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Re: A question about the translated "On the Electrodynamics of Moving Bodies"
Posted:
Mar 8, 2013 1:55 AM


On Mar 7, 11:52 am, "Paul B. Andersen" wrote: > Koobee Wublee wrote:
> > It is very necessary to understand exactly what variable means. > > Let?s reexamine the temporal transformation of the Lorentz > > transform. As the usual, any transform (Galilean, the Voigt, > > Larmor?s, or the Lorentz) is a tale of 3 points where Point #1 > > and Point #2 are observing Point #3. > > > ** dt_1 = (dt_2 + [B_12] * d[s_23] / c) / sqrt(1 ? B_12^2) > > > Or > > > ** dt_1 = (dt_2  [B_21] * d[s_23] / c) / sqrt(1 ? B_21^2) > > > Or > > > ** dt_2 = (dt_1 + [B_21] * d[s_13] / c) / sqrt(1 ? B_21^2) > > > Or > > > ** dt_2 = (dt_1  [B_12] * d[s_13] / c) / sqrt(1 ? B_12^2) > > > Where > > > ** dt_1 = Time at #3 as observed by #1 > > ** dt_2 = Time at #3 as observed by #2 > > ** [s_13] = Displacement vector from #1 to #3 > > ** [s_23] = Displacement vector from #2 to #3 > > ** [B_12] c = Velocity of #2 as observed by #1 > > ** [B_21] c = Velocity of #1 as observed by #2 > > ** [] * [] = Dot product of two vectors > > > When the direction of travel of either #1 or #2 is in parallel > > with the observed displacement segment, the above simplifies > > into the following familiar form. > > > ** dt? = (dt ? v dx / c^2) / sqrt(1 ? v^2 / c^2) > > > Where > > > ** dt? = dt_1 > > ** dt = dt_2 > > ** v^2 = B_21^2 c^2 > > ** dx = d[s_23] > > ** [B] * d[s] = sqrt(B^2 ds^2), [B] and d[s] in parallel > > > Koobee Wublee wants the discussion to stay in the form first > > mentioned since that form is more difficult for one to play > > mathemagic tricks and try to pull a fast one through the > > humanity. > > > Then, assuming the frequency is just the inverse of the time > > duration, one can then write down the relativistic Doppler > > shift as the following. > > > ** f_1 = f_2 sqrt(1 ? B_12^2) / (1 + [B_12] * [B_23]) > > > Where > > > ** f_1 = 1 / dt_1 > > ** f_2 = 1 / dt_2 > > ** [B_23] c = d[s_23] / dt_2 > > > When Point #3 is light itself, B_23^2 = 1, and ([] * []) > > becomes your cosine thing. Well, the equation above is > > obviously wrong since it predicts the exact opposite from > > the classical Doppler effect. Also, if you attempt to > > derive the Doppler effect from the Galilean transform using > > this (1/dt) thing, you will get no Doppler effect. To derive > > the classical Doppler effect, you must hold the wavelength > > invariant. > > > So, how did Einstein the nitwit, the plagiarist, and the > > liar derive the energy transformation from the Lorentz > > transform? If you don?t know, you have no right to toss > > the equation around since you have no way of controlling > > which parameter means what. <shrug> > > > Hint: Using the Lagrangian method, the following can be > > derived in which all are equivalent. <shrug> > > > ** f_1 = f_2 (1 + [B_12] * [B_23]) / sqrt(1 ? B_12^2) > > > Or > > > ** f_1 = f_2 (1  [B_21] * [B_23]) / sqrt(1 ? B_21^2) > > > Or > > > ** f_1 = f_2 sqrt(1 ? B_21^2) / (1 + [B_21] * [B_13]) > > > Or > > > ** f_1 = f_2 sqrt(1 ? B_12^2) / (1  [B_12] * [B_13]) > > Have a look at this: > http://www.gethome.no/paulba/pdf/AberrationDoppler.pdf
Yes, seen that before. Give Koobee Wublee a break. Transformation of waves? :) Can you be more professional about this? Koobee Wublee is asking you if you can derive the energy transformation for Special Relativity. <shrug>
By the way, it is Koobee Wublee who is asking you this. Why are you avoiding Koobee Wublee and tried to reply to a person who did not raise the issue? What are you scared of? Oh, the past experience dealing with Koobee Wublee still brings back nightmares. :)
> A calculation of aberration and Doppler shift between > two arbitrary frames of reference, neither of which > has to be the source frame.
Indeed, paul. What took you so long to realize the Doppler thing is basically the aberration stuff? If you don?t understand the aberration stuff is the same as the Doppler thing, how can you accuse someone else of confusing between aberration and parallax? <shrug>
> But if you wish to compare it to Einstein's derivation, > the frame called K would be the source frame.
Have you not read that paper of Einstein the nitwit, the plagiarist, and the liar? The nitwit did not offer any derivation if you have noticed. He must have plagiarized from someone else. Don?t you agree? <shrug>
Oh, by the way. You must have exhausted all the attempts to fudge a convergence in time elapses for that scenario where both twins travel with the exact same acceleration profile. Isn?t it about time to be a man of tall statue and come out to admit the Lorentz transform is just wrong? <shrug>
Koobee Wublee knows you are reading his posts. So, please reply. And we will have fun just like the old times, no? :)



