# Probability of randomly bouncing speedometer (stubborn calculus)

1. Jan 26, 2010

### Yitzach

Welcome to Intro Quantum Mechanics:
1. The problem statement, all variables and given/known data
(From previous problem) The needle on a broken car speedometer is free to swing, and bounces perfectly off the pins at either end, so that of you give it a flick it is equally likely to come to rest at any angle between 0 and pi. (Next problem) We consider the same device as the previous problem, but this time we are interested in the x-coordinate of the needle point - that is, the "shadow," or "projection," of the needle on the horizontal line. (a.) What is the probability density rho of x? (b.) Compute <x>, <x^2>, and sigma (standard deviation) for this distribution.

2. Relevant equations
Uniform distribution:$$\rho(x)=\frac{1}{b-a}$$
1. $$x=r\cos\theta$$
2. $$\frac{d}{du}\cos u=-\sin u du$$
3. $$\csc\arccos\frac{x}{r}=\frac{\left|r\right|}{\sqrt{r^2-x^2}}$$
4. $$\int\frac{dx}{\sqrt{a^2-x^2}}=\arcsin\frac{x}{r}$$
5. $$\left\langle x\right\rangle=\int^\infty_{-\infty}\bar{\psi}x\psi dx$$
6. $$\left\langle x^2\right\rangle=\int^\infty_{-\infty}\bar{\psi}x^2\psi dx$$
7. $$\sigma=\sqrt{\left\langle x^2\right\rangle-\left\langle x\right\rangle^2}$$
8. $$\int\frac{x^2dx}{a+bx^2}=\frac{x}{b}-\frac{a}{b}\int\frac{dx}{a+bx^2}$$
9. $$\int\frac{dx}{a+bx^2}=\frac{1}{2\sqrt{-ab}}\ln\frac{a+x\sqrt{-ab}}{a-x\sqrt{-ab}}$$
10. $$\int\frac{dx}{a+bx^2}=\frac{1}{\sqrt{-ab}}\tanh^{-1}\frac{x\sqrt{-ab}}{a}$$
11. $$\int\frac{x^2dx}{a+bx+cx^2}=\frac{x}{c}-\frac{b}{2c^2}\ln(a+bx+cx^2)+\frac{b^2-2ac}{2c^2}\int\frac{dx}{a+bx+c^2}$$
12. $$\int\frac{dx}{a+bx+c^2}=\frac{1}{\sqrt{-(4ac-b^2)}}\ln\frac{2cx+b-\sqrt{-(4ac-b^2)}}{2cx+b+\sqrt{-(4ac-b^2)}}$$
13. $$\tanh^{-1}=.5\ln\frac{1+x}{1-x}$$
L'Hospital's Rule:$$\lim_{x\rightarrow x_0}\frac{f(x)}{g(x)}=\lim_{x\rightarrow x_0}\frac{f'(x)}{g'(x)}$$

3. The attempt at a solution
I started out with the uniform distribution and found $$\rho(\theta)=\frac{1}{\pi}$$.
Then I found the conversions from polar r and theta to Cartesian x and for polar dtheta to Cartesian dx using 1 and 2 and manipulated them thus:
$$\arccos\frac{x}{r}=\theta$$
$$-\frac{1}{r}\csc\theta dx=d\theta$$
$$-\frac{1}{r}\csc\arccos\frac{x}{r} dx=d\theta$$.
I put that conversion in the integral of rho of theta dtheta for all space and checked the handy work with the help of 3, 4, and the fact that r is a positive real number thus:
$$\int^\pi_0\frac{1}{\pi}d\theta=\int^{-r}_r-\frac{1}{r}\csc\arccos\frac{x}{r} dx=\frac{1}{\pi}\int^r_{-r}\frac{dx}{\sqrt{r^2-x^2}}=1$$
I conclude the following from that exercise:
$$\rho(x)=\frac{1}{\pi\sqrt{r^2-x^2}}$$
I then go to find the expected/average value of x (<x>) and find it to be 0 meters from center. Makes sense as it is a symmetric distribution about x=0.
I then attempt to find <x^2> using 6, 8-13, and L'Hospital's Rule.
$$\int^\infty_{-\infty}\bar{\psi}x^2\psi dx=\int^r_{-r}\frac{1}{\pi\sqrt{r^2-x^2}}x^2\frac{1}{\pi\sqrt{r^2-x^2}}dx=\int^r_{-r}\frac{x^2dx}{\pi^2(r^2-x^2)}$$
The indefinite integral of that looks like or is related by algebra to this:
$$\int\frac{x^2dx}{r^2-x^2}=-\frac{a}{2}\ln\frac{x-r}{x+r}-x+C$$
I inevitably ran into one of two application of L'Hospital's Rule and limits in general:
$$\lim_{x\rightarrow r}r\ln\frac{x-r}{x+r}=r\ln\lim_{x\rightarrow r}\frac{x-r}{x+r}=r\ln\lim_{x\rightarrow r}\frac{1}{1}=0.$$
The other look like it except it is -r, which results in:
$$\ln-1=\pi i$$
I consistently came up with answers that made no sense. The difference between the first, third and fourth answers are if I used logarithm laws to simplify addition or not. And I think I had a few adding errors between the last two. The first answer was the most popular:
$$\left\langle x^2\right\rangle=\left\{-2\frac{r}{\pi},-2r\infty,-2\frac{r}{\pi}+\frac{r i}{2\pi},-2\frac{r}{\pi}+\frac{r i}{\pi}\right\}$$
Negative and complex answers would result in complex or imaginary standard deviations which make no sense either. My calculator tells me that my answers should be real positive numbers. But it can't tell me that for certain. Weird stuff starts happening when you use limits of integration where the points are at infinity or not included in the normal function. I found two more entries for 13 in my math tables that I haven't tried yet, but I don't expect to get better results out of them.

Last edited: Jan 26, 2010
2. Jan 26, 2010

### blkqi

Don't square the distribution!

Equation 6 is specific to 1-d quantum mechanics where dP = |ψ|^2 dx. Yet you're NOT dealing with a wavefunction. You have dP = ρ(x) dx = [π*sqrt(r^2-x^2)]^-1 dx. The expectation value is

<x^2> = ∫ x^2 dP, (all space)

Now take the integral. The integrand is even, so you have symmetry about the y-axis.

(Everything else looks good!)

Last edited: Jan 26, 2010
3. Jan 27, 2010

### Yitzach

That makes more sense, that is a mean trick to pull. And I'm know I'm not the only one who fell for it.
The resulting sigma make more sense, but not complete sense. Rho of x is defined between -r and r.
$$\sigma=\sqrt{r^2\frac{\pi}{2}-0^2}=r\sqrt{\frac{\pi}{2}}\approx1.2533r$$
How can all of the distribution and then some be within one standard deviation of the expected value of x?

4. Jan 27, 2010

### blkqi

Hmmm.. Wolfram Alpha shows me a little different answer.. Did you forget the 1/π?

This was my calculation:
Integral[2*(π*sqrt(r^2-x^2))^(-1)*x^2,{x,0,r}]

http://www.wolframalpha.com/

5. Jan 27, 2010

### Yitzach

I did drop the pi^-1. GIGO. TI-89 Titanium only knows what I tell it. Nicely caught. And that sigma makes complete sense now that it is less than one whole r.