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12 Boundary conditions in multipole techniques Ivo Severens May - PowerPoint PPT Presentation

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12 Boundary conditions in multipole techniques Ivo Severens May 7, 2002 /k 12 1. Introduction Molecular dynamics: follow the trajectories of N particles by Newtons sec- ond law: d 2 x i m i dt 2 = i , i = 1 , ..., N. /k 12 2.

  1. 12 Boundary conditions in multipole techniques Ivo Severens May 7, 2002 /k

  2. 12 1. Introduction Molecular dynamics: follow the trajectories of N particles by Newton’s sec- ond law: d 2 x i m i dt 2 = −∇ Φ i , i = 1 , ..., N. /k

  3. 12 2. Multipole method Fast multipole method N charged particles lead to N 2 pairwise interactions from Coulomb’s law, i.e. q j Φ ij = 4 πε 0 � x i − x j � , i, j = 1 , ..., N. Taylor series multipole expansions ∞ c n = f ( n ) ( a ) � c n ( z − a ) n , f ( z ) = , n ! n =0 in particular ∞ 1 | z − z 0 | = 1 1 | 1 − z 0 /z | = 1 1 � z 0 � n � � x − x 0 � = | z | · | z || | , z n =0 lead to an algorithm requiring an amount of work proportional to N to eval- uate all interactions. /k

  4. 12 Multipole method in physics Consider the electrostatic problem: ∇ 2 V ( x ) = − ρ ( x ) = 1 � ρ ( y ) ( − δ ( x − y )) dτ y . ε 0 ε 0 Its formal solution is: 1 � ρ ( y ) V ( x ) = � x − y � dτ y . 4 πε 0 Since � x − y � 2 = � x � 2 + � y � 2 − 2 ( x , y ) = � x � 2 + � y � 2 − 2 � x �� y � cos θ, we have that ∞ � n 1 1 1 1 � � y � � � x − y � = � 2 � 1 / 2 = P n (cos θ ) . � x � � x � � x � � � y � � y � � n =0 1 − 2 � x � cos θ + � x � /k

  5. 12 This leads to � ρ ( y ) ∞ � n 1 � � y � � V ( x ) = P n (cos θ ) dτ y 4 πε 0 � x � � x � n =0 � x �� � � 1 ρ ( y ) dτ y + 1 � = � x � , ρ ( y ) y dτ y + ... 4 πε 0 � x � � x � � x � � � 1 Q + 1 = � x � , p + ... , 4 πε 0 � x � | x � Q : total charge, p : electric dipole moment. Conclusion In physics, the multipole method is an integrated version of the fast multi- pole method. /k

  6. 12 3. Conducting agglomerates Theory Based on "Polarizability of conducting and dielectric agglomerates: theory and experiment" by R.C. Brown and M.A. Hemingway in Journal of Electrostatics 53 (2001) 235-254. Consider a conducting agglomerate in an external electric field E 0 : The response of a conductor to an external electric field will be such that the /k entire agglomerate has a constant electric potential.

  7. 12 q j V ijq = , E ijq = −∇ V ijq , 4 πε 0 r ij � � p j , r ij V ijp = , E ijp = −∇ V ijp . 4 πε 0 r 3 ij This yields � � V i = ( V ijq + V ijp − E 0 · r i ) , E i = ( −∇ V ijq − ∇ V ijp ) + E 0 . j � = i j � = i Furthermore, the dipole moment p , developed by a conducting sphere of radius R in a uniform electric field E 0 is given by p = 4 πε 0 R 3 E 0 . Imposed conditions: � p i = 4 πε 0 R 3 V i = V, i E i , q i = 0 . i Unknown: q i , p i , V , i.e. 4N+1. Linear equations: 4N+1. /k

  8. 12 An example In this case: E 0 = − 40 / (10 R ) e y and r 12 = r 21 = 2 R . This gives q 2 q 1 V 1 = 16 + 8 πε 0 R = V, V 2 = 24 + 8 πε 0 R = V, q 1 + q 2 = 0 , with solution q 1 = − 32 πε 0 R, q 2 = +32 πε 0 R, V = 20 . /k

  9. 12 Corrections 1 � V new • Replace the condition V i = V by dσ = V , where i 4 πR 2 ∂B ( x i ,R i ) i V new = V i + V iiq + V iip . i • Use higher order terms (quadrupole, octapole, et cetera) in the multipole expansion to model the conducting contact better. What happens at the boundary � x � = R of a conducting sphere ? ∞ � n 1 � � y � � � V ( x ) = σδ ( � y � − R ) P n (cos θ ) dτ 4 πε 0 � x � R n =0 � π ∞ σ � 2 πR 2 = P n (cos θ ) sin θdθ 4 πε 0 R 0 n =0 � 1 ∞ = σR � P n ( z ) · 1 dz 2 ε 0 − 1 n =0 ∞ = σR 2 n + 1 δ n 0 = σR 2 � /k 2 ε 0 ε 0 n =0

  10. 12 Boundary conditions First a simple case: 1 point charge q 1 opposite to a grounded plate. Then: q 1 V ( x ) � = V q 1 ( x ) = 4 πε 0 � x − x 1 � , q 1 � 1 1 � V ( x ) = V q 1 ( x ) + V q 2 ( x ) = � x − x 1 � − . /k 4 πε 0 � x − x 2 �

  11. 12 4. Image charges Consider a point charge between two grounded plates. Positive charges: − d, 2 h − d, − 2 h − d, 4 h − d, − 4 h − d, ... , − d + 2 kh , k ∈ Z . Negative charges: d, − 2 h + d, 2 h + d, − 4 h + d, 4 h + d, ... , d + 2 kh , k ∈ Z . /k

  12. 12 This leads to 1 � 1 1 � � V ( x ) = � x − ( − d + 2 kh ) e y � − . 4 πε 0 � x − ( d + 2 kh ) e y � k ∈ Z In reality: This leads to millions of calculations. /k

  13. 12 5. Conclusions • Multipole methods are often used by physicists and engineers, and give general insight into the problem. • For a multipole expansion the correct fundamental solution (function of Green) is required. • The potential cannot in general be calculated from merely the point charges. • The calculation of image charges converges very slowly for real life prob- lems. This makes the Fast Multipole Method very slow. /k

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