# fctaylor25

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# 47 Actions

 Aug19 awarded Popular Question Jul2 awarded Curious May21 revised Why does Minkowski space provide an accurate description of flat spacetime? Clarified question. May21 revised Why does Minkowski space provide an accurate description of flat spacetime? Clarified question. May21 comment Why does Minkowski space provide an accurate description of flat spacetime? @Muphrid Thanks, I see how that was unclear. I've clarified my question to reflect this. May21 revised Why does Minkowski space provide an accurate description of flat spacetime? Clarified question. May21 revised Why does Minkowski space provide an accurate description of flat spacetime? Clarified question. May21 revised Why does Minkowski space provide an accurate description of flat spacetime? Clarified question. May20 revised Why does Minkowski space provide an accurate description of flat spacetime? added 1216 characters in body May20 asked Why does Minkowski space provide an accurate description of flat spacetime? Mar18 accepted Derivation of the general Lorentz transformation Mar18 answered Derivation of the general Lorentz transformation Mar15 awarded Commentator Mar15 awarded Informed Mar15 comment Derivation of the general Lorentz transformation $U = \left(\begin{matrix} 1 & 0 \\ 0 & K^\textrm{t} \end{matrix}\right) \implies U^\textrm{t} = \left(\begin{matrix} 1 & 0 \\ 0 & K \end{matrix}\right) = \left(\begin{matrix} 1 & 0 \\ 0 & H \end{matrix}\right)$. However, I don't think that $K = H$. Secondly, I understand that you can use an intermediate basis, but I don't see the point: why not go straight back to the original basis? Thanks for helping me figure this out. Mar15 revised Derivation of the general Lorentz transformation Added a question. Mar15 asked Derivation of the general Lorentz transformation Mar15 comment Special relativity: how to prove that $g = L^t g L$? I'm glad you added the section on intuition. I was wondering how you thought of that. Mar15 comment Special relativity: how to prove that $g = L^t g L$? Thanks. I guess it would only follow if $X^\textrm{t}gX = X^\textrm{t}L^\textrm{t}gLX$. Mar15 accepted Special relativity: how to prove that $g = L^t g L$?