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Say we want to make an educated guess for critical values of time, distance and mass, where quantum gravity effects are supposed to be non-negligible. These values are given the prefix "Planck-". Now, the way that they are "derived" is to take the speed of light $c$, Planck's constant $h$ and the gravitational constant $G$. Their units contain mass [$\mathrm{kg}$], distance [$\mathrm{m}$] and time [$\mathrm{s}$]. So say we want to get the Planck distance $d_\text{Pl}$, then we just make the ansatz

$$c^A h^B G^C = d_\text{Pl}$$

where $A$,$B$,$C$ are unknown exponents that are determined by equaling the dimensions on both sides of the equation (3 unknowns: $A$,$B$,$C$ and 3 units: $\mathrm{kg}$,$\mathrm{m}$,$\mathrm{s}$). This determines $d_\text{Pl}$ uniquely.

The analogous ansatz is made for the Planck time and mass.

My question: Why do we think that this gives us a valid guess on the regime for quantum gravity?

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Possible duplicate: physics.stackexchange.com/q/20883/2451 –  Qmechanic Sep 16 '12 at 4:55
    
While that other question may be of interest to people reading this one, I don't think it's actually a duplicate. The other question doesn't have anything to do with Planck units, except to the extent that the Planck scale presents itself as a sort of natural high-energy bound for renormalization. –  David Z Sep 16 '12 at 5:17
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3 Answers 3

up vote 5 down vote accepted

Excellent question! As far as I know, there is nothing like a rigorous justification that quantum gravity effects should kick in at the Planck mass/length/time. But there is an intuitive justification, which goes something like this:

As you may know, in special relativity, many quantities can be expressed as a power series in $\frac{v^2}{c^2}$.

$$X = X_0 + X_1\frac{v^2}{c^2} + \cdots$$

When you are working with a physical situation where $\frac{v}{c} \ll 1$, you can neglect the higher-order terms in this expression. This means that $c$ is, in a sense, a characteristic speed at which relativistic effects (the higher-order terms) make a significant difference. Of course this doesn't mean you can completely ignore relativity at speeds less than $c$, but a good rule of thumb is that relativistic corrections are insignificant at speeds less than about $c/10$, which is only one order of magnitude off. In our everyday experience, eight orders of magnitude less than the speed of light, that one factor of 10 doesn't really matter that much.

A similar thing occurs in quantum mechanics, although not quite as clear-cut. Basically, quantum effects include things like interference and diffraction of wavefunctions, which become significant when the wavelength is comparable to the sizes of the objects involved. (This is a result of classical wave mechanics; I won't get into the details here.) So you can think of a quantity like $\frac{\lambda}{r} = \frac{h}{rp}$ playing a role similar to the one $\frac{v^2}{c^2}$ does in SR: it gives you an order-of-magnitude estimate of the scale at which quantum effects come into play. It's a very blunt estimate, though, because quantum corrections take many different forms - think of quantum mechanics as giving a power series in "$\frac{h}{\text{stuff}}$," not necessarily $\frac{h}{rp}$ exactly. In practice, it often turns out that $\frac{h}{rmc}$ is a better parameter to use; $\frac{h}{mc}$ is called the Compton wavelength.

Finally, the same thing can be said of general relativity, specifically the part of it that deals with strong gravitational fields. In this case the relevant parameter is $\frac{2Gm}{rc^2}$, which pops up in various exact solutions to the Einstein field equations. This means that the Schwarzschild radius $\frac{2Gm}{c^2}$ is a characteristic radius at which general relativistic effects become significant.

Now, a quantum theory of gravity should combine all of these things. So a generic quantity might be calculable in quantum gravity as a multivariable power series like this:

$$\begin{align}X &= X_{0,0,0} + X_{1,0,0}\frac{v^2}{c^2} + X_{0,1,0}\frac{h}{rmc} + X_{0,0,1}\frac{2Gm}{rc^2} + \cdots \\ &= \sum_{m,n,p} X_{m,n,p}\biggl(\frac{v^2}{c^2}\biggr)^m\biggl(\frac{h}{rmc}\biggr)^n\biggl(\frac{2Gm}{rc^2}\biggr)^p\end{align}$$

where $X_{0,0,0}$ is the classical, low-speed approximate value, and the subscripts refer to how many powers of the various correction factors are involved. Any term which has zeros for certain combinations of the subscripts, we can already calculate using existing theories. For example, calculating some quantity using special relativity gives us the power series

$$X_\text{SR} = \sum_{n} X_{n,0,0}\biggl(\frac{v^2}{c^2}\biggr)^n$$

which takes care of all the $(n,0,0)$ terms. Similarly, quantum mechanics takes care of all the $(0,n,0)$ terms, and GR all the $(0,0,n)$ terms - in fact, because GR includes special relativity, you can also get the $(m,0,n)$ terms from it. The terms with indices of the form $(m,n,0)$ come from the combination of special relativity and quantum mechanics, namely quantum field theory.

So what's left? The remaining terms, which aren't covered by existing theories, are going to be negligible unless we have a system that is:

  • very dense, $r \sim \frac{2Gm}{c^2}$
  • and very small, $r \sim \frac{h}{mc}$

Equating these conditions, we get $\frac{2Gm}{c^2} \sim \frac{h}{mc}$, or

$$m \sim \sqrt{\frac{hc}{2G}} \sim m_\text{Pl}$$

(I dropped a factor of $\sqrt{\pi}$ because we're just working in orders of magnitude), and plugging back in to either of the conditions,

$$r \sim \frac{h}{m_\text{Pl}c} \sim \ell_\text{Pl}$$

So the kinds of systems where the $(0,m,n)$ terms in this "master power series" become relevant are precisely those where the mass is on the order of the Planck mass and the size is the order of the Planck length. For calculating the behavior of these systems, we will need a theory which allows us to calculate those $(0,m,n)$ terms, and in fact all the terms with arbitrary indices - in other words, a quantum theory of gravity.

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Great answer! Thanks! –  Simon Sep 16 '12 at 10:52
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The interesting thing is that universal constant of gravitation is not even fixed constant, (why there is no accuracy of the measured value of $G$?)

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On the contrary, in current theory $G$ is constant. It is just hard to measure to more than a few digits precision. –  dmckee Oct 15 '12 at 3:05
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Because the meaning of these constants gives you that the Planck length and Planck time are limits to space, while the Planck mass is the lower-limit for a black hole (anything lighter will be called a fundamental particle).

You should understand the constants as follows: c converts from seconds to meters, it is showing that time and space is the same. You should set it to one, it is the least interesting constant of the three.

In units where c=1, G has units of length/mass, so it converts the unit of mass to a unit of length. The length associated with the mass is the black hole radius of that given mass, up to a small factor.

hbar tells you the Energy/momentum time/position uncertainty, so that the radius at which a black hole's compton wavelength is comparable to its size is the Planck length.

This is the smallest distance you can probe, since a particle of Planck energy is required to probe the Planck length, and it will produce a black hole of size the Planck length. So unlike at accelerators, where higher energies probe shorter distances, at higher energy scales than the Planck mass, you don't probe shorter distances, rather you end up making large black holes.

This means that, up to small factors, the Planck length and Planck time are the shortest distances you can measure. An application of logical positivism allows you to conclude that shorter distances are likely meaningless in quantum gravity. The Planck mass is the boundary between an elementary particle and black hole. This is dimensional analysis, so that you only trust it up to a dimensionless factor.

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You should say: shortest proper distances/times –  drake Sep 16 '12 at 8:41
    
@drake: No I shouldn't. –  Ron Maimon Sep 16 '12 at 19:52
    
Lorentz–FitzGerald contraction? Time dilation? –  drake Sep 17 '12 at 0:01
    
@drake: The measurement defines the uncertainty in discriminating between points, and the uncertainty is in position and time in specification relative to a full frame, not in proper time. The measurements of position one is imagining are not along a light-ray, these measurements cannot require me to identify a point here with a point at a nearly equal |x| and |t| value in the Andromeda galaxy. These points are only identified in a ridiculously boosted frame. The structure isn't a lattice, it needs to be compatible with relativity, and this gives the graininess different words. –  Ron Maimon Sep 17 '12 at 7:43
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@drake-- if you say "I can't discriminate between two points at small proper distance" then you are saying you can't discriminate between all the points along a light ray, no matter how far apart these points are. The invariant distance is not a topological separation distance. –  Ron Maimon Sep 17 '12 at 21:53
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