Re: 1905 Einstein's Relativity&Potential Energy

From: Androcles (Androcles_at_)
Date: 02/20/05 

Date: Sun, 20 Feb 2005 12:57:32 GMT

"Franz Heymann" <notfranz.heymann@btopenworld.com> wrote in message
news:cv9e49$93o$8@sparta.btinternet.com...
>
> "Androcles" <Androcles@ MyPlace.org> wrote in message
> news:VQERd.86127$68.55161@fe1.news.blueyonder.co.uk...
>>
>> <stephen@nomail.com> wrote in message
>> news:cv71pm$1foh$1@msunews.cl.msu.edu...
>> > In sci.physics.relativity Eric Gisse <jowr.pi@gmail.com> wrote:
>> > : Read page 2, where Ritz's theory is described. Referencing
> Ritz's
>> > : particles: "These particles were assumed to leave the charge
> with a
>> > : relative velocity c. After emission they traveled at constant
>> > velocity
>> > : (presumably with respect to any inertial system) whatever the
>> > : subsequent motion of the charges".
>> >
>> > : Oops, that sounds like SR to me. Constant velocity whatever the
>> > : subsequent motion of the charges? I thought you didn't like SR!
>> >
>> > To be fair, that does not sound like SR. To me "constant velocity
>> > whatever the subsequent motion of the charges" sounds like the
>> > particles do not change speed once they leave the charge,
> regardless
>> > if the charge changes its motion later. In other words, in the
>> > reference frame where the charge is at rest when the particle is
>> > emitted,
>> > the particle's speed is c. If the charge changes velocity, the
> charge
>> > will now have a non-zero speed in that reference frame, but the
>> > particle
>> > will still have a speed of c. In the reference frame where the
> charge
>> > is
>> > now at rest, the particle's speed will not be c. That is how I
> would
>> > read those two sentences. Unless there is a whole lot of other
> text
>> > that you snipped, it does not sound like the particle's speed
>> > will be c in all referene frames.
>> >
>> > Stephen
>>
>> Quite correct, Stephen.
>> Langevin's derivation would fail if c = (c+v)/(1+v/c) as Einstein
>> claims.
>> It doesn't fail.
>
> Perhaps I can help Androclown a little by breking the algebra into
> bite sized steps:
>
> Start with
> (c + v)/(1 + v/c)
> multiply top and bottom by c, if you can, to give
> c(c + v)/{c(1 + v/c)}
> mulyiply out the c in the denominator, to give
> c(c + v)/(c + v)
> Now divide top and bottom by (c + v) and you are left with
> c
> And the final summary of our exercise says
> therefore
> (c + v)/(1 + v/c) = c
> And just swapping left and right sides, we get
> c = (c + v)/(1 + v/c)
>
> In other words, the claim, which you said Einstein made, is in fact
> correct.
> Therefore, according to you, Langevin's derivation would fail.
>
Yes, it would, if you used the composition of velocities.
Langevin's derivation uses blue- and red-shift, one photon
moving away faster than the other.

Red doppler shift would fail, even Einstein's doppler.

Doppler's equation:
         c+v
f' = f --------
         c+u

In the absence of a medium such as aether, this reduces to

f' = f * (c+v)/c
   = f* (1+v/c)

Einstein's doppler (for phi = 0, cos(0) = 1) is Doppler's
equation divided by sqrt(1-v^2/c^2), reference
http://www.fourmilab.ch/etexts/einstein/specrel/www/
§ 7. Theory of Doppler's Principle and of Aberration

If we use c = (c+v)/(1-v/c) we then must have, clearly,
f' = f * (c/c) / sqrt( 1-v^2/c^2)
   = f / sqrt( 1-v^2/c^2)
and that is blue shift only, since sqrt(1-v^2/c^2) < 1,
producing no red shift at all.

> Get a grip on yourself.

Even the simplest algebra is way beyond your comprehension, fuckwit.
*** off.

Androcles.

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