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Talk:Counterexamples to Relativity (view source)
Revision as of 23:44, August 20, 2011
, 23:44, August 20, 2011→Open problems in cosmology
:::: Einstein's motivation for the cosmological constant was indeed a mistake. He wanted a static universe and put the cosmological constant as a repellent term to explain why, if gravity is attractive, everything isn't just lumped up together. This doesn't work however, since the equilibrium created by adding the term isn't stable. Later physicists took a new interpretation of the cosmological constant as the energy density of the vacuum. It turns out Einstein's mistake wasn't adding a cosmological constant, but not putting it on the other side of his equation with the stress–energy tensor!
:::: Einstein's motivation for the cosmological constant was indeed a mistake. He wanted a static universe and put the cosmological constant as a repellent term to explain why, if gravity is attractive, everything isn't just lumped up together. This doesn't work however, since the equilibrium created by adding the term isn't stable. Later physicists took a new interpretation of the cosmological constant as the energy density of the vacuum. It turns out Einstein's mistake wasn't adding a cosmological constant, but not putting it on the other side of his equation with the stress–energy tensor!
:::: Anyway, the cosmological constant is still used because data suggest it exists. I highly doubt cosmologists have stuck to it because of the treachery of Einstein's wife. --[[User:MatthewQ|MatthewQ]] 19:06, 20 August 2011 (EDT)
:::: Anyway, the cosmological constant is still used because data suggest it exists. I highly doubt cosmologists have stuck to it because of the treachery of Einstein's wife. --[[User:MatthewQ|MatthewQ]] 19:06, 20 August 2011 (EDT)
::::: ''''<math>H_{\mu\nu}=-8\pi T_{\mu\nu}</math>, where <math>H_{\mu\nu}</math> is what I call "Hubble tensor""''
::::: It's mentioned [[Einsteinian_gravitation#Step_8._Flatness_of_spacetime|here]] that the the "Hubble tensor" is anti-symmetric. If so, this equation doesn't make sense since the stress-energy tensor, <math>T_{\mu\nu}</math>, is symmetric. The only way for the above equation to hold is if the "Hubble Tensor" and the stress-energy tensor are both zero ( <math>H_{\mu\nu}=-8\pi T_{\mu\nu} = -8\pi T_{\nu\mu}= H_{\nu\mu} = - H_{\mu\nu}</math> and <math>H_{\mu\nu} = - H_{\mu\nu} \Rightarrow H_{\mu\nu} = 0</math>). Actually, since in the linked article the "Hubble Tensor" is defined as the antisymmetric part of Ricci tensor it is exactly zero. --[[User:MatthewQ|MatthewQ]] 19:44, 20 August 2011 (EDT)