Is the Planck length the smallest length that exists in the universe or is it the smallest length that can be observed?

Question

I have heard both that Planck length is the smallest length that there is in the universe (whatever this means) and that it is the smallest thing that can be observed because if we wanted to observe something smaller, it would require so much energy that would create a black hole (or our physics break down). So what is it, if there is a difference at all.

Answer 1

Short answer: nobody knows, but the Planck length is more numerology than physics at this point

Long answer: Suppose you are a theoretical physicist. Your work doesn't involve units, just math--you never use the fact that $c = 3 \times 10^8 m/s$, but you probably have $c$ pop up in a few different places. Since you never work with actual physical measurements, you decide to work in units with $c = 1$, and then you figure when you get to the end of the equations you'll multiply by/divide by $c$ until you get the right units. So you're doing relativity, you write $E = m$, and when you find that the speed of an object is .5 you realize it must be $.5 c$, etc. You realize that $c$ is in some sense a "natural scale" for lengths, times, speeds, etc. Fast forward, and you start noticing there are a few constants like this that give natural scales for the universe. For instance, $\hbar$ tends to characterize when quantum effects start mattering--often people say that the classical limit is the limit where $\hbar \to 0$, although it can be more subtle than that.

So, anyway, you start figuring out how to construct fundamental units this way. The speed of light gives a speed scale, but how can you get a length scale? Turns out you need to squash it together with a few other fundamental constants, and you get: $$ \ell_p = \sqrt{ \frac{\hbar G}{c^3}} $$ I encourage you to work it out; it has units of length. So that's cool! Maybe it means something important? It's REALLY small, after all--$\approx 10^{-35} m$. Maybe it's the smallest thing there is!

But let's calm down a second. What if I did this for mass, to find the "Planck mass"? I get: $$ m_p = \sqrt{\frac{\hbar c}{G}} \approx 21 \mu g $$

Ok, well, micrograms ain't huge, but to a particle physicist they're enormous. But this is hardly any sort of fundamental limit to anything. It isn't the world's smallest mass. Wikipedia claims that if a charged object had a mass this large, it would collapse--but charged point particles don't have even close to this mass, so that's kind of irrelevant.

It's not that these things are pointless--they do make math easier in a lot of cases, and they tell you how to work in these arbitrary theorists' units. But right now, there isn't a good reason in experiment or in most modern theory to believe that it means very much besides providing a scale.

Answer 2

None of the above. Though there are many speculations about the significance of the Planck length, none is proven in any currently accepted theory.

It is expected, though, that quantum gravity effects become definitely non-neglegible at the energy/distance scale set by the Planck length, so it provides a heuristic scale at which we should not expect our current theories to make accurate prediction.

Answer 3

There is a tiny bit more going on than the otherwise excellent answer by zeldrege suggests. Imagine that you wish to probe an unspecified object to examine its structure. If we use light to look at the structure of an object, we need to have its wavelength smaller than the size of the details we wish to look at. Probing an object that has a (linear) size equal to the Planck length, requires that the energy of the photon be greater than the mass of a black hole of that "size". So, a classical black hole would be formed by our energy probe, thus preventing us to see details inside the object we wish to investigate. We are lead to an apparent contradiction, which suggests an incompatibility between Relativity and Q.M.

Answer 4

The Planck length has exactly the following significance:

If a characteristic length scale $l_G$ exists for a theory that combines quantum mechanics with general-relativistic gravitation, then that scale must - by dimensional analysis - take the form

$$l_G = k \cdot \sqrt{\frac{\hbar G}{c^3}}$$

where $k$ is some arbitrary dimensionless constant. The quantity

$$l_P := \sqrt{\frac{\hbar G}{c^3}}$$

is called the Planck length. But note! Because $k$ is an arbitrary constant, there is no guarantee that $l_G = l_P$ at all! Heck, it is not necessarily the case the two are even comparable. It could be that the actual gravitational length is much smaller, or much larger. That said, given the failure to observe any quantum gravitational effects at an energy scale of about 14 TeV, which corresponds to a length scale of about 89 zm ($8.9 \times 10^{-20}\ \mathrm{m}$) by $\lambda = \frac{hc}{E}$, we can establish an experimental upper bound on $k$ of

$$k \lesssim 5.5 \times 10^{15}.$$

But other than that, we cannot say anything about whether $l_P$ as a length specifically in and of itself has any significance. The idea that space-time must discretize is actually not intrinsically required by quantum mechanics; it seems related to difficulties in quantum field theory, but the dirty secret physicists don't like to tell you is quantum field theory itself isn't even on really solid ground. That is, we do not truly have a quantum mechanical theory consistent 100% with special relativity even, that can so much as completely and rigorously treat something as simple as the hydrogen atom! Instead, the best available is a collection of mathematical "hacks" that only have very limited applicability, particular perturbation approaches that only work for low energy scattering processes, i.e. pretty much "particle accelerator experiments", and not continuously-interacting systems like $\mathrm{H}\cdot$; or else "lattice" theories that do a better job for that case, but only by rudely discretizing spacetime so as to technically break special relativity; or else very ad-hoc kinds of perturbative, hand-wavy approaches specific to only particular and narrow questions about such bound systems, such as with the famous prediction of the Lamb shift.

Because of this, I would personally contend that trying too hard to worry about quantization of general relativity at this stage is kind of "jumping the gun" - let's get (R)QFT sorted first, and perhaps the answers will become clear. After all, the problems are chiefly described as breakdown of the perturbative scattering theory, and it should not be too much of a surprise, I'd think, that if you should take what is already a patchwork, stuck-together-with-peeling-glue-and-cardboard "pile of hacks" and then try to push it even a bit hard, much less as hard as GR does by complicating spacetime far more - keep in mind that RQFT itself is necessitated by the mere introduction of spacetime (as opposed to separated space + time) - that the result is that it falls to pieces.

Answer 5

It is expected to be. However, we are so far away from experimentally observing such a tiny length that it cannot be experimentally verified.

Answer 6

I would only like to add a few things that were not mentioned in the other answers.

1. We all forget that the Planck-length comes from the Planck-constant, which comes originally from the photoelectric effect, and from Planck's discovery about the correlation between the energy and the frequency of the photon (and originally the difference between the energy levels of the emitting electron).

2. If we look at QM, we could express all particles as having a wave-function, and all particles having a probability distribution, a frequency and a wavelength as well. But let's just take photons now. If a photon has a certain frequency, it corresponds to it's wavelength too, and the smallest frequency could be (in theory) set to the Planck-length. Why? Because Em waves are information. If you think of a photon as information, then it's frequency can be measured experimentally. And it can be expressed as f=E/h. So if we have a photon (or any particle) with frequency so small, that will correspond to a wavelength that is the scale of the universe, it will not make sense any more. We are simply not able (theoretically we are, but it makes no sense) to express anything with a frequency smaller then that.

3. If the frequency cannot be smaller then that, then it makes sense to limit it so something. I understand that it is not the exact match with the Planck-length, but still there is a sense in limiting the frequency-minimum (and the wavelength maximum) even if we live in an accelerating expansion.

4. The reason that I choose photons is that they represent information. And if the wavelength of information itself cannot be larger then the scale of the universe, then it's frequency cannot be smaller then the corresponding frequency. So whether that minimal frequency is exactly the Planck-time (and so corresponds to the Planck-length and the Planck-constant) or not, is another question, especially because it changes as the universe expands.

But since information itself cannot have a smaller frequency then that, I believe it does not make sense to talk about anything (with a frequency of) smaller then that size (and since nothing can have a larger wavelength then the scale of the universe).

Of course this does not correspond to the spatial extension of particles, which, to our knowledge today in some cases are point like (and thus smaller then the Planck-length). I have no information on how we can measure experimentally something's spatial extension if this extension is smaller then the smallest possible frequency of a photon (that has the wavelength of the sale of the universe), but it would be nice to know.