Known physics defeated by simple puzzle?
in [Relativity]
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From: Inertial on 17 Feb 2010 16:12 "Ste" <ste_rose0(a)hotmail.com> wrote in message news:9760db08-176a-46dd-a823-6c34cd5d7cba(a)u9g2000yqb.googlegroups.com... > On 16 Feb, 23:21, "Inertial" <relativ...(a)rest.com> wrote: >> "Ste" <ste_ro...(a)hotmail.com> wrote in message >> >> news:e6adef4c-9c7e-421a-8a35-097ef868a9e4(a)k19g2000yqc.googlegroups.com... >> >> >> >> >> >> > On 16 Feb, 03:40, mpalenik <markpale...(a)gmail.com> wrote: >> >> On Feb 15, 6:49 pm, Ste <ste_ro...(a)hotmail.com> wrote: >> >> >> > 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? >> >> >> I'm not sure what you're trying to prove here. >> >> > I'm not trying to prove anything. I'm trying to form some sort of >> > consistent picture in my own head of how this setup behaves under >> > different transformations. >> >> >> You've brought the >> >> detector at rest with respect to the source. They're now both in the >> >> same reference frame. The only way SR says that you should observe >> >> anything "unusual" is if the source and detector are moving with >> >> respect to each other. >> >> > Perhaps. But what is confusing me is this "constant speed of light" >> > postulate. >> >> >> Translations in relativity--or in fact, even in pure mathematics--are >> >> very different things than rotations, and picking the detector up and >> >> putting it somewhere else without changing the speed is a pure >> >> translation. You don't even need relativity in this scenario, where >> >> everything is at rest with respect to everything else. >> >> > Yes, but we're going to get to the bit where relativity is required in >> > a moment. >> >> > Now, let us suppose we have two source and two detectors again: >> >> > D1 D2 D3 >> >> > S1 S2 S3 >> >> > S1 and D1 are stationary in the frame, and do not move. D2 is also >> > stationary in the frame. S2, S3, and D3 are all moving in the y+ >> > direction (i.e. same as the previous scenario) at a constant speed >> > (which is close to 'c'). Just to be sure we understand, the same setup >> > a few moments back in time would have looked like this: >> >> > D1 D2 >> >> > D3 >> >> > S1 >> >> > S2 S3 >> >> Your diagram contradicts your description .. you show in the diagram that >> S1 >> and D1 have moved apart (as have S3 and D3), but you said they do not. >> D3, >> S2 and S3 have also moved to the left. I think you need to draw your >> diagram again. > > No, I think you need to look at the diagram in a monospaced font. I did and am. I suggest YOU look at what YOU posted. Having an extra line between the D1 line and the S1 line has nothing to do with fonts.
From: Inertial on 17 Feb 2010 16:15 "Ste" <ste_rose0(a)hotmail.com> wrote in message news:9d92d181-e9c5-4562-905a-2f538a68ee09(a)j31g2000yqa.googlegroups.com... > On 17 Feb, 15:53, PD <thedraperfam...(a)gmail.com> wrote: >> On Feb 17, 9:13 am, Ste <ste_ro...(a)hotmail.com> wrote: >> >> > > > > Translations in relativity--or in fact, even in pure >> > > > > mathematics--are >> > > > > very different things than rotations, and picking the detector up >> > > > > and >> > > > > putting it somewhere else without changing the speed is a pure >> > > > > translation. You don't even need relativity in this scenario, >> > > > > where >> > > > > everything is at rest with respect to everything else. >> >> > > > Yes, but we're going to get to the bit where relativity is required >> > > > in >> > > > a moment. >> >> > > > Now, let us suppose we have two source and two detectors again: >> >> > > > D1 D2 D3 >> >> > > > S1 S2 S3 >> >> > > > S1 and D1 are stationary in the frame, and do not move. D2 is also >> > > > stationary in the frame. S2, S3, and D3 are all moving in the y+ >> > > > direction (i.e. same as the previous scenario) at a constant speed >> > > > (which is close to 'c'). Just to be sure we understand, the same >> > > > setup >> > > > a few moments back in time would have looked like this: >> >> > > > D1 D2 >> >> > > > D3 >> >> > > > S1 >> >> > > > S2 S3 >> >> > > > Now, when all sources come into line with each other (as per the >> > > > first >> > > > illustration above), a pulse is emitted towards the respective >> > > > detectors. After emission, S2 would continue towards D2, but in >> > > > reality we remove S2 from the picture before any collision (and >> > > > we've >> > > > already established that any transformation of the sources after >> > > > emission has no effect on photons already emitted). >> >> > > > Now, based on the previous scenario, I presume that in SR, D3 >> > > > receives >> > > > its pulse long after D1. However, this time, does D2 receive its >> > > > pulse >> > > > at the same time as D1? >> >> Yes, and yes. > > Ok. > > Consider yet another setup: > > > D1 D3 > > > > S1 D2 S2 D4 > > > In case it isn't clear, D1 and D2 are equidistant from S1, and D3 and > D4 are equidistant from S2. Again .. poorly 'drawn'. you show the distance S1..D2 as shorter than S2..D4. You need to take a little more care. This is the second diagram that you have mis-drawn. > The S1 group (i.e. comprising S1, D1, and D2) are always stationary in > the frame, and S1 is emitting a pulse towards both D1 and D2. There is > a similar setup for the S2 group, except that after the emission of > the pulse, the S2 group moves in the y+ direction, so that S2 is now > in the former place of D3 (again, the whole group accelerates rapidly, > and is stationary again by time of detection). > > Now, I assume D1 and D2 receive their signals simultaneously. But what > of D3 and D4? By the previous answers, I presume D4 receives it's > signal before D3? D4 will not receive the signal at all, because it is no longer in line with the direction of the signal (which went from left to right from the original location of S2.
From: PD on 17 Feb 2010 16:33 On Feb 17, 2:57 pm, Ste <ste_ro...(a)hotmail.com> wrote: > On 17 Feb, 15:53, PD <thedraperfam...(a)gmail.com> wrote: > > > > > On Feb 17, 9:13 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > > > Translations in relativity--or in fact, even in pure mathematics--are > > > > > > very different things than rotations, and picking the detector up and > > > > > > putting it somewhere else without changing the speed is a pure > > > > > > translation. You don't even need relativity in this scenario, where > > > > > > everything is at rest with respect to everything else. > > > > > > Yes, but we're going to get to the bit where relativity is required in > > > > > a moment. > > > > > > Now, let us suppose we have two source and two detectors again: > > > > > > D1 D2 D3 > > > > > > S1 S2 S3 > > > > > > S1 and D1 are stationary in the frame, and do not move. D2 is also > > > > > stationary in the frame. S2, S3, and D3 are all moving in the y+ > > > > > direction (i.e. same as the previous scenario) at a constant speed > > > > > (which is close to 'c'). Just to be sure we understand, the same setup > > > > > a few moments back in time would have looked like this: > > > > > > D1 D2 > > > > > > D3 > > > > > > S1 > > > > > > S2 S3 > > > > > > Now, when all sources come into line with each other (as per the first > > > > > illustration above), a pulse is emitted towards the respective > > > > > detectors. After emission, S2 would continue towards D2, but in > > > > > reality we remove S2 from the picture before any collision (and we've > > > > > already established that any transformation of the sources after > > > > > emission has no effect on photons already emitted). > > > > > > Now, based on the previous scenario, I presume that in SR, D3 receives > > > > > its pulse long after D1. However, this time, does D2 receive its pulse > > > > > at the same time as D1? > > > Yes, and yes. > > Ok. > > Consider yet another setup: > > D1 D3 > > S1 D2 S2 D4 > > In case it isn't clear, D1 and D2 are equidistant from S1, and D3 and > D4 are equidistant from S2. > > The S1 group (i.e. comprising S1, D1, and D2) are always stationary in > the frame, and S1 is emitting a pulse towards both D1 and D2. Let's make this a little more concrete. Let's say that S1 and S2 flash, emitting pulses that radiate in all directions, and that the detectors intercept a portion of the light from those pulses. (If you want, think of S1 and S2 as firecrackers.) > There is > a similar setup for the S2 group, except that after the emission of > the pulse, the S2 group moves in the y+ direction, so that S2 is now > in the former place of D3 (again, the whole group accelerates rapidly, > and is stationary again by time of detection). > > Now, I assume D1 and D2 receive their signals simultaneously. But what > of D3 and D4? By the previous answers, I presume D4 receives it's > signal before D3? In this frame, yes. Note, not in all inertial reference frames. :>)
From: Ste on 18 Feb 2010 09:06 On 17 Feb, 21:12, "Inertial" <relativ...(a)rest.com> wrote: > "Ste" <ste_ro...(a)hotmail.com> wrote in message > > >> > Now, let us suppose we have two source and two detectors again: > > >> > D1 D2 D3 > > >> > S1 S2 S3 > > >> > S1 and D1 are stationary in the frame, and do not move. D2 is also > >> > stationary in the frame. S2, S3, and D3 are all moving in the y+ > >> > direction (i.e. same as the previous scenario) at a constant speed > >> > (which is close to 'c'). Just to be sure we understand, the same setup > >> > a few moments back in time would have looked like this: > > >> > D1 D2 > > >> > D3 > > >> > S1 > > >> > S2 S3 > > >> Your diagram contradicts your description .. you show in the diagram that > >> S1 > >> and D1 have moved apart (as have S3 and D3), but you said they do not. > >> D3, > >> S2 and S3 have also moved to the left. I think you need to draw your > >> diagram again. > > > No, I think you need to look at the diagram in a monospaced font. > > I did and am. I suggest YOU look at what YOU posted. Having an extra line > between the D1 line and the S1 line has nothing to do with fonts. There is no "extra line". D3 is moving with S2 and S3 (and D3 should be in line with S3).
From: Ste on 18 Feb 2010 09:23
On 17 Feb, 21:15, "Inertial" <relativ...(a)rest.com> wrote: > "Ste" <ste_ro...(a)hotmail.com> wrote in message > > >> Yes, and yes. > > > Ok. > > > Consider yet another setup: > > > D1 D3 > > > S1 D2 S2 D4 > > > In case it isn't clear, D1 and D2 are equidistant from S1, and D3 and > > D4 are equidistant from S2. > > Again .. poorly 'drawn'. you show the distance S1..D2 as shorter than > S2..D4. Indeed you are correct, but in any event that is why I included the clarifying clause. > You need to take a little more care. This is the second diagram that you > have mis-drawn. I'm afraid I still can't see the problem with the first drawing. It appears exactly as intended on my screen, although your requote of it appears garbled. > > The S1 group (i.e. comprising S1, D1, and D2) are always stationary in > > the frame, and S1 is emitting a pulse towards both D1 and D2. There is > > a similar setup for the S2 group, except that after the emission of > > the pulse, the S2 group moves in the y+ direction, so that S2 is now > > in the former place of D3 (again, the whole group accelerates rapidly, > > and is stationary again by time of detection). > > > Now, I assume D1 and D2 receive their signals simultaneously. But what > > of D3 and D4? By the previous answers, I presume D4 receives it's > > signal before D3? > > D4 will not receive the signal at all, because it is no longer in line with > the direction of the signal (which went from left to right from the original > location of S2. Indeed. So let's assume that the S2 group was already moving at the time of emission, and that it continued to move. Would the pulse *then* also continue in the x+ direction (i.e horizontally across the screen) towards the location of D4 *at the time of emission*, or would it take a diagonal (or even curved) path to catch D4? (Again, the frame is that of the stationary S1 group) |