robphy said:
I think I'm the only one who has published the rotated graph paper diagrams.
I do know of two presentations by others who have used it in their slides:
Isn't this just Bondi's k-formalism visualized in a pretty clever didactical way of "rotated graph paper".
Behind this is of course the representation of vectors ##x \in \mathbb{R}^{1,1}## in terms of "light-cone coordinates", ##(\xi,\eta)=(x^0-x^1,x^0+x^1)##. The lines ##\xi=\text{const}## are the light cones for light moving in positive 1-direction, and ##\xi=\text{const}## those for light moving in negative 1 direction.
Since ##x_{\mu} x^{\mu}=(x^0)^2-(x^1)^2=\xi \eta## an Lorentz transformation, which leaves this Minkowski quadratic form invariant, is given by
$$\xi'=A \xi, \quad \eta'=\frac{1}{A} \eta.$$
This means
$$x^{\prime 0}-x^{\prime 1}=A(x^0-x^1), \quad x^{\prime 0}+x^{\prime 1}=\frac{1}{A}(x^0+x^1).$$
From this you get
$$x^{\prime 0}=\frac{1}{2} [(A+1/A) x^0+ (1/A-A) x^1], \quad x^{\prime 1}=\frac{1}{2}[(1/A-A) x^0+(1/A+A) x^1].$$
The origin of the primed system moves with velocity ##v## given by
$$\beta\frac{v}{c}=\frac{A-1/A}{A+1/A}=\frac{A^2-1}{A^2+1}.$$
i.e.,
$$A=\sqrt{\frac{1+\beta}{1-\beta}},$$
Since ##A \in \mathbb{R}## you must have ##|\beta|<1## and the transform reads, as it should
$$x^{\prime 0}=\gamma(x^0-\beta x^1), \quad x^{\prime 1}=\gamma(-\beta x^0+x^1), \quad \gamma=\frac{1}{\sqrt{1-\beta^2}}.$$
The nice thing with this representation and the "rotated graph paper" is that the area of the "diamonds" spanned by the coordinates lines ##\xi=\text{const}## and ##\eta=\text{const}## stay the same under Lorentz transformations. The consequences for the geometry of the Minkowski space in this rotated-graph-paper representation is given in detail in
@robphy 's nice paper
https://arxiv.org/abs/1111.7254