On 04/01/2010 01:12 PM, Roarty, Francis X wrote: > Stephan, > > Your reply actually answers my reply to Stevens answer. You > are saying that I will never be able to test this hypothesis because the > axis of contraction is never going to present itself to the stationary > observer – I liked your “silhouette” analogy. You mentioned this concept > of being “real” or just a trick of light is controversial. So I guess I > am just piling speculation on top of controversy when I raise the > question of stationary objects on the spatial axis where time is > occurring at different rates due to equivalence. We know time is slower > a the bottom of a gravity well due to mass while to achieve an > equivalent slowing of time without a gravity well takes high velocity on > the luminal scale like my basketball question. This high velocity > reflects the Pythagorean relationship between time and space from which > they derived Gamma (where the Y axis is a constant 300m/s ?). My > question is therefore how does space contract for “equivalent” > acceleration vs spatial acceleration? Does the gravitational vector > define the axis that contract? Would these “pancaked” basketballs > sitting in a deep gravity well be stackable such that millions of > basketballs would now fit where only a few “un-contracted” balls should fit?
The situation in a gravity well is very different from the situation where there are two inertial observes in motion with respect to each other. I'm really rusty on this (and I was never anything like an expert) so I can't give clear, detailed answers, but the following may help a little. In the case of motion-induced Fitzgerald contraction, the effect is, as I said, arguably a "trick of the light" (but it seems pretty real to me, at least in some cases). Space itself is notably not affected; only the measurements seem to change. In the case of a gravity well, space itself is unambiguously distorted. The metric tensor changes, and to deal with this you need to dive into general relativity and pseudo Riemannian geometry. To deal with the gravitational field around a planet, you can look at the Schwarzschild metric (Wiki has a page on it, for example), which will give you some feel for how the distortion "looks". But there is a very important point here, which is that there are several kinds of distortion, and they can happen together or separately. -- The "curvature" is nonzero in the presence of a gravitational field which has tidal effects. This results when there are nonzero second derivatives of the components of the metric tensor. Nearly all real gravitational fields have tidal effects. The Schwarzschild metric, for instance, is curved. -- The "connection" is nonzero when there's a gravitational gradient -- i.e., when there's a direction things fall in. This is a result of nonzero values for the the first derivatives of the components of the metric tensor. This is associated with distortion along one axis, I think, and is somewhat analogous to Fitzgerald contraction. However, note well this this has very little to do with time dilation! -- Finally, there's gravitational time dilation, which is correlated with the gravitational *potential*, *not* with the local gravitational field strength! For example, if we dig a spherical chamber in the center of a planet, there will be *no* gravitational "field" within that chamber caused by the mass of the planet. However, the gravitational potential is lower in that chamber than it is on the surface, and clocks in the chamber will run SLOWER than clocks on the surface. As another example, if we dig a spherical chamber OFF CENTER inside a *uniformly* *dense* planet, the gravitational field within that chamber caused by the mass of the planet will be almost exactly uniform. In other words, there will be no tidal effects, and it will be impossible for an observer within the chamber to determine whether he's being subjected to a gravitational field, or is simply accelerating at a uniform rate. Clocks will runs slower at the "bottom" of the chamber than at the "top". > > Regards > > Fran > > <http://www.mail-archive.com/> > > > *Re: [Vo]:checking my understanding of Lorentz contraction* > > Stephen A. Lawrence > Thu, 01 Apr 2010 06:40:36 -0700 > > > > On 03/31/2010 11:52 PM, Francis X Roarty wrote: > >> Am I correct in believing a near luminal basketball could pass through > >> the eye of a stationary needle? > >> > > > > No. The basketball is contracted fore-and-aft, but not side-to-side, as > > viewed by an observer sitting next to the needle. So, it's going to be > > too wide to fit through the needle's eye, even though it may be > > *thinner* than the needle's diameter. > > > > As Steven Johnson already said, the basketball "pancakes", but the > > pancake is flying along with one flat side to the front, so it's "splat > > time" when it hits the needle. You could also say the basketball has > > been replaced with its silhouette. > > > > It's the fact that there's no side to side contraction which leads to > > all the arguments over whether the contraction is "real". The > > fore-and-aft contraction is arguably just a "trick of the light". > > > > The fore-and-aft contraction seems to be mostly an artifact of the fact > > that time is skewed between the reference frames, and in order to > > determine how "long" something is, you need to find the locations of the > > front and back of the object "simultaneously". The definition of > > "simultaneous" turns out to be frame dependent, which is how an observer > > on the basketball and one sitting by the needle end up disagreeing about > > whether the basketball is contracted or not. > > > > Spinning disks, on the other hand, cast a rather different "light" on > > the matter, as a spinning disk which can't stretch will, in principle, > > crack as it spins up, due to the contraction of the rim. That makes the > > contraction seem rather "real". A really long train also cannot > > accelerate at more than a certain rate without breaking apart, as it > > turns out the cars at the back of the train must move faster than the > > ones in the front due to the train's contraction as it speeds up. At > > some point they'd have to break the lightspeed barrier to avoid being > > left behind by the shrinking train. That, also, makes the contraction > > seem rather "real". > > > > >
