md65536

Yes, it seems that was the only thing missing? Don't worry about errors, that's the point of the thread, where something appears paradoxical when presented in a way that each part of it makes sense, but some missing thing is hidden. You were right---at least if we're talking about events only---that if (event) A is in the middle of (events) B and C in its own frame, it's in the middle in every frame. I was wrong, it doesn't matter where where the origin of the frame is. The Lorentz transformation is a linear transformation, so it preserves the 4D linearity of those 3 events (preserving the midpoint), but not distance between them or simultaneity.

Yes, it's not correct for anybody. What I wrote earlier makes no sense, I'll try again... You have frames A, B, and C coincident at time t=0. In A's frame, B and C are equidistant from A. Let's call events BA and CA the events on B's and C's respective world lines that are at time t=0 in A's frame. Then you transformed BA and CA into B's frame and found that they're also equidistant from the event "A at proper time 0" in B's frame because of the way you put B's origin at that event. So far so good. In B's frame, at time t'=0 where you've set it up so that A is at the origin at that time, neither events BA nor CA have coordinate time t'=0. You've set it up so that they are away from the origin (x' != 0), so if you transform their time t into t' you should find that those events are not simultaneous in this frame. Therefore it's not the distances between the objects A, B, and C that you're comparing, but distances to events at different times. I guess the basic idea is that the Lorentz transformation you used applies to events, not the objects. (Note that the location of B is fixed in B's frame, so the distance to the origin is always the same and t' doesn't really matter for measuring the distance to B---I messed that up in an earlier reply---but the distance to C does depend on what time t' you measure it.)

Easy for you, maybe... with your own assumptions. But it's a good demonstration of the point of the thread's topic. You can apply SR and use its equations, then slip one little detail from one frame into to another inappropriately, and you come up with a different answer.

Yes, I made a mistake but revised it. You didn't show that the distance from A "to B" is the same as between A and C, in B's frame. You showed it for the distance from A "to the location of B at time t=0 as measured in A's frame", which is not the location of B in B's frame at time t'=0, because you have B not at the origin of B's frame. I really think you've overcomplicated it.

I still think you should start with what you understand and build up from there, but... One thing I see is that you have A at the origin, and are looking only at time t=0 where the origins of A's frame and B's coincide. Yes, distances from the origin should scale by gamma. However, that stops working for t != 0, when the 2 frames' origins no longer coincide. Another way to look at it is that in B's frame, the velocity of C is not 2 times the velocity of A, so A does not maintain its place in the middle of B and C. Also, to get things to align at this one time, you've set up everything really unconventionally. You have B not at the origin of its own frame. I think you're transforming "the starting location of B, in A's frame" into a coordinate in B's frame... but B is not actually at that coordinate at time t'=0 in B's frame! Since you have B far from the origin in its own frame, the time at that location will be transformed. So basically you're saying "In B's frame, A is in the middle of where C will be at some time in the past or future, and where B will be some time in the past or future," which I think is correct. This is too complex and confusing for me.

Looking over the thread, my opinion is that it is far too complicated, with too many aspects you haven't figured out yet, but which you would need to describe all the things you're trying to (eg. relativity of time, relativity of distance). Instead of looking at the things that don't make sense to you, why not start with at least one thing you do understand, and build off of that? I suggest the following: 1) Start with a simpler scenario. Use just 2 observers at first. The basic twin paradox is good because it is concrete; the twins start and end at the same place and can see each other's immediate age without worrying about synchronizing with a distant clock. Or if you prefer it, both observers moving inertially and setting a start event using a light signal from one to the other, is also simple enough. You could choose, but stick to one. You can add more to it later, to figure out other things. 2) Understand it first from at least ONE inertial frame of reference only, to start. Don't start trying to figure what a second observer measures, if you can't made sense of what the first observer does. 3) Write out the equations as you go. I don't see how you could understand how the numbers add up without calculating them. I bet that SR could be figured out from scratch by amateurs using just the 2 postulates, some geometry, and the timing of light signals, if you're interested... But you really have to understand more of the basics to figure out what you're trying to in this thread.

Careful, this can cause confusion. C has at least 2 reference frames. The frame you're calling "C's reference frame" is the frame where C is on the return journey to the meeting point. If that was clear to everyone in the discussion, that's fine. Edit: Nevermind me, I missed where the discussion switched to a case with no acceleration. I guess that's fine because you're basically looking at only the last phase of a twin paradox experiment. But in the full twin paradox experiment... in the C_outbound frame, B's clock is slower and B does everything that C does, later than C does. If they have a gradual acceleration phase, then when C is momentarily at rest with the "meeting point frame", B's clock has caught up and they've reached their maximum distance at exactly the same time in this frame. From then on, according to C, B's clock is ahead and anything that happens to C has happened earlier to B. So, in the case of only the inbound B and C frames, and the midpoint frame where B and C are symmetrical and always the same age, it is relativity of simultaneity that makes them not the same age in other frames (except at the meeting event). In B's frame before meeting, C is older but ageing slower.

Beside the main paradox Lorentz Jr mentioned, people will add additional confusing elements that leave out something. You personally approach it from already knowing the solution, and how to find it, and so you can quickly figure out what's missing. Before one understands that it works out mathematically and that one is as likely to find a paradox in it as in multiplication and division, it's easier to assume that some "paradoxical" complication is valid. It's like a "paradox" that shows that 1=0 by a missed mathematical detail, which can still be puzzling if it is hidden in a lot of other equations. For example, I could create paradoxes in a basic Twin paradox setup by adding something like, "now imagine that a uniform gravitational field is suddenly switched on everywhere in the universe...", and I'm sure you'll immediately see a problem in that, but if you don't already know the solution, it might be puzzling. I've seen intelligent people try to show that the asymmetry of the Twin paradox can be removed by adding some such detail, or a change in coordinates. Another thing that happens is that some rephrase it as an unresolvable debate of "why" the twin paradox happens.

Yes, and different people are arguing different things. Some are saying only that if one twin has a certain velocity and then a different velocity, there "needs" to be coordinate acceleration for that to be possible. Others will say that acceleration is the "cause" of the difference in proper time. Some say that swapping the twin with another moving in the opposite direction is not "the Twin paradox" because it describes a different experiment even if the results are the same. Others argue any number of things (they can't be meaningfully swapped, the accelerated twin physically changes, the results would be different---I can't think of any examples I think are correct). "Feeling" the acceleration or having proper acceleration shouldn't matter because it's not generally in the calculations, and the time dilation effect still happens in freefall, eg. with GPS satellites. Well that's not true. It's valid in a single reference frame (the aging of both twins can be figured out using only the inertial twin's reference frame even if the other one is never inertial) and it's valid for comparing any number of inertial reference frames (eg. composition of velocities, using triplets all leaving together and returning again, etc.). The first video says something like "SR only applies to uniform motion", and that's wrong. It generally applies only in uniformly moving reference frames, and I think that's implied by its postulates, which include that the speed of light is a constant c, and it's not constant in an accelerating reference frame. However, SR does apply to non-uniform motion of objects. For example, if the accelerating twin spends the entire trip accelerating, SR can be used to calculate its time dilation from an inertial reference frame just fine, just from its velocity.

Yes of course it's possible and common. Google says 'About 1% of supermassive black holes have an "accretion disk" of gas and dust swirling around them.' So I guess it looks like an object being pulled apart from the inside. If you want to consider some other object, it should be similar. Some "forces" give the object its shape and keep parts of the object away from the black hole, whether it's the inertia of a spinning accretion disk or chemical bonds of solid matter, and the BH rips away any parts that get too close, regardless of what holds the object together. It would look like spaghettification. The process is just gravitation, not some magical suction that vacuums everything up. For example a human on an extremely massive planet would be crushed by gravity. A black hole with similar gravitation would devour a human from the inside because the latter doesn't have the structural integrity to resist such strong pull. A small enough asteroid would barely have the strength to keep a human stuck to it, let alone damage its shape. A black hole with similar gravitation wouldn't have the strength to rip a human apart or pull more of it in.

Another thing is that its gravitational influence is similar to ordinary matter of the same mass, at a distance, so a black hole of one solar mass wandering into the solar system shouldn't be much more likely or a danger of eating us, as a star of one solar mass doing the same. Similarly a black hole created in the lab would not have enough mass to devour the earth on its own. It would only devour extremely nearby matter (probably closer than atoms are to each other*, depending on its mass of course). A BH made from Earth's matter wouldn't collapse the Earth any easier than the Earth's own mass would collapse on itself, so it would have to eat away at nearby matter until it destroyed the structural integrity of the planet before it would collapse. That said, Ton 618 has "The largest black hole ever found in the known universe" at 66 billion solar masses. "A black hole of this mass has a Schwarzschild radius of 1,300 AU (about 390 billion km in diameter) which is more than 40 times the distance from Neptune to the Sun." But it's about 18.2 billion light years away, so like you say, it's distance that keeps us safe. * Edit: I vastly overestimated that distance. Atom spacing is on the order of Angstroms (10^-10 m). A black hole with a mass of a million tonnes would have a Schwarzschild radius of about 10^-18 m, a thousandth the size of a proton! Does that mean such a BH might not even collapse a proton even if it passed right through it?

I found some ranges where you can create pieces of any area in the range, but none of them overlap at any of the values of A through E. Was that on purpose? Because, I found some other possible areas of pieces, including one that has 2 different answers. So I'll extend the puzzle with these... Cut a cube of side 2 cms into eight identical pieces such that the surface area of each piece is : F: 7 sq cms G: 6 + ✅2 sq cms H: 5 + ✅5 sq cms Then bonus: For which pairs of A through H is it possible to create 8 identical pieces with a surface area anywhere in between the two? Based on that answer, how would you cut one cube into 8 identical pieces of surface area a, and cut another cube into 8 identical pieces also of surface a but with a different shape than those from the first cube? I'm not certain that I haven't made any errors here!

You can change the area continuously for some of these cuts just by rotating the cube, so I still suspect there might be some with multiple answers.

Yes, I agree. Sorry for the confusion, I've ended up discussing what would happen in both the instantaneous and gradual acceleration cases, and the whole train stopping simultaneously only applies to the gradual case. In the instantaneous case, yes the train stretches again, because the parts of the train don't stop simultaneously in S', and the "car 0" end (the back end in S) continues moving for a short time while the other end has stopped, in the S' frame. Well it seems to work, but it's confusing. In this case of gradual acceleration, in the vfinal frame, what is observed is that we start with the train having constant velocity -v and is length contracted. Then to start the acceleration phase, the n=N end of the train begins to decelerate first, and the train begins to stretch very gradually at first. Other cars decelerate in turn until the n=0 car begins decelerating last. The n=0 car has the greatest proper acceleration (which all frames must agree on), but since it started last, it maintains a greater speed (with negative velocity) than the rest of the train in the vfinal frame during the entire acceleration phase. Then finally all cars arrive at the same velocity of 0 simultaneously, the train fully stretched to its full proper length. This does sound exactly like a time reversal of what is seen in the track frame, with everyone agreeing that the n=0 end has the higher proper acceleration. But everything I'm describing has a locally constant proper acceleration during the acceleration phase, and satisfies Born rigidity as far as I understand it. Are we talking about the same thing? Another possible source of confusion is that constant proper acceleration isn't constant coordinate acceleration in any single frame, and I may have mistook one for the other in what you wrote.

I think I figured out the basic idea in the track frame, for Born rigidity or anything else that approximately maintains the proper length of the train with non-instantaneous acceleration. In the track frame, the back of the train always accelerates at a higher rate than the front, so that as velocity increases, the train contracts. Then when acceleration stops, the back of the train stops accelerating first, when it reaches velocity v. The rest of the train is moving slower and continues to accelerate (and contract), with the front of the train being the last to stop accelerating, at which point the whole train is moving at v and is length-contracted by gamma. This is consistent with allowing the whole train to stop accelerating simultaneously in a moving frame (v's frame, if I got it right).

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". 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. 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). 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.

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?

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.

You're not still using "rapidity" as a synonym of acceleration, are you?

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, 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.)

Nothing physical is moving faster than c, it's just the timing of the local accelerations across the length of the train moving faster than c. It's only the phase velocity of this wave that's traveling faster than c. That's fine because the propagation of the wave isn't causal; the parts of the train are accelerating independently of each other. I can avoid calling it a wave to be less confusing. Yes, I've assumed "local engines" all along, that was implied by OP. I'm just saying the engines don't have to be part of the train itself. I'm trying to talk about the effects of SR without getting hung up on the difficulties of building a physical version of a thought experiment. I don't want to debate whether it's possible for 2 events to be simultaneous ie. not causally connected.

As mentioned, this isn't a rigid body and that's fine because it's not being accelerated as a rigid body, rather all the parts of the train are accelerated independently, as if it is modelled as a soup of particles. I also think it's not describing a practical impossibility, where every smallest part of the train needs to be a train engine, because the cause of the acceleration can be external. For example the train could be a metal object accelerated by a rail gun. I think it's possible that the train might not feel any stress from being stretched or compressed, it might only feel the local proper acceleration, temporal distortion, etc. This is because the acceleration "wave" crosses the train at a rate faster than c. For any particle on the train, even though the one "behind" it has started moving first, the particle has accelerated before the one behind has caught up and before any causal effect of the particle behind it has reached it. And even though the particle is moving before the one in front moves, by the time the particle gets to the location of the one in front, the one in front has already accelerated. This is true no matter how small you make the separation of the particles. (I guess this assumes that any field effects that can be felt, like EM field, are also accelerated along with the particles. Is this impossible?) On second thought... after a particle has accelerated it is now in the new inertial frame, in which the particle in front has already accelerated first! It has accelerated to rest. This gives an observer in the middle of the train (and who accelerates along with it) a bizarre account of what happens! According to her, the rear of the train begins to compress forward (not fast enough to see or feel it) while the front of the train remains stationary. Then she feels proper acceleration, then the rear of the train begins to stretch backward (but not faster than can be felt??? It seems this compression would have to be visible) while the front of the train again remains stationary! That seems surprising enough that I wonder what details I'm missing or getting wrong. Edit: Due to delay of light I think she'd see the front of the train appear to be approaching and stretched out (due to aberration of light). These are images of the front of the train before those parts accelerated, seen in the post-acceleration frame. But if it can be seen it can be felt?

https://en.wikipedia.org/wiki/Born_rigidity The "Class A" section contains an entire class of applicable motions. Instantaneous (not generally simultaneous) acceleration using the solution I proposed (different from OP's I think), does not satisfy Born rigidity, because it requires different parts of the train to accelerate at different times, in the 2 rest frames (before and after acceleration) of the train. Therefore the train's length changes in those frames, and its proper length doesn't even seem defined while accelerating. However, it should be the only solution that involves a single instantaneous acceleration at each point of the train, with which the train has the same proper length before and after the acceleration. It would seem not even Born rigidity can be satisfied with an instantaneous acceleration. However, OP's proposal never mentioned anything about a requirement of rigidity. For that matter, they neither required that the proper length of the train is the same before and after, I assumed that. 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. Then the velocity of the "wave" would have to be d/t = Lv/(L - L/gamma) = v/(1 - 1/gamma) For gamma=2, it seems intuitive that the wave would have to travel at 2v, to reach the front of the train at the same time that the back of the train traveling at v gets halfway there. With v = 3^0.5 c/2, I get a velocity of the wave equal to 3^0.5 c. Like you said this is faster than c, so it's non-causal and needs local forces to accelerate the parts of the train.

This doesn't refute the validity of Born rigidity especially in terms of kinematics. If you understood Born rigidity you'd know this thread is concerned with the "irrotational motions" class of Born rigidity, which is broken if the train turns. So bringing it up is a strawman argument. It sounds like you're trying to argue that Born rigidity can't be satisfied, by bringing up irrelevant counter-examples where it is not satisfied.