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    (**** i apologise I can not seem to edit the title, it is not a geometric distribution ! )

    q13)

    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?

    2. Relevant equations
    see above

    3. 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
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    (Original post by xfootiecrazeesarax)
    (**** i apologise I can not seem to edit the title, it is not a geometric distribution ! )

    q13)

    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?

    2. Relevant equations
    see above

    3. 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
    I'm confused as to what exactly your working is, or what is the working of the given solution. Can you put up a picture or your full working and the solution that you have.

    If I had to guess, I'd say that they are taking the modulus of a 1D Jacobian i.e. they are trying to treat a 1D case consistently with a 2D case, or something along those lines.
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    (Original post by xfootiecrazeesarax)
    ..
    If z = cos \theta, then your limits are going to be reversed after the substitution, creating an additional minus sign.

    e.g. \int_0^{\pi/2} \sin \theta d\theta, setting z = \cos \theta gives the integral

    -\int_1^0 \,dz, which is the same as \int_0^1 \,dz = 1.

    (atsruser is also correct that in general you want to use the modulus of the Jacobean)
 
 
 
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