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advanced_tools:group_theory:desitter [2018/03/24 11:08] jakobadmin [Why is it interesting?] |
advanced_tools:group_theory:desitter [2018/03/24 11:09] jakobadmin [Student] |
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C.f. http://physics.stackexchange.com/questions/299621/the-desitter-group-vs-the-poincare-group-for-a-non-zero-cosmological-constant | C.f. http://physics.stackexchange.com/questions/299621/the-desitter-group-vs-the-poincare-group-for-a-non-zero-cosmological-constant | ||
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<blockquote>Given the current picture of an increasingly expanding universe, it may very well be the case that future generations of students first learn about the de Sitter and the Newton-Hooke groups, while the Poincaré and the Galilei groups will be considered to be nothing but historical aberrations. <cite>Symmetries in Fundamental Physics by Kurt Sundermeyer</cite></blockquote> | <blockquote>Given the current picture of an increasingly expanding universe, it may very well be the case that future generations of students first learn about the de Sitter and the Newton-Hooke groups, while the Poincaré and the Galilei groups will be considered to be nothing but historical aberrations. <cite>Symmetries in Fundamental Physics by Kurt Sundermeyer</cite></blockquote> | ||
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<tabbox Student> | <tabbox Student> | ||
- | <note tip> | + | <blockquote> |
- | In this section things should be explained by analogy and with pictures and, if necessary, some formulas. | + | The group $SO(d,1)$ moves points on $dS^d$ around. We thus conclude that, just like the sphere, deSitter spacetime is maximally symmetric. So, according to the general theory of maximally symmetric spaces explained in chapter IX.6, the Riemann curvature tensor $R_{\mu\nu\lambda\sigma}$ must be equal to $(g_{\mu\lambda}g_{\nu\sigma}-g_{\mu\sigma}g_{\nu\lambda})$ up to an overall constant. [...] Then, by dimensional analysis, we must have |
- | </note> | + | $$ R_{\mu\nu\lambda\sigma} =\frac{1}{L} (g_{\mu\lambda}g_{\nu\sigma}-g_{\mu\sigma}g_{\nu\lambda}) $$ |
+ | [...] de Sitter spacetime is a solution of Einstein's field equation $R_{\mu \nu}=8\pi G\Lambda g_{\mu\nu}$, with a positive cosmological constant given by | ||
+ | $$ 8\pi G\Lambda = \frac{3}{L}$$ | ||
+ | [...] | ||
+ | Topologically, de Sitter spacetime is $R \times S^3$, with a spatial section given by $S^3$, as just explained. In contrast, we know from chapters VI.2 and VI.5 that Einstein's field equation with a positive cosmological constant leads to $ds^2=-dt^2+e^{2Ht}(dx^2+dy^2+dz^2)$ with the Hubble constant given by $H= \left( \frac{8\pi G}{3}\Lambda\right)^{1/3}$, or in terms of the de Sitter length, $H=1/L$. | ||
+ | |||
+ | <cite>page 627 in Einstein Gravity in a Nutshell - A. Zee</cite> | ||
+ | </blockquote> | ||
<tabbox Researcher> | <tabbox Researcher> |