Bounds of Integration for Random Oriented particle

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In the Stoner-Wohlfarth model, the integration for the average cosine of the angle is limited to the range from 0 to π/2 due to the symmetries of the problem, which allow for the solutions to be extrapolated across the entire parameter space. This approach simplifies calculations while still accurately representing the random orientations of uniaxial, non-interacting particles at low temperatures. The discussion highlights the significance of Fig. 4 in illustrating how these symmetries facilitate the derivation of solutions. Understanding these boundaries is crucial for grasping the behavior of the particle system. The integration limits are thus justified by the inherent symmetries in the model.
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In the Stoner-Wohlfarth model, a uniaxial, non-interacting particle is cooled to very low temperature with no exposure to an external field. Therefore, the orientation of each particle is random, if you have a group of particles. In their paper, they integrate such that:
\langle \cos (\Theta )\rangle =\int_0^{\frac{\pi }{2}} \sin (\Theta ) \cos (\Theta ) \, d\Theta

I am having a hard time understanding why they only integrate from 0 to pi over two, instead of 0 to pi. Can anyone shine any enlightenment on this?
 

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A copy of the paper can be found at http://spin.nanophys.kth.se/spin/stoner-wohlfarth.pdf There's a lot of discussion around Fig. 4 where they talk about how the symmetries of the problem allow them to reproduce the solutions everywhere in parameter space from the region ##0 \leq \theta,\phi \leq \pi/2##.
 
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fzero said:
A copy of the paper can be found at http://spin.nanophys.kth.se/spin/stoner-wohlfarth.pdf There's a lot of discussion around Fig. 4 where they talk about how the symmetries of the problem allow them to reproduce the solutions everywhere in parameter space from the region ##0 \leq \theta,\phi \leq \pi/2##.

That helped me tremendously - thank you!
 
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