Actually, because the planes fly at equivalent speeds WRT the Earth, which is a rotating frame of reference, when they get back to the geographical starting place (which has moved), they arrive at the same "local" time and according to Hafele's experimentally obtained data the clocks do not agree.
--- "Stephen A. Lawrence" <[EMAIL PROTECTED]> wrote: > > > Grimer wrote: > > At 03:50 pm 19/01/2006 -0900, Horace wrote: > >>This has not made it back to me for over 4 hours > so here goes again: > >> > >>On Jan 19, 2006, at 8:11 AM, Stephen A. Lawrence > wrote: > >> > >>><someone wrote> > >>>The speed of the flights is not a factor, either > -- the same time > >>>lag will be observed no matter how fast they go. > However, in order > >>>to keep the precision with which one needs to > keep time down to > >>>something manageable, it's important to go > quickly. If you used a > >>>ship and retraced Magellan's route instead of > using an airplane, > >>>for instance, the tiny difference in the readings > would be totally > >>>lost in the accumulated inaccuracy of the clocks > over a period of > >>>several months. > >>> > >>Interesting about the speed independence. > > > > I think one has to be careful what one means by > > speed independence here. > > Here's what we mean by that: > > Consider a rotating disk. Select a point on the > perimeter. > > Send two signals around the disk, starting from that > point, > circumnavigating the disk, and returning to that > point (which has, of > course, moved by the time the signals get back to > it). Make sure the > two signals travel at the same speed relative to the > rim of the disk. > > The signal which went around in the same direction > as the disk's > rotation will arrive back at the start _after_ the > signal which went the > other way around. The difference in the arrival > times is a function of > the rotation rate of the disk, but it is _not_ a > function of the speed > of the signal. Fast signal, slow signal, the > absolute delay between the > return of the signal on the "fast" path and the > return of the signal on > the "slow" path is the same. > > As I mentioned previously, this can be demonstrated > without the use of > any clocks, and in fact it is demonstrated all the > time. Current > generation inertial navigation systems use > ring-laser gyroscopes which > only work as a result of this effect. In a > ring-laser gyro the signal > is a a light pulse carried in a fiber optic cable, > and it travels at > roughly 3/4 C relative to the rim of the disk. The > signal speed is the > same in both directions, relative to the disk > (signal speed on a moving > body is trivial to measure, and if it weren't > invariant with respect to > the motion, moving computers would not work). The > arrival time > difference is measured by looking at interference > fringe shifts between > the counter-traversing pulses, and it's used to > determine the rate at > which the disk is turning, which datum is used by > the navigation system. > > It's sometimes claimed that the Sagnac effect is > difficult to explain in > special relativity, or that the math is a horrible > mess. That's not > true. The effect is actually pretty simple; in fact > it can be explained > in a few pictures without a (whole) lot of messy > math. See here: > > http://physicsinsights.org/sagnac_1.html > > In a nutshell, the rotation doesn't make a > difference; straighten out > the path so it's just a long straight rod that's > being traversed, and it > becomes a lot more obvious what's going on. > > > > > > In it's rotation the earth (and clocks on its > > surface) is moving in relation to the > Beta-atmosphere > > which reduces the speed of the caesium clock. > > If you go towards the setting sun then it is not > > that the clock will speed up. It is that the slow > > running will be reduced to a minumum when the > speed > > is stationary in relation to the local B-atm. > > Going round towards the rising sun slow running > will > > be increased. > > > > But the difference in speed between planes and > ships > > is small compared to light speed. If one projected > > > a caesium clock at close to the speed of light > > relative to the absolute frame of reference for > > motion then its speed would slow right down since > > mass is the reciprocal of internal closed path > > velocity (see IHM note on Beta-atm.Yahoo site). > > > > The fact that the caesium clocks rate can be > altered > > merely by flying it around the globe shows the > utter > > insanity of using it to define length. If you do, > then > > you end up with the ludicrous result that the > distance > > around the globe clockwise is different from that > around > > the globe widdershins. > > Ring-laser gyros make hardly any sense, it's true. > You're right. > However, they exist and they work. All of special > relativity has this > problem: Intuitively it's absurd. But it's born > out by an enormous > mass of experimental data. > > But there's a point you may have missed in the > "airplane" experiment. > The two aircraft don't arrive back at the starting > point at the same > moment. According to each airplane's onboard clock, > the time to go > around the world was the same -- that doesn't depend > on the direction! > And so neither does the distance the airplane > traveled. What changes is > how long it takes in Earth-minutes for the planes to > go around the world. > > At the point at which the planes meet -- which is > _NOT_ the starting > point, because they got back to the start at > different times -- they > really have traveled different distances, and their > clocks really do > show different readings. There's no contradiction > and little surprise > in that. The odd thing is that the don't get back > to the starting point > at the same time. > > > > Frank Grimer Merlyn Magickal Engineer and Technical Metaphysicist __________________________________________________ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com
