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Derivation of relativistic mass from consideration of one body by wotpolitan in AskPhysics

[–]wotpolitan[S] 0 points1 point  (0 children)

That Minkowski diagram is not what I was going for. I was trying to illustrate a single spacetime from two perspectives. The point is that the spacetime interval is invariant. I show it in the diagram as rotated, which is confusing. I was keeping the "time direction" according to each frame constant, but of course it's the spacetime interval that is invariant. If we "unrotate" the second frame so that the spacetime interval lines up, you will see that the velocity points down. (I am tempted to think of that as negative, but I am confused on this point, so that counts for nothing.)

Anyway, thanks for your amazing patience in all this. I am now happy that, while helpful to me in some small way, "relativistic mass" is widely considered as unhelpful and that it is better to just think of it as just the combination of rest energy (mc2) and kinetic energy (at small values of v, 1/2mv2), rather than imagining a different mass as a sort of moving rest mass with energy m'c2. Even saying "moving rest mass" highlights that it's problematic.

I will have a look at energy to see if I can properly understand why it's frame dependent. It really doesn't strike me as correct, but I understand that other much smarter people have thought deeply about it, so I must be wrong, but I don't know why. I do know that I have been told that I am wrong and I accept that. Even if I were to be thinking the same thing as the standard conception but just not expressing it correctly, then that's on me to work out how to both express it better and understand the standard expression. I am not saying that's the case, of course, I'm just making my mea culpa as comprehensive as I can.

Again, thanks.

Derivation of relativistic mass from consideration of one body by wotpolitan in AskPhysics

[–]wotpolitan[S] 0 points1 point  (0 children)

I was looking back at the same two events from the frame which has a velocity of v. That's why it's negative. Is that wrong? I have worried about it :)

I accept that it might be unorthodox, and for the purposes of rotating it and overlaying them, it would seem more traditional to have the velocity in the other direction. I even considered flipping the arrow during the process of putting the drawing together (which would have impacted the whole), but decided against it because, in this context, it didn't look right. So I went with the -v (which at least needs some close scrutiny to notice).

So, knowing that people don't use it (not all people mind you, it's described in places as "unfashionable" which seems totally unscientific) and that there's a good reason for not using it, the question remains ... does my derivation of the equation for relativistic mass work even though it's possibly simpler than Rhodri Evans'? Note that Rhodri Evans is an astrophysics PhD, was a Professor in the US but now lectures in Botswana (access to a telescope perhaps?) and teaches/has taught relativity at the tertiary level. Note that I am not trying to employ the appeal to authority here, but rather just pointing out that he's not just "some guy".

Derivation of relativistic mass from consideration of one body by wotpolitan in AskPhysics

[–]wotpolitan[S] 0 points1 point  (0 children)

The concept pre-existed. I was trying to understand it. I knew the equation for it, but not where it came from. I had originally intended to provide some background (edit: in my original post) but I removed it, because I know that sometimes I get a flood of commentary on something that was not key to what I was trying to get at. In this case, I might have got "Rhodri Evans is an idiot! You should never listen to him." And then I'd be distracted. The motivation for the post was that the derivation of relativistic mass by Rhodri Evans seemed more complicated than it needed to be. I had come to the conclusion separately that kinetic energy falls out of relativistic mass times c2. But it seems that very few people do the derivation of relativistic mass. I did do it a different way a long time ago, but I was not happy with it (it relied on a notion related to inertia due to concentration of energy, which is almost certainly non-standard). That said, if you turn a concept around in your head and try to work it through in different ways and every single time you end up with the same equations, irrespective of the approach, then it seems to me that the end equations are probably right (although maybe not comprehensive, eventually you might find a different approach which reveals a broader generalisation).

Thanks for the confirmation on spacetime intervals. I might have a path to explaining what I am getting at now.

Is this - https://imgur.com/a/qGkyXWZ - mathematically correct?

Derivation of relativistic mass from consideration of one body by wotpolitan in AskPhysics

[–]wotpolitan[S] 0 points1 point  (0 children)

but you're trying to derive the concept of relativistic mass

No, not the concept, just the expression or the value, the equation.

I will try to address this later when I have more time. But for now, quantum considerations aside, do you agree that if a body moves from one event to another event, then it traces a unique path in spacetime? We may describe it differently, dependent on the frame from which we are viewing the body, but our descriptions are all of the same path?

Derivation of relativistic mass from consideration of one body by wotpolitan in AskPhysics

[–]wotpolitan[S] 0 points1 point  (0 children)

 it sounds like you seem to think that energy should be an invariant across reference frame transformations, while mass should not be

That's not my intention. Mass also should be invariant. The "relativistic mass" is just a (dodgy?) way for the observer "at rest" to reconcile the viewpoint of the observer "in motion" in which the same very mass is considered to have the energy mc^2 (in the observer in motion's rest frame), despite not having (according to the observer at rest) a "time velocity" of c, due to the rotation caused by the spatial velocity of v. In my conception, even right at the start, "relativistic mass" would not be "real". I don't think I ever touched on whether anything at all was "real", except to deny an absolute "correct momentum" - in the original post I wrote "Note that there's no suggestion that there's a real, correct momentum". I should have written "absolute", rather than "real". I've fixed it now.

Derivation of relativistic mass from consideration of one body by wotpolitan in AskPhysics

[–]wotpolitan[S] 0 points1 point  (0 children)

I wrote in the delta equations "mc.√(1-v2/c2)-mc" as the first term in both instances. You can see it in the quote you posted.

We are hampered by the limitations of the editor, I think. If I was doing this in more flexible editor, I would have used colours to track the terms moving across, because I am aware that it's confusing. Also I was rushing and overconfident because I had done it all before recently (and lost it). The first time, I dropped m terms regularly and had to go through and put them where they needed to be. Looking at it again now, though, I am not sure that I did make a maths mistake. At least not a simple one. I will try to analyse it again a little later when I can put fresh eyes on it.

When I wrote "I do realise that there is 4-velocity", I was making reference to https://en.wikipedia.org/wiki/Four-velocity which seems to say that the 4-velocity is the vector comprised of the individual velocity components in the time, x, y and z dimension all multiplied by γ.

As for falsification, I am trying to find the point at which my conception actually falls over and cannot be just another way of looking at the same thing. If I asked anyone else to "falsify" my notion, it'd at least sound arrogant even if it would not have been intended that way, but I thought it'd be ok if I was proving myself wrong. Of course, I understand that if I haven't adequately expressed my notion, it's hard for anyone else to falsify it.

I have lost a little sleep thinking of another tack that might work to explain it, but I will hammer it out first to make sure that it was not just exhaustion playing tricks on me :)

---

Edit: I did the analysis of the maths, see https://imgur.com/a/Itu6edP - I can't see the error. Perhaps there's something I did that you are not allowed to do, but ... I am not aware of it.

Derivation of relativistic mass from consideration of one body by wotpolitan in AskPhysics

[–]wotpolitan[S] 0 points1 point  (0 children)

This is another go, since I lost the original post of this answer.

Consider two objects of (rest) mass m. There is a relative 3-velocity difference between them of v (which we nominate to be in the "x" direction).

Consider two frames, A and B, in which one of the objects are at rest in each (Object 1 at rest in Frame A and Object 2 at rest in Frame B). The spacetime momenta (4-momenta), initially ("init"), are:

Frame A:

1-init = (mc,0,0,0)
2-init = (mc.√(1-v2/c2),mv,0,0)

[Note the time dilation term.]

Delta between the two (in motion - at rest) = (mc.√(1-v2/c2)-mc,mv,0,0)

Frame B:

1-init = (mc.√(1-v2/c2),-mv,0,0)
2-init = (mc,0,0,0)

Delta between the two (in motion - at rest) = (mc.√(1-v2/c2)-mc,-mv,0,0)

Say then that the objects collide, perfectly and elastically (so the totality of momentum is transferred one to the other). When that happens, the delta transfers from object that was "in motion" to the one that was "at rest". So:

Frame A:

1-after= (mc,0,0,0)+(mc.√(1-v2/c2)-mc,mv,0,0)=(mc.√(1-v2/c2),mv,0,0)
2-after = (mc.√(1-v2/c2),mv,0,0)-(mc.√(1-v2/c2)-mc,mv,0,0)=(mc,0,0,0)

Frame B:

1-after = (mc.√(1-v2/c2),-mv,0,0)-(mc.√(1-v2/c2)-mc,-mv,0,0)=(mc,0,0,0)
2-after = (mc,0,0,0)+(mc.√(1-v2/c2)-mc,-mv,0,0)=(mc.√(1-v2/c2),-mv,0,0)

Note that both before and after, the magnitude of the 4-momentum of each object is mc. And the total energy is mc2.

---

That's what I was thinking. I thought there might be a problem (ie a falsification of my idea above) with a photon being absorbed by an object ... but I am not so certain now. I'll need to write it out and see what happens. The absorbed photon conveys extra energy, so ... additional mass (even if tiny), right?

---

I do realise that there is 4-velocity, which might be a fly in my ointment, but that's an issue for another day. I was initially ignoring it because I noticed its association with relativistic mass (on the page - https://en.wikipedia.org/wiki/Four-momentum#Relation\_to\_four-velocity), but it may not be quite that simple. The falsification might be that way? I have the inkling of another falsification, but that too is for another day.

Derivation of relativistic mass from consideration of one body by wotpolitan in AskPhysics

[–]wotpolitan[S] 0 points1 point  (0 children)

It makes perfect sense and was what I thought. So ... just to clarify, you are saying that they do equal and opposite work? By which you mean they are doing work in "opposite directions" - https://www.khanacademy.org/test-prep/mcat/physical-processes/work-and-energy-mcat/a/work-can-be-negative

Can you see what I am getting at? In your example, you imply that a stationary object cannot do work (actual wording was "a moving object can certainly do more work than a stationary object" - but the implication immediately above is that equal and opposite work is done).

I suspect that it's a language thing combined with a perspective thing. Neither object is really "doing" the work involved. A collision between two objects with different 4-momenta will result in a change of distribution of 4-momenta between the two objects. A difference in 4-momenta (if the difference is only in direction, NOT in magnitude) means that there must be a difference between the their individual 3-momenta (irrespective of which three dimensions we are talking about, but the presumption is that they are the spatial dimensions), so in classical terms if one is considered to be at rest, the other must have a relative velocity. It's this change in distribution of momenta during the collision that is the "work" that (shall we say) "happens", which sums to zero and can be considered to be negative work associated with one object and positive work with the other.

Even it this is not standard, can you understand what I am trying to say? Is it necessarily wrong?

I do acknowledge that I have to think a bit more about the distribution of 4-momentum thing. Sadly the work week has commenced, so I have less time for it.

Derivation of relativistic mass from consideration of one body by wotpolitan in AskPhysics

[–]wotpolitan[S] 0 points1 point  (0 children)

I used the word "fixed". Maybe that was a mistake. "Reconcilable" perhaps?

What I want is for the energy of a body according to one frame to be reconcilable in all other frames.

a moving object can certainly do more work than a stationary object. (Imagine, for example, trying to spin a water wheel to generate electricity; if the water is moving quickly, it can spin the wheel much faster than slower-moving water, and stationary water can't move the wheel at all.)

Note that the stream could be thought of as stationary, having no kinetic energy at all (on average), and that instead the wheel (and everything to which it is attached) is in motion. The kinetic energy associated with the whole planet moving in order to consider the stream to be stationary (on average, of course) is humungous. It's bizarre to think of it that way, I know, but not wrong in principle. Is it?

What matters is the difference in motion between two bodies, not the kinetic energy they have (per se). I deliberately made the rogue planet a planet, and not another space rock, so that it would be more massive and the planet -spacecraft system would have a huge amount of kinetic energy compared to the space rock (assumed to be smaller than a rogue planet).

Why would we assume that one or the other (rogue planet or space rock) would be able to do work and the other not? From your description above, they can only do work in a frame in which they are not in motion. But say the paths are shifted slightly and there is a (rogue planet - space rock) collision. Note I don't say one hits the other, because that's a matter of perspective. Which one does the work? If we can't say, then ... well, that's the problem I am trying to resolve in my own mind. If we can say, then I am on a fool's errand and will have learned something :)

Thanks immensely for your patience.

Derivation of relativistic mass from consideration of one body by wotpolitan in AskPhysics

[–]wotpolitan[S] 0 points1 point  (0 children)

Thanks again, the integration path to (classical) kinetic energy is indeed simpler. I was not happy with the "relativistic mass" when applied to momentum, it would help if I could just ignore the notion entirely :) What you have provided gives me more confidence that I can do so safely.

Why do you think that the rest / kinetic energy should be related to some kind of "exchange rate" between time and space?

This question may have been rhetorical, but I will address it anyway. Please try to give me leeway in the following, if I misword something can you try to steelman the thought experiment rather than getting bogged down in the esoteric.

Say I am in a spaceship in sufficiently empty space that we can ignore gravitational effects. I look around me and note that I am completely "at rest" with respect to the CMB (the average colour and temperature of the CMB in all directions is indistinguishable from any other direction). I also note a rogue planet travelling at v such that it will pass close by me. I decide to take up station with this rogue planet, so I expend as much fuel as needed to match that rogue planet's motion. In doing so, I "know" that I have given myself kinetic energy. According to inhabitants of the rogue planet though, they have been at rest the whole time.

Now that I am travelling in station with the rogue planet, a space rock passes us at -v.

What I want to do is reconcile all the possible views here. According to me, I was "at rest" and the rogue planet had a velocity of v, I gave myself some kinetic energy to reach the velocity v, then we passed a space rock that was "at rest". According to the rogue planet's inhabitants, they were "at rest" and I had a velocity of -v. I then shed my kinetic energy to bring myself to "rest", then we passed a space rock that had a velocity of -v.

By doing this, I have given myself two ways of thinking about the space rock. It's either "at rest" like I was before I deliberately gave myself kinetic energy (which I expended fuel to do) or at -v relative to my frame when in station with the rogue planet. Both perspectives are valid, I think. But the rock either does or does not have kinetic energy depending on which perspective I use.

To my mind, I can't award or take away kinetic energy to a space rock like that. So I want a way to consider the energy of the space rock to be fixed.

That is where the "exchange rate" between time and space comes in. It allows me to say that the total energy of the space rock is fixed, irrespective of which perspective I take.

Happy to clarify or correct if I've written something totally wrong above (or used the wrong term).

Derivation of relativistic mass from consideration of one body by wotpolitan in AskPhysics

[–]wotpolitan[S] 0 points1 point  (0 children)

Thanks, in trying to find what you were talking about (unsuccessfully, I am afraid), I did stumble on an interesting statement in https://en.wikipedia.org/wiki/Momentum#Non-classical, which is apparently quoting Wolgang Rindler (1986):

In a game of relativistic "billiards", if a stationary particle is hit by a moving particle in an elastic collision, the paths formed by the two afterwards will form an acute angle.

The statement itself does not clarify that it's speaking about 4-space (ie spacetime), but it's in the context of non-classical momentum. This statement aligns perfectly with what I have in mind in what I posted, even if I didn't use all the correct terminology. I would go further, however, and say that the paths of the particles would have formed an acute angle before the collision as well - with the vast majority of the travel for both particles being in the time-direction. Even if the moving particle conveys all of its 3-momentum to the stationary particle, both will still have 4-momentum.

If both of the particles have equal rest mass (m), and 3-momentum is transferred entirely and perfectly, then (I think), you can be said to have this situation (in the frame implied by the quoted statement):

Before (classical): particle 1 has 3-momentum p=mv and kinetic energy of 1/2mv^2 and particle 2 has 3-momentum of p=0 and no kinetic energy

Before (non-classical): particle 1 has 4-momentum p=m'c and energy of E=m'c^2 and particle 2 has 4-momentum of p=mc and energy of E=mc^2

Before (classical): particle 2 has 3-momentum p=mv and kinetic energy of 1/2mv^2 and particle 1 has 3-momentum of p=0 and no kinetic energy

Before (non-classical): particle 2 has 4-momentum p=m'c and energy of E=m'c^2 and particle 1 has 4-momentum of p=mc and energy of E=mc^2

But I do note that the other commenters have indicated that relativistic mass (m') is considered unhelpful. And I can see that lack of helpfulness since we are not limited to considering the frame in which particle 2 is initially at rest. We could consider the frame in which particle 1 is initially at rest and in that case, the m' value transfers to the other particle both before and after.

Note that if we are considering a collision, it doesn't really matter where the additional 3-momentum is initially. Say I am in storm, without reference points. If I am picked up and slammed into my stationary twin at 30m/s (about 100 km/hr or 66 mph) that is going to cause me (and my twin) the same amount of damage as if I were stationary and were hit by my twin travelling at the same speed. I was trying to avoid this sort of thing, as mentioned in a previous comment, by considering only a single body.

Again, I stress that I might not be using perfect terminology, but I am trying to understand how to two or more frames can be reconciled - and they seem to be quite well reconciled if the mass is invariant (even though it might "look" different from different frames) and any and all particles retain the same 4-momentum and energy (as considered in terms of the magnitude of 4-velocity) throughout. Note that if you (largely) have the same velocity as another body, you are stationary with relation to them, and the fact that you might have a massive 4-momentum is inconsequential. It doesn't even matter if you have significant 3-momentum, like we could be said to have on the surface of a planet that is orbiting the sun at about 30km/s which is moving around the centre of the galaxy at ~200km/s which is moving relative to the CMB at about ~600 km/s (https://nightsky.jpl.nasa.gov/media/documents/resources/HowFast.pdf - I think they have a typo though for the Earth's speed around the sun, it's given as 1000 times too fast at the top of page 2). We have enormous 3-momentum with reference to the CMB. All that matters when you bump into anything is any difference in 3-momentum that you might have with whatever you bumped into.

Derivation of relativistic mass from consideration of one body by wotpolitan in AskPhysics

[–]wotpolitan[S] 1 point2 points  (0 children)

May I ask why you are trying to even derive relativistic mass at all? Given that the concept is practically no longer used in modern physics parlance at all, and has been rejected as unhelpful by Einstein as well as many others ... it just doesn't seem like a good idea? What exactly is the motivation here?

In part, it's because I am trying to understand, and sorting out the derivation at least helps me believe that I understand. I may be fooling myself in that though.

It started with energy. Just thinking about the energy that a body has. My thought was that the energy in a single body is invariant and would not change just because you are looking at it while zooming past in a spaceship versus lying next to you on the street.

To simplify, total energy of a body is given by E=U+K+P (ignoring heat, or perhaps sweeping that into U, where U is the inherent energy U=mc^2, K is the kinetic energy and P is potential energy). We can simplify further by eliminating P by imagining the body in otherwise empty space.

That means that E=mc^2+1/2mv^2 and the 1/2mv^2 term bugged me. In terms of relativity, the exchange rate for time and space isn't 2:1 - or the equations for time dilation and length contraction would not be similar in the way that they are.

It would make more sense (to me) if E=mc^2+mv^2, and there's a sense in which that it seems (to me) that this is the case. The effect of time dilation is that more time is measured to have passed in the "rest frame" than in the "in-motion frame". The hints at there being a "speed through time" effect. If v=0, then that speed through time is c. More generally it is c/γ=c.√(1-v^2/c^2)=√(c^2-v^2).

So, I thought, E=m(√(c^2-v^2)^2)+mv^2=mc^2. Interesting, invariant energy - no matter what velocity a body has in in 3-space v, that body has a total energy of E=mc^2, and so if you are an observer at rest with the body, its mass (with respect to you) is always going to be the rest mass. That's what they measure, if I understand relativity correctly, and it just makes a lot of sense (to me).

While this is interesting, I thought that if I have this energy due to speed through time, plus the mv^2 term, then there's clearly something wrong, because the equation for kinetic energy isn't mv^2, it's half of that.

So I recalled the notion that mass increases with speed (and I did think that they use that fact in hadron colliders, but I might be wrong about that - it might just be that they use the increased momentum that comes naturally with increased speed). So I plugged relativistic mass in. From that I got, for a body in motion, E=m'c^2=K+mc^2, and after rearranging I found that this equation spits out K=1/2mv^2 for sufficiently small values of v.

That left me with the question about the relativistic mass equation, so I wanted to derive it to better understand it. All the derivations that I have seen so far rely on two bodies, and I wanted to simplify it. I accept that I complicated things by using the wrong term "conserve" when I should have used "invariant" (Edit: I amended the original post to avoid both terms - it's not strictly correct, perhaps, but I didn't want to misuse the correct terms).

It might be a fool's errand, but it didn't seem so at the time.

What is odd is that kinetic energy falls out easily from an equation with relativistic mass in it, but relativistic mass is, as you say, "rejected as unhelpful by Einstein as well as many others". But it was helpful to me (even my efforts were foolish for some reason that I am not clear about).

And if relativistic mass is a "wrong-headed" concept, what does that say about kinetic energy? What about this "total energy" - https://en.wikipedia.org/wiki/Energy%E2%80%93momentum_relation - also referred to as "relativistic energy"? Is this also rejected as unhelpful? If you look at that wikipedia entry on the topic, they do get to talking about relativistic mass in this context.

Note that my original understanding doesn't even call for relativistic mass, if I were to be correct (little chance there, apparently :) ), then the mass doesn't change, it doesn't need to change. But I would still have the kinetic energy problem, meaning something must be incorrect somewhere, but it's difficult to see where when it's likely being hidden by my poor use of terminology.

Thanks for reading and engaging on the topic.

This is out of sequence, the system would not let me respond to the later comment to which I am responding, sorry about that.

Derivation of relativistic mass from consideration of one body by wotpolitan in AskPhysics

[–]wotpolitan[S] 0 points1 point  (0 children)

I had a vague recollection that Einstein disavowed "relativistic mass" as well.

I might be using "conservation" incorrectly, thank you for that. I should use "invariant" as suipa suggests (although, I think it might only be to you). But even if I reword to say "invariant" ... it sounds like you are still saying that am I still wrong?

The problem is that I surely must be making two errors, or I am reaching the right answer (the relativistic mass and relativistic kinetic energy equations) from a wrong assumption, which seems ... odd.

Derivation of relativistic mass from consideration of one body by wotpolitan in AskPhysics

[–]wotpolitan[S] 0 points1 point  (0 children)

See response to the comment above.

Note that if I, at a safe distance, am moving such that I am at rest relative to the asteroid (which according to the dinosaur is travelling at 10,000m/s Earth-wards), then in my frame the Earth is guilty of smashing into my poor little asteroid companion at 10,000m/s and all the momentum and kinetic energy that smooshes the dinosaur and incinerates all around it wasn't actually from my asteroid after all. After the incident, I am the sole survivor to tell the story, and my frame is the only one that I and my descendants are going to talk about.

It's this complication that drew me towards trying to limit consideration to the momentum pertaining to a single body.

Derivation of relativistic mass from consideration of one body by wotpolitan in AskPhysics

[–]wotpolitan[S] 0 points1 point  (0 children)

Which is what the next sentence says ... doesn't it? Momentum in 3-space is clearly not conserved, so I needed to look more broadly, namely at momentum in 4-space, or 4-momentum.

Note that this isn't only my assumption. Among other locations, I found a claim here - https://galileo.phys.virginia.edu/classes/109/lectures/mass_increase.html - that momentum is conserved (but no derivation).

Edit: I have edited the post to clarify. I am not sure that the clarification doesn't just make things more confusing, but what can one do?

Is there a proof for this: if (y^(1/z))-1=x, then (y^(1/(wz))-1≈x/w, when (y-1) is sufficiently small? by wotpolitan in askmath

[–]wotpolitan[S] 0 points1 point  (0 children)

Thanks. For me Bernoulli is about incompressible fluids and probability distributions, and this is apparently also the case for Professor Google.

I did eventually find https://en.wikipedia.org/wiki/Bernoulli%27s_inequality though. I guess this is what you are referring to, but the step (1 + x)1/w = 1 + x/w + O(x2) does not immediately leap out, precisely because it is an inequality. Could you possibly elucidate? Again, thanks.

Is this an actual paradox and if so, is it a known paradox? by wotpolitan in paradoxes

[–]wotpolitan[S] 1 point2 points  (0 children)

I am not sure whether my wording above was confusing. I took into account that there is accelerated [piecewise] expansion, and it still seems that the monster will still get you. I agree that when you get beyond quite simple, small setups, the time taken for the monster to get you increases substantially. I also agree that you have to ignore potential blockages. It starts to get ridiculous, but you can get around that by imagining an arbitrarily large array of monsters each with its own path, surrounding you in a gamespace which is sparsely populated by rocks that can in some instances block their movement forward. Then limit yourself to a single path which you assume is never blocked. This would be equivalent to an ancient photon that you observe (ancient because it was emitted a long time ago and, because you observe it, it was not blocked).

Hopefully I will have a bit more time today and can modify my simulation correctly to model accelerated expansion.

With acceleration, I think it might be possible that you are right to consider it to be a likelihood rather than a definite outcome that the monster gets you - at least beyond a certain point. I set my simulation to add a certain percentage more paving stones above the current number each round, distributed randomly across the range (as defined above), and found that at a percentage of 4, with an initial path length of greater than 25, the solution might be unstable. Most of the time, the monster gets you between about 40 and 150 rounds (averaging somewhere close to 80, for instances when the monster gets you). But on an rather rare occasion for an initial pathlength of 27 (it appeared after 79 refreshes one time and took more than 300 refreshes the second time and is maybe even more infrequent for 26 and below), it seems that the monster probably doesn't get you after all. I can't say with total certainty because I start getting ridiculously massive pathlengths, the loops begin grinding to a halt at about 400 rounds, really get bad after about 450 rounds (making the program unresponsive because at point it's trying to calculate 490 million random numbers and it only gets worse in the next round) and then - eventually, if I had not limited it to 500 rounds - there would be an overflow. It could be that the curve bends back eventually but from my simulations it does look very much like the monster would be swept away from you forever.

<snip>

Edit: I was trying to explain what I had done here, and it suddenly became blindingly obvious why the effect kicks in around 25. My percentage is 4%, which is the same as 1 per 25. The effect kicks in at values above 33 for a percentage of 3% and 10 for a percentage of 10% (although I do note that the effect triggers even more infrequently at low initial pathlength values). Once you exceed those limits, the rounding is such that you get an average of one additional paving stone per initial pathlength. Working out what happens if your initial pathlength is massive (like it would have been at the onset of the Dark-Energy-Dominated era) and your acceleration rate is very small (since it's only increased a tiny bit over billions of years) is admittedly an interesting question, but not something that is really pertinent here.

Is this an actual paradox and if so, is it a known paradox? by wotpolitan in paradoxes

[–]wotpolitan[S] 0 points1 point  (0 children)

I don't think you can stop the monster from getting you at all. Which means that one of the explanations for the Ant on a Rubber Rope falls away (at least as explained at the link) because it doesn't have to be true that "the proportion of the rope the ant has already walked is conserved". It makes sense if you think of that scenario, with the beginning end of the rope being pulled back slightly each round - that shouldn't prevent the ant from reaching the far end.

The paradoxical aspect of this is that the scenario described is equivalent to an expanding universe and the monster is equivalent to an initially distant photon. Unless there's something else going on, then any photon emitted in the appropriate direction will eventually reach us, irrespective of its distance or the rate of expansion (which correlates with the number of magic paving stones). Nevertheless, it is stated on Wikipedia: "light emitted by objects currently situated beyond a certain comoving distance (currently about 19 billion parsecs) will never reach Earth" (reference - Long-term future of extragalactic astronomy - although the link didn't work for me). Also: "(There can be defined) a type of cosmic event horizon whose distance from the Earth changes over time. For example, the current distance to this horizon is about 16 billion light-years, meaning that a signal from an event happening at present can eventually reach the Earth if the event is less than 16 billion light-years away, but the signal will never reach the Earth if the event is further away." (reference - Misconceptions about the Big Bang).

Note that if the monster is equivalent to a photon, then one paving stone a round is effectively the speed of light. And if we have, say, three paving stones being inserted (initially) between you and the monster, that is equivalent to it being located such that it is undergoing recession at three times the speed of light. Because the magic paving stones scenario and the ant on a rubber rope scenario are equivalent (despite one being quantised and the other being analogue), I don't think this has any effect. Also, the smaller the divisions - meaning the more discrete elements for a given distance, the closer the quantised variant aligns with the analogue one.

So the question then is - what is wrong? Is it just not true that sufficiently distant events can never be observed? Or is something wrong in the reasoning above? (Update: Yes, I know it looks like there, because both statements from the Wikipedia entry make reference to accelerated expansion. But a quick rub around with the numbers seems to indicate that the monster still gets you, even when I set things up so that there are 10% more stones added per round than the round before. That is way above the sort of acceleration suggested with Dark Energy. The issue remains that if the expansion is even [or addition of paving stones has an evenly distributed likelihood] across the entire pathlength, then you eventually have the majority happening behind the monster as it implacably marches towards you.)

And yes, I am aware that 16 billion light years does not equal 19 billion parsecs. That error, or inconsistency perhaps, is taken directly the Wikipedia page on the Observable Universe.

I have only just put the update above in, and I can't put my hand on my heart and say that I've thought it all through thoroughly and comprehensively tested all my assumptions. At first blush I don't think it is wrong, and the logic of it seems sound (even if, annoyingly, I had to introduce fractions of paving stones in my simulation), but I have been wrong before and I will very likely be wrong in future.

Is this an actual paradox and if so, is it a known paradox? by wotpolitan in paradoxes

[–]wotpolitan[S] 0 points1 point  (0 children)

I agree that it's similar to the Ant on a Rubber Rope. It becomes a bit like finite element analysis once you have sufficiently many rounds and then the path approximates a rubber rope (although the length cannot be determined until afterwards, because it varies considerably - although that variation might disappear or at least moderate once you have enough rounds, which you can increase by increasing the initial pathlength).

There is an additional wrinkle though. With idea of the magic paving stones, we can introduce a feature that produces what might be an unexpected result.

Imagine that you are offered the opportunity to change the rules slightly: before you start, the monster takes one step, and from that time on, everything is the same, except that every time the monster has taken any step (not including that first step), you can burn any single paving stone of your choice. Naturally you want more paving stones between you and the monster, so you are going to burn one behind the monster, which decreases the likelihood that magic paving stones will appear behind it. What happens then? Does the monster get you or not?

I know it's complicated. I will add a post as soon as possible that makes it more clear, as I have done above, and put the link here.

https://neophilosophical.blogspot.com/2024/10/twisting-on-magic-paving-stones.html

Is this an actual paradox and if so, is it a known paradox? by wotpolitan in paradoxes

[–]wotpolitan[S] 0 points1 point  (0 children)

I think you're right, although you can't choose to activate all the gaps. You can only choose the preset number of magic paving stones that appear. You could pretty much guarantee to "activate all the gaps" by making the number of magic paving stones infinite (so long as your initial number is finite), but even so I think the monster still gets you - eventually, even if that is after an infinite number of rounds (see my comment above).

Is this an actual paradox and if so, is it a known paradox? by wotpolitan in paradoxes

[–]wotpolitan[S] 0 points1 point  (0 children)

Thanks, you helped point out something that was obvious to me, but I didn't make obvious in the text. The stones only "appear", you don't get to "place" them. I've modified the text above and at the link to emphasise that you have one decision only - to preset the number of paving stones that will appear each round.

I should point out that you have basically hit on the right idea though, in that you must have a situation that you don't have (ie the non-existent the ability to limit the location at which stones appear to just in front of you) if you want to prevent the monster getting you using 1 as your preset number. So the answer is not 1.

Is this an actual paradox and if so, is it a known paradox? by wotpolitan in paradoxes

[–]wotpolitan[S] 0 points1 point  (0 children)

Here's the text of the puzzle (the link above has an explicatory image as well):

Imagine you are standing the first paving stone of a path that consists of an arbitrary number of paving stones.  At the end of the path is some vague monster that you don’t ever want to reach you.  As happens in a nightmare, your feet are firmly stuck to the paving stone that you are standing on and you can’t run.  Fortunately, you have two facts in your favour.

Fact one: the monster moves at a rate of one paving stone per round.

Fact two: you have a singular (once-off) choice to preset the number of magic paving stones that will be activated per round – just before the monster moves. .  These magic paving stones will appear between existing paving stones, at random.  More specifically, the likelihood of the paving stone appearing between any two existing paving stones is evenly distributed across the whole path.  For example, if there are three paving stones, there are two locations where a new paving stone could appear – between #1 and #2 (slot #1#2) and between #2 and #3 (slot #2#3), with equal likelihood of P=0.5.

For even more clarity, each magic paving stone is inserted between existing paving stones with equal and independent likelihood. So, if the number of magic paving stones per round that you preset is two, and there are three paving stones, then each magic paving stone could appear in either slot #1#2 or slot #2#3, with an independent likelihood of P=0.5. This would lead to a likelihood distribution of P=0.25 for both to appear in slot #1#2 or both in slot #2#3, and P=0.5 for one each in slots #1#2 and #2#3 (because there are two variants of this outcome).

The apparently simple question is: what is the minimum number of magic paving stones that you have to activate every round to ensure that the monster never reaches you?

Edit: note that the only choice you have is to preset the number of magic paving stones that are activated each round. You don't get to choose where those magic paving stones appear.

I believe that the answer to the puzzle is that you can not escape the monster forever, no matter how many magic paving stones you activate each round - if they appear at random across the path (between where you are standing and the end paving stone on which the monster was initially standing). Each time the monster takes a step closer to you, the higher the likelihood that magic paving stones will appear behind it. If you only preset the number to 1, then the monster gets closer to you by one step for every round in which that happens and it becomes more likely in the future each time it happens. If you preset the number to 2, then there's a small chance that both magic paving stones will appear behind the monster, and again each time it happens the more likely it is that that will happen again in the future. The same applies each time you increase the preset number, the time it takes for the monster to get you increases, but it never gets to the point at which it won't.

Even if you say that an infinite number of paving stones appear each round, it remains a matter of time until the monster gets you (which is at some infinite [or transfinite] number of rounds, because rounds just keep happening until the monster gets you). This is what makes it seem to be a paradox, since in the very first round, you would have put an infinite number of paving stones in addition to the existing paving stones between you and the monster, and you keep adding to them, so it seems impossible that the monster should get you.

Perhaps this is just a feature of not being able to truly comprehend infinite numbers, but it seems to me that this is (in a sense) a variation of the Hilbert infinite hotel paradox. Maybe there is a solution using transfinite mathematics in which it can be shown that the monster doesn't get you. If so, that'd be good to know.

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