- #1

binbagsss

- 1,265

- 11

## Homework Statement

I have probabilty density ##\rho(\theta,\phi)=\frac{1}{4\pi} sin \theta d\theta d\phi##

The question is let ##\theta## be measured with respect to the z-axis, find the probability ##p_{z}(z) dz ## that the particle lies between ##z## and ##z+dz##?

## Homework Equations

see above

## The Attempt at a Solution

My method:

##z=cos \theta ##

##\frac{d\theta}{dz}= \frac{-1}{sin\theta}##

##\int^{2\pi}_{0} d\phi \int \frac{1}{4\pi} sin \theta dz . \frac{d\theta}{dz} ##

= ##2\pi \frac{-1}{4\pi} = - \frac{1}{2} ##

This is wrong by a minus sign and the solutions instead do:

##P_{z}(z)=...=p(\theta) |\frac{d\theta}{dz}|= 1/2##?

(the '...' being the integration over ##\phi##. )

MY QUESTION:

I don't understand where or why the modulus signs come from in the transformation variables : ##|\frac{d\theta}{dz}|##?

Many thanks