PDA

View Full Version : When "Milkdromeda" happens.



Buttercup
2015-Sep-29, 08:39 PM
So I was watching some episodes (Sci Channel) of "How the Universe Works." The eventual merger of our galaxy and Andromeda was mentioned. Including speculation that despite the big event, no two stars will likely ever collide during all that. That surprised me. Yes, I know stars are very distant from each other. But still...

Noclevername
2015-Sep-29, 08:54 PM
So I was watching some episodes (Sci Channel) of "How the Universe Works." The eventual merger of our galaxy and Andromeda was mentioned. Including speculation that despite the big event, no two stars will likely ever collide during all that. That surprised me. Yes, I know stars are very distant from each other. But still...

Stars and bodies of similar size tend to go into partial or full orbits around each other if they get near enough. They are almost always flung out into new trajectories, always curved. Almost nothing goes in a straight line at that scale.

StupendousMan
2015-Sep-29, 09:01 PM
Stars in the local neighborhood are separated from each other by very very roughly 5 million times their typical radius.

So, imagine that the United States is emptied of people and buildings and animals and trees, and all that remains are tennis balls. Each tennis ball is placed about 5 million times its radius away from the nearest other balls ... or very very roughly 100 miles apart. The entire US has maybe 500 tennis balls, spaced out evenly around the country.

China, likewise, has been emptied of people and trees and animals, and it, too, has tennis balls sitting here and there, spaced 100 miles apart. Picture another 500 tennis balls placed hither and thither.

Now, continental drift carries the North American plate toward the Asian plate. North America collides with China, and through the magic of the imagination, drifts right through it! What are the chances that 1 of the 500 tennis balls in the US collides with 1 of the 500 tennis balls in China?

Buttercup
2015-Sep-29, 09:04 PM
Stars in the local neighborhood are separated from each other by very very roughly 5 million times their typical radius.

So, imagine that the United States is emptied of people and buildings and animals and trees, and all that remains are tennis balls. Each tennis ball is placed about 5 million times its radius away from the nearest other balls ... or very very roughly 100 miles apart. The entire US has maybe 500 tennis balls, spaced out evenly around the country.

China, likewise, has been emptied of people and trees and animals, and it, too, has tennis balls sitting here and there, spaced 100 miles apart. Picture another 500 tennis balls placed hither and thither.

Now, continental drift carries the North American plate toward the Asian plate. North America collides with China, and through the magic of the imagination, drifts right through it! What are the chances that 1 of the 500 tennis balls in the US collides with 1 of the 500 tennis balls in China?

Um...17? :D

Joking. I see your point (and Noclevername's, too).

ngc3314
2015-Sep-29, 09:17 PM
Maybe not zero, but only a few. There's a back-of-the-envelope calculation to show this. Take 400 billion stars in each galaxy, and let them be distributed randomly (but with uniform surface density) across disks of radius 20 kiloparsecs. That gives one star per 0.003 square parsecs when summed all the way through each galaxy's disk perpendicular to its plane. Even a red giant star has a cross-section of only about 2E-11 square parsecs, so the expected number of star-star collisions if the galaxies pass face-on through each other is <4E11 x 2E-11 ~10 ( an upper limit since only a minority of stars are this large). This simple multiplication neglects important details such as the bulges having more stars but being smaller targets, and a merger involving more than one passage for parts of the galaxies, but captures the gist of why galaxy mergers produce so few direct collisions between the two sets of stars.

Buttercup
2015-Sep-29, 09:21 PM
Maybe not zero, but only a few. There's a back-of-the-envelope calculation to show this. Take 400 billion stars in each galaxy, and let them be distributed randomly (but with uniform surface density) across disks of radius 20 kiloparsecs. That gives one star per 0.003 square parsecs when summed all the way through each galaxy's disk perpendicular to its plane. Even a red giant star has a cross-section of only about 2E-11 square parsecs, so the expected number of star-star collisions if the galaxies pass face-on through each other is <4E11 x 2E-11 ~10 ( an upper limit since only a minority of stars are this large). This simple multiplication neglects important details such as the bulges having more stars but being smaller targets, and a merger involving more than one passage for parts of the galaxies, but captures the gist of why galaxy mergers produce so few direct collisions between the two sets of stars.

Thank you.

Of course, I'm not the least bit qualified to contradict anything; but for all those stars of two big galaxies, I'd expect a few collisions. But I'm not qualified...

Jeff Root
2015-Sep-30, 12:06 AM
You can expect a few. If there turn out to be none, though,
I won't be surprised, because I don't have the patience to
wait that long. If there are no collisions in the first couple
of million of years, I'll quit paying attention.

-- Jeff, in Minneapolis

Noclevername
2015-Sep-30, 01:04 AM
Galactic mergers are like NASCAR racing; most people only watch it for the crashes.

George
2015-Sep-30, 05:29 PM
Maybe not zero, but only a few. There's a back-of-the-envelope calculation to show this. Take 400 billion stars in each galaxy, and let them be distributed randomly (but with uniform surface density) across disks of radius 20 kiloparsecs. That gives one star per 0.003 square parsecs when summed all the way through each galaxy's disk perpendicular to its plane. Even a red giant star has a cross-section of only about 2E-11 square parsecs, so the expected number of star-star collisions if the galaxies pass face-on through each other is <4E11 x 2E-11 ~10 ( an upper limit since only a minority of stars are this large). This simple multiplication neglects important details such as the bulges having more stars but being smaller targets, and a merger involving more than one passage for parts of the galaxies, but captures the gist of why galaxy mergers produce so few direct collisions between the two sets of stars.
That's interesting. Wouldn't potential colliders minimize their actual impact due to mutual gravitational interaction? Perhaps many or most of these 10, or whatever, become shredders more than impactors, or would their gravity actually serve to improve their impact odds?

Without looking to see the rotation differences and inclination between us, wouldn't these be important to the odds as well? In other words, are the disk rotations such that the outer stars of the disk will find themselves in head-on traffic? Perhaps tidal tearing will minimize this if in fact we (our disks) are so rotationally oriented.

kzb
2015-Oct-05, 12:29 PM
The current numbers are an approach velocity of 110 km/s and transverse velocity of 17 km/s.

Presumably the approach velocity will increase as the two galaxies fall together under gravity.

But then both galaxy disks are rotating. About 240 km/s for the milky way and 255 km/s for M31.

I don't know what the angle of collision will be and whether the rotation directions will be the same or opposite. But surely it is safe to say the relative velocities of stellar systems from M31 will be in the 100's of km/s. Far higher than relative velocities in our stellar neighbourhood currently. I imagine most stars would just shoot by, unless they approach very closely.

I realise I am stretching the scope of this thread, but if you want to calculate the effect on things in general, you don't just need to consider the cross-section of the stars themselves. The cross section for causing planetary system disruption is much larger, in one recent paper this was estimated as 100AU as a ball-park figure.

Here is a paper which argues the two galaxies have approached closely on a previous occasion (55kpc) and the structures we see now can be explained by this past encounter.

Local Group timing in Milgromian dynamics. A past Milky Way-Andromeda encounter at z>0.8

http://arxiv.org/abs/1306.6628

Lastly, I hope someone comes up with a better name for our new galaxy when the time comes !

antoniseb
2015-Oct-05, 01:13 PM
... Local Group timing in Milgromian dynamics. A past Milky Way-Andromeda encounter at z>0.8 ...

Just as a heads up, the cited paper can be used for first order estimates on velocities and collision likelihood, and such, but as it is based on MoND, and not GR, it will have a few things built in which are ATM ideas.

publiusr
2015-Oct-09, 09:42 PM
I seem to remember how early galaxy mergers can spawn new stars--but we look to get cheated, as our two galaxies are supposed to be more worked over.

jamesabrown
2015-Oct-10, 01:44 AM
It seems that within both galaxies there are a small but non-zero number of stellar collisions, aren't there?

If so, then I would guess that the number of stellar collisions that occur when two similar-sized galaxies merge would only go up by a factor of two.

Jeff Root
2015-Oct-10, 09:00 AM
jamesabrown,

The relative speeds of the stars is a factor, so the collision
rate (collisions per unit time) would likely more than double
in any region of overlap where the star density is doubled.

-- Jeff, in Minneapolis

grapes
2015-Oct-10, 03:07 PM
Stars in the local neighborhood are separated from each other by very very roughly 5 million times their typical radius.

So, imagine that the United States is emptied of people and buildings and animals and trees, and all that remains are tennis balls. Each tennis ball is placed about 5 million times its radius away from the nearest other balls ... or very very roughly 100 miles apart. The entire US has maybe 500 tennis balls, spaced out evenly around the country.

China, likewise, has been emptied of people and trees and animals, and it, too, has tennis balls sitting here and there, spaced 100 miles apart. Picture another 500 tennis balls placed hither and thither.

Now, continental drift carries the North American plate toward the Asian plate. North America collides with China, and through the magic of the imagination, drifts right through it! What are the chances that 1 of the 500 tennis balls in the US collides with 1 of the 500 tennis balls in China?
Using plate tectonics to throw them at each other is slow.

Let's make them smaller, then gather them into a ball, and throw them at each other.

Let's see, instead of a 7cm tennis ball, reduce the 500 objects to 1mm, then throw them at each other. The groups would only be . . . Let's see, fifty miles wide. OK that's still too big to throw at each other.

Noclevername
2015-Oct-10, 06:20 PM
Would the star density double? I vaguely recall that the stars, in the arms at least, would be pulled further apart by gravity as the galaxies approach each other. Does that affect star density in the cores as well?

Jetlack
2015-Oct-10, 06:37 PM
Thank you.

Of course, I'm not the least bit qualified to contradict anything; but for all those stars of two big galaxies, I'd expect a few collisions. But I'm not qualified...

I would imagine, even without direct collisions its going to cause chaos in some planetary systems where a star passes by too closely.