- #1

dRic2

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Let's immagine a collision of two point particles in which one can be considered at rest and suppose the scattering process viewed in center of mass frame is isotropic. Then the probability of one particles to be scattered in one direction is ##p(\Omega) = \frac 1 {4 \pi}##. Let us suppose then the process is also symmetric w.r.t. the polar angle ##\phi##. Then

$$p(\Omega) d \Omega = \frac 1 {4 \pi} \sin \theta d\theta 2 \pi = \frac 1 2 \sin \theta d \theta = p( \theta) d\theta$$

so

$$p( \theta) = \frac 1 2 sin \theta$$

This probability doesn't seem so "isotropic" to me tough. In fact according to this result the particle is more likely to be scattered in an orthogonal direction. But isotropic means that every direction has the same probability to be "chosen".

What did I got wrong ?