Physics, 30.08.2019 17:30 meth77

When you have two einstein solids (a and b), the probability of a certain state = (\omega_a*\omega_b)/\omega_tot.
can someone explain why we have to multiply \omega _a and \omega _b and why doesn't just one of these two suffice?
for example: n_a=12, n_b=18, q_a=10, q_b=50.
what is the probability of q_a=10?
can't i just calculate \omega _a/\omega_tot, since having \omega _a, automatically means \omega _b is defined (and has no influence on the probability anymore)?

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When you have two einstein solids (a and b), the probability of a certain state = (\omega_a*\omega_b...