PeterDonis said:
...you know that free particles still have to have straight worldlines, and that light still has to have speed ##c##. And you know that the coordinate transformation has to be linear. So you do have information to work from.
This info is used to derive the Lorentz transformation - in the ##(t,x)\to(t',x')## case, it's ##\mathcal{L}=\begin{bmatrix}\gamma & \beta\gamma\\\beta\gamma & \gamma\end{bmatrix}## (sign of off-diagonal terms depends on relative velocity so let's consider them positive in this case).
PeterDonis said:
Orthogonality is a coordinate-independent property, so a pair of vectors that are orthogonal in one chart will be orthogonal in any chart.
Now
this is important info and something I'm embarrassed to admit that I didn't know. I thought if orthogonality is coordinate-independent as you said, then maybe even the inner product is (googling confirmed this). So if we have a couple of vectors ##\mathbf{a},\mathbf{b}## with representations ##[ a ]_{\mathcal{A}}, [ b ]_{\mathcal{A}}## in the initial coordinate chart ##\mathcal{A}## (and since we know that ##\eta=\text{diag}(-1,1)## is the metric
representation in ##\mathcal{A}## by construction), their inner product is ##[ a ]^T_{\mathcal{A}}\eta[ b ]_{\mathcal{A}}##.
Let ##\eta'## be the metric
representation in the new coordinate chart ##\mathcal{B}## (the Minkowski metric doesn't change - its representation changes - something I realized after reading your post and googling related stuff). The representation of ##\mathbf{a}## in ##\mathcal{B}## will be ##[ a ]_{\mathcal{B}}=\mathcal{L}[ a ]_{\mathcal{A}}## (similarly for ##\mathbf{b}##). Then
$$[ a ]^T_{\mathcal{B}}\eta'[ b ]_{\mathcal{B}}=[ a ]^T_{\mathcal{A}}\eta[ b ]_{\mathcal{A}}=[ a ]^T_{\mathcal{A}}\mathcal{L}^T(\mathcal{L}^T)^{-1}\eta\mathcal{L}^{-1}\mathcal{L}[ b ]_{\mathcal{A}}
=[ a ]^T_{\mathcal{B}}(\mathcal{L}^{-1})^T\eta\mathcal{L}^{-1}[ b ]_{\mathcal{B}}
\\\implies \eta'=(\mathcal{L}^{-1})^T\eta\mathcal{L}^{-1}=\eta$$
So this was the missing link. And now since the
metric representation is the same even in the new coordinate chart, and since the representation of the
new basis vectors in the new chart ##\mathcal{B}## is ##[1,0]## and ##[0,1]##, their inner product is zero.
PeterDonis said:
It's a good start, but you spoke in general terms. There is a specific requirement for an IRF that requires isotropy.
I confess I don't know about that requirement. As far as I've read, isotropy is justified loosely by the fact that the orientation of the relative velocity vector between IRFs can be arbitrary. If there's an even stronger/more specific justification, I'm lost.
It's a chance for all of us (especially me haha) to clarify concepts. I want to go into deep conceptual waters sooner or later for sure. I might be getting ahead of myself at times and that might slow me down, but I guess that's a downside of self-teaching.
)
