Henri Wilson wrote:
> On 12 Apr 2006 05:01:24 -0700, "George Dishman" <[email protected]>
> >Henri Wilson wrote:
> >> On 11 Apr 2006 04:53:58 -0700, "George Dishman" <[email protected]>
> >> wrote:
> >> >> > Of course, that's what we have been talking about
> >> >> > all the time. The energy of a _single_ photon is
> >> >> >
> >> >> > E = hv
> >> The correct equation is E = h(c+v)/lambda.
> >No, it is what I wrote, the energy imparted in the
> >phooelectric effect depends on the frequency at
> >the target.
> ....where nu is the arrival rate of wavecrests....
> which = (c+v)/lambda
Wrong, in your terminology c+v is the speed at which
the light was emitted relative to the target but you also
claim "extinction" changes the speed after it has left
the target. What you could say is that the energy in
the frame of the source is E = hc/lambda since the
speed relative to the source is c. That also tells you
how much energy is removed from the source so you
would retain the law for black body radiation which is
what led Planck to the concept in the first place.
> >> Frequency is inferred as the arrival rate of 'wavecrests', (c+v)/lambda
> >For a grating that's true. For a radio receiver it isn't,
> >the receiver responds to the time varying voltage
> >produced by the field at the antenna so we measure
> >frequency and infer wavelength. A 3m signal arriving
> >at 1.5c would produce the same voltage pattern from
> >the antenna as a 2m signal travelling at c.
> A radio signal is a wave created in a 'of photons'. I say it isn't the same as
> a single photon emitted during a charge energy transition.
I don't care what you say, the topic is what ballistic
> >> >> However, ONE photon does NOT have a frequency when considered
> >> >> independently.
> >> >
> >> >Get a clue Jim, the whole point is that light comes in little
> >> >packets, and the energy of a packet is proportional to the
> >> >frequency of that packet. For one photon:
> >> >
> >> > photon energy = Planck's Constant * photon frequency
> >> ..but, George, we say 'frequency' is a function of relative velocity.
> >Jim doesn't think one photon has a frequency _or_ a
> >wavelength, he seems to think wavelength is the
> >distance between them and frequency is the rate
> >at which they arrive.
> Well in a way that is correct. Jim was using the bullet analogy to ilustrate
> wavelength. I say photons have an intrinsic wavelength that opertates in the
> same way that Jim had in mind..
What Jim is saying now is that the distance between the
bullets is wavelength and that photons cannot possibly
have an intrinsic frequency or wavelength. He's off on a
path of his own.
> >> Photon energy as we know it is not intrinsic to the photon.....although it
> >> must
> >> also possess some amount of intrinsic energy to keep itself going.
> >> I have posed the question before, "does doppler shifted light have the same
> >> P.E. effect as non-shifted light of the same incident wavelength?"
> >A simpler way to resolve that is to note that some
> >types of detectors have a lower cutoff where the
> >photon energy is insuffient to create an effect and
> >that cutoff doesn't change if you coat the surface
> >with a material of a different RI. The wavelength
> >changes, the frequency doesn't and the energy
> >of the photon stays the same too.
> Yes , energy appears to be the crucial factor in the PE effect....but if we
> could test it with light from a fast moving source, in which case wavelength
> remains the same, we might learn something.
> I guess the Mossbaeur effect is somewhat similar to this.
I think the problem is that you have assumed you could
take the existing understanding of photons and graft it
into ballistic theory which is a classical wave theory
without too much trouble. It isn't that simple and I think
you need to understand balistic theory first and then see
if you can quantise it. For example saying that E=h.nu at
the source and that then the energy in the source frame
is conserved might be a starting point from which you
could try to work out a relation between energy, frequency
and wavelength at the target. Certainly what you have at the
moment has problems which is why you are having to say
differnt methods of creating EM signals result in different