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As a first approximation, I don't see how a wavelength of less than 2 Planck distances could exist. The question is: Are there any other limits that would come into play before that?

For example:

  • Would the energy density cause the photon to turn into a black hole or something like that?
  • Would the energy of the photon exceed the total mass+energy of the universe?
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  • $\begingroup$ In order to create a photon, You need some exited object which has the energy to radiate such a photon. $\endgroup$
    – Georg
    Nov 1, 2011 at 16:15
  • $\begingroup$ @Georg, the question isn't about creating the photon. OTOH, it's interesting to consider how you could do that (or how you could pump an exiting photon to that high an energy level). $\endgroup$
    – BCS
    Nov 1, 2011 at 18:00
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    $\begingroup$ My comment was thought to make You think of the fact, that the two reasons You name as possible limits would apply for that "emitter" first. And: of course the most energetic precursor is the upper limit for a photon. Photons dont emerge "from heaven" like Manna. $\endgroup$
    – Georg
    Nov 1, 2011 at 18:04

2 Answers 2

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The principle of relativity guarantees that the energy of a particle may always be boosted to a higher value, e.g. by looking at the same situation from a different inertial system. All the situations with 1 particle and arbitrary allowed energy (any number not smaller than the rest mass times $c^2$: the rest mass of the photon is zero) are physically equivalent.

That's why the wavelength (which is linked to the inverse momentum) of a photon, or any other particle, may be arbitrarily short, whether it's shorter than the Planck length or not. You can't produce a black hole just from one particle because it's fast. You only produce a black hole if a sufficient amount of mass is concentrated within the Schwarzschild radius from the center-of-mass reference frame.

There's a lot of misconceptions in popular science literature about the Planck length as the "minimum distance". The Planck length is only the minimum allowed distance of "proper distances measured in the rest/otherwise-natural frames" i.e. distances within a hypothetical nearly static object, measured at rest. But the wavelength associated with an arbitrary particle is just some difference of coordinates according to any frame and this quantity can't be constrained because of the principle of relativity.

So the answer to both questions of yours is a resounding No:

  1. No, a single particle with a vanishing or low rest mass can never turn into a black hole, regardless of the high energy, high momentum, and corresponding high frequency or short wavelength. You need to collide at least 2 particles of Planckian energies to produce a black hole. What matters is the center-of-mass energy (which is also zero for a single photon).

  2. No, a photon (or any other particle) whose wavelength is comparable to the Planck length carries the energy equal to the Planck energy which is $c^2$ times the Planck mass. The Planck mass is just 10 micrograms or so, extremely below the mass of the Universe. ;-) It's, in fact, 100 times lighter than a mosquito. It's a big energy if you concentrate it to a single particle – which is what particle physicists usually want to do (in their minds) with the Planck energy. But it is a negligible energy relatively to the latent energy of the macroscopic objects and surely the Universe as well.

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  • $\begingroup$ Note that with regards to point 1, The energy density I'm referring to is not that cause by relativistic speeds (if a photon had such, it would be infinitely heavy) but rather just the energy of the photon it's self. For example, what would happen if a single photon was emitted using the total matter/energy conversion of an entire super massive black hole (or an equivalent mass)? $\endgroup$
    – BCS
    Nov 1, 2011 at 15:34
  • $\begingroup$ @BCS: you can't emit just one photon from a black hole, but if a supermassive black hole would decay into two photons going in opposite directions, both photons would be indistinguishable from highly boosted supermassive black holes themselves, unless you chase them with impossible speed. All boosted matter is eventually indistinguishable from a boosted black hole experimentally for stationary observers. $\endgroup$
    – Ron Maimon
    Nov 1, 2011 at 19:07
  • $\begingroup$ I specifically left unspecified how the photon is generated and even what matter source is used to power the experiment. The only things I specified is the total amount of energy dumped into it. $\endgroup$
    – BCS
    Nov 1, 2011 at 23:01
  • $\begingroup$ Dear @Ron, I agree that the momentum conservation prohibits the emission of just one photon from a black hole; otherwise, nope. A photon is always distinguishable from a black hole. Its rest mass is zero; the rest mass of any black hole is always greater than the Planck mass (times a numerical constant of order one). $\endgroup$
    – Luboš Motl
    Nov 3, 2011 at 10:07
  • $\begingroup$ Dear @BCS, it is completely irrelevant how a photon was generated. Whatever the circumstances of the birth were, you may always boost the whole arrangement and get a photon of a shorter wavelength than before. There is no lower bound on the allowed wavelength. If you are uncertain what is allowed with "large energy densities", "short wavelengths" etc. and if it looks like a conflict of many infinities, try to find a reference frame in which the situation doesn't look too extreme. That's what relativity allows you to do. You will see that a single photon is never "less or more extreme". $\endgroup$
    – Luboš Motl
    Nov 3, 2011 at 10:10
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A possible (short) non-existence argument for a minimum wavelength:

Observer A transmits a photon of "minimum wavelength" towards observer B who is approaching observer A. Due to the relative motion of B w.r.t. A, the photon will be blue shifted to a wavelength shorter than the "minimum". Contradiction.

q.e.d.

The only possible counter argument to this (on the off chance one exists) would amount to that the proximity of the observer's matter interacts with the photon to cause it to shed energy (e.g. via some hypothetical phenomenon akin to Cherenkov radiation). Even that however would only limit the minimum that can be observed in a given reference frame rather than an absolute minimum for it.

OTOH, if the energy gets high enough, the photon could only be observed briefly or indirectly as the very bright flash caused by the direct observer being totally obliterated.

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  • $\begingroup$ You are doing 'magic thinking' . 10 observers 10 realities ? $\endgroup$
    – Helder Velez
    Nov 3, 2011 at 2:13
  • $\begingroup$ (cont) Once you release the photon there it goes unperturbed until catched by one and only one observer. This observer will perceive it differently according to its motion, but to the photon nothing happens. The measure and only the measure change. $\endgroup$
    – Helder Velez
    Nov 3, 2011 at 2:26
  • $\begingroup$ @HelderVelez, good point: Close only counts in Horseshoes, Hand Grenades, and Tactical Nuclear Weapons. (OTOH that last one has an interesting relevance here.) $\endgroup$
    – BCS
    Nov 3, 2011 at 16:13

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