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One of the oldest problems in astronomy, which dates back to to the time of Kepler, is the problem of development in infinite trigonometric series of the "equation of the center" - to express the difference between true anomaly $v$ and mean anomaly $M$ as a trigonometric series of integer multiples of mean anomaly $M$, with coefficients of the series that are functions of the eccentricity $\epsilon$:

$$v - M = \sum_{n=1}^\infty \frac{{1}}{{n}}C_n \sin(nM)$$

where $C_n = f(\epsilon)$. The problem of determination of this trigonometric series is essentially a modern equivalent of the ancient method of epicycles - this was a model of the celestial positions of the planets which used a series of "wheels" ("epicycles") with different radiuses and angular speeds to trace the location of the planet. The coefficient $\frac{C_n}{n}$ is essentially the radius of the $n$th wheel, while it's angular frequency is $\frac{nM}{t}$ ($t$ is the time of a given mean anomaly measurement $M$).

Restating the problem in a more formal way, it amounts to finding the following integral (Fourier coefficients):

$$C_n = \frac{1}{\pi} Re \left[ \int_{-\pi}^{+\pi} \left( \frac{dv}{dM} \right) e^{inM} dM \right] = \frac{1}{\pi} Re \left[ \int_{-\pi}^{+\pi} \left( \frac{dv}{dM}\cdot \frac{dM}{dE} \right) e^{inM} dE \right]$$

Since $\frac{dv}{dM}$ can be expressed through the eccentric anomaly $E$ and the eccentricity $\epsilon$ by the kepler equation, this problem is closely related to Kepler equation. Since Kepler wrote that Kepler equation cannot be solved a priori (on account of the different nature of the arc and the sine), this was historically a very challenging problem.

C.F. Gauss (unpublished, 1805), F. Carlini (in 1817) and P.S. Laplace (in 1827) solved this problem, and as a by-product discovered an interesting phenomenon that the trigonometric series diverges (it doesn't converge to $v-M$ ) for values of the eccentricities above approximately $0.662$, a value which is now called "Laplace limit". Carlini determined the trigonometric series by a very remarkable method that, according to several sources, anticipated the WKB approximation of quantum mechanics (Carlini's work was popularized by C.G.J. Jacobi in 1849), while Gauss used a different method which was apparently based on his ideas on complex analysis. Gauss, and later the mathematician Wilhelm Scheibner (Scheibner in 1856 - Ueber die asymptotischen Werthe der Coefficienten in den nach der mittleren Anomalie vorgenommenen Entwickelungen, here is a link to his article), applied a certain "imaginary transformation" connecting the eccentric anomaly $E$ and the eccentricity $\epsilon$:

$$E = i\cdot \log \cot \left(\frac{{\varphi}}{{2}} \right) + \epsilon$$

where $\varphi$ is an angle defined as $\sin(\varphi) = \epsilon$.

Now, I have many misunderstandings concerning this substitution - $E$ and $\varphi,\epsilon$ are well defined and real-sizes, so how does the imaginary unit $i = \sqrt{-1}$ enters the equation? I understand that this has something to do with contour integration in the complex plane - according to one source the Fourier integral (which I wrote before) can be transformed into a contour integral in complex variable $z$ along the circular contour $|z| = 1$ by the substitution $z = e^{iE}$. But the whole picture is very unclear to me.

Therefore, my question is:

  • Can anyone who is familiar with the theory of Kepler equation explain the meaning of this substitution? and how it enables to find the coefficients of the Fourier series for $v-M$?

Side remarks:

  • The story of the center equation is a very convoluted one, and I'm well aware that the short background I gave in this question is not enough, so if anyone wants me to give more useful information, or add references and links, just keep me know and I'll update my posted question.
  • I have already posted a more comprehensive background to this question in this HSM post, so anyone who wants more information can find it there.
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  • $\begingroup$ Why are you torturing yourself? There is an easy solution to what I call "Doctor, it hurts when I do this" problem. (A patient goes to see his doctor and says "Doctor, it hurts when I do this", and then hits himself in the face.) The solution is "Don't do that then." Reading old technical papers, any paper that involves math that predates the computer age is akin to hitting yourself in the face. So don't do that then. $\endgroup$ – David Hammen Aug 26 '20 at 8:16
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    $\begingroup$ In this case, the solution is to solve Kepler's equation for the eccentric anomaly (algorithms galore, but Newton Raphson works quite nicely with an initial guess of pi), then solve for the true anomaly (easy, as every modern computer language has atan and a sqrt functions), and then compute the difference between the true anomaly and the mean anomaly. $\endgroup$ – David Hammen Aug 26 '20 at 8:19
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    $\begingroup$ I am somewhat suspicious about that formula for the eccentric anomaly. The eccentric anomaly varies as the object moves around its orbit, while the formula and description you give only provide a dependency on the eccentricity and not on orbital phase. Are you sure $\sin \varphi = \epsilon$? $\endgroup$ – user24157 Aug 26 '20 at 8:52
  • $\begingroup$ @antispinwards- you might be right (after all, this is what i also don't understand about this substitution), but on page 420 here - gdz.sub.uni-goettingen.de/download/pdf/PPN236018647/… (this is volume 10-1 of Gauss's collected works) - Gauss defines the eccentricity $\epsilon$ to be $sin\varphi$. $\endgroup$ – user2554 Aug 26 '20 at 10:55
  • $\begingroup$ In addition, on p. 446 in the link i just gave you appears an explanation (which i don't really understand) how using contour integration in the complex plane leads exactly to the substitution used by Gauss and later by Scheibner. If you need to translate it from german to english, i recommend viewing the same volume not through a pdf excerpt but through this website gdz.sub.uni-goettingen.de/id/PPN236018647?tify={%22panX%22:0.522,%22panY%22:0.613,%22view%22:%22thumbnails%22,%22zoom%22:0.889} , which also enables to do Google translate. $\endgroup$ – user2554 Aug 26 '20 at 11:10

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