Probability that a magnetic dipole is oriented with theta

  • #1
1.
the problem goes like this :
The energy of interaction of a classical magnetic dipole with the magnetic field B is given by
E = −μ·B.
The sum over microstates becomes an integral over all directions of μ. The direction of μ
in three dimensions is given by the angles θ and φ of a spherical coordinate system
The integral is over the solid angle element
= sin θdθdφ. In this coordinate system
the energy of the dipole is given by E = −μB cos θ.
Choose spherical coordinates and show that the probability p(θ, φ)dθdφ that the dipole is
between the angles θ and + dθ and φ and φ + dφ is given by

p(θ, φ)dθdφ = (e^(μB cos θ) sin(θ) dθ dφ)/z



2. Homework Equations


z = ∫∫ e^(μB cos θ) sin(θ) dθ dφ .


The Attempt at a Solution


i have no idea what to do , and i tried all i know
i know that the boltzmann distribution gives you the probability that a particle has an energy is :
e^([/B]μB cos θ)/∫e^(μB cos θ) , but how do i integrate the spherical coordinates i don t know . please help me and thank you .
 
Last edited:

Answers and Replies

  • #2
this is the best that i could do
the probability that the dipole between x and dx is
dp(x) = (1 / Z ) e^(μ(x)B cos θ) dx = dx because we assume that B is parallel to z so μ(x) . B = 0
dp(y) =
(1 / Z ) e^(μ(y)B cos θ) dy = dy
dp(z) = (1 / Z )e^(μ(z)B cos θ) dz
d^3 p( x,y,z)= (1 / Z )e^(μ(z)B cos ) dz dy dx = (1 / Z )e^(μ(z)B cos θ) dv
d^3 p( r,
θ,)
= (1 / Z )e^(μ(z)B cos θ) rd²r dθ dφ

is this true ? or am i making horrible mistakes ?
 

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