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Phys 215A Hw 1, solutions
each problem is 20 points, please format your results as
a weaker proof: ππΊ
ππ₯=
ππΊ
ππ¦=
ππΊ
ππ= 0 βΌ πΊ ππ πππππππ§ππ ππ πππ₯ππππ§ππ
ππΊ
ππ= 0 implies πΉ = 0 , the constraint is satisfied
ππΊ
ππ₯=
ππΊ
ππ¦= 0 implies βπ + π βπΉ = 0 this means, locally at the solution point:
the gradient of field π(π₯, π¦) and πΉ(π₯, π¦) are parallel βπ // βπΉ.
ππ
ππ₯= 0
π(π₯) is minimized
or maximized
Landau Book, Page 35
or the gradient of π(π₯, π¦) is perpendicular to the constraint line (or surface) βπ(π₯, π¦) β₯ { πΉ(π₯, π¦) = 0}
Therefore, for under constraint πΉ(π₯(π ), π¦(π )) = 0 , the derivative is zero ππ(π₯(π ),π¦(π ))
ππ = 0
~ the solution is a local extreme.
Landau Book, 3rd Edition, Page 36, 37:
Examples:
convection This is an example, where the system is not in equilibrium. But you can still maximize the entropy, using some method. And you will see, there is a relative macroscopic motion.
Heat death of the universe? when our universe is in equilibrium no macroscopic motions
Page 43
We can write down the answer using Boltzmann distribution: πΈ(π, π) = πππ
β π
ππ΅π+0
πβ
πππ΅π+1
= ππ
π
πππ΅π+1
The distribution π(πΈ) βΌ πβ
πΈ
ππ΅π can be derived from the principle of maximum entropy.
However, we can also solve this problem, directly using the principle of maximum entropy.
Suppose the total energy of the system is πΈ, π =πΈ
π
Numbers of microscopic states = Ways to put π indistinguishable particles in π boxes
Ξ© = πΆππ
Using the Stirling's approximation
π
ππ΅= log Ξ© β π log π β π log π β (π β π) log(π β π)
Taking the derivative:
π
ππlog Ξ© = β log π + log(π β π) = log
π β π
π= log(
π
πβ 1)
Replacing log Ξ© and π with π and πΈ
π
ππ΅
ππ
ππΈ= log(
ππ
πΈβ 1)
ππ
ππΈ=
1
π
π
ππ΅π= log(
ππ
πΈβ 1)
Finally:
πΈ(π) = ππ
π
πππ΅π + 1
The total energy is maximized in the limit π β β, πΈmax =ππ
2 π(π π‘ππ‘π 1) = π(π π‘ππ‘π 2) = 50%
ππ΅π
π
πΈ
ππ
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