# Why does the closest approach of star S2 to Sgr A* not appear to be near the focus of its elliptical orbit?

Phys.org's Very Large Telescope sees star dance around supermassive black hole, proves Einstein right links to several ESO videos, including The star S2 makes a close approach to the black hole at the centre of the Milky Way.

In the screen shot below, the location of Sgr A* is nowhere near a focus of the elliptical orbit.

In addition to the excellent accepted answer I have written this answer to Why does Earth not appear to be at the focus of TESS' elliptical orbit in this video? and so I know that looks can be deceiving depending on one's perspective.

But I still don't really understand why the offset can look so bad here. For the video did they put the observer (the "camera") at a fairly close distance and thereby distort the perspective, or even if you choose the distance to be infinite can elliptical orbits appear to not have their central body at their focus, or at least along their axis?

Is it possible to state the orbital elements of S2 in such a way and in some coordinate system so that I can plot them in 3D and then try to reproduce this effect?

The paper in question is now on the arXiv here: "Detection of the Schwarzschild precession in the orbit of the star S2 near the Galactic centre massive black hole". This gives the following orbital elements (Table E.1):

a = 125.058 mas
e = 0.884649
i = 134.567°
ω = 66.263°
Ω = 228.171°


The orbital elements $$i$$, $$\omega$$ and $$\Omega$$ are essentially Euler angles describing the orbit's orientation in space. Start with the orbit in the $$xy$$-plane, with the periapse pointing towards $$+x$$. Then rotate about the $$z$$-axis by $$\omega$$, rotate about the $$x$$-axis by $$i$$, then rotate about the $$z$$-axis again by $$\Omega$$.

The usual convention I've seen is to have the $$x$$-axis pointing North (positive declination) and the $$y$$-axis pointing Eastwards (positive right ascension).

Putting it all together, here's some Python code to plot the orbit:

import numpy as np
import matplotlib.pyplot as plt

N = 1800

# generate eccentric anomaly values
Evals = np.linspace(-np.pi, np.pi, N*2+1)

# orbital elements
a = 125.058
e = 0.884649

b = a * np.sqrt(1-e*e)

# orbit in xy-plane with periastron towards +x
xvals = a*(np.cos(Evals) - e)
yvals = b*np.sin(Evals)

# rotate about z-axis by ω
xvals2 = xvals*np.cos(omega) - yvals*np.sin(omega)
yvals2 = xvals*np.sin(omega) + yvals*np.cos(omega)

# rotate about x-axis by i
xvals3 = xvals2
yvals3 = yvals2*np.cos(i)

# rotate about z-axis by Ω
xvals4 = xvals3*np.cos(Omega) - yvals3*np.sin(Omega)
yvals4 = xvals3*np.sin(Omega) + yvals3*np.cos(Omega)

# plot the orbit - note that y is RA and x is Dec
plt.plot(yvals4, xvals4)

# plot the black hole
plt.plot(0, 0, marker='o')

# plot the position of pericentre
plt.plot(yvals4[N], xvals4[N], marker='o')

# plot the line of apsides
plt.plot([yvals4[0], yvals4[N]], [xvals4[0], xvals4[N]], linestyle='--')

# plot the closest point in projected separation
proj_sep = np.sqrt(xvals4*xvals4 + yvals4*yvals4)
min_args = np.argmin(proj_sep)
plt.plot(yvals4[min_args], xvals4[min_args], marker='o')

# RA increases to the left
plt.gca().invert_xaxis()

plt.xlabel('ΔRA (mas)')
plt.ylabel('ΔDec (mas)')

plt.show()


This results in the following sky-projected orbit, where the orange dot is the black hole, the green dot is the position at pericentre, and the purple dot is the position at closest projected separation:

This matches the shape of the orbit shown in the paper's Figure 1. Now to compare that to the video screenshot, my best effort at matching the displayed orbits gives the following:

While not a perfect match, it looks as though they might be depicting the closest approach in projected separation, rather than the closest approach in 3D space.

• This is an amazing and thorough answer, thank you very much! – uhoh Apr 17 at 3:03

The orbital elements are on wikipedia:

$$e=0.884\ a=0.125'',\ i=134^\circ,\, \Omega=228^\circ$$

(At an assumed distance of 8kpc, $$0.125'' = 1000au$$)

It is the inclination that means that the black hole is not at the focus of the projected ellipse.

Imagine a circular orbit, with a central body. If viewed from a distant but high inclination, the orbit is projected as an ellipse, with the central body at the centre of the ellipse (not the focus)

A distant projection of an ellipse is another ellipse, but the projection of the focus is not the focus of the projection.

Below you can see another example. The ellipse is in the grey plane, with foci at A and B. But the apparent ellipse has a completely different axis, and the focus is nowhere near the axis. These image have been generate with geogebra and no perspective has been used.

• I've put a couple of images, from geogebra (it doesn't do any fancy camera at finite distance) to show how the focus and axis don't commute with projection. I haven't tried to get the exact same perspective as S2, but it illustrates the principle – James K Apr 16 at 10:30
• This is excellent, thank you! – uhoh Apr 16 at 11:46
• Ah, I've often wondered whether about distant projections of ellipses are still ellipses. Thanks. – notovny Apr 16 at 12:13
• Yes, I think you can show that an affine projection of a quadratic form is still quadratic. – James K Apr 16 at 12:30