# How to get Planck length

I know that what Planck length equals to.

1. The first question is, how do you get the formula $$\ell_P~=~\sqrt\frac{\hbar G}{c^3}$$ that describes the Planck length?

2. The second question is, will any length shorter than the Planck length be inaccessible? If so, what is the reason behind this?

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Hi user2346! For future reference, we prefer that you ask each separate question in a separate post. –  David Z May 21 '12 at 20:35

The expression $(\hbar G/c^3)^{1/2}$ is the unique product of powers of $\hbar, G,c$, three most universal dimensionful constants, that has the unit of length. Because the constants $\hbar, G,c$ describe the fundamental processes of quantum mechanics, gravity, and special relativity, respectively, the length scale obtained in this way expresses the typical length scale of processes that depend on relativistic quantum gravity.

The formula and the value were already known to Max Planck more than 100 years ago, that's why they're called Planck units.

Unless there are very large or strangely warped extra dimensions in our spacetime, the Planck length is the minimum length scale that may be assigned the usual physical and geometric interpretation. (And even if there are subtleties coming from large or warped extra dimensions, the minimum length scale that makes sense – which could be different from $10^{-35}$ meters, however – may still be called a higher-dimensional Planck length and is calculated by analogous formulae which must, however, use the relevant Newton's constant that applies to a higher-dimensional world.) The Planck length's special role may be expressed by many related definitions, for example:

• The Planck length is the radius of the smallest black hole that (marginally) obeys the laws of general relativity. Note that if the black hole radius is $R=(\hbar G/c^3)^{1/2}$, the black hole mass is obtained from $R=2GM/c^2$ i.e. $M=c^2/G\cdot (\hbar G/c^3)^{1/2} = (\hbar c/G)^{1/2}$ which is the same thing as the Compton wavelength $\lambda = h/Mc = hG/c^3 (\hbar G/c^3)^{-1/2}$ of the same object, up to numerical factors such as $2$ and $\pi$. The time it takes for such a black hole to evaporate by the Hawking radiation is also equal to the Planck time i.e. Planck length divided by the speed of light. Smaller (lighter) black holes don't behave as black holes at all; they are elementary particles (and the lifetime shorter than the Planck time is a sign that you can't trust general relativity for such supertiny objects). Larger black holes than the Planck length increasingly behave as long-lived black holes that we know from astrophysics.

• The Planck length is the distance at which the quantum uncertainty of the distance becomes of order 100 percent, up to a coefficient of order one. This may be calculated by various approximate calculations rooted in quantum field theory – expectation values of $(\delta x)^2$ coming from quantum fluctuations of the metric tensor; higher-derivative corrections to the Einstein-Hilbert action; nonlocal phenomena, and so on.

The unusual corrections to the geometry, including nonlocal phenomena, become so strong at distances that are formally shorter than the Planck length that it doesn't make sense to consider any shorter distances. The usual rules of geometry would break down over there. The Planck length or so is also the shortest distance scale that can be probed by accelerators, even in principle. If one were increasing the energy of protons at the LHC and picked a collider of the radius comparable to the Universe, the wavelength of the protons would be getting shorter inversely proportionally to the protons' energy. However, once the protons' center-of-mass energy reaches the Planck scale, one starts to produce the "minimal black holes" mentioned above. A subsequent increase of the energy will end up with larger black holes that have a worse resolution, not better. So the Planck length is the minimum distance one may probe.

It's important to mention that we're talking about the internal architecture of particles and objects. Many other quantities that have units of length may be much shorter than the Planck length. For example, the photon's wavelength may obviously be arbitrarily short: any photon may always be boosted, as special relativity guarantees, so that its wavelength gets even shorter.

Lots of things (insights from thousands of papers by some of the world's best physicists) are known about the Planck scale physics, especially some qualitative features of it, regardless of the experimental inaccessibility of that realm.

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According to which established, experimentally verified theory can one assert that''once the protons' center-of-mass energy reaches the Planck scale, one starts to produce the "minimal black holes" mentioned above''? Where is a proof of the assertion that ''the usual rules of geometry would break down'' at distances shorter than the Planck scale? How,in the absence of a consistent theory of quantum gravity, can one prove that ''the Planck length is the radius of the smallest black hole that (marginally) obeys the laws of general relativity''? –  Arnold Neumaier May 21 '12 at 16:18
Dear Arnold, according to which established, experimentally verified theory can one assert what I did? The theory you're looking for is known as general relativity. One may prove that with a sufficient concentration of energy in a small volume such as one I described, one inevitably forms black holes. This has been known since the singularity theorems due to Penrose and Hawking from the 1970s. Also, for radii larger than the Planck length, one may show that the corrections to GR are small so that the conclusion is unchanged. –  Luboš Motl May 21 '12 at 16:33
One may prove that distances shorter than the Planck scale fail to obey the laws of geometry in many independent ways, from semiclassical GR to individual full-fledged consistent descriptions of string/M-theory, from AdS/CFT to Matrix theory. –  Luboš Motl May 21 '12 at 16:34
Try e.g. page 3 of Joseph Polchinski's "String Theory" or the first chapter of any other basic textbook about the subject. –  Luboš Motl May 21 '12 at 18:24
Let me just mention that those of us who have actually studied the subject – after the equivalent of those 1,000 pages and maybe a bit earlier – not only know that classical GR becomes invalid near the Planck scale, which one understands from page 3, but we also have a working knowledge of the actual correct physics that replaces it. I don't quite understand the logic of "arguing against" my proofs organized by someone who not only confesses not to have the knowledge of those 1,000 pages – but who seems boldly ignorant even about the "first three pages" of introductions to the subject. –  Luboš Motl May 22 '12 at 5:21

Using fundamental physical constants, try to construct an expression which has a length unit.
So using dimensional analysis, we have:

• $G = m^3 \cdot kg^{-1} \cdot s^{-2}$
• $c = m \cdot s^{-1}$
• and $\hbar = J \cdot s = kg \cdot m^2 \cdot s^{-1}$.

Than we are to construct length $l = m$ in the following way: $$l = G^a c^b \hbar^d = m^{3a + b+d} \cdot kg^{-a+d} \cdot s^{-2a-b-d} \equiv m$$ It's equivalent to the following system of equations $$\begin{cases} 3a+b+2d & = 1 \\-a+d & = 0 \\-2a-b-d & = 0 \end{cases}$$ And the only solution is just what we call now Planck's length.

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The formula is obtained by dimensional analysis. Up to a constant dimensionless factor, the given expression is the only one of dimension length that one can make of the fundamental constants $\hbar$, $c$, and $G$.

Discussions about the physical significance of the Planck length have no experimental (and too little theoretical) support, so that your second question cannot be answered (except speculatively).

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Comment to the answer (v1): The sentence [...]the given expression is the only dimensionless one[...] seems to be a typo, since $\ell_P$ is not dimensionless, but has dimension of length. –  Qmechanic May 22 '12 at 12:20
@Qmechanic: Thanks, I corrected it. –  Arnold Neumaier May 22 '12 at 14:05

I must agree with Lubos (except for the exception he makes regarding Photons, since SR is the wrong tool to use and GR doesn't let Photons stand out either) that it's theoretically very well established that Planck's scale sets a point beyond which new physics should happen and string theory gives one possible form this new physics might take.

Forgetting about strings, other than Blackhole arguments, one can appeal to the modern RG framework to claim any renormalizable but not asymptotically-free field theory at low energies (like Standard Model) signals the existence of a UV scale beyond which a new field theory must get replaced. The Planck's scale is the only relevant scale we know that might possibly be the candidate for a gravitational qft. Look at Delamotte's "A hint of renormalization" for a clear description of this point.

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It's clear to me that if there are only a finite number of (length) scales in the theory, that one can expect something to happen in field theoretic considerations w.r.t. to that unit (say Planck scale here $\ell_P$). However, what makes $\ell_P$ more fundamental than $2\ell_P$, how can one conclude that a particular numeric value has any significance without a good theory one that level? –  NikolajK May 22 '12 at 14:23
Just based on general considerations, the transition doesn't even have to be sharp I guess (it will be sharp only if some symmetry is spontaneously broken and I'm ignorant of whether or not that must be the case for a renormalizable low-energy theory to appear). So I agree with you, nothing is so special about Planck's length before considering a specific consistent QG. –  Arash May 22 '12 at 15:48

This is an answer to the part of the question about why smaller scales are inaccessible.

Particle physicists are in the business of measuring things at very small distances. To do this, they have to use particles with wavelengths comparable to the distance scale they're trying to probe, and they have to collide those particles with the thing they're trying to probe.

However, something goes wrong if you keep trying to make the wavelength $\lambda$ shorter and shorter. Although accelerating a particle to ultrarelativistic speed doesn't make it into a black hole (after all, in its own frame it's at rest), the collision with the object being probed can create a black hole, and it will do so, roughly speaking, if the energy $E$ is equivalent to an $mc^2$ for which the Schwarzschild radius $2Gm/c^2$ is smaller than the $\lambda\sim hc/E$. (This is not rigorous, since it's really the stress-energy tensor that matters, not the energy, but it's good enough for an order-of-magnitude estimate.) Solving for $\lambda$, we get something on the order of the Planck length.

If you make the wavelength shorter than the Planck length, you're making the energy higher. The collision then produces a larger black hole, which means you're not probing smaller scales, you're probing larger ones.

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