Known physics defeated by simple puzzle?
in [Relativity]
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From: Ste on 15 Feb 2010 18:43 On 15 Feb, 23:26, "Inertial" <relativ...(a)rest.com> wrote: > "mpalenik" <markpale...(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. Ok then, "along the y-axis", if we assume that the x- is left and right across the screen (right is +), y- is up and down the screen (up is +), and z- is in and out the screen (out being 'toward the viewer', and out being +), although I'm sure at some point someone is going to want further clarification.
From: Ste on 15 Feb 2010 18:49 On 15 Feb, 23:05, "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. So do we have a consensus on the matter then? The pulse has an independent existence from its source so that, once emitted, a translation of the actual source does not translate the 'apparent' source of a pulse already in flight?
From: Inertial on 15 Feb 2010 19:05 "Ste" <ste_rose0(a)hotmail.com> wrote in message news:e5455758-949e-4255-a0e6-d6c872b6c7d7(a)q29g2000yqn.googlegroups.com... > On 15 Feb, 23:05, "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. > > So do we have a consensus on the matter then? I think so .. now that we've cleared up what you meant. It certainly doesn't even begin to come close to defeating known physics. > The pulse has an > independent existence from its source so that, once emitted, a > translation of the actual source does not translate the 'apparent' > source of a pulse already in flight? Of course .. there's only one theory going around trying to salvage emission theory that says reflection works so that light somehow remembers the speed of its source, and when it reflects off a moving mirror, rather than reflecting at the incident speed, or even being re-emitted at c, it adjusts its speed so that it remains as c wrt the source frame (ie it may speed up or slow down as required to keep its speed as c in that frame). I don't think any rational physicists take that 'theory' seriously. Certainly in mainstream physics, and even the sensible (but refuted) theories like emission theory, once light leaves a source, the source speed has no effect on the light speed. SR even goes as far as saying that the source speed has no effect on the light speed at all (as do basic ether theories). And that is supported by experimental evidence.
From: Inertial on 15 Feb 2010 19:06 "Ste" <ste_rose0(a)hotmail.com> wrote in message news:7e282ca9-964c-4db5-9032-ecd72386d3e2(a)g11g2000yqe.googlegroups.com... > 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? Once we all work out what your really meant .. and which of your two scenarios we are talking about. its not your puzzle that is 'defeating' us ... its your confusing presentation of it.
From: kenseto on 15 Feb 2010 21:06
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? I assume the following: |_______________| | S1 D2 | ^ | | /|\ | | | | | V | | |__D1______S2___| During acceleration D1 will detect blue shift and D2 will detect red shift. After acceleration D1 and D2 will detect the same freequency. Ken Seto | |