A constant speed of light in all reference frames? Surely you can't be serious.
in [Relativity]
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From: mpalenik on 4 Mar 2010 13:07 On Mar 4, 1:06 pm, Ste <ste_ro...(a)hotmail.com> wrote: > On 4 Mar, 17:49, mpalenik <markpale...(a)gmail.com> wrote: > > > > > On Mar 4, 12:46 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > On Mar 4, 11:17 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > On 4 Mar, 16:49, mpalenik <markpale...(a)gmail.com> wrote: > > > > > > On Mar 4, 11:45 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > > > On 4 Mar, 16:32, mpalenik <markpale...(a)gmail.com> wrote: > > > > > > > > On Mar 4, 11:28 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > > > > > On 4 Mar, 16:20, mpalenik <markpale...(a)gmail.com> wrote: > > > > > > > > > > On Mar 4, 10:31 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > > > > > > > On 4 Mar, 13:40, mpalenik <markpale...(a)gmail.com> wrote: > > > > > > > > > > > > On Mar 4, 3:12 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > > > > > > > > > On 3 Mar, 20:01, mpalenik <markpale...(a)gmail.com> wrote: > > > > > > > > > > > > > > On Mar 3, 12:52 pm, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > > > > > > > > > > > No. In SR, clocks *appear* to run slower as you are increasing your > > > > > > > > > > > > > > distance from the clock. The effect is entirely apparent in SR. > > > > > > > > > > > > > > You must just go through the entire thread and not pay any attention > > > > > > > > > > > > > to what anybody says. Ever. > > > > > > > > > > > > > > 1) What you've stated above is not an effect of SR. It is an effect > > > > > > > > > > > > > of propagation delay, which was used to calculate c from the motion of > > > > > > > > > > > > > the moons of jupiter hundreds of years ago. > > > > > > > > > > > > > Ok. > > > > > > > > > > > > > > 2) If you were to move TOWARD the clock, it would appear to run > > > > > > > > > > > > > faster. But SR says nothing about whether you are moving toward or > > > > > > > > > > > > > away from an object. > > > > > > > > > > > > > <suspicious eyebrow raised> Ok. > > > > > > > > > > > > > > 3) The amount that the clock would appear to slow down is DIFFERENT > > > > > > > > > > > > > from the amount that SR predicts the clock *actually* slows down > > > > > > > > > > > > > Really? I'm growing increasingly suspicious. In what way does SR > > > > > > > > > > > > predict the "actual" slowdown, as opposed to the "apparent" slowdown? > > > > > > > > > > > > And for example, if we racked up the value of 'c' to near infinity, > > > > > > > > > > > > would SR still predict an "actual" slowdown, even though the > > > > > > > > > > > > propagation delays would approach zero? > > > > > > > > > > > > With what you have described, I checked just to be sure, even though I > > > > > > > > > > > was already pretty sure what the answer would be, the time you read > > > > > > > > > > > moving away the clock would be: > > > > > > > > > > > > t2 = t - (x+vt)/c = t(1-v/c) - x > > > > > > > > > > > > and when you move toward the clock > > > > > > > > > > > > t2 = t + (x+vt)/c = t(1+v/c) + x > > > > > > > > > > > > so moving away from the clock: > > > > > > > > > > > dt2/dt = 1-v/c > > > > > > > > > > > and toward > > > > > > > > > > > dt2/dt = 1-v/c > > > > > > > > > > > > Special relativity predicts that the moving clock will always slow > > > > > > > > > > > down as > > > > > > > > > > > dt2/dt = sqrt(1-v^2/c^2) > > > > > > > > > > > > What you *measure* is a combination of the actual slow down predicted > > > > > > > > > > > by SR (sqrt(1-v^2/c^2) and whatever changes occur due to propagation > > > > > > > > > > > delays (which depend on the direction of motion). > > > > > > > > > > > Ok. So let us suppose that we take two clocks. Separate them by a > > > > > > > > > > certain distance, synchronise them when they are both stationary, and > > > > > > > > > > then accelerate them both towards each other (and just before they > > > > > > > > > > collide, we bring them stationary again). Are you seriously saying > > > > > > > > > > that both clocks report that the other clock has slowed down, even > > > > > > > > > > though they have both undergone symmetrical processes? Because there > > > > > > > > > > is obviously a contradiction there. > > > > > > > > > > Yes, that is correct. Both will report a slow down. And in fact, > > > > > > > > > which ever one breaks the inertial frame to match speed with the other > > > > > > > > > is the one that will be "wrong". This is still within the realm of > > > > > > > > > SR, not GR. > > > > > > > > > What if they both "break the inertial frame"? > > > > > > > > Then whichever frame they both accelerate into will be the one that > > > > > > > has measured the "correct" time dilation. > > > > > > > So in other words, the clocks will register the same time, but will > > > > > > have slowed in some "absolute sense"? > > > > > > Yes--assuming they both accelerated by the same amount (that is to > > > > > say, assuming they both broke the inertial frame in a symmetric way). > > > > > Otherwise, they will register different times. > > > > > Agreed. > > > > > So let's explore an extension of this scenario. Let's say you have two > > > > clocks, and you accelerate both of them up to a common speed, and > > > > after they have travelled a certain distance, you turn them around and > > > > return them to the starting point. The only difference is that one > > > > clock goes a certain distance, and the other clock goes twice that > > > > distance, but they *both* have the same acceleration profile - the > > > > only difference is that one clock spends more time travelling on > > > > inertia. > > > > > Obviously, one clock will return to the starting point earlier than > > > > the other. But when both have returned, are their times still in > > > > agreement with each other, or have they changed? > > > > Agreement. Both of them will agree, but will be showing a time earlier > > > than a third clock that was left behind at the starting point. > > > Wait, maybe I'm confused by Ste's setup. Didn't he say that one > > travels twice as far as the other? But then he also says that you > > turn them both around and return them to the start after traveling a > > certain distance. Have they moved different distances in his scenario > > or not? > > One has travelled twice as far as the other, but both at the same > speed, with the effect that the clock that has travelled the shorter > distance returns to the start point before the other clock. In that case, what I said is correct. The two clocks have the same acceleration profile They both accelerate up to speed v One clock travels at speed v twice as long as the other when returned to rest, the two clocks will display different times.
From: mpalenik on 4 Mar 2010 13:10 On Mar 4, 1:03 pm, JT <jonas.thornv...(a)hotmail.com> wrote: > On 4 mar, 18:49, mpalenik <markpale...(a)gmail.com> wrote: > > > > > On Mar 4, 12:46 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > On Mar 4, 11:17 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > On 4 Mar, 16:49, mpalenik <markpale...(a)gmail.com> wrote: > > > > > > On Mar 4, 11:45 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > > > On 4 Mar, 16:32, mpalenik <markpale...(a)gmail.com> wrote: > > > > > > > > On Mar 4, 11:28 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > > > > > On 4 Mar, 16:20, mpalenik <markpale...(a)gmail.com> wrote: > > > > > > > > > > On Mar 4, 10:31 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > > > > > > > On 4 Mar, 13:40, mpalenik <markpale...(a)gmail.com> wrote: > > > > > > > > > > > > On Mar 4, 3:12 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > > > > > > > > > On 3 Mar, 20:01, mpalenik <markpale...(a)gmail.com> wrote: > > > > > > > > > > > > > > On Mar 3, 12:52 pm, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > > > > > > > > > > > No. In SR, clocks *appear* to run slower as you are increasing your > > > > > > > > > > > > > > distance from the clock. The effect is entirely apparent in SR. > > > > > > > > > > > > > > You must just go through the entire thread and not pay any attention > > > > > > > > > > > > > to what anybody says. Ever. > > > > > > > > > > > > > > 1) What you've stated above is not an effect of SR. It is an effect > > > > > > > > > > > > > of propagation delay, which was used to calculate c from the motion of > > > > > > > > > > > > > the moons of jupiter hundreds of years ago. > > > > > > > > > > > > > Ok. > > > > > > > > > > > > > > 2) If you were to move TOWARD the clock, it would appear to run > > > > > > > > > > > > > faster. But SR says nothing about whether you are moving toward or > > > > > > > > > > > > > away from an object. > > > > > > > > > > > > > <suspicious eyebrow raised> Ok. > > > > > > > > > > > > > > 3) The amount that the clock would appear to slow down is DIFFERENT > > > > > > > > > > > > > from the amount that SR predicts the clock *actually* slows down > > > > > > > > > > > > > Really? I'm growing increasingly suspicious. In what way does SR > > > > > > > > > > > > predict the "actual" slowdown, as opposed to the "apparent" slowdown? > > > > > > > > > > > > And for example, if we racked up the value of 'c' to near infinity, > > > > > > > > > > > > would SR still predict an "actual" slowdown, even though the > > > > > > > > > > > > propagation delays would approach zero? > > > > > > > > > > > > With what you have described, I checked just to be sure, even though I > > > > > > > > > > > was already pretty sure what the answer would be, the time you read > > > > > > > > > > > moving away the clock would be: > > > > > > > > > > > > t2 = t - (x+vt)/c = t(1-v/c) - x > > > > > > > > > > > > and when you move toward the clock > > > > > > > > > > > > t2 = t + (x+vt)/c = t(1+v/c) + x > > > > > > > > > > > > so moving away from the clock: > > > > > > > > > > > dt2/dt = 1-v/c > > > > > > > > > > > and toward > > > > > > > > > > > dt2/dt = 1-v/c > > > > > > > > > > > > Special relativity predicts that the moving clock will always slow > > > > > > > > > > > down as > > > > > > > > > > > dt2/dt = sqrt(1-v^2/c^2) > > > > > > > > > > > > What you *measure* is a combination of the actual slow down predicted > > > > > > > > > > > by SR (sqrt(1-v^2/c^2) and whatever changes occur due to propagation > > > > > > > > > > > delays (which depend on the direction of motion). > > > > > > > > > > > Ok. So let us suppose that we take two clocks. Separate them by a > > > > > > > > > > certain distance, synchronise them when they are both stationary, and > > > > > > > > > > then accelerate them both towards each other (and just before they > > > > > > > > > > collide, we bring them stationary again). Are you seriously saying > > > > > > > > > > that both clocks report that the other clock has slowed down, even > > > > > > > > > > though they have both undergone symmetrical processes? Because there > > > > > > > > > > is obviously a contradiction there. > > > > > > > > > > Yes, that is correct. Both will report a slow down. And in fact, > > > > > > > > > which ever one breaks the inertial frame to match speed with the other > > > > > > > > > is the one that will be "wrong". This is still within the realm of > > > > > > > > > SR, not GR. > > > > > > > > > What if they both "break the inertial frame"? > > > > > > > > Then whichever frame they both accelerate into will be the one that > > > > > > > has measured the "correct" time dilation. > > > > > > > So in other words, the clocks will register the same time, but will > > > > > > have slowed in some "absolute sense"? > > > > > > Yes--assuming they both accelerated by the same amount (that is to > > > > > say, assuming they both broke the inertial frame in a symmetric way). > > > > > Otherwise, they will register different times. > > > > > Agreed. > > > > > So let's explore an extension of this scenario. Let's say you have two > > > > clocks, and you accelerate both of them up to a common speed, and > > > > after they have travelled a certain distance, you turn them around and > > > > return them to the starting point. The only difference is that one > > > > clock goes a certain distance, and the other clock goes twice that > > > > distance, but they *both* have the same acceleration profile - the > > > > only difference is that one clock spends more time travelling on > > > > inertia. > > > > > Obviously, one clock will return to the starting point earlier than > > > > the other. But when both have returned, are their times still in > > > > agreement with each other, or have they changed? > > > > Agreement. Both of them will agree, but will be showing a time earlier > > > than a third clock that was left behind at the starting point. > > > Wait, maybe I'm confused by Ste's setup. Didn't he say that one > > travels twice as far as the other? But then he also says that you > > turn them both around and return them to the start after traveling a > > certain distance. Have they moved different distances in his scenario > > or not?- Dölj citerad text - > > > - Visa citerad text - > > lol you framejumping grasshoppers have just have no idea what is > ***REALLY*** going on have you. Don't forget u can always use the > fudgefactor. > > JT You figured me out! Damn it. I guess the days of the lie are over. Pretty soon the physicists absolute control over government, politics, and economics will come to an end. Damn you for uncovering our secret. Damn you all to hell!
From: PD on 4 Mar 2010 13:12 On Mar 4, 11:49 am, mpalenik <markpale...(a)gmail.com> wrote: > On Mar 4, 12:46 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > > On Mar 4, 11:17 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > On 4 Mar, 16:49, mpalenik <markpale...(a)gmail.com> wrote: > > > > > On Mar 4, 11:45 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > > On 4 Mar, 16:32, mpalenik <markpale...(a)gmail.com> wrote: > > > > > > > On Mar 4, 11:28 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > > > > On 4 Mar, 16:20, mpalenik <markpale...(a)gmail.com> wrote: > > > > > > > > > On Mar 4, 10:31 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > > > > > > On 4 Mar, 13:40, mpalenik <markpale...(a)gmail.com> wrote: > > > > > > > > > > > On Mar 4, 3:12 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > > > > > > > > On 3 Mar, 20:01, mpalenik <markpale...(a)gmail.com> wrote: > > > > > > > > > > > > > On Mar 3, 12:52 pm, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > > > > > > > > > > No. In SR, clocks *appear* to run slower as you are increasing your > > > > > > > > > > > > > distance from the clock. The effect is entirely apparent in SR. > > > > > > > > > > > > > You must just go through the entire thread and not pay any attention > > > > > > > > > > > > to what anybody says. Ever. > > > > > > > > > > > > > 1) What you've stated above is not an effect of SR. It is an effect > > > > > > > > > > > > of propagation delay, which was used to calculate c from the motion of > > > > > > > > > > > > the moons of jupiter hundreds of years ago. > > > > > > > > > > > > Ok. > > > > > > > > > > > > > 2) If you were to move TOWARD the clock, it would appear to run > > > > > > > > > > > > faster. But SR says nothing about whether you are moving toward or > > > > > > > > > > > > away from an object. > > > > > > > > > > > > <suspicious eyebrow raised> Ok. > > > > > > > > > > > > > 3) The amount that the clock would appear to slow down is DIFFERENT > > > > > > > > > > > > from the amount that SR predicts the clock *actually* slows down > > > > > > > > > > > > Really? I'm growing increasingly suspicious. In what way does SR > > > > > > > > > > > predict the "actual" slowdown, as opposed to the "apparent" slowdown? > > > > > > > > > > > And for example, if we racked up the value of 'c' to near infinity, > > > > > > > > > > > would SR still predict an "actual" slowdown, even though the > > > > > > > > > > > propagation delays would approach zero? > > > > > > > > > > > With what you have described, I checked just to be sure, even though I > > > > > > > > > > was already pretty sure what the answer would be, the time you read > > > > > > > > > > moving away the clock would be: > > > > > > > > > > > t2 = t - (x+vt)/c = t(1-v/c) - x > > > > > > > > > > > and when you move toward the clock > > > > > > > > > > > t2 = t + (x+vt)/c = t(1+v/c) + x > > > > > > > > > > > so moving away from the clock: > > > > > > > > > > dt2/dt = 1-v/c > > > > > > > > > > and toward > > > > > > > > > > dt2/dt = 1-v/c > > > > > > > > > > > Special relativity predicts that the moving clock will always slow > > > > > > > > > > down as > > > > > > > > > > dt2/dt = sqrt(1-v^2/c^2) > > > > > > > > > > > What you *measure* is a combination of the actual slow down predicted > > > > > > > > > > by SR (sqrt(1-v^2/c^2) and whatever changes occur due to propagation > > > > > > > > > > delays (which depend on the direction of motion). > > > > > > > > > > Ok. So let us suppose that we take two clocks. Separate them by a > > > > > > > > > certain distance, synchronise them when they are both stationary, and > > > > > > > > > then accelerate them both towards each other (and just before they > > > > > > > > > collide, we bring them stationary again). Are you seriously saying > > > > > > > > > that both clocks report that the other clock has slowed down, even > > > > > > > > > though they have both undergone symmetrical processes? Because there > > > > > > > > > is obviously a contradiction there. > > > > > > > > > Yes, that is correct. Both will report a slow down. And in fact, > > > > > > > > which ever one breaks the inertial frame to match speed with the other > > > > > > > > is the one that will be "wrong". This is still within the realm of > > > > > > > > SR, not GR. > > > > > > > > What if they both "break the inertial frame"? > > > > > > > Then whichever frame they both accelerate into will be the one that > > > > > > has measured the "correct" time dilation. > > > > > > So in other words, the clocks will register the same time, but will > > > > > have slowed in some "absolute sense"? > > > > > Yes--assuming they both accelerated by the same amount (that is to > > > > say, assuming they both broke the inertial frame in a symmetric way). > > > > Otherwise, they will register different times. > > > > Agreed. > > > > So let's explore an extension of this scenario. Let's say you have two > > > clocks, and you accelerate both of them up to a common speed, and > > > after they have travelled a certain distance, you turn them around and > > > return them to the starting point. The only difference is that one > > > clock goes a certain distance, and the other clock goes twice that > > > distance, but they *both* have the same acceleration profile - the > > > only difference is that one clock spends more time travelling on > > > inertia. > > > > Obviously, one clock will return to the starting point earlier than > > > the other. But when both have returned, are their times still in > > > agreement with each other, or have they changed? > > > Agreement. Both of them will agree, but will be showing a time earlier > > than a third clock that was left behind at the starting point. > > Wait, maybe I'm confused by Ste's setup. Didn't he say that one > travels twice as far as the other? But then he also says that you > turn them both around and return them to the start after traveling a > certain distance. Have they moved different distances in his scenario > or not? My understanding is that both clocks accelerate up to speed v, but one goes for x kilometers and the other goes for 2x kilometers before turning around and returning at speed v. Naturally, one arrives back in half the time it takes the other one. Oh, I see the issue. When the first clock comes back, it has to stop, and therefore its rate will be different than when traveling. Yes, I agree now, the two clocks will read different times. PD
From: PD on 4 Mar 2010 13:12 On Mar 4, 12:04 pm, Ste <ste_ro...(a)hotmail.com> wrote: > On 4 Mar, 17:46, PD <thedraperfam...(a)gmail.com> wrote: > > > > > On Mar 4, 11:17 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > On 4 Mar, 16:49, mpalenik <markpale...(a)gmail.com> wrote: > > > > > On Mar 4, 11:45 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > > On 4 Mar, 16:32, mpalenik <markpale...(a)gmail.com> wrote: > > > > > > > On Mar 4, 11:28 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > > > > On 4 Mar, 16:20, mpalenik <markpale...(a)gmail.com> wrote: > > > > > > > > > On Mar 4, 10:31 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > > > > > > On 4 Mar, 13:40, mpalenik <markpale...(a)gmail.com> wrote: > > > > > > > > > > > On Mar 4, 3:12 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > > > > > > > > On 3 Mar, 20:01, mpalenik <markpale...(a)gmail.com> wrote: > > > > > > > > > > > > > On Mar 3, 12:52 pm, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > > > > > > > > > > No. In SR, clocks *appear* to run slower as you are increasing your > > > > > > > > > > > > > distance from the clock. The effect is entirely apparent in SR. > > > > > > > > > > > > > You must just go through the entire thread and not pay any attention > > > > > > > > > > > > to what anybody says. Ever. > > > > > > > > > > > > > 1) What you've stated above is not an effect of SR. It is an effect > > > > > > > > > > > > of propagation delay, which was used to calculate c from the motion of > > > > > > > > > > > > the moons of jupiter hundreds of years ago. > > > > > > > > > > > > Ok. > > > > > > > > > > > > > 2) If you were to move TOWARD the clock, it would appear to run > > > > > > > > > > > > faster. But SR says nothing about whether you are moving toward or > > > > > > > > > > > > away from an object. > > > > > > > > > > > > <suspicious eyebrow raised> Ok. > > > > > > > > > > > > > 3) The amount that the clock would appear to slow down is DIFFERENT > > > > > > > > > > > > from the amount that SR predicts the clock *actually* slows down > > > > > > > > > > > > Really? I'm growing increasingly suspicious. In what way does SR > > > > > > > > > > > predict the "actual" slowdown, as opposed to the "apparent" slowdown? > > > > > > > > > > > And for example, if we racked up the value of 'c' to near infinity, > > > > > > > > > > > would SR still predict an "actual" slowdown, even though the > > > > > > > > > > > propagation delays would approach zero? > > > > > > > > > > > With what you have described, I checked just to be sure, even though I > > > > > > > > > > was already pretty sure what the answer would be, the time you read > > > > > > > > > > moving away the clock would be: > > > > > > > > > > > t2 = t - (x+vt)/c = t(1-v/c) - x > > > > > > > > > > > and when you move toward the clock > > > > > > > > > > > t2 = t + (x+vt)/c = t(1+v/c) + x > > > > > > > > > > > so moving away from the clock: > > > > > > > > > > dt2/dt = 1-v/c > > > > > > > > > > and toward > > > > > > > > > > dt2/dt = 1-v/c > > > > > > > > > > > Special relativity predicts that the moving clock will always slow > > > > > > > > > > down as > > > > > > > > > > dt2/dt = sqrt(1-v^2/c^2) > > > > > > > > > > > What you *measure* is a combination of the actual slow down predicted > > > > > > > > > > by SR (sqrt(1-v^2/c^2) and whatever changes occur due to propagation > > > > > > > > > > delays (which depend on the direction of motion). > > > > > > > > > > Ok. So let us suppose that we take two clocks. Separate them by a > > > > > > > > > certain distance, synchronise them when they are both stationary, and > > > > > > > > > then accelerate them both towards each other (and just before they > > > > > > > > > collide, we bring them stationary again). Are you seriously saying > > > > > > > > > that both clocks report that the other clock has slowed down, even > > > > > > > > > though they have both undergone symmetrical processes? Because there > > > > > > > > > is obviously a contradiction there. > > > > > > > > > Yes, that is correct. Both will report a slow down. And in fact, > > > > > > > > which ever one breaks the inertial frame to match speed with the other > > > > > > > > is the one that will be "wrong". This is still within the realm of > > > > > > > > SR, not GR. > > > > > > > > What if they both "break the inertial frame"? > > > > > > > Then whichever frame they both accelerate into will be the one that > > > > > > has measured the "correct" time dilation. > > > > > > So in other words, the clocks will register the same time, but will > > > > > have slowed in some "absolute sense"? > > > > > Yes--assuming they both accelerated by the same amount (that is to > > > > say, assuming they both broke the inertial frame in a symmetric way). > > > > Otherwise, they will register different times. > > > > Agreed. > > > > So let's explore an extension of this scenario. Let's say you have two > > > clocks, and you accelerate both of them up to a common speed, and > > > after they have travelled a certain distance, you turn them around and > > > return them to the starting point. The only difference is that one > > > clock goes a certain distance, and the other clock goes twice that > > > distance, but they *both* have the same acceleration profile - the > > > only difference is that one clock spends more time travelling on > > > inertia. > > > > Obviously, one clock will return to the starting point earlier than > > > the other. But when both have returned, are their times still in > > > agreement with each other, or have they changed? > > > Agreement. Both of them will agree, but will be showing a time earlier > > than a third clock that was left behind at the starting point. > > Oh dear. Mark contends otherwise. Right. I misunderstood. He's right. I was wrong.
From: Ste on 4 Mar 2010 13:12
On 4 Mar, 17:20, mpalenik <markpale...(a)gmail.com> wrote: > On Mar 4, 12:02 pm, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > > > On 4 Mar, 16:29, mpalenik <markpale...(a)gmail.com> wrote: > > > > On Mar 4, 10:31 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > On 4 Mar, 13:40, mpalenik <markpale...(a)gmail.com> wrote: > > > > > > On Mar 4, 3:12 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > > > On 3 Mar, 20:01, mpalenik <markpale...(a)gmail.com> wrote: > > > > > > > > On Mar 3, 12:52 pm, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > > > > > No. In SR, clocks *appear* to run slower as you are increasing your > > > > > > > > distance from the clock. The effect is entirely apparent in SR. > > > > > > > > You must just go through the entire thread and not pay any attention > > > > > > > to what anybody says. Ever. > > > > > > > > 1) What you've stated above is not an effect of SR. It is an effect > > > > > > > of propagation delay, which was used to calculate c from the motion of > > > > > > > the moons of jupiter hundreds of years ago. > > > > > > > Ok. > > > > > > > > 2) If you were to move TOWARD the clock, it would appear to run > > > > > > > faster. But SR says nothing about whether you are moving toward or > > > > > > > away from an object. > > > > > > > <suspicious eyebrow raised> Ok. > > > > > > > > 3) The amount that the clock would appear to slow down is DIFFERENT > > > > > > > from the amount that SR predicts the clock *actually* slows down > > > > > > > Really? I'm growing increasingly suspicious. In what way does SR > > > > > > predict the "actual" slowdown, as opposed to the "apparent" slowdown? > > > > > > And for example, if we racked up the value of 'c' to near infinity, > > > > > > would SR still predict an "actual" slowdown, even though the > > > > > > propagation delays would approach zero? > > > > > > With what you have described, I checked just to be sure, even though I > > > > > was already pretty sure what the answer would be, the time you read > > > > > moving away the clock would be: > > > > > > t2 = t - (x+vt)/c = t(1-v/c) - x > > > > > > and when you move toward the clock > > > > > > t2 = t + (x+vt)/c = t(1+v/c) + x > > > > > > so moving away from the clock: > > > > > dt2/dt = 1-v/c > > > > > and toward > > > > > dt2/dt = 1-v/c > > > > > > Special relativity predicts that the moving clock will always slow > > > > > down as > > > > > dt2/dt = sqrt(1-v^2/c^2) > > > > > > What you *measure* is a combination of the actual slow down predicted > > > > > by SR (sqrt(1-v^2/c^2) and whatever changes occur due to propagation > > > > > delays (which depend on the direction of motion). > > > > > Ok. So let us suppose that we take two clocks. Separate them by a > > > > certain distance, synchronise them when they are both stationary, and > > > > then accelerate them both towards each other (and just before they > > > > collide, we bring them stationary again). Are you seriously saying > > > > that both clocks report that the other clock has slowed down, even > > > > though they have both undergone symmetrical processes? Because there > > > > is obviously a contradiction there. > > > > I don't know what happened to this thread, or why google registers > > > several different copies of this thread, each with only a few messages > > > in it, but I'm going to describe the thought experiment I posted > > > before in greater detail. > > > > In order for the ^ to receive the pulse of light from both emitters at > > > the same time and half way between the two emitters, the light must be > > > emitted BEFORE the ^ is half way between the two emitters. > > > Indeed. > > > > From the stationary frame, the two emitters emit light at the same > > > time and the ^ moves vertically until it is half way between the two > > > of them, when it receives both pulses. > > > Ok. > > > > But from the moving frame, the ^ sees the emitters emit light when it > > > is not half way between them, and then it sees the emitters move until > > > he is exactly half way between them. However, the speed of light is > > > the same in all frames, and unaffected by the motion of the emitters. > > > > So the the emitter emit while one is closer to him than the other. > > > The fact that they are moving does not affect the way light > > > propagates. If the light signal from the closer one reaches him at > > > the same time as the light signal from the farther one, that means the > > > farther one must have emitted first. The fact that he is half way > > > between the emitters at the end doesn't matter, because in his frame, > > > it is the emitters that are moving, which does not affect the speed of > > > light. > > > This seems implausible. You cannot possibly (that is, physically) have > > the situation you describe, in the way you describe it. The receiver, > > when it receives the pulse of light, cannot possibly be in more than > > one position relative to the two sources. > > It isn't. > > > It either receives when the > > sources are equidistant, or it receives when one source is closer. > > The sources are equidistant. > > > It > > cannot receive when it is equistant *and* when one source is closer, > > for that would be an obvious contradiction. > > What you don't understand is the fact that the sources were not > equidistant when they emitted the pulse. Indeed, but it's neither here nor there where the sources where at the time of emission. The question is where the sources are at the time of reception, and at that time they must be equidistant if both pulses were emitted at the same time. No matter what frame you look at it from, the sources are not equidistant at the time of emission. |