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seen Jul 8 at 18:00

Aug
19
awarded  Popular Question
Jul
2
awarded  Curious
May
21
revised Why does Minkowski space provide an accurate description of flat spacetime?
Clarified question.
May
21
revised Why does Minkowski space provide an accurate description of flat spacetime?
Clarified question.
May
21
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.
May
21
revised Why does Minkowski space provide an accurate description of flat spacetime?
Clarified question.
May
21
revised Why does Minkowski space provide an accurate description of flat spacetime?
Clarified question.
May
21
revised Why does Minkowski space provide an accurate description of flat spacetime?
Clarified question.
May
20
revised Why does Minkowski space provide an accurate description of flat spacetime?
added 1216 characters in body
May
20
asked Why does Minkowski space provide an accurate description of flat spacetime?
Mar
18
accepted Derivation of the general Lorentz transformation
Mar
18
answered Derivation of the general Lorentz transformation
Mar
15
awarded  Commentator
Mar
15
awarded  Informed
Mar
15
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.
Mar
15
revised Derivation of the general Lorentz transformation
Added a question.
Mar
15
asked Derivation of the general Lorentz transformation
Mar
15
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.
Mar
15
comment Special relativity: how to prove that $g = L^t g L$?
Thanks. I guess it would only follow if $X^\textrm{t}gX = X^\textrm{t}L^\textrm{t}gLX$.
Mar
15
accepted Special relativity: how to prove that $g = L^t g L$?