1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A problem on calculus in Griffiths' book

  1. Sep 27, 2016 #1
    upload_2016-9-27_22-5-42.png
    I can't understand the solution to Problem 1.4(a). The solution is the following:
    upload_2016-9-27_22-10-18.png
    What puzzles me is that ρ(θ)dθ=ρ(x)dx ? Why are they equal?
     
  2. jcsd
  3. Sep 27, 2016 #2

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    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
    $$\tilde{P}(x) = \left |\frac{\mathrm{d} \theta}{\mathrm{d} x} \right| P(\theta)=\frac{1}{\pi \sqrt{r^2-x^2}}.$$
     
  4. Sep 27, 2016 #3
    Why should a probability distribution satisfy |dθ|P(θ)=|dx|~P(x) ? That's really my question. Thank you!
     

    Attached Files:

  5. Sep 27, 2016 #4

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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.
     
  6. Sep 27, 2016 #5

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    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!
     
  7. Sep 27, 2016 #6

    TSny

    User Avatar
    Homework Helper
    Gold Member

    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
    upload_2016-9-27_11-17-25.png
     
  8. Sep 27, 2016 #7

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
  9. Sep 27, 2016 #8
    This explanation really helps me understanding the physical meaning of that, thank a lot!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: A problem on calculus in Griffiths' book
  1. Griffiths Problem 5.38 (Replies: 6)

  2. Griffiths Problem 5.41 (Replies: 4)

Loading...