md65536

I won't admit it but let's say that it's true. It's still not important because it doesn't have a measurable (detectable) effect. If it does, what is the difference between measurements when the "hidden" acceleration is neglected, vs. when it is included? I have not seen anyone include acceleration in any equation in this thread. In the typical twin paradox with the traveler using one clock, change of velocity is real and important because there is a difference between measurements when the acceleration is applied (the twin returns home) vs. not (the twin continues out into space). What difference occurs if your "hidden" effects are not considered? Is it possible that you're inventing a hidden effect to explain something that you don't understand? Only partly, and I made the mistake described in your link (http://math.ucr.edu/home/baez/physics/Relativity/SR/clock.html) of thinking that the equations of SR prove the postulate is true, when really applying the equations of SR, when acceleration is involved, assumes that the postulate is true. If the postulate was not true, the accepted calculations of the twin paradox effect according to SR would be wrong, while SR could still correctly predict the outcome of the experiment in post #1, which would then not be equivalent to a twin paradox setup.

This is an excellent analysis demonstrating correct application of the principle of relativity with respect to the relativistic Doppler effect, with a small error corrected by Janus, in which you're describing the traveler's point of view using the Earth's clock (two years each way) instead of the traveler's own clock (1.732 years each way).

It's just the best literal interpretation that I could come up with. It doesn't matter much if it's right because I already know it wasn't the intended interpretation. The word "next" in "the next second" is relative to "every second". Every day I go to sleep, and wake up in the next day. Hopefully there are multiple next days.

You've misunderstood what I said in 74. I agree with Iggy's assessment. In any post of mine that refers to C, it is treated as an inertial observer and there is no acceleration from v_B to v_C nor any equivalent. If you speak of C's history, I am assuming that C has remained inertial for all relevant history.

Now that everyone seems to be in agreement, perhaps someone can help me figure out the error, or "hidden trick" in my original premise. Please tell me the first statement here that you disagree with: 1) In a twin paradox using constant outbound v = sqrt(3)/2 ~= 0.866, "instant" turnaround, and inbound v = -v, where the traveling twin B ages 1 year proper time on each of the two legs of the trip, inertial twin A will age 4 years between B's departure and B's return (gamma = 2). 2) This works regardless of the method of travel of B. If B is on an outbound train, and then "jumps" onto a different inbound train, instantly accelerating and experiencing infinite proper acceleration during its frame switch, this properly implements the twin paradox experiment described in (1). 3) A clock on the train will tick at the same rate as B's clock while B is at rest on that train. So, a clock on the outbound train will age one year while B is on it, and a clock on the inbound train will age one year while B is on it. 4) Before and after B jumps trains, it will have a different relative simultaneity relative to A. Before, B calculates that A has aged 0.5 years since leaving, and after, B calculates that A has aged 3.5 years since leaving. In negligible proper time of B, B has calculated a change of 3 years of A's proper time. 5) The relative time depends on distance and velocity. So while B is at rest on the first train, B and the train agree on the relative simultaneity with respect to A, ie. just before the trains pass, both agree that A has aged 0.5 years. Just after the trains pass, and B is at rest on the inbound train, B and the inbound train agree that A has aged 3.5 years since B's leaving. To see this is true, according to (3) both B and train will age 1 year until arriving at A, both will calculate that A ages 0.5 years during that time, and both will agree on A's age (4 years) when they arrive. 6) The relative simultaneity of events at A as measured by the trains is the same whether or not B is on the train. If B misses the jump, the outbound train still ages 1 year before passing the inbound train, and the inbound train still ages 1 year before arriving at A, and A has still aged 4 years between trains, even though B may be stuck on the outbound train or splatted somewhere in space. 7) If B stays at rest on the outbound train, no object accelerates at the passing of the outbound and inbound train. 8) Any observer can calculate the relative simultaneities of the various objects, and all observers agree on the proper time or aging of the trains and of A and B between any of the events they pass. 9) Therefore the effects of the twin paradox are evident in objects that follow a part of the same path as the traveling twin, and the full effect is evident in a set of objects that cumulatively follow the full path of the traveling twin, even if no single object undergoes acceleration. 10) Acceleration is still important to implement the twin paradox experiment in flat spacetime using only one clock per twin.

Climbing 3 feet implies an upward movement of 3 feet (cumulative, not necessarily contiguous or constant), and slipping 2 feet implies a downward movement of 2 feet. It makes no sense to consider these as simultaneous. Ascending 1 foot while slipping is still a climb of 1 foot; why would anyone say the monkey slipped 2 feet if it never lowered in height?, what would the 2 feet mean? Edit: I think I opened a can of worms bringing up interpretations. Sorry. You could say "The monkey climbed 3 feet while its hands/feet slipped (slid downward) 2 feet (across the surface of the pole)" and that makes sense, but it's not what the puzzle says.

It seems to be three vs three in this thread. I'll wait before replying to your counter argument, hoping that an expert will weigh in, because I don't think I have any credibility left. What do you make of this line from the conclusion in your link: "It is important to point out, however, that appealing to General Relativity is not necessary to resolve the paradox, as demonstrated above."?

I agree with this answer and a very literal (mis?)interpretation of the puzzle: The monkey is climbing 3 feet "every" second, it doesn't say it stops climbing for alternating seconds (though that's probably what was meant). It also doesn't imply constant velocity. Reading the puzzle to mean that the monkey climbs 3 feet sometime over every second, and slips 2 feet sometime over every second except the first, it could make it in as little as 26 seconds if it completes the final 3 feet before any slipping in that second, otherwise as long as 28 seconds if it slips before the climb in the final seconds.

Are you sure about that? The "paradox" of the twin paradox is simply time dilation. Time dilation occurs between any 2 relatively moving clocks, even ones that are each inertial. It's a reciprocal relationship, so they (or anyone) can't agree that one actually ticks slower than the other unless they have a way to universally establish it (enter the same inertial frame and compare at agreed upon events, or come together at a single event) which the twin paradox experiment accomplishes using a turn-around, with acceleration. Two clocks on 2 different paths between the same 2 events will generally measure a difference in proper time. Both paths can't be inertial if they're different, in flat spacetime. ALL I'M SAYING is that a single clock measures the same proper time along a given path, as would 2 clocks with the path split into 2 parts with each part measured separately by one of the two clocks. This does not disagree with SR, in fact SR assumes that it is true! Any two clocks, side-by-side in an inertial frame, will both tick at one second per second. Contrary to xyzt's belief, no clock jumps in proper time. Your "obvious and well established fact" is neither obvious nor established nor a fact, but if you have a reference to back it up I'll check it out... I could be wrong.

Oh brother. I HAVE searched for it. I can't find it. I've been told that it's "hidden". So someone show it to me! But not with nonsense, hand-waving, or even with math if it backs up what I'm claiming. But Iggy's right, opinion---anybody's---counts for nothing, it has to be backed up.

In response to "You're right, the experiment would be equivalent to a corresponding twin paradox even with different speeds.": In the experiment I have B and C traveling at the same speed relative to A, only for the sake of simplicity, but it would work at any speeds. A traveling astronaut would age less by a predictable amount even if the return trip was faster or slower. (Note: In post #1 I say B "passes clock C traveling in the opposite direction at the same speed" but I don't say for whom it is the "same". In the twin paradox it's A, and that's probably clear because if it was interpreted to mean the same speed according to B, any speed at all suffices! It would just have to be fast enough that A and C eventually meet: v_BC would have to be greater than v_AB... if it was interpreted to mean "B passes C at the same relative speed that it passed A" then A and C would be at rest. But as long as A and C meet, the twin paradox effect is still demonstrated. But actually on second thought nevermind this note!, by saying B is traveling at v it is clearly referring to the frame where B is moving, that is A's. ) Clocks don't skip. What is the theoretical basis for predicting such a thing??? Is there a skip in the A clock at the AB event? Observer A resets its clock at the AB event, does that bother you? In the experiment that I described, I have 3 inertial clocks each passing the other 2 clocks in 2 respective events. Why do you think that event BC is special?

I can't change it but I prefer my previously suggested: "Acceleration is not required to physically measure the time dilation effect demonstrated in the twin paradox."

I admit the claim in the title is false. You're right, the experiment would be equivalent to a corresponding twin paradox even with different speeds.

The experiment in #201 is exactly the same as the experiment in #1, with some additional unimportant details given in each. The clock C in #1 is set at BC, but setting a clock does not affect proper time. C still ages 2 years at a rate of one year per yer whether or not it sets its clock midway. Setting a clock doesn't make it jump in age.

Everything you quoted looks good to me. As mentioned before the thread title could be replaced. It would be useful to include a statement that the equivalence assumes the clock postulate.

Yyyyyyyyup. I've modified my experiment to suit your counter-arguments, without changing the relevant conditions or the outcome. There is no discontinuity of time for clock C at event BC. There is no discontinuity in proper time according to any clock in the experiment. There is no discontinuity of proper time experienced by any clock ever, within the domain of SR.

Not according to SR. Using A's frame of reference and the original values, A ages 4 years between AB and AC, while B ages 2 years, and C ages 2 years. The B and C times are coordinate times, or a proper time for B between AB and B_end where B_end is simultaneous with AC according to A, and a proper time for C between C_start and AC where C_start is simultaneous with AB according to A. Both B and C age at the same rate according to A, because v and -v give the same value of gamma. B and C meet halfway through the experiment according to A (or B or C) using these newly defined events. Thus B and C age 1 year---proper time---before meeting, and each ages 1 year after (using these events). In the original post I was ignoring as irrelevant the aging of C before BC, and the aging of B after, however these can be calculated consistently with everything else. It is irrelevant because all I need to know is the proper time for B between AB and BC, and the proper time for C between BC and AC, which can be compared to the proper time for A between AB and AC. Each describes a path from AB to AC, which can meaningfully be compared. No hidden acceleration.

Post #185 agrees with the twin paradox, and the experiment described in post #1. There is no mention of acceleration in post #185.

Post #73 defines a proper time at A according to F_C, which is different from the times at A used in more recent posts. The equations of #73 might be correct equations for the wrong events... events which do not correspond to events described in a twin paradox nor the experiment described in this thread. This is the incorrect part. This calculation would work for tau_B, as in post #185 (which is what I considered "corrected"). Or it would work for tau_A if d was a rest length in F_C's frame. Right!

Alright I think that with your corrected equations we all might be coming to agreement? I'll incorporate info from the 189 replies, and change the claim made in the title of the thread. As-is it doesn't make sense, because the twin paradox by name involves just 2 clocks, and one clock must not be inertial thus acceleration is important. So instead, a better title might be: The time dilation measured in the twin paradox, in the case where the traveling twin's path can be separated into several independent inertial sections, can also be measured using a set of independent inertial clocks, one for each section, assuming that the clock postulate holds. Or... to stick to the point: Acceleration is not required to physically measure the time dilation effect demonstrated in the twin paradox.

But 5 > 3.75/6, so you're saying that according to A, B's clock ticks faster? Nevermind, I can't read... Okay, I think I agree with this math. Earlier the equation was [math]\tau_A=\frac{d \sqrt{1-(v/c)^2}}{v}[/math], incorrectly calculated for A instead of B, which is what I disagreed with.

I can see what you're doing here, and I agree. I can't see what xyzt is doing, by always referring to proper times without defining them, or explaining what events they span. The formula d/v *sqrt(1-(v/c)^2) calculates the proper time of a muon passing through events "enter atmosphere" and "hit Earth" where d is the (rest) distance between those two events. Setting t=d/v, then T_u = t*sqrt(1-(v/c)^2). From the muon's perspective, setting t to the local time that the muon experiences passing through those two events, then the proper time on Earth of two events that are simultaneous ACCORDING TO THE MUON is also T_A = t*sqrt(1-(v/c)^2) using a different value of t than before. However this is fairly meaningless. Time t is not equal to d/v, because the distance (height of atmosphere) is length contracted according to the muon. At very high v, the height of the atmosphere is small for a muon, and the proper time of crossing it is small, and the time that elapses on Earth is tiny, and on Earth the tiny proper time between those events is meaningless, because the events don't describe anything relevant to the Earth observer. They're events on Earth that are simultaneous according to another observer (the muon) with the muon entering the atmosphere, and it hitting the Earth, respectively---the latter event can be agreed on by all. I think that xyzt's confusion is in calculating "proper times" using remote events, thus not bothering to define events using any specific simultaneity (assuming any will do?), thus calculating different "proper times" depending on whose clock is used for the calculations, and THEN ignoring the problems of simultaneity by saying "proper time is invariant", as if the proper time at A between any two events must be the same as the proper time at A between any other 2 events, which is wrong. However, I can only guess that that is the error, because no details, no specific events are given. The events are always implied and they keep changing in xyzt's counter examples.

Equations are meaningless if improperly applied.

I thought that d was the distance that C travels, according to A. In post #164 you use d "as measured by the Earth (A)". If C is moving at v relative to A, it will close the distance d in a time of d/v, ie. the proper time that elapses at A while (according to A) C closes that distance. (Edited for clarity) Using an example with numbers would clarify what you mean (maybe let you see where your equations are not working out)...

Okay. And you've done that for the proper time C measures between an event at C that is simultaneous with AB according to A (I think) and should have found that T_C = T_A / gamma. You've also done it for the proper time C measures between an event at C that is simultaneous with AB according to F, and you found T_C = T_A. No problem there. And we've done that for the proper time C measures between BC and AC, and also for the proper time B measures between AB and BC, which are the identical calculations done in a corresponding twin paradox experiment. No problem. These are all okay because they're different proper times measured between different events.