# Can a photon have a wavelength less than the planck length?

Would the photon effectively have no wavelength?

• No offense , but most questions on this site show at least some research on the part of the person asking the question. It would greatly assist your question, (on this or any other topic) if you were able to cite experimental results, rather than "what if" type questions. So, in this case, have you evidence, for example, of delays in light speed over cosmological distances due to interactions on the scale you mention? – user108787 Aug 11 '16 at 23:57
• Have a look at Is the Planck length the smallest length that exists in the universe or is it the smallest length that can be observed? and its linked questions. There is no significance to the Planck length in the currently accepted theories. – ACuriousMind Aug 12 '16 at 9:18
• One close vote as "not clear what you're asking" and two as "primarily opinion-based." Neither makes any sense. The question is both crystal clear and indisputably about a matter of fact. – Nathaniel Aug 12 '16 at 11:12
• It is a very clear and concise question, some people have bothered to answer it. Courtesy costs you nothing. – Promissory Estoppel Aug 12 '16 at 13:28

I am going to weigh in on this. In one sense we can say that we really do not know. Since you mentioned experimental evidence, we have no experimental evidence of anything at the Planck energy $E~=~\sqrt{c^5/G\hbar}$. The wave length of a particle at this energy would be equal to the Planck length. The Planck length is computed by equating the wavelength of a particle at rest with the circumference of the event horizon. The $4$-momentum $P^\mu~=~(mc,~0,~0,~0)$ with the deBroglie type equation $p\lambda~=~h$, we equation $\lambda$ equated to $2\pi$ times the Schwarzschild radius $r~=~2GM/c^2$ the rest is algebra and you get $\ell_p~=~\sqrt{G\hbar/c^3}$

So what does it mean for a particle to have a wavelength shorter than $\ell_p$? It means the particle is in a region smaller than the even horizon of the smallest quantum unit of black hole. There is nothing immediate that says this can't happen. What we do hypothesize is that this unit of black hole represents the smallest region one can locate a qubit. 't Hooft and Susskind formulated holography initially by looking at event horizons according to units of Planck areas that can hold a qubit of information.

So suppose you have photons or any quantum field with an arbitrary spectrum of energy or frequencies. The energy in a Fourier sum that exceeds the Planck energy is not able to hold a qubit of information. In other words, these probably do not play any physically meaningful role and can be removed. This follows in some sense with the idea that the Planck scale is sort of the ultimate renormalization cut-off in QFT or quantum gravity. Of course as yet the details of this are not yet certain.

• Thank you. I guess the problem is we cannot measure at this microscopic level as of yet. It would be interesting to see how this may be done, along with re-running the double slit experiment with particles with a wavelength of the Planck length (or lower), particularly with newer experiments like the quantum eraser etc. It would be interesting to see if the frequency effectively reached a limit and became a constant such as 0 or 1. – Promissory Estoppel Aug 12 '16 at 2:17
• @PromissoryEstoppel Did you miss the part where the answer notes that you would get a black hole? That's something very observable and expected to be seen in e.g. particle collider experiments. It's pointless to consider such a photon since it wouldn't really be a photon long before it got anywhere near as energetic. There's nothing special about the Planck length, it's not a "size limit" of the universe - it's just yet another unit that tries to avoid (directly or indirectly) using anthropocentric units like feet. It has no significance in cosmology or physics to the best of our knowledge. – Luaan Aug 12 '16 at 12:15
• Actually, I read it quite clearly and it implied that the creation of a black hole is a possibility, not a certainty. – Promissory Estoppel Aug 12 '16 at 13:27

If there was a minimum wavelength, you just could increase your own velocity in direction of the photon to still make it smaller.

Since the relative velocity to the photon must always be c the only thing that can increase then is the frequency, so you always can get a still smaller wavelength just by increasing the doppler effect.

Because of that there is no minimum wavelength. If you could reach c (which you can't) the wavelength would be zero, but since you can get arbitrarily close to c the wavelength can also get arbitrarily close to zero.

• +1 for good, highly relevant physcical insight, but I don't think it answers the OPs question because you need to assume that classical SR holds down to arbitrarily small length scales, something which we don't know nor (I believe as a lay reader) even believe. See my answer where I talk about a couple of variations on classical relativity that allow for both an invariant velocity and an invariant length scale. – Selene Routley Aug 12 '16 at 3:17
• It will be interesting if there turns out to be a maximum frequency allowed before the photon itself suffers gravitational collapse. How would this work with Doppler blue shift? – Jus12 Jan 29 '18 at 20:07

The short answer is we simply do not know: this hypothesis is utterly beyond anything that we can either test experimentally or reason about with a widely accepted theory.

Симон Тыран's Answer is a good, concise exposition showing what classical, relativistic reasoning has to say about this. But I don't believe this answers the question because you have to make the assumption that classical relativity works down to an arbitrarily small scale and that's something we neither know nor (I get the impression as a lay reader) even believe.

Moreover, even without full quantum gravity, one can formulate classical theories with both an invariant velocity (as in the $c$ of special relativity) and and invariant length scale. Examples of such theories are "Doubly Special Relativity" and also de Sitter Invariant Special Relativity wherein the symmetry group $SO(1,\,4)$ is a supergroup of the Lorentz group and is the same as the symmetry group of de Sitter space, a highly symmetric vacuum solution of the Einstein field equations. In such a universe one would have a natural, invariant length scale that can be used to invalidate the classical special relativistic reasoning that there always exists an inertial observer whom a wavelength is arbitrarily small for.

• Of course, even though it's far beyond anything we could test experimentally, it's highly unlikely that 1 planck length would be significant in any way. Even if there is a hard limit, there's no reason to believe that 1 plack length would be that limit, just like there's no reason to believe that 1 planck mass is the smallest possible mass of something, or that nothing can move slower than 1 m/s. It's not like e.g. 0 K which actually does have physical significance. – Luaan Aug 12 '16 at 12:19
• @Luaan That's a good point that the Planck length itself is unlikely to be the length scale where things go weird. There's nothing in doubly special relativity nor de Sitter invariant relativity that links the invariant length to anything we know - they're just possible toy theories to show that invariant lengths can sit with relativistic principles. Or at least that's my understanding. – Selene Routley Aug 12 '16 at 13:14

We don't know if massless particles may have a wavelength beyond planck length, but there are some significative indices (including the ones provided in the other answers) against limitation of wavelength. In particular, according to Wikipedia,

There is currently no proven physical significance of the Planck length; it is, however, a topic of theoretical research.

This research is referring in particular to the discreteness of spacetime and not to wavelengths.

The possibility to increase the relative velocity of an observer is an essential argument, and it is important to note that the spacetime interval of massless particles is zero. A zero interval means that there is no place for a n y kind of wavelength, even not for a wavelength smaller than Planck length. However, contrary to space intervals, the spacetime interval does not correspond to any observer but to a (hypothetical, non existent) observer moving at speed of light. You can now assume an observer moving very close to speed of light, and as far as we know today there is no limit, that means the observer will never reach speed of light, but he may come arbitrarily close to it, and this observer would observe a wavelength which may be arbitrarily small, that means arbitrarily close to the zero length of the zero spacetime interval of the massless particle.

Now, the speed of the observer is a space interval per time interval, and by this, your question transforms into the question if admitted wavelengths are discrete, in particular if space and/ or time are discrete. But as I mentioned above, mainstream today considers space and time to be continuous.