Physics Asked on October 29, 2021
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?
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 Einstein-Hilbert action for gravity and expand around a flat metric $g_{munu} = eta_{munu} + h_{munu}$, where we can interpret $h_{munu}$ 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 $$mathcal{L}_{EH} sim frac{1}{2} partial hpartial h+ frac{1}{m_P} hpartial h partial h + frac{1}{m_P^2} h^2partial h partial h + ldots $$ (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 non-renormalizable effects become important and all the quantum corrections and higher order terms render the above Lagrangian equation useless, and we require a new description.
Answered by 4xion on October 29, 2021
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 point-like 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 point-like particles is of order one Planck mass multiplied by $c^2$, then we need a quantum gravity type of theory to describe the process.
Answered by Andrew Steane on October 29, 2021
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 size-wise, 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).
Answered by Rivers McForge on October 29, 2021
Get help from others!
Recent Answers
Recent Questions
© 2024 TransWikia.com. All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP