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In a major assignment I am to determine the semi-major axis of an elliptic orbit for the star S2 around Sagittarius A*. I found some data that I have used to fit the points to an ellipse - however the equation I get is in terms of

$ax^2 + bxy + cy^2 + dx + ey + f = 0$

rather than

$ \frac{x^2}{a^2} + \frac{y^2}{ b^2}= 1$.

Can any of you smart people teach me how to 'translate' it, or if it's even necessary in order to determine the semi-major axis of the ellipse?

Emilio Novati
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William Randrup
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  • Where is your data relative too, Sagittarius A*? – Paul Dec 12 '18 at 13:00
  • Yes. It's measured in arc seconds of declination and right ascension relative to SgrA*. – William Randrup Dec 12 '18 at 13:05
  • Well, the first problem is that in the second equation, it's assumed that the ellipse is located at the origin of the coordinate system. Therefore, you have to move it to the right location first ... – Matti P. Dec 12 '18 at 13:05
  • I'm not very adept when it comes to ellipses, but I think it is located at the origin of the coordinate system as (0,0) is located at one of the foci. – William Randrup Dec 12 '18 at 13:08
  • Another problem is that declination and right ascension are spherical coordinates and are measured in different units, so there's a question of whether "one arc-second" measures the same apparent distance in both directions. Also, it seems we are not looking "straight down" on S2's orbit but rather view it from an oblique angle, which means the major axis of its path on the celestial sphere may be very different from the major axis of the actual orbit. The distance between two points on the orbit depends on the change in distance from us as well as the apparent angle. – David K Dec 12 '18 at 13:33
  • Altogether, if you want a scientifically meaningful answer, this is far more complicated than just fitting a major axis to some $x,y$ coordinates in a standard Cartesian plane. – David K Dec 12 '18 at 13:36
  • There is a lot written on this site about deriving the parameters of an ellipse from its equation, mostly in two dimensions but sometimes in three (which I think you want). You could start here: https://math.stackexchange.com/questions/1217796/compute-center-axes-and-rotation-from-equation-of-ellipse – David K Dec 12 '18 at 13:40
  • I took the data from the original publication in nature where it is already applied to an orbital fit. Therefore it seems that they have already done the dirty work of making it so the orbit is indeed viewed "straight down" upon to quote you, David K. However I'm not entirely sure how they managed the fact that an arc-second does not measure the exact same distance. I will try this method first, and see if I get the same results as them - otherwise I will try something else, but thank you for your insights either way! – William Randrup Dec 12 '18 at 14:47
  • Please refer to another question here. – Ng Chung Tak Dec 12 '18 at 19:06

1 Answers1

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Think of the ellipse as a quadratic form: $$[x,y,1]\begin{bmatrix}a&b/2&d/2\\b/2&c&e/2\\d/2&e/2&-f \end{bmatrix}\begin{bmatrix}x\\y\\1 \end{bmatrix}=0 $$. Lets call the matrix in the middle $Q_0$ and then $[x,y,1]Q_0 \begin{bmatrix}x\\y\\1 \end{bmatrix}=0 $

Since you are allowed translate and rotate the coordinate system, you can push a 2d "rigid motion" matrix $E$ such that $[x,y,1] E^T Q_1 E\begin{bmatrix}x\\y\\1 \end{bmatrix}=0 $ and you can look for such and $E$ such that $Q_1=\begin{bmatrix} a'&0&0\\0&b'&0\\0&0&-1 \end{bmatrix}$

The general form of $E$ is $\begin{bmatrix}\cos(\theta)&\sin(\theta)&p_x \\-\sin(\theta)&\cos(\theta)&p_y\\ 0&0&1 \end{bmatrix}$

user619894
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  • In such issues (attraction by a heavy mass), it is preferable to use polar representation $p=p_0/(1+e \cos(\theta))$ where the center is the center of mass (https://math.stackexchange.com/q/2389034) – Jean Marie Dec 12 '18 at 20:48
  • Why is it preferable? – user619894 Dec 13 '18 at 05:36
  • if you look for example to the way the 3 Kepler's laws are derived from the (Newton's) inverse square law of gravitation, you will see that it's this form which is used. See for example formula $r(\theta)=...$ on page 6 of this document http://www.math.utk.edu/~freire/teaching/fall2006/m142f06NewtonKepler.pdf – Jean Marie Dec 13 '18 at 07:25
  • In fact, I wanted to make this comment to the OP, not to you... – Jean Marie Dec 13 '18 at 07:26
  • I think that for regression fitting purposes, it is in fact more convenient to keep the $x,y$ Cartesian form, but to to each his own. – user619894 Dec 13 '18 at 10:11