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I was reading about Minkowski diagrams in Morin's Introduction to Classical Mechanics, pg.537, and was not able to understand a particular ratio.

He showed that a point $(x',ct')=(0,1)$ is at a distance $\gamma\sqrt{1+\frac{v^2}{c^2}}$ from the origin whose derivation made sense. So he next shows the relation between one $ct'$ unit and one $ct$ unit as: $$\frac{one \ ct'\ unit}{one \ ct\ unit}=\sqrt{\frac{1+\beta^2}{1-\beta^2}}$$

I believe this is the scaling factor, so from this when $\beta\to1$, that is when it approaches the speed of light the scaling factor is infinite, what does this mean intuitively?

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On a position vs time graph (a Spacetime diagram), the point $(x', ct')=(0,1)$ lies on a hyperbola centered at the origin and physically represents "one tick" of an inertial astronaut's watch, where the astronaut travels with velocity $v=\beta c$.

That point (event P) on the hyperbola can be written as $$(\gamma v/c, \gamma) =\left(\frac{\beta}{\sqrt{1-\beta^2}}, \frac{1}{\sqrt{1-\beta^2}}\right) =(\sinh\theta,\cosh\theta),$$ where $\beta$ is the slope with respect to the vertical (time axis running upward) and $\theta$ is the rapidity (the Minkowski angle [Minkowski arc-length on the hyperbola divided by the radius, twice the sector-area divided by the square-of-the-radius]).

So, as $\beta \rightarrow 1$, the event P moves further up the hyperbola.
These are the "first ticks" for a sequence of faster inertial observers.

The expression $$\frac{one \ ct'\ unit \mbox{ drawn}}{one \ ct\ unit \mbox{ drawn}}=\sqrt{\frac{1+\beta^2}{1-\beta^2}}$$ is the ratio of the Euclidean-lengths of the segments (each representing "one tick") if you use a Euclidean ruler on your diagram,
since one "$ct$" unit drawn has Euclidean-length $\sqrt{(0)^2+(1)^2}=1$
and one "$ct'$" unit drawn has Euclidean-length $\displaystyle\sqrt{\left(\frac{\beta}{\sqrt{1-\beta^2}}\right)^2+\left(\frac{1}{\sqrt{1-\beta^2}}\right)^2}=\sqrt{\frac{\beta^2+1^2}{{1-\beta^2}}}$.
That ratio describes what the diagram looks like.
So, as $\beta \rightarrow 1$, the drawn length of a worldline-segment from the origin to P gets very large (as P moves up the hyperbola).


Here's the case for $\beta=3/5$:
robphy-segments-06


However, the physical ratio $$\frac{one \ ct'\ unit \mbox{ on the $t'$-watch}}{one \ ct\ unit \mbox{ on the $t$-watch}}=1$$ is the ratio of the Minkowski-lengths of the segments (each representing "one tick") if you use a Minkowski ruler [i.e. a watch] on your diagram,
since one "$ct$" unit on the $t$-watch has Minkowski-length $\sqrt{-(0)^2+(1)^2}=1$
and one "$ct'$" unit on the $t'$-watch has Minkowski-length $\displaystyle\sqrt{-\left(\frac{\beta}{\sqrt{1-\beta^2}}\right)^2+\left(\frac{1}{\sqrt{1-\beta^2}}\right)^2}=1$.
That ratio describes that their watches are using the same standard of elapsed time.
So, as $\beta \rightarrow 1$, the elapsed time on a watch along the worldline-segment from the origin to P remains constant at 1 (as P moves up the hyperbola).

A Minkowski-ruler can be visualized as a "causal diamond" (the intersection of the future of the origin event and the past of the event P), since the diamond-areas are equal for all the causal diamonds with diagonal along $OP$, where $P$ is on the hyperbola.

Here's the case for $\beta=3/5$:
robphy-causalDiamond-06

Here's the case for $\beta=4/5$:
robphy-causalDiamond-08


Presumably, Morin will talk about "time-dilation", which can be interpreted as the ratio of the vertical-components of these segments:

The expression $$\frac{\mbox{vertical-component of } one \ ct'\ unit \mbox{ drawn or timed}} {\mbox{vertical-component of }one \ ct\ unit \mbox{ drawn or timed}}=\sqrt{\frac{1}{1-\beta^2}}=\gamma=\cosh\theta$$ is the time-dilation factor.
The stationary watch measures
the elapsed time of "one tick of the moving watch"
to be equal to "$\gamma$ ticks of the stationary watch".
So, as $\beta \rightarrow 1$, that ratio grows large (as P moves up the hyperbola).

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