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?

[EDIT] Ok, I figured out a way to describe the scenario in which I'm interested. Let's do it like Einstein would using a thought experiment.

Say you were in orbit around a black hole with a size of 10 solar masses. The exact size is not that important but for this scenario we need the point where spaghettification happens is inside the event horizon so objects that fall through the event horizon will stay intact in that immediate region of space. Let's also say you brought a flashlight with you and thinking about Einstein and the speed of light you realize you can do an experiment simply by dropping the flashlight into the black hole. Now, there is something special about the flashlight in that it has the quality that allows it to always point away from the black hole and it emits a stream of individual photons. As the flashlight falls through space it continuously emits photons at the speed of light but since the curvature of space is redshifting the photons, we could graph the changing frequency of the light coming from the flashlight as a function of distance from the event horizon using the X axis to represent distance from the event horizon and the Y-axis being the photon's frequency. There is likely to be a limit at X=0 but what does that tell us about the curve of space there. Do photons get stuck at the event horizon and just build up over the lifetime of the hole? When the hole gets larger as it gobbles up more nearby mass, the event horizon also gets larger but what happens to those photons that were caught at the previous horizon? This may or may not be related to recent talk about the event horizon being a hologram that stores the inbound information but it seems to me that there is still something intriguing about the region near the horizon that can be used to investigate some profound physical concepts.

In summary, when David Z says,

"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 ϵ that will actually give you an infinite wavelength (zero frequency).

If you were to go all the way down to ϵ=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."

For the photon emitted right on the event horizon, yes it may be stuck from a null geodesic point of view but what can we say about its wavelength/frequency? If it is stuck, does that mean the frequency is zero? If the frequency can be zero wouldn't that make the energy of the photon zero? If that is somehow impossible, then are we saying that there may be some natural limit involved that hasn't been contemplated?

  • 4
    $\begingroup$ Redshift corresponds to an increase in wavelength, not a decrease. The wavelength never gets to zero by increasing! $\endgroup$
    – Ted Bunn
    Commented Jun 29, 2011 at 17:25
  • $\begingroup$ Oops, will have to edit. Meant zero frequency. $\endgroup$ Commented Jun 29, 2011 at 21:53
  • $\begingroup$ @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? $\endgroup$
    – anna v
    Commented Oct 28, 2011 at 15:06
  • $\begingroup$ In reference to the comment by @annav the electromagnetic field surrounding a charged black hole is a "fossil field", it doesn't have to escape the event horizon. That's also the case for its gravitational field. See math.ucr.edu/home/baez/physics/Relativity/BlackHoles/… for details. $\endgroup$
    – PM 2Ring
    Commented Apr 9, 2020 at 7:02
  • $\begingroup$ @PM2Ring thanks, Since 2011 I have been introduced to the concept of virtual particle building the static field at the limits. $\endgroup$
    – anna v
    Commented Apr 9, 2020 at 8:29

2 Answers 2


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.

  • $\begingroup$ 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? $\endgroup$ Commented Jun 29, 2011 at 19:09
  • $\begingroup$ 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. $\endgroup$ Commented Jun 29, 2011 at 20:41
  • $\begingroup$ 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 $\endgroup$ Commented Jun 29, 2011 at 20:45
  • $\begingroup$ trajectories [wave vectors] that are orthogonal to the rotational symmetry of the black hole.) $\endgroup$ Commented Jun 29, 2011 at 20:46
  • $\begingroup$ @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. $\endgroup$
    – anna v
    Commented Oct 28, 2011 at 15:48

No. The photon escapes the gravity well and gets red shifted as it climbs out or it falls in.

To be stuck on the event horizon is to be captured by it.

Photons don't experience time. But to freeze a photon you would need to stop time. Maybe the wavelength can be drawn out to light years and the frequency could be longer than the age of the universe...


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