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:(adsbygoogle = window.adsbygoogle || []).push({});

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.

- First, "the
energy carried by a photon is related to it's frequency"^{http://en.wikipedia.org/wiki/Photon#Physical_properties"}

- 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.
- 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.
- We can derive mass of a photon from the following:

- Photons are have m
_{0}of zero and speed c- Relativistically, momentum p = m
_{0}[itex]\gamma[/itex]c and energy E = m_{T}c^{2}- m
_{0}[itex]\gamma[/itex] is total mass m_{T}, which is m_{rel}+ m_{0}. 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, orE / c^{2}= m_{T}= m_{rel}- And gravity well can be approximated as F = G m
_{T}M / r^{2}- 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.- and recall m
_{T}= E c^{-2}- We're interested in red shift, and thus frequency. E = h [itex]\omega[/itex] / (2[itex]\pi[/itex]), m = h/(2[itex]\pi[/itex]c
^{2}) [itex]\omega[/itex], dm/dt = h/(2[itex]\pi[/itex]c^{2}) d[itex]\omega[/itex]/dt- combining: F = G m
_{T}M / r^{2}->

([STRIKE]c[/STRIKE])( [STRIKE]h/(2[itex]\pi[/itex][/STRIKE][STRIKE]c^{2}[/STRIKE]) d[itex]\omega[/itex]/[STRIKE]dt[/STRIKE] ) ([STRIKE]dt[/STRIKE]/dr [STRIKE]c[/STRIKE]) = (G M / r^{2}) ([STRIKE]h/(2[itex]\pi[/itex][/STRIKE]c^{2}) [itex]\omega[/itex]) ->

d[itex]\omega[/itex]/dr - [itex]\omega[/itex] GMr^{-2}c^{-2}= 0- Solving the differential:
[itex]\omega[/itex] = [itex]\omega[/itex]_{0}e^{GM/c2(1/r-1/r0)}

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

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# Understanding Red Shift

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