# User talk:86.160.170.7

Dear 86.160.170.7 I have moved your contribution to the Maier-Saupe mean field model page here because, although it looks interesting, it is currently not of a presentable standard for an article contribution. It would be wonderful if you could work a little to tidy and spell check your contribution and then re-insert it on the Maier-Saupe page. SklogWiki uses LaTeX math markup for equations. If you are unfamiliar with LaTeX please let me know and I will be more than happy to lend a hand. All the best -- **Carl McBride** (talk) 14:47, 9 April 2009 (CEST)

Maier and Saupe Theory
Aim is to calculate S as function of T.
Maier and Saupe (1960) anisotropic attraction
Onsager (1949) anisotropic repulsion
We will look at MS theory and then consider its strengths and weaknesses.
(i)
Simplest attractive interaction between two polarizable rods. Instantaneous dipole interacts with induced dipole.
⎟⎠⎞⎜⎝⎛−=21cos23)(),(12212121212ββrurU
β12
2
1
r12
(ii)
Too difficult to consider interaction of every molecule with every other molecule so we construct an average potential energy function that one molecule feels due to immersion in a sea of other similar molecules. Mean field approximation. )(cos)(21cos23)(1)()(21cos23)(222222iiiiiiiiPVASUVASUVUSUUββββββββ−=⎟⎠⎞⎜⎝⎛−−=∝∝⎟⎠⎞⎜⎝⎛−−∝
i.e. potential proportional to cos squared of angle
and order parameter
and density squared
n
βi
A defines strength of potential. Ignores fluctuations and SRO
(iv)
Now calculate orientational distribution function: angle. azimuthal theis and director theand axis longmolecular ebetween th anglepolar theis wheresin where1)(020)()(αβαβββππββddeZeZfiiTkUTkUiBiiBii∫∫−−==
(v)
The order parameter can now be calculated using the method outlined in lecture 2. The order parameter is just the average value of )(cos2β
P. That is)cos(2βPS=.
In more detail… αβββαββββαββββππππππddTkVASPddPTkVASPddPfSBBsin)(cosexpsin)(cos)(cosexpsin)(cos)(020220220220220∫∫∫∫∫∫⎟⎟⎠⎞⎜⎜⎝⎛⎟⎟⎠⎞⎜⎜⎝⎛==
This is just an equation with S on both sides. It is tricky to solve because S appears within an integral but solutions can be found using the following method. ()()ATkmVSTkVASmdxmxdxxmxSdxTkVxASPdxxPTkVxASPSdxTkVxASPdxxPTkVxASPSBBBBBB221022102102221022112221122or where21expexp23)(exp)()(exp)(exp)()(exp==−=∴⎟⎟⎠⎞⎜⎜⎝⎛⎟⎟⎠⎞⎜⎜⎝⎛=⎟⎟⎠⎞⎜⎜⎝⎛⎟⎟⎠⎞⎜⎜⎝⎛=∫∫∫∫∫∫−−
(vi)
Two simultaneous equations in S and m. Integral may be done numerically and equations solved “graphically”. Slope of straight line is proportional to T.
The results are:
Tmax
0
T
S
m
low T
high T
1
-0.5
S
)/(22284.02maxBkVAT×=
At high T (ie T > Tmax) there is only 1 solution, S = 0
At low T (ie T < Tmax) there are 3 self consistent solutions. S=0, S>0 and S<0
(vii)
Which one has lowest free energy? The one with lowest free energy! Use : Helmholtz free energy = energy – T × entropy Σ
−=TU
Recall from Statistical Mechanics:
Probability of the system being in a state with energy Er : 1 and ==Σ−rrTkErPZePBr mean energy of system, UrrrEPΣ= entropy of system, rrrBPPklnΣ−=Σ
(viii)
Average energy of a molecule : )(cosS using and function,on distributi over the averagean represent where)(cos)(cos)(2222222ββββPVASPVASPVASUUii=−=−=−== Energy of phase of N molecules: 2221VASNU−= (Note the half)
(ix)
Entropy of a molecule is –kB times average of ln(distribution): ZkTUfkBiiBiln)(ln+=−=Σβ from (iv) Entropy of N average molecules: ⎟⎟⎠⎞⎜⎜⎝⎛−=−=Σ−=+−=+=Σ=ΣZTkVASNZTNkVASNTUFZNkTVASNZNkTUNNBBBBiiln21ln21lnln222222 Unfortunately, this too must be evaluated numerically: For each value of S find m then calculate Z
ddmPZβββππsin))(cosexp( where0202∫∫−= Iit turns out that for BkVAT222019.0< the positive S solution has the lowest free energy. Hence BNIT22019.0=
1
-0.5
T
0.43
0
Tmax
S
S decreases steadily as T is increased until it suddenly drops to zero at TNI .
(x)
TNI is less than TMAX so have first order transition.
43.0)(22019.02==NIBNITSkVATA reasonable value compared with experiment.
1-K Joules 5.3Joules 5.3=ΔΣ×=ΔNININITU Much weaker than crystal to liquid transition
Strong angle dependant attraction (large A) increases TNI.
Dilution (increasing V) decreases TNI.
Why does it work? We have neglected the shape completely but it seems to give reasonable values.

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