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5. Getting the Physics Right

5.1. An Analytic Recipe

Alice: Hi Bob! Well, are you ready to derive the distribution function of Plummer's model from first principles?

Bob: If we're really going to get to the bottom of this, I insist on using . Let's roll up our sleeves then. Here is the, by now very familiar, potential-density pair for our Plummer's model:

Substituting the right-hand side of the first equation in the second equation, we find:

Now we have a choice: we can use this expression either in eq. (52) or in eq. (53). If I were to guess, the sheer fact that Binney and Tremaine have added the second equation would suggest that that one is the easiest to work with; otherwise they would have limited themselves to the first one. They don't seem to add anything more than necessary.

Alice: Unlike our style of writing.

Bob: I'd say! It's a good thing that we are writing this for our students . . .

Alice: . . . and for ourselves . . .

Bob: . . . and for ourselves, yes, but not for our colleagues. I certainly wouldn't want to show them how many small steps it takes me to derive these types of equations.

Alice: But don't forget, we are planning to put this up on the web.

Bob: Ah, yes, I'd forgotten about that already. Oh, well, we can just call each page `class notes', and that will chase our colleagues away.

Alice: I'm not so sure, but I don't want to discourage you. Let's move on!

Bob: Yes, let's. As I was saying, my bet is on eq. (53). Let's write it down again here, and then substitute with eq. (55):

The second term is zero, and the expression in the denominator invites us to introduce the following change of variables:

5.2. The Full Form

Well, I guess I was not as rusty in these kind of manipulations as I had thought I was.

Alice: You shouldn't, since you did your undergraduate studies a lot more recently than I did!

Bob: Yet it seems so long ago! I guess everything is relative.

Alice: At least the power came out correctly, as expected for a polytrope of index . A while ago you wrote down an expression you got from some book, where was that, ah, eq. (37). Let's write it here again:

Bob: Since , we got exactly what was ordered. It was a bit of work, but I must admit, it is fun to derive things from scratch. But wait, what is that factor doing there? That one is missing from what I just derived.

Alice: It was the mass of a single star, in our star cluster in which all masses are the same. The book you picked your expression for the distribution function from, must have defined as the number density of the stars. If you multiply both sides with , you get on the right-hand side what you found. This then must be the mass density of the stars in phase space, namely the number density multiplied by the mass of a single star.

Bob: That must be the answer.

Alice: You may remember that I did not protest when you decided to do your derivation purely in dimensionless units. As a reward, I insist on you writing the final answer with all the 's, 's, and 's put back in.

Bob: That's easy. Here it is:

But don't ask me to prove it to you with hats or tildes or a substantial part of the Greek alphabet!

Alice: I won't. But since that is how I would do it, I'm curious to know what you just did. How did you figure this out?

5.3. Dimensional Analysis

Bob: Well, I used dimensional analysis, but not in a very systematic way, just substituting what seems to work in the quickest way. If we denote the unknown combination of factors by , we can start by writing

where I have dropped all numerical factors, since they won't make any difference as far as dimensional analysis is concerned. Since you just pointed out that we have computed the mass density in phase space, we know that the expression

has the dimension of mass, something we can write as


Since energy is proportional to mass times velocity squared, think , our , the energy per unit mass, has a dimension of velocity squared, so

Before you protest that I should use only physical dimensions of [mass], [length] and [time] . . .

Alice: . . . which is what I would have done . . .

Bob: . . . I want to point out that I have a freedom of choice, as long as my three units are not degenerate. I find it more convenient to work with [mass], [length] and [velocity].

Alice: Come to think of it, that makes sense, since the last two reflect the two different spaces of which phase space is the product, and mass is what we have been sprinkling into phase space.

Bob: I guess. You always find a way to suggest more abstract reasons for what I am doing intuitively. It just has been my experience that velocity is a more useful quantity than time in this kind of dimensional analysis.

Alice: I think I'll follow your suggestion, from now on. Learn something new every day!

5.4. Getting The Details Right

Bob: Where was I? Ah, yes, I started with

and using the last two expression above, I can find that has the following dimension:

Now the crucial point is that only can help us in giving us a velocity dimension. Velocity involves time, since its dimension is length over time, and neither nor have a time component.

The dimension of follows from the fact that potential energy must have the same dimension as kinetic energy:

for whatever 's, 's, 's, etc. you care to use. This implies:

So this tells us that we can factor out as follows:

Since the only mass we have is , and the only length we have is , we have to conclude that

This then means that our dimensionless expression, that we derived with the help of Binney and Tremaine:

has to be expanded in the following way, to make it again dimensionally correct:

5.5. From Implicit to Explicit

Alice: This is indeed what you wrote down before, eq. (59).

Bob: I must say, I'm surprised at how many words and lines of equations I have to write down to make it all explicit. I did it mostly in my head, with a few scribbles here and there to provide some help.

Alice: That's because you've grown so familiar by now with these kinds of manipulations. But I definitely think it is a good idea to show your students how you do this, if only once.

Bob: Don't you think they will figure it out for themselves, sooner or later.

Alice: Probably later. I like the idea of being a catalyst, speeding up the process of letting them discover things. And by making your implicit stream of thoughts explicit, tedious as it may seem, we may be doing them a real favor.

Besides, to be really honest, I think we can still learn a lot as well. I had no idea that your approach would be so different from mine, and I can see the advantage of your way of thinking, at least in some cases such as these. It would definitely have taken me longer to derive your result in my way.

Bob: It is interesting, isn't it, that normally we don't talk very much about how we do these kind of little, or not so little, derivations. And I must admit, I too have learned quite a bit from the way you approach these problems. Besides, it is more fun to struggle with all this together.

Well, I think we've struggled enough now. Let's call it a day.

Alice: call it a day? But we haven't yet checked the truth behind the Abel integral transform, whether that way of inverting the equation was correct. We have just taken Binney and Tremaine's word for it!

Bob: I'd forgotten all about that. Well, hmm. Do you really insist on checking it all the way?

Alice: I do.

Bob: Okay, okay. Let's get it over with. Can I look at that page, where they talk about Mr. Abel and his transforms? You would at least expect them to give us a hint.

Alice: They do give a hint: go read appendix 1.B.4.

5.6. Abel Integral Transforms

Bob: So they do. Here it is. They start with the following two lines.



I like that terse style: only "Let . . then . .", what a difference from our constant chattering!

Alice: It is a good move of them to reduce the more complex equations (52) and (53) to these simplified forms. But let us first check whether the "Let . . then . ." claim, if true, will solve our problem. If it really does, we then have to verify the claim itself, to make sure that the problem really has been solved.

Let us start with eq. (62). For this to be equal to our previous equation (51):

We have to use the following dictionary, with their appendix notation on the left and their main text notation on the right:

Bob: Now that's what I call confusing. They use symbols for two very different functions, in their two languages. We'll have to be careful with this dictionary.

Alice: But I thought you didn't like hats and tildes, and preferred to use the same symbol for different things?

Bob: Only if they have the same physical meaning. Here we're talking mathematics. I would have prefered that they would have used instead of in the appendix. Oh well, small point, as long as we're careful.

Alice: Now let us assume that their eq. (65) indeed provides the solution for eq. (62). We then have to translate eq. (62) back into main text quantities, using the dictionary in the other direction. Here we go:

Bob: Indeed, these are exactly the eqs. (
52) and (53) that we set out to prove. Great! Now can we start testing my code?

Alice: Ho! Not so quick. All we have done is verify that Binney and Tremaine's claim in the appendix leads to their claim in the main text. Now we have to prove the claim they make in their appendix.
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