Known physics defeated by simple puzzle?
in [Relativity]
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From: dlzc on 15 Feb 2010 22:30 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. You chose the thread title, I didn't. You are trolling. David A. Smith
From: mpalenik on 15 Feb 2010 22:33 On Feb 15, 6:37 pm, Ste <ste_ro...(a)hotmail.com> wrote: > 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? We all agree, based on the way you've described it, they arrive at different times. It was only because I misunderstood the setup that I thought they would arrive at the same time. The direction of motion is very important here. But we all agree.
From: eric gisse on 15 Feb 2010 22:38 Ste 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). So open up a textbook and read it, rather than rudely demand to be taught for free. http://math.ucr.edu/home/baez/physics/Administrivia/rel_booklist.html Go away until you have read one or two.
From: mpalenik on 15 Feb 2010 22:40 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. 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. 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.
From: paparios on 16 Feb 2010 11:08
On 16 feb, 10:46, Ste <ste_ro...(a)hotmail.com> wrote: > On 16 Feb, 03:40, mpalenik <markpale...(a)gmail.com> wrote: > > 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? Your lack of basic knowledge in math shows clearly in the way you try to describe your *puzzles*. You should have realized by now the importance of providing some geometrical information, like the location (x,y) of your sources and detectors and, more general, the (x,y,z,t) coordinates of each event. Let me try to illustrate this with another simple puzzle that you should be able to answer. y | | _ ___|_________________________x K *********-> D O v _ Let us assume an object O, with a proper length of value L=x2-x1= 1 km, is moving along the x axis of a frame of reference K, towards the positive values of x. Let us assume the speed of the object on frame K is constant and equal to v=0.6c=180000 km/sec. Assume there is a laser gate D, through which the object will pass. The gate consists of a laser light illuminating a photodetector. So when the object goes through the laser gate, it will interrupt the laser light signal, as seen in a fast oscilloscope (with a response time of around 10 nanoseconds, short enough to be ignored). The question is: what would be the signal seen in the oscilloscope screen? Is it true that it would be a pulse signal of duration around 4.44 microseconds? Miguel Rios |