Thomsons principle and Rayleighs monotonicity law Martti Meri 1 - PowerPoint PPT Presentation

T-79.7001 Postgraduate Course in Theoretical Computer Science - Spring 2006 Thomson’s principle and Rayleigh’s monotonicity law Martti Meri 1

• Currents minimize energy dissipation • Conservation of energy • Thomson’s principle • Rayleigh’s monotonicity law • A probabilistic explanation of the Monotonicity law • A Markov chain proof of the Monotonicity law 2

Currents minimize energy dissipation i 2 � � E = xy R xy = i xy ( v x − v y ) (1) x,y x,y j a + j b = x j x = y j xy = � � � x 1 x,y ( j x,y − j y,x ) = 0 � 2 x,y ( w x − w y ) j xy = � x ( w x y j xy ) − � y ( w y x j xy ) = � � � w a y j ay + w b y j by − w a x j xa − w b x j xb = � � � � w a j a + w b j b − w a ( − j a ) − w b ( − j b ) = 2( w a − w b ) j a 3

Conservation of energy ( w a − w b ) j a = 1 � ( w x − w y ) j xy (2) 2 x,y v a i a = 1 ( v x − v y ) i xy = 1 i 2 � � xy R x,y (3) 2 2 x,y x,y R eff = v a /i a a R eff = 1 i 2 i 2 � xy R x,y (4) 2 x,y energy dissipated by unit current flow is just R eff 4

Thomson’s principle If i is the unit flow from a to b determined by Kirchhoff’s x,y i 2 xy R xy minimizes Laws, then the energy dissipation � x,y j 2 xy R xy among all unit flows j the energy dissipation � from a to b . d x,y = j x,y − i x,y . d is a flow from a to b with d a = x d a,x = 1 − 1 = 0. � x,y j 2 x,y ( i xy + d x,y ) 2 R xy = xy R xy = � � x,y i 2 x,y d 2 xy R xy + 2 x,y i xy R xy d x,y + xy R xy = � � � x,y i 2 x,y d 2 xy R xy + 2 x,y ( v x − v y ) d x,y + xy R xy setting � � � w = v and j = d and recalling (2.) the middle term is 4( v a − v b ) d a = 0 5

x,y j 2 x,y i 2 x,y d 2 x,y i 2 xy R xy = xy R xy + xy R xy ≥ xy R xy . � � � � 6

Rayleigh’s monotonicity law If the resistances of a circuit are increased, the effective resistance R eff between any two points can only increase. If the are decreased, it can only decrease. i unit current from a to b with the values R x,y and j unit current from a to b with the values R x,y so that R x,y ≥ R x,y . R eff = 1 xy R x,y ≥ 1 x,y j 2 x,y j 2 xy R x,y ≥ � � 2 2 1 x,y i 2 xy R x,y = R eff � 2 In the last inequation Thomson’s principle used. 7

A probabilistic explanation of the Monotonicity law Assume v r > v s and mark u x expected number of times in x and u xy expected number of crossing from x to y . C rs u rs = u r P rs = u r = v r C rs C r C sr u sr = u s P sr = u s = v s C sr C s Since C rs = C sr and recalling assumption we have u rs > u sr 8

Effect on escape probability esc = p esc + ( v r − v s ) d ( ǫ ) where Claim. p ( ǫ ) d ( ǫ ) = u ( ǫ ) C r + ǫ − u ( ǫ ) ǫ ǫ C s + ǫ denotes the expected net number r s of times the walker crosses from r to s Proof. Fortune at x is v x is the probability of returning to a before reaching b in the absence of bridge rs . 1 · (1 − p ( ǫ ) esc ) + 0 · p ( ǫ ) esc = 1 − p ( ǫ ) esc lost amount when stepping away from r. � � � C rx ǫ ǫ v r − C r + ǫ v x + C r + ǫ v s = ( v r − v s ) x C r + ǫ total amount expected to loose ǫ C s + ǫ = p esc + ( v r − v s ) d ( ǫ ) ǫ p esc + ( v r − v s ) C r + ǫ + ( v s − v r ) for small ǫ 9

u ( ǫ ) u ( ǫ ) C s + ǫ = u r C r − u s C s = v r C a − v s C r + ǫ − r s C a meaning that, p ( ǫ ) esc − p esc = ( v r − v s ) 2 ǫ C a monotonicity law p ( ǫ ) esc ≥ p esc ≥ 0 10

A Markov chain proof of the Monotonicity law P is ergodic Markov chain associated with an electric network. ˆ Bridge ǫ from r to s . P ( ǫ ) rs = C rs + ǫ C rs P rs = C r + ǫ C r + ǫ ˆ P ( ǫ ) ǫ rr = 0 P rr = C r + ǫ Q ( ǫ ) = ˆ Q + hk (5) ǫ ǫ � T � h = 0 , . . . , 0 , ( C r + ǫ ) , 0 , . . . 0 , ( − C s + ǫ ) , 0 , . . . , 0 k = (0 , . . . , 0 , − 1 , 0 , . . . , 0 , 1 , 0 , . . . , 0) 11

N ( ǫ ) = ˆ N + c ( ˆ Nh )( k ˆ N ) (6) � ˆ ˆ N i,r ǫ N i,s ǫ N ( ǫ ) i,j = ˆ � ( ˆ N sj − ˆ N i,j ǫ + N rj ) (7) C r + ǫ − C s + ǫ 1 c = (8) ˆ ˆ ˆ ˆ N r,r ǫ N s,r ǫ N s,s ǫ N r,s ǫ 1 + C r + ǫ + C r + ǫ − C s + ǫ − C s + ǫ � B xb = N x,y P y,b (9) y since P ( ǫ ) x,b = ˆ P x,b 12

ˆ ˆ N x,r ǫ N x,s ǫ B ( ǫ ) xb = ˆ C s + ǫ )( ˆ B sb − ˆ B xb + c ( B rb ) (10) C r + ǫ − a,x = ˆ since P ( ǫ ) P a,x u r ǫ ˆ u s ǫ ˆ C s + ǫ )( ˆ B sb − ˆ p ( ǫ ) esc = ˆ p esc + c ( B rb ) (11) C r + ǫ − ˆ ˆ B x,a B x,a u x C x = C a = ˆ ˆ C a esc = p esc + ǫc ( ˆ B sb − ˆ p ( ǫ ) B rb ) 2 (12) C a 13