Known physics defeated by simple puzzle?
in [Relativity]
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From: Ste on 14 Feb 2010 14:42 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?
From: Darwin123 on 14 Feb 2010 14:51 On Feb 14, 2:42 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?
From: Darwin123 on 14 Feb 2010 15:13 On Feb 14, 2:42 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? The answer to your problem is actually very easy. The signals won't reach either detector. The path of the pulses, in the accelerated reference frame, will be curved downward. So the pulses will fall below the the detectors. By the time the light reaches the detector plane, the light beams will have fallen below the level of the detectors. The "measured" speed of light in this case is irrelevant. Since the light travel a curved path, it will fall below the detectors regardless of the "speed of light." If the acceleration itself is an impulse, the detectors and sources will reach a constant velocity close to the speed of light and then continue sailing in that frame. However, the light still will have curved downward durring the acceleration. So the light will still pass below the detectors. Even if the acceleration is short, it passes below the detectors. So the answer is quite obvious, and doesn't really involve relativity. In the inertial frames, light travels in a straight line. However, the detector moves out of the way before the pulse reaches it. In the accelerated frame, The light moves in a curved path that falls below the detectors. It is as though the light beam fell. According to both frames, inertial and accelerated, the light passed below the detectors. I think you had in mind a different problem. In general relativity, the speed of light and its path really do change with gravitational potential or with acceleration. The rules are different for accelerated frames. The constancy of the speed of light is strictly valid only for inertial frames, meaning the detectors are not accelerating. So you can't present a problem with accelerated detectors, invoke the constancy of the speed of light, and then honestly claim that you found a paradox.
From: William Hughes on 14 Feb 2010 16:15 On Feb 14, 3:42 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? What's more think of running with a long pole, fast enough so it is foreshortened enough to fit in a barn. An observer on the roof closes the in door and opens the out door simultaneously. But you see the barn foreshortened, not the pole ... Those who do not study the FAQ are condemned to repeat it. - William Hughes
From: Inertial on 14 Feb 2010 17:48
"Ste" <ste_rose0(a)hotmail.com> wrote in message news:929ce032-2892-4258-a004-fe497de345ee(a)k41g2000yqm.googlegroups.com... Oh dear .. another armchair physicist trying to show physics is wrong based on his own lack of understanding > No takers for this simple question then? Eh? You only just asked it. > Consider this setup: > > S1 D2 > > > D1 S2 Fine > 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. Fine > 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) Do they all accelerate the same acceleration profile and tart accelerating at the same time? If so, it would rip the device apart. > 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, They don't receive them at all. Gees. > or does one receive its signal before the other? And are the signals > identical, or do they suffer from Doppler shifting, etc? |