Timeline for Minkowski space-time

Current License: CC BY-SA 3.0

9 events

when toggle format	what		by	license	comment
Jan 2, 2016 at 17:42	history	edited	Omar Nagib	CC BY-SA 3.0	deleted 5 characters in body
Dec 30, 2015 at 22:06	comment	added	Gyro Gearloose		I see, $\Lambda^T=\Lambda$, so $\eta=\Lambda^T\eta\Lambda=\Lambda\eta\Lambda$. If $v$ is an eigenvector of $\Lambda$ with $v\Lambda=\alpha v$ then $v\Lambda\eta\Lambda=v\eta$ and $v\Lambda\eta\Lambda=\alpha v\eta\Lambda=v\eta$. Thus $v\eta$ is an eigenvector of $\Lambda$ and thus $\eta$ maps eigenvectors onto eigenvectors of $\Lambda$. So, up to scaling (or permutation), not much of a choice.
Dec 30, 2015 at 21:49	comment	added	Omar Nagib		@GyroGearloose But the physics is the same if you choose any other $n$.
Dec 30, 2015 at 21:48	comment	added	Omar Nagib		@GyroGearloose Indeed $\mathbf \Lambda^T \eta \mathbf\Lambda=\eta$ has no unique solution, its general solution is given by $$\eta = \begin{bmatrix}n & 0\\0 & -n\end{bmatrix}$$ for all $n$ that is real. The solution $n=0$ is excluded for being trivial, since any arbitrary transformation matrix $A$ satisfy it, therefore the zero matrix metric does not uniquely specify Lorentz transformation. Actually you can choose any $n$(except $0$), but then your inner product will be given by $n(x^2-t^2)$. Obviously $n=1$ is the most natural and convenient to work with.
Dec 30, 2015 at 18:01	history	edited	Omar Nagib	CC BY-SA 3.0	added 1 character in body
Dec 30, 2015 at 18:00	vote	accept	FUUNK1000
Dec 30, 2015 at 17:25	history	edited	Omar Nagib	CC BY-SA 3.0	added 650 characters in body
Dec 30, 2015 at 17:23	comment	added	Gyro Gearloose		How exactly do you solve $\mathbf \Lambda^T \eta \mathbf\Lambda=\eta$? I am stuck on that, because your $\eta$ can be scaled and so is not the only solution. Can't see at the moment why there could not be even other solutions.
Dec 30, 2015 at 17:04	history	answered	Omar Nagib	CC BY-SA 3.0