Is Length Contraction in SR physical??
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From: mpalenik on 6 Feb 2010 23:14 On Feb 6, 9:35 pm, Ste <ste_ro...(a)hotmail.com> wrote: > On 6 Feb, 10:55, mpalenik <markpale...(a)gmail.com> wrote: > > > > > > > On Feb 6, 5:32 am, mpalenik <markpale...(a)gmail.com> wrote: > > > > On Feb 6, 5:01 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > On 6 Feb, 09:23, mpalenik <markpale...(a)gmail.com> wrote: > > > > > > On Feb 6, 3:43 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > > > Yes, with a bit of jiggerypokery, I think I have drawn it correctly, > > > > > > and indeed, it does become shorter in the x' orientation as against > > > > > > the x axis. > > > > > > Just to be sure we're on the same page, you should have something like > > > > > this (I added the circle, which represents the proper length of the > > > > > ladder):http://s424.photobucket.com/albums/pp327/mpalenik/?action=view¤... > > > > > Yes, that's basically what I had but without the circle (I was just > > > > using a line on the x-axis to represent the ladder). > > > > The ladder lies along the x' axis. The x' axis is the rest frame of > > > the ladder. > > > > > > Keep in mind this is in *non-Minkowski* spacetime, which means it will > > > > > produce the oppose results that relativity produces (length expansions > > > > > instead of contractions, reversed breaking of simultanaity). > > > > > Ah right, because the way I looked at it, if you were using the x-axis > > > > of the diagram to measure, then the "width" of anything on the x'-axis > > > > had actually reduced when viewed from the x-axis. > > > > That's because you're doing it wrong. You can only measure the parts > > > of the ladder that are intersecting with the regular x axis. The end > > > points of the ladder are marked by dots in my picture. You'll notice > > > that the ladder intersects the x axis but at different values of t'. > > > Let me try to clarify--different parts of the ladder intersect the x > > axis at different values of t'. It traces out a solid rectangle in > > spacetime as it moves. > > Indeed. > > > You can only measure along the x axis at a given t the parts of the > > ladder that intersect the x axis at that given t. > > Of course. Because t=0 represents what you're visually observing in > the present. Right, so if you see a length contraction in my picture, it means that you're trying to observe parts of the rod that aren't at t=0. Look at the intersection with the volume the rod sweeps out with t=0. This intersection is *larger* than the length of the rod (you can see it extends slightly past the edge of the circle). > > > The two ends of the > > ladder intersect the x axis at *different* values of t'. This > > corresponds to different *times* in the ladder's frame. Look at where > > the t' lines extended from the endpoints of the ladder intersect the x > > axis. You'll see the line they intersect to make a line that is > > *longer* than the length of the ladder. This has nothing to do with > > optical illusions. This has to do with the amount of space the ladder > > traces out along the x axis as it moves through spacetime. > > I'm afraid I disagree with this interpretation. I'm reminded of > Einstein's quip that the "only thing that gets in the way of thinking, > is education". > This is not an interpretation. This is you using the diagram incorrectly. We haven't even gotten to the interpretation yet. You need to look at the diagram the right way first before we can interpret anything. This is not a subjective description that I'm talking about. If you look at the intersection of the 2D volume the ladder traces out with the t=0 line (the x axis) you will see this intersection has a greater length than the ladder itself.
From: mpalenik on 6 Feb 2010 23:18 On Feb 6, 9:35 pm, Ste <ste_ro...(a)hotmail.com> wrote: > > Trust me, the interpretation you are putting on this is *not > warranted* by the observational evidence.- Hide quoted text - > I deliberately set up a situation that works the opposite of the way the world actually works to avoid having to explain Minkowski spacetime to you. It's not *supposed* to match observational evidence. Once we agree on how to interpret this diagram, I will have you repeat the same exercise in Minkowski spacetime (assuming I can get you to understand what Minkowski spacetime is), where you will see that it *does* match observation. But first, we need to agree on how this diagram works. In this diagram, you do, indeed see length expansion instead of length contraction. See my previous post.
From: mpalenik on 6 Feb 2010 23:30 On Feb 6, 9:35 pm, Ste <ste_ro...(a)hotmail.com> wrote: > > Of course. Because t=0 represents what you're visually observing in > the present. > Since you at least agree with this, I made up another picture: http://s424.photobucket.com/albums/pp327/mpalenik/?action=view¤t=non-minkowski2.gif Remember, this is NOT in Minkowski spacetime, so the results we get will be the opposite of those predicted by relativity. Once we can agree on this, we can move on to repeating this picture in Minkowski spacetime. The gray represents the volume that the ladder traces out in spacetime as it moves. The blue line represents the length of the ladder in the ladder's moving frame. The green line represents the intersection of the moving ladder with t=0. Note that the blue line is LONGER than the green line (if it were the same length or shorter, it would not go outside of the circle I drew).
From: Ste on 6 Feb 2010 23:40 On 7 Feb, 03:43, "Peter Webb" <webbfam...(a)DIESPAMDIEoptusnet.com.au> wrote: > "Ste" <ste_ro...(a)hotmail.com> wrote in message > > news:86583d1c-3ce6-4d27-b43b-8463952c1d02(a)z41g2000yqz.googlegroups.com... > On 6 Feb, 10:29, mpalenik <markpale...(a)gmail.com> wrote: > > > On Feb 6, 5:01 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > On 6 Feb, 09:23, mpalenik <markpale...(a)gmail.com> wrote: > > > > But as I'm saying to you, I think you're confusing an optical effect > > > with a physical effect. That is, you're confusing *appearances* with > > > concrete reality. > > > What part of this picture do you think is optical? It's *geometrical* > > it doesn't have anything to do with what you can visibly see. > > Don't you realise that SR is about the behaviour of *light* - that is, > EMR? And SR describes how *observations* made by way of *light* change > in response to physical circumstances? > > ___________________________ > No. SR says the length is contracted. It is. Doesn't matter if you measure > its length with light waves, neutrinos, or a metre ruler. Have you actually tried measuring it with a meter ruler, and *without* using light? Remember, using your eyes is out of the question - you have to find a way of measuring that does not involve photons. > > > > An important point, however, is that neither one of the sets of axes > > > > (either x,t or x',t') is inherantly better than the other. Neither > > > > set is more correct than the other. > > > > Indeed. > > > How can you agree with that and claim that the differences in > > measurement are optical? > > Because *that* is what SR is all about - it is about describing the > behaviour of *light*. > > _______________________ > No. Its not. It describes the behaviour of any particle or object capable of > carrying information (causality). The *only* thing that SR is *not* required > for is light, as the length transforms for ligh waves were already known to > be the Lorentz transformations from solving Maxwell's equations. "Information" is a poorly defined term, and once again, SR does not describe the effects of anything but EMR, and I believe classical mechanics takes care of all physical reality that is not mediated by EMR. > > The differences in measurement are due to > > the different coordinate systems. It has nothing to do with what you > > *see* it has to do with how you make your measurements. > > It has *everything* to do with what you *see*. > > _______________________________ > What is observed. You don't have to "see it" at all; Einstein's transforms > work in the dark just as well. Do they now? What experimental evidence do you have to show that SR "works" where photons are not involved? I assume you *do* have evidence, don't you? > > When one > > observer measures length, he measures along the x axis. When another > > observer measures length, he measures along the x' axis. The proper > > length of the object doesn't change, but the measured length has > > nothing to do with optics, visibility, or propagation delays. > > Of course the measured length has *everything* to do with optics. How > do you think we usually carry out measurements? > > _________________________________ > Lots of ways. Normally I measure length by puting a ruler next to an object. And by looking at the readings with your eyes no doubt - so we've gone back to a measurement mediated by light? > You put a ruler next to the ladder, you will see it has contracted exactly > per Einstein. But in fact we don't see that. > And why do you think the ladder appears the correct size for an > observer in the x' frame? > > ____________________________ > Without knowing what the x' frame is, I would hazard a guess and say the > ladder is stationary in that frame. Indeed, and are not wondering why the ladder seems just as big when it is measured optically in that frame?
From: Ste on 7 Feb 2010 00:33
On 7 Feb, 03:54, PD <thedraperfam...(a)gmail.com> wrote: > On Feb 6, 8:52 pm, Ste <ste_ro...(a)hotmail.com> wrote: > > > > The only thing that is required is to note at the detector X or the > > > detector Y whether the signals from the events arrive at the same time > > > or at different times. This is a point decision. It is a yes or no > > > question. "Signal from A just arrived at X. Did signal from B arrive > > > at X at the same time? Yes or no." > > > If detection is instantaneous (i.e. if a photon is absorbed > > instantaneously), then it is possible for A and B to be simultaneous > > according to both X and Y. However, if detection is not instantaneous, > > then it is *not* possible. > > I didn't say "according to both X and Y". What I said in fact was the > opposite. Please reread. > What I did say is that X is *right* in concluding that A and B are > simultaneous, based on the procedure we established as reliable. But the procedure isn't reliable! I've said that repeatedly. > Also, Y is *right* in concluding that A and B are not simultaneous, > based on the same reliable procedure. The procedure is not reliable! > But yes, photons are absorbed instantaneously, or at least MUCH faster > than the propagation delay for the signals to arrive. Indeed. > > > If the answer is yes, and if we ALSO know that the distance from X to > > > A is the same as the distance from X to B (which we can check later if > > > we wish), and if we ALSO know that the signal speeds from A and B are > > > the same (which we can check later if we wish), then we KNOW the > > > events A and B were simultaneous, even though they happened some time > > > ago. Likewise, if the answer is no, then we KNOW the events A and B > > > were not simultaneous, even though they happened some time ago. > > > As I say, there is a third way here: the answer is "no", and we know > > that the events were indeed simultaneous. > > How would you derive from the procedure we agreed upon, where the > answer to the relevant condition is "no", that the events were > nevertheless simultaneous? We didn't agree on the procedure. I said from the very beginning that it was not reliable because it fails to take into account the fact that detection is not an instantaneous process > On what basis would you come to that conclusion. Because as I've said, the absorbtion/detection of the photon requires a time interval, and although X and Y can be equidistant from both events *at a single instant*, they cannot both maintain equidistance *over an interval of time*. > > > Do you agree that those are the right conclusions, based on the yes or > > > no question above, and given that the other conditions can be > > > established? > > > No. I think your mistake is in assuming that both the photon and > > detector have an absolutely zero diameter (and therefore detection > > occurs as soon as the surface of the zero-diameter objects touch). In > > reality, nothing in space will have a diameter of zero. > > I don't know why you think diameter has anything to do with it. Note > the size of the distance between A and X and between B and X. If a > detector is 1.5 mm across, do you think this is going to be a dominant > effect? No, I'm talking about the diameter of the photon and the atom. As I say, the visualisation I have is somewhat like two bubbles in water, and clearly if they are forced together so that they become one bubble, that is not an instantaneous process. Certainly, the bubbles do not merge merely at the first instant their (idealised spherical) surfaces touch - they must be actually forced together until their surface tension breaks. Having thought about it for a moment however, I realise that my previous argument might not be wholly relevant or may be speculative (although it embodies some likely factors that may be relevant in the real world). You can theoretically (in particular, in the absence of gravity) bring about absorbtion of photons from both events simultaneously for both A and B, assuming that the photons and the atoms maintain a constant speed as they impact and are absorbed by the atom. However, I've formulated a slightly different argument based on the same principle of detection requiring a time interval. Once detection has taken place, a signal must then be sent that tells the detection- system to check whether the events are an equal amount of distance away. By time *this* signal is being acted upon, the events are no longer equidistant from the observer. As I say though, the ultimate argument with this scenario Paul is that equidistance is not maintained over the measurement interval, and so each detector is subject to non-symmetric factors that influence the measurement. > The signal starts at the location of the > event, where it leaves a mark, and the signal stops at the location of > the detector, where there is also a mark. Then we can certainly > measure the distance between the marks at our leisure, no? Indeed. But real world measurements do not happen this way. |