Difference between revisions of "Planck scales and fundamental constants"

Let's analyze the methods for measuring of considered quantities to answer this quenstion. $E$ and $E/c^2$ are measured by the same procedure, for example, using a calorimeter, but frequency is measured in a fundamentally different way, for example, using a clock. Consequently, the expression $\omega = E/\hbar$ contains the relation between two different types of measurements, while expression $e = E/c^2$ does not. This metrological difference has mainly historical origin. Recall, that modern chronometers are based on the measurement of difference between energies of atomic levels.

This is impossible, since momentum $\vec p$ is a vector, while $e$ and $m$ are scalars in 3--dimensional space.

$l_{_{Pl}} = \sqrt {{G\hbar} \over {c^3}}=1.6 \times 10^{-35}\mbox{m} $, $t_{_{Pl}} = \sqrt{{G\hbar } \over {c^5 }}=5.38 \times 10^{-44}s$, $M_{_{Pl}} = \sqrt {{\hbar c} \over G}=2.18 \times 10^{-8} \mbox{kg}$, $T_{_{Pl}} = \sqrt {{\hbar c^5} \over {Gk_B^2}} = 1.4\cdot 10^{32}K$, $\rho _{_{Pl}} = {{c^5 } \over {G^2 \hbar }}=5.17 \times 10^{96} \mbox{kg/m}^3$

$10^{35}$

$r_g={{2GM_{_{Pl}}} \over {c^2}} = {\hbar \over M_{_{Pl}}c} = l_{_{Pl}}$

Quantum gravity effects can be neglected in the case when the Compton wavelength $\lambda _c = \frac{\hbar }[[:Template:Mc]]$ of the particle is much greater than the Schwarzschild radius $r_s = \frac[[:Template:2mG]][[:Template:C^2]]$. Parameter that defines the role of quantum gravity,\[\frac{\lambda _c }{r_s } \approx \frac{m_{Pl}^2}{m^2}\] For example, for an electron \[\frac{\lambda _c }{r_s} \approx 10^{45}\] so quantum gravity effects are completely negligible. The same is true for all other Standard Model particles.

$M_{Pl} = 1.42 \cdot 10^{32} \,K = 6.2 \cdot 10^{32}\mbox{\it cm}^{ - 1} = 1.8 \cdot 10^{43} s^{ - 1} $.

$G = 7.4243\times 10^{-29}~\mbox{cm/g}$.

Since the Coulomb force in $D$--dimensional space is $F_e \propto e^2 r^{-(D - 1)}$, the dimensionality of charge depends on the dimension of space, therefore $[\alpha ] = \left[ {{{e^2 } / {\hbar c}}} \right] = L^{D - 3}$.

Recall, that dimensionalities of ($e, G$) depend on the dimension of space. The required dimensionless combination is $$\alpha^{(D)} = e^{D - 1} \hbar ^{2 - D} c^{D - 4} G^{{{3 - D} \over 2}}. $$ Gravitational constant $G$ doesn't enter this relation only for $D=3$ so that $\alpha^{(3)} = {{e^2 } \over {\hbar c}}$.

$T^{GU} \approx 10^{28}~K$.