What's your guys' take on Moving Dimensions Theory? Vs. String Theory / LQG?
in [Relativity]
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From: Androcles on 8 Nov 2009 15:00 "dre" <drelliot(a)gmail.com> wrote in message news:53e8273e-a6e6-4818-8f6a-9c7094a1cfa3(a)m33g2000pri.googlegroups.com... Can anyone refute any of these proofs? "Simple, logical proofs of MDT: MDT PROOF#1: Relativity tells us that a timeless, ageless photon remains in one place in the fourth dimension. ========================================== Photons travel from A to B. Proof by reductio-ad-absurdum. Dead on arrival, drivellot.
From: BURT on 8 Nov 2009 15:05 On Nov 8, 12:00 pm, "Androcles" <Headmas...(a)Hogwarts.physics_p> wrote: > "dre" <drell...(a)gmail.com> wrote in message > > news:53e8273e-a6e6-4818-8f6a-9c7094a1cfa3(a)m33g2000pri.googlegroups.com... > > Can anyone refute any of these proofs? > > "Simple, logical proofs of MDT: > > MDT PROOF#1: Relativity tells us that a timeless, ageless photon > remains in one place in the fourth dimension. > > ========================================== > Photons travel from A to B. > Proof by reductio-ad-absurdum. > Dead on arrival, drivellot. Everything is moving. Everything is flowing. And the aether can be still. Mitch Raemsch
From: cjcountess on 8 Nov 2009 18:02 > MDT PROOF#1: Relativity tells us that a timeless, ageless photon > remains in one place in the fourth dimension. Does relativity suggest this, beccause photons cycle as they move at c through space, and although I agreed that they in a sense surf, to be more accurate as I wass in an earlier post, if we looked at all photons as just surfing, we could say that all photons travel at same speed through space, and that higher frequency photons emerge more rapidly but at the same speed as lower frequency photons, sort of like a machine gun that fires more bullets in same amount of time, but with same constant velocity, and they would have same energy, mass. and momentum. But they actualy cycle as though they were orbiting something, which could be considered a string by analogy, or a constant mass, = to (h=c). And so equation E=hf/c^2 is exactly equal to F=mv/r^2 and F=Mm/r^2 on quantum level. Sense higher cycles per time unit, translate into higher kinetic energy, mass. and momentum, it is as if they were moving faster than photons of lower frequency, and if not in the linier direction, at least in the angular direction, as opposed to being still and surfing Quantum mechanics tells > us that a photon propagates as a spherically-symmetric expanding > wavefront at the velocity of c. Ergo, the fourth dimension must be > expanding relative to the three spatial dimensions at the rate of c, > in a spherically-symmetric manner. How do you reconcil this statement, which seems to contridict the first, with the first? For the first time in the history of relativity, > change has been wedded to the fundamental fabric of spacetime in MDT. This is also not correct but I am not here to nit=pic, still I want to know why you make statement I questioned Conrad J Countess
From: Ahmed Ouahi, Architect on 8 Nov 2009 18:02 Whatsoever, as along that matter, anything is going a just fine, as it would remind something which it remains as a crucial as a primordial matter, which is the time matter, or at least an interval of that time, which it would take to an approipriate combination, along that matter all along... However, that an inteval time, which it would even take an appropriate formula, first of all, which it would be as an interval which it would be an equal matter to a 2pi square root(L/g), as along that matter the L, would be the lenght of an appropriate moving and also the g... Therefore, along that matter, would be an acceleration suitable along the gravity on the shell of the earth, whether already we do know the constant 2pi, would be an equal or more to 6.28, which it should be closed to an appropriate value as the I, something which makes anything to turn along any approximation, all along... For the time being, that any determination along that matter would remain depending on the length L, and the acceleration g, to attempt to a time, which it would be allowed its a possible combination a just along the square root of the L/g, simply as that, and this what is all about, a definitely as a matter a fact... -- Ahmed Ouahi, Architect Best Regards! "cjcountess" <cjcountess(a)yahoo.com> kirjoitti viestiss�:b27003ec-5d27-48e5-8008-a3e48abb8539(a)u20g2000vbq.googlegroups.com... > Ahmed, > If we use the electron mass, time frequency cycle, and length, which > are equal to c^2, according to my own geometrical interpretation of > E=mc^2 = E=mc^circled and are the real natural Planck units which are > also equal to (c^2 = h/2pi = G), you get different results than if you > use (c = h/2pi = G), which gives length of 1.6.10-35m, time of 10-43s, > and mass which matches no known particle in nature. > > So it depends on which Planck units one uses, the ones built on (c,h/ > 2pi,G) which get you this complication or (c^2 = G = h/2pi) for which > we have a natural object as it manifestation which is electron and a > simple conversion factor (c^2 as c^circled = to h/2pi = hx2pi because > at this (1/1=1x1=1)= G) > > If we use the electron mass length and time as basic units I am not > sure what the measurements will be but I bet they would be more > correct that using what are traditionally considered Planck units > which as I said are the equating of (c, h/2pi, G), when it should > simply be (c^2 = h/2pi = G) or simply c^2 the Universe age length and > time will be a multiple of c^2. > > http://graham.main.nc.us/~bhammel/PHYS/planckunits.htm > > http://math.ucr.edu/home/baez/planck/node2.html > > http://abyss.uoregon.edu/~js/glossary/planck_time.html > > http://graham.main.nc.us/~bhammel/PHYS/planckmass.html > > Conrad J Countess
From: cjcountess on 8 Nov 2009 19:24
2pi square root(L/g), as long as (c x 2 pi) = (c^2) as in my geometrical interpretation of (E=mc^2) = (E=mc^circled) = (h x 2pi) = (c x 2 pi). This would make (2pi = c), in this special case, and just as c is sqrt of (c^2), so would (2pi). This would also = (G as L/T^2), as (c^2), is the ultimate (L/T^2), and also energy in circular motion, as (c in liniear direction x c in 90 degree angular direction) creating a (90 degree arc), which if constant creates a circle, and balence of centrifugal and centripital forces. It would also = Einstien and Minkowsky's, (c x sqrt-1) or (ct x sqrt-1) as I see not enough difference to invaldate this equality. (E=mc^2) = (E=mc^circled) = (ct x sqrt -1) and (c = sqrt -1) also (c^2) = (h / 2 pi) = (c / 2pi) for momentum and wavelengh being inversly proportional to it = (h x 2pi) = (c x 2 pi) Conrad J Countess |