# md65536

Senior Members

2032

4

## Posts posted by md65536

### A straight forward Geometrical question !

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!

### A straight forward Geometrical question !

Spoiler

B. Cut like layer cake into 2 layers, then cut top with an X, corner to corner.

C. Cut like a pizza into 8 slices, starting corner to corner.

D. Cut like layer cake into 8 layers.

E. Cut into 4 layers, and then each layer into a wedge, cutting from one edge to the farthest opposite edge.

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.

### A straight forward Geometrical question !

Spoiler

A. would be 8 1x1x1 cm cubes. That should be the smallest possible surface area for 8 pieces.

Anything bigger can be done by cutting 4 1x1x2 cm pieces, and then cutting a symmetric comb joint. There are multiple answers. Do your solutions result in the fewest possible faces, or only convex pieces? Or some extra symmetry? Or min number of planar cuts?

### Neutral simultaneity for two frames.

12 hours ago, DimaMazin said:

I think in moving frame the train should stretch again. Therefore forward part of the train stoped earlier in S'.Because forward part of the train (in S) is backward part of the train in S'. Maybe you are correct but then you should correctly remake the model.

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.

12 hours ago, Lorentz Jr said:

Exactly. And therefore the ending solution in the vfinal frame must be the same as the beginning solution in the track frame, except the n's are reversed (n=0 at the front) and the formula applies to negative values of t (relative to the stopping time). Now write down that solution (it's in the "primed" frame) and transform it into the track frame to see what it looks like! 😋

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.

### Neutral simultaneity for two frames.

16 hours ago, md65536 said:

what I just described doesn't seem to make sense from the tracks frame.

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).

### Neutral simultaneity for two frames.

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.

### Neutral simultaneity for two frames.

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?

### Neutral simultaneity for two frames.

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.

### Neutral simultaneity for two frames.

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?

### Neutral simultaneity for two frames.

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.)

### Neutral simultaneity for two frames.

21 minutes 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?

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.

### Neutral simultaneity for two frames.

On 12/10/2022 at 8:52 AM, DimaMazin said:

Local forces accelerate the train. Therefore no force travels faster than c.

Why velosity of wave of acceleration can be faster than c? Because local forces don't need to wait when light crosses already contracted part of the train.

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?

### Neutral simultaneity for two frames.

1 hour ago, Mordred said:

I have yet to see a treatment where Born rigidity holds true. Care to provide one ?

Lol course we can also examine Ehrenfests treatment

The "Class A" section contains an entire class of applicable motions.

Quote

Born rigidity is satisfied if the orthogonal spacetime distance between infinitesimally separated curves or worldlines is constant,[7] or equivalently, if the length of the rigid body in momentary co-moving inertial frames measured by standard measuring rods (i.e. the proper length) is constant and is therefore subjected to Lorentz contraction in relatively moving frames.[8] Born rigidity is a constraint on the motion of an extended body, achieved by careful application of forces to different parts of the body. A body rigid in itself would violate special relativity, as its speed of sound would be infinite.

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.

15 hours ago, DimaMazin said:

Because velosity of the wave of the acceleration = (gamma +1)v/gamma

For example  gamma=2

v=31/2c/2

velosity of wave acceleration=3*31/2c/4

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.

### Neutral simultaneity for two frames.

5 hours ago, Mordred said:

Sure I can go with that

Here is the Born Rigidity examination

From this article it should be clear just how tricky any rigid rod examination can get

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.

### Neutral simultaneity for two frames.

4 hours ago, Lorentz Jr said:

Each car stretches out separately. If two adjacent cars get longer, the car in front has to move forward or the rear one has to move back.

I've thought about it more and I still think my first reply is correct. The issue is, if there's an instant change in velocity, does length contraction apply to all of the velocities in between as if it accelerates through them all in an instant, and I say it doesn't. SR doesn't predict the effects of acceleration, rather it is an assumption that acceleration doesn't have an effect, only velocity itself does (the "clock postulate"). SR neither says that the train would survive the acceleration, nor that it wouldn't. It says the train would be contracted at the beginning velocity, and also at the end, but not what strains would happen to the train in the zero length of time in between. So, whether the train can accelerate instantly or not is an assumption we have to make, not one we can derive from SR itself. If the train parts can instantly accelerate from 0 to v, then I think my answer works, and if not then it doesn't, and SR doesn't change that.

But yes, if there's any frame in which a train is moving and then all parts of it simultaneously stop, SR says its new proper length is smaller than it was, so it physically must get squished. Going from -u to +u instantaneously doesn't have a moment when it is at rest, so SR does not by itself say it will behave as if at rest in that case.

### Neutral simultaneity for two frames.

2 hours ago, Mordred said:

How much GR have you studied ? Under GR coordinate time is the time at each event. The proper time follows the worldline.

What I stated earlier stands

I agree this is irrelevant nonsense and you're incorrectly applying definitions that have no bearing on the issue.

### Neutral simultaneity for two frames.

34 minutes ago, studiot said:

This problem is a classic example of forgetting that both SR and GR are point function theories.

Difficulties can easily arise if you try to apply either to 'bodies' or systems that are too large to be considered as 'point partcles'.

It's not a problem if we consider only the the kinematics of the system, and treat the train as a set of particles, and consider only as many particles as needed. Usually one at the front of the train and one at the back is enough for most things. The solution to Bell's spaceship paradox is the same if you consider one large metal ship, or 2 small ships with an imaginary string between them. Einstein used trains in thought experiments without problems.

Considering only kinematics, the forces on the train and communication between the parts is irrelevant. Things like lengths and speeds are what's important.

3 hours ago, Lorentz Jr said:

It will start out contracted and then get squished halfway through because it loses its contraction. That's what I meant by complications.

Accelerating all at once is just the limit of accelerating quickly. It doesn't mean nothing happens in between.

I think this is wrong but I'm not certain.

If it squishes (like, physically crumples) in that one frame, that must happen in all frames. I don't see how that's possible in the moving train's frame, where the train is simply stretching out from length-contracted to rest length. Are you saying that what I described will crush the train with high v, or that I made some other mistake?

Also, if the train "loses contraction" only for an instant, the effects of the "squishing" would be causal, and there's no time for the effects to propagate.

Also, the "squishing" seems to imply that you can't accelerate an object without either stretching or compression strain on the object (you can only minimize it with gradual acceleration), do you agree? However, Born rigid acceleration is possible without spacial strain.

### Neutral simultaneity for two frames.

3 hours ago, Lorentz Jr said:

Accelerating all of the cars in a train instantly in the ground's reference frame would result in a chain of length-contracted cars separated by "uncontracted" distances. The couplings between the cars would all break.

But that's not the ground frame. There are also complications between -u and +u, so the couplings would have to be flexible. Either that or each car would have to be pre-programmed with its own specific acceleration schedule. In the ground frame, the back end of the train has to initially accelerate more quickly than the front end, because the distances between the cars need to shrink along with the cars' lengths.

Right, the parts of the train aren't accelerated simultaneously in the ground frame, only the in-between frame. The parts are each accelerated instantly but not simultaneously, in other frames. In the ground frame, the train starts at its proper length and ends up length-contracted; the rear of the train must accelerate first. In the post-acceleration train's frame, the train starts length-contracted and ends at rest at its proper length; the front of the train must accelerate first.

This isn't a question of practicality, it's mathematical. If an object can be made to accelerate all at once from -u to +u, it doesn't get pulled apart (in any frame) or physically squished, just length-contracted. If you want to debate whether a real train can do this, you should probably find out what v is before deciding if it can handle it, because in practical cases v is small enough that length-contraction is negligible.

### Neutral simultaneity for two frames.

You can do it if all the parts of the train can change speed instantly.

Say you have a train of proper length L, at rest on some tracks, and instantly accelerate each part of it to velocity v so that it has a proper length of L in its new frame. There's a frame of reference in between those two, where the tracks + rest train are moving in one direction at some velocity -u, and after the train accelerates, it's moving in the opposite direction but same speed +u, so that the length contraction factor before and after are the same. This is the frame in which all parts of the train would accelerate simultaneously. You can find u using the composition of velocity formula, so that v is the composition of u and u.

Is that what you're talking about, and does this match your result?

I suppose that if you did this repeatedly using small accelerations, in limit form you'd get Born rigid motion?

### Is "positionary-temporal" uncertainty built into spacetime?

On 11/10/2022 at 11:53 PM, Markus Hanke said:

What does this do physically? Well, having a difference rather than a sum enables you to make simultaneous changes to the space part and the time part in equal but opposite measure, without affecting the overall separation in any way. So you can trade a decrease in space for an increase in time (or vice versa), and still end up with the same overall separation.

And that’s exactly what happens in Special Relativity - for example, if you are looking at a clock passing you by at relativistic speeds, you’ll find that the clock is time-dilated (meaning it takes longer for the clock’s hands to move, from your point of reference), while at the same time the clock itself will be length-contracted in its direction of motion, so its size becomes shorter (again, from your point of reference). So in this scenario, and from your point of reference, “space is traded for time”, in a manner of speaking. This happens in equal but opposite measures - the decrease in size is by the same factor as is the increase in time - which is why the ratio between them remains the same always, which physically means that the speed of light is always the same in any inertial frame.

I'm trying to make sense of this and can't. I think you have it backwards.

s^2 is the spacetime interval between 2 events. I don't think that equation can directly represent length-contraction of an object, because the length of an object in a reference frame is the spatial distance between 2 simultaneous (in that frame) events. The spacetime interval is the measure between the same 2 events in different frames, while length contraction compares 2 different sets of events in different frames. I don't think you can put 2 lengths of an object (ie. proper and contracted) into the equation and get the same s^2, or in other words I don't see how you can express "longer t and shorter x" with that equation.

The interval being invariant implies that the space and time coordinates actually change in the same way, not opposite. That means that in one frame, if 2 events are a certain time apart and distance apart, in another frame where they're a longer time apart, they're also a greater distance apart.

I always confused that idea with the way that length contraction seems to say the opposite of that, so I'll just go through an example to show how relativity of simultaneity clears up the confusion. Say you have a train passing through a tunnel so that in the train's frame, the train is exactly the length of the contracted tunnel. The events are the front of the train passing the exit of the tunnel, and the rear of the train passing the entrance of the tunnel, simultaneously. The spacetime interval has (delta)t=0 and x=the proper length of the train = contracted length of tunnel, so s^2 is negative, a spacelike interval. In the tunnel frame, the tunnel is longer than it is in the train frame, while the train is length-contracted and fits entirely inside the tunnel. s^2 is the same since it's invariant, with x'=proper length of tunnel > x, but now t' is also larger than t because the train's rear enters the tunnel before the front of the train leaves. If I were to confuse things, I might say "let x' be the contracted length of the train, and it is smaller while t' is larger", but those statements are talking about 2 different spacetime intervals.

### What is gamma factor of object, which is falling into black hole?

23 hours ago, DimaMazin said:

If they fall directly toward each other then what mass would be after such collision?

There are additional replies better than I can give here: https://astronomy.stackexchange.com/questions/32445/head-on-collision-of-two-black-holes

23 hours ago, DimaMazin said:

Why creation of gravitational waves ,before their collision, takes only their KE and no significant mass?

Why creation of gravitational bubble,at the collision, takes only their mass and no significant KE?

If energy is lost from a system (in its CoM frame), the system loses mass. The binary system is losing mass equivalent to the gravitational wave energy radiated away, before the collision.

If an individual component of the system doesn't radiate energy itself, that component don't lose rest mass.

Radiating gravitational waves can reduce angular momentum, so the remnant BH should be decreasing its angular momentum during merger and ringdown.

After they're merged, the system only has the one object in it. Your question is basically asking about the components of the system before the merger, and the system as a whole after, that's the main difference. Also, I think the energy radiated while merging is vastly greater than that radiated during the inspiral phase.

### What is gamma factor of object, which is falling into black hole?

13 hours ago, DimaMazin said:

These are inspiral orbiting black holes. Yes, I see the numbers in the wiki add up to zero kinetic energy included. However if you look at the references eg. [4], the numbers don't add up exactly, and the uncertainties in the estimates of the black hole masses are so huge that it's not possible to give an exact figure for other energy in the system from the masses. The equation describing this would be more like, remnant BH mass Mf = m1+m2+everything_else_ignored - radiated_energy/c^2. It's likely that whatever is ignored is small on a scale of solar masses.

These aren't two masses falling directly toward each other, they come together because their orbits decay due to energy lost to gravitational waves. The kinetic energy is being radiated away as they approach. When they merge they'd generally have net angular momentum, resulting in a spinning black hole. The rotational energy of a black hole contributes to its mass (https://en.wikipedia.org/wiki/Kerr_metric#Mass_of_rotational_energy). Certainly then, some kinetic energy of the orbiting black holes contributed to the mass of the resulting black hole. In this case, at about 8 solar masses of energy radiated, much much more than any energy not accounted for is lost to gravitational waves.

### What is gamma factor of object, which is falling into black hole?

14 hours ago, DimaMazin said:

Then the rest mass after collision should be = m1 + m2 - energy of gravitational bubble /c2 +KE/c2

But really mass=m1 +m2 - energy of gravitational bubble /c2 + 0   Why so?

What are you basing this on? You're saying it should be one thing but is "really" another, where do you get the latter from?

If you apply your equations to quarks forming a proton, it really does include the KE. From https://en.wikipedia.org/wiki/Proton "The mass of a proton is about 80–100 times greater than the sum of the rest masses of its three valence quarks, while the gluons have zero rest mass. [...] The rest mass of a proton is, thus, the invariant mass of the system of moving quarks and gluons that make up the particle, and, in such systems, even the energy of massless particles is still measured as part of the rest mass of the system."

In the case of a black hole you can't measure or assert any internal motion, but the externally measurable rest mass is still there and the energy is still conserved.

### Minkowski space and geometric intuition

11 hours ago, Markus Hanke said:

So the correct question would be why inertial frames are related via hyperbolic rotations in spacetime - that’s a very valid question, but it isn’t one that any of our present theories can answer. So to make a long story short, we don’t have an explanation of why this happens, only a description of it. That’s not the same thing at all.

It's easy to explain the hyperbolic rotation relation in terms of other things. If you assume that the speed of light is constant, and consider a path of light c*tau long in one frame, then consider it in another frame (as commonly done with a light-clock at rest and moving, in thought-experiments used to explain time dilation), chosen to form a right triangle with a sides x and c*tau and hypotenuse c*t (right? am I messing up the details?). Then by the Pythagorean Theorem, the value of x^2 - (ct)^2 is constant for a given tau. That equation happens to describe a hyperbola. You don't need to explain why it's described by a hyperbolic rotation, if you instead explain why the speed of light is constant and why Pythagoras' theorem holds. "Hyperbolic rotation" is just an equivalent mathematical description of the relationship. But likewise, a constant speed of light and Pythagorean theorem could also be just descriptions of what's happening, equivalent to some other things. Maybe there's some simplest explanation of what's "really" happening, or maybe they're all just equivalent descriptions of what we observe.

I think we have the same point, that you don't need to explain a given description of a phenomenon to explain the phenomenon itself. Likewise, you don't need things to "really" be rotating for a mathematical equation for rotation to be a valid description.

### Using entanglement is not forbidden by relativity??

21 hours ago, geordief said:

Is this true ?Have there only been statistical observations and never a one to one observation?

As far as I know (which is not much and hopefully someone else can explain or correct me), whether an experiment is measuring the distribution of many particles or a single entangled pair doesn't matter much because the results are consistent. However, even with a single observation, I think there is always an inherent randomness to them (at least where probability wave functions are involved), which makes drawing a conclusion from it probabilistic. A single photon doesn't show a diffraction pattern in a double-slit experiment, it just follows the probability distribution of the pattern.

For example say you have an experiment to see if some action maintains the entanglement of a pair of particles, or breaks it. Suppose a measured event is expected to happen 70% of the time for entangled particles, and 50% of the time for random particles. If you observe a single pair and the event happens, were they entangled, or was it random chance? If the event doesn't happen, were they not entangled, or a random entangled outlier?

My under-educated impression is, anything you do to make a result 100% certain ("collapsed wave function") removes the interesting effects you only see with the interaction of probability wave functions.

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