1000 Solved Problems in Modern Physics

(Tina Meador) #1

80 1 Mathematical Physics


N!→


2 πNNNe−N

x!→


2 πxxxe−x

(N−x)!→


2 π(N−x)(N−x)N−xe−N+x

B(x)→f(x)=


N

2 πx(N−x)

pxqN−xNN
xx(N−x)N−x

=


N

2 πx(N−x)

(

Np
x

)x(
Nq
(N−x)

)N−x

Letδdenote the deviation ofxfrom the expected valueNp, that is

δ=x−Np

thenN−x=N−Np−δ=Nq−δ

f(x)=

[

2 πNpq

(

1 +

δ
Np

)(

1 −

δ
Nq

)]− 1 / 2

(

1 +

δ
Np

)−(Np+δ)
·

(

1 −

δ
Nq

)−(Nq−δ)

Assume that|δ|Npqso that




δ
Np




∣1 and





δ
Nq




∣^1

The first bracket reduces to (2πNpq)−

(^12)


. Take logeon both sides.


ln

[

f(x)(2πNpq)

12 ]

=−(Np+δ)ln

(

1 +

δ
Np

)

−(Nq−δ)ln

(

1 −

δ
Nq

)

=−(Np+δ)

(

δ
Np


δ^2
2 N^2 p^2

+

δ^3
3 N^3 p^3

−···

)

−(Nq−δ)

(


δ
Nq


δ^2
2 N^2 q^2


δ^3
3 N^3 q^3

−···

)

=−

δ^2
2 Npq


δ^3 (p^2 −q^2 )
6 N^2 p^2 q^2

Neglect theδ^3 term and puttingσ=


Npqandσ=x−Np=x ̄.

f(x)=

1

σ


2 π

exp

[

−(x− ̄x)^2
2 σ^2

]

(Normal or Gaussian distribution)
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