63
$\begingroup$

Is the net angular momentum vector of our solar system pointing in roughly the same direction as the Milky Way galaxy's net angular momentum vector? If yes or no, is that common for most stars in the galaxy?

$\endgroup$
0

3 Answers 3

73
$\begingroup$

There is no alignment between the Sun or the Solar System's net angular momentum and the "spin axis" of the Galaxy. Think for a moment about whether the line of the ecliptic (which marks the "equatorial line" of the Solar System) and the Milky Way (which roughly marks the plane of the Galaxy) are lined up? If this were so, then you would always see the planets (Jupiter, Mars, etc.) projected against the Milky Way. In fact, the spin axes of the Solar System and Galaxy are inclined at an angle of 63 degrees with respect to each other (see cartoon below - note, the Solar System is not drawn to scale compared with the Galaxy!).

Spin alignment of the Galaxy and the Solar System

We do not know much about the alignments of other solar systems. Both the Doppler shift discovery method and the transit discovery method have a rotational ambiguity about the plane of the exoplanets' orbits. In other words, if we were to observe a transiting planet, we know that the inclination is close to 90 degrees to the line of sight, but we could rotate the system around our line of sight by any angle, and would see the same observational signatures.

The general assumption is that there is no relationship between the angular momentum directions of stars (and their planetary systems) and the Galaxy. Turbulence in molecular clouds on relatively small scales compared with the dimensions of the Milky way randomises the angular momentum vectors of collapsing prestellar cores. A possible alignment mechanism could come about through the threading of giant molecular clouds by the Galactic magnetic field.

If we knew what fraction of stars had close-in, potentially transiting planets, we could use the numbers of detected transiting exoplanets in the Kepler field to say whether that number was consistent with random orientations or not. Alternatively, if we had another Kepler field pointing in a different Galactic direction, but with similar sensitivity to the original Kepler field, then the relative numbers of detected transiting planets in the two fields might tell us of any non-random orientations. For example, if orbital planes were all aligned with the Galactic plane, then no transits would be seen for any star viewed out of the Galactic plane. (I think this extreme possibility can already be ruled out.)

$\endgroup$
10
  • $\begingroup$ I also wonder if anyone has looked at this thermodynamically. The (non-bulge, non-halo) stars in a spiral constitute a relatively cold dynamical population. But there are excursions from planar orbits, and there are varying eccentricities, so the temperature is positive. How cold would it have to be to not excite the internal degrees of freedom in planetary systems (assuming everything was coupled), given that our planetary system's angular momentum is more than 12 orders of magnitude smaller than the Sun's orbital angular momentum? $\endgroup$
    – user10851
    Commented Sep 7, 2015 at 17:49
  • $\begingroup$ However what? I'd need to review the content and find the videos again to make a good answer. This is less than "link only" since I don't recall the details. $\endgroup$
    – JDługosz
    Commented Sep 8, 2015 at 6:07
  • $\begingroup$ @JDlugosz If you find something, it would also be an answer to my question that I wasted 50 bounty points on... physics.stackexchange.com/questions/148268/… $\endgroup$
    – ProfRob
    Commented Sep 8, 2015 at 6:12
  • 1
    $\begingroup$ @JDlugosz I do see what you are getting at though. But isn't the main Kepler field more or less in the Galactic plane? So you would have to argue that the fraction of stars showing transits was the same as in some other control direction. But there is no equivalent second Kepler field with similar sensitivity. $\endgroup$
    – ProfRob
    Commented Sep 8, 2015 at 6:18
  • 6
    $\begingroup$ "note, the solar system is not drawn to scale compared with the Galaxy" ...or compared with itself, for that matter. $\endgroup$ Commented Sep 8, 2015 at 18:25
12
$\begingroup$

Various data from Kepler and stellar modeling allows the inclination of the orbit to be determined. Several SETI Seminar videos goes into this in some detail. Whether the eclipse cuts a little bit or straight through the middle of the star and the statistics that indicate complete misses are consistent with random orientations without enough data to see if they are biased with the galaxy rotation.

This talk tells us stars are oriented randomly, and they can tell which way it's pointing and which (different) way the planet's orbits are.

Enter image description here

Also text here

To measure the rotation of the star in Kepler-56, the authors looked at oscillation modes due to a combination of pressure and gravity waves inside the star. Certain pressure-dominated and gravity-dominated modes will split in a unique way depending on the inclination of the star’s rotation axis. The authors model the splitting of six modes to derive the star’s inclination. Because the planets transit, their inclination angle is very close to 90 degrees. Therefore, any angle different from 90 degrees for the star’s inclination is a direct measure of obliquity.

The advent of steroseismology via other instruments and exploration as a new viable technique will give more varied data as to whether we're looking at the side or top of a star, or whatever degree in between. But existing spectra (a mature field) already provide that: more Doppler broadening if you are looking at the side, none if looking down at the pole.

The transit also causes a spectral shift as it shades the limb rotating toward or away from us, and the rate of rotation varies with latitude. So these measurements (I can't find a video that goes over such detailed geometry) give more.

This paper talks about inclination, but I could not find a SETI Seminar video that went over the data analysis that explains how you can infer inclinations.

Estimates of the inclination of a transiting planet come from the impact parameter b which is the projected distance between the center of the planet at mid-transit and the center of the star, in units of the star’s radius.

They know whether the planetary system is edge-on, just clipping the side of the star's disk and thus barely transiting, or something in between. Most of the time it misses completely, but the statistics of the ones we see and the number of those we don't see are presumed to be random, which is necessary to figure out the percentage of stars with planets (at all) even though we only see those that are approximately edge on. Since the amount of tilt that will provide a visible transit varies with the size of the star, size of the planet, and distance of orbit, the statistics can be diced rather well to look for any sign that the tilt is not truly random.

$\endgroup$
3
  • $\begingroup$ You need to develop your argument more clearly. All of this refers to determining the inclination of stars and their planetary systems. This is not the same as knowing the orbital plane. You can rotate the entire system through 360 degrees and get the same observational signatures. $\endgroup$
    – ProfRob
    Commented Sep 8, 2015 at 9:58
  • $\begingroup$ Yes, there is an ambiguity around the line of sight. It would be suspicious if they were all the same inclination but presumed different angles relative to our particular sight line! If they were not uniformly distributed in the ways we can tell, I'm sure there would be some effort to see what the absolute plane was to see what everything was biased toward. $\endgroup$
    – JDługosz
    Commented Sep 8, 2015 at 10:40
  • $\begingroup$ Really not following this at all. The talk you reference has no bearing on this problem (at least the 5 minute segment I watched) and simply determines the inclination of a star and then shows that the planets in this system are misaligned with the stellar rotation axis. Ditto the other sources you refer to. They merely determine inclination. $\endgroup$
    – ProfRob
    Commented Sep 9, 2015 at 6:24
3
$\begingroup$

Let's look closer to home.

Axial tilt gives the axial tilt of the more familiar bodies in the Solar System. An axial tilt of greater than 90 degrees implies the body is rotating backwards.

So we see Venus, with little axial tilt, rotating very slowly backwards (due to a tidal resonance with Earth) and Uranus and Pluto with pronounced tilts exceeding 90 degrees.

All the other bodies in the list rotate in a prograde direction, within 27 degrees of the orbital plane / ecliptic.

So in the solar system at least, most bodies rotate in the same direction as the sun. This suggests they were formed from the same accretion disk that formed the sun, which was a mass of gas and dust swirling in a particular direction.

I suspect that the same is more or less true of the stars /solar systems in our galaxy. The galaxy clearly has angular momentum in a particular direction, and matter gathered from it to form a star might be expected to swirl in the same way.

However we must not overlook the case of Venus, whose near-perfect reversal of rotation is due to tidal effects. Matter orbiting closer to the centre of the galaxy overtakes matter that is orbiting further out. This might lead to retrograde solar systems being formed.

In general, if there is a rule, there will be many exceptions. It's like the common idea that the Coriolis force makes water swirl anticlockwise down the plughole in the Northern hemisphere and anticlockwise in the Southern: it may have an influence, but there are lots of other factors (for example, which way the water was swirling in the sink before the plug was pulled), which means the effect is largely irrelevant.

$\endgroup$
7
  • 1
    $\begingroup$ What do tidal effects hqve to do with Venus's rotation axis? $\endgroup$
    – JDługosz
    Commented Sep 8, 2015 at 2:24
  • $\begingroup$ @JDługosz en.wikipedia.org/wiki/Tidal_locking $\endgroup$ Commented Sep 8, 2015 at 5:20
  • $\begingroup$ Crazy. $\endgroup$
    – JDługosz
    Commented Sep 8, 2015 at 5:35
  • 1
    $\begingroup$ Re: "true of the galaxies in the solar system" Don't you mean it the other way around? $\endgroup$ Commented Sep 8, 2015 at 18:54
  • 1
    $\begingroup$ (Accidentally hit enter, edit timed out) Due to the vastly lower density of the Galaxy, there is much less interaction between the angular momentum of solar systems and the interstellar medium compared to the interaction of planets with the accretion disk, so the above is not a good analogy. $\endgroup$ Commented Sep 9, 2015 at 4:56

Not the answer you're looking for? Browse other questions tagged or ask your own question.