## Saturday, March 20, 2010

### Landau: Mechanics, 3rd Ed., p. 32

On the page mentioned above in the book mentioned above, Landau makes the following claim, in the context of talking about motion in a central field; that is, the motion of two particles, where the only forces between the two particles are directed on the straight line joining the two particles.

Such cases are exceptional, however, and when the form of $U(r)$ is arbitrary the angle $\Delta\phi$ is not a rational fraction of $2\pi$. In general, therefore, the path of a particle executing a finite motion is not closed. It passes through the minimum and maximum distances an infinity of times, and after infinite time it covers the entire annulus between the two bounding circles. The path shown in Fig. 9 is an example.

A note to the reader: Fig. 9 looks very much like the graph on the front of this book.

Now, in classical mechanics, we have no tunneling or other means of teleportation. This implies that the particle must continuously traverse whatever path it is on. We should also note that a particle is like a mathematical point - it has no extension.

Now, mathematically, Landau is saying that there is a continuous bijection from the half-infinite line $[0,\infty)$ to the annulus $\overline{B}(0,r_{\text{max}})\setminus B(0,r_{\text{min}})$. Is this possible? The annulus obviously requires two continuous variables to locate a point in its interior. However, given the path of the particle as a function of time, we need only specify one variable (time) in order to locate its position.

So help me out, you topologists. Is there a homeomorphism between these two sets? I could believe that the particle's path is dense in the annulus, but I'm not sure about traversing every point.

In Christ.

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At 3/20/2010 08:07:00 PM ,  Adrian C. Keister said...

Well, I partly answered my own question. The annulus, as I've described it, is compact, because it is a closed and bounded set in $R^{n}$. Hence, any continuous function defined on it is bounded. Therefore, there cannot be an onto function from the annulus to the interval $[0,\infty)$, since the latter is unbounded. Ergo, any candidate for a continuous bijection from $[0,\infty)$ to the annulus cannot have a continuous bijective inverse. Therefore, by the definition of homeomorphism, there can be no homeomorphism between the annulus and the half-infinite line $[0,\infty)$.

So that answers that. I suppose the next question would be this: does the proven lack of a homeomorphism guarantee that, therefore, this particle cannot trace out the entire annulus?

At 5/25/2010 02:52:00 AM ,  Jacqueline said...

So I have NO idea what you're talking about, but I loved what you wrote on Gene Veith's blog about mathematics and classical Christian education.

My question: When are you going to write a math curriculum for home educators like me? Make it quick, please, my boys are 10 and 4.

In Christ,
Jacqueline

At 5/25/2010 04:48:00 AM ,  Adrian C. Keister said...

Jacqueline:

For good or ill, it's probably going to be a good while before I really start writing it. I need real classroom experience to help out, especially classroom experience in the lower grades.

In the meantime, I hope you find something that works well for your sons.

In Christ.

At 5/29/2010 11:43:00 PM ,  Brian Fulton said...

Greetings old friend.

I hope you and Susan and Hans are doing well.

You are correct in your skepticism. The path does not traverse every point of the annulus.

I am not sure how well the explanation will come across in this format, but here it goes.

I'm going to let "<=" stand for "less than or equal to" and
">=" stand for "greater than or equal to".

First, from page 31 we have

d\phi = M*dt / (m*r^2) <= M*dt / (m*r_min^2).

This gives

dt >= (m*r_min^2)*d\phi / M.

This is important because we now know that each change of 2*pi in the value of \phi takes up at least k units of time, where k = [2*pi*(m*r_min^2) / M] is a quantity greater than 0.

Now, consider a fixed ray emanating outward from the origin at an angle \theta.
This ray will cross both circles r = r_min and r = r_max. If the path of the particle is to cover the entire annulus, then every point on the ray which satisfies
r_min <= r <= r_max must be a point of intersection with the path.

However, the time intervals [0, k) , [k, 2k) , [2k, 3k), [3k, 4k) , ...... each contain at most one point of intersection between the path and the ray. This is because the polar angle \phi of the particle changes by at most 2*pi in each of the above intervals and also (bottom of p. 31) the polar angle is always increasing.

Therefore, the set of intersection points on the ray is at most countable. This tells us that not every point on the ray which satisfies r_min <= r <= r_max is touched by the path. Hence, the path does not completely fill up the annulus.

At 5/31/2010 12:41:00 PM ,  Adrian C. Keister said...

Brian,

That's another good way to look at it. Your way is perhaps a bit more concrete than mine; concrete, constructive approaches definitely have their advantages! I think your approach also goes further than mine. Well done, doc.