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\title[Volume of a spherical trirectangular tetrahedron]{\large The volume of a spherical trirectangular tetrahedron}


\author[N.V. ABROSIMOV, B.P. Bayzakova]{{\bf N.V. Abrosimov\protect\orcidicon{0000-0001-5429-6589}, B.P. Bayzakova}}



\address{Nikolay Vladimirovich Abrosimov  % Контактные данные всех авторов и место работы указываются только на английском языке
\newline\hphantom{iii} Regional Scientific and Educational Mathematical Center, 
\newline\hphantom{iii} Tomsk State University,
\newline\hphantom{iii} pr. Lenina, 36,
\newline\hphantom{iii} 634050, Tomsk, Russia\medskip
\newline\hphantom{iii} Sobolev Institute of Mathematics,
\newline\hphantom{iii} pr. Koptyuga, 4,
\newline\hphantom{iii} 630090, Novosibirsk, Russia}%
\email{\textcolor{blue}{abrosimov@math.nsc.ru}}%

\address{Bibizada Polatovna Bayzakova  % Контактные данные всех авторов и место работы указываются только на английском языке
	\newline\hphantom{iii} Novosibirsk State University,
	\newline\hphantom{iii} str. Pirogova, 1,
	\newline\hphantom{iii} 630090, Novosibirsk, Russia}%
\email{\textcolor{blue}{b.baizakova@g.nsu.ru}}%

\thanks{\sc Abrosimov, N.V.,
The volume of a trirectangular spherical tetrahedron}
\thanks{\copyright \ 2024 Abrosimov N.V., Bayzakova B.P}
\thanks{\rm This work was supported by the Ministry of Science and Higher Education of Russia (agreement No. 075-02-2024-1437)}
\thanks{\it Received November, 8, 2024, Published December, 31, 2024}%


\semrtop \vspace{1cm}
\maketitle {\small

\vspace{-12pt}
\centerline{{\it Communicated by} {\sc P.P. Petrov}
%({\it filled by Editorial board})
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\bigskip
\bigskip

\sloppy
\begin{quote}
\noindent{\bf Abstract:} We consider a three-parameter family of tetrahedra in the spherical space, which three edges at one vertex are pairwise orthogonal. We establish conditions for the existence of such tetrahedra in spherical space. We obtain formulae for its dihedral angles and volume in terms of edge lengthes. \medskip

\noindent{\bf Keywords:} trirectangular tetrahedron, spherical space, dihedral angle, volume, Coxeter tetrahedron.
 \end{quote}
}

\bigskip
\sloppy

\section{Introduction}
Calculating the volume of polyhedra is a classical problem, well-known since the time of Euclid. This problem remains relevant even today, in particular because the volume of the fundamental polyhedron is the main geometric invariant of a three-dimensional manifold. 

Detailed overview of the results on calculating the volumes of polyhedra in spaces of constant curvature can be found in \cite{NA}. 

Volume formulas for non-Euclidean tetrahedra in some important special cases have been known since the time of N.~Lobachevsky, J.~Bolyai and L.~Schl\"{a}fli (see, e.g., \cite{NA} and \cite{M}). In particular, Schl\"{a}fli found the volume of the orthoscheme in $\mathbb{S}^3$ \cite{LS}. \emph{Orthoscheme} is an $n$-dimensional simplex in which there exists a path along the edges $(v_0,v_1), (v_1,v_2), \ldots, (v_{n-1},v_n),$ connecting all vertices, and any two edges of the path are perpendicular to each other. A three-dimensional orthoscheme is also called a \emph{birectangular tetrahedron}. By definition, a three-dimensional orthoscheme has three right dihedral angles (in Fig.~\ref{ortoscheme} the corresponding edges are highlighted with colored bold lines) and three others, which we will call \emph{essential dihedral angles} (in Fig.~\ref{ortoscheme} they are designated as $A, B, D$).\vspace{-5pt}
\begin{figure}[h]
	\centering
	\includegraphics[scale=0.23]{sphortoscheme.png}\vspace{-5pt}
	\caption{Ortoscheme with essential dihedral angles $A,B,D$}
	\label{ortoscheme}
\end{figure}\vspace{-10pt}

\begin{theorem}[Schl\"{a}fli, 1858]
	Let $T$ be a spherical orthoscheme with essential dihedral angles $A, B, D.$ Then its volume $V=V(T)$ can be calculated by the following formula       \begin{equation*}
		V=\frac{1}{4}\,S(A,B,D), \,\textit{where}
	\end{equation*}
	\begin{multline*}
		S\left(\frac{\pi}{2}-x, y, \frac{\pi}{2}-z\right)=\widehat{S}(x, y, z)=\\
		=\sum_{m=1}^{\infty}\left(\frac{D-\sin (x) \sin (z)}{D+\sin (x) \sin (z)}\right)^m \frac{\cos (2 m x)-\cos (2 m y)+\cos (2 m z-1)}{m^2}-\\
		-x^2+y^2-z^2 \vspace{-35pt} 
	\end{multline*} 
	and $D=\sqrt{\cos^2{x}\cos^2{z}-\cos^2{y}}.$
\end{theorem}

In the work of J.~Murakami \cite{JM} formulae for the volume of an arbitrary spherical tetrahedron in terms of dihedral angles and edge lengths are proposed. However, each of them contains 16 multi-valued complex dilogarithms of implicit functions, which makes them difficult to use. The purpose of this work is to obtain an explicit integral formula for the volume of a sufficiently wide three-parameter family of spherical tetrahedra, which will be convenient to use for further calculations and the study of the asymptotic behavior of the volume.

In this paper, the object of study is a \emph{trirectangular tetrahedron}, that is, a tetrahedron that has three pairwise orthogonal edges with a common vertex. In the work \cite{AS} a formula for the volume of a three-rectangular tetrahedron in hyperbolic space was obtained. %(Fig.~\ref{hypertetrahedron}).
%\begin{figure}[h]
%	\centering
%	\includegraphics[scale=0.4]{Hypertetra.png}\vspace{-15pt}
%	\caption{Hyperbolic trirectangular tetrahedron $T(\ell_1,\ell_2,\ell_3)$}
%	\label{hypertetrahedron}
%\end{figure}

\begin{theorem}[Abrosimov, Stepanishchev, 2023]
	Let $T=T(\ell_1,\ell_2,\ell_3)$ be a trirectangular hyperbolic tetrahedron given by lengths of pairwise orthogonal edges $\ell_1,\ell_2,\ell_3$ at a common vertex. Then the volume $V = V (T) $ can be found by the formula 
	\begin{multline*}
		V=\\
        \frac{1}{2}\int\limits_0^{\tanh{\ell_1}}\int\limits_0^\frac{\tanh{\ell_2}(\tanh{\ell_1} - x)}{\tanh{\ell_1}}\hspace{-15pt}\left[\frac{1}{1 - x^2 - y^2} + \frac{1}{x(x - 2x_0) + y(y - 2y_0) - e^{2\ell_3}}\right]dx\,dy,
	\end{multline*}
	where $x_0 = \dfrac{1-e^{2\,\ell_3}}{2\tanh{\ell_1}}$ and $y_0 = \dfrac{1-e^{2\,\ell_3}}{2\tanh{\ell_2}}$.
\end{theorem}


In the present paper we consider a trirectangular tetrahedron in spherical space. For this purpose, we introduce some notations and definitions.
Consider 3-dimensional sphere
\begin{equation*}
	\mathbb{S}^3=\{(x_1,x_2,x_3,x_4)\in \mathbb{R}^4 :x_1^2+x_2^2+x_3^2+x_4^2=1\}.
\end{equation*}

The \emph{distance} between two points $v_1, v_2\in \mathbb{S}^3$ is the angle (in radians) between their radius vectors $v_1$ and $v_2$ in $\mathbb{R}^4$. Thus, the scalar product in $\mathbb{R}^4$ induces a metric in $\mathbb{S}^3$.
%$$<v_1,v_2>=\sum_{j=1}^4(v_1)^j\cdot(v_2)^j,$$ где $(v_i)^j-$ $j$-$\text{я}$ координата вектора $v_i.$ 
We have 
$$\cos a=<v_1,v_2>,$$
where $a$ is a distance between $v_1$ and $v_2.$

\emph{Geodesics} are lines along which the shortest distances are achieved. Geodesics on a sphere are the so-called \emph{great circles}, that is, circles whose center coincides with the center of the sphere.

A \emph{spherical tetrahedron} is a convex hull (in the sense of the spherical metric) of four points in $\mathbb{S}^3$, called the \emph{vertices of the tetrahedron}.

In the book \cite{AVS} one can find more detailed definitions, as well as the derivation of some classical formulas of spherical geometry.






\section{Conditions for the existence of a trirectangular tetrahedron in $\mathbb{S}^3$}
Consider a standard basis in $\mathbb{R}^4$
\begin{align*}
    e_1&=(1,0,0,0),\\
    e_2&=(0,1,0,0),\\
    e_3&=(0,0,1,0),\\
    e_4&=(0,0,0,1).
\end{align*}

Let $P_{ij}$ denotes the plane formed by vectors $e_i$ and $e_j$. Then the circle $S_{ij}:=P_{ij}\cap \mathbb{S}^3$ is a geodesic.

\begin{figure}[h]
    \centering
    \includegraphics[scale=0.23]{sphertrirectetr.png}\vspace{-5pt}
    \caption{Spherical trirectangular tetrahedron $T(a,b,c)$}
    \label{sphertetrahedron}
  \end{figure}

Consider an arbitrary trirectangular tetrahedron $T = T(a, b, c)$ in $\mathbb{S}^3$ given by the lengths of its three pairwise orthogonal edges $a, b, c$ with a common vertex (Fig.~\ref{sphertetrahedron}). Without loss of generality, we will assume that this vertex is placed at the point $v_1 = (1, 0, 0, 0)$, and the three orthogonal edges $v_1v_2, v_1v_3, v_1v_4$ are directed along the geodesics $S_{12}, S_{13}, S_{14},$ lying in the coordinate planes $P_{12}, P_{13}, P_{14} \subset \mathbb{R}^4$. Then the vertices of tetrahedron $T (a, b, c)$ have the following coordinates
\begin{equation}\label{coor_ver}
    \begin{aligned}
        v_1&=(\,\,\,\,\,\,\,\,\,1,\,\,\,\,\,\,\,\,\,\,0,\,\,\,\,\,\,\,\,\;0,\,\,\,\,\,\,\,\,\,\,0),\\
        v_2&=(\cos{a},\;\sin{a},\,\,\,\,\,\,\,\,\;0,\,\,\,\,\,\,\,\,\;0),\\
        v_3&=(\cos{b},\,\,\,\,\,\,\,\,\,\;0,\;\sin{b},\,\,\,\,\,\,\,\,\;0),\\
        v_4&=(\cos{c},\,\,\,\,\,\,\,\;\;0,\,\,\,\,\,\,\,\;\;0,\;\sin{c}).
    \end{aligned}
\end{equation}

We have constructed a model of an arbitrary trirectangular tetrahedron $T(a, b, c)$ in $\mathbb{S}^3$ and thus proved the following theorem.

\begin{theorem}\label{existence}
    A trirectangular tetrahedron $T(a,b,c)$ exists in $\mathbb{S}^3$ for arbitrary $a, b, c \in (0,\pi)$. 
\end{theorem}

\begin{remark}
    For $a, b$ or $c=0$, tetrahedron $T(a, b, c)$ in $\mathbb{S}^3$ loses its dimension (i.e. degenerates into a triangle, segment, or point). For $a, b$ or $c=\pi$, tetrahedron $T(a, b, c)$ degenerates into a rectangular ``spindle'', i.e. a body with two vertices, at each of which three orthogonal edges meet.
\end{remark}
\begin{proof}
    Indeed, let $a = 0$, then according to equalities (\ref{coor_ver}), the vertex $v_2 = (1, 0, 0, 0)$ coincides with the vertex $v_1$. In this case, the tetrahedron loses one dimension and degenerates into a triangle. 

    On the other hand, if $a=\pi$, then according to equalities (\ref{coor_ver}), the vertex $v_2=(-1,0,0,0)\in\mathbb{S}^3$ is diametrically opposite to vertex $v_1$. In this case, the edge $v_1v_2$ is a half of the great circle $S_{12}$. Then the two edges $v_1v_3$ and $v_3v_2$ form a half of the great circle $S_{13}$. The remaining two edges $v_1v_4$ and $v_4v_2$ form a half of the great circle $S_{14}$. Thus, tetrahedron $T(a, b, c)$ is a rectangular ``spindle'' with diametrically opposite vertices $v_1=(1,0,0,0)$, $v_2=(-1,0,0,0)\in\mathbb{S}^3$, at each of which three orthogonal edges meet.
\end{proof}


\section{Dihedral angles in terms of edge lengths}
\begin{theorem}\label{angles}
    Let $T(a,b,c)$ be a spherical trirectangular tetrahedron given by the lengths of its three pairwise orthogonal edges $a, b, c$ with a common vertex. Then the dihedral angles opposite to these edges can be found by the formulas
    \begin{equation*}
        \begin{aligned}
            \cos{D}&=\frac{\cos{a}\sin{b} \sin{c}}{\sqrt{R}}, \\
            \cos{E}&=\frac{\sin{a}\cos{b} \sin{c}}{\sqrt{R}},\\
            \cos{F}&=\frac{\sin{a}\sin{b} \cos{c}}{\sqrt{R}},
        \end{aligned}
    \end{equation*}
where $R=1-\cos^2{a}\cos^2{b}-\cos^2{a}\cos^2{c}-\cos^2{b}\cos^2{c}+2\cos^2{a}\cos^2{b}\cos^2{c}.$
\end{theorem}
\begin{proof}
    Consider a section of tetrahedron $T(a,b,c)$ by a sphere centered at vertex $v_2$ (Fig.~\ref{sectiontetr}).
    \begin{figure}[h]
        \centering
        \includegraphics[scale=0.23]{sectetr.png}
        \caption{Section of tetrahedron $T(a,b,c)$ by a sphere centered at vertex $v_2$}
        \label{sectiontetr}
    \end{figure}

    As a section we get a spherical right triangle (Fig.~\ref{sectriangle}).\newpage
    \begin{figure}[h]
        \centering
        \includegraphics[scale=0.23]{sectriangle.png}\vspace{-5pt}
        \caption{Spherical triangle as a section of tetrahedron $T(a,b,c)$}
        \label{sectriangle}
    \end{figure}

    By the first spherical cosine law for this triangle we have
    \begin{equation*}
        \cos{E}=\frac{\cos{x_3} \cos{x_1}\cos{x_2}}{\sin{x_1}\sin{x_2}}.
    \end{equation*}

    Replacing in the last equality $\cos{x_2}=\cos{x_1} \cos{x_3}\,$ according to the spherical Pythagorean theorem, we obtain
    \begin{equation}\label{cosE}
        \cos{E}=\frac{\cos{x_3} \sin{x_1}}{\sin{x_2}}.
    \end{equation}
    
    Let us write the spherical Pythagorean theorem for three right angled faces of tetrahedron $T(a,b,c)$
    \begin{equation}\label{Pifagor}
        \cos{d}=\cos{b}\cos{c},\quad \cos{e}=\cos{a}\cos{c}, \quad \cos{f}=\cos{a}\cos{b}.
    \end{equation}

    Consider the lower face $v_1v_2v_3$ of tetrahedron $T(a,b,c)$. By the spherical law of sines in this right triangle we have
    \begin{equation*}
        \sin{x_3}=\frac{\sin{b}}{\sin{f}}=\frac{\sin{b}}{\sqrt{1-\cos^2{f}}}. 
    \end{equation*}

    Substituting here $\cos{f}$ from (\ref{Pifagor}), we get
    \begin{equation*}
        \sin{x_3}=\frac{\sin{b}}{\sqrt{1-\cos^2{a}\cos^2{b}}}. 
    \end{equation*}
    
    Hence,
    \begin{equation}\label{cosx3}
        \cos{x_3}=\frac{\sin{a}\cos{b}}{\sqrt{1-\cos^2{a}\cos^2{b}}}.
    \end{equation}.

    By the first spherical cosine law for the lateral face $v_1v_2v_4$ we have
    \begin{equation*}
        \cos{x_1}=\frac{\cos{c}-\cos{a}\cos{e}}{\sin{a}\sin{e}}.
    \end{equation*}

    Substituting here $\cos{e}$ from (\ref{Pifagor}), we get
    \begin{equation}\label{sinx1}
        \cos{x_1}=\frac{\cos{c}\,\sin{a}}{\sqrt{1-\cos^2{a}\cos^2{c}}}.
    \end{equation}

    Hence,
    \begin{equation*}
        \sin{x_1}=\frac{\sin{c}}{\sqrt{1-\cos^2{a}\cos^2{c}}}.
    \end{equation*}

    Now consider the face $v_2v_3v_4$ of tetrahedron $T(a,b,c)$. The first spherical cosine law gives
    \begin{equation*}
        \cos{x_2}=\frac{\cos{d}-\cos{e}\cos{f}}{\sin{e}\,\sin{f}}.
    \end{equation*}

    Substituting $\cos{d}, \cos{e}, \cos{f}$ from (\ref{Pifagor}), by straightforward calculations we obtain
    \begin{equation*}
        \cos{x_2}=\frac{\sin^2{a}\cos{b}\cos{c}}{\sqrt{(1-\cos^2{a}\cos^2{b})(1-\cos^2{a}\cos^2{c})}}.
    \end{equation*}

    Hence, 
    \begin{multline}\label{sinx2}
        \sin{x_2}=\\
        \sqrt{\frac{1-\cos^2{a}\cos^2{b}-\cos^2{a}\cos^2{c}-\cos^2{b}\cos^2{c}+2\cos^2{a}\cos^2{b}\cos^2{c}}{(1-\cos^2{a}\cos^2{b})(1-\cos^2{a}\cos^2{c})}}.
    \end{multline}

    It remains to substitute the found expressions (\ref{cosx3}), (\ref{sinx1}), (\ref{sinx2}) into the relation (\ref{cosE}). Then we obtain
    \begin{multline*}
        \cos{E}=\\
        \frac{\sin{a}\cos{b}\sin{c}}{\sqrt{1-\cos^2{a}\cos^2{b}-\cos^2{a}\cos^2{c}-\cos^2{b}\cos^2{c}+2\cos^2{a}\cos^2{b}\cos^2{c}}}.
    \end{multline*}

    From the last formula one can also obtain the corresponding expressions for $\cos{D}, \cos{F}$ by renaming the edges.
\end{proof}



\section{Edge lengths in terms of dihedral angles}
\begin{theorem}\label{edges}
    Let $T$ be a spherical trirectangular tetrahedron. Then the lengths of pairwise orthogonal edges $a, b, c$ of $\,T$ can be expressed in terms of  dihedral angles $D, E, F$ opposite to those edges by the formulas
    \begin{equation*}
        \begin{aligned}
            \cos{a}&=\frac{\cos{D}}{\sqrt{\sin^2{E}-\cos^2{F}}},\quad \cos{d}=\frac{\cos{E}\cos{F}}{\sqrt{(\sin^2{D}-\cos^2{F})(\sin^2{E}-\cos^2{D})}},\\
            \cos{b}&=\frac{\cos{E}}{\sqrt{\sin^2{D}-\cos^2{F}}},\quad \cos{e}=\frac{\cos{D}\cos{F}}{\sqrt{(\sin^2{E}-\cos^2{F})(\sin^2{E}-\cos^2{D})}},\\
            \cos{c}&=\frac{\cos{F}}{\sqrt{\sin^2{E}-\cos^2{D}}},\quad \cos{f}=\frac{\cos{D}\cos{E}}{\sqrt{(\sin^2{E}-\cos^2{F})(\sin^2{D}-\cos^2{F})}}.
        \end{aligned}
    \end{equation*}
\end{theorem}

\begin{proof}
    Consider sections of tetrahedron $T$ by three spheres centered at vertices $v_2, v_3, v_4$ (Fig.~\ref{3sectiontetr}).\newpage 
    \begin{figure}[h]
        \centering
        \includegraphics[scale=0.23]{3sectiontetr.png}\vspace{-5pt}
        \caption{Sections of tetrahedron $T$ by a spheres centered at vertices $v_2, v_3, v_4$}
        \label{3sectiontetr}
    \end{figure}

    As these sections we get three spherical right triangles (Fig.~\ref{3sectriangle}).
    \begin{figure}[h]
        \centering
        \includegraphics[scale=0.23]{3sectriangle.png}\vspace{-5pt}
        \caption{Spherical triangles as sections of tetrahedron $T$}
        \label{3sectriangle}
    \end{figure}

    By the second spherical cosine law for these right triangles we have
    \begin{equation}\label{cosx}
        \begin{aligned}
            \cos{x_1}&=\frac{\cos{F}}{\sin{E}},\quad \cos{x_3}=\frac{\cos{E}}{\sin{F}}, \\
            \cos{x_4}&=\frac{\cos{D}}{\sin{F}},\quad \cos{x_6}=\frac{\cos{F}}{\sin{D}}, \\
            \cos{x_7}&=\frac{\cos{D}}{\sin{E}},\quad \cos{x_9}=\frac{\cos{E}}{\sin{D}}.
        \end{aligned}
    \end{equation}

    By the second spherical cosine law for each of the three right angled faces of tetrahedron $T$ we have
    \begin{equation}\label{cosabc}
        \begin{aligned}
            \cos{a}=\frac{\cos{x_7}}{\sin{x_1}}=\frac{\cos{x_7}}{\sqrt{1-\cos^2{x_1}}}, \\
            \cos{b}=\frac{\cos{x_9}}{\sin{x_6}}=\frac{\cos{x_9}}{\sqrt{1-\cos^2{x_6}}}, \\
            \cos{c}=\frac{\cos{x_1}}{\sin{x_7}}=\frac{\cos{x_1}}{\sqrt{1-\cos^2{x_7}}}.
        \end{aligned}
    \end{equation} 

    We substitute $\cos{x_i}$ in equalities (\ref{cosabc}) by expressions (\ref{cosx}) and obtain 
    \begin{equation*}
        \begin{aligned}
            \cos{a}&=\frac{\cos{D}}{\sqrt{\sin^2{E}-\cos^2{F}}},\\
            \cos{b}&=\frac{\cos{E}}{\sqrt{\sin^2{D}-\cos^2{F}}},\\
            \cos{c}&=\frac{\cos{F}}{\sqrt{\sin^2{E}-\cos^2{D}}}.
        \end{aligned}
    \end{equation*}

    Then we apply spherical Pythagorean theorem for right angled faces of tetrahedron $T$ and use the latter formulas for $\cos{a}, \cos{b}, \cos{c}$ to get the remaining formulas for $\cos{d}, \cos{e}, \cos{f}$.
\end{proof}




\section{Volume of a trirectangular tetrahedron in $\mathbb{S}^3$}
\begin{theorem}\label{volume}
    Let $T=T(a, b, c)$ be a spherical trirectangular tetrahedron given by the lengths of its pairwise orthogonal edges at a common vertex, equal to $a, b, c$. Then the volume $V=V(T)$ can be found by the formula
    \begin{multline*}
        V=\int\limits_{0}^{a}\frac{1}{2R}\left[\frac{\arccos{(\cos{b}\cos{c})}(1-\cos^2{b}\cos^2{c})\sin{\alpha}\sin{b}\sin{c}}{\sqrt{R-\cos^2{\alpha}\sin^2{b}\sin^2{c}}}-\right.\\
        -\cos{\alpha}(R-\sin^2{\alpha}(\cos^2{b}+\cos^2{c}-2\cos^2{b}\cos^2{c}))\cdot\\
        \left.\cdot\left(\frac{\arccos{(\cos{\alpha}\cos{c})}\cos{b}\sin{c}}{\sqrt{R-\sin^2{\alpha}\cos^2{b}\sin^2{c}}}
        +\frac{\arccos{(\cos{\alpha}\cos{b})}\sin{b}\cos{c}}{\sqrt{R-\sin^2{\alpha}\sin^2{b}\cos^2{c}}}\right)\right] d\alpha,
    \end{multline*}
	where $R=1-\cos^2{\alpha}\cos^2{b}-\cos^2{\alpha}\cos^2{c}-\cos^2{b}\cos^2{c}+2\cos^2{\alpha}\cos^2{b}\cos^2{c}.$
\end{theorem}
\begin{proof}
    Consider a spherical trirectangular tetrahedron $T(a,b,c)$ with given edge lengths $a,b,c.$ According to Theorem~\ref{existence}, the domain of existence of such a tetrahedron has the form $\Omega=\{(a,b,c)\in \mathbb{R}^3 : a,b,c\in [0,\pi]\}$ (Fig.~\ref{existencedomain}).
    \begin{figure}[h]
        \centering
        \includegraphics[scale=0.23]{existdom.png}
        \caption{Existence domain $\Omega$ of a spherical trirectangular tetrahedron $T(a,b,c)$}
        \label{existencedomain}
    \end{figure}

    At the boundary ${a=0}$ of the existence domain $\Omega$, the tetrahedron loses its dimension, degenerating into a triangle, and its volume wanishes.

    Let $D,E,F$ denote the dihedral angles opposite to the edges $a,b,c,$ respectively. According to Theorem~\ref{angles}, the dihedral angles are uniquely determined by the lengths of the edges $a,b,c.$ Differentiating the volume as a complex function of $a$, we obtain
    \begin{equation}\label{vol}
            \frac{\partial{V}}{\partial{a}}=\frac{\partial{V}}{\partial{D}}\cdot \frac{\partial{D}}{\partial{a}}+\frac{\partial{V}}{\partial{E}}\cdot \frac{\partial{E}}{\partial{a}}+\frac{\partial{V}}{\partial{F}}\cdot \frac{\partial{F}}{\partial{a}},
        \end{equation}
        since $\displaystyle A=B=C=\frac{\pi}{2}$ and $\displaystyle\frac{\partial{A}}{\partial{a}}=\frac{\partial{B}}{\partial{a}}=\frac{\partial{C}}{\partial{a}}=0.$\vspace{5pt}

    Recal the well-known Schl\"{a}fli equation \cite{Sch1858-60}
    \begin{equation*}
        dV=\frac12\sum_{\theta} l_{\theta}\,d\theta,
    \end{equation*}
    where the sum is taken over all edges of the tetrahedron, $l_{\theta}$ denotes the length of the edge and $\theta$ is the dihedral angle along it. Therefore,
    \begin{equation}\label{PDLeng}
        \frac{\partial{V}}{\partial{D}}=\frac{d}{2},\quad \frac{\partial{V}}{\partial{E}}=\frac{e}{2},\quad \frac{\partial{V}}{\partial{F}}=\frac{f}{2},
    \end{equation}
    where $d=\arccos(\cos{b}\cos{c}),\, e=\arccos(\cos{a}\cos{c}),\, f=\arccos(\cos{a}\cos{b}),$ according to (\ref{Pifagor}).

    Note that
    \begin{equation*}
        \frac{\partial{\cos{D}}}{\partial{a}}=-\sin{D}\,\frac{\partial{D}}{\partial{a}}\quad\textrm{or, equivlently,}\quad\frac{\partial{D}}{\partial{a}}=\frac{-1}{\sqrt{1-\cos^2{D}}}\cdot\frac{\partial{\cos{D}}}{\partial{a}}.
    \end{equation*}

    Using Theorem~\ref{angles}, by straightforward calculations we obtain
    \begin{equation}\label{pD}
        \frac{\partial{D}}{\partial{a}}=\frac{(1-\cos^2{b}\cos^2{c})\sin{a}\sin{b}\sin{c}}{R\sqrt{R-\cos^2{a}\sin^2{b}\sin^2{c}}},
    \end{equation}
    where $R=1-\cos^2{a}\cos^2{b}-\cos^2{a}\cos^2{c}-\cos^2{b}\cos^2{c}+2\cos^2{a}\cos^2{b}\cos^2{c}.$

    Likewise,
    \begin{equation}\label{pEF}
        \begin{aligned}
            \frac{\partial{E}}{\partial{a}}&=
            \frac{-(R-\sin^2{a}(\cos^2{b}+\cos^2{c}-2\cos^2{b}\cos^2{c}))\cos{a}\cos{b}\sin{c}}
            {R\sqrt{R-\sin^2{a}\cos^2{b}\sin^2{c}}},\\
            \frac{\partial{F}}{\partial{a}}&=
            \frac{-(R-\sin^2{a}(\cos^2{b}+\cos^2{c}-2\cos^2{b}\cos^2{c}))\cos{a}\sin{b}\cos{c}}
            {R\sqrt{R-\sin^2{a}\sin^2{b}\cos^2{c}}}.
        \end{aligned}
    \end{equation}
    
    Integral of a differential form
    \begin{equation}\label{df}
        dV=\frac{\partial{V}}{\partial{a}}da+\frac{\partial{V}}{\partial{b}}db+\frac{\partial{V}}{\partial{c}}dc
    \end{equation}
    does not depend on the choice of the integration path in $\Omega$ connecting two fixed points. Since $V=0$ on the boundary of $\Omega$ at $a=0$, then by the Newton--Leibniz formula, the volume of tetrahedron $T(a,b,c)$ is equal to the integral of the form (\ref{df}) over any piecewise smooth curve $\gamma\subset\Omega$ with origin $(0,b,c)$ and end at the point with coordinates $(a,b,c).$ We will integrate over a straight line segment with origin $(0,b,c)$ and end $(a,b,c),$ since in this case $b,c$ are constant and the differential form (\ref{df}) takes the form
    \begin{equation*}
        \displaystyle dV=\frac{\partial{V}}{\partial{a}}da. 
    \end{equation*}
    It remains to integrate both parts of the equation (\ref{vol}) with respect to the variable $a$.

    Substituting the obtained derivatives (\ref{PDLeng}), (\ref{pD}) and (\ref{pEF}) into equation (\ref{vol}) and integrating it along the straight line segment with origin $(0,b,c)$ and end $(a,b,c),$ we obtain the desired formula for the volume. To distinguish the length of the edge $a$ and the integration variable, we denote the latter by $\alpha.$
\end{proof}


\section{Coxeter trirectangular tetrahedra in $\mathbb{S}^3$}
Let $T$ be a trirectangular tetrahedron in $\mathbb{S}^3$. Consider sections of tetrahedron $T$ by three spheres centered at vertices $v_2, v_3, v_4$ (Fig.~\ref{3sectiontetr}).\vspace{-5pt}
    \begin{figure}[h]
        \centering
        \includegraphics[scale=0.23]{3sectiontetr.png}\vspace{-5pt}
        \caption{Sections of tetrahedron $T$ by a spheres centered at vertices $v_2, v_3, v_4$}
        \label{3sectiontetr}
    \end{figure}

    As these sections we get three spherical right triangles (Fig.~\ref{3sectriangle}).\vspace{-10pt}
    \begin{figure}[h]
        \centering
        \includegraphics[scale=0.23]{3sectriangle2.png}\vspace{-5pt}
        \caption{Spherical triangles as sections of tetrahedron $T$}
        \label{3sectriangle}
    \end{figure}

Since the sum of angles in a spherical triangle is greater than $\pi$, then the dihedral angles of tetrahedron $T$ satisfy the conditions
\begin{eqnarray}\label{system}
	\begin{cases}
		E+F>\displaystyle\frac{\pi}{2}, \\[7pt]        
		D+F>\displaystyle\frac{\pi}{2}, \\[7pt]
		D+E>\displaystyle\frac{\pi}{2}.
	\end{cases}
\end{eqnarray}

\emph{Coxeter tetrahedron} is a tetrahedron whose dihedral angles are of the form $\displaystyle\frac{\pi}{n},$ where $n\geq 2$ is an integer. The list of spherical Coxeter tetrahedra was constructed by H.S.M.~Coxeter \cite{Cox}. He shown that there are 11 types of Coxeter tetrahedra in $\mathbb{S}^3$. In this section we will show that exactly 5 of these types belong to the family of trirectangular tetrahedra. We will calculate their volumes using Theorem~\ref{volume} in order to check our formula.

Let $D=\displaystyle\frac{\pi}{\ell}, E=\displaystyle\frac{\pi}{m}, F=\displaystyle\frac{\pi}{n},$ where $l,m,n\geq 2$ are integers. Solving the system of inequalities (\ref{system}) in integers $\ell, m, n$ we get the complete family of Coxeter trirectangular tetrahedra in $\mathbb{S}^3$ (see Table~\ref{examplestable} below).

Here we should note that conditions (\ref{system}) are necessary, but not sufficient for existence of a trirectangular tetrahedron $T$ in $\mathbb{S}^3$.  That is why we check the existence criterion for each of the solutions in addition. The existance criterion for arbitrary tetrahedron in $\mathbb{S}^3$ in terms of dihedral angles was given by F.~Luo \cite{Luo}.

\begin{theorem}[F.~Luo, 1997]
    The Gram matrix $G=[-\cos{\theta_{ij}}]$ of a spherical $n$-simplex is symmetric, positive definite with the diagonal entries equal to 1. Conversely, any positive definite symmetric matrix with diagonal entries equal to 1 is the Gram matrix of a spherical $n$-simplex unique up to isometry.
\end{theorem}

A {\em Gram matrix} $G$ of tetrahedron $T$ is defined by its dihedral angles as follows 
$$
G=[-\cos\theta_{ij}]_{i,j=1,2,3,4}=
\left(%
\begin{array}{cccc}
1 & -\cos{A} & -\cos{B} & -\cos{C} \\
-\cos{A} & 1 & -\cos{F} & -\cos{E} \\
-\cos{B} & -\cos{F} & 1 & -\cos{D} \\
-\cos{C} & -\cos{E} & -\cos{D} & 1 \\
\end{array}%
\right),
$$ 
where $\theta_{ij}$ denotes a dihedral angle along the edge $v_iv_j$, we assume here that $-\cos\theta_{ii}=1$.

This allow us to exclude example~7 in Table~\ref{examplestable} with angles $D=E=\displaystyle\frac{\pi}{3}, F=\displaystyle\frac{\pi}{5}$ since the determinant of corresponding Gram matrix is negative.

According to Theorem~\ref{edges}, we calculate the cosines of edge lengths $a, b, c$ for the given dihedral angles $D, E, F$. Then we use Theorem~\ref{volume} to calculate the volumes.

Example~1 in Table~\ref{examplestable} represents the infinite family of Coxeter trirectangular tetrahedra in $\mathbb{S}^3$. The next simple formula follows from Theorem~\ref{volume} in this particular case.
\begin{corollary}
    The volume of a Coxeter trirectangular tetrahedron $T$ in $\mathbb{S}^3$ with dihedral angles $\left(\displaystyle\frac{\pi}{2},\frac{\pi}{2},\frac{\pi}{2},\frac{\pi}{2},\frac{\pi}{2},\frac{\pi}{n}\right), n\geq 2,$ is equal $V(T)=\displaystyle\frac{\pi^2}{4n}$.
\end{corollary}

\begin{proof}
    Without loss of generality we set 
    $$A=B=C=D=E=\displaystyle\frac{\pi}{2},\quad F=\frac{\pi}{n}.$$ 

    Then by Theorem~\ref{edges} we have
    $$\cos{a}=\cos{b}=0,\quad \cos{c}=\cos{\frac{\pi}{n}}.$$
    
    Substituting the latter values into the volume formula from Theorem~\ref{volume}, we find
    \begin{equation*}
        V(T)=\frac{\pi\sin{\frac{\pi}{n}}}{4}\int\limits_{0}^{\frac{\pi}{2}}\frac{d\alpha}{1+\cos{\frac{\pi}{n}}\cos{\alpha}}=\frac{\pi\sin{\frac{\pi}{n}}}{4}\cdot\frac{\pi}{n\sin{\frac{\pi}{n}}}=\frac{\pi^2}{4n}.
    \end{equation*}
\end{proof}

\begin{table}[h]
\renewcommand{\arraystretch}{3}
\centering
\begin{tabular}{|c|c|c|c|c|c|c|c|c|}
\hline
\textit{№} & \textit{D} & \textit{E} & \textit{F} & $\cos{a}$ & $\cos{b}$ & $\cos{c}$ & \textit{Existence} & \textit{Volume} \\[2pt] \hline
1 & $\displaystyle\frac{\pi}{2}$ & $\displaystyle\frac{\pi}{2}$ & $\displaystyle\frac{\pi}{n}$ & $0$ & $0$ & $\displaystyle\cos{\frac{\pi}{n}}$ & exist for $n\geq 2$ & $\displaystyle\frac{\pi^2}{4n}$      \\
2 & $\displaystyle\frac{\pi}{2}$ & $\displaystyle\frac{\pi}{3}$ & $\displaystyle\frac{\pi}{3}$ & $0$ & $\displaystyle\frac{1}{\sqrt{3}}$ & $\displaystyle\frac{1}{\sqrt{3}}$ & exists & $\displaystyle\frac{\pi^2}{24}$      \\
3 & $\displaystyle\frac{\pi}{2}$ & $\displaystyle\frac{\pi}{3}$ & $\displaystyle\frac{\pi}{4}$ & $0$ & $\displaystyle\frac{1}{\sqrt{2}}$ & $\displaystyle\sqrt{\frac{2}{3}}$ & exists & $\displaystyle\frac{\pi^2}{48}$      \\
4 & $\displaystyle\frac{\pi}{2}$ & $\displaystyle\frac{\pi}{3}$ & $\displaystyle\frac{\pi}{5}$ & $0$ & $\displaystyle\sqrt{\frac{5+\sqrt{5}}{10}}$ & $\displaystyle\sqrt{\frac{3+\sqrt{5}}{6}}$ & exists & $\displaystyle\frac{\pi^2}{120}$      \\
5 & $\displaystyle\frac{\pi}{3}$ & $\displaystyle\frac{\pi}{3}$ & $\displaystyle\frac{\pi}{3}$ & $\displaystyle\frac{1}{\sqrt{2}}$ & $\displaystyle\frac{1}{\sqrt{2}}$ & $\displaystyle\frac{1}{\sqrt{2}}$ & exists & $\displaystyle\frac{\pi^2}{96}$      \\
6 & $\displaystyle\frac{\pi}{3}$ & $\displaystyle\frac{\pi}{3}$ & $\displaystyle\frac{\pi}{4}$ & $1$ & $1$ & $1$ & degenerates & $0$ \\
7 & $\displaystyle\frac{\pi}{3}$ & $\displaystyle\frac{\pi}{3}$ & $\displaystyle\frac{\pi}{5}$ & --- & --- & --- & doesn't exist & --- \\[5pt] \hline
\end{tabular}\vspace{10pt}
\caption{Coxeter trirectangular tetrahedra in $\mathbb{S}^3$}
\label{examplestable}
\renewcommand{\arraystretch}{1}
\end{table}

In example~6 we see that $\cos{a}=\cos{b}=\cos{c}=1$, which means that tetrahedron $T(a,b,c)$ in this case degenerates into a single point.

The volumes in the last column of Table~\ref{examplestable} coincide with well known volumes of respected Coxeter tetrahedra in $\mathbb{S}^3$ (see, e.g., \cite{Kol}, Table~1).

\bigskip
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\end{document}
