Known physics defeated by simple puzzle?
in [Relativity]
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From: mpalenik on 15 Feb 2010 18:15 On Feb 15, 6:05 pm, "Inertial" <relativ...(a)rest.com> wrote: > "mpalenik" <markpale...(a)gmail.com> wrote in message > > news:90a5859a-7eaf-41ef-bda5-133028662cf0(a)h17g2000vbd.googlegroups.com... > > > > > On Feb 15, 4:09 pm, PD <thedraperfam...(a)gmail.com> wrote: > >> 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. > > > Yes, this is the revised picture. For some reason, I was thinking the > > motion was supposed to be perpendicular to the plane of the apparatus, > > but when I thought about it later, I specifically remembered a post > > from Ste where he said this was not the case (that he meant "up", as > > in "up on the screen"). This is how I understand the description as > > well. > > Its much simpler now. The pulses arrive at different times, because the > detectors have been moved (one closer to where pulses was emitted, the other > further way from it). And no doppler as the detectors are at rest again by > the time the pulses gets to it. I agree. When I gave my first answer (that the light arrives simultaneously to both detectors), I was still thinking the accelleration was supposed to be perpendicular to the plane of the setup, but that, apparently, is not what he was describing.
From: Inertial on 15 Feb 2010 18:26 "mpalenik" <markpalenik(a)gmail.com> wrote in message news:6bc53230-fe29-471d-a694-ab2f404f8d64(a)t42g2000vbt.googlegroups.com... > On Feb 15, 6:05 pm, "Inertial" <relativ...(a)rest.com> wrote: >> "mpalenik" <markpale...(a)gmail.com> wrote in message >> >> news:90a5859a-7eaf-41ef-bda5-133028662cf0(a)h17g2000vbd.googlegroups.com... >> >> >> >> > On Feb 15, 4:09 pm, PD <thedraperfam...(a)gmail.com> wrote: >> >> 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. >> >> > Yes, this is the revised picture. For some reason, I was thinking the >> > motion was supposed to be perpendicular to the plane of the apparatus, >> > but when I thought about it later, I specifically remembered a post >> > from Ste where he said this was not the case (that he meant "up", as >> > in "up on the screen"). This is how I understand the description as >> > well. >> >> Its much simpler now. The pulses arrive at different times, because the >> detectors have been moved (one closer to where pulses was emitted, the >> other >> further way from it). And no doppler as the detectors are at rest again >> by >> the time the pulses gets to it. > > I agree. When I gave my first answer (that the light arrives > simultaneously to both detectors), I was still thinking the > accelleration was supposed to be perpendicular to the plane of the > setup, but that, apparently, is not what he was describing. That's the problem when one uses words like 'up' that are ambiguous .. and give support to the reason why physics uses mathematics as its 'language' .. it is concise and far less ambiguous. A reason why Ste should learn more math so he can understand the language of physics.
From: Ste on 15 Feb 2010 18:37 On 15 Feb, 19:59, mpalenik <markpale...(a)gmail.com> wrote: > 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. Hell fire! Can't anyone make up their minds?
From: Ste on 15 Feb 2010 18:39 On 15 Feb, 21:36, "Inertial" <relativ...(a)rest.com> wrote: > "PD" <thedraperfam...(a)gmail.com> wrote in message > > news:70fd5e10-013a-474a-9b0f-ad6fcdd107b3(a)f15g2000yqe.googlegroups.com... > > > > > > > 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. > > Well, the original post says: > ==== > 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 > ==== > So for most of this thread, the example we're dealing with definitely does > NOT at rest in the initial frame of reference .. its going at 'near' the > speed of light. > > But then in this branch of the thread, Ste has changed the scenario so that > the device accelerates to (I assume) near the speed of light and then > decelerates back to being at rest, all while the light is in transit from S1 > to D1 and S2 to D2. Indeed. > Unless one is looking at a tree view of the thread, or expands out all the > >>'s one cannot easily tell which example one is dealing with now (or > necessarily know that there WAS a second scenario introduced). Agreed. This perhaps has led to some confusion.
From: Ste on 15 Feb 2010 18:40
On 15 Feb, 22:42, dlzc <dl...(a)cox.net> wrote: > Dear Inertial: > > On Feb 15, 2:36 pm, "Inertial" <relativ...(a)rest.com> wrote: > > > > > > > "PD" <thedraperfam...(a)gmail.com> wrote in message > > >news:70fd5e10-013a-474a-9b0f-ad6fcdd107b3(a)f15g2000yqe.googlegroups.com.... > > > On Feb 15, 2:41 pm,dlzc<dl...(a)cox.net> wrote: > > >> 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). > > > > 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. > > Yes. > > > > Then, after emission, they are all briefly accelerated > > > in the same direction and in the plane of the > > > apparatus, > > Yes. > > > > and then brought back to rest in the initial reference > > > frame. > > I don't see that he said that in his initial post. > > > > In this case, the whole apparatus has been simply > > > displaced while the photons are in transit. > > > Well, the original post says: > > ==== > > 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 > > ==== > > So for most of this thread, the example we're dealing > > with definitely does NOT at rest in the initial frame of > > reference .. > > Just the initial light pulses started when everything was at rest. > > > its going at 'near' the speed of light. > > > But then in this branch of the thread, Ste has changed > > the scenario so that the device accelerates to (I > > assume) near the speed of light and then decelerates > > back to being at rest, all while the light is in transit > > from S1 to D1 and S2 to D2. > > Typical crank behavior. They don't like nearing the answer, so they > change the question. > > > Unless one is looking at a tree view of the thread, or > > expands out all the >>'s one cannot easily tell which > > example one is dealing with now (or necessarily know > > that there WAS a second scenario introduced). > > Review Ste's original post. S1, S2, D1, D2 were all at rest at the > instant of emission. They were all in some uniform motion before > detection at either D1 or D2. > > I usually refuse to respond to them when they change questions in mid- > stream. They aren't looking for answers, only nibbles at the bait. Rubbish. I'm looking for some fairly simple answers, so that I can get an idea of how this bloody thing works (relativity, I mean). |