2}[/math] This should be [math]\displaystyle{-\frac{d\nu}{dn} = k_1 \nu^2}[/math]. And a reminder that this isn't the whole solution. It's only the beginning, when the train is just getting started. Edited December 17, 2022 by Lorentz Jr Link to comment Share on other sites More sharing options...

md65536 Posted December 17, 2022 Share Posted December 17, 2022 43 minutes ago, Lorentz Jr said: And a reminder that this isn't the whole solution. It's only the beginning, when the train is just getting started. I don't understand, is this an interpretation of OP's stated problem, or are you rejecting instantaneous acceleration and substituting your own problem? If the latter, why not just apply Born rigidity? Is that what you're trying to do? If so, why approximate, and why not allow constant proper acceleration at each part or car of the train? Are you treating the train differently because it's made up of cars, and if so why would a solution for Born rigidity that works on the train as a whole, not work for the individual cars? Link to comment Share on other sites More sharing options...

Lorentz Jr Posted December 17, 2022 Share Posted December 17, 2022 (edited) 55 minutes ago, md65536 said: I don't understand, is this an interpretation of OP's stated problem, or are you rejecting instantaneous acceleration and substituting your own problem? Instantaneous acceleration is a solution, not a problem. I take the problem as wanting to accelerate the train quickly and the idea of instantaneous acceleration as the OP's attempt to solve it. Or, if you take the problem more specifically to be wanting to accelerate the cars instantaneously, then yes, I'm saying that's obviously impossible, and I'm showing why it's impossible and providing a solution to a more realistic problem. As soon as a car accelerates, it would crash into the one ahead of it if that one isn't also accelerating. 55 minutes ago, md65536 said: If the latter, why not just apply Born rigidity? Is that what you're trying to do? I'm not sure what "apply Born rigidity" means. It's defined mathematically, but it can't be maintained by objects that are made of ordinary materials and being forcefully pulled from ahead or pushed from behind. I only mentioned it because eliminating stresses between the cars is mathematically equivalent, at least if the cars aren't stressed too much internally by their propulsion systems. 55 minutes ago, md65536 said: If so, why approximate, and why not allow constant proper acceleration at each part or car of the train? That wouldn't work. At t=0, the cars are all stationary, so the formula I posted is the proper acceleration of each car at that moment, and the formula shows that they have to be different from each other. And then, as the train approaches its target speed, the front cars have to keep accelerating while the back cars are already up to speed. That's an interesting suggestion though. I suppose it might turn out to be approximately true for some parts of the train's trajectory when the accelerations are all converted to each car's reference frame. I couldn't say for sure though. I seriously doubt there's an exact analytic solution to the problem. 55 minutes ago, md65536 said: Are you treating the train differently because it's made up of cars, and if so why would a solution for Born rigidity that works on the train as a whole, not work for the individual cars? I'm still not sure what you mean. A "solution for Born rigidity" is only defined mathematically. It's physically impossible for ordinary materials that are subjected to high accelerations by external sources. The point is that a solution that works for individual cars doesn't work for the train as a whole. Compression in the front part of the car can compensate for stretching in the back part, and the car is short enough that those effects occur almost instantaneously. So the proper length of the car can stay relatively constant without too much trouble, but there's no way that can happen to a train with only engines at the ends. Edited December 17, 2022 by Lorentz Jr Link to comment Share on other sites More sharing options...

DimaMazin Posted December 17, 2022 Author Share Posted December 17, 2022 (edited) If instantaneous acceleration does not exist then acceleration cannot exist at all. We try to use magic acceleration to search key for development of SR. I see no sence to comlicate the problem by gradual acceleration and then to go back by the complex way to get the same. Edited December 17, 2022 by DimaMazin Link to comment Share on other sites More sharing options...

Mordred Posted December 17, 2022 Share Posted December 17, 2022 (edited) 1 hour ago, DimaMazin said: If instantaneous acceleration does not exist then acceleration cannot exist at all. We try to use magic acceleration to search key for development of SR. I see no sence to comlicate the problem by gradual acceleration and then to go back by the complex way to get the same. There is an instantaneous acceleration treatment by applying the four acceleration equations. A specific equation that describes this is \[\alpha=\gamma^3 a\frac{1}{(1-u^2/c^2}\frac{du}{dT}\] where \[\alpha\] is the proper acceleration for objects with mass The large T is specifying coordinate time to be more obvious. u here is the instantaneous velocity. you can further simply that equation by applying motion on the Minkoskii hyperbolic curve the above equation leads to which simplifies to \[g^4/c^2\] \[x'^2-ct^2=x^2\]. the equation above works for both forms of acceleration via change in velocity or direction. This equation has been used in Born rigidity examination as well. An interesting consequence of relativity is the observer effects. Place an observer at a static location your classic rest frame. The train has length so he's going to observe different parts of the train at different angles. Even if we only consider the observer along the x axis on top of the train he will observe a different length front to rear. The approximate point of simultaneity of signals received from the front and rear would be the center of the train. The only way to preserve that simultaneity from any two equidistance points either in the x+ or x- direction the length contraction must occur in a symmetric fashion from that observer point of view. In a linear acceleration case that isn't too hard if you allow some mechanism that the entirety of the bus gains speed. however once the train starts to turn your going to lose simultaneity from that same location. At least I don't know of any solution where you won't. treating simultaneity in terms of signals received by an Observer Edited December 17, 2022 by Mordred Link to comment Share on other sites More sharing options...

DimaMazin Posted December 17, 2022 Author Share Posted December 17, 2022 On 12/11/2022 at 9:57 AM, md65536 said: The way I figure it, if you have the back of the train accelerate at time 0, the rest of the train accelerates over time until the front of the train finally accelerates when the length of the train has become L/gamma in the initial frame, where L is its original proper length. By that time, the back of the train has traveled a distance of (L - L/gamma) at velocity v. Therefore the time when the front of the train accelerates would be t=d/v = (L - L/gamma)/v. Now I think it is beter to define equation for non-simultaneity t=(L-L/gamma)/v=(gamma-1)L/(gamma_{*}v) then the equation of non-simultaneity is t = t_{0} + (gamma-1)x/(gamma*v) then we should translate the non-simultaneity of frame S in frame S'. Link to comment Share on other sites More sharing options...

md65536 Posted December 18, 2022 Share Posted December 18, 2022 (edited) 18 hours ago, Lorentz Jr said: Or, if you take the problem more specifically to be wanting to accelerate the cars instantaneously, then yes, I'm saying that's obviously impossible, and I'm showing why it's impossible and providing a solution to a more realistic problem. As soon as a car accelerates, it would crash into the one ahead of it if that one isn't also accelerating. The result I get is that as soon as a car accelerates to its new speed, it is in a different inertial frame in which the car in front has accelerated first (or rather decelerated to rest in this frame), and is already out of the way. I argued before that I thought it wouldn't feel any stress due to causality, but now I'm sure that's wrong. Still, I don't see a result from SR that is theoretically impossible for every type of "train". 18 hours ago, Lorentz Jr said: That wouldn't work. At t=0, the cars are all stationary, so the formula I posted is the proper acceleration of each car at that moment, and the formula shows that they have to be different from each other. And then, as the train approaches its target speed, the front cars have to keep accelerating while the back cars are already up to speed. I mean that there's a solution where any individual part (car) of the train has a constant proper acceleration over time, not that all parts have the same proper acceleration. But yes, if the train is meant to reach a specific velocity relative to the tracks, and then stop accelerating, there should be a solution that uses a set of constant proper accelerations, beginning simultaneously in the momentary rest frame of the train, and stopping simultaneously in another momentary rest frame of the train. Such a rest frame should exist because Born rigidity maintains the proper length of the train. Or another way to look at it, the train stopping at its final velocity should be simultaneous in that frame, because the process should be time-reversible. Ie. when the back car is "already up to speed", all of the cars are. The cars had different proper accelerations for different proper times, because they started simultaneously in one frame, and ended simultaneously in another. However... the more I think about it the less I'm sure I have that figured out correctly, because what I just described doesn't seem to make sense from the tracks frame. Maybe an acceleration phase and a separate deceleration phase are both needed, not just a stop to the acceleration. I'll have to think about that more because 2 different answers make sense in 2 different frames, so for sure I'm wrong somewhere. 18 hours ago, Lorentz Jr said: The point is that a solution that works for individual cars doesn't work for the train as a whole. Compression in the front part of the car can compensate for stretching in the back part, and the car is short enough that those effects occur almost instantaneously. So the proper length of the car can stay relatively constant without too much trouble, but there's no way that can happen to a train with only engines at the ends. Yes, the same solution works. The point of Born rigidity is that all of the points or parts of the object are accelerated individually, it never specifies "only two engines" unless the train consists of only those 2 points (such as 2 rockets with an imaginary string between them). Besides, if your solution works for cars of length L_0, then it would work for a train with one engine and a length of L_0. Either your solution works on the train as a whole with certain restrictions, or it fails for cars when there are no restrictions (unless you make n infinitely high and L_0 infinitesimal, but then you're just describing a solid rod where every part of it is accelerated locally). 15 hours ago, DimaMazin said: Now I think it is beter to define equation for non-simultaneity t=(L-L/gamma)/v=(gamma-1)L/(gamma_{*}v) Interesting that the second form is what you posted in the very first post, and the first form I posted later, and they are the same but I didn't realize it. We were talking about the same thing all along. Edited December 18, 2022 by md65536 Link to comment Share on other sites More sharing options...

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