Homework Help: A problem on calculus in Griffiths' book

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1. Sep 27, 2016

Tspirit

I can't understand the solution to Problem 1.4(a). The solution is the following:

What puzzles me is that ρ(θ)dθ=ρ(x)dx ? Why are they equal?

2. Sep 27, 2016

vanhees71

Obviously Griffiths in problem 1.4 has not given a precise definition. So the only way is to guess from the solution (I don't like his QM textbook so much, because of all these little imprecisions, although I've not read the book myself but extrapolate this from the questions here in the forum). He takes usual polar coordinates for the speedometer with the angle restricted to $\theta \in [0,\pi]$. Then the $x$ coordinate of the needle hand is
$$x=r \cos \theta.$$
The probability distribution for the angle is uniform, i.e.,
$$P(\theta)=\frac{1}{\pi}.$$
Now he wants the probability for the projection $x$, which is in the range $x \in [-r,r]$. You have to remember that this is a probability distribution, i.e., you must have
$$|\mathrm{d} \theta| P(\theta) = |\mathrm{d} x| \tilde{P}(x).$$
Now you have
$$|\mathrm{d} x|=r \sin \theta=\sqrt{r^2-x^2}.$$
Thus we have
$$\tilde{P}(x) = \left |\frac{\mathrm{d} \theta}{\mathrm{d} x} \right| P(\theta)=\frac{1}{\pi \sqrt{r^2-x^2}}.$$

3. Sep 27, 2016

Tspirit

Why should a probability distribution satisfy |dθ|P(θ)=|dx|~P(x) ? That's really my question. Thank you!

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4. Sep 27, 2016

PeroK

What he means is that if you take a small interval (a physical interval), then there is a definite probability that the needle lies in that interval, independent of coordinates.

If you express this probability in polar coordinates you get $\rho (\theta) d\theta$ and if you express this probability in cartesian coordinates you get $\rho (x) dx$ and, therefore, they must be equal.

What you must be careful of is that when you set up your problem, $d \theta$ and $dx$ do indeed cover the same physical interval.

5. Sep 27, 2016

vanhees71

Yes, and it must be a one-to-one-mapping (more precisely a diffeomorphism).

PS: I'm sorry that I gave the complete answer, but when I answered, this was posted to the QT forum not the homework forum!

6. Sep 27, 2016

TSny

It helps me to think of it this way (similar to PeroK's reply). The only way for the shadow to be between x and x + dx is for the pointer to be between the corresponding θ and θ + dθ as shown below (and vice versa). So, the probability that the shadow is between x and x + dx equals the probability that the pointer is between the corresponding values θ and θ + dθ. The values of dθ and dx that correspond to each other are given by dx = -r sinθ dθ

7. Sep 27, 2016

Ray Vickson

Part of the problem is Griffith's sloppy use of the same symbol $\rho$ to stand for two different things in the same problem.

To understand what is happening, start by avoiding such symbols at the start. The basic principle is that if a single event has a single probability value. If you represent the same event in two different ways (using two different variables) then the "manner" of expressing the probability in the two variables must lead to the same numerical answer. We have $x = r \cos(\theta)$ and $x + \Delta x = r \cos(\theta + \Delta \theta)$, where $r > 0$ is constant. To first order in small increments, we have $\cos(\theta + \Delta \theta) = \cos(\theta) - \sin(\theta) \Delta \theta$, so the events $(x \to x + \Delta x)$ in $x$-space and $( r \cos(\theta) \to r \cos(\theta) - r\sin(\theta) \Delta \theta )$ in $\theta$-space are the same event, represented in two different ways. But, being the same event they have the same probability, so the probability that the $x$-projection lies between $x$ and $x + \Delta x$ is the probability that the angle lies between $\theta$ and $\theta + \Delta \theta$. Note that if $\Delta x > 0$ we have $\Delta \theta < 0$, so the length of the small $\theta$-interval is $|\Delta \theta|$, not $\Delta \theta$. This means that the probability we seek is $\frac{1}{\pi}|\Delta \theta|$. We have
$$\Delta x = - r \sin(\theta) \Delta \theta = - y \Delta \theta = -\sqrt{r^2-x^2} \Delta \theta,$$
so that
$$|\Delta \theta| = \frac{1}{\sqrt{r^2-x^2}} \Delta x.$$

Thus, the probability that the x-projection lies between $x$ and $x + \Delta x$ is
$$P ( x \to x + \Delta x) = \frac{1}{\pi \sqrt{r^2-x^2}} \Delta x$$.
That means that the so-called probability density of $x$ is
$$f(x) = \frac{1}{\pi \sqrt{r^2-x^2}}$$

In a nutshell, $f(x) |dx|$ and $\rho(\theta) |d \theta|$ are equal because they are just two different representations of the same probability.

8. Sep 27, 2016

Tspirit

This explanation really helps me understanding the physical meaning of that, thank a lot!