Why is the Planck length considered fundamental, but not the Planck mass?
Solution 1
From the perspective of particle physics, you are correct, the Planck length and Planck mass are essentially equivalent concepts: the Planck mass describes a (very high) energy scale ($\sim 10^{19}$ GeV) at which new physics must emerge, just as the Planck length entails a (very short) length scale beyond which we need a new description. If we set $\hbar=c=1$ (which are really just conversion factors between units) we see that they are inverses of each other, $m_P=1/l_P$.
More precisely, if we take the EinsteinHilbert action for gravity and expand around a flat metric $g_{\mu\nu} = \eta_{\mu\nu} + h_{\mu\nu}$, where we can interpret $h_{\mu\nu}$ as the graviton field, the resulting action will have an infinite number of higher order terms suppressed by powers of the Planck mass. Roughly, we have $$\mathcal{L}_{EH} \sim \frac{1}{2} \partial h\partial h+ \frac{1}{m_P} h\partial h \partial h + \frac{1}{m_P^2} h^2\partial h \partial h + \ldots $$ (as well as terms from higher derivative corrections, which are also higher order in $1/m_P$). So we have predictive control at energies scales much less than $m_P$, where the infinite number of higher order terms can be ignored. But once we reach the Planck scale (i.e. energy scales of $m_P$ or length scales of $l_P$) the nonrenormalizable effects become important and all the quantum corrections and higher order terms render the above Lagrangian equation useless, and we require a new description.
Solution 2
The Planck mass makes the units work out "nicely" in a lot of equations, sort of like radians are a very "natural" unit of angle measure, or $e \approx 2.71828...$ is a very "natural" base for exponential functions and logarithms.
But sizewise, the Planck mass isn't anything special. Wikipedia says a flea egg weighs about one Planck mass; so, it is possible to have masses much smaller than the Planck mass.
Mass isn't "quantized" in the sense that every object has mass an integer multiple of the Planck mass, the way electric charge is "quantized" in the sense that every object has electric charge an integer multiple of the charge on an electron (or, if you prefer, the charge on a quark).
Solution 3
You haven't understood the concepts here quite right I think. It is not that ordinary physics cannot describe things happening over small distances (Planck length for example), it is rather a question of interaction energies between pointlike entities such as quarks and electrons. Even Newtonian physics can describe an ordinary ball moving through a distance of one Planck length. But if a process is characterised in its dynamics by very short distances, then quantum theory will be needed.
The Planck mass is important in that if the collision energy between pointlike particles is of order one Planck mass multiplied by $c^2$, then we need a quantum gravity type of theory to describe the process.
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korni1990
Updated on August 01, 2022Comments

korni1990 5 months
The planck length is considered by many to be a lower bound of the scale where new physics should appear to account for quantum gravity. The reasoning behind, as far as I understand, is that $l_{P}=\sqrt{\dfrac{\hbar G}{c^3}}$ consists of the fundamental constants of gravity and relativistic quantum mechanics.
By the same argument $m_{P}=\sqrt{\dfrac{\hbar c}{G}}$ should be equally important, no?
What am I missing?

robert bristowjohnson over 2 yearsTo make them all more "fundamental", $G$ and $k_\mathrm{e}$ should both be normalized to $\frac{1}{4\pi}$. This, along with normalizing the speed of propagation $c$ to $1$ will result in the characteristic impedance of free space set to $1$ for both EM and gravitational radiation.


4xion over 2 yearsSeems not quite right to say that $m_p$ isn't anything special as it corresponds to the scale at which the nonrenormalizable nature of gravity becomes important. Another minor comment: electric charge appears quantized but (unless we observe monopoles) there's no guarantee of this.

Henry over 2 yearsThe Planck mass does not in itself need a new physics (plenty of living creatures are larger and many others are smaller). But a black hole with the Planck mass may need a new physics.

Rivers McForge over 2 years@4xion Right, I’m using “special” in the narrow sense of “not a unit of quantization” here. My impression is also that the change in physics from mass scales above the Planck mass to below the Planck mass is far less drastic than the change in physics from length scales above the Planck length to below the Planck length.

Bob Stein over 2 years@4xion did I correct that typo correctly? unless > useless

Rick over 2 yearsWhat creature is smaller than a Planck length

golf69 over 2 years@Rick not the Planck length, but the Planck mass (~10^19u)

4xion over 2 years@Henry Such a mass is spread out over a huge distance (on the scales relevant for particle physics). But you’re missing the point, the electrons, photons, etc all around you are usually interacting at energies far far below the Planck scale. But once you consider interactions at these high energies, you’ll need new physics.

Henry over 2 years4xion You are saying that a Planck mass concentrated at a single point (or within a Planck length to meet @Rick's question) needs new physics; I do not disagree, but that is not a question of mass alone but one of density. Similarly it is easy enough to have explosions orders of magnitude larger or smaller than the Planck energy, but you only need new physics when this is sufficiently concentrated.

4xion over 2 yearsI don't disagree, I just think thinking about concentration is a bit misleading, especially in QFT where we have interactions which are local and pointlike. It's the energy of these local interactions that's important. The energies of the explosions you mention are just summing up lots of interactions at very low energies (compared to the Planck scale), which is not really what we mean by Planck scale physics. But I do agree that the idea of a local interaction is problematic at high energies

4xion over 2 yearsIt's also maybe worth mentioning that the above Lagrangian also breaks down at very large gravitational field values of $h$, which is just the statement that you need a new description in extremely high curvature regimes (e.g. for those in a black hole)

4xion over 2 years@BobStein re typo: good catch, thanks!