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 11:20 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.
From: Ste on 4 Mar 2010 11:24 On 4 Mar, 15:54, PD <thedraperfam...(a)gmail.com> wrote: > On Mar 4, 1:03 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > This is what fundamentally sets apart things like creationism from > > > > science. Whatever other hoops creationism manages to jump through, it > > > > will never jump through the hoop of naturalism, and that is what > > > > *fundamentally* sets it apart from science. > > > > And also FUNDAMENTALLY distinguishes science from religion. Thanks. > > > Agreed, but then religion in general never claimed to be science, > > Agreed! And so science is not a religion in the same fashion. No, but neither did one religion ever claim to be the other. > > and > > traditional religion is almost immediately identifiable by its > > supernaturalism. Creationism is different in that it actually claims > > to be scientific in some essential respects. > > Ah, yes, but as has been demonstrated even to layfolk (Dover v > Kitsmiller), this is an unsupportable claim. I agree. I'm glad you brought up that case. I just reviewed the judgment quickly, and apparently the court agrees that the defining essence of science is naturalism.
From: Ste on 4 Mar 2010 11:28 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"?
From: mpalenik on 4 Mar 2010 11:29 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. 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. 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.
From: mpalenik on 4 Mar 2010 11:32
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. |