Redshifted photon emission vs redshifted Sun-Earth photon transport

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• Mickey1
Mickey1
TL;DR Summary
Is a photon from the Sun towards Earth red-shifted already at emission?
I am considering the magnitude of the gravitational redshift and I look at the process of a photon leaving an atom from the Sun. I am asking whether the processes in the atom, viewed as a clock, would lead us to conclude that the emitted photon, at the time of emission, would itself be red-shifted, and then whether the journey out of the Sun´s gravitational field would redshift it once more, or alternatively if these two effects are one and the same?

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Doppler shift is a phenomenon to do with the relation between the states of motion of the source and receiver (and the light path, in general). A light pulse is neither redshifted nor blueshifted in of itself. So your question is incomplete. You need to specify who is measuring its frequency and how.

That said, your actual question is straightforward. Gravitational redshift and gravitational time dilation are one and the same thing. A local measurement of the light pulse frequency near its point of emission will show it to have a frequency matching that of the atomic transition that created it. A distant observer watching through a telescope would conclude that the atom, instrumentation, and light pulse were all time dilated by the same factor.

vanhees71, lomidrevo and PeroK
Mentor
I am asking whether the processes in the atom, viewed as a clock, would lead us to conclude that the emitted photon, at the time of emission, would itself be red-shifted, and then whether the journey out of the Sun´s gravitational field would redshift it once more, or alternatively if these two effects are one and the same?

"Redshift" is not a property of a photon by itself. It is a relationship between the photon and the device that receives it. So it makes no sense to ask if the photon itself is "redshifted". It only makes sense to ask whether a particular receiver device will measure the photon to be redshifted. As @Ibix says, a receiver device on the Sun, right next to where the photon is emitted, will not measure it to be redshifted; but a receiver device far away from the Sun will.

vanhees71, Imager, Ibix and 1 other person
You can find a detailed discussion in On the Interpretation of the Redshift in a Static Gravitational Field

Eugene.

To my understanding, the paper must be wrong in the lab-frame (but I am not 100 percent sure):
paper said:
On the other hand, the energy (frequency) of the photon is conserved as it propagates in a static gravitational field
...
Clearly,in the laboratory system there is no room for the interpretation in which the photon loses its energy when working against the gravitational field.
Source:
https://arxiv.org/abs/physics/9907017

Reason: The energy of a photon is frame-dependent. In the (accelerated) lab frame, a photon (or electro-magnetic wave puls) must loose kinetic energy, while moving up in the gravitational field. Else, one could build a perpetual motion machine: Consider two masses, a mass ##A## on the floor and a mass ##B## at the ceiling. If mass ##A## emits the photon, it looses a mass of ##\Delta E/c^2##. If mass ##B## would gain the same amount of mass (for example heat energy) by absorbing the photon (or electromagnetic wave pulse), then additional energy could be generated by letting a small fraction ##\Delta E/c^2## of this mass fall down to gain the additional kinetic energy. Then you would have on the floor more energy than at the beginning.

Mentor
the paper must be wrong in the lab-frame

Yes, in the sense that it assumes that its interpretation of "energy" is the only viable one, which is not the case.

The paper is using "energy" to denote energy at infinity, which is indeed a constant of geodesic motion in any stationary spacetime; so on this interpretation it is true that the photon does not and cannot change energy as it propagates.

However, you are describing a different use of the term "energy", which is perfectly valid. See further comments below.

The energy of a photon is frame-dependent.

More precisely, it is dependent on the spacetime relationship between the emitter and the receiver. You can actually construct a scalar invariant that describes the energy of a photon as measured by a particular emitter or receiver: it's just the inner product of the photon's 4-momentum with the receiver's 4-velocity. And if we pick an emitter and receiver at different altitudes in the gravitational field, and a photon propagates between them, the photon's 4-momentum stays the same (for the same reason that energy at infinity is a constant of the motion) but the receiver's 4-velocity is different from the emitter's 4-velocity, so the energy measured by the receiver will be different than the energy measured by the emitter.

Your particular description of this process is frame-dependent, yes; it requires you to choose a local inertial frame that contains the emitter, the receiver, and the photon. But the invariant I described above can be constructed for any emitter and receiver, even if they are much too far apart to both fit within the same local inertial frame (in which case your description could not even be applied).

Sagittarius A-Star, vanhees71 and ChrisVer
Your particular description of this process is frame-dependent, yes; it requires you to choose a local inertial frame that contains the emitter, the receiver, and the photon.
I actually described it in an accelerated frame (also the paper talked in my quote of the "laboratory system"). If I had choosen an inertial frame (for example the frame of a nearby free falling elevator cabin), then the photon would fly with constant kinetic energy. The red-shifted reception at the ceiling could then be described as a relativistic Doppler effect only.

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vanhees71
Mentor
I actually described it in an accelerated frame

Yes, agreed, but it's still a local accelerated frame. It can't cover a region of spacetime large enough for the "acceleration due to the Earth's gravity" to vary significantly in magnitude or direction.

If I had choosen an inertial frame (for example the frame of a nearby free falling elevator cabin), then the photon would fly with constant kinetic energy. The red-shifted reception at the ceiling could then be described as a relativistic Doppler effect only.

Yes, agreed.

You can actually construct a scalar invariant that describes the energy of a photon as measured by a particular emitter or receiver: it's just the inner product of the photon's 4-momentum with the receiver's 4-velocity.
This is interesting and shows, that, according to RT, the energy of a photon transfoms by the same factor as it's frequency, without making use of the QT-formula ##W=h*f##.

In an inertial frame, in which the sender is at rest at least at the point in time, when the photon is sent-out, the energy ##E## of the photon can be transformed in the following way to the receiver:

##E' = \mathbf {P_{photon}} \cdot \mathbf {U_{receiver}} = \begin {pmatrix} E/c \\ \vec {p} \end {pmatrix} \cdot \begin {pmatrix} c \\ \vec {v} \end {pmatrix} \gamma = (E - \vec{p} \cdot \vec{v}) \gamma = (E - p * v * \cos{\varphi}) \gamma##

with ##\varphi## = angle between ##\vec{p}## and ##\vec{v}## in the source-frame. With ##p = E/c ## for photons follows:

##E' = E \gamma (1 - \frac{v}{c} * \cos{\varphi})##.

That is the same factor as in the relativistic doppler shift of the frequency (see Eq. 7):
https://en.wikipedia.org/wiki/Relativistic_Doppler_effect#Einstein_Doppler_shift_equation

Mentor
In an inertial frame, in which the sender is at rest at least at the point in time, when the photon is sent-out, the energy ##E## of the photon can be transformed in the following way:

##E' = \mathbf {P_{photon}} \cdot \mathbf {U_{receiver}} = \begin {pmatrix} E/c \\ \vec {p} \end {pmatrix} \cdot \begin {pmatrix} c \\ \vec {v} \end {pmatrix} \gamma = (E - \vec{p} \cdot \vec{v}) \gamma = (E - p * v * \cos{\varphi}) \gamma##

Note that this calculation assumes a global inertial frame, which requires flat spacetime. More precisely, it assumes that the components of the photon's 4-momentum remain unchanged as they are parallel transported along the photon's worldline from the event of emission to the event of reception. In a general curved spacetime, this will not be the case; but you can still do the calculation, you just have to add an extra computation of what happens to the photon's 4-momentum during the parallel transport from the emission event to the reception event. This is what you would do, for example, to calculate the observed redshift of photons from a distant galaxy in an expanding universe.

Sagittarius A-Star
$$hf = e U$$