In either frame of reference, we compute coordinates and velocities of:

• the actual particles $\BBa$ and $\BAa$ ($\act$ for actual),
  representing the actual positions of Bob and Alice respectively;
• the measured particles $\BBs$ and $\BAs$ ($\sim$ for simultaneous),
  representing what Bob and Alice can measure of each other;
• the apparent particles $\BBv$ and $\BAv$ ($\vis$ for visible),
  representing what Bob and Alice can observe of each other.

Physical import

The space-time diagrams make concrete the clock hypothesis, i.e. proper time as absolute time, recovering absolute time as the fundamental dynamic variable, and with it the relativistic symmetry of the frame perspectives.

The diagrams also show that, relative to any frame of reference, any particle moves into the future and in space at a speed that is faster than measured in that frame (i.e. in a plane of simultaneity) by exactly its Lorentz factor. And this already and essentially is time travel.

In particular, the 4-velocity of light in any frame of reference is (scalarly) infinite in all components, its magnitude being exactly the measured light speed times its Lorentz factor (see below for the formal details).

Formal setup

In Special Relativity in $1\+1$ dimensions,
with generic coordinates $\ovP \~ \vec{t^\vex{\P}}{x^\vex{\P}}\!$,
and velocities $\ovv^\vex{\P} \~ \vec{v_t^\vex{\P}}{v_x^\vex{\P}}\!$.

Given $\c \ino{0}{\infty}$ the measured speed of light.
Given $\T \ino{0}{\infty}$ the proper time period.

For $v \inc{0}{1}\c$ a (constant) measured frame speed
  (i.e., WLOG, we are only considering the case $v\ge\!0$ in the unprimed frame).
For $\t \inc{0}{1}\T$ a proper time (by the clock).

Let $\b \= \b\!_v \is v\/\c \inc{0}{1}\!$.
Let $\gb \is 1/{\small \sqr{1\-\b^2}} \inc{1}{\gc}\!$.
Let $\epb \is 1/{\small (1\+\b)} \inc{1/2}{1}\!$.
Let $\enb \is 1/{\small (1\-\b)} \inc{1}{2\,\gc^2}\!$.
where $\epb\;\enb \= \gb^2$.
  (Note that, instead of saying $\gb \inc{1}{\infty}$, which is undefined, we keep track of $\gb$ throughout as a scale factor, so we can keep our intervals closed and have a uniform and finitely representable treatment for all $v$. In fact, we make explicit the range of all coordinates to facilitate delimiting the space-time diagrams.)

Diagramatically, for any fixed velocity $\uhat{v}$ and point $\uhat{\ovP} \~ \vec{\uhat{t}}{\uhat{x}}$:

• The world line through $\uhat{\ovP}$ is the straight line $\uhat{v} (t \- \uhat{t}) \= x \- \uhat{x}$,
  which can be written as $\uhat{\b\!} (t \- \uhat{t}) \= (x \- \uhat{x})/\c$.
• Reciprocally, the simultaneity plane at $\uhat{\ovP}$ is $t \- \uhat{t} \= \uhat{\b\!} (x \- \uhat{x})/\c$.
• And, similarly, the light cones at $\uhat{\ovP}$ are $t \- \uhat{t} \= \pmf (x \- \uhat{x})/\c$.

Finally, for reference, the form of our Lorentz transformations is:
$\pri{t} \= \gb\, (t \- \v x/\c^2)$
$\pri{x} \= \gb\, (x \- \v t)$

Formal derivation

We parametrize all formulas, besides implicitly on the (measured) relative speed, on the one and only (universal) proper time.

Frame $\B$

This is the frame of reference in which Bob is at rest at the origin of space.

We derive coordinates in this frame, whence velocities, by imposing the relevant physical constraints, then by plain geometry.

Frame $\B$ :  Particle $\BBa$

This represents Bob's actual space-time position.

We derive coordinates of $\BBa$ in this frame by imposing the constraints that Bob is at rest at the origin of space, hence $x \= 0$, and that Bob's proper distance from the origin of space-time is $|\t|$, hence by the metric $|\t| \= {\small \sqr{{t}^2 \- x^2/\c^2}} \= {\small \sqr{{t}^2}} \= |t|$, whence, with the direction of $t$ in accordance with that of $\t$, $t \= \t$.

In summary, we get:
$\pBBa(\t) \~ \vec{1}{0}\t$
$\ovv^\vex{\BBa}(\t) \~ \vec{1}{0}$

with coordinates as derived above:
$t^\vex{\BBa}(\t) \= \t \quad\, \inc{0}{1}\T \;\;\; \subseteq \ivc{0}{1}\T$
$x^\vex{\BBa}(\t) \= 0 \quad \inc{0}{0}\c\T \; \subseteq \ivc{0}{0}\c\T$

whence, by taking derivatives w.r.t. $\t$, the velocity.

Also, since Bob's measured speed is $0 \= x^\vex{\BBa} / t^\vex{\BBa}$:

• Bob's world line is the vertical line through $\pBBa$,
  i.e. $0\, (t \- t^\vex{\BBa}) \= (x \- x^\vex{\BBa})/\c$, whence $x \= 0$.
  
• Bob's simultaneity plane is the horizontal line through $\pBBa$,
  i.e. $t \- t^\vex{\BBa} \= 0\, (x \- x^\vex{\BBa})/\c$, whence $t \= \t$.
  
• Bob's light cones are the lines with slope $\pmf\! 1$ through $\pBBa$,
  i.e. $t \- t^\vex{\BBa} = \pmf (x \- x^\vex{\BBa})/\c$, whence $t \= \t \pmf x/\c$.
  

Frame $\B$ :  Particle $\BAa$

This represents Alice's actual space-time position.

We derive coordinates of $\BAa$ in this frame by imposing the constraints that Alice's measured speed is $v$, hence $x \= \v t$, and that Alice's proper distance from the origin of space-time is $|\t|$, hence by the metric $|\t| \= {\small \sqr{{t}^2 \- x^2/\c^2}} \= {\small \sqr{{t}^2 \- v^2\,t^2/\c^2}} \= |t|\/\gb$, whence, with the direction of $t$ in accordance with that of $\t$, $t \= \gbt$ and $x \= \v\gbt$.

In summary, we get:
$\pBAa(\t) \~ \vec{1}{v}\gbt$
$\ovv^\vex{\BAa}(\t) \~ \vec{1}{v}\gb$

with coordinates as derived above:
$t^\vex{\BAa}(\t) \= \gbt \quad\quad \inc{0}{\gb}\T \quad\;\; \subseteq \ivc{0}{\gb}\T$
$x^\vex{\BAa}(\t) \= \v\gbt \quad\, \inc{0}{\b\gb}\c\T \; \subseteq \ivc{0}{\gb}\c\T$

whence, by taking derivatives w.r.t. $\t$, the velocity.

Also, since Alice's measured speed is $v \= x^\vex{\BAa} / t^\vex{\BAa}$:

• Alice's world line is the line with slope $1/v \= 1/(\c\b)$ through $\pBAa$,
  i.e. $t \= \gbt \+ (x \- \v\gbt)/v$, whence $t \= x\/(\c\b)$   (or $x \= 0\,$ if $v \= 0$).
  
• Alice's simultaneity plane is the line with slope $\b/\c$ through $\pBAa$,
  i.e. $t \= \gbt \+ \b (x \- \v\gbt)/\c$, whence $t \= \t/\gb \+ \b x/\c$.
  
• Alice's light cones are the lines with slope $\pm\! 1$ through $\pBAa$,
  i.e. $t \= \gbt \pmf (x \- \v\gbt)/\c$, whence $t \= \exb\t/\gb \pmf x/\c$.
  

Frame $\B$ :  Particle $\BAs$

This represents what Bob measures of Alice: here he finds her time-dilated and length-contracted.

We derive coordinates of $\BAs$ in this frame by intersecting Alice's world line, $t \= x\/v$, with Bob's simultaneity plane, $t \= \t$, whence $x \= \v\t$.

In summary, we get:
$\pBAs(\t) \~ \vec{1}{v}\t$
$\ovv^\vex{\BAs}(\t) \~ \vec{1}{v}$

with coordinates as derived above:
$t^\vex{\BAs}(\t) \= \t \quad\quad \inc{0}{1}\T \quad\, \subseteq \ivc{0}{1}\T$
$x^\vex{\BAs}(\t) \= \v\t \quad\; \inc{0}{\b}\c\T \; \subseteq \ivc{0}{1}\c\T$

whence, by taking derivatives w.r.t. $\t$, the velocity.

Frame $\B$ :  Particle $\BBs$

This represents what Alice measures of Bob: here she finds him time-dilated and length-contracted.

We derive coordinates of $\BBs$ in this frame by intersecting Bob's world line, $x \= 0$, with Alice's simultaneity plane, $t \= \t/\gb \+ \b x/\c$, whence $t \= \t/\gb$.

In summary, we get:
$\pBBs(\t) \~ \vec{1}{0}\t/\gb$
$\ovv^\vex{\BBs}(\t) \~ \vec{1}{0}/\gb$

with coordinates as derived above:
$t^\vex{\BBs}(\t) \= \t/\gb \quad \inc{0}{1/\gb}\T \; \subseteq \ivc{0}{1/\gb}\T$
$x^\vex{\BBs}(\t) \= 0 \quad\quad\; \inc{0}{0}\c\T \quad\, \subseteq \ivc{0}{0}\c\T$

whence, by taking derivatives w.r.t. $\t$, the velocity.

Frame $\B$ :  Particle $\BAv$

This represents what to Bob appears of Alice, i.e. by just observing her from his position.

We derive coordinates of $\BAv$ in this frame by intersecting Alice's world line, $t \= x\/v$, with Bob's negative light cone, $t \= \t \- x/\c$, whence $t \= \epb\t$ and $x \= \v\epb\t$.

In summary, we get:
$\pBAv(\t) \~ \vec{1}{v}\epb\t$
$\ovv^\vex{\BAv}(\t) \~ \vec{1}{v}\epb$

with coordinates as derived above:
$t^\vex{\BAv}(\t) \= \epb\t \quad\quad \inc{0}{\epb}\T \quad\;\; \subseteq \ivc{0}{1}\T$
$x^\vex{\BAv}(\t) \= \v\epb\t \quad\, \inc{0}{\b\epb}\c\T \; \subseteq \ivc{0}{1/2}\c\T$

whence, by taking derivatives w.r.t. $\t$, the velocity.

Frame $\B$ :  Particle $\BBv$

This represents what to Alice appears of Bob, i.e. by just observing him from her position.

We derive coordinates of $\BBv$ in this frame by intersecting Bob's world line, $x \= 0$, with Alice's positive light cone, $t \= \epb\t/\gb \+ x/\c$, whence $t \= \epb\t/\gb$.

In summary, we get:
$\pBBv(\t) \~ \vec{1}{0}\epb\t/\gb$
$\ovv^\vex{\BBv}(\t) \~ \vec{1}{0}\epb\/\gb$

with coordinates as derived above:
$t^\vex{\BBv}(\t) \= \epb\t/\gb \quad \inc{0}{\epb\/\gb}\T \; \subseteq \ivc{0}{1}\T$
$x^\vex{\BBv}(\t) \= 0 \quad\quad\quad\;\, \inc{0}{0}\c\T \quad\quad \subseteq \ivc{0}{0}\c\T$

whence, by taking derivatives w.r.t. $\t$, the velocity.

Frame $\B$ :  Diagram boundaries

Extrapolating, coordinate boundaries in frame $\B$ are:
$t \in \ivc{0}{\gb}\T \quad\;\;\, \subseteq \ivc{0}{\gb}\T$
$x \in \ivc{0}{\b\gb}\c\T \; \subseteq \ivc{0}{\gb}\c\T$

Frame $\A$

This is the frame of reference in which Alice is at rest at the origin of space.

We derive coordinates in this frame by Lorentz-transforming by speed $v$ the corresponding coordinates in frame $\B$.

We double-check that the results obtained are symmetric to those in frame $\B$, i.e. they are the same modulo swapping any $\pri{\A}$, resp. $\pri{\B}$ particles with the corresponding $\B$, resp. $\A$ particles, and inverting the sign of the space coordinates in all places.

Frame $\A$ :  Particle $\AAa$

Given that $\pBAa(\t) \~ \vec{1}{v}\gbt$, by transforming coordinates we get:
$\pri{t}^\vex{\AAa}(\t) \= \gb \left( \gbt \- v^2\,\gbt/\c^2 \right) \= \t$
$\pri{x}^\vex{\AAa}(\t) \= \gb \left( \v\gbt \- \v\gbt \right) \= 0$

Whence $\pAAa(\t) \~ \vec{1}{0}\t$, symmetric to $\pBBa$ as expected.

Frame $\A$ :  Particle $\ABa$

Given that $\pBBa(\t) \~ \vec{1}{0}\t$, by transforming coordinates we get:
$\pri{t}^\vex{\ABa}(\t) \= \gb \left( \t \- 0 \right) \= \gbt$
$\pri{x}^\vex{\ABa}(\t) \= \gb \left( 0 \- \v\t \right) \= \!-\!\!\v\gbt$

Whence $\pABa(\t) \~ \vec{1}{-\!v}\gbt$, symmetric to $\pBAa$ as expected.

Frame $\A$ :  Particle $\ABs$

Given that $\pBBs(\t) \~ \vec{1}{0}\t/\gb$, by transforming coordinates we get:
$\pri{t}^\vex{\ABs}(\t) \= \gb \left( \t/\gb \- 0 \right) \= \t$
$\pri{x}^\vex{\ABs}(\t) \= \gb \left( 0 \- \v\t/\gb \right) \= \!-\!\!\v\t$

Whence $\pABs(\t) \~ \vec{1}{-\!v}\t$, symmetric to $\pBAs$ as expected.

Frame $\A$ :  Particle $\AAs$

Given that $\pBAs(\t) \~ \vec{1}{v}\t$, by transforming coordinates we get:
$\pri{t}^\vex{\ABa}(\t) \= \gb \left( \t \- v^2\,\t/\c^2 \right) \= \t/\gb$
$\pri{x}^\vex{\ABa}(\t) \= \gb \left( \v\t \- \v\t \right) \= 0$

Whence $\pAAs(\t) \~ \vec{1}{0}\t/\gb$, symmetric to $\pBBs$ as expected.

Frame $\A$ :  Particle $\ABv$

Given that $\pBBv(\t) \~ \vec{1}{0}\epb\t/\gb$, by transforming coordinates we get:
$\pri{t}^\vex{\ABa}(\t) \= \gb \left( \epb\t/\gb \- 0 \right) \= \epb\t$
$\pri{x}^\vex{\ABa}(\t) \= \gb \left( 0 \- \v\epb\t/\gb \right) \= \!-\!\!\v\epb\t$

Whence $\pABv(\t) \~ \vec{1}{-\!v}\epb\t$, symmetric to $\pBAv$ as expected.

Frame $\A$ :  Particle $\AAv$

Given that $\pBAv(\t) \~ \vec{1}{v}\epb\t$, by transforming coordinates we get:
$\pri{t}^\vex{\ABa}(\t) \= \gb \left( \epb\t \- v^2\,\epb\t/\c^2 \right) \= \epb\t/\gb$
$\pri{x}^\vex{\ABa}(\t) \= \gb \left( \v\epb\t \- \v\epb\t \right) \= 0$

Whence $\pAAv(\t) \~ \vec{1}{0}\epb\t/\gb$, symmetric to $\pBBv$ as expected.

Frame $\A$ :  Diagram boundaries

Symmetrically to frame $\B$, coordinate boundaries in frame $\A$ are:
$\pri{t} \in \ivc{0}{\gb}\T \quad\quad\:\, \subseteq \ivc{0}{\gb}\T$
$\pri{x} \in \ivc{-\!\b\gb}{0}\c\T \; \subseteq \ivc{-\!\gb}{0}\c\T$

That completes our calculations.