Known physics defeated by simple puzzle?
in [Relativity]
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From: Inertial on 15 Feb 2010 16:36 "PD" <thedraperfamily(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. 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).
From: dlzc on 15 Feb 2010 17:42 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. David A. Smith
From: Inertial on 15 Feb 2010 17:48 "dlzc" <dlzc1(a)cox.net> wrote in message news:b91787b3-c543-49fa-a199-16ec17d4eb28(a)m27g2000prl.googlegroups.com... > 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 did .. I quoted it above > 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. > > David A. Smith
From: mpalenik on 15 Feb 2010 17:53 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.
From: Inertial on 15 Feb 2010 18:05
"mpalenik" <markpalenik(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. |