Any coordinate system in GR?
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From: Sorcerer on 29 Aug 2006 11:12 "Daryl McCullough" <stevendaryl3016(a)yahoo.com> wrote in message news:ed1k7a01hib(a)drn.newsguy.com... | Koobee Wublee says... | | [stuff deleted] No, he didn't say that at all, you did. You relativists are so shitheaded you are compelled to snip what you have no answer for. Androcles
From: JanPB on 29 Aug 2006 11:43 Sorcerer wrote: > "Koobee Wublee" <koobee.wublee(a)gmail.com> wrote in message > news:1156826811.550173.60410(a)b28g2000cwb.googlegroups.com... > | After a few days of fastening these cows, it is time for slaughter. > > ROFL! > According to Bielawski, Relativity works because > > > An error in Relativity "would be like Stephen Hawking dividing by zero or > something equally trivial." -- Bielawski. > > It's WAY too simple-minded.-- Bielawski. > > "would have been caught immediately by the AdP reviewer." -- Bielawski. > > It is boring to find out how far Bielawski can be pushed into > irrelevancies and nonsense. Just to keep it incontext - I meant specifically mathematical errors in special relativity (i.e., the type of errors you complained about). These errors are out. -- Jan Bielawski
From: JanPB on 29 Aug 2006 11:50 Sorcerer wrote: > "Daryl McCullough" <stevendaryl3016(a)yahoo.com> wrote in message > news:ed1k7a01hib(a)drn.newsguy.com... > | Koobee Wublee says... > | > | [stuff deleted] > > > No, he didn't say that at all, you did. > > You relativists are so shitheaded you are compelled > to snip what you have no answer for. Why not scroll up the screen then? :-) -- Jan Bielawski
From: Sorcerer on 29 Aug 2006 12:12 "JanPB" <filmart(a)gmail.com> wrote in message news:1156866628.917737.216100(a)m73g2000cwd.googlegroups.com... | Sorcerer wrote: | > "Daryl McCullough" <stevendaryl3016(a)yahoo.com> wrote in message | > news:ed1k7a01hib(a)drn.newsguy.com... | > | Koobee Wublee says... | > | | > | [stuff deleted] | > | > | > No, he didn't say that at all, you did. | > | > You relativists are so shitheaded you are compelled | > to snip what you have no answer for. | | Why not scroll up the screen then? :-) | | -- | Jan Bielawski Such trite comment displays your troll nature and fanatical lunatic belief in division by zero. http://www.androcles01.pwp.blueyonder.co.uk/DominoEffect.GIF Any so-called "metric" containing a reversing velocity has to be the product of lunacy, all you guys are debating is how many angels can dance on the head of a pin. Androcles.
From: carlip-nospam on 29 Aug 2006 13:56
In sci.physics Koobee Wublee <koobee.wublee(a)gmail.com> wrote: > These two gentlemen fail to understand what the metric actually > represents. It is thus necessary to explain what the metric really > represents. Given the ubiquitously familiar Schwarzschild metric as > described in the very second equation, dt, dr, and dO (longitudinal > and latitudinal information) are the observed quantities. No, they aren't. As a simple example, compare the Schwarzschild metric in isotropic coordinates and Schwarzschild coordinates (see http://relativity.livingreviews.org/open?pubNo=lrr-2000-5&page=node9.html for the explicit expressions). Both describe the same spacetime, and both have a radial coordinate called r. But at a given point, the numerical value of r_Schwazrschild will be different from the value of r_isotropic. Clearly they can't both be "the observed quantities." So how do you propose to decide which one is "right"? (Note: an answer of the form, "I saw the Schwarzschild form first" is not passing.) For a *physical* example of the difference, see Bodenner and Will, "Deflection of light to second order: A tool for illustrating principles of general relativity," American Journal of Physics, Vol. 71, No. 8, pp. 770รข??773, August 2003. You will learn that the coordinate values for the deflection of light by the Sun differ depending on which "r" you use, and that the correct result can only be obtained by using "measurable, coordinate-independent quantities." [...] > In the meantime, the Voodoo mathematical minds still need to answer and > address the following. > "Koobee Wublee" <koobee.wublee(a)gmail.com> wrote in message > news:1156798554.868899.215910(a)m73g2000cwd.googlegroups.com... >> With the following spacetime, >> ds^2 = c^2 dt^2 / (1 + r / K) - (1 + r / K) (K^4 / r^4) dr^2 - (K^4 / >> r^2) (1 + r / K)^2 dO^2 >> How do you specify a distance 2 AU from the center of the sun? First of all, you can't use this metric to answer that question -- the Sun is not a vacuum, and the metric you've written down is only valid in empty space. To answer the question as stated, you need first to give an interior metric that "glues" on to the exterior metric you've written down. The answer will depend on the details of that interior metric. You can, however, answer the question, "How do you specify a distance 2 AU from the edge of the Sun." To answer that, you need to take the interval ds at fixed time and fixed angle (so dt=0 and dO=0) and integrate ds from the edge of the Sun to an arbitrary value r. That integral will give you the physical ("proper") distance D, as measured by an observer at rest. Inverting the answer, you can find r as a function of D. Take that expression and set D = 2 AU. This is fairly basic general relativity. It's true that beginners sometimes get coordinate differences and physical quantities mixed up, but if you take a decent course in the subject, or actually read a decent book (and work through the exercises, etc.), you should be able to correct that kind of mistake pretty quickly. (For example, see the discussion on page 191 of Weinberg's text about the problem of whether a prediction "really refers to an objective physical measurement or whether it has folded into it arbitrary subjective elements dependent on our choice of coordinate system"; or, for the meaning of Schwarzschild coordinates, Wald's careful discussion in section 6.1 of his text. And by all means, read the paper by Bodenner and Will that I cited above, which is about almost precisely this issue.) Steve Carlip |