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Derivatives of Newtonian gravity.

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Higher order derivatives for N-body simulations Keigo Nitadori July 3, 2014 1 Derivatives of power functions Consider y = xn , (1) with time dependent x and y, and constant n. From lny =n lnx, (2) y y =n x x , (3) we have 0 =n xy x y, 0 =n xy + (n 1) x y x y, 0 =n ... xy + (2n 1) x y + (n 2) x y x ... y , 0 =n .... x y + (3n 1) ... x y + (3n 3) x y + (n 3) x ... y x .... y . (4) Here, [ ] = d dt [ ] and we also note [ ](n) = dn dtn [ ]. From these, higher order derivatives of y are recursively available in y =[n xy]/x, y =[n xy + (n 1) x y]/x ... y =[n ... xy + (2n 1) x y + (n 2) x y]/x, .... y =[n .... x y + (3n 1) ... x y + (3n 3) x y + (n 3) x ... y ]/x, (5) and so on. A general form for k 1 is y(k) = 1 x k1 i=0 Ck,ix(ki) y(i) , (6) 1 with coecients Ck,0 =n Ck,k = 1 Ck,i =Ck1,i1 + Ck1,i = k 1 i n k 1 i 1 (1 i k 1). (7) 2 Derivatives of gravitational force Now we focus on the time derivatives of the gravitational force, f = m r r 3 . (8) Let s = (r r), and q = s3/2 , (9) then f =mqr, f =mq r + q q r , f =mq r + 2 q q r + q q r , ... f =mq ... r + 3 q q r + 3 q q r + ... q q r , (10) and f (k) =m k i=0 k i q(i) r(ki) =mq k i=0 k i q(i) q r(ki) . (11) For the derivatives of q, by using n = 3 2, we have q =s3/2 , q q = 3 s s 2 , q q = 3 s s 2 + 5 3 s 2 q q ... q q = 3 s ... s 2 + 8 3 s 2 q q + 7 3 s 2 q q . (12) The coecients multiplied by 6 are: 2 @ @ @k i 0 1 2 3 4 5 1 3 (2) 2 3 5 (2) 3 3 8 7 (2) 4 3 11 15 9 (2) 5 3 14 26 24 11 (2) Derivatives of s have simple form in, s =(r r), s =2(r r), s =2(r r) + 2( r r), ... s =2(r ... r ) + 6( r r), s(4) =2(r r(4) ) + 8( r ... r ) + 6( r r), s(5) =2(r r(5) ) + 10( r r(4) ) + 20( r ... r ), (13) and a general form is s(k) = k i=0 k i (r(i) r(ki) ) = 2 (k1)/2 i=0 k i (r(i) r(ki) ) (k is odd) 2 (k2)/2 i=0 k i (r(i) r(ki) ) + k k/2 (r(k/2) r(k/2) ) (k is even) . (14) 3 Another approach Le Guyader (1993) took a slightly dierent approach. For the derivatives of r = r , r2 = (r r), (15) r r = (r r), (16) and after (k 1) times dierentiations, k1 i=0 k 1 i r(i) r(ki) = k1 i=0 k 1 i (r(i) r(ki) ). (17) Finally, we have r(k) = 1 r (r r(k) ) + k1 i=1 k 1 i (r(i) r(ki) ) r(i) r(ki) , (18) 3 for k 2. For the derivatives of q = r3, from r q = 3rq, (19) after dierentiating (k 1) times, k1 i=0 k 1 i r(i) q(ki) = 3 k1 i=0 k 1 i r(ki) q(i) . (20) Thus, q(k) = 1 r 3r(k) q + k1 i=1 k 1 i 3r(ki) q(i) + r(i) q(ki) , (21) for k 2. References Le Guyader, C. 1993, A&A, 272, 687. http://adsabs.harvard.edu/abs/1993A&A... 272..687L 4