Radial Distribution of Points over the Area of a Circle

[SaintsTheMeta]

This is a tiny part of a presentation I am giving Friday, any and all help is appreciated.

Homework Statement

Suppose we have a circle centered on O. We are looking for the distribution of the points generated by the following method:

We choose a random radius of the circle, and then choose points along that radius with a uniform distribution. i.e. for any arbitrary radius of the circle, we will have a uniform distribution of points $f(r)=\frac{1}{R}$.

My question is how do we extrapolate this to the area of a circle, to account for all "random radiuses?"

ET Jaynes gives a solution: $f(r)=\frac{1}{2πRr}$
here: http://www.google.com/url?sa=t&rct=...OQeRMbfgQ&sig2=goMBcl6Cr-jKjCMYz11G_g&cad=rja
page 5, last sentence before the section 4 header.

Homework Equations

$f(r)=\frac{1}{2πRr}$

The Attempt at a Solution

This solution seems logical. But I need to be able to give a FORMAL proof of where this comes from. I just CAN'T seem to understand exactly where this equation comes from.

My thinking is it is a uniform distribution over circumference $f(r)=\frac{1}{2\pi r}$ multiplied with the uniform distribution of each radius $\frac{1}{R}$

Or, we could integrate $\frac{1}{R}$ over angles 0 to 2pi which would give what we want, but Whyyyy

Can anyone help me PRECISELY understand this? This is the kind of thing that is almost always "hand waved" in physics, and while it seems logical, formalism is difficult for me after my studies focused on physics.

Thank you!

edit: My best explanation at this point would be a series of "rings" with circumference 2∏r, then multiplied with the uniform distribution 1/R. But still, this is not logically sound, argh... Why would it be 1 over 2piR?? And what could allow you to just multiply these expressions together??

 

[haruspex]

You mention midpoints, but there's no reference to such in the construction.
(The pdf link crashed Firefox.)

 

[SaintsTheMeta]

You mention midpoints, but there's no reference to such in the construction.
(The pdf link crashed Firefox.)

sorry, I meant just points. Updated in first post..

Will post any alternate link to Jaynes' paper I can find, for now I have updated the OP to a new link from Google, which at least doesn't crash my Firefox..
http://books.google.com/books?id=3f...v=onepage&q=jaynes well posed problem&f=false page 139 has the same page I cited in OP, but the previous 2 pages are removed. Best I could find other than the original link in OP

In case I was unclear, it's pretty simple in theory. Every radius of the circle has a uniform distribution of points on it. ie if we let x=the distance along the radius from the center of the circle O, then we have the pdf:

f(x)=1/R when x is an element of [0,R], else =0

If we then consider infinitely many radii with distributions as such, according to ET Jaynes we see a distribution function over the entire circle of 1/2∏Rr

 

[SaintsTheMeta]

BUMP

I'm thinking I need to somehow integrate 1/R over the circumference of infinitesimal rings to get an equation:

normalized to equal 1 of course...
1=f(r)*(2pi*r)(R)

But how and why could I do this? What would allow me to integrate this way?

Thank you!

 

[haruspex]

I don't see why it's so hard. You want a pdf over the circle which is only a function of r, so f = f(r). The probability of picking a point in an area element dA = r dr dθ is f(r)dA. For a given θ, that means the probability of picking a radius in the range (r, r+dr) is f(r)rdr. Since you need that to be a uniform distribution, f(r) is proportional to 1/r.

 

[SaintsTheMeta]

I don't see why it's so hard. You want a pdf over the circle which is only a function of r, so f = f(r). The probability of picking a point in an area element dA = r dr dθ is f(r)dA. For a given θ, that means the probability of picking a radius in the range (r, r+dr) is f(r)rdr. Since you need that to be a uniform distribution, f(r) is proportional to 1/r.

yes that is understood... but how do we go from f(r)~1/r to f(r)~1/(2πRr) ??

if we set up an integral normalized to 1 with limits of integration are 0 to R and 0 to 2π, we would have
1 = ∫∫f(r) * r * dr *dθ ... add in a term of 1/r and we have
1 = ∫∫f(r) * 1/r * r * dr *dθ ...
1 = ∫∫f(r) dr *dθ ...
1 = f(r) * R * 2π
which gives f(r)=1/2πR ... missing a term of r (!)
how do we end up with a dependence on both 1/r AND 1/R ??
again, I need to arrive at this solution: $f(r)=\frac{1}{2πRr}$

Thanks a million if you or anyone can help clear this up by tomorrow morning :)

 

[haruspex]

1 = ∫∫f(r) * r * dr *dθ ... add in a term of 1/r and we have
1 = ∫∫f(r) * 1/r * r * dr *dθ ...

No, that's redefining f. You mean f(r) = k/r, 1 = ∫∫f(r) * r * dr *dθ = ∫∫k * dr *dθ = 2πRk. So k = 1/2πR, f(r) = 1/2πRr

 

[SaintsTheMeta]

No, that's redefining f. You mean f(r) = k/r, 1 = ∫∫f(r) * r * dr *dθ = ∫∫k * dr *dθ = 2πRk. So k = 1/2πR, f(r) = 1/2πRr

Wow thank you! I don't know why I got so confused by that... totally over-thought it..

Amazing how much my mathematics skills have left me in just 1.5 years of not using them...