Understanding Red Shift: A Derivation from Classical and Relativistic Physics

[FireStorm000]

This is another one of those concepts where I understand what happens at the macroscopic level, but don't understand the why, or rather what is happening at the fundamental level. Now my understanding of QM has improved a fair amount in the last week thanks to some help from ZapperZ, but I'm still having trouble moving from a relativistic understanding of the photon to a QM one. Below is a derivation of redshift based on some classical and relativistic physics:

1. First, "the energy carried by a photon is related to it's frequency"^{http://en.wikipedia.org/wiki/Photon#Physical_properties"}
   
   
2. Second, energy is conserved. What this means, at least to my understand is that the photon must loose energy as it moves out of the potential well created by a massive object.
3. From a classical perspective, gravity acts on an object's total mass? (Not sure about this one). Total mass includes relativistic mass, so if true, gravity acts on photons.
4. We can derive mass of a photon from the following:
   • Photons are have m₀ of zero and speed c
   • Relativistically, momentum p = m₀\gammac and energy E = m_Tc²
   • m₀\gamma is total mass m_T, which is m_rel + m₀. That gives us the substitution m_T = m_rel for the photon due to the identity property of addition.
   • a little algebra on the definitions of relativistic p and E gives E / c = p
   • the above means that, for photons, energy is proportional to momentum, rather than the classical relation.
   • E / c = m_rel c, or E / c² = m_T = m_rel
5. And gravity well can be approximated as F = G m_T M / r²
6. realizing that force is the change in momentum F = dp/dt; Recall that v = dr/dt, dt/dr c = 1. dp/dt = d/dt(mc) = dm/dt c.
7. and recall m_T = E c^-2
8. We're interested in red shift, and thus frequency. E = h \omega / (2\pi), m = h/(2\pic²) \omega, dm/dt = h/(2\pic²) d\omega/dt
9. combining: F = G m_T M / r² ->
   ([STRIKE]c[/STRIKE])( [STRIKE]h/(2\pi[/STRIKE][STRIKE]c²[/STRIKE]) d\omega/[STRIKE]dt[/STRIKE] ) ([STRIKE]dt[/STRIKE]/dr [STRIKE]c[/STRIKE]) = (G M / r²) ([STRIKE]h/(2\pi[/STRIKE]c²) \omega) ->
   d\omega/dr - \omega GMr^-2c^-2 = 0
10. Solving the differential: \omega = \omega₀e^{GM/c2(1/r-1/r₀)}

As per that equation, a photon leaving the surface of the sun and traveling to 1au is red-shifted to .9999978905 of its original frequency, or a change of -2.1095E-4% of the original. One could also see that a photon emitted from a singularity is red-shifted to a frequency of zero. On the extreme, a photon from the accretion disk of a black hole of 1000 solar masses from 1 Earth radius moving to infinity red-shifts to .791386 or about a 21% decrease of its original wavelength. A photon falling into Earth's well blue-shifts 7.058E-8% of the initial frequency, an (I should think) barley detectable amount.

So, are my results correct, and how would one do the equivalent from the quantum perspective, using the wave-function model for light?

 

[Naty1]

Um, I forgot the question.

But the first five or siz equations seem fine...did not look further...

I've never seen a quantum physics explanation for redshift...a quick search did not turn up anything interesting:

A special relativistic redshift formula (and its classical approximation) can be used to calculate the redshift of a nearby object when spacetime is flat. However, many cases such as black holes and Big Bang cosmology require that redshifts be calculated using general relativity. Special relativistic, gravitational, and cosmological redshifts can be understood under the umbrella of frame transformation laws. There exist other physical processes that can lead to a shift in the frequency of electromagnetic radiation, including scattering and optical effects; however, the resulting changes are distinguishable from true redshift and not generally referred as such (see section on physical optics and radiative transfer).

http://en.wikipedia.org/wiki/Red_shift

 

[Ken G]

The way I would put it is, nothing is happening to the photon at all, so there's no "micro physics" to understand there in the first place. Instead, something is happening to the observer, or there is some difference in the environment of the observer relative to an observer where the photon was emitted, and that difference is causing them to interpret observations of the photon differently. That pretty much holds for all relativistic effects. Remember also that "causes" in physics are rarely unique, they often depend on the point of view taken in answering the question. (For example, what is the microphysics that causes two charges to attract? Or is the observation that they attract the cause of our having dreamt up the microphysics?)