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Newsgroups: sci.physics.relativity, sci.physics, sci.astro, sci.physics.particle, sci.math
From: glird <gl...@aol.com>
Date: Thu, 5 Nov 2009 17:57:23 -0800 (PST)
Local: Fri, Nov 6 2009 1:57 am
Subject: Re: Call for a Paradigm Shift in Fundamental Physics
On Nov 5, 10:02 am, GSS wrote:
> < I hope you are aware that any shrinkage or a strain field in the region and that all strain fields are subjected to certain physical constraints like continuity of associated displacements and equilibrium of associated stresses. > In his 1904 paper Lorentz showed that the opposite is correct, i.e. > < Consider a steel rod of length L laid along X-axis of a stationary reference frame K. Suppose there are n 'witches' (W1, W2, W3, ..., Wn)flying along the X-axis at uniform velocities of V1, V2, V3,...,Vn respectively. If we assume that the length L of the steel rod will actually become L1 for witch W1, L2 for W2, L3 for W3 and Ln for the witch Wn, will you call it Witchcraft or Relativity? > No; I would call it defective semantics. (The length of a stationary rod won't "actually become" a function of which witch is looking at it; it will only APPEAR to be deformed as measured with the help of esynched clocks of each witch's system. > < Now consider two point A and B fixed on the surface of earth and separated by distance D. Let us position two identical atomic clocks at A and B and ensure their absolute synchronization. When we send a laser pulse from location A to B, we can arrange to record the up-link pulse propagation time Tu from the instantaneous transmission and reception time readout of the atomic clocks A and B respectively. Similarly we can record the down-link time Td for the pulse propagation from B to A. IF we assume that Earth is a stationary system and that A and B are > < As per Relativity, the up-link signal propagation time Tu is SUPPOSED to be equal to the down-link signal propagation time Td in any stationary reference frame when the two clocks A and B are stationary in that reference frame (Tu = Td). > The two times WOULD be equal if the clocks were at rest in any > < But when the two clocks are moving along AB with a common velocity U, the up-link and down-link signal propagation times will no longer be equal (Tu <> Td). > That's right, although it is wrong. (If you find that ambiguous, note that so is your "absolute synchronization". If you use Einstein's method of setting clocks then stationary clocks WOULD measure Tu as identical to Td. But if you esynch clocks of a moving system, i.e. set them to MEASURE the speed of light as identical in any and all directions, they will do so even though the actual times are not equal. > < However, when the two clocks A and B are SIMULTANEOUSLY at rest in the local or Lab "Simultaneously" as measured by who, and how? > frame and in motion in the BCRF and the Galactic reference frames, the up-link and down-link signal propagation times Tu and Td will be required to be simultaneously equal and unequal at the same time. If you mean "at the same instant", then yes; Tu will be equal to Td as plotted by an esynched moving system even though they won't be equal as plotted by a differently moving system. Indeed that is the entire meaning of Einstein's long equation immediately prior to his setting x' "infinitesimally small". > < If you can make two physical measurements Tu and Td to be equal and unequal at the same time, will you call it Relativity or Witchcraft? > Yes. And No. (It depends on what you mean by "at the 'same time'" :-). glird You must Sign in before you can post messages.
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