dkv Posted August 14, 2008 Author Report Posted August 14, 2008 Sometimes I wonder why the momentum needs to be conserved in virtual interactions. Just as the energy can be violated for a short time,similarly momentum can be violated for a short distances... If two virtual particles come close to each other then they can "gain" extra momentum. Quote
CraigD Posted August 14, 2008 Report Posted August 14, 2008 Obviously I disagree with Craig on this point. in the surface-resting box, the measured acceleration is slightly greater near the floor of the box than near the ceiling, while in the spacecraft-born box, they are the same. Thought experiments along these lines go by Born rigidity or “Bell's spaceship paradox” and it is more completely reveled using Rindler observers in Rindler coordinates. Accelerating a rod evenly in SR requires more force applied to the trailing edge than the leading edge. It’s explained in the link to accelerated coordinates above.I’ve read and understand the resolution of Bell’s spaceship paradox that show that the top of a rocket-propelled box (or, in the usual terms of the paradox, the lead spaceship), and while it does contradict my overly-approximate claim that the acceleration near the floor of the spacecraft-born box is the same as the acceleration near the ceiling, it doesn’t contradict the first part of the claim... one can in principle determine if one is in a box resting on the surface of a planet or in a spaceship accelerating at exactly the planet’s surface acceleration of gravity …as illustrated by the following thought experiment: A box of height h is resting on a platform distance r from the center of a body with nearly all of its mass M concentrated in a small core. The acceleration of gravity at the bottom of the box, then, is [math] a_0= \frac{MG}{r^2}[/math], at the top of the box, [math] a_{h,M,r} = \frac{MG}{(r+h)^2} = \frac{MG}{r^2+2rh+h^2} [/math] Another box of the same height h is resting on a platform distance 2r from the center of a body with nearly all of its mass 4M concentrated in a small core. The acceleration of gravity at the bottom of the box, then, is [math] a_0 = \frac{4MG}{(2r)^2} = \frac{MG}{r^2}[/math], at the top of the box, [math] a_{h,4M,2r}= \frac{4MG}{(2r+h)^2} = \frac{MG}{r^2+rh+(\frac{h}2)^2}[/math] The acceleration [math]a(h)[/math] of a point at a given height [math]h[/math] from the floor of the box (“a point on the rod” in formula (27) of this paper),[math]a(h) = \frac{a_0}{1+\frac{ha_0}{c^2}}[/math], where [math]a_0[/math] is the acceleration of a point on the floor,is has only 2 terms, [math]h[/math] and [math]a_0[/math]. While its possible to chose [math]h[/math], [math]M[/math] and [math]r[/math] such that [math]a(h) = a_{h,M,r}[/math], because [math]a_{h,M,r} \not= a_{h,4M,2r}[/math], this choice won’t work for both boxes-on-a-platform, even though both measure equal accelerations at the floor [math]a_0[/math]. I’ve a suspicion I’m missing some critical “GR length contraction” that would refute my claim, but have not been able to discover such a thing. So, in short, it appears to me that, with the exception of a single [math]h[/math] for a given [math]M[/math] and [math]r[/math] pair, it’s possible in principle to determine if one is in a box resting on the surface of a planet or in a spaceship accelerating at exactly the planet’s surface acceleration of gravity. :shrug: Quote
dkv Posted August 15, 2008 Author Report Posted August 15, 2008 Equivalence principle applies only with respect to a weak Gravitational field. However this is debatable. We consider Equilavence principle to be true for an instant then it becomes possible to treat the instant as unifromly accelerating at all points... What was Einstein's principle of Equivalence?*by JOHN NORTON is a good guide on such matters. pdf can be found on the web. there are many assumptions... Quote
modest Posted August 15, 2008 Report Posted August 15, 2008 While its possible to chose [math]h[/math], [math]M[/math] and [math]r[/math] such that [math]a(h) = a_{h,M,r}[/math], because [math]a_{h,M,r} \not= a_{h,4M,2r}[/math], this choice won’t work for both boxes-on-a-platform, even though both measure equal accelerations at the floor [math]a_0[/math]. I agree. While both functions decrease hyperbolically with h - there is variability with gravity as to the shape of the curve which isn't there with two accelerating observers keeping constant distance. It's notable that when deriving gravitational length contraction or time dilation it is assumed that g is constant over r such as is done here. Which makes sense because it's really the difference in potential between the points that cause those things. In both gravity and a frame of constant acceleration there's difference in potential with h - regardless of the effects of born rigidity. ~modest Quote
dkv Posted August 16, 2008 Author Report Posted August 16, 2008 Suppose an object is kept at distance x then as the light is send to measure the distance of object at x the light penetrates the object some distance y before reflecting back. (gamma rays can pass thru the object)Therefore the actual distance measured becomes (x+y) and the time taken depends on the material (2x/c+ t') where t' is the time taken to cover the distance 2y. Inside the material we dont know what is the speed of light because the atomic transitions take time. Therefore it is obvious that SR deals with very special light which immediately reflects back without interacting with source. Which again shows that speed of light is not constant. (inside the object virtual interactions are also possible because the vaccum near a charged proton is not empty)Quantum mechanics doesnt allow such precise measurements. Had the plank's constant been little more than h it would not have been possible to measure the coordinates at all. Clearly SR is a very special theory which carries no intrinsic reasons to be true.If SR is not true then GR must also be incorrect. In a different thread I have already proved that the SR doesnt reduce to modified Galilean transformations.Thats why I think that SR is an inconsistent theory. Quote
CraigD Posted August 16, 2008 Report Posted August 16, 2008 Suppose an object is kept at distance x then as the light is send to measure the distance of object at x the light penetrates the object some distance y before reflecting back. (gamma rays can pass thru the object)Therefore the actual distance measured becomes (x+y) and the time taken depends on the material (2x/c+ t') where t' is the time taken to cover the distance 2y.This is true even of low energy photons such as in the visible light band. A good demonstration of this are various reflective metals, which, when deposited of hammered into thin films or foils, are nearly transparent.Therefore it is obvious that SR deals with very special light which immediately reflects back without interacting with source.Although many of the thought experiments intended to help people understand special relativity - a well-known one that comes to mind involving reflection is the “light clock” illustration of time dilation – these descriptions must be understood to be idealized describes the behavior of light predicted by the theory. Any physical experiment will involve imprecision such as precisely the position in a metal reflector of the electron shell that actually reflects a specific measured photon. The limited accuracy of a particular physical experiment, however, does not invalidate the predictions of a particular theoretical prediction, be it special relativity or some other. Experimental predictions must account for the inaccuracy of their instruments, whether on the scale of atoms (around [math]10^{-10} \,\mbox{m}[/math]) or the much smaller scale of quantum mechanical uncertainty(around [math]10^{-35} \,\mbox{m}[/math]) .Which again shows that speed of light is not constant.Stating that observational precision is limited – that is, that any given experimental measurement of the speed of light varies – is not equivalent to showing that the speed of light in vacuum c is not constant. An experiment to showing that the c is not constant in some predictable way – such as being different for light of different frequencies - is not hard to design. For example, one could repeat a LLRE with lasers of different frequencies, predicting a difference in the round-trip times. As first page finds of a google search of the terms ‘“lunar laser” wavelength’ shows, over the more than 36 year history of these experiments, this has in fact been done to very high precision (even accounting for the refractive effects of the earths atmosphere) without finding any variation in c. Given the large body of evidence that c is constant, I don’t think the claim that it is not is supported. Quote
dkv Posted August 17, 2008 Author Report Posted August 17, 2008 Lunar laser or no laser the fact remains that speed of light need not be constant. There is a precise physical reason for it. No matter how idealized the theoretical model is the practical physics does not accept it. Measurement changes the state of measured. No matter how much but the change is definitely there.Moreover I think that SR tranformations are wrong. Quote
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