Known physics defeated by simple puzzle?
in [Relativity]
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From: Ste on 15 Feb 2010 14:01 On 15 Feb, 18:27, mpalenik <markpale...(a)gmail.com> wrote: > On Feb 15, 1:06 pm, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > > > On 15 Feb, 16:53, PD <thedraperfam...(a)gmail.com> wrote: > > > > On Feb 15, 10:05 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > On 15 Feb, 15:18, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > On Feb 14, 1:42 pm, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > > > No takers for this simple question then? > > > > > > > Consider this setup: > > > > > > > S1 D2 > > > > > > > D1 S2 > > > > > > > We've got sources S1 and S2, paired with detectors D1 and D2. They're > > > > > > all mechanically connected, so that a movement in one of them > > > > > > produces > > > > > > a movement in all the others - in other words, their relative > > > > > > distances are always maintained. Each source is transmitting a > > > > > > regular > > > > > > pulse of light to its counterpart detector (so S1 is transmitting to > > > > > > D1, etc.), and both sources are transmitting simultaneously with each > > > > > > other. > > > > > > > Now, we calculate that a pulse has just been emitted from both > > > > > > sources, and we suddenly accelerate the whole setup "upwards" (i.e. > > > > > > relative to how it's oriented on the page now) to near the speed of > > > > > > light, and we complete this acceleration before the signals reach > > > > > > either detector. > > > > > > > Now, do both detectors *still* receive their signals simultaneously, > > > > > > or does one receive its signal before the other? And are the signals > > > > > > identical, or do they suffer from Doppler shifting, etc? > > > > > > In what frame do you want the answer? > > > > > The frame into which the set-up is accelerating, or the frame from > > > > > which the set-up is accelerating? > > > > > The answer is dependent on that choice. > > > > > The answer will be according to an observer that is attached to the > > > > setup and is equidistant from both detectors. And remember, the > > > > acceleration is complete before the detection occurs. > > > > OK, first of all, this is not an inertial reference frame, because it > > > is accelerating. > > > This is the reason why I gave you two choices of inertial reference > > > frame, neither of which is the noninertial reference frame you've > > > chosen. > > > I can certainly give you an answer for the noninertial reference frame > > > you've chosen: they do not arrive at the same time, and yes, they are > > > frequency/wavelength shifted. > > > Actually Paul, I just want to add another clause to this. What happens > > if, still before the detection occurs, the whole setup decelerates > > back to the same velocity as it started with. So the whole setup has > > moved, but at the time of the detection, the setup is not moving > > relative to its original frame. I assume that they still arrive at > > different times, but does the Doppler shifting remain?- Hide quoted text - > > If you bring it back to the original reference frame--that is, back to > the original velocity that it was moving at when the emitters emitted, > before the dectors detect the pulse, there will be no doppler shift > and both detectors will recieve the pulses simultaneously. I find that somewhat hard to believe (which is not to say I necessarily disbelieve it). So you're saying that the time it takes for a photon to move between source and detector, depends only on the relative distance between the source and detector, and not on any movement from a notional "start point"?
From: PD on 15 Feb 2010 14:20 On Feb 15, 12:52 pm, PD <thedraperfam...(a)gmail.com> wrote: > On Feb 15, 12:06 pm, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > On 15 Feb, 16:53, PD <thedraperfam...(a)gmail.com> wrote: > > > > On Feb 15, 10:05 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > On 15 Feb, 15:18, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > On Feb 14, 1:42 pm, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > > > No takers for this simple question then? > > > > > > > Consider this setup: > > > > > > > S1 D2 > > > > > > > D1 S2 > > > > > > > We've got sources S1 and S2, paired with detectors D1 and D2. They're > > > > > > all mechanically connected, so that a movement in one of them > > > > > > produces > > > > > > a movement in all the others - in other words, their relative > > > > > > distances are always maintained. Each source is transmitting a > > > > > > regular > > > > > > pulse of light to its counterpart detector (so S1 is transmitting to > > > > > > D1, etc.), and both sources are transmitting simultaneously with each > > > > > > other. > > > > > > > Now, we calculate that a pulse has just been emitted from both > > > > > > sources, and we suddenly accelerate the whole setup "upwards" (i.e. > > > > > > relative to how it's oriented on the page now) to near the speed of > > > > > > light, and we complete this acceleration before the signals reach > > > > > > either detector. > > > > > > > Now, do both detectors *still* receive their signals simultaneously, > > > > > > or does one receive its signal before the other? And are the signals > > > > > > identical, or do they suffer from Doppler shifting, etc? > > > > > > In what frame do you want the answer? > > > > > The frame into which the set-up is accelerating, or the frame from > > > > > which the set-up is accelerating? > > > > > The answer is dependent on that choice. > > > > > The answer will be according to an observer that is attached to the > > > > setup and is equidistant from both detectors. And remember, the > > > > acceleration is complete before the detection occurs. > > > > OK, first of all, this is not an inertial reference frame, because it > > > is accelerating. > > > This is the reason why I gave you two choices of inertial reference > > > frame, neither of which is the noninertial reference frame you've > > > chosen. > > > I can certainly give you an answer for the noninertial reference frame > > > you've chosen: they do not arrive at the same time, and yes, they are > > > frequency/wavelength shifted. > > > Actually Paul, I just want to add another clause to this. What happens > > if, still before the detection occurs, the whole setup decelerates > > back to the same velocity as it started with. So the whole setup has > > moved, but at the time of the detection, the setup is not moving > > relative to its original frame. I assume that they still arrive at > > different times, but does the Doppler shifting remain? > > No, they arrive at the same time. All that has happened is that you've > *displaced* the apparatus while the photons are in flight. There is > also no longer any frequency/wavelength shifting. I'm sorry, I misrepresented the diagram. They would arrive at different times but there would be no red/blue-shifting.
From: mpalenik on 15 Feb 2010 14:59 On Feb 15, 2:20 pm, PD <thedraperfam...(a)gmail.com> wrote: > On Feb 15, 12:52 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > > On Feb 15, 12:06 pm, Ste <ste_ro...(a)hotmail.com> wrote: > > > > On 15 Feb, 16:53, PD <thedraperfam...(a)gmail.com> wrote: > > > > > On Feb 15, 10:05 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > > On 15 Feb, 15:18, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > > On Feb 14, 1:42 pm, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > > > > No takers for this simple question then? > > > > > > > > Consider this setup: > > > > > > > > S1 D2 > > > > > > > > D1 S2 > > > > > > > > We've got sources S1 and S2, paired with detectors D1 and D2. They're > > > > > > > all mechanically connected, so that a movement in one of them > > > > > > > produces > > > > > > > a movement in all the others - in other words, their relative > > > > > > > distances are always maintained. Each source is transmitting a > > > > > > > regular > > > > > > > pulse of light to its counterpart detector (so S1 is transmitting to > > > > > > > D1, etc.), and both sources are transmitting simultaneously with each > > > > > > > other. > > > > > > > > Now, we calculate that a pulse has just been emitted from both > > > > > > > sources, and we suddenly accelerate the whole setup "upwards" (i.e. > > > > > > > relative to how it's oriented on the page now) to near the speed of > > > > > > > light, and we complete this acceleration before the signals reach > > > > > > > either detector. > > > > > > > > Now, do both detectors *still* receive their signals simultaneously, > > > > > > > or does one receive its signal before the other? And are the signals > > > > > > > identical, or do they suffer from Doppler shifting, etc? > > > > > > > In what frame do you want the answer? > > > > > > The frame into which the set-up is accelerating, or the frame from > > > > > > which the set-up is accelerating? > > > > > > The answer is dependent on that choice. > > > > > > The answer will be according to an observer that is attached to the > > > > > setup and is equidistant from both detectors. And remember, the > > > > > acceleration is complete before the detection occurs. > > > > > OK, first of all, this is not an inertial reference frame, because it > > > > is accelerating. > > > > This is the reason why I gave you two choices of inertial reference > > > > frame, neither of which is the noninertial reference frame you've > > > > chosen. > > > > I can certainly give you an answer for the noninertial reference frame > > > > you've chosen: they do not arrive at the same time, and yes, they are > > > > frequency/wavelength shifted. > > > > Actually Paul, I just want to add another clause to this. What happens > > > if, still before the detection occurs, the whole setup decelerates > > > back to the same velocity as it started with. So the whole setup has > > > moved, but at the time of the detection, the setup is not moving > > > relative to its original frame. I assume that they still arrive at > > > different times, but does the Doppler shifting remain? > > > No, they arrive at the same time. All that has happened is that you've > > *displaced* the apparatus while the photons are in flight. There is > > also no longer any frequency/wavelength shifting. > > I'm sorry, I misrepresented the diagram. They would arrive at > different times but there would be no red/blue-shifting. I had gathered from what I read that the motion was perpendicular to the plane of the setup, so if not, they would arrive at different times.
From: dlzc on 15 Feb 2010 15:41 Dear PD: On Feb 15, 12:20 pm, PD <thedraperfam...(a)gmail.com> wrote: .... > I'm sorry, I misrepresented the diagram. They would arrive at > different times but there would be no red/blue-shifting. The emitters are in a rest frame, and the detectors are (later) in a frame with +v and -v. Better think again. Same situation would obtain the S1 and D2 were stationary, D1 and D2 were moving at v, and the pulses were sent through some form of clock synchronization (no more complex than obtaining "rigid" geometry). David A. Smith
From: PD on 15 Feb 2010 16:09
On Feb 15, 2:41 pm, dlzc <dl...(a)cox.net> wrote: > Dear PD: > > On Feb 15, 12:20 pm, PD <thedraperfam...(a)gmail.com> wrote: > ... > > > I'm sorry, I misrepresented the diagram. They would arrive at > > different times but there would be no red/blue-shifting. > > The emitters are in a rest frame, and the detectors are (later) in a > frame with +v and -v. Better think again. > > Same situation would obtain the S1 and D2 were stationary, D1 and D2 > were moving at v, and the pulses were sent through some form of clock > synchronization (no more complex than obtaining "rigid" geometry). > > David A. Smith Perhaps I don't understand the (revised) picture, either. My understanding was that S1, D1, S2 and D2 are initially all at rest in a common reference frame. Then, after emission, they are all briefly accelerated in the same direction and in the plane of the apparatus, and then brought back to rest in the initial reference frame. In this case, the whole apparatus has been simply displaced while the photons are in transit. |