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Is there any lower or upper limit on the energy of a photon? i.e. does the mathematical framework we currently use to study photons blow up when a photon surpasses a certain upper limit of energy? (or the same on the opposite side?)

My thoughts: If I let the energy of a photon tend to infinity, its wavelength would be tending to zero, and since it is thought that we cannot really distinguish things when they are on the scale of a Planck Length, would the photon have its maximum energy when its wavelength is equal to the Planck length? (Feel free to correct me, I feel I might not be.) and for the opposite end of the spectrum a photon with the least energy would have zero wavelength, implying no photon, which is a trivial case.

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    $\begingroup$ very interesting! I heard about Plank length but never paid attention. I even don't remember in which connection I heard of it. Can you tell me what is Plank length? $\endgroup$ – Sofia Jan 17 '15 at 15:01
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    $\begingroup$ See ACuriousMind's answer here. The Planck length isn't that important. $\endgroup$ – HDE 226868 Jan 17 '15 at 15:08
  • $\begingroup$ Its supposed to be the base unit for length in the system of units created by Planck. The value of the Planck Length is derived from h, c and G, which are assumed to be fundamental constants of the universe. math.ucr.edu/home/baez/planck/node2.html you should read this. If the mass of something is the Planck mass, then its Schwarzchild Radius is equal to its De Broglie Wavelength. $\endgroup$ – Hritik Narayan Jan 17 '15 at 15:08
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    $\begingroup$ @HDE 226868 - ACuriousMind was just saying the Planck length has no special significance in current quantum field theories, which are incompatible with general relativity - most physicists working on quantum gravity would probably bet that it will be very important in such a theory, though I don't know (and ACuriousMind didn't address) what current attempts at such a theory like superstring theory would say about the implications for the energy of individual photons. $\endgroup$ – Hypnosifl Jan 17 '15 at 15:58
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    $\begingroup$ BTW, for a good not-too-technical introduction to why it's expected that the Planck scale is where general relativity's predictions will become completely inaccurate and quantum gravity will be needed instead, see the paper Six easy roads to the Planck scale. $\endgroup$ – Hypnosifl Jan 17 '15 at 20:26
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One of the main problems of Quantum Gravity is that Quantum Mechanics (in broad sense, including QFT) holds for arbitrary energies, i.e. there is no structural inner bounds to its validity nor there are known systems for which QM fails. In other terms, in the framework of QM, a collison between particles having Planckian energy has nothing special. On the other hand, at very high energy General Relativity holds and, as far it is known, its consequences must be taken in account. In particular, it is a theorem by Schoen and Yau that "[w]hen enough matter is condensed in a small region, gravitational effects will be strong enough to cause collapse and a black hole will be formed" (quoting the abstract of their article). So, there is a limit on the energy of a photon? Strictly speaking, I don't know. There is no physical evidence, as far as I know, for saying yes or no. If the answer was in the affermative however, QM has to be modified, in order to include an "inner blow up" at very high energies. Notice that one can't say for sure that a QG theory exists nor even that it is a physical necessity. The only persuading physical reason I know of to say that a more comprehensive theory ought to exist is that the cosmic microwave background, a specific GR-object, follows a black-body curve, a specific QM-behaviour.

At the opposite extreme, "soft" photons would not create any difficulty of this sort (provided that GR is "switched off", i.e., has no effects, otherwise picking sufficiently small regions the situation is in principle the same than above); interestingly, this limit is "fussy" for electrons, in the sense that, since electrons are very light but not massless, it is difficult to succeed in keeping them at nonrelativistic regime and check whether they behave according to nonrelativistic QM. Obviously, photons of too high energy are rather unuseful as probes, since annoying effects such as couple creation arise and most measurements become unsensible.

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    $\begingroup$ How can a QG theory "not exist" in principle? GR exists, QFT exists, there must be something in-between. It may appear to be e.g. string theory, or something not yet invented. $\endgroup$ – Ruslan Jan 17 '15 at 17:39
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    $\begingroup$ Why "there must be"? Rember that in physics such a statement must be justified on the physical ground. It could equally well be that in nature GR and QM are always well separated for some strange reason. Of course, it is natural to think that an overlap must be, all of us think so, but we can't take this as a physical motivation: only experimental results are. Theoretical physics aims to describe nature, i.e., experiments, not our picture of nature. $\endgroup$ – Federico Jan 17 '15 at 17:44
  • $\begingroup$ But how can it not exist? Surely there must be a way to predict phenomena in between GR and QM, at least in some way, why not? $\endgroup$ – Ruslan Jan 17 '15 at 20:33
  • $\begingroup$ @Ruslan It's not like there is a continuum between GR and QM -- their domains of applicability are rather orthogonal. $\endgroup$ – user10851 Jan 17 '15 at 20:36
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    $\begingroup$ Unprendictability is not same as existence of a theory. To be clear, QFT is only known to exists as a perturbative theory in 4 dimensions. So far, no one can say whether it exists or not in a nonperturbative sense. However, this is not a serious drawback in physics, since we have a set of rules thanks to which we can perform calculations, verified with the highest accuracy. This is a drawback for our common "unified" vision of the world. We feel be compelled to find a more complete theory, but actually this is physical unnecessary until we find an experimental result that disagree with QFT. $\endgroup$ – Federico Jan 17 '15 at 23:20
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If a photon is too energetic then it might create a black hole according to general relativity. This poses a serious problem when trying to define the process of measuring something with arbitrary accuracy. Some heuristic considerations, based on putting the Heinsenberg's uncertainty principle together with classical General Relativity lead to an estimate of the accuracy with which you can measure some of the coordinates of an event in space-time by sending, say, photons with a certain energy distribution. After this threshold an event horizon is formed and no light can get back to the observer, so the measuring process loses its operational meaning. This has led Doplicher, Fredenhagen and Roberts to postulate that the coordinates of local orthogonal frames should be replaced by (unbounded) self-adjoint operators. For more details on this, a nice review on the subject of Quantum Space-time has just appeared on the arxiv. Briefly, commutation relations among the coordinates are assumed, $$[q^\mu,q^\nu] = i\lambda_P^2 Q^{\mu\nu}$$ and they depend on a characteristic scale given by Planck's length $\lambda_P$, and this captures the above considerations, since one can then retrieve an uncertainty principle for coordinates which is in good accordance with what it is expected from the above heuristic considerations, that is not all coordinates of an event in space-time can be measured simultaneously with arbitrary accuracy.

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    $\begingroup$ except that in the standard model the photon is an elementary particle, a point in space, so no coordinates. These would apply for measuring interferences, which is a different question. The frequency of the photon is just the energy divided by h $\endgroup$ – anna v Jan 17 '15 at 15:38
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    $\begingroup$ last time i checked, points had coordinates...but as you say, photons are elementary particles in the sense that they are quantum objects, so the notion of position is ill-defined in this case. But localisation can still take place, even if between certain bounds... $\endgroup$ – Phoenix87 Jan 17 '15 at 15:42
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    $\begingroup$ @Phoenix87 I don't know much about general relativity, could you given an intuitive explanation on why a photon that is too energetic (or how energetic is too energetic?) might create a black hole? $\endgroup$ – M. Zeng Jan 17 '15 at 16:10
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    $\begingroup$ In general relativity, the "fabric" of space-time is governed by the energy-matter distribution. Under certain conditions (cf. en.wikipedia.org/wiki/…) on such distribution, the solutions of Einstein's equations might exhibit some singularities which can be interpreted as event horizons, or black holes. Even light cannot escape the gravity of such objects (that's why they're called black holes), so the "light" sent to measure an event might create a Planckian black hole and remain trapped in its inside. $\endgroup$ – Phoenix87 Jan 17 '15 at 16:16
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    $\begingroup$ @annav of course a 0 dimensional object is nonetheless a quantum object and has an associated wave packet. And it's the wavepacket who has the characteristic size, which can be used somehow to (statistically) measure distances. $\endgroup$ – Ruslan Jan 17 '15 at 17:37
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Edited because I can't read.

The Greisen–Zatsepin–Kuzmin (GZK) limit is one particular theoretical limit involving high-energy protons. Beyond this energy limit, protons scatter too much with the cosmic microwave background radiation to travel any significant distance towards Earth.

Now, in principle, photons should also interact with the CMB, scattering of two photons has already been experimentally confirmed. There are reference frames in which CMB photons have (nearly) arbitrarily high energies, and so one would expect pair production from extremely high energy gamma rays. According to this source the threshold energy is about $10^{15}\,\mathrm{eV}$. A quick search for the term "gamma ray opacity" brings up a lot of interesting papers also, including this one by Dweck and Krennrich , which deals with gamma ray scattering off of extragalactic background light.

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    $\begingroup$ No, the GZK limit is for cosmic rays (in general, these would be extremely energetic protons), not photons. This is even stated in the very first sentence of the link you provide. $\endgroup$ – Kyle Kanos Jan 17 '15 at 19:35
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    $\begingroup$ I need to change my prescription then :) $\endgroup$ – lionelbrits Jan 17 '15 at 19:36
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    $\begingroup$ While interesting, the scattering of high-energy photons with CMB photons doesn't seem to address the question at all: is there a maximum photon energy? (whether it could travel through outer space is another issue) $\endgroup$ – Tim S. Jan 17 '15 at 22:22
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I would say no, since the energy of a photon depends on the reference frame, so if a photon beam is directed toward you, the energy of the photons in your reference frame will depends on the relative speed between you and the photon source. When this speed tend toward $c$ then the energy of each photon will tend toward $+\infty$

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  • $\begingroup$ Isn't the relative speed between you and a photon always the speed of light? $\endgroup$ – Noah Jan 18 '15 at 4:41
  • $\begingroup$ @Noah: It is, but the behavior described here is known as "blue shift". Those photons do have higher frequencies and therefore higher energies. We see the opposite (red shift, lower frequency/energy) in galaxies which are moving away from us. $\endgroup$ – MSalters Jan 18 '15 at 10:55
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    $\begingroup$ I understand the Doppler effect and blue shift. The measured energy of a photon depends on the speed of the photon source compared to the receiver; the speed of the photon (through empty space) relative to the receiver is always c. $\endgroup$ – Noah Jan 18 '15 at 13:50
  • $\begingroup$ Yes, that what I have written, and that's why without the CMB to interact with there is no theoretical limit on the energy of a photon $\endgroup$ – agemO Jan 18 '15 at 19:32
  • $\begingroup$ Why is this not the accepted answer? $\endgroup$ – pela Apr 1 at 21:17
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Yes, you could say a photon could gain enough energy to be bound by its own gravitational field, this class of particle is known as the Geon, invented by John A. Wheeler. That energy would correspond to the Planck energy - but its very speculative and such a particle is probably not stable.

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protected by ACuriousMind Oct 25 '15 at 18:01

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