Known physics defeated by simple puzzle?
in [Relativity]
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From: Ste on 6 Apr 2010 13:11 On 6 Apr, 16:42, PD <thedraperfam...(a)gmail.com> wrote: > On Apr 5, 7:15 pm, Ste <ste_ro...(a)hotmail.com> wrote: > > > On 5 Apr, 22:57, PD <thedraperfam...(a)gmail.com> wrote: > > > > On Apr 5, 4:29 pm, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > I do not see how this can work, > > > > unless the clocks themselves fall out of synchronisation (and hence > > > > are not actually measuring the same periods of time as each other). > > > > The clocks' synchronization can be (and is) checked both before and > > > after the operations described above and verified to still hold, so > > > your supposition of what must have happened is ruled out. > > > Not necessarily. If both slowed in the middle of the operation, then > > they would still be synchronised with each at the end, but it would > > utterly confound any calculation about simultaneity, unless you also > > knew by how much the other clock had slowed and when. > > > To give you an example, if I have two clocks stationary relative to > > each other and ten light-seconds apart, and I suddenly slow both > > clocks to half-speed (in other words, ticking once for every two > > previous ticks), then according to each clock, the other one speeds up > > to double speed, > > Why does the other speed up to double speed, when it is also > identically slowed? It doesn't speed up. But it *appears* to because the distant ticks now come in at a frequency double that of local ticks. You do accept, don't you, that (for a short time) the ticks come in twice as fast from the distant clock when both clocks are slowed down? > > because for a period of 10 seconds (the time it takes > > for the change of rate to propagate), there is only one "local" tick > > for every 2 "received" ticks. > > And at the end there is a similar and compensating two local ticks for > every one received ticks. No. There is no occasion, in this example, on which the local clock ticks twice as fast as the distant clock. To give you the actual numbers, first we'll start with what each clock "really" reads (i.e. if information of their display travelled instantly). Local Distant 00:01 00:01 00:02 00:02 00:03 00:03 00:04 00:04 00:05 00:05 00:06 00:06 00:07 00:07 00:08 00:08 00:09 00:09 00:10 00:10 00:11 00:11 Now for the apparent readings (i.e. what an observer will actually see, when the clocks start going): Local Distant 00:01 00:00 00:02 00:00 00:03 00:00 00:04 00:00 00:05 00:00 00:06 00:00 00:07 00:00 00:08 00:00 00:09 00:00 00:10 00:00 00:11 00:01 Obviously, the explanation for this is that it takes 10 seconds for each first tick to propagate to the other clock, so the local clock reads 00:11 when the distant clock first turns over to 00:01. But we can reason backwards from this, and say, since it takes 10 seconds to reach us, the distant clock must "really" read 00:11, which happens to be the same as our own. Next, we show what happens "really" when we slow both clocks down at 00:20: Local Distant 00:20 00:20 00:21 00:21 00:22 00:22 00:23 00:23 00:24 00:24 00:25 00:25 And the apparent readings: Local Distant 00:20 00:10 00:21 00:12 00:22 00:14 00:23 00:16 00:24 00:18 00:25 00:20 And if we use the same reasoning again, that it takes 10 seconds for the distant clock to propagate its display, then by that logic, at 00:25 local, the distant clock "really" reads 00:30 (because it is always apparently 10 seconds behind itself). But obviously, that isn't true, because in fact the speed of light is now travelling twice as fast - the real reading of the distant clock is 00:25, the same as the local clock. Both clocks are always in synchronisation, both always tick simultaneously. For reference, the apparent figures going forward are these: Local Distant 00:26 00:21 00:27 00:22 00:28 00:23 00:29 00:24 00:30 00:25 And if we speed the clocks up again, obviously the whole thing happens in reverse. The local clock spings ahead of the distant clock. Obviously, this doesn't begin to capture the complexity of SR, where the clocks are *not* stationary, and so the calculation would be made even more complex. But it does show that interesting things can happen to your perception of time (and your calculations about what is happening at a distant location, and when) if your own clock is falling victim to a *real* relativistic slowing (which is, of course, totally imperceptible in itself, because *everything* slows down at a fundamental level). > Furthermore, recall that there is no TIMING that is done by any clock > during the signal propagation at all. The only decision that has to be > made is whether the signals are received at the receiver at the same > time or at different times. This is not something that is done with > the need of a clock. When you look with your eyes at two flashes of > light, you do not need to refer to a clock to decide whether the > flashes were received at the same time or at different times. You can > see it with your plain eyes. If you want to use electronics, you do it > with a coincidence counter, not with a clock. I'm not saying that you need clocks Paul. I'm giving you a particular hypothetical scenario, where the regular interval of the ticking allows us to explore how our reasoning about what is happening at a distant location, and when, can be distorted by "illusions". > The only other bit of information you need is whether the signal > propagation time is identical. Again for this, you do not need a > clock. You need a measuring stick to measure the distance of flight of > the signals, and you need an independent measurement of the velocity > of the signal. The velocity measurement can be done at leisure, either > before the entire experiment or after or both. The only presumption > that is made is that the speed of the signal propagation did not > happen to change during the one time you were receiving signals from > the two events. Since signal propagation speed is governed by constant > laws of nature and are not subject to random whim, this is a > justifiable presumption -- especially since you can repeat that speed > measurement any number of times at random moments and confirm that the > speed never varies. I'm unclear whether 'c' really is constant. It seems reasonably certain to say that it is constant within any frame (that is, constant between any two stationary objects), but I'm less clear about whether it is constant between frames (i.e. between objects moving at high relative speeds). But anyway, this is beyond my current level of comprehension. > Again, there is NO CLOCK needed to determine simultaneity. I never said there was. We just happen to be using clocks here, because they provide a regular inverval.
From: Ste on 6 Apr 2010 13:14 On 6 Apr, 17:02, PD <thedraperfam...(a)gmail.com> wrote: > On Apr 6, 10:39 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > > > On 6 Apr, 16:28, PD <thedraperfam...(a)gmail.com> wrote: > > > > On Apr 5, 6:41 pm, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > On 5 Apr, 22:10, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > On Apr 5, 3:47 pm, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > > > I'm happy to make concessions, but I have to be able to ask questions > > > > > > and get meaningful answers. The ladder in the barn is a prime example. > > > > > > I'm told "the ladder contracts to fit in the barn". > > > > > > It contracts to fit the barn in the rest frame of the barn, which is > > > > > why the ladder makes no marks on the barn doors when the doors are > > > > > shut at the same time. In the rest frame of the ladder, the ladder > > > > > does not contract and indeed does not fit inside the barn at all. In > > > > > the rest frame of the ladder, the reason why there are no marks on the > > > > > barn doors when they are shut is that they were not shut at the same > > > > > time in this frame. > > > > > This logic is easily defeated Paul, because if we contracted the > > > > ladder's length *just* enough so that it marked the door in the barn > > > > frame (in other words, the ladder has contracted just enough to manage > > > > an interference fit with both doors shut), then this cannot be > > > > accounted for in the ladder frame (because, in the ladder frame, if > > > > the ladder is *even larger* relative to the barn than when it started, > > > > then the ladder could not possibly mark the doors in the same way). > > > > I'm not sure what the fuss is. The observation is that the doors are > > > shut and open without striking the pole, and this is true in both > > > reference frames examined (as well as any other inertial reference > > > frame). The account in the ladder frame is, however, not because the > > > ladder fits inside the barn. It is exactly as I described above. Why > > > is this difficult? > > > It isn't difficult for me. I can explain it, in its entirety, in terms > > of a "visual effect" and the careful timing of the doors. It's other > > people here who keep insisting that it is not a visual effect, and > > hence the fuss. > > I don't believe it's a "visual effect" at all. The doors are closed > and opened by a common trigger, electronically. > > Is it your claim that the doors are "really" closed and opened > simultaneously, but they only visually appear to close and open > nonsimultaneously in the ladder frame? Then how does the 80 foot > ladder fit into the 40 ft barn without the doors striking the ladder, > if the doors are "really" closed simultaneously? I would contend that the ladder never does fit, and even though it would *appear* observationally to be small enough to fit, any attempt to actually shut the doors while the ladder is inside will produce catastrophe.
From: paparios on 6 Apr 2010 13:22 On 6 abr, 12:28, Ste <ste_ro...(a)hotmail.com> wrote: > On 6 Apr, 05:54, "Inertial" <relativ...(a)rest.com> wrote: > > > > A thing cannot fit and at > > > the same time not fit, > > > yes it can.. just as it can be travelling at 100km/hr and 50 km/hr at the > > same time. > > But that's not true. Again, hidden in the meaning here is 100km/h > *relative to what*. An object cannot travel at *both* 100km/h *and* 50 > km/hr *relative to me*. > > And by the same token, if we are talking about the relationship > between the barn and the ladder, it cannot fit and not fit. > Again you are misunderstanding the purpose of using these inertial frames of reference on SRT. SRT says that: First postulate The laws by which the states of physical systems undergo change are not affected, whether these changes of state be referred to the one or the other of two systems in uniform translatory motion relative to each other. Second postulate - Light in vacuum propagates with the speed c (a fixed constant, independent of direction) in at least one system of inertial coordinates (the "stationary system"), regardless of the state of motion of the light source. So in the pole and barn gedanken, we have a pole moving quite fast and two doors that close and then open (also quite fast). The relevant EVENTS are the closing and opening of the barn doors and their effect on the moving pole. If the closing of one door produces the pole crushing that door (generating a megaton size explosion), that EVENT has (logically) to be seen, heard and measured as an explosion in every single frame of reference, to satisfy the first postulate. If the pole fits inside the barn, allowing both doors of the barn to close simultaneously and with the pole never touching any of the doors (as seen and measured on the barn frame of reference), that does not mean all of the other frames of reference will see, hear or measure the same (that is, that the pole is inside the barn with the doors closed). What other frames will see and measure will be the pole moving and the barn doors closing and opening at some time. The effects of these conditions have to be consistent with the first observation, that is, the pole and the barn never touch each other and the barn doors close and open. The rest (at v=0 with respect to the barn frame of reference) dimension of the pole is 20 meters and the barn is 10 meters. If the pole is moving at v=0.9c, its length as measured in the barn frame of reference is of 8.73 meters, that is the pole is contracted in that frame. In the rest frame of the pole, it is 20 meters long while the barn is moving approaching it at v=-0.9c, which means the barn is contracted to 4.37 meters on the pole frame of reference. So the gedanken shows that for a very short time, on the barn frame of reference, the barn doors can be closed at the same time with the whole contracted pole inside the barn. The gedanken also shows that from the pole frame of reference, the barn doors close and open at different times which allows the 20 meters pole to pass through the 4.37 meters barn without touching the barn doors. Miguel Rios
From: PD on 6 Apr 2010 13:28 On Apr 6, 12:14 pm, Ste <ste_ro...(a)hotmail.com> wrote: > On 6 Apr, 17:02, PD <thedraperfam...(a)gmail.com> wrote: > > > > > On Apr 6, 10:39 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > On 6 Apr, 16:28, PD <thedraperfam...(a)gmail.com> wrote: > > > > > On Apr 5, 6:41 pm, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > > On 5 Apr, 22:10, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > > On Apr 5, 3:47 pm, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > > > > I'm happy to make concessions, but I have to be able to ask questions > > > > > > > and get meaningful answers. The ladder in the barn is a prime example. > > > > > > > I'm told "the ladder contracts to fit in the barn". > > > > > > > It contracts to fit the barn in the rest frame of the barn, which is > > > > > > why the ladder makes no marks on the barn doors when the doors are > > > > > > shut at the same time. In the rest frame of the ladder, the ladder > > > > > > does not contract and indeed does not fit inside the barn at all. In > > > > > > the rest frame of the ladder, the reason why there are no marks on the > > > > > > barn doors when they are shut is that they were not shut at the same > > > > > > time in this frame. > > > > > > This logic is easily defeated Paul, because if we contracted the > > > > > ladder's length *just* enough so that it marked the door in the barn > > > > > frame (in other words, the ladder has contracted just enough to manage > > > > > an interference fit with both doors shut), then this cannot be > > > > > accounted for in the ladder frame (because, in the ladder frame, if > > > > > the ladder is *even larger* relative to the barn than when it started, > > > > > then the ladder could not possibly mark the doors in the same way). > > > > > I'm not sure what the fuss is. The observation is that the doors are > > > > shut and open without striking the pole, and this is true in both > > > > reference frames examined (as well as any other inertial reference > > > > frame). The account in the ladder frame is, however, not because the > > > > ladder fits inside the barn. It is exactly as I described above. Why > > > > is this difficult? > > > > It isn't difficult for me. I can explain it, in its entirety, in terms > > > of a "visual effect" and the careful timing of the doors. It's other > > > people here who keep insisting that it is not a visual effect, and > > > hence the fuss. > > > I don't believe it's a "visual effect" at all. The doors are closed > > and opened by a common trigger, electronically. > > > Is it your claim that the doors are "really" closed and opened > > simultaneously, but they only visually appear to close and open > > nonsimultaneously in the ladder frame? Then how does the 80 foot > > ladder fit into the 40 ft barn without the doors striking the ladder, > > if the doors are "really" closed simultaneously? > > I would contend that the ladder never does fit, and even though it > would *appear* observationally to be small enough to fit, any attempt > to actually shut the doors while the ladder is inside will produce > catastrophe. This is inconsistent with experiment. As I mentioned before, the first thing one must do in the case of an observational science, is to not deny what is actually seen in experiment, just because you do not immediately understand how that outcome could arise. As indicated above, the doors are closed and then opened by a single trigger and equal-length, exchangeable cables. There is no catastrophe observed. PD
From: PD on 6 Apr 2010 13:33
On Apr 6, 12:11 pm, Ste <ste_ro...(a)hotmail.com> wrote: > On 6 Apr, 16:42, PD <thedraperfam...(a)gmail.com> wrote: > > > > > On Apr 5, 7:15 pm, Ste <ste_ro...(a)hotmail.com> wrote: > > > > On 5 Apr, 22:57, PD <thedraperfam...(a)gmail.com> wrote: > > > > > On Apr 5, 4:29 pm, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > > I do not see how this can work, > > > > > unless the clocks themselves fall out of synchronisation (and hence > > > > > are not actually measuring the same periods of time as each other). > > > > > The clocks' synchronization can be (and is) checked both before and > > > > after the operations described above and verified to still hold, so > > > > your supposition of what must have happened is ruled out. > > > > Not necessarily. If both slowed in the middle of the operation, then > > > they would still be synchronised with each at the end, but it would > > > utterly confound any calculation about simultaneity, unless you also > > > knew by how much the other clock had slowed and when. > > > > To give you an example, if I have two clocks stationary relative to > > > each other and ten light-seconds apart, and I suddenly slow both > > > clocks to half-speed (in other words, ticking once for every two > > > previous ticks), then according to each clock, the other one speeds up > > > to double speed, > > > Why does the other speed up to double speed, when it is also > > identically slowed? > > It doesn't speed up. But it *appears* to because the distant ticks now > come in at a frequency double that of local ticks. > > You do accept, don't you, that (for a short time) the ticks come in > twice as fast from the distant clock when both clocks are slowed down? > > > > because for a period of 10 seconds (the time it takes > > > for the change of rate to propagate), there is only one "local" tick > > > for every 2 "received" ticks. > > > And at the end there is a similar and compensating two local ticks for > > every one received ticks. > > No. There is no occasion, in this example, on which the local clock > ticks twice as fast as the distant clock. To give you the actual > numbers, first we'll start with what each clock "really" reads (i.e. > if information of their display travelled instantly). > > Local Distant > 00:01 00:01 > 00:02 00:02 > 00:03 00:03 > 00:04 00:04 > 00:05 00:05 > 00:06 00:06 > 00:07 00:07 > 00:08 00:08 > 00:09 00:09 > 00:10 00:10 > 00:11 00:11 > > Now for the apparent readings (i.e. what an observer will actually > see, when the clocks start going): > > Local Distant > 00:01 00:00 > 00:02 00:00 > 00:03 00:00 > 00:04 00:00 > 00:05 00:00 > 00:06 00:00 > 00:07 00:00 > 00:08 00:00 > 00:09 00:00 > 00:10 00:00 > 00:11 00:01 > > Obviously, the explanation for this is that it takes 10 seconds for > each first tick to propagate to the other clock, so the local clock > reads 00:11 when the distant clock first turns over to 00:01. But we > can reason backwards from this, and say, since it takes 10 seconds to > reach us, the distant clock must "really" read 00:11, which happens to > be the same as our own. > > Next, we show what happens "really" when we slow both clocks down at > 00:20: > > Local Distant > 00:20 00:20 > 00:21 00:21 > 00:22 00:22 > 00:23 00:23 > 00:24 00:24 > 00:25 00:25 > > And the apparent readings: > > Local Distant > 00:20 00:10 > 00:21 00:12 > 00:22 00:14 > 00:23 00:16 > 00:24 00:18 > 00:25 00:20 > > And if we use the same reasoning again, that it takes 10 seconds for > the distant clock to propagate its display, then by that logic, at > 00:25 local, the distant clock "really" reads 00:30 (because it is > always apparently 10 seconds behind itself). But obviously, that isn't > true, because in fact the speed of light is now travelling twice as > fast - the real reading of the distant clock is 00:25, the same as the > local clock. Both clocks are always in synchronisation, both always > tick simultaneously. > > For reference, the apparent figures going forward are these: > > Local Distant > 00:26 00:21 > 00:27 00:22 > 00:28 00:23 > 00:29 00:24 > 00:30 00:25 > > And if we speed the clocks up again, obviously the whole thing happens > in reverse. The local clock spings ahead of the distant clock. > > Obviously, this doesn't begin to capture the complexity of SR, where > the clocks are *not* stationary, and so the calculation would be made > even more complex. But it does show that interesting things can happen > to your perception of time (and your calculations about what is > happening at a distant location, and when) if your own clock is > falling victim to a *real* relativistic slowing (which is, of course, > totally imperceptible in itself, because *everything* slows down at a > fundamental level). > > > Furthermore, recall that there is no TIMING that is done by any clock > > during the signal propagation at all. The only decision that has to be > > made is whether the signals are received at the receiver at the same > > time or at different times. This is not something that is done with > > the need of a clock. When you look with your eyes at two flashes of > > light, you do not need to refer to a clock to decide whether the > > flashes were received at the same time or at different times. You can > > see it with your plain eyes. If you want to use electronics, you do it > > with a coincidence counter, not with a clock. > > I'm not saying that you need clocks Paul. I'm giving you a particular > hypothetical scenario, where the regular interval of the ticking > allows us to explore how our reasoning about what is happening at a > distant location, and when, can be distorted by "illusions". Of course. I'm not denying the existence of illusions. However, since there is no relevance to the ladder and barn puzzle, and there is no relevance to the classification of the simultaneity of two events, I'm hard pressed to see value in the strawman. Thank you, however, for all your carefully wasted effort. > > > The only other bit of information you need is whether the signal > > propagation time is identical. Again for this, you do not need a > > clock. You need a measuring stick to measure the distance of flight of > > the signals, and you need an independent measurement of the velocity > > of the signal. The velocity measurement can be done at leisure, either > > before the entire experiment or after or both. The only presumption > > that is made is that the speed of the signal propagation did not > > happen to change during the one time you were receiving signals from > > the two events. Since signal propagation speed is governed by constant > > laws of nature and are not subject to random whim, this is a > > justifiable presumption -- especially since you can repeat that speed > > measurement any number of times at random moments and confirm that the > > speed never varies. > > I'm unclear whether 'c' really is constant. It seems reasonably > certain to say that it is constant within any frame (that is, constant > between any two stationary objects), but I'm less clear about whether > it is constant between frames (i.e. between objects moving at high > relative speeds). But anyway, this is beyond my current level of > comprehension. You've mangled the language quite a bit. Perhaps you should be more careful about understanding the meaning of words like "frame" until you know what they mean in physics context. Constant in an inertial reference frame does not mean as measured between two stationary objects, nor does "between frames" mean as measured between two moving objects. > > > Again, there is NO CLOCK needed to determine simultaneity. > > I never said there was. We just happen to be using clocks here, > because they provide a regular inverval. |