# Did Gauss's expression for the differential of the hyperbolic volume of the tetrahedron agree with later results?

My question is mainly about getting explanation to Gauss's third note (note III) in his writing "Zur Astralgeometrie", (see Cubierung der Tetraeder, pages 228-229), and about placing it in the right historical context. In this note Gauss is writing the following:

"In the tetrahedron 1234 (1,2,3,4 are it's vertexes), whose faces 124 and 134 are orthogonal, denote the volume as $$\Delta,$$ then it holds that: $$\partial \Delta = -24.\partial341,$$ and if the face angles at vertex 3 are constant, then the following also holds:

$$\alpha\alpha\cdot cotg^2341 - \beta\beta\cdot (tgi.24/i)^2 = 1$$

when: $$\alpha = cotg431,\quad \beta = cotg 234.$$

Two things i didn't understand about Gauss's notation immediately came to my mind:

• What does the notation 24 point "something" mean? does the point mean product? or something else?
• What is the "i" that appears in his formulas? how is it defined?

The questions i asked are of great interest to me since I think the problem of determining the volume of the non-euclidean tetrahedron really represents the peak of Gauss's work on hyperbolic geometry, and I really want to get a formal statement of his result, and to understand how is it compared with later results of Lobachevsky and Schlafli.

Update: recently i became aware of a certain interconnection between this fragment (CUBIRUNG DER TETRAEDER) and a second fragment of this kind - "Volumenbestimmungen in der Nichteuklidischen Geometrie" (Gauss's werke, volume 8, p.232-233). This fragment was found among the pages of Gauss's copy of Lobachevsky's 1840 publication (these two fragments are the only ones that deal with the probelm of hyperbolic tetrahedron). As a matter of fact, Stackel mentions, in his commentary of Gauss's "cubirung der tetraeder", that Gauss used the results of the second fragment implicitly when he derived his first displayed formula from p. 228 ($$\partial \Delta = -\frac{{1}}{{2}}(24)d(341)$$) - he used the two formulas:

• $$d(341) = -sinh(14)\cdot d(13)$$
• $$\partial \Delta = \frac{{1}}{{2}}d(13)\cdot (24)\cdot sinh(14)$$

to set up this relation ($$\partial \Delta = -\frac{{1}}{{2}}(24)d(341)$$). The second of these formulas was also mentioned in his later fragment, and though this fragment was written later (Gauss wrote this since the publication of Lobachevsky captured his attention), it seems that he was aware of it earlier.

I made this update because maybe dechipering Gauss's second fragment might serve as a breakthrough in understanding what Gauss knew on hyperbolic tetrahedrons.

• According to the remarks by Staeckel, $\partial\Delta$ represents the change in volume from an infinitely small extension of the edge 31 and maintaining the right angles of the sides of the tetrahedron. Likewise $\partial\,3\,4\,1$ is the change in area of the side, which Staeckel denotes $d(3\,4\,1)$. Staeckel also refers to a formula on p. 233, which he uses to derive Gauss's formula. – Michael E2 Dec 10 '17 at 14:29
• Michael E2 - do you know to read german? Can you translate Stackel's commentary and answer my question? – user2554 Dec 10 '17 at 16:12
• I never studied German, but I know other languages. With a dictionary I can make it through math in German. I skip things that are grammatically challenging unless I cannot figure out the math. Translating it would take a bit of work and I'd probably make some blunders. I don't have time right now to make the effort. – Michael E2 Dec 10 '17 at 16:33

(Edited). My German is not too good, but the dot clearly stands for multiplication, and $i=\sqrt{−1}$ so $$\frac{\mathrm{tg}i.24}{i}$$ is what we call the hyperbolic tangent today, $\tanh(24)$. Symbol probably $\partial$ means the differential.

If you can read German, read the comments below the note: Stackel explains what is going on there. In particular he says that Gauss missed the multiple of $1/2$ in his variation formula (his first displayed formula). The formula describes the variation of the volume under the conditions that the following angles 142, 342, 312, 314 are right. Under these conditions he derives Gauss's second displayed formula, (written by Stackel in the bottom line on p. 228), and then differentiates it to obtain Gauss's first formula, with extra factor $1/2$.

• @user1554: You missed the minus sign in the formula for $\partial\Delta$, and the assumption that the three angles that I listed must be right. – Alexandre Eremenko Dec 11 '17 at 0:02
• O.k thanks. And now for the last part of my question: did Gauss's statement agree with the later results on volume of orthoscheme tetrahedron? in particular how can it be compared with the results of Lobachevsky and Schlafli? – user2554 Dec 11 '17 at 6:52

Gauss's procedure does imply Bolyai's result on the volume of orthoscheme tetrahedron, as i'll show here. However, Gauss's result is a little bit more limited than Bolyai, since in Gauss's tetrahedron 4 of the 12 face angles of the tetrahedron are right, while Bolyai refers to a slightly more general tetrahedron whose only 3 face angles ar right. For the sake of consistency, we denote the angles 431, 234, and 214 as $$\alpha$$, $$\beta$$ and $$\gamma$$, respectively. Now lets look at the link of vertex 3 of the tetrahedron: it's a spherical triangle whose two edges lengths are $$\alpha$$ and $$\beta$$ and one angle is $$\gamma$$ (it is the dihedral angle of edge 31 and it's also equal to $$\gamma$$). In addition the sides $$\alpha, \beta$$ of this spherical triangle are orthogonal to each other. Therefore, by a combination of the spherical sine theorem and the spherical pythagoras theorem, we get:

$$\frac{{sin(arccos(cos\alpha\cdot cos\beta))}}{{sin 90}} = \frac {{sin\beta}}{{sin\gamma}}$$, or:

$$sin\gamma = \frac {{sin\beta}}{{\sqrt{{1 - (cos\alpha \cdot cos\beta)^2}}}}$$

Now, Gauss's procedure for the calculation of the volume leads to the following integral:

$$\Delta = \frac {{tan\beta}}{{2 tan \gamma}}\int_{0}^{c}\frac {{x sinh(x) dx}}{{(cosh^2(x) - 1 + \frac {{cosh^2(x)}}{{sin^2\alpha cot^2\beta}})\sqrt{{\frac{{cosh^2x}}{{cos^2\beta}} - 1}} }}$$.

Now, the left factor of the denominator $$cosh^2(x)(1 + \frac{{1}}{{sin^2\alpha \ cot^2\beta}})-1$$, is exactly equal to $$cosh^2(x)\cdot \frac{{1}}{{cos^2\gamma}}-1$$, because subtitution of $$cos\gamma = \sqrt {1 - \frac {{sin^2\beta}}{{1-(cos\alpha\cdot cos\beta)^2}}}$$ in this expression gives the previous one.

Concluding remarks:

$$Vol(T) = \frac {{tan\beta}}{{2 tan \gamma}}\int_{0}^{c}\frac {{x sinh(x) dx}}{{(\frac {{cosh^2(x)}}{{cos^2\gamma}} - 1)\sqrt{{\frac{{cosh^2x}}{{cos^2\beta}} - 1}} }}$$

can be seen as treating a slightly more general case then the one treated by Gauss (note: the differences in notation between the Bolyai integral in the presentation and Gauss's integral are just due the different symbols of the angles 431, 234, and 214).

However, for the case treated by Gauss, his formulas are absolutely correct. He should also be given credit for the identification of the calculation of the orthoscheme tetrahedron as the basis for volume formulas of general tetrahedrons (without right angles). In one of his letters, he refered to those calculations of volumes as "die jungle" - i guess he refered to the extremely complicated integrals that arise in the attempts to the decompose the general tetrahedron into orthoscemes (this problem was only solved very recently).

• It's still necessary to understand how Gauss arrived at the formula $$\partial \Delta = -\frac{{1}}{{2}}(24)d(341)$$ (he missed the factor $$\frac {{1}}{{2}}$$ at the first attempt); the second formula from his note can be derived with relative ease. I think dechipering Gauss's second fragment might serve as a clue for understanding Gauss's reasoning.