|   |   | 
| Line 5: | Line 5: | 
|  | where <math>\theta</math> is the angle between the molecular symmetry axis and the [[director]]. <math>\overline P_2</math> |  | where <math>\theta</math> is the angle between the molecular symmetry axis and the [[director]]. <math>\overline P_2</math> | 
|  | is the [[Order parameters | uniaxial order parameter]]. |  | is the [[Order parameters | uniaxial order parameter]]. | 
|  | Maier and Saupe Theory
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|  | Aim is to calculate S as function of T.
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|  | Maier and Saupe (1960) anisotropic attraction
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|  | Onsager (1949) anisotropic repulsion
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|  | We will look at MS theory and then consider its strengths and weaknesses.
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|  | (i)
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|  | Simplest attractive interaction between two polarizable rods. Instantaneous dipole interacts with induced dipole.
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|  | ⎟⎠⎞⎜⎝⎛−=21cos23)(),(12212121212ββrurU
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|  | β12
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|  | 2
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|  | 1
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|  | r12
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|  | (ii)
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|  | 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−=⎟⎠⎞⎜⎝⎛−−=∝∝⎟⎠⎞⎜⎝⎛−−∝
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|  | i.e. potential proportional to cos squared of angle
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|  | and order parameter
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|  | and density squared
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|  | n
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|  | βi
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|  | A defines strength of potential. Ignores fluctuations and SRO
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|  | (iv)
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|  | Now calculate orientational distribution function: angle. azimuthal theis and director theand axis longmolecular ebetween th anglepolar theis wheresin where1)(020)()(αβαβββππββddeZeZfiiTkUTkUiBiiBii∫∫−−==
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|  | (v)
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|  | The order parameter can now be calculated using the method outlined in lecture 2. The order parameter is just the average value of )(cos2β
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|  | P. That is)cos(2βPS=.
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|  | In more detail… αβββαββββαββββππππππddTkVASPddPTkVASPddPfSBBsin)(cosexpsin)(cos)(cosexpsin)(cos)(020220220220220∫∫∫∫∫∫⎟⎟⎠⎞⎜⎜⎝⎛⎟⎟⎠⎞⎜⎜⎝⎛==
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|  | 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==−=∴⎟⎟⎠⎞⎜⎜⎝⎛⎟⎟⎠⎞⎜⎜⎝⎛=⎟⎟⎠⎞⎜⎜⎝⎛⎟⎟⎠⎞⎜⎜⎝⎛=∫∫∫∫∫∫−−
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|  | (vi)
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|  | Two simultaneous equations in S and m. Integral may be done numerically and equations solved “graphically”. Slope of straight line is proportional to T.
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|  | The results are:
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|  | Tmax
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|  | 0
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|  | T
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|  | S
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|  | m
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|  | low T
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|  | high T
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|  | 1
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|  | -0.5
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|  | S
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|  | )/(22284.02maxBkVAT×=
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|  | At high T (ie T > Tmax) there is only 1 solution, S = 0
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|  | At low T (ie T < Tmax) there are 3 self consistent solutions. S=0, S>0 and S<0
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|  | (vii)
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|  | Which one has lowest free energy? The one with lowest free energy! Use : Helmholtz free energy = energy – T × entropy Σ
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|  | −=TU
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|  | Recall from Statistical Mechanics:
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|  | 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
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|  | (viii)
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|  | 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)
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|  | (ix)
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|  | 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
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|  | ddmPZβββππsin))(cosexp( where0202∫∫−= Iit turns out that for BkVAT222019.0< the positive S solution has the lowest free energy. Hence BNIT22019.0=
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|  | 1
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|  | -0.5
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|  | T
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|  | 0.43
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|  | 0
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|  | Tmax
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|  | S
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|  | S decreases steadily as T is increased until it suddenly drops to zero at TNI .
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|  | (x)
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|  | TNI is less than TMAX so have first order transition.
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|  | 43.0)(22019.02==NIBNITSkVATA reasonable value compared with experiment.
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|  | 1-K Joules 5.3Joules 5.3=ΔΣ×=ΔNININITU Much weaker than crystal to liquid transition
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|  | Strong angle dependant attraction (large A) increases TNI.
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|  | Dilution (increasing V) decreases TNI.
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|  | Why does it work? We have neglected the shape completely but it seems to give reasonable values.
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|  | 
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|  | 
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|  | ==References== |  | ==References== | 
|  | #W. Maier and A. Saupe "EINE EINFACHE MOLEKULARE THEORIE DES NEMATISCHEN KRISTALLINFLUSSIGEN ZUSTANDES", Zeitschrift für Naturforschung A '''13'''  pp. 564- (1958) |  | #W. Maier and A. Saupe "EINE EINFACHE MOLEKULARE THEORIE DES NEMATISCHEN KRISTALLINFLUSSIGEN ZUSTANDES", Zeitschrift für Naturforschung A '''13'''  pp. 564- (1958) |