Value of this 'Science' Channel as quick intro

[MikeGomez]

An off-topic discussion about videos by @ScienceClic has been deleted and he will not be able to post again in this thread.

I thought this is thread is a discussion about a video by @ScienceClic. Could you please clarify what exactly is off topic here?

Specifically I was interested to hear what @D.S.Beyer finds to be dubious in another video that he mentions which visualizes 4D space-time .

 

[berkeman]

I thought this is thread is a discussion about a video by @ScienceClic. Could you please clarify what exactly is off topic here?

This is still under discussion by the Mentors. More in a bit...

 

[vanhees71]

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:

• https://web.archive.org/web/20131107065620/http://www.physics.pomona.edu/sixideas/VisTalk.pdf (see pg 24 - 50 - 66.)
• https://www.einsteinianphysics.com/iit-docs/Time%20in%20SRT%20and%20GRT.pdf (pg 15, 19)
• [forthcoming?] appendix in a revised edition of short relativity text

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

 

[D.S.Beyer]

I thought this is thread is a discussion about a video by @ScienceClic. Could you please clarify what exactly is off topic here?

Specifically I was interested to hear what @D.S.Beyer finds to be dubious in another video that he mentions which visualizes 4D space-time .

I'm going to save that for another thread since this thread is ...uh... in 'flux'.
I'll make sure to remember to @ you, cause it's going to be fun.