Is Length Contraction in SR physical??
in [Relativity]
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From: Ste on 6 Feb 2010 04:43 On 6 Feb, 08:10, mpalenik <markpale...(a)gmail.com> wrote: > On Feb 5, 11:15 pm, Ste <ste_ro...(a)hotmail.com> wrote: > > > > Length contraction must follow for logical consitancy based on other > > > measurements. > > > "Logical consistency" is a far cry from "experimental evidence". > > If you don't think the universe has to be logically consistant, then > the entire framework of physics, and science in general, falls apart. I agree the universe has to be logically consistent - I'm a determinist, after all. The point is that most theories have some element of unspoken assumptions, and so there is a lot of room for "logical conclusion" that is neither central to the theory, nor consistent with reality. I think the most spectacular unspoken assumption is that SR describes "physical reality", as opposed to merely describing the behaviour of EMR. > > > If the speed of light is constant in every reference > > > frame, > > > And it is my contention that it isn't relatively constant, but merely > > *appears* to be so. It is my contention that the speed of light is > > constant only with reference to an absolute frame, and that the time > > dilation effects are due to classical-mechanical effects on the > > velocity of photons. > > This isn't consistant with what is observed. > > First of all: > > Time dilation isn't measured on photons. I don't even know how you > could claim to do that (other than, in a sense, by measuring red > shifts, which has been done, but that's not generally what people mean > when they say time dilation has been measured). On the contrary. I've seen the transverse doppler shift deployed as evidence for "real" time dilation. > It has been measured > with actual clocks, which, after moving, can be shown to register less > time having passed, once they are returned to rest, in a manner > consistant with relativity. It has been measured through the half > life of high energy particles, which changes in a way consistant with > relativity. No, you mean it's been measured with atomic clocks, which incorporate simple counters. But I have a theory that explains that behaviour physically. > Second of all: > > The speed of light has been measured in multiple different ways, from > observing the moons of jupiter to bouncing light beams back and forth > along mountain tops, to interferometry, to frequency and wavelength > measurements. In all cases, it comes out the same. I'm saying 'c' is invariant relative to an absolute frame of reference. Neither the Earth nor Jupiter moves at a significant fraction of the speed of light. Remember, I'm not challenging relativity per se - I'm explaining the physical origins of its effects. > Third of all: > > There are many other observations, such as particle energies, which > obey relativistic physics and are inconsistant with Newtonian > mechanics. I think you'll find there *is* a credible classical mechanical explanation. > > > length contraction must necesessarily follow, as we've > > > described it to you. That's how the original derivation of all of > > > special relativity worked. You apply logic to the two postulates of > > > relativity and see what it necessitates for consistancy. If the speed > > > of light is constant, E=mc^2 logically follows (or rather, E^2 = > > > p^2c^2 + m^2c^4), as does length contraction, time dilation, and the > > > differences is simultanaity. Can you admit this is the case? So > > > while length contraction may not have been experimentally measured > > > directly, what has been measured? > > > But what if, for argument's sake, the speed of light is constant *only > > relative to an absolute frame*? > > First of all, this is demonstrably false based on measurements. No it isn't. The measurements are consistent with my argument. If you have any particular measurements in mind, then let's discuss them. > Second of all, even if you choose not to believe those measurements, > it would still negate the relation E^2 = p^2c^2 + m^2c^4. That > expression is a direct consequence of the consistancy of the speed of > light in every frame. Without it, you return to the Newtonian > expression E = 1/2mv^2. The speed of light *is* constant. But it is constant relative to the absolute frame. And I can explain to you why the only currently observed effect of this is that time appears to slow down.
From: Ste on 6 Feb 2010 05:01 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). > 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. > The dashed lines represent the motion of the ends of the ladder. The > solid line labeled t' represents the motion of the center of the > ladder. > > The moving (ladder's) frame of reference is represented by x' (the > ladder's spacial axis) and t' (the ladder's time axis). If we want to > find the location of any point, we can use either x and t or x' and > t'. Both are equally valid. However, distances measured along the x > axis will not agree with distances measured along the x' axis and > times measured along the t axis will not agree with times measured > along the t' axis. Indeed. But even on this allegedly expansionary diagram, if one measures up from the x-axis, the distance between the points has actually *reduced*, not increased. > > > Two quick comments, though -- which end of the ladder is in the > > > future > > > and which is in the past is the *opposite* of how it would be in > > > Minkowski spacetime. Also, in this diagram, you will see length > > > *expansions*, whereas in Minkowski spacetime, you get length > > > *contractions*. > > > > Please try to actually draw this out. If you get confused, let me > > > know. Once you completely understand this, we can move on to modify > > > it to work in Minkowski spacetime. > > > I'd be interested to learn more about that. > > We have to get things to make sense to you in this first framework > before doing that. It's not required but understanding Minkowski > spacetime is another layer on top of what's required to understand > this picture, so I think it's best to understand this picture first. > > > > > But I will say is that I'm quite sure that this is not inconsistent > > with my views. It is true that, according to SR, the *apparent length* > > at any one *instant* is shorter. But that is *not* a shortening of the > > physical length. SR describes what you *observe*, not what is > > physically happening. > > The proper length of the ladder is unchanged in relativity but the > amount of space the ladder takes up changes because the ladder gets > rotated in time, almost exactly like in the picture I've shown you. > It's exactly like the rotation analogy with fitting the ladder into a > shorter barn by rotating it that many of us here have been trying to > describe to you. 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. > 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.
From: mpalenik on 6 Feb 2010 05:29 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. > > > 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.- Hide quoted text - How can you agree with that and claim that the differences in measurement are optical? 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. 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.
From: mpalenik on 6 Feb 2010 05:32 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'.
From: mpalenik on 6 Feb 2010 05:55
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. 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. 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. |