Einstein's equation is $$G_{\mu\nu} + \Lambda g_{\mu\nu} = {8 \pi G \over c^4} T_{\mu\nu}$$ where $G_{\mu\nu} = R_{\mu\nu} - (1/2)g_{\mu\nu}\,R$ is the Einstein tensor, which combines the Ricci tensor, the scalar curvature and the metric tensor, $\Lambda$ is the cosmological constant, $T_{\mu\nu}$ is energy-momentum tensor of matter, $\pi$ is the number, $c$ is the speed of light, $G$ is Newton's gravitational constant. This equation can be written as $$\frac{1}{4\pi}(G_{\mu\nu} + \Lambda g_{\mu\nu}) = 2\left ({G\over c^3}\right )\left ({1\over c}\,T_{\mu\nu}\right )$$ where $\left ({1\over c}\,T_{\mu\nu}\right )$ is the density of the energy-momentum of matter; $({G/c^3})$ is also in the formula for the Planck length: $\ell_P=\sqrt {(G/c^3)\hbar}$. In the derivation of his equations, Einstein suggested that physical spacetime is Riemannian, ie curved. A small domain of it is approximately flat spacetime. For any tensor field $N_{\mu\nu...}$ value $N_{\mu\nu...}\sqrt{-g}$ we may call a tensor density, where $g$ is the determinant of the metric tensor $g_{\mu\nu}$. The integral $\int N_{\mu\nu...}\sqrt{-g}\,d^4x$ is a tensor if the domain of integration is small. It is not a tensor if the domain of integration is not small, because it then consists of a sum of tensors located at different points and it does not transform in any simple way under a transformation of coordinates [Dirac]. Here we consider only small domains. This is also true for the integration over the three-dimensional hypersurface $S^{\nu}$. Thus, Einstein's equations for small spacetime domain can be integrated by the three-dimensional hypersurface $S^{\nu}$. Have $$\frac{1}{4\pi}\int\left (G_{\mu\nu} + \Lambda g_{\mu\nu}\right )\sqrt{-g}\,dS^{\nu} = {2G \over c^3} \int \left (\frac{1}{c} T_{\mu\nu}\right )\sqrt{-g}\,dS^{\nu}$$ Since integrable spacetime ''domain'' is small, we obtain the tensor equation $$R_{\mu}=\frac{2G}{c^3}P_{\mu}$$ where $P_{\mu}=\frac{1}{c}\int T_{\mu\nu}\sqrt{-g}\,dS^{\nu}$ is the 4-momentum of matter, $R_{\mu}=\frac{1}{4\pi}\int\left (G_{\mu\nu}+\Lambda g_{\mu\nu}\right )\sqrt{-g}\,dS^{\nu}$ is the radius of curvature small domain. Since $P_{\mu}=mc\,U_{\mu}$ then $$R_{\mu}=\frac{2G}{c^3}mc\,U_{\mu}=r_s\,U_{\mu}$$ where $r_s$ is the Schwarzschild radius, $U_ {\mu}$ is the 4-speed, $m$ is the gravitational mass. This record reveals the physical meaning of $R_{\mu}$. Here $R_{\mu}R^{\mu}=r_s^2$ (compare $dx_{\mu}dx^{\mu}=dS^2$) There is a similarity between the obtained tensor equation and the expression for the gravitational radius of the body. Indeed, for static spherically symmetric field and static distribution of matter have $U_{0}=1, U_i=0 \,(i=1,2,3)$. In this case we obtain $$R_0= \frac{2G}{c^3}mc\,U_0=\frac{2G\,m}{c^2}=r_s$$ In a small area of spacetime is almost flat and this equation can be written in the operator form $$\hat R_{\mu}=\frac{2G}{c^3}\hat P_{\mu}=\frac{2G}{c^3}(-i\hbar )\frac{\partial}{\partial\,x^{\mu}}=-2i\,\ell^2_{P}\frac{\partial}{\partial\,x^{\mu}}$$ where $\hbar $ is the Dirac constant. Then commutator operators $\hat R_{\mu}$ and $\hat x_{\mu}$ is $$[\hat R_{\mu},\hat x_{\mu}]=-2i\ell^2_{P}$$ From here follow the specified uncertainty relations $$\Delta R_{\mu}\Delta x_{\mu}\ge\ell^2_{P}$$ Substituting the values of $R_{\mu}=\frac{2G}{c^3}m\,c\,U_{\mu}$ and $\ell^2_{P}=\frac{\hbar\,G}{c^3}$ and cutting right and left of the same symbols, we obtain the Heisenberg uncertainty principle $$\Delta P_{\mu}\Delta x_{\mu}=\Delta (mc\,U_{\mu})\Delta x_{\mu}\ge\frac{\hbar}{2}$$ Note that now, according to the equation $R_{\mu}=({2G}/{c^3})\,P_{\mu}$, together with the expressions for the energy-momentum quantum $P_{\mu}=\hbar\, k_{\mu}$ valid expressions for the quantum spacetime curvature $R_{\mu}=\ell^2_P\, k_{\mu}$ (but not quantum spacetime), where $k_{\mu}$ - the wave 4-vector. That is, the curvature of spacetime is quantized, but the quantization step is extremely small. This can serve as a basis for building a quantum theory of gravity. In the particular case of a static spherically symmetric field and static distribution of matter $U_{0}=1, U_i=0 \,(i=1,2,3)$ and have remained $$\Delta R_{0}\Delta x_{0}=\Delta r_s\Delta r\ge\ell^2_{P}$$ where $r_s$ is the Schwarzschild radius, $r$ is radial coordinate. Here $R_0=r_s$ and $x_0=c\,t=r $, since the matter moves with velocity of light in the Planck scale. Last uncertainty relation allows make us some estimates of the equations of general relativity at the Planck scale. For example, the equation for the invariant interval $dS$ in the Schwarzschild solution has the form $$dS^2=\left( 1-\frac{r_s}{r}\right)c^2dt^2-\frac{dr^2}{1-{r_s}/{r}}-r^2(d\Omega^2+\sin^2\Omega d\varphi^2)$$ Substitute according to the uncertainty relations $r_s\approx\ell^2_P/r$. We obtain $$dS^2\approx\left( 1-\frac{\ell^2_{P}}{r^2}\right)c^2dt^2-\frac{dr^2}{1-{\ell^2_{P}}/{r^2}}-r^2(d\Omega^2+\sin^2\Omega d\varphi^2)$$ It is seen that at the Planck scale $r=\ell_P$ spacetime metric is bounded below by the Planck length, and on this scale, there are real and virtual Planckian black holes.

Question: What is wrong here?


closed as off-topic by Ryan Unger, Kyle Kanos, ACuriousMind, CuriousOne, Gert Mar 23 '16 at 1:53

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    $\begingroup$ There is a completely unjustified switch here right in the middle of the "derivation": You write $P_\mu\propto\partial_\mu$ and suddenly consider $P$ and $x$ as operators, but you never fixed a Hilbert space upon which $P_\mu = \partial_\mu$ is supposed to hold, and seem completely unaware that $P_\mu \propto \partial_\mu$ is not how quantum field theory works. Also, you pull $U_\mu$ and $m$ out of nowhere without saying the four-velocity or mass of what they're supposed to be. What's the actual physics question here? We're not a site to review your work. $\endgroup$ – ACuriousMind Mar 22 '16 at 17:38
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    $\begingroup$ What is Dirac constant??? $\endgroup$ – Les Adieux Mar 22 '16 at 17:38
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    $\begingroup$ Dirac constant is $h/2\pi$ $\endgroup$ – Alexander Klimets Mar 22 '16 at 17:56
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    $\begingroup$ I'm voting to close this question as off-topic because you haven't defined most of the terms and this seems to be an advanced numerology problem, not physics. $\endgroup$ – Ryan Unger Mar 22 '16 at 20:20