Known physics defeated by simple puzzle?
in [Relativity]
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From: Ste on 16 Feb 2010 06:35 On 16 Feb, 03:30, dlzc <dl...(a)cox.net> wrote: > On Feb 15, 4:40 pm, Ste <ste_ro...(a)hotmail.com> wrote: > ... > > > 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). > > Lying does not become you. I know. That's why I tend not to lie. > You chose the thread title, I didn't. Yes, sarcastically, because on the other thread all I had got (at the time of starting this new thread) was a lecture from Peter Webb. > You are trolling. That old broken record. Exactly what would you expect to see from a "non-troll"?
From: Peter Webb on 16 Feb 2010 07:29 > You are trolling. That old broken record. Exactly what would you expect to see from a "non-troll"? _________________________ *You* are a broken record. In answer to your question: 1. A genuine attempt to learn. 2. Shame at your ignorance of high school maths, not pride 3. A bit of courtesy to people who are trying to help You clearly are a troll. See, got me to respond!
From: Ste on 16 Feb 2010 08:46 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 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?
From: PD on 16 Feb 2010 09:31 On Feb 16, 7:46 am, Ste <ste_ro...(a)hotmail.com> wrote: > 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. Another terminology issue here. The setup as described is undergoing an acceleration or two. It therefore changes from being at rest in one reference frame to being at rest in another reference frame. "Transformations", as in the context Galilean transformations or Lorentz transformations, is the relationship between the descriptions of this system in those reference frames, but it does not mean the physical process of the acceleration of the system itself. Do you understand the terminology? > > > 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. What is confusing about it? Keep in mind that the constancy of the speed of light is something that is experimentally, unambiguously confirmed. > > > 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? The answer again depends on which reference frame you want the answer in. I am guessing, but need confirmation, that the answer you want is in the frame in which S1, D1, and D2 continue to be at rest?
From: dlzc on 16 Feb 2010 14:04
Dear Ste: On Feb 16, 6:46 am, Ste <ste_ro...(a)hotmail.com> wrote: > On 16 Feb, 03:40, mpalenik <markpale...(a)gmail.com> wrote: .... > > 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. This postulate is superfluous. The constancy of the speed of light is the result of Maxwell's equations (the laws of physics that are the same, regardless of source speed), and has good agreement with reality, with the proviso that only the two-way speed of light can ever be measured. David A. Smith |