…prove that this cloud will generally have a non-zero net angular momentum. This explains why spinning galaxies and spinning solar systems is the norm than the exception. Select any suitable “distribution of velocities” for this purpose, although Maxwellian would be nice.
As a practical example, when interstellar gases collapses gravitationally to form a star, it invariably forms not only a spinning star but a spinning outlying disc as well, which ultimately forms planets generally all orbiting in the same direction, and even individually spinning on their own axes in the same direction.
Update:
This is a spinoff (yeah, pun intended) from FGR!’s Y!A question:
http://answers.yahoo.com/question/index?qid=2014…
Update 2:
Feel free to make any simplifying assumptions, such as all the particles have the same point mass.
Update 3:
Let’s reframe the question in this way: We know that spinning galaxies and spinning solar systems (with spinning stars and planets) are extremely commonplace. Why?
Update 4:
murgh17, in fact, this is a difficult problem, and I don’t blame you if you can’t fit the full explanation here. I was just hoping that someone could offer a reasonable explanation, as you have started to. But the key is that any minute departure from zero average momentum at the periphery of the cloud will be greatly magnified, as you have pointed out.
Update 5:
This is the danger of making an oversimplified assumption that, given that this is using a frame of reference coincident with the center of gravity, such a cloud of particles would necessarily have EXACTLY cancelled out net angular momentum.
Update 6:
I will leave this open for additional suggestions as to the most reasonable (or economical, a la Occam’s Razor) explanation for this phenomenon.
Update 7:
Here’s the kernel of this matter: If we start with a cloud of particles of a given distribution of velocity, adjusted so that the center of mass is stationary, when you work out the net angular momentum, you’ll actually end up with a Gaussian distribution with a FINITE Q, not infinite. The hasty conclusion too many people make is that this Q is “infinite for all practical purposes”. No, it is not. Statisticaly speaking, the odds is that it is not exactly 0, and however tiny it is, it does and will make its presence felt when the cloud finally does gravitationally collapse significantly.
Update 8:
As a crude analogy, if I toss a perfectly fair coin uncountable trillions of times, while the ratio between heads and tails will approach exactly one, the difference does not nowhere approach 0.
Update 9:
FGR, the whole point of this discussion, which was inspired by your previous Y!A question, is that we really don’t need to appeal to “strange physics” in order to explain the prevalence of spinning galaxies and solar systems. In a nutshell, given any collection of nearby particles or masses, the odds that it’d have EXACT 0 net angular momentum is very small. Of course, when all of this is finally worked out, there’s going to be an inverse relation between the fine-graininess of the “cloud” and the net angular momentum. In other words, if we had a perfectly uniform gas cloud, like inside of a balloon at rest, the net isn’t much. On the other hand, a swarm of asteriods is likely to have a significant net angular momentum. And as you have pointed out, even the anistropy of the early universe, ESPECIALLY after gravitational collapse, would be a pretty fecund genesis of these relatively minor local net angular momentum variations, which leads to nice spinning galaxies and solar systems.
Update 10:
I’m going to have to close this out, because I’m going to be gone for a while. I’ll accept the friendly differences of opinion among the Answerers here, because I realize that with an IDEAL ball of gas one is not going to find a significant net angular momentum. At the same time, I did offer a lot of latittude in treating this problem, and we at least all do agree that it does require at least a bit of “chunkiness” to generate this statistical outcome.
Update 11:
The only thing that is clearly not true is that, given any passive collection of particles wtih a stationary center of mass, necessarily it also has 0 net angular momentum, or, worse, will not continue to have a stationary center of mass!
Update 12:
FGR, your mention of fractals is noteworthy, because even with the Navier-Stokes equation which describes fluid dynamics, although many of the solutions are laminar, many others are chaotic involving eddies in a fractal pattern. It’s not hard to imagine that a generalized gravitational collapse of a vast extended “cloud” could break up into local eddies, merely because it’s a possible (and very likely) solution.
Update 13:
Again, because time has run out for me (I won’t be back for at least a week), unless I can get internet access somewhere else, I’m not able to address all of the different answers in here. And so I’ll be closing out this Y!A question shortly.
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L = m v r, where mass is the mass of the particle, v is its velocity, and r is its distance. This is a simplified model, not using partition functions. The rotational frequency f = v / 2Pi r. Plugging this back in to angular momentum, we get L = 2Pi m f r^2. Thus as a cloud of spinning gases collapses, its net angular momentum, or sum of angular momentums from each particle in the cloud, actually gets amplified by the sum of f1/f2 = (r2/r1)^2. The friction between the rotating gas molecules is what actually forms the disk. I have a statistical mechanics proof as well, but I do not know how to attach latex documents, sorry. Let me know if you would like to see it, or know a way to link it to this answer.
Here is a link to a scholarly article that does a numerical calculation:
http://www.springerlink.com/content/k0178p43227731…
I also found the article for free, let me know if it is incomplete. I did not compare the full version to mine.
I hope I understood your question correctly…
Say the cloud has mass M and center of mass positioned at C. You can divide the cloud into two sections, each with mass M/2… the division can be made along any open line (or any open surface, for a 3-d cloud). The centers of mass A(t) and B(t) of the two sections can move, but they are subject to the condition
(A(t)+B(t))/2 = C (constant)
from this equation (or just by thinking about it) you see that A, B, and C are collinear. Taking the time derivative:
A'(t) + B'(t) = 0
which is just the result that the two sub-clouds must have equal and opposite velocities — their momenta must sum to zero to keep point C stationary. Otherwise there is no restriction on the direction of the velocities, since the particles are moving randomly and no direction is preferred. There will be some probability distribution for the magnitude of the velocities, based on the particle speed distribution, but that’s not important.
Now, getting to angular momentum: if the velocities of the sub-clouds are directed along the line ACB then the whole system has L=0 (vectors CA and CB are parallel to A’ and B’ so the cross-products vanish). However, if A'(t) has a component orthogonal to A(t) there will be finite angular momentum in the whole cloud.
So I would say there isn’t any reason the cloud should not have finite angular momentum: as you put it, the cloud generally has a non-zero L, except in the special case that the momenta are directed radially outward.
As an aside, I don’t think this really explains the prevalence of spinning structures in outer space. They spin because they started off with angular momentum. Similarly, if the cloud in this problem has angular momentum, the “reason” is that the system started off with it. This problem simply demonstrates that it’s possible for a system of randomly moving particles to have finite angular momentum… kind of surprising, since I would have expected that no direction is preferred when the particles are moving randomly. But I guess the point is that the velocities are distributed randomly in the following sense: a particle chosen at random from the whole ensemble is equally likely to be moving in any direction; however, (if the cloud has non-zero L) particle velocity is correlated with position in the cloud, so that particles chosen at random from a specified region in the cloud are *not* moving randomly.
I think that your assertion is not true. If all the particles in the cloud have velocities drawn from the same distribution, the net angular momentum (about the center of mass) should in fact be zero.
Consider a reference frame moving with the center of mass. The average velocity is now zero, by definition. If particles on the left (for the sake of argument) have velocities drawn from the same distribution as particles on the right, then the clumps of particles will both have average velocity of zero, giving zero angular momentum on average.
The resolution to this problem is that the _average_ angular momentum will be zero. The angular momenta of all the clouds we create in this manner will have a distribution that is centered about zero, but none will have exactly zero. If we have a cloud that has an angular momentum slightly different than zero, it will be pulled together by gravity, so that all the radii of the particles from the center of mass will decrease. The angular momentum will stay the same, so the angular velocity of all these particles will increase. Now our slightly nonzero angular momentum has formed a rotating cloud. One could argue that this doesn’t imply that everything should rotate about the same axis, but a friction-like process combined with gravity could probably smush things into a single pancake.
Also, any clouds with really small angular momenta would contract to very small rotating clouds, which would become rotating black holes. If a galaxy is going to persist for very long, it must be rotating fast enough to avoid this collapse.
“Given a cloud of particles of random distribution of velocities and stationary center of
gravity?”
This is immediately impossible. The center of mass is a function of the positions and masses of the particles that compose the system. If the velocities of particles that compose the system are random the center of gravity will be constantly shifting in all practical cases. Also as someone has already said that the residual angular momentum comes from initial perturbations prior to the gravitational collapse. Since the angular momentum is always conserved if all the particles prior to collapse have exactly zero angular momentum then I think the resulting celestial body will also not be spinning. Though of course it has almost zero probability of occurring in real cases.
Let me add the following since you are closing:
First there is a simple vector analysis showing collapse
onto the “equatorial plane” of a slowly rotating ball of gas.
I believe your question is not concerned with this.
Given the current state of the universe via the known conveyor of
information, the probability that any large (or small) dynamic gas cloud has
no rotation (zero net angular momentum vector) is essentially zero.
From this and barring a few mysteries (like DM) , the physics and
thermodynamics of what follows is known. In fairness to First Grade, there
is an problem with the non-Keplerian rotation curve of the milky way. The
“flatness” of the curve could be caused by unseen matter, but we are
inferring it’s existence mostly via EM Doppler data which may be in error. If so,
the amount of unseen matter may drop to the point where Kepler may
no longer roll over in his grave. I think we all agree that the probability
of a cloud rotation seed is essentially one. I’ll close with an interesting
question and thought experiment I have. (Initially, I thought your question would
lead to this) :
Would a large, cold, isolated, homogeneous gas cloud (not a mathematical cloud of point particles) with Maxwellian distribution of velocity and ZERO net angular momentum contract under the action of known acceleration vectors and end up as a rotating disk(s) with essentially parallel angular momentum vectors?
Without proof, my tentative answer is yes> when I consider one theoretical property of the universe in tandem with QM: There is a known transporter of energy and potentially net angular momentum (not spin) out of the cloud as it collapses along its rotational axis. Proving this is another matter.
Jeez, we got to go back to the Big Bang (BB) to answer this one.
[Notwithstanding, I really like the answers above me].
Our present view of the big bang is that space began expanding from a singularity. The expansion of space between various points in the universe was greater then the speed of light. The consequence of this proposition is that you would get a very uneven distribution of matter. This is because of the lack of “mixing.” If we look north and we look south to the edge of the CMBR, those two points have not been in contact with each other since the moment of the BB. The random local Boltzmann distributions would cause severe lumpiness and we would have a universe littered with black holes.
So astrophysicists invented “inflation.” This is a super-rapid expansion in the first second of the universe’s existence spread everything out to make the universe relatively smooth instead of lumpy.
Next we will skip ahead to the CMBR, which represents the de-ionization of the of the plasma gas making up the early universe. Once the free electrons were coupled with the ionized atoms changing the plasma to a gas, light could move freely about the universe.
However, the CMBR is not completely smooth. It was not isotropic, but anisotropic. There were pockets of cooler and hotter gases. In reality since the coupling occurs at a uniform temperature (~50K), the cool pockets represent where the plasma cooled sooner and the density was less and are farther away in time, and the hotter pockets cooled later, the density greater, and are closer to us in time.
The anisotropic nature of the CMBR limits the amount of time over which “inflation” happened, otherwise we would have a smooth universe. It also set the seeds for the genesis of the large scale features of the universe and the galaxies themselves.
My layman’s mind likes to think of this as a series of fractuals (this is where the important math resides, and which I — lacking a supercomputer in may head — am going to pass on trying to terse out) representing the great voids and galactic superclusters, the local group/galactic clusters, the large galaxies, the dwarf galaxies and the globular clusters, and finally the open star clusters, the stars and the planets
Like a whirlpool, it is the very subtle initial angular momentum around the center of gravity which causes a much faster velocity as the galactic cluster/galaxy/star/planet collapses towards its center. Remember that there is a lot of empty space between the stars and there is even more between the galaxies. Matter has fallen from a long ways away to make it into the star/galaxy. (I have heard it said that “nothing ever falls straight into a black hole” — similarly nothing falls straight into a galaxy). Very little initial angular velocity can result in a fast rotating disk as the matter collapses.
Notwithstanding your assertion that there is a stationary center of gravity, the gravitational field will not be uniform. You will have outside gravitational sources such as nearby masses which are also collapsing, you will also have internal sources that may not be centered on the center of gravity/mass. All of these create a non uniform gravitational field. (you can consider these to be “tidal effects”).
There are, of course, objects like globular clusters which show little if any condensing into a disk. Also elliptical galaxies which are formed as a consequence of collisions between multiple galaxies no longer have a galactic plane. (Maybe after many billions of years they would, but stars do not collapse quickly, only gas and dust).
As for the math, the guy at the top did a pretty good job and for the really interesting stuff you’ll need a supercomputer.
**********Addendum************
I was all set to add that this can be simplified into an n-body problem where you seed your galactic cloud with a set of randomly placed masses within the cloud, and each has a random velocity.
But, I see where you are going. You need to find the net velocity perpendicular to the center of mass and with respect to the eventual galactic plane.
Rough estimate would simply to use velocity space and take the average velocity x average radius and divide by the square root of 3x N- particles. The √ 3 represents the component of the velocity perpendicular to the center of mass and projected in the galactic plane. The √ N represents a random walk in velocity space by an H2 molecule.
L = m* v.i * r.i * N / √ (3N)
If you started with a very large radius and you collapsed the gas cloud you would eventual cause the velocity perpendicular to the center of mass increase in proportion to the decrease in radius.
But since angular momentum is conserved but the other velocities cancel out:
L.final = L.initial
m* v.f * r.f *N = m* v.i * r.i * N/√(3N)
v.f = v.i *r.i/r.f *1/√(3N)
where v.f represents the average final orbital velocity.
Given the huge number for N.
N = ~10^9 stars x 10^30 kg/star x 3×10^26 molecules H2/kg
N=~ 3×10^65 particles
The net angular momentum will be very, very small unless you have very, very large initial velocities.
For example, lets use are sun as a prototypical star. Its velocity is 2 km/s and radius is 20,0 light years. If we assume that it originally collapsed from a distance roughly 50x its current distance from the galactic center.
v.i = v.f *r.f/r.i * √(3N)
v.i = 2 km/s * 1/50 * √ (3*3×10^65)
v.i = 3.8×10^35 m/s
Oops, that is a little fast.
It is far, far better to model using an anisotropic universe. This gives you a lot more initial pop. You need cold matter that has turbulence and an angular momentum rather than hot particles that give you a net angular momentum by chance alone.
********Addendum 2*********
My result above is a gross back of the envelope estimation. But in contemplating it, I realized that large average velocity (which is temperature) supports the big bang.
Even though you start out with a boltzmann distribution of matter and energy, you have discontinuities caused by a very hot, expanding universe. Now that everything has cooled, those discontinuities have resulted in local gravitational cells that have net angular momentum even thought there is no net universal angular momentum.
Of course, my calculations are far too crude for anything but providing a little insight.
Here’s my two cents, for what it’s worth.
Consider a static cloud, with the equivalent to drift velocity zero everywhere, but with thermal velocity well defined by a probability distribution connected with temperature, in such a way that instantaneously we may say we are describing the cloud at the exact right moment it’s total mass starts driving collapse to the center.
Mathematically, corresponding to matter being continuous, if at any given element of volume you compute the expectation value for angular momentum, because the velocity probability distribution should be isotropic, you get a zero value. You may try to proceed computing the probability that average angular momentum deviates from the expectation value. To make sure I’m not sticking my foot into my mouth, how would you do that? In a perfect (continuous) mathematical model, this doesn’t seem to make any sense. For any amount of mass in the element of volume having velocity +v, you have the exact same amount of mass having velocity canceling out in the integration, meaning that the probability of deviating from the expectation value is zero.
However matter is the cloud is not continuous. In any element of volume there exists a finite number of particles that though large, will not ever completely cancel each other out (or the likelihood of that happening is very low). The net angular momentum will be residual, noise, in any small volume. But consider that low density increases its value, in the lines that one particle is net worth, two may partially cancel a bit, three even more so, and so on and so forth. The more particles you have, the more they approach the “mathematical ideal” case. Consider additionally that the longer is the position vector of that element of volume, meaning the outskirts of the gas cloud, that can extend up to hundreds of AU or even a few ly, the larger is the angular momentum obtained from incomplete statistical cancellation of velocities. Elements of volume having minute residual angular momentum at their center, will be amplified by low density, far away position vectors relative to the center of mass of the cloud.
Now, just give me two or three minutes to work the mathematics of this.
It seems to me that in large clouds, and stellar systems proto clouds are very large indeed, perhaps the expectation value for angular momentum will be zero, but the probability that in an actual cloud it deviates from this value may even be larger than the probability that its value is zero. No joking. Just notice that the expectation value is given by trying gas clouds an infinite number of times, but we are concerned with the probability for one cloud. In the infinite set of gas clouds, each having a given +L will cancel with another cloud having -L, independently of the likelihood of |L|. Does this make sense, anyone? Imagine the probability distribution for L is given by a U shaped curve. Expectation value is still zero.
Of course, any initial asymmetry of the mass distribution in the cloud guarantees that angular momentum does not depend alone of the residual, finite number of particles, effect I just described. And again the outlaying, faraway parts of it play a larger relative role.
The truly neat part of this would be if we could from the mathematics of this extract probability distributions for planets, their masses and composition. Starting with a cloud with a given mix of heavy elements and hydrogen will we ever be able to compute the probability of Jupiters and Earths, either close to the star or a bit farther out?
I’ll be back to you in a NY minute. I just have some paper scribbling to do. This time Y!A will beat ArXiv.
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[EDIT:] Summarizing what I said above:
1) In a “physically ideal” perfectly symmetrical sphere with no internal currents, what we would call drift velocities, an adamant zero everywhere, and with only thermal velocities, it is possible that the probability of large angular momentum is better than the unlikely case that it cancels overall to zero. This would need some mathematical legs to walk on.
2) When a gas cloud that may be a few light-years across collapses to the sizes of a Solar System that potential huge angular momentum is conserved, but of course you know this.
3) Therefore, I submit, the probability of having spinning structures everywhere (with the exception of elliptical galaxies that may be the result of dynamics posterior to cloud collapse) is huge considering that you never get perfectly spherical gas clouds to begin with, and the collapse of even slightly asymmetric clouds will result in something spinning.
Apart the fact that I am throwing item 1) that at this point I am unable to substantiate mathematically, I am not sure Scythian, I understand the clarification you made on the question. Would you clarify the clarification, if you may?
Well the text of your question does say “prove.” Roger to that, …
You’re kidding, right?
__________________________
[EDIT 2:] I am a bit puzzled by your statement that the difference between heads and tails in an infinite toss does not approach zero.
Given that probability is defined by taking the number of tries to infinity, I would think that
lim (H – D) = lim H – lim D
unless there is some technical non-intuitive aspect in the mathematics of this.