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** Alice**: Now I understand what your code is doing, except for a few crucial
lines. First there is the one-liner about choosing the distance
between a new star and the center of the star cluster:

radius = 1.0 / sqrt( rand ** (-2.0/3.0) - 1.0)

Can you tell me what this expression means, and how it is derived? It must somehow be related to the density distribution , which you have already derived from the potential. How exactly do Aarseth, Henon and Wielen use the density for particle sprinkling?

** Bob**: They describe their technique in a few words, and I had to read
those words carefully, and do some head scratching, to figure out what
it meant. But as always, once you see it, it is really easy. Let me
try to summarize it in my own words.

First we introduce the cumulative mass distribution

which is the amount of mass contained within the star cluster inside a distance from the center. When we create a new star, and place it at radius , that star will see an amount of mass of the cluster at positions closer to the center than its own position, and therefore an amount of mass at positions further from the center.In other words, it will see a fraction of the total mass inside its radial position. Now that fraction could be anything between 0 and 1. It will be 0 if the particle is placed exactly in the center, and it will approach 1 if the particle is placed very far away, reaching 1 when the particle is placed at infinity.

The ranking of each particle, in terms of the enclosed mass, is random and uniform in the mass fraction. In other words, will be a random value between 0 and 1, with each value equally likely.

So here is the idea: spin a random number generator in order to obtain a a random number , with , and we consider that to be the fractional mass contained within the radius of particle . So all we know is that , but what we need to know is itself. So the procedure is to invert (18) to obtain a function , and then life is simple: .

** Alice**: That sounds straightforward. Can you show me the expressions
you found for and ?

** Bob**: I simply took their expressions. They use a system of units in which
the total mass * M*, the gravitational constant * G* and the structural length
scale * a* that we used above are all unity. The mass enclosed within
a radius * r* then becomes:

radius = 1.0 / sqrt( rand ** (-2.0/3.0) - 1.0)

** Alice**: Hmm. You didn't check whether they had done their math
correctly?

** Bob**: No need to. This is a paper by Aarseth, Henon, and Wielen.
Besides, it is thirty years old and has been cited zillions of times
by others. I'm sure this is a result that can be trusted.

** Alice**: I don't like to accept things on trust, no matter what the
authority may be behind it. Not that I expect them to be wrong,
I agree that that would be highly improbable. Still, I would feel
much better if we derive the results ourselves. Besides, if we work
it out now, we can both use those notes when we have to teach it in
class to the students. Better still, we just put it into the material
we prepare for them on the web.

** Bob**: Okay, if you like. Your turn, though, I already derived the
density. Do you want to use a package from symbolic integration?
Differentiation is easy enough by hand, but I must admit, I'm a bit
rusty in my integration.

** Alice**: So am I, and that is a really good reason to do it with pen and
paper, tempting as it is to use a symbolic package. Okay. I'll start
with the density you derived:

which gives us:

This gives us:

** Bob**: Well, if you are rusty in your integrations, then I don't know what
to call myself. Nice job! It is always surprising to me how the result
of that type of calculation can come out in such a simple form.

** Alice**: There probably is a good physical reason for it to be this simple.
Let's think. I started with density, something that you had found by
differentiation, and then I integrated the product of the density and
the geometric opening angle factor of . Apart from
that factor, integration and differentiation would have canceled. Pity.

** Bob**: Hey, wait a minute. I found the density by integrating alright, but
in the following way, using Poisson's equation:

** Alice**: It does . . . Hey, I could have started there! I could have written:

** Alice**: Not only that, here is the physical meaning we were looking for!
You also mentioned that the gradient of the potential is the gravitational
force, apart from a minus sign. So what this equation is telling us, is
simply that the physical force is proportional to mass and inversely
proportional to the radius squared: Newton's gravity! We could have
started that way. The magnitude of the force on a particle with mass
at distance from the center is of course:

and also equal to:

since everywhere.

** Bob**: You're right. If we would have started with those two lines, we
could have written

right away. And with the expression I wrote down yesterday,

this would have given us:

** Alice**: Quite a bit faster than my juggling of integrals! We could have
used a healthy dose of physical intuition, before embarking on that lengthy
computation.

** Bob**: Now that we have found the radius dependence of the cumulative mass,
we only have to invert that relationship, to get the dependence of cumulative
mass on radius. That shouldn't be too hard. However, I'm getting tired of
carrying along the factors and which we
may as well consider to be the physical units used for measuring
and , so that the latter are used as
dimensionless parameters. Setting , we can write:

** Alice**: But you'd better explain your students how you can restore the
correct factors of and , otherwise they
will think that you were cutting corners.

** Bob**: Good point. It takes a while to learn to think in terms of
dimensionless quantities, and to transform easily and confidently
between those and the corresponding physical quantities. In this
case, I can just point out that the dimensionless quantity
has to be replaced by , and the
dimensionless quantity has to be replaced by
. We then get:

** Bob**: Yes, students do indeed ask such questions! It just means that they
have to practice more with simple examples, until it comes to them naturally.

** Alice**: Well, that is not really answering the question. My answer would be
to introduce two different sets of symbols, to remove the confusion between the
physical quantities and the dimensionless quantities.

** Bob**: I won't stop you!

** Alice**: If you define

** Bob**: Yes, that is full mathematical rigor. But of course, in practice,
you wouldn't go to such extravagance. This is like what you were pushing
for earlier, with your request of putting hats on all kind of variables,
just because their mathematical functional form changed. Since I'm a
physicist, I prefer to change notation only if the physical meaning
of the variables change.

** Alice**: I must admit, I often do use this type of switching of variables,
along the lines of what I just illustrated. I can see that you're comfortable
with omitting that step, and that is mostly a matter of taste, I guess.
Still, you can't insist or wish that your students have the same quirks or
abilities as you. So I suggest we at least add my derivation as well.

** Bob**: In that case I insist that we leave my derivation in too, for those
younger versions of me who already got the physics, and don't want to accrete
unnecessary mathematical niceties.

** Alice**: So be it.

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