Супергравитация - Темная Энергия

Last modified by Alexey Popov on 2021/12/22 11:30

$(\Delta - k^{2})\varphi =4\pi G(\rho_u - \rho)$

$\varphi = - G\frac{M}{R}e^{-k\cdot R} + \frac{4}{3}\pi G\rho_u R^2 e^{-k\cdot R} +C$

$H_0^2 = \frac{4}{3}\pi G\rho_u$

$k = \frac{c^2}{2GM_u} = \frac{H_0}{c}$

$-k^{2}+{\frac {\omega ^{2}}{c^{2}}}={\frac {c^{4}}{4G^2M_u^2}}$

$\Delta \phi +{\frac {\partial }{\partial t}}\left({\frac {1}{c^{2}}}{\frac {\partial \phi }{\partial t}}+\nabla \cdot \mathbf {A} \right)-\left({\frac {H_0}{c }}\right)^{2}\phi =4\pi G(\rho_u - \rho)$

$\Delta \mathbf {A} -\nabla \left({\frac {1}{c^{2}}}{\frac {\partial \phi }{\partial t}}+\nabla \cdot \mathbf {A} \right)-\left({\frac {H_0}{c }}\right)^{2}\mathbf {A} =4\pi \frac{G}{c^2}(\rho_u - \rho)\mathbf v$

Темная Материя -

$\Delta \mathbf a = \frac{G H_0}{c^2}\rho \mathbf v$

$\nabla \cdot \Delta \mathbf a = -\frac{G H_0^2}{c^2}\rho_u$

$\Delta \mathbf a \frac{ \mathbf S}{V} = \frac{G H_0^2}{c^2}\rho_u$

$\mathbf a = -\frac{1}{3}\frac{G H_0^2}{c^2}M_u$

$\mathbf s = - G \frac{M R_u}{R^3}H_0^2$

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Created by Alexey Popov on 2021/12/18 17:40

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