FR Bending of Light -- Proof
in [Relativity]
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From: Phil Bouchard on 27 Jan 2010 17:47 Sam Wormley wrote: > > Velocity is defined as dr/dt, Phil! > http://scienceworld.wolfram.com/physics/Velocity.html > http://en.wikipedia.org/wiki/Velocity > > Didn't you study mathematics once, or did you just make > that up! Thanks for the correction, I posted too quickly.
From: Androcles on 27 Jan 2010 17:48 "Phil Bouchard" <phil(a)fornux.com> wrote in message news:4b60c20f$1(a)news.x-privat.org... > Androcles wrote: >> >> Oh look, we are playing Jeopardy! >> "I don't know." - Phil Bouchard. >> How did mechanical energy get into the conversation? >> >> Illogical idiot alert! > > You got me there, I can't answer that! You just saved Einstein! >> Ok, so PE can be negative. >> Now the rock falls and the PE is converted to KE. >> >> What is the KE of 1/2 m_rock * v^2 when v is negative? > > KE = 1/2 m_rock * v^2 > > KE will always be positive. "Negative PE converts to positive KE." - Phil Bouchard. Illogical idiot alert!
From: Androcles on 27 Jan 2010 18:10 "Phil Bouchard" <phil(a)fornux.com> wrote in message news:4b60c2a3$1(a)news.x-privat.org... > Sam Wormley wrote: >> >> Velocity is defined as dr/dt, Phil! >> http://scienceworld.wolfram.com/physics/Velocity.html >> http://en.wikipedia.org/wiki/Velocity >> >> Didn't you study mathematics once, or did you just make >> that up! > > Thanks for the correction, I posted too quickly. Idiot alert! Wormley should state that velocity is defined as upsilon = d(xi)/d(tau) in the "moving frame" and v = dx/dt in the "stationary frame". Didn't he NOT study relativity once, or did he just make that up! (oops... ...up?) Didn't he NOT study English punctuation once, or did he just invent his own?
From: glird on 28 Jan 2010 16:09 On Jan 27, 5:44 pm, Sam Wormley <sworml...(a)gmail.com> wrote: > Velocity is defined as dr/dt, Phil! > http://scienceworld.wolfram.com/physics/Velocity.html I looked there and then at http://scienceworld.wolfram.com/physics/Wavenumber.html and found this: "There are unfortunately two different definitions of the wavenumber. French (1971, p. 214) uses the definition (1) k = 1/ gamma, where gamma is the wavelength. However, as French notes, it is more common in theoretical physics to use the definition (2) k = 2 pi/ gamma". In the quantum of energy in a photon, the length per wave is equal to 2 pi r, where r is the length of a radius of a circle. Let r = 1 cm for now. Then gamma = 2pi cm; and Eq 1 says the wavenumber k = 1/gamma = 1/2pi = .1591549 cm. HOWEVER, if we let r = .0001 cm then Eq 1 says the wavenumber k = 1/gamma = 1/(2pi x .0001) = 1591.5494 cm and Eq 2 says k = 2pi/(2pi x gamma = 2 pi/(2pi x .0001) = 1/.0001) = 10,000 cm. Either way, why should the "wavenumber" be a function of an unknown value of r; but if r is stipulated it remains a constant regardless of how many waves there might be in a given photon? Why should the "wave number" of the 4th wave in a series of 500 be the same as that of the 44th and the 53nd and all of them? As to Sam's "Velocity is defined as dr/dt, Phil!"; Phil had said: "If v = m/s and s is reduced then v will increase." In his equation, m denotes "meters" and s denotes "seconds". In Sam's equation, dr denotes a length and dt an interval of time. Since the unit of length is a meter and the unit of time is a second, v = m/s = meters per second is either identical or equivalent to v = dr/dt = meters per second. glird
From: spudnik on 29 Jan 2010 14:29
there are no photons. see, you say taht r is "the radius of a circle," which seems to be a confusion of the notion of a "plane wave," whose radius is ideally infinite; sibstitute diameter-not-in-the-plane. > Either way, why should the "wavenumber" be a function of an unknown > value of r; but if r is stipulated it remains a constant regardless of > how many waves there might be in a given photon? Why should the "wave > number" of the 4th wave in a series of 500 be the same as that of the > 44th and the 53nd and all of them? thus: I would say that the LHC or its omnipotent caretakeers had to take a few pages out of its flip-book -- just rip them right out & shred, like the Royal Astronomer would try to do to yours, if you were to question his Reality Bumpersticker. > comparable reason in the known physics of space and time to dismiss it > as an illusion? I know of none. The only stimulus is a negative one. --les OEuvres! http://wlym.com |