OPTIMUM EXPRESSION FOR COMPUTATION OF THE GRAVITY FIELD OF A POLYHEDRAL BODY WITH LINEARLY INCREASING DENSITY 1

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1 OPTIMUM EXPRESSION FOR COMPUTATION OF THE GRAVITY FIEL OF A POLYHERAL BOY WITH LINEARLY INCREASING ENSITY 1 V. POHÁNKA2 Abstract The formla for the comptation of the gravity field of a polyhedral body with density linearly depending on some coordinate is derived and transformed in the optimm form for nmerical calclation. Introdction The optimm expression for the gravity field of a homogeneos polyhedral body was derived in Pohánka 1988 hereafter denoted as HPB. However, in many cases the difference density within the body varies in a more or less simple way with the position. The most sal case is the linear dependence on the depth; therefore, it is desirable to derive an exact formla for the gravity field of a polyhedral body with density linearly depending on some coordinate. erivation of the Formla We follow here the derivation of the formla for the gravity field presented in HPB. If denotes the interior of the polyhedral body and ρr is the difference density within the body, then the gravity potential of the body at the point r is V r = κ dτ ρr r r, where κ is the gravitational constant and dτ is the volme element at the point r. The intensity of the gravity field is Er = r V r = κ dτ ρr r = κ dτ ρr r r r + κ In or case the density can be expressed as r r = κ dτ ρr 1 r r r = dτ 1 r r r ρr. 2 1 ρr = ρ 0 + ρ 1 r 3 ths ρ 0 is the vale of density in a sitably chosen origin of coordinate system and ρ 1 is the gradient of density and we have Er = κ dτ ρ 0 + ρ r 1 r r + κ dτ ρ 1 r r r. 4 1 Received Janary Slovak Academy of Sciences, Geophysical Institte, Bratislava, úbravská cesta 9, Slovak Repblic 1

2 Similarly as in HPB, we express the integrals on the rhs of 4 as srface integrals over the srface S of the body sing the the Gass theorem. If fr is a vector fnction with integrable gradient in the domain, it holds dτ r fr = dσ fs, 5 S where dσ is the srface element at the point s on the srface S oriented otwards from domain. Accordingly, if fr is a scalar fnction with integrable gradient in the domain, it holds dτ r fr = dσ fs. 6 S In order to transform the second integral on the rhs of 4, we se the identity r r r r r = 2 r r ; ths formla 4 reads Er = κ dτ ρ 0 + ρ r 1 r r + κ dτ r r r r 2 r r ρ 1 7 and according to 5 and 6 we obtain Er = κ dσ ρ 0 + ρ 1 s S s + κ dσ s r r S 2 s r ρ 1. 8 Now we write the integrals on the rhs of 8 as a sm of integrals over the sides S k of the polyhedron 1 k K. At any point of the k-th side we have dσ = n k dσ, where n k is the nit normal vector pointing otwards from this side and dσ is the scalar srface element; ths K Er = κ dσ k=1 Sk ρ 0 + ρ n 1 s K k s + κ dσ n k r k=1 S k K [ = κ dσ ρ0 + ρ n 1 r k k=1 S k s + ρ 1 s r r s n k r = κ K k=1 where F k r = dσ 1 S k s r, s r 2 s r ρ 1 = s ] r 2 s r ρ 1 = [n k ρ 0 + ρ 1 r F k r + ρ 1 G k r 1 ] 2 n k G k r ρ 1, 9 10 G k r = dσ S k s r s r. 11 Similarly as in HPB, we express srface integrals in fnctions F k r and G k r as line integrals along the closed crve L k, which is the bondary of the side S k. This can be done sing the 2 Gass theorem which we express in the following form: if fr is a vector fnction with integrable gradient in the domain S k i.e. for r = s and n k fs = 0 i.e. vector fs lies in the plane of the side S k, it holds dσ s fs = dξ fl, 12 S k L k 2

3 where dξ is the linear element at the point l of the crve L k vector dξ is orthogonal to the crve L k and vector n k, and it is oriented otwards from the domain S k. For correctness we note that we do not explicitly se an operator of the 2 gradient: operator r represents always the 3 gradient and expression s fs is in fact a limit of r fr for r s. Nevertheless, this expression does not contain the derivative with respect to coordinate perpendiclar to the side S k ths it is in fact a 2 divergence, as n k fs = 0 here it is important that n k is a constant vector. In the case that fr = gr v, where v is a constant vector, gr is a tensor fnction with integrable gradient in the domain S k and n k gs = 0 i.e. vector gs v lies in the plane of the side S k, we get from 12 dσ s gs v = dσ s gs v = dξ gl v = dξ gl v. S k S k L k L k As this holds for any constant vector v, we obtain the formla dσ s gs = dξ gl 13 S k L k note that it is not necessary that tensor gs is a 2 tensor, ths it need not be gs n k = 0. For explanation, if, v are vectors and g is a tensor, g and g are vectors obtained by the scalar mltiplication of tensor g with vector from the left and right, respectively, and g v is a scalar obtained by the scalar mltiplication of tensor g with vector from the left and vector v from the right the order of mltiplications is not sbstantial. For the fnction F k r it was shown in this way in HPB that F k r = dξ l r L k l r + n k l r ; 14 ths we need to transform only the srface integral in G k r. We are looking for a tensor fnction g k r sch that r g k r = r r r r, n k g k r = Note that it is necessary that these conditions hold only in the domain S k ; however, for simplicity we reqire that they hold in the whole space as this does not represent any loss of generality. We constrct tensor g k r from the nit tensor I and vectors r r and n k sing the scalar and tensor prodct of vectors. We denote the tensor prodct of vectors and v in this order simply as v this tensor is a dyad composed from vectors and v. We first decompose vector r r into two components, r r which is parallel to the side S k and r r perpendiclar to the side S k, r r = Z k n k, Z k = n k r r, r r = r r r r, 16 and denote ρ k = r r, z k = Z k. 17 Then the most general form of tensor g k r constrcted as mentioned above and satisfying the second condition of 15 is g k r = I n k n k g 0 ρ k, Z k + r r r r g 1 ρ k, Z k + n k g 2 ρ k, Z k. 18 3

4 We have and ths r r r = I n k n k, r r r = 2, r g i ρ k, Z k = r r ρ k g i ρ k, Z k + n k g i ρ k, Z k, ρ k Z k r r r g i ρ k, Z k = ρ k ρ k g i ρ k, Z k, r g k r =I n k n k r g 0 ρ k, Z k + + r r r r r g 1 ρ k, Z k + n k g 2 ρ k, Z k + + r r r r r g 1 ρ k, Z k + =r r 1 ρ k + r g 1 ρ k, Z k r r + r g 2 ρ k, Z k n k = = r r g 0 ρ k, Z k + 2 r r ρ k ρ g 1 ρ k, Z k + n k g 2 ρ k, Z k + k + r r g 1 ρ k, Z k + r r ρ k g 1 ρ k, Z k + n k ρ k g 2 ρ k, Z k = ρ k ρ k g 0 ρ k, Z k + 3 g 1 ρ k, Z k + ρ k g 1 ρ k, Z k + ρ k ρ k + n k 2 g 2 ρ k, Z k + ρ k ρ k g 2 ρ k, Z k Inserting in the first condition of 15 and sing the eqality r r r r = r r + Z k n k ρ 2 k + z2 k we obtain eqations 1 ρ k ρ k g 0 ρ k, Z k + 3 g 1 ρ k, Z k + ρ k 2 g 2 ρ k, Z k + ρ k ρ k g 2 ρ k, Z k = Ths we have ρ g 0ρ, Z + 1 ρ ρ ρ3 g 1 ρ, Z =. ρ k g 1 ρ k, Z k = Z k ρ 2 k + z2 k. 1, ρ 2 k + z2 k ρ ρ 2 + z 2, 19 ρ ρ ρ2 g 2 ρ, Z = Z ρ 2 + z, 2 20 where z = Z. In order to se eqality 13 we reqire that tensor g k r is bonded for r r; this implies that g 0 ρ, Z, ρ 2 g 1 ρ, Z and ρ g 2 ρ, Z have to be bonded for ρ 0+. Then the single soltion of 20 is g 2 ρ, Z = Z ρ ρ z 2 z = Z ρ 2 + z 2 + z 21 4

5 note that it is always z 0; in the case of 19 we are free to choose one of the fnctions g 0 ρ, Z, g 1 ρ, Z. In view of 13, 15 and 18 we get from 11 G k r= dσ s g k s = S k = dξ L k dξ g k l = L k ] [I n k n k g 0 ρ k, Z k + l r l r g 1 ρ k, Z k + n k g 2 ρ k, Z k. 22 In or case L k is a closed broken line composed from Lk line segments L k,l 1 l Lk; we assme that these segments are nmbered in the direct sense conter-clockwise viewed from the otside of the body. Vertices of the side S k are nmbered in the same order: end points of segment L k,l are denoted as a k,l and a k,l+1 ths a k,lk+1 = a k,1. For every segment L k,l we define two nit vectors: vector µ k,l, which is parallel with the segment and has the same orientation µ k,l = a k,l+1 a k,l d k,l, 23 where d k,l is the length of the segment L k,l d k,l = a k,l+1 a k,l, 24 and vector ν k,l lying in the plane of the side S k, perpendiclar to the segment L k,l and directed otwards from the side S k, which reads ν k,l = µ k,l n k. 25 Then we have on the segment L k,l and l = a k,l + µ k,l ξ 0 ξ d k,l 26 dξ = ν k,l dξ. 27 Frther we denote and k,l r = µ k,l a k,l r, v k,l r = k,l r + d k,l, w k,l r = ν k,l a k,l r; 28 Z k r = n k a k,1 r, z k r = Z k r, 29 As the qantity n k l r = n k a k,l r does not depend on ξ nor on the index l, this qantity is eqal to Z k r on any segment of S k ; ths we have according to 16 and 17 on the segment L k,l Z k = Z k r, z k = z k r, l r = k,l r + ξ µ k,l + w k,l r ν k,l, ρ k = ρ k,l r + ξ, w k,l r, ρξ, w = ξ 2 + w

6 Now we can write 22 in the form G k r = = = = Lk l=1 Lk l=1 Lk l=1 If we denote we obtain where Lk l=1 dk,l 0 dk,l 0 vk,l r k,l r L k,l dξ ] [I n k n k g 0 ρ k, Z k + l r l r g 1 ρ k, Z k + n k g 2 ρ k, Z k = [ ] dξ ν k,l g 0 ρ k, Z k + ν k,l l r l r g 1 ρ k, Z k + n k g 2 ρ k, Z k = dξ [ ν k,l g 0 ρ k,l r + ξ, w k,l r, Z k r + dξ Φ 1, v, w, Z = Φ 2, v, w, Z = Φ 3, v, w, Z = G k r= Lk l=1 Using 21 we get + w k,l r k,l r + ξ µ k,l + w k,l r ν k,l g 1 ρ k,l r + ξ, w k,l r, Z k r + ] + n k g 2 ρ k,l r + ξ, w k,l r, Z k r = [ ν k,l g 0 ρξ, w k,l r, Z k r + v v v + w k,l r ξ µ k,l + w k,l r ν k,l g 1 ρξ, w k,l r, Z k r + dξ w ξ g 1 ρξ, w, Z, dξ ] + n k g 2 ρξ, w k,l r, Z k r. 31 g 0 ρξ, w, Z + w 2 g 1 ρξ, w, Z, dξ w g 2 ρξ, w, Z, 32 [Φ 1 k,l r, v k,l r, w k,l r, Z k r µ k,l + Φ 2 k,l r, v k,l r, w k,l r, Z k r ν k,l + ] + Φ 3 k,l r, v k,l r, w k,l r, Z k r n k. 33 Φ 3, v, w, Z = Z Φ, v, w, Z 34 Φ, v, w, Z = v dξ w ξ 2 + w 2 + z 2 + z, 35 z = Z and ths Φ, v, w, Z = Φ, v, w, z. In HPB this fnction is denoted as φ, v, w, z and it was shown that 14 can be expressed as F k r = Lk l=1 Φ k,l r, v k,l r, w k,l r, z k r. 36 6

7 Explicit formla for the fnction Φ, v, w, z reads where Φ, v, w, z = w L, v, w, z + 2 z A, v, w, z, 37 L, v, w, z = ln v 2 + w 2 + z 2 + v 2 + w 2 + z 2 +, 38 A, v, w, z = arctan 2 w v T, v, w, z 2 v T, v, w, z z, 39 T, v, w, z = 2 + w 2 + z 2 + v 2 + w 2 + z Integral in the first formla of 32 can be removed if we pt g 1 ρ, Z = 1 ρ hρ, Z; 41 ρ then we obtain sing the last formla of 30 Φ 1, v, w, Z = v Inserting 41 in the condition 19 we get ρ g 0ρ, Z + 1 ρ dξ w hρv, ξ hρξ, w, Z = w w, Z hρ, w, Z. 42 ρ ρ2 ρ hρ, Z = ρ ρ 2 + z 2, what can be written in the form g 0 ρ, Z + ρ ρ ρ hρ, Z = ρ ρ 2 + z 2 43 note that eqations are not changed if we replace hρ, Z by hρ, Z + cz, where cz is a fnction of Z. Ths we have g 0 ρ, Z + ρ ρ hρ, Z = ρ 2 + z 2, 44 as the additional integration term of the form cz on the rhs can be removed if we replace hρ, Z by hρ, Z + cz. Then we have g 0 ρ, Z + w 2 g 1 ρ, Z = ρ 2 + z 2 hρ, Z ρ2 w 2 ρ sing the last formla of 30 we get g 0 ρξ, w, Z + w 2 g 1 ρξ, w, Z = and from the second formla of 32 we obtain hρ, Z, ρ ξ 2 + w 2 + z 2 ξ hρξ, w, Z, ξ v Φ 2, v, w, Z = dξ ξ 2 + w 2 + z 2 ξ ξ hρξ, w, Z = v = dξ ξ 2 + w 2 + z 2 v hρv, w, Z hρ, w, Z. 45 7

8 We can easily calclate the indefinite integral dξ ξ 2 + w 2 + z 2 = 1 ξ ξ w 2 + z 2 + w 2 + z 2 ln ξ 2 + w 2 + z 2 + ξ, ths the integral on the rhs of 45 is v dξ and we finally have ξ 2 + w 2 + z 2 = 1 v v w 2 + z w 2 + z 2 + w 2 + z 2 L, v, w, z Φ 2, v, w, Z = 1 2 v v 2 + w 2 + z w 2 + z 2 + w 2 + z 2 L, v, w, z v hρv, w, Z hρ, w, Z. 47 Ths we have expressed fnctions Φ 1, v, w, Z and Φ 2, v, w, Z formlae 42 and 47 in terms of fnction hρ, Z, which has to satisfy only the condition that hρ, Z and ρ hρ, Z/ρ have to be bonded for ρ 0+. Now we are able to determine the form of fnction hρ, Z in sch a way that the expression 33 for the fnction G k r wold be the simplest possible. We first pt hρ, Z = 0 48 and from 42 and 47 we obtain Φ 1, v, w, Z = 0, Φ 2, v, w, Z = 1 2 v v 2 + w 2 + z w 2 + z 2 + w 2 + z 2 L, v, w, z. 49 Now we can try to find still simpler expression for these fnctions. It can be easily shown that the term containing fnction L, v, w, z cannot be removed from the expression of fnction Φ 2, v, w, Z by any choice of fnction hρ, Z: if it wold be for some fnction fρ, Z v fρv, w, Z fρ, w, Z = 1 2 w2 + z 2 L, v, w, z, then inserting = v we wold get fρv, w, Z = w2 + z 2 v ln 2 + w 2 + z 2 + v 4 v v 2 + w 2 + z 2 v ; however, neither the argment of logarithm nor the term in front of it can be expressed as a fnction of ρv, w and Z. This means that fnction Φ 2, v, w, Z has to contain the logarithm term and that it is not reasonable to consider any fnction hρ, Z containing any logarithmic or other transcendental fnction. We can still try to remove or simplify the first term on the rhs of 47: if we pt 46 hρ, Z = 1 ρρ, z, 2 50 we obtain Φ 1, v, w, Z = 1 2 w v 2 + w 2 + z w 2 + z 2, Φ 2, v, w, Z = 1 2 w2 + z 2 L, v, w, z. 51 8

9 It is clear that there are no simpler expressions for the fnctions Φ 1, v, w, Z and Φ 2, v, w, Z than 49 or 51 note also that in both cases Φ 1, v, w, Z = Φ 1, v, w, z and Φ 2, v, w, Z = Φ 2, v, w, z. It has to be noted that the expression of fnction G k r corresponding to formlae 33 and 49 was obtained in a slightly different way and in context of an other problem by Ivan 1996 athor is indebted to the anonymos reviewer for this information. Now we can present the reslting formla for the intensity of the gravity field Er: from 9, 33, 34 and 36 we get Er = κ K Lk k=1 l=1 [ ρ 0 + ρ 1 r + ρ 1 n k Z k r Φ k,l r, v k,l r, w k,l r, z k r + + ρ 1 µ k,l Φ 1 k,l r, v k,l r, w k,l r, z k r + + ρ 1 ν k,l Φ 2 k,l r, v k,l r, w k,l r, z k r n k 1 ] 2 Z kr Φ k,l r, v k,l r, w k,l r, z k r ρ 1, 52 where fnctions Φ, v, w, z, Φ 1, v, w, z, Φ 2, v, w, z are given by 37 and 49 or 51. Nmerical Aspects Now we modify the obtained formla for nmerical calclation, closely following HPB and Pohánka 1990: we aim to eliminate any ndefined operations e.g. expressions of the type 0/0 and to improve the accracy of calclation. For the fnction Φ, v, w, z this is already done in the mentioned articles: we express fnction L, v, w, z in the form note that we have always v sign = signv : sign signv : L, v, w, z = signv ln v 2 + w 2 + z 2 + v 2 + w 2 + z 2 +, L, v, w, z = ln v 2 + w 2 + z 2 + v 2 + w 2 + z 2 + w 2 + z 2, 53 and replace fnction Φ, v, w, z given by 37 in the formla 52 by the fnction Φ, v, w, z, ε defined as Φ, v, w, z, ε = w L, v, w, z + ε + 2 z A, v, w, z + ε, 54 where the positive parameter ε has a dimension of length and it is mch smaller than the characteristic dimension of the polyhedron. In the case of fnctions Φ 1, v, w, z and Φ 2, v, w, z we first write fnction L, v, w, z in the form 53 and rewrite the other terms according to eqalities v 2 + w 2 + z w 2 + z 2 = v v + T, v, w, z, v v 2 + w 2 + z w 2 + z 2 = 1 2 v v + 2 T, v, w, z + T, v, w, z 55 9

10 in order to decrease the error at points far away from the body. Then we replace fnctions Φ 1, v, w, z, Φ 2, v, w, z in the formla 52 by fnctions Φ 1, v, w, z, ε, Φ 2, v, w, z, ε, respectively, where Φ 1, v, w, z, ε = 0, Φ 2, v, w, z, ε = 1 4 v v + 2 T, v, w, z + ε + T, v, w, z + ε w2 + z 2 L, v, w, z + ε, 56 in the first case formla 49 and Φ 1, v, w, z, ε = v v + w 2 T, v, w, z + ε, Φ 2, v, w, z, ε = 1 2 w2 + z 2 L, v, w, z + ε, 57 in the second case formla 51. Then the expressions 54, 56 and 57 are well defined for v, z 0 and they are sitable for the nmerical calclation. Finally we compare the two variants 56 and 57 with respect to the speed of calclation. Expression 57 is a little simpler than 56, as it contains two additions less than the other. However, by the calclation of expression 52 sing 56 we need neither to calclate the qantity ρ 1 µ k,l ths perform three mltiplications and two additions, nor to mltiply it by a nonzero qantity and add to the other terms. Therefore the most simple variant for the nmerical calclation is that given by formlae 52, 54 and 56. Algorithm of Calclation Let s repeat briefly the optimm algorithm for the calclation of the gravity field of a polyhedral body with linearly increasing density we follow here closely the algorithm presented in HPB. The inpt parameters of the body are: vale of the density at the origin of coordinates ρ 0 note that this origin can lie otside the body, gradient of the density ρ 1 ths density at the point r within the body is given by ρr = ρ 0 + ρ 1 r, nmber of polyhedron sides K, for every side designated by the nmber k, 1 k K, the nmber of edges Lk eqal to the nmber of vertices, for every vertex of the k-th side designated by the nmber l, 1 l Lk, its radis-vector a k,l the vertices are ordered in the direct sense viewed from the exterior of the body and a k,lk+1 = a k,1. The l-th edge is defined by its end points a k,l and a k,l+1. The last inpt parameter is the radis-vector r of the point of calclation of the gravity field. In the calclation we proceed as follows: 1. For each side i.e. for each k we find the following qantities: a For each edge i.e. for each l its length d k,l = a k,l+1 a k,l, and the nit vector µ k,l = a k,l+1 a k,l d k,l. 10

11 b The nit normal vector n k pointing otwards from the side see HPB where n k = N k N k, N k = Lk 1 l=2 a k,l a k,1 a k,l+1 a k,1. c For each edge the nit vector ν k,l = µ k,l n k. None of the qantities mentioned in this paragraph depend on r. 2. For the given point r we calclate the following qantities: a For each k Z k = n k a k,1 r, z k = Z k. b For each k and l k,l = µ k,l a k,l r, v k,l = k,l + d k,l, w k,l = ν k,l a k,l r. 3. For some choice of the small positive parameter ε, the aproximation of the intensity of the gravity field Er, ε is given by Er, ε = κ K Lk k=1 l=1 [ ρ 0 + ρ 1 r + ρ 1 n k Z k Φ k,l, v k,l, w k,l, z k, ε + + ρ 1 ν k,l Φ 2 k,l, v k,l, w k,l, z k, ε n k 1 2 Z k Φ k,l, v k,l, w k,l, z k, ε ρ 1 ], 58 where fnctions Φ, v, w, z, ε, Φ 2, v, w, z, ε can be calclated as follows. For the given nmbers, v v = + d, d > 0, w, z z 0, we make the following consective steps: a z ε = z + ε, b W 2 = w 2 + z 2, W 2 ε = w 2 + z 2 ε, c U ε = d T ε = U ε + V ε, 2 + Wε 2, V ε = v 2 + Wε 2, e sign = signv : L ε = signv ln V ε + v U ε +, 11

12 sign signv : A ε = arctan L ε = ln V ε + v U ε + Wε 2, 2 w d T ε + d T ε d + 2 T ε z ε, f Φ, v, w, z, ε = w L ε + 2 z A ε, Φ 2, v, w, z, ε = 1 4 d v + 2 T ε + T ε W 2 L ε. Conclding Remarks We have derived a formla for the comptation of the gravity field of a polyhedral body with linearly increasing density. This formla has the following properties: 1 it is maximally simple, 2 it is valid for every point of space, and 3 it needs no special attention for the points near or on the srface of the body. However, when compared with the formla for a polyhedral body with constant density, there has to be paid an attention for the points very far from the body. This is becase of the behavior of the terms of the sm on the rhs of formla 52: these terms behave for large distances from the body as polynomials of coordinates of second order, compared with polynomials of the first order in the case of constant density. Therefore it is necessary to se by the nmerical calclation higher precision expression of real nmbers. Acknowledgement. The athor is gratefl to the Slovak Grant agency for science grant No. 2/1063 for the partial spport of this work. References Ivan, M Polyhedral approximations in physical geodesy. Jornal of Geodesy 70, Pohánka, V Optimm expression for comptation of the gravity field of a homogeneos polyhedral body. Geophysical Prospecting 36, Pohánka, V Reply to comment by M. Ivan. Geophysical Prospecting 38,