Fuss With Time

Why is the $x$ axis in the post "Thought Experiment Gone Wrong" dated 19 Nov 2017 moving at one unit per unit time?

Because the origin is defined as a point.  A line is a point moving.  In this case arbitrarily at one unit per unit time and the axis goes on forever.  A short line is a point moved, given its velocity, $v$

$L=\int^{t}_0{v}dt=\sqrt{\left(\int^{t}_{0}{v(x,\,t)}dt\right)^2+\left(\int^{t}_{0}{v(y,\,t)}dt\right)^2}$


where $v(y,\,t)$ is the $y$ component of $v$ and $v(x,\,t)$ is the $x$ component of $v$.  $v$ is a vector integrated over time.  With,

$\cfrac{d y}{d t}=a$

$\cfrac{d x}{d t}=b$

and so,

$\cfrac{d y}{d x}=\cfrac{a}{b}$

with

$(x_1,\,y_1)$  as the starting point,

$L=\sqrt{\left(\int^{x_2}_{x_1}{v(x,\,t)\cfrac{1}{b}}dx\right)^2+\left(\int^{y_2}_{y_1}{v(y,\,t)\cfrac{1}{a}}dy\right)^2}$

$(x_2,\,y_2)$ is the end point after time $t$, and $y_2=\cfrac{a}{b}(x_2-x_1)+y_1$ and

$x_1=b.t$

Over one unit time,

$x_2=b + x_1$

and

$y_2=a + y_1$

but,

$v(y,\,t)=\cfrac{d y}{d t}=a$

$v(x,\,t)=\cfrac{d x}{d t}=b$

so,

$L=\sqrt{\left(\int^{x_2}_{x_1}{b\cfrac{1}{b}}dx\right)^2+\left(\int^{y_2}_{y_1}{a\cfrac{1}{a}}dy\right)^2}$

$L=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt{b^2+a^2}$

which is what is expected.  Why the fuss?  We started with the origin as a point, the rest of the graph is with this point as the lowest denominator.  A infinite line is a point with a velocity; a finite length on the graph is a point with a velocity after a finite integral over an arbitrary time interval.

Both length and the axes have a direction and are vectors.  A curve is a point with changing velocity.  Intersections are paths that crosses irrespective of time.  But a collision is in both space and time.

Still what's the fuss?  The time dimension enters into the discussion because the issue at hand is space and time in spacetime diagrams.