Derivation of SR formula involving an absolute interval I am working with the web book Reflections on Relativity.
I am having trouble deriving the equation of the absolute interval between the intersection points of two worldlines with a hyperbola.
Here is an image of that page:

I was thinking that the "1/4" exponents of the velocity factors are typos and should actually be "1/2"s. If anyone can assist me in deriving this formula it would be greatly appreciated.
 A: The "1/4" exponents are correct.
Here is a derivation of those formulas using the rapidity $\theta$, 
where the relative-velocity is $v=\tanh\theta$, the time-dilation factor is $\gamma=\frac{1}{\sqrt{1-v^2}}=\cosh\theta$, and
(the key part of this derivation...) the Doppler factor $k=\sqrt{\frac{1+v}{1-v}}=\exp\theta$.
Note: $\cosh\theta=\frac{k+k^{-1}}{2}$ and 
$\sinh\theta=\frac{k-k^{-1}}{2}$.

short answer
The rapidity (spacelike-arclength along the unit-hyperbola (of timelike-radius $1$) is
$$\theta_{\small{A\ to\ B}}=\frac{1}{2}\ln\left(\frac{1+v_{BA}}{1-v_{BA}}\right)=\ln k.$$
The spacelike arc $AB$ (along that unit hyperbola) has size
$$s_{\small{arc\ A\ to\ B}}=(1)\ \theta_{\small{A\ to\ B}}.$$
The spacelike chord $AB$ (which is the base of an isosceles triangle with unit legs) has size
\begin{align*}
s_{\small{chord\ A\ to\ B}}
&=2 (1)\sinh\left(\frac{1}{2}  \theta_{\small{A\ to\ B}} \right)\\
&=2 (1) \left(\frac{ \exp\left(\frac{1}{2}  \theta_{\small{A\ to\ B}} \right)-\exp\left(-\frac{1}{2}  \theta_{\small{A\ to\ B}} \right)}{2}\right)\\
&=2 (1) \left(\frac{\sqrt{k}-\sqrt{k^{-1}}}{2} \right)\\
&=(1) \left(\sqrt{k}-\sqrt{k^{-1}} \right)\\
&=(1) \left(\left(\frac{1+v_{BA}}{1-v_{BA}}\right)^{1/4}-\left(\frac{1-v_{BA}}{1+v_{BA}}\right)^{1/4} \right)\quad\checkmark\\
\end{align*}
where $\left(\frac{1}{2}  \theta_{\small{A\ to\ B}} \right)$ is the (rapidity-)angle-bisector of $\theta_{\small{A\ to\ B}}$.
Note: 
$\frac{\theta_{\small{A\ to\ B}}}{2}=\frac{1}{2}\ln k=\ln\sqrt{k}=\ln\left(\frac{1+v_{BA}}{1-v_{BA}}\right)^{1/4}$.

The spacetime diagrams below might be helpful.


*

*the first diagram is used to derive the relationship between the Doppler factor $k$ and the hyperbolic-trig functions of the rapidity $\theta$ (using a radar method inspired by the Bondi k-calculus)

*the second diagram is drawn in the bisector frame between the two observers OA and OB. Note the isosceles triangle (with base AB and unit legs) can be decomposed into two Minkowski-right triangles with angle $\frac{\theta_{\small{A\ to\ B}}}{2}$.


The red lines (the chord AB [where the legs OA and OB are radii] ]and the Ray from O to the midpoint of AB) are Minkowski-orthogonal to each other.


(If necessary, I can elaborate on the derivations.)

answers to comments
In response to the OP’s comment,
in the first diagram, the (t,x) coordinates are as follows: $A=(1,0)$ and $B=(\cosh\theta,\sinh\theta)$.
The square-interval is 
$s^2_{\small{chord\ A\ to\ B}}=
(\cosh\theta-1)^2-(\sinh\theta-0)^2=2-2\cosh\theta=-4\sinh^2(\theta/2) \quad\checkmark$

 This agrees with geometrical argument using the definitions of $k$ and $\theta$ above.

In response to the OP’s second comment,
we repeat the calculation in terms of the Doppler factor $$k=\left(\frac{1+v_{BA}}{1-v_{BA}}\right)^{1/2}=\exp\theta$$
(which is what the "Reflections on Relativity" document was using implicitly)
without explicitly using hyperbolic-trig functions of the rapidity $\theta$.
The spacelike arc $AB$ (along that unit hyperbola [with radius 1]) has size
$$s_{\small{arc\ A\ to\ B}}=(1)\ 
\theta_{\small{A\ to\ B}}=(1)\ \frac{1}{2}\ln\left(\frac{1+v_{BA}}{1-v_{BA}}\right)=(1)\ln k.$$
The spacelike chord $AB$ (which is the base of an isosceles triangle with unit legs) has size given by the square-root of the square-interval from 
$A=\left(1,0\right)$ to $B=\left(\frac{k+k^{-1}}{2},\frac{k-k^{-1}}{2}\right)$


*

*B’s coordinates in terms of $k$ is from a radar measurement made by observer OA
where $\frac{k+k^{-1}}{2}T_B$ is the elapsed time of event B assigned by OA
and $\frac{k-k^{-1}}{2}T_B$ is the apparent distance of event B assigned by OA

*This may be a useful formula: Since $k=\sqrt{\frac{1+v}{1-v}}$, then $v=\frac{k^2-1}{k^2+1}$. 
So, $\gamma=\frac{1}{\sqrt{1-v^2}}=\frac{1}{\sqrt{1-\left(\frac{k^2-1}{k^2+1}\right)^2}}=\left(\frac{k^2+1}{2k}\right)=\frac{k+k^{-1}}{2}$,

and $\gamma v=\left(\frac{k^2+1}{2k}\right) \frac{k^2-1}{k^2+1}=\frac{k^2-1}{2k}=\frac{k-k^{-1}}{2}$.


So the square-interval is:
\begin{align*}
s^2_{\small{chord\ A\ to\ B}}
&=(t_B-t_A)^2-(x_B-x_A)^2\\
&=\left(\frac{k+k^{-1}}{2}-1\right)^2-\left(\frac{k-k^{-1}}{2}\right)^2\\
&=\left(\frac{k+k^{-1}-2}{2}\right)^2-\left(\frac{k-k^{-1}}{2}\right)^2\\
&=\frac{1}{4}\left( \left(k+k^{-1}-2\right)^2-\left(k-k^{-1}\right)^2 \right)\\
&=\frac{1}{4}\left( 
\left(\vphantom{\frac12}\left(k+k^{-1}-2\right)+\left(k-k^{-1}\right)\right) 
\left(\vphantom{\frac12}\left(k+k^{-1}-2\right)-\left(k-k^{-1}\right)\right) 
\right)\\
&=\frac{1}{4}
\left(2k-2\right) \left(2k^{-1}-2\right) \\
&=\frac{1}{4}
\left(4-4k-4k^{-1}+4\right) \\
&=
-\left(k-2+k^{-1}\right) \\
&=
-\left(\sqrt{k}-\sqrt{k^{-1}}\right)^2 \\
s_{\small{chord\ A\ to\ B}}
&=\sqrt{k}-\sqrt{k^{-1}} \\
&=\left(\left(\frac{1+v_{BA}}{1-v_{BA}}\right)^{1/4}-\left(\frac{1-v_{BA}}{1+v_{BA}}\right)^{1/4} \right)\quad \checkmark\\
\end{align*}
Of course, you can write $k$ algebraically in terms of $v_{AB}$... but you might be lost as how to simplify efficiently.


*

*While the velocity $v$ may seem for familiar and physical, the Doppler factor $k$ is more natural for relativity because $k$ and its reciprocal $k^{-1}$ are eigenvalues of a Lorentz boost transformation, associated with the lightlike eigenvectors [encoding the invariance of the speed of light]. (This is related to the "null coordinates" mentioned in the "Reflections on Relativity" document.)

*Since velocity composition in terms of the Doppler factor is multiplicative ($k_{CA}=k_{CB}k_{BA}$), using $k=\exp\theta$ we find that velocity composition in terms of rapidity is additive ($\theta_{CA}=\theta_{CB}+\theta_{BA}$). 

*Using hyperbolic-trig functions, $v_{CA}=\tanh\theta_{CA}= \tanh(\theta_{CB}+\theta_{BA})=\frac{\tanh(\theta_{CB})+\tanh(\theta_{BA})}{1+\tanh(\theta_{CB})\tanh(\theta_{BA})}=\frac{v_{CB}+v_{BA}}{1+v_{CB}v_{BA}}$ shows that velocity composition is non-additive in terms of velocity. In other words, the velocity $v$ is not optimal for exploiting the symmetries of special relativity (just like the slope of a line is not optimal for exploiting the symmetries of Euclidean geometry).


I hope this is now an acceptable answer.
