Origin of the atttractive part of Lennard-Jones potential
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1 Origin of the atttractive part of Lennard-Jones potential Masahiro Yamamoto Department of Chemistry, Konan University, Kobe, , Japan (Dated: February 7, 06) A. Origin of the attractive part of L-J potential Now we condsider two hydrogen atoms. If the two atoms are located far away, the Hamiltonian H 0 and the wavefunction Ψ of the system is given by two independent H atoms, h m ( + ) e e 4πɛ 0 r 4πɛ 0 r () Ψ ψ (r )ψ (r ) R nl (r )Y lm (ˆr )R nl (r )Y lm (ˆr ) nlm nlm () H 0 The ground state Ψ 0 is given by If the two H atoms get closer, they interact each other and the potential e e + + z r r r FIG. : van der Waals interaction between two hydrogen atom V (perturbation) is given by V e 4πɛ 0 r e 4πɛ 0 r + r e 4πɛ 0 r + r + e 4πɛ 0 r + r + r (3) If we assume r >> r, r, and vector r is in the z direction. r (0, 0, z). If we consider the order up to r (x /r ) because the (/r)(z /r) order terms are cancelled out. If we use ( + x) / x/ + 3x /8... r r (x x ) + (y y ) + (z z ) ] / r z zz + x + y + z r + zz + x + y + z r r + r r ( zz + x + y + z r + 3z r ) r ] / ] / r ( zz + x + y + z r + 3z r ) r r + r r zz + zz + x x x + x + y y y + y + z z z + z r + 3 V e 4πɛ 0 r ( + zz + x + y + z r 3z + zz + x + y + z r 3z + zz + zz + x x x + x + y y y + y + z z z + z r + 3 e 4πɛ 0 r 3 (x x + y y + z z 3z z ) (z z ) r ] z + z z z r ) e 4πɛ 0 r 3 (x x + y y z z ) (4) This is the instantaneous dipole-dipole interaction /(4πɛ 0 r 3 ) µ µ 3( µ ˆr)( µ ˆr)]. Here µ e(r ), µ e(r ). Using perturbation theory the energy change is given by E n E n (0) + () n + () n +... (5) () n n V n V nn (6)
2 () n k n V nk E (0) n E (0) k (7) For ground state n 0, E (0) 0 ɛ 0 (atom) + ɛ 0 (atom) and () e n 0 4πɛ 0 r 3 (x x + y y z z ) 0 e 4πɛ 0 r 3, 0, 0(atom), 0, 0(atom) (x x + y y z z ), 0, 0(atom), 0, 0(atom) e 4πɛ 0 r 3, 0, 0(atom) x, 0, 0(atom), 0, 0(atom) x, 0, 0(atom) (8) In the last equation the integral of the product of even, odd, even functions becomes zero. and the second order () 0 V 0k ɛ 0 (atom) + ɛ 0 (atom) ɛ k (atom) + ɛ k (atom) k 0 e 4 (4πɛ 0 ) r 6 k 0 0 x x + y y z z k k x x + y y z z 0 ɛ 0 (atom) + ɛ 0 (atom) ɛ k (atom) + ɛ k (atom) (9) Using the relation k k k k 0 k k, and we assume that the denominator is constant and ɛ 0 (atom) + ɛ 0 (atom) ɛ k (atom) + ɛ k (atom) (I + I ), here I, I is the ionization energy of atom and. () e 4 0 (4πɛ 0 ) r 6 0 (x x + y y z z ) 0 I + I e 4 (4πɛ 0 ) r 6 0(atom) x I + I 0(atom) 0(atom) x 0(atom) + 0(atom) y 0(atom) 0(atom) y 0(atom) + 4 0(atom) z 0(atom) 0(atom) z 0(atom) ] In the same perturbation theory the atomic polarizability is given by α e k 0 0 z k k z 0 E (0) 0 E (0) k e I 0 z 0 (0) Here the electric field is in z-direction. Then we can get 0 z 0 0 x 0 0 y 0 (αi)/(e ) With this relation () e 4 α I α I 0 (4πɛ 0 ) r 6 I + I e e + α ] I α I e e + 4α I α I e e 3 α α I I (4πɛ 0 ) r 6 () I + I This equation was derived by F. London (Z. Phys. 63, 45-79,930). Since the vdw interaction can not be estimated from usual density functional theory( because we should consider excitation and dynamical dipole-dipole interaction), some semiempirical correction based on Eq.() is reported. (e.g. F. Ortmann et al. PRB, 73, 050, 006) B. LJ potential The final result give the attractive part of the Lennard-Jones potential, φ(r) 4ɛ(σ/r) (σ/r) 6 ]. In the mixed system one can use the following convention(lorentz-berthelot mixing rule) σ ij σ i + σ j () ɛ ij ɛ i ɛ j (3)
3 In the case of σ ij we can understand the rule from the view point of hard-sphere. In the case of ɛ ij we should consider Eq.() α i 4ɛ i σi 6 3 I i (4πɛ 0 ), 4ɛ jσj 6 3 α j (4πɛ 0 ) I j 3 ɛ ij ɛ i ɛ j 3 α i α j Ii I j 6 (4πɛ 0 ) σ 3 i σ3 j (4) From the final equation the approximation of Eq.(3) may be not bad. If one consider the deviation form the standard mixing rule, the modified Lorentz-Berthelot mixing rule can be used. ξ is the deviation parameter. ɛ ij ξ ɛ i ɛ j, σ ij σ i + σ j (5) FIG. : from peattiea/research main.html The Gecko (???? jcan stick to the wall. (both on the hydrophilic and hydrophobic surace and even dead gacko can stick!!) They have billion contacting point per foot and interact via van der Waals force!!. The total adhesive stress from Autumn s force measurements is 576 kpa (5.68 atm). Kellar Autumn, Yiching A. Liang, S. Tonia Hsieh, Wolfgang Zesch, Wai Pang Chan, Thomas W. Lenny, Ronald Fearing and Robert J. Full (000) Adhesive force of a single gecko foot-hair, Nature, 405(6787) , and Proc Natl Acad Sci U S A , 5-6. Evidence for van der Waals adhesion in gecko setae. Autumn K, Sitti M, Liang YA, Peattie AM, Hansen WR, Sponberg S, Kenny TW, Fearing R, Israelachvili JN, Full RJ.] Recently Huber et al. showed that the evidence for capillary force contribution to gecko adhesion from single spatula AFM force measurment. (PNAS, 0, , 005) The pull-off force is increased lineary (from 7 nn to nn) with the humidity for hydrophilic flat surface and the the increase is not so much (from 6 nn to 9 nn) for hydrophobic surface. At the humidity around 90 % the monolayer of water is adsorbed on the hydrophilic surface. C. Kittel s derivation of the attractive part of LJ potential: dynamical effect Now we consider the two neutral atom is separated by R. At a certain moment the atoms induce dipole moment in each other and the induced dipole moments cause an attractive interaction between the atoms. We assume the two atoms are identical linear harmonic oscillators and only electrons move in the x-direction. When the displacement of electrons are x and x, the force constants are K, and the momenta of the oscillators are p and p The hamiltonian of the system without the interaction between the induced dipole moments is H 0 m p + Kx + m p + Kx (6)
4 4 FIG. 3: capillary force x x R FIG. 4: Kittel s explanation on the vdw attractive potential (ISSP book) Here m is the mass of electron and K mω 0. The energy of the ground state is the double of the zero point energ The interaction H between the dipoles becomes H e 4πɛ 0 R }{{} ++ e E 0 hω 0 hω 0 (7) e + 4πɛ 0 (R + x ) 4πɛ 0 (R + x x ) 4πɛ 0 (R x ) }{{}}{{}}{{} + + Here e is the elementary charge and ɛ 0 is the permittivity of free space. If we assume that R x, x. 4πɛ 0 H e R ( x ) R R + x R + x x + (x x ) ] R R R ( + x R R + x ) R (9) R x + (x x ) x ] x x R 3 (0) Here we used the approximation /( ± x) x + (/)x +... The total hamiltonian becomes If we use the symmetric and anti-symmetric coordinate H H 0 + H m p + Kx + m p + Kx x x R 3 () and also use the symmetric and anti-symmetric momenta e (8) x s (x + x ), x a (x x ) () x (x s + x a ), x (x s x a ) (3) p s (p + p ), p a (p p ) (4)
5 p (p s + p a ), p (p s p a ) (5) 5 The total hamiltonian becomes H m p s + ) ] (K e 4πɛ 0 R 3 x s + m p a + ) ] (K + e 4πɛ 0 R 3 x a (6) The two vibrational frequencies are found for this interacting system ) / ( ω (K ± e K e /m] )] / 4πɛ 0 R 3 ± m 4πɛ 0 KR 3 (7) e ω 0 ( ) / ± 4πɛ 0 KR 3 (8) If assume that (e )/(4πɛ 0 KR 3 ), we can use ( ± x) / ± (/)x (/8)x +... ω ω 0 ± ( ) e 4πɛ 0 KR 3 ( ) e 8 4πɛ 0 KR ] (9) The zero point energy change is E hω 0 + ( e 4πɛ 0 KR 3 ( ) e hω 0 8 4πɛ 0 KR 3 ) ( ) e ] 8 4πɛ 0 KR 3 + hω 0 ( ) e 4πɛ 0 KR 3 ( ) e ] 8 4πɛ 0 KR 3 hω 0 hω 0 e 4 3π ɛ 0 K R 6 A R 6 (30) The attractive interaction is proportional to R 6 and A include the dynamics of the oscillator through ω 0
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