Known physics defeated by simple puzzle?
in [Relativity]
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From: Inertial on 14 Feb 2010 22:25 "nuny(a)bid.nes" <alien8752(a)gmail.com> wrote in message news:09e1ab4f-8d10-49b7-89bd-9838a44eb404(a)a16g2000pre.googlegroups.com... > On Feb 14, 3:55 pm, "Inertial" <relativ...(a)rest.com> wrote: >> "n...(a)bid.nes" <alien8...(a)gmail.com> wrote in message >> >> news:05d6a5e5-9996-4a4b-8bf9-a3c93108da51(a)x10g2000prk.googlegroups.com... >> >> >> >> > On Feb 14, 11:42 am, Ste <ste_ro...(a)hotmail.com> wrote: >> >> 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. > > Sorry, my question should have read: > >> > By "upwards", do you mean within the page along the direction >> > parallel to a line drawn from D1 to S1 (S2 to D2), i. e. that way: >> > >> > ^ >> > | >> > >> > ...or out of the screen towards me? >> >> >> 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? >> >> > If the former, both receive their signals simultaneously. Doppler >> > applies. >> >> I don't agree. As D1 is accelerating in the opposite direction to the >> light, and D2 in the same direction as the light .. the light would hit >> D1 >> first, > > That would be correct if light velocity depended on the velocity of > the emitter, but it doesn't. True that it doesn't, but false that that is required for the arrival times to be different. > Light is always observed to travel at c > no matter the relative state of motion of detector and emitter. No .. only when that motion is inertial. > When > an emitter/detector don't move WRT each other, but are accelerated as > a unit, Which would tear them apart. > they see Doppler shift during acceleration and deceleration. Yes > When they are moving at a *constant velocity* the signal is seen at > it's at rest wavelength. Only if it was emitted at that constant velocity > The device is an accelerometer, not a speedometer. I didn't say it was. > If it were, it > would demonstrate the existence of an absolute frame of reference. No .. it demosntrates that acceleration is 'absolute' (in that all inertial frames agree whether a given object is accelerating or not) > I suspect that's the OPs point. I don't think he has one > If it is, I suggest he build the > apparatus and see; it isn't complicated. No matter how he yanks it > around (hundreds, even thousands of gs of *nondestructive* > acceleration is not hard to achieve with small, well-built > electronics) I predict he will not observe a timing change with > acceleration. I would be very interested if he did. He would. have a look at a sagnac device, that gives fringe shifts due to change in arrival time of light around an accelerating path. >> for the same reason that light would not hit the detectors at all in >> the case where the device was accelerating out of the page. > > That's because if the acceleration is fast enough the light simply > misses the detector; Exactly > in the first case the light has nowhere to go > *but* the detector. Yeup > It's still an "accelerometer" of sorts, but go-no go rather than > analog. Yeup >> Mind you, during accelerating the notion of 'first' is not all that clear >> .. >> according to which frame of reference are we talking? But the experiment >> does claim that the acceleration starts after the pulses leave the >> sources >> and stops before the light reaches either detector .. so that really >> gives >> us a choice of two reasonable frames to consider for deciding which is >> first >> .. the original rest frame and the final rest frame. > > I was being generous and assuming the emitters were triggered by, > and the detectors compared by, mechanisms equidistant from both, You misunderstand. Distance doesn't matter here (you would account for that). Its whether or not they are simultaneous. > tied > to the frame holding everything together. The signal paths from the > trigger to the emitters, and the path from the detectors to the > comparator, *and* the timing signal from the trigger, must be the same > lengths (or suitable delays introduced to make their signal travel > times identical). Or you account for it. But that is not the point. > Notice the frame, being made of matter (since there isn't anything > else to work with) is held together by electron-mediated bonds, which > can't respond faster than disturbances in their mutual electromagnetic > field can travel, IOW light speed. This is essentially the same > "puzzle" the OP posed. If lightspeed bonds independent of velocity > can't hold the structure together there's no puzzle. Eh? >> > If the latter, neither detector registers a hit *until the setup >> > reaches a constant velocity*. >> >> Yeup.
From: Owen Jacobson on 15 Feb 2010 02:35 Subject: Known physics defeated by simple puzzle? If you're serious about actually learning something about physics, as opposed to being serious about winding people up, don't take this approach. It discourages others from taking you seriously. We already have enough "physics is wrong!" threads kicking around. If you don't understand what modern physical theories (in this case, relativity) predict for a given experiment, that's fine, and asking about it is not unwise; however, presupposing that they offer no prediction or that the prediction is illogical or inconsistent is only going to make you look foolish to the people who could help you understand. Conversely, if you don't understand what happens for a given experimental setup, your best bet is to /run the experiment/ -- that's how physical theories are (in)validated, after all. -o
From: Ste on 15 Feb 2010 06:14 On 14 Feb, 20:13, Darwin123 <drosen0...(a)yahoo.com> wrote: > 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? > > The answer to your problem is actually very easy. The signals > won't reach either detector. The path of the pulses, in the > accelerated reference frame, will be curved downward. So the pulses > will fall below the the detectors. By the time the light reaches the > detector plane, the light beams will have fallen below the level of > the detectors. I can see there has been a misunderstanding. When I said "upwards", I didn't mean "above the page in 3D", I meant "the top of the page in 2D". In other words, "the heading of the page", so that D1 and S2 head towards S1 and D2, and S1 and D2 head towards where I wrote "consider this setup". > The rules are different for accelerated frames. The constancy > of the speed of light is strictly valid only for inertial frames, > meaning the detectors are not accelerating. So you can't present a > problem with accelerated detectors, invoke the constancy of the speed > of light, and then honestly claim that you found a paradox. I'm not necessarily claiming to have "found a paradox". I'm asking a relatively straightforward question that if the setup moves while the photon is in "mid-air", what is the outcome?
From: Ste on 15 Feb 2010 06:19 On 14 Feb, 21:15, William Hughes <wpihug...(a)hotmail.com> wrote: > On Feb 14, 3:42 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? > > What's more think of running with a long pole, fast enough > so it is foreshortened enough to fit in a barn. An observer > on the roof closes the in door and opens the out door > simultaneously. But you see the barn foreshortened, not the > pole ... > > Those who do not study the FAQ are condemned to repeat > it. Let's forget the pole and barn paradox because it introduced too much argument. This setup is simple, in that it involves a setup that undergoes a straightforward translation along a single dimension, except that one detector moves towards the source (as it was before the translation), and the other moves away.
From: Ste on 15 Feb 2010 06:22
On 14 Feb, 22:48, "Inertial" <relativ...(a)rest.com> wrote: > > > 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) > > Do they all accelerate the same acceleration profile and tart accelerating > at the same time? If so, it would rip the device apart. Forget about the acceleration profile or the physical nature of the device. All it requires is that an acceleration initiates after emission at the sources, and before reception at the detector. The fact that it would be physically impossible to achieve this setup is the same as how virtually every other gedanken is impossible to achieve in practice (what with ladders being accelerated into garages at near the speed of light, etc). |