GENERAL I ARTICLE The Archimedes Principle and Gauss's Divergence Theorem Subhashis Nag This article explores the connection between the Ar chimedes principle in physics and Gauss's divergence theorem in mathematics. Introduction Recently, while teaching a university course in vector-calcu lus, I remembered and revisited the business of deducing Subhashis Nag received Archimedes' famous principle in hydrostatics from Gauss's his BSc(Hons) from divergence theorem (which relates certain surface integrals Calcutta University and to volume integrals).(A well-known anecdote relates that PhD from Cornell Archimedes -. having felt fluid pressure on his body (as I University. Subsequently presume we all do) while immersed in a bathtub - under he has worked at many stood it all in a flash, jumped out of the tmb and emerged on institutions in India and the street in a state of nature, shouting 'Eureka!'.) There abroad, and is presently a Professor at the Institute is indeed a very elegant and deep connection between the of Mathematical Sciences, Gauss divergence theorem and Archimedes' principle about Chennai. His research the net effect due to pressure on any immersed solid. That interests centre around study has generated the present article. complex analytic geometry and its intimate relation The main point for us will be to analyze the resultant force with mathematical physics as well as the resultant torque (= turning moment) due to via 'string theory'. the entire system of forces arising from fluid pressure acting on the boundary surface of any three-dimensional body im mersed in a homogeneous fluid at rest. (As we shall remark at the end of our calculations, the fluid need not even have been homogeneous for a rather general form of Archimedes' principle to be valid.) I shall start by describing the resultant effect of an arbitrary system of forces acting at arbitrary points in R 3. The math ematically minded reader rnay find the following discussion of three-dimensional statics independently useful. Systems of Forces in R 3 Let any finite (non-empty) set of force vectors Fi,l ::; i ::; -18-------------------------------~-----------R-E-S-O-N-A-N-C-E--I-N-o-ve-m-b-e-r-1-9-9-a GENERAL j ARTICLE N, be assigned as acting at arbitrary points of application ~ E R 3, in three-dimensional Euclidean space. For non triviality, we may assume each Fi to be non-zero. The precise points of application are actually irrelevant as long as the line of action, say Ai, for each force Fi is specified. (N amely, Ai is the unique line passing through Pi in direc tion Fi; indeed Pi may be chosen arbitrarily on Ai in what follows.) Associated to any such system of forces, we consider two fundamental vectors: The resultant-force vector: (1) Torque-vector with respect to an arbitrary origin 0: (2) Clearly, Torq( 0) is independent of the particular positions of the ~ along the Ai. (Note: The X above denotes the usual vector-product ('cross-product') for vectors in R3 (See any .) college-level mechanics text, e.g. French [2].) Now, if we change the origin 0 to a different point 0' we immediately deduce the following evident, but very basic, relationship: R Torq(O) - Torq(O') = OCY x (3) Thus, given the resultant force R, the vector Torq(O) deter mines uniquely the torque-vector Torq( 0'), computed with respect to any other choice of origin. The discussion above shows that it makes sense to make the following definition: Mechanically Equivalent Systems: Two given systems of forces in R3: {System 1 : Fi along lines Ai}, and {System 2 : G along lines !Vj}, are said to be mechanically equivalent if j -E-S-O-N-A-N-C-E-I-N-o-ve-m-b-e-r-19-9-8----------~------------------------------'9 R GENERAL I ARTICLE the corresponding resultant-force vectors are equal, and also the corresponding torque-vectors (computed with respect to any origin) are equal. Remarks: We note for future use two evident yet important remarks: (1) Torq(Q), as a vector-valued function of the point" Q, R - is constant on lines parallel to the resultant-force as is obvious from equation (3). (2) There exists Q E R3 such that the torque-vector com puted with respect toQ is zero, i.e, Torq( Q) = 0, if and only if the force system is equivalent to a single force acting along some line in R3. The point Q can be chosen arbitrarily on this line. What Force Systems can be Reduced to a Single Force Only? We inquire: under what circumstances can we get a mechanically equivalent system consisting of just one force R (assumed nOll-zero) acting along some specific line, say A, in R3? By the remark number (2) above, we see that we must be able to choose Q E R3 so that Torq(Q) = O. The equation (3) then implies that we must be able to solve for the vector OQ from the equation: Torq(O) = OQ x R (4) The equation shows, of course, (recall the defining properties of the cross-product), that Torq(O) must be orthogonal to R. Consequently the necessary condition for a solution Q to exist is that: R· Torq(O) = 0 (5) A simple calculation now shows that if (5) is valid then the solutions Q to (4) lie along the following line: ......, 1 OQ = -->-(R-I x Torq(O)) + tR......, , any real t (6) IRI2 The torque vanishes when computed with respect to any point on this particular line. Consequently, subject to con dition (5), the equation (6) represents the line A along which -20-----------------------------~-----------RE-S-O-N-A-N-C-E-I-N-o-ve-m-b-e-r-19-9-a GENERAL I ARTICLE Ii acts, and there is then no residual torque or any other forces. Algorithm for Reducing an Arbitrary Force System: Our analysis demonstrates that, given an arbitrary system of forces acting along specified lines in R3, we obtain a me chanically equivalent reduced system by following the simple step-by-step algorithm below: :Ii First step: Calculate and Torq(O) (the latter w.r.t. an arbitrary origin 0) Case 1: If R = 0, then equation (3) shows that the torque vector is independent of the origin. Thus the system reduces to a pure 'couple', (representable (in infinitely many ways) as a pair of equal and opposite forces acting along some pair of parallel lines.) Case 2: If 11 is non-zero then check the dot-product: Ii· Torq(O) Case 2a: If:R· Torq( 0) ::::: 0: then get a single force it as the reduced equivalent system, acting along the line whose equation was exhibited in (6) above. Case 2b: R.Torq(O) non-zero: This is the generic case. it A reduced equivalent system must consist of the force (say applied at origin 0), together with a couple representing the torque Torq(0 ). Interestingly enough, we shall demonstrate in a later section that the system of forces arising from fluid pressure acting on any immersed surface S in a homogeneous fluid, always reduces to Case 2a above! Namely, there is no turning mo ment left over. That is part of the beauty of Archimedes' principle - (although this matter is seldom explained in a school or college course). Some Remarks and a Question: Here is a query that the mathematically minded among you may enjoy investigating. In the generic Case 2b, we saw that we can never choose Q so that Torq( Q) vanishes. What then is the set of points -RE-S-O-N-A-N-C-E-\-N-o-v-em--be-f-'-9-98-----------~-------------------------------~ GENERAL I ARTICLE Q E R3 such that the vector Torq(Q) achieves its minimum possible magnit1.tde? This is an instructive calculus problem of optimization. The reader will discover that this 'best' set of torque-centers Q comprises either a line, A , or a plane, n. In either case it is clear that the line or plane must be parallel to it (since, by Remark (1) after equation (3), we know that Torq( Q) stays constant along lines parallel to 11). Prove that one gets a line or a plane consisting of 'minimizing' centers Q according as the rank of the following symmetric matrix, QuadCR), is 2 or 1, (respectively). 2 + 2 r2 r3 -T1T2 -T1T3 QUad(R) 2+ -T2r1 r3 r12 -T2T3 + -r3r1 -r3r2 112 T22 where 11 = (r1' '2, r3). Note that the above matrix, whose entries are quadratic expressions in the components of the resultant it, always has determinant zero (check this iden tity!). Its rank is therefore always strictly less than three. R::::: Rank 0 is being ignored because that entails 0, (which we disallowed for Case 2b). Gauss's Divergence Theorem Let S be a smooth closed (= compact) surface in R 3 enclos ing (Le., bounding) a three-dimensional region V We shall av s. write = We let n(x, y, z) denote the unit normal pointing outward at the general point (x,y,z) E S. Let G be any vector field on V U S, possessing continuous partial derivatives on an open set containing VUS. Then the well-known Gauss divergence theorem states: Jis JJ I v G ii dS = div(G) dV (7) Notation: Recall that for a vector field G we set div( G) ::::: \7 G and curl(G) = \J x G where \J = ia/ax + ja/ay + ica/oz. Therefore, div(G) = ap/ax + oQ/By + BR/8z, if G = iP + 3Q + kR. We shall also denote partial derivatives by subscripts, where convenient. Finally note that i, j, k ~-- ---------------------------~-----------RE-S-O-N-A-N-C-E-l-N-o-ve-m-b-e-r-1-99-a denote the usual orthonormal set of uni t basis vectors in the coordinate directions. It is important to note that the divergence theorem relates the integral of a derivative of G (namely div(G» over a three-dimensional region V to the surface integral (the 'flux through the surface S') of the vector field G itself over the boundary surface. The double integral on the left side is with respect to the element of surface area on S, whereas the triple integral on the right is with respect to the volume element in R3. The proof can, in fact, be reduced to several judicious ap plications of the standard fundamental theorem of calculus (which asserts that integration is the inverse process to dif ferentiation). Karl Friedrich Gauss (1777-1855) discovered the above theorem while engaged in his research on electro statics. The books by R Courant (a classic calculus text), and that by M Spivak, listed in the Suggested Reading, are good places to look for the divergence theorem. In Chapter V, section 5 of his text Courant indeed explains - very briefly - the idea that is behind what we are presently working out; also Spivak's last exercise (no. 5-36) on the last page of the book talks about the idea being presented in this article, (but, in both cases without mention of the possible turning effects arising from pressure). The Resultant of Pressure Forces: A Generalized Archimedes' Principle Let a solid region V (a compact region of R3 with non empty interior), having a smooth boundary surface BV = S, be immersed in a tank of liquid. We fix the xy plane to be the horizontal surface of the liquid, with the positive z-axis pointing upwards. Thus the fluid region is the lower half space: {(x, y, z) E R3 : z :S OJ. (Note Figure 1) :From basic physics we recall that the pressure at any point P(x, y, z), which is force per unit area, appears due to the weight of the column of liquid above that point. This force always acts in the direction that is normal to any element --ES-O-N-A-N-C-E-I--N-ov-e-m-b-er--19-9-8-----------~-------------------------------~ R GENERAL I ARTICLE of surface that sits there at P( x, y, z) (precisely because a liquid in equilibrium cannot support any shearing stresses) and its magnitude is independent of the direction of the normal to the area element. Note on units: We shall identify the units of weight with the units of mass in what follows, i.e., we measure unit magni tude of force to be the gravitational force acting on one unit of mass. This eliminates the g factor in our equations. Let p denote the density of the fluid. (Note:The perceptive reader will notice that in what follows, Theorem Eureka.R will go through for even a general inhomogeneous fluid (i.e. the density p could as well be a function of (x, y, z) varying from point to point), while Theorem Eureka.Torq. will go through for certain inhomogeneous fluids - for instance, if the density is a function of z alone.) The fluid pressure, call it p,(x,y,z), at any point (x,y,z), is obtained, as explained, from the weight of (an infinitesimally thin) vertical column of liquid standing above that point. Consequently: l(X'Y,O) p,(x, y, z) = pdz (8) (x,y,z) the line integral being along the vertical line segment. In a homogeneous liquid, one has simply p,(x, y, z) = -pz. The General Mathematical Problem: Forgetting all the physics now, the chief mathematical problem we propose is to analyze the system of forces arising from a pressure of magnitude p, per unit area, acting in the inward normal direction all over any immersed surface S. This system of forces is therefore obtained by partitioning the entire surface S into tiny area elements, typically denoted b,.S, with b,.S containing the point P(x, y, z) E S; the force acting at P on the area element b,.S is of magnitude equal to p,(P)D.S in the direction of the inward normal to the surface element ~S . Namely, Force at P on surface-element~S = -p,(x, y, z)D.S n(x, y, z) (9) -24-----------------------------~-----------R-ES-O-N-A-N-C-E--1 -N-ov-e-m-b-e-r-19-9-a GENERAL I ARTICLE (The minus sign arises because we defined :fie P) as denoting the outward unit normal to S at P.) The mathematical 'A rchimedes problem' is to now find the resultant fo.rce and the resultant torque due to these forces acting over the entire closed surface S. (Of course, we must in our calculations, proceed to the limit - as the partitioning of S becomes finer and finer - in the time-tested tradition of integral calculus.) R: R The Resultant Force is the vector sum of ~he forces (9) over all the area elements in the partitioning of S. Pass ing from Riemann sum to integral (replacing ~S by dS), we deduce that the three components of Rare: J Is R· i = - /-l(x, y, z)i· odS (10i) i Is Ii· j = - /-leX, y, z)j . adS (lOj) JI s R· k = - /-leX, y, z)k ftdS (lOk) Bu t the above three are precisely the surface 'flux' integrals for the following three vector fields, respectively: where the G are vector fields defined in the entire fluid i region {z :::; O}. Therefore, the Gauss divergence theorem (7) implies that the resultant force is: R=iJ J IvdiV(G1)dV+ iv JJ kJ JI v j div(G2)dV + div(G3)dV (12) But clearly we have: div(G3) = -o/-l/8z = p (see the defini tion (8) of /-l). Consequently we have proved a basic result: -E-S-O-N-A-N-C-E-I-N-o-v-em-b-e-r-1-99-8-----------~-----------------------------2-5 R GENERAL I ARTICLE R, Theorem Eureka.R. The resultant-force which is the vector sum of all the pre.ssure forces acting on the compact immersed body V is: k The component i.s thus exactly equal to the weight of the fluid displaced by the im,m,ersed body, acting in the upward direction. The i and J components vani.sh .since the pre.s.sure depends only on depth (i.e., /-Lx = /-Ly = 0). I surely need not remind the reader that the assertion above k regarding the component is exactly what they teach you in school physics as Archimedes' principfe. Analysis of the Resultant Torque: As we have learnt earlier, the resultant-force is not by any means the entire story about any given force system in 3-space. We need to understand also the torque-vector Torq( 0), and thence obtain a reduced mechanically equivalent system. That is the goal we now pursue. Let r(x, y, z) = xi + yJ + zk denote the position vector OP for P(x, y, z) E S. Since the force on the element .6.S at P is -/-L(x, y, z).6.Sft(x, y, z) (see (9)), we observe that: is J Torq( 0) = - r x (/-Lft)dS (13) When decomposed into components we get: is J Torq( 0) i = f x (/-Li) ftdS, (14) and, of course, two other similar expressions for the J and k components. (Each integrand is a 'scalar triple product',) The Gaus.s divergence theorem again applie.s immediately to each of these three integrals, and we obtain: JJ I v Torq(O) i = div(f' x (/-Li))dV (15i) -26----------------------------~~-----------R-ES-O-N-A-N-C-E--1 -N-ov-e-m-b-e-r-19-9-a GENERAL I ART'CLE JJ l Torq(O) . J = div(r x (J1.J))dV (15j) Torq(O) k = J J'l div(r x (JLk))dV (15k) We are now admirably positioned to state and prove the basic and final theorem: Theorem Eureka.T orq. The resultant torque-vector about the origin, owing to all the pressure forces on the surface of the immersed body V is k J J l[YJLx - xJLyjdV Since JLx = JLy = 0 throughout the fluid region, the torque simplifies to: Torq(O) = iJ J lypdV - j J J l xpdV (16) 11 But the resultant force (vide Theorem Eureka.R) acts vertically upward, whereas the torque vector lies in the hor :R izontal plane - namely is orthogonal to Torq(O). The system of forces arising from pressure therefore falls under Case 2a of the first section. Consequently, the entire system of forces is equivalent to a single upward force equal to the weight of the fluid dispLaced - acting (as we will prove be low) in a verlicalline passing through the center of mass 01 the region V of fluid displaced. Proof: We need simply calculate the divergences of each of the vector fields appearing on the right sides of equations (15i,j,k). One obtains: diver x (JLi)) == YP + zJLy diver x (JL3)) = -xp - ZJ-Lx div(r x (JLk)) = -XJLy + YJLx -E-S-O-N-A-N-CE--I-N-ov-e-m-b-er-1-9-98-----------~---------------------------2-1 R