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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 necessarysome 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> ​
advanced_tools/group_theory/desitter.txt · Last modified: 2018/03/24 11:10 by jakobadmin