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As a photon leaves a strong gravitational field, it loses energy and redshifts. Is the exchange in potential energy of a photon characterized by energy quanta?

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No. A photon of a given frequency $f$ is exactly one quantum of energy. An electromagnetic wave has a total energy given by $E_\text{total} = \langle N\rangle hf$, where $\langle N\rangle$ is the number of photons in the wave.

When an EM wave escapes from a gravitational potential well (or falls into one), it's the energy of each individual photon that changes; in other words, the unit in which energy is quantized gets smaller or larger. The number of units (number of photons) stays the same.

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This doesn't feel completely satisfactory. One can imagine that interaction between two "gravitationally charged" objects happens by exchange of gravitons in the very same way that interaction between electrically charged objects happens by exchange of photons. –  Marek Feb 18 '11 at 23:43
    
@Marek: hm, okay, good point. In true quantum gravity, I suppose the frequency changes might be quantized. But I don't know that we have a well-accepted way of calculating that quantization, and I think my answer is valid as long as you keep GR classical. –  David Z Feb 18 '11 at 23:46
    
@David: sure, in classical GR it's fine. As for that true quantum gravity: string theory would be a bit overkill here, I think :) Just quantize linearized GR on a Schwarzschild background. –  Marek Feb 18 '11 at 23:55
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what I meant was, is the given frequency of a photon recognized as a value that can take any value such that there is no minimum discrete increment in which the energy of a photon can be gained or lost? –  HAL Feb 19 '11 at 4:51
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@HAL: well, that question is hard to answer. Currently our theories are based on particle exchange and energy can be transferred continuously in this picture (because the intermediate particles can have any energy too). Nevertheless, it might be that for some reason this picture is modified and energy will be discrete at very small scales. In any case, one would need full quantum gravity as David said. –  Marek Feb 19 '11 at 9:52
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