1. The problem statement, all variables and given/known data Consider a polycrystalline sample of Subscript[CaSO, 4]\[CenterDot]2 Subscript[H, 2]O in an external magnetic field Overscript[B, \[RightVector]] in the z direction. The internal magnetic field (in the z direction) produced at the position of a given proton in the Subscript[H, 2]O molecule by the neighboring proton is given by ∂/a^3 ( 3 cos^2 \θ -1 ) if the spin of this neighboring proton points along the applied field. It is given by -(∂/a^3)( 3 cos^2 θ-1 ) if this neighboring spin points in a direction opposite to the applied field. Here, \[Mu] is the magnetic moment of the proton and a is the distance between the two protons, while \[Theta] denotes the angle between the line joining the two protons and the z axis. In this sample of randomly-oriented crystals, the neighboring proton is equally likely to be located anywhere on the sphere of radius a surrounding the given proton. 2. Relevant equations da=a^2sinθdθdδ b=\[∂]/a^3 ( 3 cos^2θ -1 ) 3. The attempt at a solution da=a^2sinθdθdδ The probability of finding p\[Theta]d(\[Theta]) is p(θ)dθ=[a^2sinθdθ∫dδ/(4∏a^2) The itnregral being from 0 to 2∏ The integral deserves some explanation. As the external proton orbits around the molecule the electric field will be constant throught it because the external magnetic field is everywhere the same around 2Pi simplifying gives us pθdθ=sinθdθ/2 We are looking for w(b)db as our result so we write down the equation. The graph showing this distribution is given on a seperate page. We are uncertain about the probability of locating the external proton at radius a and angle θ. We do know that θ andθ+dθ lie somehwere withing b and b+db. Adding up all the values of b that have θ through θ+dθ is given by w(b)db=2pθdθ*absvalue(db/dθ) We look at the plot of b vs θ to see that db intersects \[Theta] twice. Substituting P(θ)d\θ] =sinθdθ/2 into w(b)db and we get the following. w(b)db=2sinθd/2*absvalue(db/dθ) w(b)db=sinθdθ*absvalue(db/dθ) Now we need to find db. Start by writing what we know about b. b=\[Mu]/a^3 ( 3 cos^2 θ -1 ) To find db we take the derivative with respect to θ. d(b/dθ) =-6a^-3*∂(cosθsinθ) substitute this into w(b)db w(b)db=6sinθdθ*a^-3∂(cosθsinθ) We dropped the minus sign because we take the absolute value of db/dθ. Simply further w(b)db= sinθ^2cosθ dθ 6a^-3 The answer key has this in terms of area, which is something I am having difficulty with. Reif solves the area as A=1/2sqrt(a^3/3∂) I start out with da=a^2sinθdθdδ I take the double integral of both sides ∫∫da=a^2∫∫sinθdθdδ A^2=a^2sinθdθδ A=(2a^2sinθdθδ)^1/2 Substitute in pθdθ=sinθdθ/2 A=(4a^2Pθdθδ) This is where I get stuck.