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In reading another question about gravity's effects on a photon, I wondered if it were possible for a photon to ever be redshifted to zero wavelength.

I know that black holes have a gravity field strong enough to keep light from escaping but what about the affect on the photon that is escaping. Say we have a single photon escaping on a path directly away from a non-spinning singularity. Also conjecture that we can release a single photon at various distances from the singularity (still on a trajectory directly away) so as to select the strength of the gravity field it must traverse. I know the distance has already been calculated (the event horizon I believe) but what of the photon itself?

Is there a point where the amount of redshifting (e.g. gravitational) causes the wavelegth of the photon to become infinitely large? This answer mentions the change in potential and kinetic energy. Is it possible for all of a photon's kinetic energy be transformed to potential energy?

What are the implications, both if it were possible and/or impossible? More importantly if impossible, what makes it impossible? This would beg the question, what is the smallest amount of energy possible (or quanta) for a photon? If that minima-photon attempted to escape from within a gravity field, it can't slow down so what does it do? As the wavelength increases approaching infinity, so must it's frequency decrease to approaching zero.

[EDIT] Thanks to @david-z for an excellent answer but I was hoping for more depth for the case where the photon is emitted at the precise event horizon. He shows that the photon is indeed 'stuck' there but I'm interested in more of the math describing such a photon. If such a photon has a zero wavelength then what else does it imply about the physical existence of the photon? If a photon has zero wavelength then how can you calculate it's energy? Using an equation from @david-z 's answer to the above question,

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

If the number of photons $\langle N\rangle$ = 1 and the wavelength is zero, it doesn't matter what the frequency is, the total energy goes to zero. If the total energy of the photon is zero then can we claim that the photon must not exist? I am making the statement in order to be corrected but also so readers understand the core of the question that holds my interest.

I'm thinking about it like in calculus where a curve has no value at a specific point either due to an asymptote, a hole (like caused by division by zero), or other type of discontinuity. Considering that a photon can only have the velocity of c, what can we say about the velocity of a photon caught at the event horizon? Does it become undefined or is there more exotic physics that need to be brought in to describe the situation? I realize that this may end up as a meaningless question but am open to why it is a meaningless question.

Basically, what weirdness happens when we force a photon to achieve zero wavelength?

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Redshift corresponds to an increase in wavelength, not a decrease. The wavelength never gets to zero by increasing! – Ted Bunn Jun 29 '11 at 17:25
Oops, will have to edit. Meant zero frequency. – Kelly S. French Jun 29 '11 at 21:53
@Ted Bunn I am curious how would one describe a charged black hole, if there exists such a beast? Would there not be an electric field radiating out, which in a sense is a 0 frequency limit? – anna v Oct 28 '11 at 15:06

1 Answer 1

Ted's comment is correct, redshift makes the wavelength increase. But if you actually meant zero frequency (which corresponds to infinite wavelength), then it's almost possible, if you emit a photon from just outside the event horizon of a black hole.

Specifically, suppose the photon is emitted from a small coordinate distance $\epsilon R_s$ ($\epsilon \ll 1$) outside the event horizon. According to Wikipedia, the redshift of that photon once it reaches an infinite distance away from the hole would be

$$\lim_{r\to +\infty}z(r) = \frac{1}{\sqrt{1-\left(\frac{R_s}{R^*}\right)}}-1 = \frac{1}{\sqrt{1-\left(\frac{R_s}{R_s(1 + \epsilon)}\right)}}-1 \approx \sqrt{\frac{1}{\epsilon}}$$

Redshift relates to wavelength as $z = \frac{\lambda_o - \lambda_e}{\lambda_e}$ ($\lambda_o$ is the observed wavelength, $\lambda_e$ is the wavelength at emission), so given that $\epsilon$ is very small,

$$\lambda_o \approx \frac{\lambda_e}{\sqrt{\epsilon}}$$

In other words, by emitting a photon close enough to the event horizon, you can arrange for it to be redshifted to as large a positive wavelength (and thus as small a frequency) as you want. But there's no value of $\epsilon$ that will actually give you an infinite wavelength (zero frequency).

If you were to go all the way down to $\epsilon = 0$, i.e. emit the photon from right on the event horizon, then it would just be stuck there, since there are no outgoing null geodesics.

*Everything here applies only to Schwarzschild black holes, although I don't know of anything that would allow a photon to be redshifted to infinite wavelength from any other kind of black hole either.

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Is there any physical process for which photons are emitted with a population that limits to infinity as the frequency goes to zero? Since the energy is proportional to frequency there are easily mathematical forms that can accomplish this. Whether or not something in the universe creates such a thing is a different question. I see no Newtonian argument that we can't make infinitely many photons in a finite space, but the quantum effects become proportionally more significant. What would be some consequences of that? – Alan Rominger Jun 29 '11 at 19:09
Regarding your * comment: gravitational redshift in classical GR is defined relative to a time-like congruence. In the static black hole, the static killing vector field induces a natural time-like congruence. Without this "naturality", we would have difficult discerning how much of the red-shift is due to gravitational effects and how much it is due to Doppler effect. For a spinning black hole, things becomes interesting: due to the frame dragging effects of the ergosphere, the stationary killing field is not time-like close to the event horizon. – Willie Wong Jun 29 '11 at 20:41
So it is not a priori clear how one gets a preferred global family of time-like congruences that "rules out Doppler shift" effects. So it is a bit hard to say what the gravitational red-shift is for distant observers. On the other hand, a local/infinitesimal version of gravitational redshift survives, if near the Kerr black hole you use the Hawking vector field instead of the stationary vector field to define your preferred time-like congruence. (This also means that a notion of gravitational redshift can be defined relative to the stationary killing vector field, if you restrict to photon – Willie Wong Jun 29 '11 at 20:45
trajectories [wave vectors] that are orthogonal to the rotational symmetry of the black hole.) – Willie Wong Jun 29 '11 at 20:46
@David I just realized this is an old question brought current because of an edit. Could you have a look at my comment/question on the original question, since Ted Bunn is not entering frequently. – anna v Oct 28 '11 at 15:48

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