can you set up a FLRW-model in which the Hubble sphere has a constant comoving radius? what would be needed for such a construction?

bonus question: for which $n$ is the equality $\mathrm{i}^{n^2}=\pm\mathrm{i}^n$ with $\mathrm{i^2}=-1$ fulfilled?

The de Sitter model is such a model.

ReplyDeleteIn this model, but not in general, the Hubble sphere coincides with the event horizon. This confused some people until Rindler cleared it up in his paper in the fifties, but some younger dudes apparently haven't read his paper and are confused today.

answer to the bonus question: like the last week's case, it's best to split up the proof into 4 cases: $n=4m+q$, $q=0,1,2,3$ with the respective squares following from the binomial formula: $n^2=16m^2+8q+q^2$. then, $\mathrm{i}^{n^2}=+\mathrm{i}^n$ is fulfilled for $q=0,1$ and $\mathrm{i}^{n^2}=-\mathrm{i}^n$ for $q=2,3$.

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