Slopes of lines in Lorentz transformation

Question

Came across this paragraph reading Philosophy Of Physics by Tim Maudlin.

”In our Euclidean space-time diagrams, the trajectories of comoving clocks are related to their equal-t slices in the following way: as the trajectories of the clocks tip over in the diagram, the associated equal-t slices tip up so that the light rays always split the difference between the clock trajectories and the equal-$t$ slices. In other words, if the slope of the master clock in the diagram is $s$, the slope of the equal-$t$ surface for that clock's time coordinate is $1/s$. As a consequence, the angle in the diagram between a light ray and a clock trajectory always equals the angle between the light ray and the clock's equal-t slice. This is merely a consequence of the conventions used in drawing the diagrams."

Couldn't understand this paragraph about how and why the slope should be s and 1/s and the significance of it

Can someone please explain it?

Answer 1

This elaborates on the answer by @JEB.

We follow the usual convention that time runs upwards on a spacetime diagram, and that coordinates are labeled (t,x,y,z) and units are chosen so that light-rays are drawn at 45-degrees. We restrict to one spatial dimension.

Consider the unit hyperbola $t^2-x^2=1$ (where the tips of the 4-velocities [thought of a radius-vectors] would lie, representing one tick for that inertial observer's worldline)--this is the "unit-circle" in Minkowski spacetime geometry.

For an observer with spatial velocity $s$, the tip of the 4-velocity would be on the unit-hyperbola at $(T,X)$, where $s=X/T$. This is a slope, relative to the [vertical] t-axis. The tangent line to the unit-hyperbola [the "unit-circle"] has slope $1/s$. This tangent line represents a set of simultaneous ("equal-t") events for that moving observer.

• [Implicit differentiation of $t^2-x^2=1$ yields $\frac{dx}{dt}=\frac{t}{x}$. Thus, the slope of the tangent line is the reciprocal of the slope of the radius. For a unit circle in Euclidean geometry, implicit differentiation of $t^2+x^2=1$ yields $\frac{dx}{dt}=\frac{-t}{x}$. Thus, in Euclidean geometry, the slope of the tangent line is the minus the reciprocal of the slope of the radius.]

To visualize this, visit my visualization https://www.desmos.com/calculator/awgqxtkqcc where time runs to the right like the usual position-vs-time graph [instead of upwards]. Play around with the E-slider.

By the way, the unit-hyperbola $t^2-x^2=1$ can be represented, using a different set of variables [called light-cone coordinates], by $$u=1/v$$ where $u=t+x$ and $v=t-x$. Note that $1=t^2-x^2=uv$.