>>>>Sorry to jump into discussion, but there are some absolutely wrong ideas that you borrowed from people not enough educated in physics.
>>>
>>>The person I cited has studied GR far more than you or I. I included references to his lecture notes from the University of Chicago and a link to his book on the subject. I think he's more than adequately educated in physics. You can follow the links at the bottom of this message to verify this.
>>
>>And it is unfortunate to the students he teached.
>>If you consider it is important, I am PhD in Physics and Applied Math
>
>Cool. What fields specifically did you study in physics?

Laser Physics.

>
>
>>>>Redshifting means that the observer sees the light frequency shifting when source of light and observer move RELATIVE to each other. Nothing to do with photon energy, which is still h * C / L, where C is the (absolute) light speed, L is the light wave length, and h is the Plank constant. The very same observer will not see any changes in L.
>>>
>>>That's not exactly correct, AFAIK. The observer will see the wavelength increase. c = fL, so when the frequency (f) drops, since c stays the same, the wavelength (L) gets larger.
>>
>>And you cannot do it here with f, because f is relative measure, but not the actual frequency of the light.
>
>Maybe you can teach me a thing or two about this.
>
>I don't "get" the frequency of photons. I "get" the frequency of radiation, it is the rate over time in which a wave reaches you between wavelengths.
>
>But that would imply that more than one wave-packet is required.
>
>So how is there an "actual frequency" of a single photon?
>

If you consider waves, then it is understandable what is frequency or wavelength. From our everyday life: you can feel waves while swimming if their length is bigger than the size of your body. The huge ship cannot "feel" small waves. Similar to that, if you are going to measure the property of the light: if the sensor size is less than the wavelength, than you can measure the wave properties, otherwise you need to consider particles.

>I had just assumed that there really isn't, and that the frequency of photon is really just a measure of the energy being radiated from a distant electromagnetic interaction.

Yes, if you do not consider quantities measured in the observer's frame, but in the absolute frame.

>
>In that case, how is the relative measure different than the actual frequency?
>

Depends on what do you measure. Let us, say, for Doppler Radar instruments that emits light to the moving object and catches the reflected light, you may use the formula:

f_change = 2 * V / L

where L is the wavelength of light source, and v is the speed of object relative to the instrument. F_change is positive or negative depending in what direction the object is moving.


>
>>>I've been discussing these ideas in moderated usenet groups for a bit now. According to most people, in expansion the wavelength DOES increase, contrary to what you've claimed above:
>>>
>>>"Note that a redshift IS an increase in wavelength."
>>>http://groups-beta.google.com/group/sci.physics.research/msg/bbb44a7740ff43e0
>>
>>I am not surprised having in mind the quality of math and physics education system in USA.
>
>:-)
>
>I'm a bit more cynical in that the state of physics at the moment (relativity + Big Bang + copenhagen interpretation of QM) has way too much wrong to have a coherent picture.

And it is OK about "coherent picture", because the more we learn the more we understand where our ideas/theories are applicable, and where they are not.
Therefore I would say that the state of physics is approaching new hights, but the level of physics (school) education in USA is not comparable with what I know in Russia and/or in Europe.