Re: single photon through glass
- From: ajiko <ajiko2004@xxxxxxxxx>
- Date: Wed, 2 Jul 2008 02:01:26 +0000 (UTC)
"Dirk Bruere at NeoPax" <dirk.bruere@xxxxxxxxx> wrote in message
news:6csf6qF3ilvjfU2@xxxxxxxxxxxxxxxxxxxxx
ajiko wrote:I would say the default is that an event is not localized. We can localize
"Dirk Bruere at NeoPax" <dirk.bruere@xxxxxxxxx> wrote in message
news:6cn68gF3h05caU1@xxxxxxxxxxxxxxxxxxxxx
ajiko wrote:
Thanks. Forgot about that Feynman book. I have it and just re-read it.So the photon interacting with the glass atoms does not constitute a
The photon takes all paths through the glass. Some of the paths
include one
or more scattering. Each scattering results in a 90 degree phase shift.
When
the paths are merged with the phase shifts included, the effect is
equivalent to
a slower velocity.
It was the inclusion of the scatterings in the merging process that is
confusing.
I thought that event type things were supposed to make the field
collapse
and the process start over.
measurement that would localise the photon?
--
Dirk
That's a core point for looking at this. I'm still not sure. Feynman'
setup uses a monochromatic source. His other examples were almost
always single photons. Not the case for the glass
reflection/transmission example. There was a reflection analysis
(grating) that required the originating
time at the source for the photon to vary. This implies that multiple
source photons went into the calculation for a single destination
photon. The transmission analysis is getting more confusing each time I
review
it. I'm wanting to build the correct model of QED equations. I feel I'm
almost ready to get started. It feels like a photon scattering in glass
is:
1) an absorption by a glass molecule
2) a glass molecule in an excited state for a time
3) a re-emission at the same energy
The glass molecule holds the excited state for a quarter cycle. I'm
wondering what determines the duration of an excited state. It seems it
must be associated with the ability to complete another interaction. The
interaction is virtual in the sense that it gets merged with all
variations. The only measured interaction is the initial emission to
final absoption. I would say no. The photon is never localised within
the glass.
That would appear to be obvious, but *why*?
And how many atoms might one expect the photon to interact with (whatever
that means)? Think fibre optic.
--
Dirk
http://www.transcendence.me.uk/ - Transcendence UK
Remote Viewing classes in London
the initial emission and the final detection only because the devices have
locations. Everything inbetween...
Feynman uses the number 0.2 as the amplitude for an interaction.
0.2^2 as the probability. I think that Feynman left off all the cascading
variations to keep it simple. I think every path gets two variations with
every molecule it encounters - a scattering and a no scattering -
with each scattering variation actually a multitude of variations. One
variation for each scattering direction.
To keep it "simple", can we assume that all the side variations cancel out
and we can calculate from only the straight forward and straight
backward variations? They usually are assumed to cancel, but yet again
there is another "why?".
--Ned
.
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- From: Dirk Bruere at NeoPax
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