md65536

Physical laws have certain properties, and you can have maths that have the same properties, but you can also have other maths that don't, including abstract mathematics that aren't based on physical things. For example, you can mathematically take apart an orange and rearrange it into two oranges identical to the original. https://en.wikipedia.org/wiki/Banach–Tarski_paradox The maths aren't restricted by the physical law of conservation of mass.

And B's "owner" doesn't see its clock slow down, because its clock doesn't slow down in its frame of reference. Not because "the clock owner is affected by time dilation to the same extent as his clock", because what the owner measures is in its own frame of reference where it is not affected by the time dilation A measures. I suppose the "both B and its owner are affected by the same time dilation" could be used to show that A agrees that observers in B's frame of reference would not experience a difference in time among their own frame's clocks, but that's all described in A's frame. However, there are people who are trying to learn relativity who get stuck on ideas like "B really 'actually' does slow down, it's just that B's owner doesn't notice", because they're not thinking about the effects with respect to different specific frames of reference.

Sounds good. Yes, I was assuming the Earth approximates an inertial frame, as with the basic twin paradox setup. The twins don't have the same turnaround point. They travel at different speeds, but start together and end together. They travel different distances as measured from Earth. Each twin measures the distance that they travel as length contracted by a respective Lorentz factor, and I've contrived these length-contracted distances to be the same, ie. the distances they each separately measure. They would turn around simultaneously according to the Earth frame. Definitely running them on the same x-axis is a good idea. Over the whole trip, the faster twin ages half the slower one, but I don't think the Lorentz factor of one twin moving relative to the other is equal to 2 during the outbound or inbound leg. Each twin measures the other's clock as slower during the inertial sections. I think that each twin, just before its turnaround, says the other twin has not yet turned around, and just after turnaround it says the other twin turned around some time ago. It seems that would be the case in any situation where the twins turn around simultaneously in the Earth frame, whatever their speeds. "In physics, the twin paradox is a thought experiment in special relativity involving identical twins, one of whom makes a journey into space in a high-speed rocket and returns home to find that the twin who remained on Earth has aged more." Do you agree? Are you going to argue both that the twin paradox requires an inertial twin, and that the Earth twin is not an inertial twin? I don't need to go out of my way to point out that non-inertial frames are involved. If any two twins separate and then re-unite, in SR, at least one of the twins is non-inertial. We can all agree they're not paradoxes, I should have put the word in quotes. I assumed understanding the resolution of the "paradox", my puzzle isn't intended be a new paradox that seems to not make sense.

Then assume a negligible turnaround time. Or a giant circle, all turn around time (but constant speed), the answer's the same. This problem doesn't involve calculations involving acceleration. The maths don't get any more complicated than the Lorentz factor. Treating it like a triplet problem is probably the easiest way to solve it. Is this too ambiguous? If a twin makes a round-trip journey that is 2 light years as measured on Earth, with constant gamma=2, is it unambiguous to say that the twin measured traveling a distance of one light year? Or do you have to technically include "one light year relative to Earth"? Or is this really mixing frames, and if using the twin's frame to measure distance, it doesn't travel at all, you'd have to say that the Earth traveled one light year?

I never said constant velocity, I said constant speed *; swansont already pointed that out. GR is not needed for this, the basic twin paradox is an SR problem and so is this variation. You can assume flat spacetime and ignore any complications not mentioned in the problem. I didn't think it was such a difficult problem. Assume the simplest paths that won't change the answer. I think the simplest is probably "straight line away, straight line back, negligible turn-around time". * That wasn't meant as a trick to the problem, just a simplification of the description. I usually make that simplification automatically because it's burned into my brain that the direction doesn't matter. I was about to say, "I should have said constant velocity outbound and constant with the same magnitude back", but on second thought, I think it's more important to not get stuck thinking that a straight-line path somehow matters.

I disagree. If the "stay at home twin" moves around, even at high speed (but slower enough than the other twin), the different aging still happens. Ceasing to be inertial doesn't make the effect go away. Further, if any two worldlines intersect at two separate events, the proper time along the two paths will generally be different, and there is differential aging. There are variations to the twin paradox, but if "twin paradox" means only the case where one twin is inertial, then instead of "twin paradox" I mean the seemingly paradoxical different aging of twins that depart and meet again, having traveled at different speeds, where each measures the other's clock as ticking slower than their own when erroneously neglecting relativity of simultaneity. No, that's not the intent. The two twins could each travel in a giant circle in empty space, or they could leave in a straight line and return in a straight line with a sharp turn around or even come to a stop if they spend negligible time accelerating, their direction doesn't matter, only speed, and a path that will let them meet again. Well of course they all have valid reference frames. If you want to do it with inertial frames, each traveling twin would require a minimum of two. Earth is used as a reference for the requirement that their speeds are constant, because their speeds are not constant relative to each other all the time, in other reference frames. I choose that to make the description and calculations as simple as possible. You could have the twins leave from one event and meet up at some other event anywhere, it wouldn't have to be at the same place... except of course that you could join the two events with an inertial observer, in whose frame the events are at the same place. However, no physical observer actually needs to inertially intersect the two events in order for the twins' different aging to occur. Direction doesn't matter, the different aging can be calculated from time and relative speed alone.

The twin paradox doesn't require an inertial twin. Suppose two twins left Earth at the same time and returned at the same time, each traveling a different but constant speed relative to Earth. Whom does your intuition say traveled a longer distance, the twin who ages more, or less? Don't read the following puzzle if you want to think about it first.

That's not a useful way to put it. Only time relative to another observer is affected (the coordinate time of the clock according to the observer). The clock itself is not affected by time dilation and measures "proper time", at 1 second per second. Your statement can lead to thinking like, if a neutron star (as an observer) rotates rapidly around another that it's colliding with 10 billion light years from here, then my clock slows down, but I don't notice because all clocks on Earth slow down. The real reason I don't notice any time dilation is that another observer's motion and its measure of my clock in its reference frame, doesn't affect time as measured in my own frame of reference.

I think you might be misunderstanding the principle of relativity. It's not that measurements of a single experiment, observed from different frames of reference, will make the same measurements. It's that the experiment, performed in each of the different frames of reference, will have the same measurements within those frames. Obviously, many measurements will be "relative" to the observer. In Galilean relativity, things like relative speeds will be different depending on inertial frame. In special relativity, things like lengths and times will also be different. In the example mentioned, you don't have a person drop a ball while standing on the ground, and measure it from the ground and from a moving airplane. You drop the ball on the ground and measure it from the ground, and you drop the ball on the plane, and measure it on the plane. If both adequately approximate an inertial frame and have the same gravity, both experiments behave the same. A more general example, if you had two trains, each at rest in a different perfectly inertial frame, there would be no experiment that could be performed that could identify that one train is at rest and another is in motion except relative to a frame of reference, and so there is no concept of universal motion or rest.

Not "should", you must take it into account or you'll get a completely wrong answer and very wrong intuition of it. They'd see it moving at about 99.97% c, assuming they're moving in the same relative direction. If you want to talk about "close approximation to c" usually you would talk about "approaching c" and use mathematical limits for equations where v=c fails. "Approximately c" can be misleading, because that is completely different from "exactly c". For example if something is moving inertially at .9999c relative to you, it still has a reference frame where it is stationary, and it does not see itself moving at some high fraction of c, but rather sees light behaving no differently than you do. In the above example, if an object is moving at .9997 c, ie. "very close to c", and another object accelerates another .5c away from both you and the object, that other object is now traveling about .9999c away from you. In that sense, yes an object that is moving with speed very close to c varies in speed less than it does in frames of reference in which it is slower. But that's not what invariant means, and since an object traveling at .9999c is traveling at 0c in another reference frame, it is not at all invariant. Maybe you could say something like "As v approaches c, v approaches being invariant among frames that are all moving at non-relativistic speeds relative to each other", but then you should see how unhelpful such a thing is to understanding relativity, because you avoid relativity here by avoiding relativistic observers.

I just realized the topic was split, but I wanted to sum it up to bring this back to the original topic anyway. Saying that it is "harder" to accelerate an object that is moving fast (as measured from some reference frame), can instead be said more precisely: A certain change in velocity of an object in a frame in which it is moving fast, requires a greater change in velocity (ie. higher rate of acceleration or acceleration for longer time) in a frame in which it is moving slowly, because the difference in the velocities is not the same in different frames of reference.

Can you clarify "same accelerations"? It's confusing because a given acceleration wouldn't be the same in the two frames. Do you mean the same proper acceleration, which would be measured as relatively minor in the muon frame? Otherwise suppose the "quite considerable" acceleration is of 0.001 c, then accelerating that much in the muon's frame would be measured as an acceleration of 0.001 c (ie. quite considerable, not relatively minor). Ugh, there's always complicated and simple ways to look at anything in relativity. Complicated: Say you are at rest relative to Earth, and are able to easily accelerate to 0.001 c relative to Earth. While at rest on Earth, you are also traveling at say .999 c toward a muon. If you want that "same acceleration" of 0.001 c relative to the muon: You can't accelerate 0.001 c toward it (measured in its frame), but you can accelerate away (or decelerate) so that you're traveling at .998 c toward it. If you did this starting from at rest on Earth, you'd accelerate 333.556 times as much relative to the Earth (using more than that many times as much energy?), so you're now traveling at 0.333556 c relative to Earth (determined using composition of velocities formula). Simple: When you're at rest you have a rest frame, and your mass is the same regardless of the relative velocities of other moving objects. In each of their rest frames, you're the one that's moving. You feel the same regardless of their motion (you can't distinguish between being at rest and being in a moving inertial frame), and proper acceleration is equally easy no matter which inertial frame you're in; the faster you go, the same it always feels. The difference is that a large change in velocity in your frame may be small in another frame, due to the way velocities add in special relativity, so you have to accelerate greater and greater amounts to achieve some specific acceleration in the other frame. (Is this simplification valid? Can the change in total energy of an object as v approaches c, be found using composition of velocities formula, or is there something more to it?)

If you were the body and accelerating yourself, do you know how this would feel or what you would measure? For example, if you're on a rocket accelerating away from Earth at ever-increasing speeds. I think that understanding that would help understand the issue. (Hint: a simple rocket model implies constant proper acceleration, and is probably an easier answer. If instead you suppose constant acceleration relative to Earth you'd measure different things. You could also imagine accelerating in short rocket burns and then describe what you measure in between; what differences do you notice between subsequent burns?) Another question that might help understanding: How would you describe your body's total energy (or "relativistic mass") in the frame of a muon generated by a cosmic ray in the atmosphere, relative to which your speed is over .999 c? In what way is it hard to accelerate yourself relative to it? This isn't a trick question.

Changed my mind! No, I think the interpretation of "James knows only a minus d" is the right one. There are no numbers with a-d=7 that have a unique product. Therefore he knows that James doesn't know the answer. I think 8221 might work in the end? There was the idea that 4421 or 4222 would then also work. However, a-d=3 can be eliminated by 5552 (a unique product), and a-d=2 by 3111. If 0's are not allowed, then 8211 fits all the criteria. However I end up with 3 working solutions: 8211, 9881, and 9981 all seem to work.

The solution also works if Jack/P1 doesn't know a and d but just d-a, so the puzzle could be interpreted either way. The less information Jack is given, the harder it is for us to reason about it, but the easier it is to eliminate possibilities once we figure it out, because for Jack to be certain that James/P2 doesn't know the answer implies that the info Jack actually has must be substantial. If d<=4, then we know that d-a>4, eliminating possibilities like 3222 earlier.

Because not all of the frames will agree on the simultaneity of events. You definitely can define a "now" and you can do it however you want to (a foliation of spacetime would make sensible instants of time throughout space, but the problem is you can foliate it in many different ways, so essentially your "now" would be completely arbitrary). To be useful, you would want to be able to say that all the events in your "now" are simultaneous. Someone in a different inertial frame would not agree that those events are simultaneous, and would have no reason to accept your personal definition of "now".

On second thought I think what I crossed out is true, and possibly equivalent to what I implemented.

I think I've solved it and the answer surprised me. I'll restate the puzzle as I solved it: There's some number with digits abcd, with 0<=d<=c<=b<=a<=9. Person P1 knows a and d. P2 knows a*b*c*d. P3 knows a+b+c+d. P1 says "A) I don't know the digits but B) neither does P2". P2 says "C) I don't know the digits but D) neither does P3." P2 says "D) I know that P3 doesn't know the digits, and C) I don't know the digits either." --- I believe this order is important because P3 can know the answer while the statement is being given. P3 says "E) I didn't know the digits but F) now I do." It does not seem to need to say more but it could say now P1 knows or now everyone knows. Solution: A) a and d must be different or it would know. B) The range must contain some number with ambiguous factorization. For example, 9*1=3*3. If the range contains 1 and 9, or just 3, then it is still good. Do this for all possibilities and you find the range must contain 0, 2, 3, or 4. If it contains 1 we already know it must also contain either 0 or 2. So, eliminate all possibilities where d>=5. C) Eliminate all remaining possibilities that have a unique product. I think that a key element here is the concept of common knowledge. At each step, everyone can eliminate any possibilities that are based on what was said, but not necessarily the possibilities that the others could eliminate. Anyway, that might not be important yet... D) P2 can eliminate all possibilities that add up to a unique sum. E,F) Here is the key to the puzzle and where most of the possibilities are eliminated. P3 can also eliminate all possibilities that add up to a unique sum. However, since it now knows what the answer is, it must be that it was possible for P3 to assume that P2 knew the digits. The answer must be a combination whose sum is not unique sum, but where all other combinations that add up to the same sum have a unique product. Edit: No wait, that's not what I implemented! I looked for combinations that P2 knew had a unique sum *among* only possibilities that have no unique product, but which P3 did not know had a unique sum. Very confusing! I'm not sure I got that right or implemented it right, but I end up with only one possibility left and that is

Intuition would be to minimize non-diagonal segments, so AjkC (However it's easy to add bad diagonal paths that would break the strategy.)

If the solution were 9xy1, the product would be the same as xy33, so Jack could claim that James doesn't know the digits. (Then, 9621 adding up to 18, and 9431 adding up to 17, both with products 108, might be possible reasons why John can eliminate 4333.)

What about 6332? Its product is also 108. It's the only such one that adds up to 14. If John supposes the answer is 4333, then James would have considered 6332 as a possible solution, and wouldn't have been certain that John wouldn't know the answer??? Is that enough to rule out John thinking it's 4333? Or would we have to also show that every other possibility that James might suppose that John might suppose, doesn't have 2 or more combinations that add up to 14? But wait... 8411 adds up to 14... So John considering 4333 as an answer might have considered that James knew both 6332 and 8411 add up to 14, so could say that John didn't know the answer, meaning that 4333 is still a possible answer to John?

Yes. If the range is known to be [3, 4], then Jack cannot know that James doesn't know the answer, as he states. If the range is [3, 4] then Jack knows that James must know the answer. The only way that Jack can assert that James doesn't know the answer is if the range includes factors that can multiply to the same product in different ways, eg. 9*1 or 3*3 (requires a range of at least [1, 9]) or 8*2 or 4*4 (range at least [2, 8]). First ruling out 0, the smallest range for which this works is [1, 4]. *We* know this much after Jack's statement, so everyone else (John, James) are able to know this after his statement too, so John can rule it out. No wait... my line of reasoning is wrong and might not go anywhere useful...

Yes that makes sense. That's what happens at speeds low enough to neglect relativistic effects. https://en.wikipedia.org/wiki/Einstein_synchronisation Clock synchronization can be "achieved by "slowly" transporting a third clock from clock 1 to clock 2, in the limit of vanishing transport velocity." Then you'd say these clocks measure the same time. Worse than conflated and confused, time dilation was completely ignored, and relativity was presented over simplistically. They cared about getting Einstein's hair right; the science wasn't important. I think it's terrible, because a lot of people base their understanding of things on fiction, then go on to be a politician who decides NASA's budget or what to do about climate change etc., all the while perpetuating "the science wasn't important." (Rant cont.) Fictional science is probably fine, but writers throw in real concepts or jargon to be more interesting or sound legitimate or whatever. How many people could earnestly debate multi-world theories, without ever taking a physics course? Where do they get their information?

What do you mean by time reference? In the situation described, with no one stopping, each observer would measure the other's clock ticking slower, in addition to seeing it tick much slower due to delay of light.

No, Insignificance gets relativity wrong. However the quote above was cut off right before it gets bad. https://www.springfieldspringfield.co.uk/movie_script.php?movie=insignificance I bolded the main wrong part, and strikethrough'd ... yuck. The character is only talking about delay of light and has neglected time dilation entirely, but explains it as if it's time dilation. In the striked part, the character is describing differential ageing or total time, and describing it as a differential rate of time (which is constant at a fixed relative velocity).