Wednesday, December 26, 2018

Solutions To Statistical Mechanics Problems (1)

1) Quantify the magnetic energy for the system at time t(o) compared to time t(o) + t, if the magnetic energy of one spin magnet can be written:

M = - m B cos Θ 

where m  is the magnetic moment (-eL/2m, L = 1) and assume Θ = +/- π, and B = 0.1T. 


For the elementary spin magnets in the system, the orientation of the up spins is such that  Θ =  90 deg, so   cos Θ  =  1.  Given the orientation we may use for the elementary spin magnetic moment:  m s z  =   L  (given the spin is aligned with z-axis)

where:     u B = 9.27 x 10-24 J/T  is the Bohr magneton

Then the magnetic energy difference for the spin states at   t(o) compared with  t(o) + t, can be written:

D M =  12 m s z B cos Θ    =   {12  ( 9.27 x 10-24 J/T)  (0.1T)}  =  1.1  x 10 -23 J

2)  Say that S = log (g) determines the entropy for a  simple statistical mechanical system, where g denotes the number of accessible states. Then estimate S for the 2D ice crystal model - including any errors that might enter.


In the Ising model case applied to the ice crystals, the energy is the sum of interaction of all adjacent squares. The contribution  of each pair of neighbors depends only on their colors. Since the total number of white squares is fixed, it is proportional to the total length of the contours separating white from blue.    Thus:

Energy = 2 × Length of contours

The key to successful solution here is to note that white squares save energy by clumping, but low energy does not automatically imply high probability (i.e. associated with the highest entropy state).  Then, as an approximation, we take the clumped contours (if readers have performed the contour-drawing exercise) to be higher order (analogous to the spin ups in the ferromagnetic system) and non-clumped "singletons" to be lower order,analogous to spin downs.

From the contour sketch there are 4 clumped states, and 5 non-clumped or singleton (lower order) states. Thus, there are 5 accessible (higher entropy) states for the system.


S =  log (g)  =  log (5)  = 1.6

Approximately. (Remember, 'log' here refers to base e log, not base 10!)

Obviously, errors could enter if the contours (yielding clumped, non-clumped crystals)   are incorrectly drawn.

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