Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

Topic: A question about the translated "On the Electrodynamics of Moving Bodies"
Replies: 3   Last Post: Mar 13, 2013 11:16 PM

 Messages: [ Previous | Next ]
 Koobee Wublee Posts: 1,417 Registered: 2/21/06
Re: A question about the translated "On the Electrodynamics of Moving Bodies"
Posted: Mar 13, 2013 2:01 AM

Paul,

So, you think you understand gravitational red shift, eh? Here is the
issue. As the ever-so-humble Koobee Wublee who has managed to kick
Paul?s ass so hard in these recent encounters understands, the
frequency transformation (please, be more professional and no more
wave transforms, OK?) is not a simple tale of (1/dt) according to the
Lorentz transform. According to the spacetime geometry described by
the Schwarzschild metric, the only way to predict a gravitational red
resolve this inconsistency. <shrug>

In the meantime, since Paul has refused to publish the results
concerning the twins? paradox where both twins do travel away and
reunite with the same acceleration profile, the following definitely
would dash all hopes among the Einstein Dingleberries in trusting and
believing in a divine resolution to the paradox itself. As Paul has
demonstrated, it is a piece of cake to fudge the results in the
asymmetrical case where only one twin travels (experiencing all the
accelerations) away. The symmetrical case of the traveling twins
proves to be much more elusive for the Einstein Dingleberries to
brainstorm through. Naturally, Paul has found it to be a little bit
challenging to fudge the results. That is why he remains ever so
impotent when confronted with such excited revelation. <shrug>

The twins twins Paul Draper and Paul Andersen leave the earth at the
same time with instruction telling each when to stop accelerating and
when to start decelerating/accelerating to eventually reunite between
Paul^2 --- the exact same acceleration profile. Given an arbitrary
time period where there is no acceleration between these two buffoons,
the mutual time dilation should relentlessly build up according to the
Lorentz transform. Since the time period of the building up of mutual
time dilation is arbitrary, there is no possible brainstorming that
can fudge the results towards the mathemaGical realm of resolving the

<CHECKMATE>

Koobee Wublee will take lack of an answer as so. So, happy
brainstorming, Paul, and don?t forget to chase after the chickens in
your neck of the woods. :-)

Ahahahaha...,
Koobee Wublee

Reference:

On Mar 7, 12:20 am, Koobee Wublee <koobee.wub...@gmail.com> wrote:
> On Mar 6, 10:54 am, "Paul B. Andersen" <some...@somewhere.no> wrote:
>

> > Remember, phi is the angle observed in the source frame,
> > so you have to put yourself at the source.
> > So what do you see?
> > Lets use the compass.

>
> > S --------- 90
> > \ phi
> > \
> > \
> > \
> > O -> v
> > | \
> > 180 160

>
> It is very necessary to understand exactly what variable means. Let?s
> re-examine 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])

Date Subject Author
3/13/13 Koobee Wublee
3/13/13 Brian Q. Hutchings