🚀 go-pugleaf

On Nov 20, 5:30 pm, Tom Roberts <tjroberts...@sbcglobal.net> wrote in
sci.physics.relativity:
> bz wrote:
> >TomRoberts<tjroberts...@sbcglobal.net> wrote in
> >news:TFE%i.22137$lD6.20414@newssvr27.news.prodigy.net:
> > bz wrote:
> >>> As I have pointed out to Henri many times, (and perhaps to you),
> >>> 'extinction' requires a suspension of the laws of thermodynamics
> >> This is simply not true. Optical extinction is what makes eyeglasses and
> >> other lenses work.
>
> > I am afraid that my understanding of optical extinction is different from
> > yours. [quote
> >http://www.edpsciences.org/articles/epjd/abs/1999/06/d8252/d8252.html] The
> > optical extinction rate of the particle embedded in an absorbing host is
> > defined as a rate of local energy losses caused by the particle
> > (absorption in the particle volume and scattering by the particle-matrix
> > interface) referenced to the matrix background. [unquote]
>
> > There is nothing about energy gain from the medium.
>
> The difficulty is that there are two different theoretical contexts
> being discussed, with no clear demarcation between them. They are:
>   A) the modern theory of classical electrodynamics, including SR
>   B) some generic emission or ballistic theory involving c+v
>
> Your quote is using (A) but most of my statements were related to (B).
> Some of my statements, like the one above about eyeglasses, apply to
> both (A) and (B).
>
> > The light leaving the medium has LESS energy (some is absorbed and
> > scattered) than the light that entered.
>
> One must carefully distinguish the energy of individual photons from the
> energy (intensity) of the entire light beam.
>
> For theoretical context (A):
>
> The energy of individual photons is proportional to their frequency. As
> photons propagate between mediums their frequency does not change
> (continuity of fields at the boundary), so neither does their energy.
>
> The total energy of a light beam is related to its intensity, and as a
> light beam propagates between mediums some is reflected at the boundary,
> and some is absorbed in the medium, so the total energy of the beam
> decreases.
>
> For theoretical context (B):
>
> It's not clear that photons are appropriate; certainly (B) does not
> include QED (which is the theory that defines photons). So I'll say
> "light packet" in context (B) where I would say "photon" in context (A).
>
> If individual packets of light have energy related to their speed, then
> as light propagates from a more dense medium to a less dense medium, the
> speed of the packets must increase, so some "magic" method must be
> invented to provide such an increase in the energy of the light packets.
> If packets of light have energy proportional to their frequency and
> independent of their speed, then no "magic" is required and the
> situation is the same as for context (A).
>
> In any case the intensity and total energy of the beam is reduced by the
> medium, as in (A).
>
> As the implicit goal of ballistic theories is to relate light to
> Newtonian mechanics, it is not clear how a light packet could have an
> energy independent of its speed, in which case such "magic" is indeed
> required.

The hypnotist advances a red herring, startles, sings "Divine
Einstein" and goes into convulsions, followed by his zombies. Roberts
Roberts since the constancy/inconstancy of the speed of photons is at
stake, your first task is to show how exactly the speed of photons
varies with the gravitational potential. You have already started
this:

Tom Roberts wrote in sci.physics.relativity:
> Pentcho Valev wrote:
> > CAN THE SPEED OF LIGHT EXCEED 300000 km/s IN A GRAVITATIONAL FIELD?
> Sure, depending on the physical conditions of the measurement. It can
> also be less than "300000 km/s" (by which I assume you really mean the
> standard value for c). And this can happen even for an accelerated
> observer in a region without any significant gravitation (e.g. in
> Minkowski spacetime).
> Tom Roberts tjroberts@lucent.com

Roberts Roberts, as the speed of photons varies with the gravitational
potential, does it do so in accordance with Einstein's 1911 equation
c'=c(1+V/c^2) or would you offer another equation? (Note that any
equation should be consistent with the gravitational redshift factor
1+V/c^2 confirmed by Pound and Rebka). Give the correct equation
Roberts Roberts!

Then Roberts Roberts you have a second task: by applying Einstein's
equivalence principle, you will have to convert Einstein's 1911
equation c'=c(1+V/c^2) into c'=c+v, where v is the relative speed of
the light source and the observer, in the absence of a gravitational
field. If c'=c(1+V/c^2) is correct, then c'=c+v is correct and vice
versa. Do you agree Roberts Roberts?

And that would be all Roberts Roberts. These two tasks will
demonstrate that you are, if not the cleverest hypnotist in Einstein
criminal cult (John Norton is cleverer than you), at least the most
honest hypnotist in Einstein criminal cult. Sounds good doesn't it.

Pentcho Valev