On 6 Apr 2006 01:35:05 -0700, "George Dishman" <[email protected]>
>Henri Wilson wrote:
>> On 5 Apr 2006 03:30:32 -0700, "George Dishman" <[email protected]>
>> >[email protected] wrote:
>> >> Henri Wilson wrote:
>> >> > On Mon, 3 Apr 2006 23:39:08 +0100, "George Dishman"
>> >> > <[email protected]>
>> >> > wrote:
>> >> >
>> >> > >
snip agreed stuff.
>> >> DO the photons deliver MORE ENERGY?????????
>> >There is another effect called Doppler shift which you
>> >are forgetting which also operates and it does change
>> >the received frequency (colour) of each photon. I've
>> >been trying to teaching you what causes Doppler effect
>> >based on your riflemen for weeks but you insist on
>> >changing the subject all the time.
>> George, Jim is NOT forgetting about doppler shift.
>No? Then why is he now talking about "the number
>of photons arriving at a target per time" as if that
>was the frequency that Doppler affected.
Ah! I'm not sure if that's what he meant or not.
Now this is becoming interesting.
How do you define 'photon frequency'.
>> >The energy of an individual photon is proportional to the
>> >received frequency (whether the source is moving or not)
>> >so the change of frequency caused by Doppler produces
>> >a change of energy.
>> I think Jim knows that George.
>I think he knows most of the pieces but can't put
>them together into a coherent picture.
>> >> (the answer being yes, George is desperate to avoid the comparison)
>> >The answer being that the energy is proportional to
>> >the frequency, Jim is desperate to pretend we hadn't
>> >already addressed the comparison. Here is a reprise:
>> >You suggested the analogy of a burst of gunfire to
>> >represent a photon and I addressed that a few days
>> >ago as I had promised when you finally gave an answer,
>> >the correct one as it happens, to my question about
>> >scenario (3).
>> >As an analogy for a photons of different colour, bursts
>> >of gunfire hitting the target at different rates (photon
>> >frequency) do damage (photon energy) proportional to
>> >the rate, therefore bursts must be of equal duration at
>> >the target.
>> >That means bursts from a high frequency source contain
>> >more bullets than a burst from a low frequency source
>> >and in the analogy it is the increased number of bullets
>> >in the burst that causes the increased amount of damage.
>> >Now where do you want to go with this Jim, or are you
>> >again going to pretend I didn't answer?
>> I've had enough...I'm getting out!
>Sensible move, how about spending a little time
>working out that modulation frequency problem
>instead so that we can test your program.
It is done.