# Reference request: a differential equation in elementary geometry

15 hours and four up-votes but not a word from anybody. That's the result of this question to stackexchange.

My question is where the following differential equation arises naturally and where it appears in the literature: $$\frac{d\alpha}{\sin\alpha} = \frac{d\beta}{\sin\beta}. \tag 1$$

Villarceau's theorem says that a plane bitangent to a torus (embedded in the usual way in $\mathbb R^3$) intersects the torus in two circles. I have found that as a point with longitude $\alpha$ and latitude $\beta$ moves along one of the Villarceau circles, equation $(1)$ holds. One could say Villarceau's circles are solutions to this differential equation. The equation is invariant under permutation of the two variables. I have found the geometric meaning of that symmetry.

So I'm wondering whether this equation has appeared in print or folklore in the context either of that bit of geometry or of anything else?

PS: The solution to $(1)$ can be defined implicitly by $$\tan\frac\beta2=m\tan\frac\alpha2$$ or by $$\cos\beta=\frac{\cos\alpha+n}{n\cos\alpha+1}.$$

• I think p. 40 of A Treatise on Plane and Spherical Trigonometry By William Chauvenet is related to this question. – user62675 May 2 '14 at 23:42
• @SanathDevalapurkar : Interesting. (Actually it's page 240.) None of these is exactly the one I had, and none is asserted to be related to Villarceau circles. I don't recall having come across the term "differential variation" before. – Michael Hardy May 3 '14 at 2:01
• Thanks for the typo correction. I referred to that book because I remembered seeing something reminiscent to the formula written above before in that book. If that reference didn't really help, I apologize. – user62675 May 3 '14 at 2:04
• It might help; I'm not sure yet..... – Michael Hardy May 3 '14 at 2:06
• If I map each circle of the torus stereographically to the line, then by your P.S. the equation becomes the equation for a straight line through the origin. – Austen May 4 '14 at 8:29

Let us consider two light rays of the same frequency in the $x$-$y$ plane forming the angles $\alpha$ and $\alpha+d\alpha$ with the axis $x$. The four-momenta of the corresponding photons are $$p_1=\left (\frac{E}{c}, \frac{E}{c}\cos{\alpha}, \frac{E}{c}\sin{\alpha} \right ),\;\;\;\mathrm{and}\;\;\; p_2=\left (\frac{E}{c}, \frac{E}{c}\cos{(\alpha+d\alpha)}, \frac{E}{c}\sin{(\alpha+d\alpha)}\right ).$$ From the invariance of the scalar product $$p_1\cdot p_2=\frac{E^2}{c^2}(1-\cos{(d\alpha)})\approx \frac{1}{2} \frac{E^2}{c^2}(d\alpha)^2,$$ we get $$E\,d\alpha=E^\prime\,d\alpha^\prime, \tag 1$$ where the primed quantities are measured in the inertial frame $S^\prime$ moving with some velocity $V$ along the $x$-axis. On the other hand, transverse components of the 3-momentum is not changed under the Lorentz transformations and, therefore, $$\frac{E}{c}\sin{\alpha}=\frac{E^\prime}{c}\sin{\alpha^\prime}.\tag 2$$ Combining (1) and (2), we get $$\frac{d\alpha}{\sin{\alpha}}=\frac{d\alpha^\prime}{\sin{\alpha^\prime}}. \tag 3$$ Hence this differential equation describes the (relativistic) aberration of light. An integration constant $n$ in its solution $$\cos{\alpha}=\frac{\cos{\alpha^\prime}+n}{1+n\,\cos{\alpha^\prime}}\tag 4$$ can be fixed in the following way. Let $\alpha^\prime=\pi/2$, then it is evident that the $x$-component of the photon's velocity $c_x=c\,\cos{\alpha}$ in the original "stationary" frame equals to the $S^\prime$ frames velocity $V$. on the other hand, from (4) in this case we have $\cos{\alpha}=n$. Therefore, $c\,n=V$ and $n=V/c$.
• I'm still digesting this. I did some edits to add \tag in four places and to change "cos" to "\cos", but at one point where I'd have written it differently I did nothing. That is where you wrote $n=\beta=V/c$. I'd have written $n=\cos\beta=V/c$. – Michael Hardy May 10 '14 at 22:10
• In fact $n=V/c$. It follows from $c_x=c\cos{\alpha}=V$ ($c_x=V$ because in the frame $S^\prime$ the photon is emitted at right angle $\alpha^\prime=\pi/2$ so that $c_x^\prime=c\cos{\alpha^\prime}=0$). $\beta$ is often used in special relativity to mean $V/c$, it is not an angle. – Zurab Silagadze May 12 '14 at 3:18
• I wasn't suggesting that $n$ differs from $V/c$. I wrote "$n=\cos\beta=V/c$", which clearly implies $n=V/c$. I was suggesting that one should not use the two letters $n$ and $\beta$ to refer to the same thing, but rather use $\beta$ to refer to the arccosine of $n=V/c$. ${}\qquad{}$ – Michael Hardy May 12 '14 at 3:46
• I agree that $\beta$ can lead to a confusion, so I excluded it from my answer. – Zurab Silagadze May 14 '14 at 5:31