Let L be a Lorentz transformation in one dimension of space and one of time. Let V and W be the two world lines of light traveling in the positive and negative directions. Then V and W are eigenspaces of L (constancy of speed of light). If F is the linear map taking V into W and vice versa, with FF=1, then FLF is the inverse of L and det(F) = 1, which implies det(L) = 1. If (t1,t2) are coordinates with respect to V and W (i.e. for a point p, light travels t1 along V and t2 along W to get to p.), then L is represented by the 2x2 matrix diag(u,1/u), where u and 1/u are the eigenvalues of V and W.
2 Answers
No not really. The Lorentz transformation can be written in the form of a hyperbolic rotation \begin{align} \left( \begin{array}{cc} \cosh (w) & \sinh (w) \\ \sinh (w) & \cosh (w) \\ \end{array} \right) \end{align} where $\cosh(w)=\gamma$, $\sinh(w)=-(v/c)\gamma= -\beta/\gamma$, with $\gamma=1/\sqrt{1-\beta^2}$. Your derivation is just one where $u=e^{w}$. Moreover, unless you can connect the eigenvalues to the parameter $\beta=v/c$, it's not terribly useful to have everything in terms of $u$.
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Sorry for reading too quickly. That derivation looks fine, but it's not what I would really call particularly new. Your derivation centers on several assumptions:
- Lorentz transformations are linear, so they can be represented by matrices
- The speed of light is the same in both directions, and the inverse of a boost is a boost with the same speed in the opposite direction (space is isotropic and boosts compose with one another)
- The speed of light is the same in all reference frames
These are the fundamental assumptions baked into all derivations of the Lorentz transformations, in one way or another. Your derivation is different from most that I see because you're comfortable jumping straight into a new coordinate system in which the Lorentz transformation matrix is diagonal. In that sense, it is new to me, but it is ultimately just the typical one viewed in rotated coordinates.
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