Тёмная сила вместо тёмной материи

Last modified by Alexey Popov on 2019/10/02 18:30

$\nabla \cdot \mathbf g = G\rho$

$\mathbf g = -G \frac{M \mathbf R}{R^3}$

$\mathbf g = - \nabla \varphi - \frac{\partial \mathbf A}{\partial t} - \nabla \times \mathbf M$

$\mathbf g = - \nabla \varphi - \frac{\partial \mathbf A}{\partial t}$

$\varphi$

$\mathbf A$

$\mathbf M$

$\varphi = G \frac{M}{R}$

$\mathbf A = \frac{1}{c^2}G \frac{M \mathbf v}{R}$

$\mathbf M = \frac{1}{c^2}G\frac{M}{R}(\mathbf r \times \mathbf a)$

$\mathbf g \sim \frac{\text{1} }{R}$

$\mathbf g = - \nabla \varphi - \frac{\partial \mathbf A}{\partial t} - \nabla \times \mathbf M = - G \frac{M}{R^2} -\frac{1}{c^2}G \frac{\mathbf v}{R} \frac{\partial M}{\partial t} - \frac{1}{c^2}G \, \nabla \times\frac{M}{R}(\mathbf r \times \mathbf a)$

$\mathbf g = - \nabla \varphi - \frac{\partial \mathbf A}{\partial t} = - G \frac{M}{R^2} -\frac{1}{c^2}G \frac{\mathbf v}{R} \frac{\partial M}{\partial t}$

$\boldsymbol{\omega} = \nabla \times \mathbf A - \frac{1}{c^2}\frac{\partial \mathbf M}{\partial t} -\frac{1}{c^2} \nabla \xi$

$\xi = \frac{1}{c}G \frac{M (\mathbf v \cdot \mathbf v)}{R}$

$\mathbf g R= - \frac{1}{c^2}{G \mathbf v} \frac{\partial M}{\partial t} = \text{k}$

$\nabla \cdot \mathbf M + \frac{1}{c^2}\frac{\partial \xi}{\partial t} = 0$

$\nabla \cdot \mathbf A + \frac{1}{c^2}\frac{\partial \varphi}{\partial t} = 0$

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Created by Alexey Popov on 2019/10/02 13:56

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