md65536

In case anyone reading is confused about whether the effects are "real", there's no argument here from anyone that they're not. They're real. "It can't be both" (a ruler can't be .5% shorter and 3% shorter) can be understood to mean, "It can't be both in any single frame of reference." Because length and time are relative, it is perfectly reasonable for example for one observer to say "Clock A is slowed but clock B isn't; ruler X is .5% shorter" and another to say "Both clocks A and B are slowed; ruler X is 3% shorter", and another to say "Clock B is slowed but clock A isn't; ruler X is not length-contracted." All of those are real physically measurable quantities made by individual observers, none of them need to be interpreted by another observer. It is not reasonable to say a rocket is length-contracted to 90% in all frames but the other frames measure it differently because their rulers are also length-contracted differently.

Because time dilation is real. And yet, the atomic clock recorded time at one second per second, never slowing down or "being affected" by time dilation in its own frame. This is all consistent with relativity. Human interpretation??? What do you mean by that? The length contraction is just as real as time dilation. What does it have to do with humans? Do you need the humans' frame to describe the proper time measured by the muon? Does the humans' frame have some special status?

In the scientists frame, the process is really slowing down. In the muon's frame, the process is really NOT slowing down, but the height of the atmosphere is really length contracted, meaning it can pass through it in a shorter proper time. The unhelpful part is trying to describe what is "really" happening in terms of only one frame of reference (like the scientists'), because many people think that different frames' measurements include one "real" measurement and just how it "falsely appears" in other frames. And once again, what the muon observes, it measures in its own frame, where the process really does not slow down. The muon doesn't have to transform its measurements to the Earth frame (or any other), to say "the process really slowed down but my clock is also slowed down." It can just say, "My clock did not slow down."

Perfect! You've also answered the bonus question. d=v*t' (or v*tau) measured in the traveler's frame. Half the proper time during the trip means double the speed, to keep the distances the same. Half the proper time also means double gamma. So 2v_a = v_b, and 2/sqrt(1-v_a^2) = 1/sqrt(1-v_b^2), solve and you get 1/sqrt(5), 2/sqrt(5), which is what you got. The red curve shows something I'd hoped might be intuitive.

You could set it up so that in the "lab" frame, the light is emitted from one point along the circumference, passes through the center, and hits a detector on the opposite side, all in a straight line. This is true no matter what frame the light source is in (it would just be pointed differently in different frames, due to "aberration of light"). If the detector was rotating around the circumference, it wouldn't be hit directly opposite the emitter, in the rotating frame. The observer moving around the circumference would see the same beam of light aimed "backward" of center, curve "forward" (in the direction the observer is moving) so that it hits the center, and continue curving so that it hits the circumference a bit "closer" along the circumference rather than directly across from it. All observers would agree on this: if the light is emitted as the observer passes by, the rotating frame turns during the time it takes the light to travel across the diameter, and "directly opposite the observer in the rotating frame" has moved on by the time the light reaches the other side. There's probably a better way to describe it. You can figure out a lot by describing things in terms of events (happening at one time and place, so all observers agree that it happens (like if a particular beam passes through the center, it does so in all frames)), adding extra measuring tools (like a marked disk that turns with the centrifuge), replacing the accelerating observer with an inertial one that shares a momentary inertial frame at an event, etc.

Sounds like "begging the question" fallacy. You're assuming that the maths are based on physical processes, and concluding that the paradox has a physical resolution.

Local would have to be a lot smaller than that, eg. "measured at the center of the centrifuge" might work. If you're moving fast enough for it to not be negligible, and you send light in a path around the edge of the centrifuge, then an observer moving around the centrifuge will measure that light will take more time to make a round trip (from observer back to observer) in the direction the observer is going, and less than usual in the opposite direction. Both path lengths are measured to be the same in either direction. You can confirm the timing in the "outside" inertial frame: by the time the light has made a full circle in the outside frame, the observer has moved on from its original position, and light must make more than a full circle when sent in the same direction the observer is moving, and less in the opposite direction. However, this is not really a valid measure of the speed of light. You could call it the "coordinate speed" of light, and it's been argued on these forums that it's a meaningless measure. I'd say it'd be like making individual measurements in different momentary inertial frames of the revolving observer; you can get similar invalid measures of the speed of light if you switch between inertial frames in SR without properly accounting for the switch.

I think it's reversible, but it's not 1 + 1 = 1, because adding oranges together conserves oranges. The cloning operation isn't adding. Same with adding finite natural numbers (quantity is conserved), which have properties corresponding to everyday objects. Adding volumes of water conserves volume, but adding it to a black hole doesn't. Different mathematical objects have different properties and correspond to different physical things (some not at all). In this case it's not about "1", but really about infinity, because decomposing the orange involves splitting it into an infinite number of points, and you can add two infinities together and end up with one infinity. So even natural numbers, which have properties I would guess are based on our understanding of the physical world, have properties not necessarily restricted by physical laws.

And if a neutron star 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. Because, the distant star's reference frame is just as valid as any other. Okay maybe that's useful, maybe I should have said I don't think it's helpful. I don't think we're helping anyone who is struggling to understand relativity by stating things this way. My clock doesn't slow down in my reference frame and my ruler doesn't length-contract in my inertial reference frame. It is not a case of it happening without me noticing, unless you want to argue that it merely "appears normal" because I don't see the measurements that could be made in another reference frame. There are so many caveats to explain, when it all could have been said so much clearer and intuitively.

Yes, that all sounds good. I definitely like the idea of switching clocks at the turnaround, because then you're dealing with their different frames of reference instead of thinking of something "happening" to the twin or single clock. Political correctness I suppose is used to avoid arguments about the role of acceleration in the setup. But, politics can be avoided too if we just say "assume that the clock postulate holds", which is a given. http://math.ucr.edu/home/baez/physics/Relativity/SR/clock.html However, no one has yet taken a stab at the puzzle, by calculating the speeds of the two twins or clocks, given that they both measured (in their respective frames) having traveled the same distance relative to Earth, and that one aged twice as much as the other over the trip.

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.