Phys 215A Hw 1, solutions

each problem is 20 points, please format your results as

a weaker proof: 𝜕𝐺

𝜕𝑥=

𝜕𝐺

𝜕𝑦=

𝜕𝐺

𝜕𝜆= 0 ∼ 𝐺 𝑖𝑠 𝑚𝑖𝑛𝑖𝑚𝑖𝑧𝑒𝑑 𝑜𝑟 𝑚𝑎𝑥𝑖𝑚𝑖𝑧𝑒𝑑

𝜕𝐺

𝜕𝜆= 0 implies 𝐹 = 0 , the constraint is satisfied

𝜕𝐺

𝜕𝑥=

𝜕𝐺

𝜕𝑦= 0 implies ∇𝑓 + 𝜆 ∇𝐹 = 0 this means, locally at the solution point:

the gradient of field 𝑓(𝑥, 𝑦) and 𝐹(𝑥, 𝑦) are parallel ∇𝑓 // ∇𝐹.

𝑑𝑔

𝑑𝑥= 0

𝑔(𝑥) is minimized

or maximized

Landau Book, Page 35

or the gradient of 𝑓(𝑥, 𝑦) is perpendicular to the constraint line (or surface) ∇𝑓(𝑥, 𝑦) ⊥ { 𝐹(𝑥, 𝑦) = 0}

Therefore, for under constraint 𝐹(𝑥(𝑠), 𝑦(𝑠)) = 0 , the derivative is zero 𝑑𝑓(𝑥(𝑠),𝑦(𝑠))

𝑑𝑠= 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: 𝐸(𝜖, 𝑇) = 𝑁𝜖𝑒

− 𝜖

𝑘𝐵𝑇+0

𝑒−

𝜖𝑘𝐵𝑇+1

= 𝑁𝜖

𝑒

𝜖𝑘𝐵𝑇+1

The distribution 𝑝(𝐸) ∼ 𝑒−

𝐸

𝑘𝐵𝑇 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 𝐸, 𝑀 =𝐸

𝜖

Numbers of microscopic states = Ways to put 𝑀 indistinguishable particles in 𝑁 boxes

Ω = 𝐶𝑁𝑀

Using the Stirling's approximation

𝑆

𝑘𝐵= log Ω ≈ 𝑁 log 𝑁 − 𝑀 log 𝑀 − (𝑁 − 𝑀) log(𝑁 − 𝑀)

Taking the derivative:

𝑑

𝑑𝑀log Ω = − log 𝑀 + log(𝑁 − 𝑀) = log

𝑁 − 𝑀

𝑀= log(

𝑁

𝑀− 1)

Replacing log Ω and 𝑀 with 𝑆 and 𝐸

𝜖

𝑘𝐵

𝑑𝑆

𝑑𝐸= log(

𝜖𝑁

𝐸− 1)

𝑑𝑆

𝑑𝐸=

1

𝑇

𝜖

𝑘𝐵𝑇= log(

𝜖𝑁

𝐸− 1)

Finally:

𝐸(𝑇) = 𝑁𝜖

𝑒

𝜖𝑘𝐵𝑇 + 1

The total energy is maximized in the limit 𝑇 → ∞, 𝐸max =𝑁𝜖

2 𝑃(𝑠𝑡𝑎𝑡𝑒 1) = 𝑃(𝑠𝑡𝑎𝑡𝑒 2) = 50%

𝑘𝐵𝑇

𝜖

𝐸

𝜖𝑁