On Vectors and Tensors, Expressed in Cartesian Coordinates

CHAPTER1 On Vectors and Tensors, Expressed n Cartesan Coordnates It s not enough, to characterze a vector as somethng that has magntude and drecton. Frst, we ll look at somethng wth magntude and drecton, that s not a vector. Second, we ll look at a smlar example of somethng that s a vector, and we ll explore some of ts propertes. Then we ll gve a formal defnton of a cartesan vector that s, a vector whose components we choose to analyse n cartesan coordnates. The defnton can easly be generalzed to cartesan tensors. Fnally on ths subject, we ll explore some basc propertes of cartesan tensors, showng how they extend the propertes of vectors. As examples, we ll use the stran tensor, the stress tensor, and (brefly) the nerta tensor. Fgure 1.1 shows the outcome of a couple of rotatons, appled frst n one sequence, then n the other sequence. We see that f we add the second rotaton to the frst rotaton, the result s dfferent from addng the frst rotaton to the second rotaton. So, fnte rotatons do not commute. They each have magntude (the angle through whch the object s rotated), and drecton (the axs of rotaton). But Frst rotaton + Second rotaton Second rotaton + Frst rotaton. However, nfntesmal rotatons, and angular velocty, truly are vectors. To make ths pont, we can use ntutve deas about dsplacement (whch s a vector). Later, we ll come back to the formal defnton of a vector. What s angular velocty? We defne ω as rotaton about an axs (defned by unt vector l, say) wth angular rate d dt, where s the angle through whch the lne PQ(see Fgure 1.2) moves wth respect to some reference poston (the poston of PQ at the reference tme). So ω = d dt l, and = fnte angle = (t). 1

2 Chapter 1 / ON VECTORS AND TENSORS, EXPRESSED IN CARTESIAN COORDINATES Start Frst rotaton Second rotaton Start Second rotaton Frst rotaton "Frst rotaton" o = rotate 90 about a vertcal axs o "Second rotaton" = rotate 90 about a horzontal axs nto the page FIGURE 1.1 A block s shown after varous rotatons. Applyng the frst rotaton and then the second, gves a dfferent result from applyng the second rotaton and then the frst. To prove that angular velocty ω s a vector, we begn by notng that the nfntesmal rotaton n tme dt s ωdt = ld. Durng the tme nterval dt (thnk of ths as δt, then allow δt 0), the dsplacement of P has ampltude QP tmes d. Ths ampltude s r sn θd, and the drecton s perpendcular to r and l,.e. dsplacement = (ω r)dt. Dsplacements add vectorally. Consder two smultaneous angular veloctes ω 1 and ω 2. Then the total dsplacement (of partcle P n tme dt)s (ω 1 r)dt + (ω 2 r)dt = (ω 1 + ω 2 ) rdt for all r. We can nterpret the rght-hand sde as a statement that the total angular velocty s ω 1 + ω 2. If we reversed the order, then the angular velocty would be the same sum, ω 2 + ω 1, assocated wth the same dsplacement, so ω 2 + ω 1 = ω 1 + ω 2. So: sometmes enttes wth magntude and drecton obey the basc commutatve rule that A 2 + A 1 = A 1 + A 2, and sometmes they do not. What then s a vector? It s an entty that n practce s studed quanttatvely n terms

ON VECTORS AND TENSORS, EXPRESSED IN CARTESIAN COORDINATES 3 P r O θ Q l A rgd object s rotatng about an axs through the the fxed pont O. FIGURE 1.2 P s a pont fxed n a rgd body that rotates wth angular velocty ω about an axs through O. The pont Q les at the foot of the perpendcular from P onto the rotaton axs. 3' 3 2' 2 V 1' O 1 FIGURE 1.3 Here s shown a vector V together wth an orgnal cartesan coordnate system havng axes Ox 1 x 2 x 3 (abbrevated to O1, O2, O3). Also shown s another cartesan coordnate system wth the same orgn, havng axes Ox 1 x 2 x 3. Each system s a set of mutually orthogonal axes. of ts components. In cartesans a vector V s expressed n terms of ts components by V = V 1ˆx 1 + V 2ˆx 2 + V 3ˆx 3 (1.1) where ˆx s the unt vector n the drecton of the -axs. An alternatve way of wrtng equaton (1.1) s V = (V 1, V 2, V 3 ), and sometmes just the symbol V. Then V 1 = V ˆx 1 and n general V = V ˆx. Thus, when wrtng just V, we often leave understood (a) the fact that we are consderng all three components ( = 1, 2, or 3); and (b) the fact that these partcular components are assocated wth a partcular set of cartesan coordnates. What then s the sgnfcance of workng wth a dfferent set of cartesan coordnate axes? We shall have a dfferent set of components of a gven vector, V. See Fgure 1.3 for an llustraton of two dfferent cartesan coordnate systems. What then are the new components of V?

4 Chapter 1 / ON VECTORS AND TENSORS, EXPRESSED IN CARTESIAN COORDINATES We now have V = V 1ˆx 1 + V 2ˆx 2 + V 3ˆx 3 where ˆx 1 s a unt vector n the new x j drecton. So the new components are V j. Another way to wrte the last equaton s V = (V 1, V 2, V 3 ), whch s another expresson of the same vector V, ths tme n terms of ts components n the new coordnate system. Then (a thrd way to state the same dea), V j = V ˆx j. (1.2) We can relate the new components to the old components, by substtutng from (1.1) nto (1.2), so that where 3 V j = (V 1ˆx 1 + V 2ˆx 2 + V 1ˆx 3 ) ˆx j = l j V l j = ˆx ˆx j. Snce l j s the dot product ot two unt vectors, t s equal to the cosne of the angle between ˆx and ˆx j ; that s, the cosne of the angle between the orgnal x axs and the new x j axs. The l j are often called drecton cosnes. In general, l j l j : they are not symmetrc, because l j s the cosne of the angle between the x j -axs and the x -axs, and n general ths angle s ndependent of the angle between x - and x j-axes. But we don t havetobe concerned about the order of the axes, n the sense that cos( θ) = cos θ so that l j s also the cosne of the angle between the x j axs and the x axs. At last we are n a poston to make an mportant defnton. We say that V s a cartesan vector f ts components V j n a new cartesan system are obtaned from ts components V n the prevously specfed system by the rule =1 3 V j = l j V. (1.3) =1 Ths defnton ndcates that the vector V has meanng, ndependent of any cartesan coordnate system. When we express V n terms of ts components, then they wll be dfferent n dfferent coordnate systems; and those components transform accordng to the rule (1.3). Ths rule s the defnng property of a cartesan vector. It s tme now to ntroduce the Ensten summaton conventon whch s smple to state, but whose utlty can be apprecated only wth practce. Accordng to ths conventon, we don t bother to wrte the summaton for equatons such as (1.3) whch have a par of repeated ndces. Thus, wth ths conventon, (1.3) s wrtten V j = l jv (1.4)

ON VECTORS AND TENSORS, EXPRESSED IN CARTESIAN COORDINATES 5 and the presence of the 3 =1 s flagged by the once-repeated subscrpt. Even though we don t bother to wrte t, we must not forget that ths unstated summaton s stll requred over such repeated subscrpts. [Lookng ahead, we shall fnd that the rule (1.4) can be generalzed for enttes called second-order cartesan tensors, symbolzed by A, wth cartesan coordnates that dffer n the new and orgnal systems. The defnng property of such a tensor s that ts components n dfferent coordnate systems obey the relatonshp A jl = A kl j l kl.] As an example of the summaton conventon, we can wrte the scalar product of two vectors a and b as a b = a b. The requred summaton over n the above equaton, s (accordng to the summaton conventon) sgnalled by the repeated subscrpt. Note that the repeated subscrpt could be any symbol. For example we could replace by p, and wrte a b = a p b p. Because t doesn t matter what symbol we use for the repeated subscrpt n the summaton conventon, or p here s called a dummy subscrpt. Any symbol could be used (as long as t s repeated). The Ensten summaton conventon s wdely used together wth symbols δ j and ε jk defned as follows: and δ j = 0 for j, and δ j = 1 for = j; (1.5) ε jk = 0 f any of, j, k are equal, otherwse ε 123 = ε 312 = ε 231 = ε 213 = ε 321 = ε 132 = 1. (1.6) Note that ε jk s unchanged n value f we make an even permutaton of subscrpts (such as 123 312), and changes sgn for an odd permutaton (such as 123 213). The most mportant propertes of the symbols n (1.5) and (1.6) are then δ j a j = a, ε jk a j b k = (a b), (1.7) for any vectors a and b; and the symbols are lnked by the property ε jk ε lmn = δ l δ jl δ kl δ m δ jm δ km δ n δ jn δ kn (1.8) from whch t follows that ε jk ε lm = δ jl δ km δ jm δ kl. (1.9)

6 Chapter 1 / ON VECTORS AND TENSORS, EXPRESSED IN CARTESIAN COORDINATES To prove (1.8), note that f any par of (, j, k) or any par of (l, m, n) are equal, then the left-hand sde and rght-hand sde are both zero. (A determnant wth a par of equal rows or a par of equal columns s zero.) If (, j, k) = (l, m, n) = (1, 2, 3), then the left-hand sde and rght-hand sde are both 1 because ε 123 ε 123 = 1 and 1 0 0 0 1 0 = 1. (1.10) 0 0 1 Any other example of (1.8) where the left-hand sde s not zero, wll requre that the subscrpts (, j, k) be an even or an odd permutaton of (1, 2, 3), and smlarly for the subscrpts (k, l, m), gvng a value (for the left-hand sde) equal ether to 1 or to 1. But the same type of permutaton of (, j, k) or (l, m, n) (whether even or odd) wll also apply to columns or to rows of (1.10), gvng ether 1 (for a net even permutaton) or 1 (for a net odd permutaton), and agan the left-hand sde of (1.8) equals the rght-hand sde. Because of the frst of the relatons gven n (1.7), δ j s sometmes called the substtuton symbol or substtuton tensor. In recognton of ts orgnator t also called the Kronecker delta. ε jk s usually called the alternatng tensor. (1.9) follows from (1.8), recognzng that we need to allow for the summaton over. Thus δ δ j δ k ε jk ε lm = δ l δ jl δ kl δ m δ jm δ km = δ (δ jl δ km δ jm δ kl ) δ j (δ l δ km δ m δ kl ) + δ k (δ l δ jm δ m δ jl ), but here δ s not equal to 1 (whch s what most people who are unfamlar wth the summaton conventon mght thnk at frst). Rather, δ = 3 =1 δ = δ 11 + δ 22 + δ 33 = 3. Usng ths result, and the substtuton property of the Kronecker delta functon (the frst of the relatons n (1.7)), we fnd ε jk ε lm = 3(δ jl δ km δ jm δ kl ) (δ jl δ km δ jm δ kl ) + (δ kl δ jm δ km δ jl ), whch smplfes to (1.9) after combnng equal terms. As we should expect, the subscrpt does not appear n the rght-hand sde. 1.1 Tensors Tensors generalze many of the concepts descrbed above for vectors. In ths Secton we shall look at tensors of stress and stran, showng n each case how they relate a par of vectors. We shall develop () () () the physcal deas behnd a partcular tensor (for example, stress or stran); the notaton (for example, for the cartesan components of a tensor); a way to thnk conceptually of a tensor, that avods dependence on any partcular choce of coordnate system; and

1.1 Tensors 7 n δ F = T δs S x δs S s an nternal surface, nsde a medum wthn whch stresses are actng. δs s a part of the surface S. x s the pont at the center of δ S. FIGURE 1.4 The defnton of tracton T actng at a pont across the nternal surface S wth normal n (a unt vector). The choce of sgn s such that tracton s a pullng force. Pushng s n the opposte drecton, so for a flud medum, the pressure would be n T. (v) the formal defnton of a tensor (analogous to the defnton of a vector based on (1.3) or (1.4)). To analyze the nternal forces actng mutually between adjacent partcles wthn a contnuum, we use the concepts of tracton and stress tensor. Tracton s a vector, beng the force actng per unt area across an nternal surface wthn the contnuum, and quantfes the contact force (per unt area) wth whch partcles on one sde of the surface act upon partcles on the other sde. For a gven pont of the nternal surface, tracton s defned (see Fg. 1.4) by consderng the nfntesmal force δf actng across an nfntesmal area δs of the surface, and takng the lmt of δf/δs as δs 0. Wth a unt normal n to the surface S, the conventon s adopted that δf has the drecton of force due to materal on the sde to whch n ponts and actng upon materal on the sde from whch n s pontng; the resultng tracton s denoted as T(n).IfδF acts n the drecton shown n Fg. 1.4, tracton s a pullng force, opposte to a pushng force such as pressure. Thus, n a flud, the (scalar) pressure s n T(n). For a sold, shearng forces can act across nternal surfaces, and so T need not be parallel to n. Furthermore, the magntude and drecton of tracton depend on the orentaton of the surface element δs across whch contact forces are taken (whereas pressure at a pont n a flud s the same n all drectons). To apprecate ths orentaton-dependence of tracton at a pont, consder a pont P, as shown n Fgure 1.5, on the exteror surface of a house. For an element of area on the surface of the wall at P, the tracton T(n 1 ) s zero (neglectng atmospherc pressure and wnds); but for a horzontal element of area wthn the wall at P, the tracton T(n 2 ) may be large (and negatve). Because T can vary from place to place, as well as wth orentaton of the underlyng element of area needed to defne tracton, T s separately a functon of x and n. So we wrte T = T(x, n). At a gven poston x, the stress tensor s a devce that tells us how T depends upon n. But before we nvestgate ths dependence, we frst see what happens f n changes sgn.

8 Chapter 1 / ON VECTORS AND TENSORS, EXPRESSED IN CARTESIAN COORDINATES n 2 n 1 P FIGURE 1.5 T(n 1 ) T(n 2 ). The tracton vector n general s dfferent for dfferent orentatons of the area across wth the tracton s actng. T(n) n n T( n) FIGURE 1.6 A dsk wth parallel faces. The normals to opposte faces have the same drecton but opposte sgn. By consderng a small dsk-shaped volume (Fgure 1.6) whose opposte faces have opposte normals n and n, we must have a balance of forces T( n) = T(n) (1.11) otherwse the dsk would have nfnte acceleraton, n the lmt as ts volume shrnks down to zero. (There s neglgble effect from the edges as they are so much smaller than the flat faces.) In a smlar fashon we can examne the balance of forces on a small tetrahedron that has three of ts four faces wthn the planes of a cartesan coordnate system, as shown n Fgure 1.7. The oblque (fourth) face of the tetrahedron has normal n (a unt vector), and by projectng area ABC onto each of the coordnate planes, we fnd the followng relaton between areas:

1.1 Tensors 9 3 2 n C B T(n) O A 1 T( xˆ 2 ) xˆ 2 FIGURE 1.7 The small tetrahedron OABC has three of ts faces n the coordnate planes, wth outward normals ˆx j ( j = 1, 2, 3) (only one of whch s shown here, j = 2), and the fourth face has normal n. (OBC, OCA, OAB) = ABC (n 1, n 2, n 3 ). (1.12) There are four forces actng on the tetrahedron, one for each face. Thus, face OBC has the outward normal gven by the unt vector ˆx 1 = ( 1, 0, 0). Ths face s pulled by the tracton T( ˆx 1 ), and hence by the force T( ˆx 1 ) tmes area OBC (remember, tracton s force per unt area). The balance of forces then requres that T(n) ABC + T( ˆx 1 ) OBC + T( ˆx 2 ) OCA + T( ˆx 3 ) OAB= 0. (1.13) (If the rght-hand sde were not zero, we would get nfnte acceleraton n the lmt as the tetrahedron shrnks down to the pont O.) Usng the two equatons (1.11) (1.13), t follows that T(n) = T(ˆx 1 )n 1 + T(ˆx 2 )n 2 + T(ˆx 3 )n 3 = T(ˆx j )n j (usng the summaton conventon). (1.14) If we now defne τ kl = T l (ˆx k ), (1.15) then T (n) = τ j n j. (1.16) If we can show that τ j = τ j, then T = τ j n. (1.17)

10 Chapter 1 / ON VECTORS AND TENSORS, EXPRESSED IN CARTESIAN COORDINATES Ths equaton gves a smple rule by whch the components of the tracton vector, T, are gven as a lnear combnaton of the components of the normal vector n j. The nne symbols τ j are the cartesan components of a tensor, namely, the stress tensor. Frst, we ll show that ndeed τ j = τ j. Second, we ll show that the symbols τ j specfy a surface whch s ndependent of our partcular choce of coordnate axes. 1.1.1 SYMMETRY OF THE STRESS TENSOR To see why the τ j are symmetrc, we can look n some detal at a partcular example, namely τ 21 and τ 12. They quantfy components of the tractons T (force per unt area) actng on the faces of a small cube wth sdes of length δx 1, δx 2, δx 3 as shown n Fgure 1.8. The force actng on the top face of the cube s tracton area, whch s T(ˆx 2 )δx 1 δx 2. And on the opposte face the force s T( x ˆ 2 )δx 1 δx 2. From these two faces, what s the strength of the resultng couple that tends to make the cube rotate about the x 3 axs? The x 2 and x 3 components of T have no relevance here (they are assocated wth the tendency to rotate about dfferent axes) only the x 1 component of T, whch s τ 21. Fgure 1.8b shown the resultng couple, and t s τ 21 δx 1 δx 2 δx 3 n the negatve x 3 drecton. When we look at the tractons actng on the rght and left faces, as shown n Fgure 1.8c, the couple that results (see Fg. 1.8d) s τ 12 δx 1 δx 2 δx 3 n the postve x 3 drecton. No other couple s actng n the x 3 drecton, so the two couples we have obtaned must be equal and opposte, otherwse the cube would spn up wth ncreasng angular velocty. It follows that τ 21 = τ 12. By smlar arguments requrng no net couple about the x 1 or x 2 drectons, we fnd τ 32 = τ 23 and τ 31 = τ 13. So n general, τ j = τ j and we have proven the symmetry requred to obtan (1.17) from (1.16). 1.1.2 NORMAL STRESS AND SHEAR STRESS Fgure 1.9 shows two components of a tracton vector T, one n the normal drecton, and the other n the drecton parallel to the surface across whch T acts. The normal stress, σ n, s gven by σ n = component of tracton n the normal drecton = T n = T j n j = τ j n n j. (1.18) Note that normal stress, as defned here, s a scalar. But t s assocated wth a partcular drecton, and the vector σ n n s often used for the normal stress (as shown n Fgure 1.9). The shear stress s the component of T n the plane of the surface across whch tracton s actng, so t may be evaluated by subtractng the normal stress σ n n from T tself: shear stress = T (T n) n. (1.19)

1.1 Tensors 11 2 xˆ 2 (a) T(x ˆ 2) (b) area = δx 1 δx 3 τ 21 δx 2 δx 1 δx 3 1 dstance = δx 2 τ 21 3 T( x ˆ 2 ) xˆ 2 2 (c) (d) τ 12 xˆ 1 δx 2 δx 1 δx 3 T(x ˆ 1) xˆ 1 1 dstance = δx 1 area = δx 2 δx 3 T( x ˆ 1 ) τ 12 3 FIGURE 1.8 A small cube s shown, wth an analyss of the couple tendng to make the cube spn about the Ox 3 axs. Thus, n (a) s shown the tractons actng on the top and bottom faces. In (b) s shown the couple assocated wth these tractons whch wll tend to make the cube rotate about the axs nto the plane of the page the couple conssts of a par of forces τ 21 δx 1 δx 3 separated by a dstance δx 2. In (c) s shown the tractons actng on the rght and left faces, and n (d) the assocated couple tendng to make the cube rotate about an axs out of the plane of the page.

12 Chapter 1 / ON VECTORS AND TENSORS, EXPRESSED IN CARTESIAN COORDINATES T(n) normal stress = (T(n).n) n shear stress = T (T(n).n) n FIGURE 1.9 The tracton T s shown resolved nto ts normal and shear components. n P O normal to Σ at P Σ FIGURE 1.10 s a quadrc surface centered on the coordnate orgn O. P s the pont on at the place where a lne drawn from the orgn n the drecton of n (a unt vector) meets the quadrc surface. The normal to at P s shown, and n general t wll be n a drecton dfferent from the drecton of n. It s unfortunate that the word stress s used for scalar quanttes as n (1.18), for vector quanttes as n (1.19), and for tensor quanttes as n (1.15). In a flud wth no vscosty, the scalar normal stress s equal to pressure (but wth opposte sgn, n our conventon where tracton s postve f t s pullng). And n such a flud, the shear stresses are all zero. 1.1.3 A QUADRIC SURFACE ASSOCIATED WITH THE STRESS TENSOR The surface gven by the equaton τ j x x j = constant (1.20) has propertes that are ndependent of the coordnate system used to defne components x and τ j. s a three-dmensonal surface, ether a spherod or a hyperbolod (whch may have one or two separate surfaces). Because (1.20) has no lnear terms (terms n x only), s a surface whch s symmetrc about the orgn O, as shown n Fgure 1.10. (If x s a pont on the surface, then so s the pont x.) If P s at poston x, then

1.1 Tensors 13 BOX 1.1 On fndng the drecton of a normal to a surface If s the surface composed of ponts x whch satsfy the equaton F(x) = c for some constant c, then consder a pont x + δx whch also les on. Then F(x + δx) = c, so But F(x + δx) F(x) = 0. F F(x + δx) = F(x) + δx x = F(x) + δx F. (Taylor seres expanson) Therefore, δx F = 0 for all drectons δx such that x and x + δx le n the surface. Snce δx s tangent to at the pont x, and δx F = 0 for all such tangents, ths last equanton shows that F has the defnng property of the normal to at x (t s perpendcular to all the tangents at x). We conclude that the normal to the surface F(x) = c s parallel to F. vector OP = x = OP n (1.21) and the length OP s related to the normal stress. Ths follows, because constant = τ j x x j = τ j OP n OP n j Hence, = OP 2 τ j n n j = OP 2 σ n from (1.18). σ n 1 OP 2. (1.22) Ths s a geometrcal property of, whch depends only the shape of the quadrc surface, not ts absolute sze. Another such geometrcal property s assocated wth the normal to at P. We can use Box 1.1 to see that ths normal s n the drecton whose -th component s

14 Chapter 1 / ON VECTORS AND TENSORS, EXPRESSED IN CARTESIAN COORDINATES x k x l (τ kl x k x l ) = τ kl x l + τ kl x k x x x (chan rule) = τ kl δ k x l + τ kl x k δ l (the partal dervatves are Kronecker delta functons) = τ l x l + τ k x k (usng the substtuton property (1.7)) = 2τ j x j (usng symmetry of the stress tensor, and changng dummy subscrpts) It follows that = 2OP τ j n j (from (1.21)) = 2OP T (from (1.17)). the normal to at P s parallel to the tracton vector, T. (1.23) The results (1.22) and (1.23) are the two key geometrcal features of the quadrc surface representng the underlyng tensor τ, and they are ndependent of any coordnate system. Gong back to Fgures 1.4 and 1.9, the quadrc surface shown n Fgure 1.10 s a geometrcal devce for obtanng the drecton of the tracton vector, and the way n whch normal stress vares as the unt normal n vares (specfyng the orentaton of an area element across whch the tracton acts). We can for example see geometrcally that there are three specal drectons n, for whch the tracton T(x) s parallel to n and hence (for these n drectons) the shear stress vanshes. To obtan the same result algebracally, we note that these specal values of n are such that T n and so, usng (1.17), τ j n j = λn. (1.24) As dscussed n Box 1.2, ths last equaton has solutons (for n), but only f λ takes on one of three specal values (egenvalues); and the three resultng values of n (one for each λ value) are mutually orthogonal. 1.1.4 FORMAL DEFINITION OF A SECOND ORDER CARTESIAN TENSOR If two cartesan coordnate systems Ox 1 x 2 x 3 and Ox 1 x 2 x 3 are related to each other as shown n Fgure 1.3, wth drecton cosnes defned by l j = ˆx ˆx j, then the entty A s a second order cartesan tensor f ts components A k n the Ox 1 x 2 x 3 system and A jl n the system are related to each other by Ox 1 x 2 x 3 A jl = A kl j l kl. (1.25) It s left as an exercse (n Problem 1.1) to obtan the reverse transformaton, and to show that the assocated quadrc surface A j x x j = c s gven also by A j x x j = c. The tensor A s sotropc f ts components are the same n the orgnal and the transformed coordnate systems. It follows geometrcally that the quadrc surface of such a tensor s smply a sphere, and that A j = A j = constant δ j. The stress tensor n an nvscd flud s sotropc (Problem 1.4 asks you to prove ths).

1.1 Tensors 15 BOX 1.2 On solutons of A x = λx where A s a symmetrc real matrx Here we brefly revew the man propertes of the set of lnear equatons A j x j = λx (where A j = A j ; = 1,...,n; and j = 1,...,n). (1) These are n equatons n n unknowns. In the present chapter, usually n = 3. Equaton (1) s fundamentally dfferent from the standard lnear problem A x = b, or A j x = b, (2) whch has a straghtforward soluton provded det A 0, for then the nverse matrx exsts and x = A 1 b. Note that for equaton (1) there s no restrcton on the absolute sze of x: f x s a soluton of (1), then so s cx for any constant c. But f x s a soluton of (2), then n general cx wll not be a soluton. Equaton (1) has the trval soluton x = 0, but we are nterested n non-trval solutons, for whch at least one component of x s non-zero. It follows that equaton (1) represents a set of n scalar relatons between n 1 varables, for example x j /x 1 ( j = 2,...,n). If x 1 = 0 then we can nstead dvde (1) by some other component of x. We can use the Kronecker delta functon to wrte (1) as (A j λδ j )x j = 0, or as (A λi) x = 0, (1, agan) where I s the dentty matrx, I j = δ j, wth 1 at every entry on the dagonal and 0 everywhere else. If det (A λi) 0, then the nverse (A λi) 1 exsts and the only soluton of (1) s x = (A λi) 1 0, gvng x = 0. It follows that the only way we can obtan non-trval solutons of (1), s to requre that det (A λi) = 0. (3) Equaton (3) s an n-th order polynomal n λ, and n general t has n solutons, λ (α) (α = 1,...,n), called egenvalues of the matrx A. For each egenvalue, we have an assocated egenvector x (α) satsfyng A x (α) = λ (α) x (α) (not summed over α). (4) These egenvectors (there are n of them, snce α = 1,...,n) are the vector solutons of (1), so they have the specal property that A x (α) s n the same drecton as the vector x (α) tself. Equaton (1) mposes no contrant on the absolute sze of solutons such as x (α), and we are free to normalze these solutons f we wsh. A common choce s to make x (α) x (α) = 1. The fnal mportant property of the egenvectors, s that any two egenvectors correspondng to two dfferent egenvalues must be orthogonal. To prove ths, we consder two egenvalues λ (α) and λ (β) wth correspondng egenvectors x (α) and x (β), and λ (α) λ (β). Then A j x (α) j = λ (α) x (α) and A j x (β) j = λ (β) x (β). (5)

16 Chapter 1 / ON VECTORS AND TENSORS, EXPRESSED IN CARTESIAN COORDINATES BOX 1.2 (contnued) We can multply both sdes of the frst of equatons (5) by x (β) and sum over, and multply both sdes of the second of equatons (5) by x (α) and sum over, gvng and A j x (α) j A j x (β) j x (β) x (α) = λ (α) x (α) x (β) = λ (α) x (α) x (β) (6a) = λ (β) x (α) x (β) = λ (β) x (α) x (β). (6b) Note here that we are usng the Ensten summaton conventon n the usual way for repeated subscrpts, but we are not usng t for repeated superscrpts α and β. [The summaton conventon does not work naturally wth such superscrpts, because one sde of equatons such as (4) or (6a) has repeated superscrpts, and the other sde has the superscrpt n just one place.] Because A s a symmetrc matrx, A j x (β) j x (α) = A j x (β) j x (α). But we can use any symbol for repeated subscrpts, and n partcular we can exchange the symbols and j n ths last expresson, so that A j x (β) j x (α) = A j x (β) j x (α) = A j x (β) j x (α) = A j x (α) x (β) j. Hence, the lefthand sde of (6a) equals the left-hand sde of (6b). Subtractng (6b) from (6a), we see that (λ (α) λ (β) ) x (α) x (β) = 0. But the frst of these factors s not zero, snce we took the egenvalues to be dfferent (λ (α) λ (β) ). It follows that the second factor must be zero, x (α) x (β) = 0, and hence that the egenvectors correspondng to dfferent egenvalues are orthogonal. The best way to become famlar wth the egenvalue/egenvector propertes descrbed above, s to work through the detals of a few examples (such as Problem 1.2). The above formal defnton of a tensor s rarely useful drectly, as a way to demonstrate that an entty (suspected of beng a tensor) s n fact a tensor. For ths purpose, we usually rely nstead upon the so-called quotent rule, whch states that f the relatonshp T k A k = B s true s all coordnate systems, where A k and B are the components of vectors A and B, then the T k are components of a second order cartesan tensor provded the components of A can all be vared separately. The statement that the relatonshp T k A k = B s true s all coordnate systems means that T jk A l = B j as well as T ka k = B. To show that the quotent rule means the formal defnton s satsfed, note that T jk A l = B j = B l j snce B s a vector = T k A k l j usng T k A k = B = T k A l l kll j from the reverse transform A k = A l l kl see Problem 1.1. Hence

1.2 The stran tensor 17 Q 0 δx P0 u(x + δx) u(x) P Q x O FIGURE 1.11 A lne element δx s shown wth ends at P 0 and Q 0, and also after the neghborhood of P 0 has undergone deformaton. The new poston of the lne element s from P to Q. P 0 s at x, Q 0 s at x + δx, P s at x + u and Q s at x + δx + u(x + δx). (T jl T kl j l kl )A l = 0. (1.26) Ths s a set of three scalar equatons ( j = 1, 2, or 3). But snce they are also true as the components A l are vared ndependently, t follow that the coeffcents of the A l (l = 1, 2, or 3) must all vansh. So T jl = T kl j l kl, and T satsfes the formal defnton of a tensor (compare wth (1.25)). The entty we have been callng the stress tensor, τ, satsfes the quotent rule (see equaton (1.17), n whch the components of τ relate the tracton vector and n). It s for ths reason that ndeed we are justfed n referrng to τ as a second order cartesan tensor. 1.2 The stran tensor Two dfferent methods are wdely used to descrbe the motons and the mechancs of moton n a contnuum. These are the Lagrangan descrpton, whch emphaszes the study of a partcular partcle that s specfed by ts orgnal poston at some reference tme, and the Euleran descrpton, whch emphaszes the study of whatever partcle happens to occupy a partcular spatal locaton. Note that a sesmogram s the record of moton of a partcular part of the Earth (namely, the partcles to whch the sesmometer was attached durng nstallaton), so t s drectly a record of Lagrangan moton. A pressure gauge attached to the sea floor also provdes a Lagrangan record, as does a neutrally buoyant gauge that s free to move n the water. But a gauge that s fxed to the sea floor and measurng propertes (such as velocty, temperature, opacty) of the water flowng by, provdes an Euleran record. We use the term dsplacement, regarded as a functon of space and tme, and wrtten as u = u(x, t), to denote the vector dstance of a partcle at tme t from the poston x that t occupes at some reference tme t 0, often taken as t = 0. Snce x does not change wth tme, t follows that the partcle velocty s u/ t and that the partcle acceleraton s 2 u/ t 2. To analyze the dstorton of a medum, whether t be sold or flud, elastc or nelastc,

18 Chapter 1 / ON VECTORS AND TENSORS, EXPRESSED IN CARTESIAN COORDINATES we use the stran tensor. If a partcle ntally at poston x s moved to poston x + u, then the relaton u = u(x) s used to descrbe the dsplacement feld. To examne the dstorton of the part of the medum that was ntally n the vcnty of x, we need to know the new poston of the partcle that was ntally at x + δx, where δx s a small lne-element. Ths new poston (see Fgure 1.11) s x + δx + u(x + δx). Any dstorton s lable to change the relatve poston of the ends of the lne-element δx. If ths change s δu, then δx + δu s the new vector lne-element, and by wrtng down the dfference between ts end ponts we obtan δx + δu = x + δx + u(x + δx) (x + u(x)). Snce δx s arbtrarly small, we can expand u(x + δx)asu + (δx )u plus neglgble terms of order δx 2. It follows that δu s related to gradents of u and to the orgnal lneelement δx va δu = (δx )u, or δu = u x j δx j. (1.27) However, we do not need all of the nne ndependent components of u x j to specfy true dstorton n the vcnty of x, snce part of the moton s due merely to an nfntesmal rgd-body rotaton of the neghborhood of x. We shall wrte δu = u ( δx j = 1 u x 2 + u ) ( j δx j + 1 u j x j x 2 u ) j δx j x j x and use the result gven n Problem 1.8 to nterpret the last term of the above equaton. Introducng the notaton u, j for u x j, we then see that (1.27) can be rewrtten as δu = 1 2 (u, j + u j, )δx j + 1 2 (curl u δx), (1.28) and the rgd-body rotaton s of amount 1 2curl u. The nterpretaton of the last term n (1.28) as a rgd-body rotaton s vald f u, j 1. If dsplacement gradents were not nfntesmal n the sense of ths nequalty, then we would nstead have to analyze the contrbuton to δu from a fnte rotaton a much more dffcult matter, snce fnte rotatons do not commute and cannot be expressed as vectors. In terms of the nfntesmal stran tensor, defned to have components e j 1 2 (u, j + u j, ), (1.29) the effect of true dstorton on any lne-element δx s to change the relatve poston of ts end ponts by a dsplacement whose -th component s e j δx j. By the quotent rule dscussed n Secton 1.1.4, t follows that e s ndeed a second order cartesan tensor. Rotaton does not affect the length of the element, and the new length s

1.2 The stran tensor 19 δx + δu = δx δx + 2δu δx (neglectng δu δu) = δx δx + 2e j δx δx j (from (1.28), and usng (curl u δx) δx = 0) = δx (1 + e j ν ν j ) (to frst order, f e j 1), where ν s the unt vector δx/ δx. It follows that the extensonal stran of a lne-element orgnally n the ν drecton, whch we defne to be s gven by e(ν) = change n length orgnal length = PQ P 0Q 0 P 0 Q 0, e(ν) = e j ν ν j. (1.30) Stran s a dmensonless quantty. The durnal sold Earth tde leads to strans of about 10 7. Strans of about 10 11 due to sesmc surface waves from dstant small earthquakes can routnely be measured wth modern nstruments. 1.2.1 THE STRAIN QUADRIC We can defne the surface e j x x j = constant. If t s referred to the axes of symmetry as coordnate axes, ths quadrc surface becomes E 1 x1 2 + E 2x2 2 + E 3x3 2 = constant and the stran tensor components become E 1 0 0 e = 0 E 2 0. 0 0 E 3 The E ( = 1, 2, or 3), called prncpal strans, are egenvalues of the matrx e 11 e 12 3 13 e = e 12 e 22 3 23 e 13 e 23 3 33 (ths matrx s dfferent for dfferent cartesan coordnate systems, but ts egenvalues are the same). Fgure 1.12 shows an example of the stran quadrc, n the case that the E do not all have the same sgn. In ths case, s a hyperbolod. The fgure capton descrbes the two key geometrcal propertes of, analogous to the results shown prevously for the stress quadrc (see (1.22) and (1.23)). The frst property, e(ν) OP 2, follows from takng x = OPν and substtutng for the components of x n e j x x j = constant. The second property, concernng the drecton of the dsplacement due to dstorton, follows from showng that the drecton of the normal at P s parallel to (e j x x j ), whch has -th component 2e j x j = 2OPe j ν j. In terms of the lne-element δx, ths -th component s proportonal to e j δx j, and ndeed ths s n the dsplacement drecton due to the dstorton (see comments followng (1.29)). The physcal nterpretaton of the axes of symmetry of, s that these are the three specal (mutually orthogonal) drectons n whch an orgnal lne-element s merely

20 Chapter 1 / ON VECTORS AND TENSORS, EXPRESSED IN CARTESIAN COORDINATES prncpal axs #2 prncpal axs #1 O ν P Σ FIGURE 1.12 A hyperbolod s shown here, as an example of a stran quadrc. The thrd prncpal axs s perpendcular to the frst two (out of the page). For a lne-element n the ν drecton, extensonal stran s nversely proportonal to OP 2. The dsplacement of the end of the lne-element, due to dstorton, s n the drecton of the normal at P. shortened or lengthened by the deformaton. For these specal drectons, lne-elements are not subject to any shearng motons. The only rotaton, s that whch apples to the whole neghborhood of the lne-element as a rgd body rotaton (dstnct from deformaton). 1.3 Some smple examples of stress and stran Examples are () Compactng sedments, whch shrnk n the vertcal drecton but stay the same n the horzontal drecton. Thnk of them as beng n a large tank (see Fgure 1.13). If x 1 and x 2 are horzontal drectons wth x 3 vertcal, then the stran tensor has components 0 0 0 e = 0 0 0. 0 0 e 33 Compresson at depth, from the weght of those above, leads to e 33 < 0. But although the stran tensor has only one non-zero component, the stress tensor has non-zero values of τ 11,τ 22,τ 33. If the sedments lack strength, then the stress wll essentally be the same as f the materal were composed of a flud (lackng any ablty to resst shearng forces). In ths case, the stress at depth x 3 s due solely to pressure generated by the overburden. There are no shearng forces, and P 0 0 x3 τ = 0 P 0 where P = ρg dx 3. 0 0 0 P

1.3 Some smple examples of stress and stran 21 x 3 FIGURE 1.13 Sedments, compressed vertcally. () A wre that s stretched n the x 1 drecton wll shrnk n the perpendcular drectons x 2 and x 3. The stran tensor s e 11 0 0 e = 0 e 22 0, wth e 11 > 0 and e 22 = e 33 < 0. 0 0 e 33 The stress tensor has only one non-zero component: τ 11 0 0 τ = 0 0 0. 0 0 0 () We use the term pure shear to descrbe the type of deformaton shown n Fgure 1.14, n whch a small rectangle s subjected to the stress feld 0 τ 12 0 τ = τ 12 0 0. 0 0 0 In ths deformaton, the pont B has moved to the rght from ts orgnal poston, by an amount u 1 δx 2, and the pont A has moved up an amount u 2 δx 1. In the case x 2 x 1 of pure shear, there s no rgd body rotaton. So curl u = 0 whch means here that u 1 = u 2. The lne element OB 0 has been rotated over to the rght (clockwse) x 2 x 1 by an angle gven by u 1 δx 2 δx 2 = u 1. The lne element OA 0 has been rotated x 2 x 2 antclockwse by the angle u 2 δx 1 δx 1 = u 2 = u 1 snce there s no rgd body x 1 x 1 x 2 rotaton, and the stran tensor components are 0 e 12 0 e = e 21 0 0. 0 0 0 Always, e j = e j. And for pure shear, e 12 = e 21 = u 1 = u 2. x 2 x 1 The reducton n the orgnal π/2 angle at O (Fgure 1.14) s 2e 12. The rato between stress and ths angle reducton,

22 Chapter 1 / ON VECTORS AND TENSORS, EXPRESSED IN CARTESIAN COORDINATES x 2 orgnal shape x 2 deformed shape (pure shear) B0 B δx2 A O δx1 A 0 x1 O x 1 FIGURE 1.14 A small rectangular prsm s sheared wth no change n area. The shearng stresses τ 12 and τ 21 (= τ 12 ) are the same as those shown n Fgure 1.8b and 1.8d. x 2 x 2 angle = 2 e 12 angle = e 12 δx2 angle = e 12 O δx1 x1 O x 1 smple shear = pure shear as shown above, plus a clockwse rotaton of the whole block through an angle e12 FIGURE 1.15 A smple shear s shown on the left, wth u 1 x 2, so that an orgnal rectangle n the Ox 1 x 2 -plane s now sheared wth moton only n the x 1 -drecton. Ths moton s equvalent to a pure shear followed by a clockwse rgd-body rotaton. The angle reducton at the orgn s 2e 12 n both cases. Rgdty s the rato between shear stress and the angle reducton. For a vscous materal, the materal contnues to deform as long as the shear stress s appled, and vscosty s defned as the rato between shear stress and the rate of angle reducton. µ = τ 12 2e 12, (1.31) (v) s known as the rgdty. Consder a deformaton n whch the dsplacement s u = (u 1, 0, 0) and u 1 depends on x only va the x 2 -component. An example s shown n Fgure 1.15, and ths stuaton s called a smple shear. It can be regarded as a pure shear descrbed by e 12 = e 21 = 1 ( u 1, plus a rgd rotaton of amount 1 2 x 2 curl u = 0, 0, 1 ) u 1. 2 2 x 2 The defnton of rgdty s easer to understand wth a smple shearng deformaton. It s the shearng stress τ 21 dvded by the dsplacement gradent u 1 (equal x 2 to the angle reducton as shown n Fgure 1.15).

1.4 Relatons between stress and stran 23 extenson (stran) stran hardenng Rubber onset of yeldng Steel stran weakenng force (tenson, stress) FIGURE 1.16 For some sold materals, stran grows n proporton to stress at low stress, but the lnearty s lost and stran ether ncreases more slowly wth ncreasng stress (for example, rubber), or ncreases more quckly (for example, steel). At hgh enough stress, the materal breaks. Ths s a schematc dagram: some solds break at extensonal strans as small as 10 3, and some forms of rubber mantan lnearty even for strans of order 1. All of the above examples n ths subsecton are elastc examples. Stress and stran go to zero together, f τ j = c jkl e kl, and there s no tme-dependence. Real materals can be qute dfferent, exhbtng vscous behavor, or stran hardenng, or a tendency to yeld at hgh values of appled stress (see Fgure 1.16). We take up ths subject next, wth emphass on solds. 1.4 Relatons between stress and stran A medum s sad to be elastc f t possesses a natural state (n whch strans and stresses are zero) to whch t wll revert when appled forces are removed. Under the nfluence of appled loads, stress and stran wll change together, and the relaton between them, called the consttutve relaton, s an mportant characterstc of the medum. Over 300 years ago, Robert Hooke summarzed the relatonshp today known as Hooke s Law between the extenson of a sprng, and the force actng to cause the extenson. He concluded expermentally that force extenson. The constant of proportonalty here s often called a modulus, M (say), and then force = M extenson. (Hooke s Law appeared orgnally as an anagram, cenosssttuv, of the Latn phrase ut tenso, sc vs meanng as the extenson, so the force. Some scentfc personaltes are very strange.) The modern generalzaton of Hooke s law s that each component of the stress tensor s a lnear combnaton of all components of the stran tensor,.e., that there exst constants c jkl such that τ j = c jpq e pq. (1.32) A materal that obeys the consttutve relaton (1.32) s sad to be lnearly elastc. The

24 Chapter 1 / ON VECTORS AND TENSORS, EXPRESSED IN CARTESIAN COORDINATES quanttes c jkl are components of a fourth-order tensor, and have the symmetres and It s also true from a thermodynamc argument that c jpq = c jpq (due to τ j = τ j ) (1.33) c jqp = c jpq (due to e qp = e pq ). (1.34) c pq j = c jpq, (1.35) for a materal n whch the energy of deformaton (assocated wth tensors τ and e) does not depend on the tme hstory of how the deformaton was acqured. The c jkl are ndependent of stran, whch s why they are sometmes called elastc constants, although they are varyng functons of poston n the Earth. In general, the symmetres (1.33), (1.34), and (1.35) reduce the number of ndependent components n c jkl from 81 to 21. There s consderable smplfcaton n the case of an sotropc medum, snce c must be sotropc. It can be shown that the most general sotropc fourth-order tensor, havng the symmetres of c, has the form c jkl = λδ j δ kl + µ(δ k δ jl + δ l δ jk ). (1.36) Ths nvolves only two ndependent constants, λ and µ, known as the Lamé modul. Substtutng from (1.36) nto (1.32), we see that the stress stran relaton becomes τ j = λδ j e kk + 2µe j (1.37) n sotropc elastc meda. If we consder only shearng stresses and shearng strans, usng (1.37) wth j, then τ j = 2µe j and µ here s just the rgdty we ntroduced earler, n (1.31). For many materals, the Lamé modul λ and µ (sometmes called Lamé constants) are approxmately equal. They can be used to generate a number of other constants that characterze the propertes of materal whch are subjected to partcular types of stran and stress, for example the stretched wre dscussed as tem () of Secton 1.3. For that example, (1.37) gves τ 11 = λ(e 11 + e 22 + e 33 ) + 2µe 11 0 = λ(e 11 + e 22 + e 33 ) + 2µe 22 0 = λ(e 11 + e 22 + e 33 ) + 2µe 33. (1.38) By comparng the second and thrd of these equatons, we see that e 22 = e 33 ; and then from these same two equatons t follows that λe 11 + 2(λ + µ)e 22 = 0. Posson s rato s defned as ν = shrnkng stran stretchng stran

Problems 25 n ths example of smple stretchng, so ν = e 22 λ = e 11 2(λ + µ) 1 4 (1.39) f λ and µ are approxmately equal (whch s the case for many common materals). Another mportant class of materals s that for whch shearng stresses lead to flow. Such materals are not elastc. We ntroduce the devatorc stress as the dfference between τ j and the average of the prncpal stresses, whch s 1 3 τ kk and whch we can symbolze by P. (In an nvscd or perfect flud, P wth ths defnton s smply the scalar pressure.) The vscosty of a materal that can flow, such as syrup (wth low vscosty) or the Earth s mantle (wth hgh vscosty, and sgnfcant flow occurrng only over many mllons of years), s the constant of proportonalty between the devatorc stress and twce the stran rate. Thus, the vscosty ν s gven by τ j + Pδ j = 2νė j. (1.40) A factor 2 appears here, for the same reason that a factor 2 appears n the defnton of rgdty (see (1.31)): both rgdty and vscosty are better apprecated for smple shearng motons, than for pure shearng. For a vscous materal, smple shearng due to applcaton of τ 21 as shown n Fgure 1.15 results n angle reducton of an orgnal rectangle at the rate u 1 x 2 = 2ė 12. Vscosty s shear stress dvded by the spatal gradent of partcle velocty. Suggestons for Further Readng Menke, Wllam, and Dallas Abbott. Geophyscal Theory, pp 41 50 for basc propertes of tensors, pp 237 252 for propertes of stran and stress tensors. New York: Columba Unversty Press, 1990. Problems 1.1 Consder the two sets of cartesan axes shown n Fgure 1.3, wth l j as the drecton cosne of the angle between Ox and Ox j. Thus, the unt vector x 1 expressed n Ox 1 x 2 x 3 coordnates has the components (l 11, l 21, l 31 ). a) Show n general that l j l k = δ jk and also that l j l kj = δ k. b) Hence show that ( l 1) j = l j. c) We know from (1.4) that a vector V havng components V n the Ox 1 x 2 x 3 system has components n the Ox 1 x 2 x 3 system gven by V j = l jv. Show that the orgnal components are gven n terms of the transformed components by V k = l kj V j. d) For two vectors a and b wth components defned n both of the two coordnate systems, show that a b = a j b j.

26 Chapter 1 / ON VECTORS AND TENSORS, EXPRESSED IN CARTESIAN COORDINATES e) We know from Secton 1.1.4 that components of A n the two coordnate systems are related by A jl = A kl j l kl. Show that A jl l pjl ql = A pq and hence that the reverse transformaton can be wrtten as A k = A jl l jl kl. f) The quadrc surface assocated wth A s A j x x j = c n the Ox 1 x 2 x 3 system. Show that ths same surface s A j x x j = c n the Ox 1 x 2 x 3 system. 1.2 Fnd the egenvalues and egenvectors of the followng matrces: A = ( 3 1 1 3 ), A = ( 5 2 2 5 ), A = ( 0.11 0.48 0.48 0.39 ), and ( ) 144p + 25q 60(p q) A = 60(p q) 25p + 144q (normalze the egenvectors, so that x (α) x (α) = 1). 1.3 Consder a stress tensor τ whose components n a partcular cartesan coordnate system are 3 1 0 τ = 1 3 0. 0 0 3 a) What are the prncpal stresses σ 1,σ 2,σ 3 n ths coordnate system (order them so that σ 1 <σ 2 <σ 3 )? And what are the correspondng three mutually orthogonal unt vector drectons ν 1,ν 2,ν 3? b) Evaluate the system of drecton cosnes l 11 l 12 l 13 l 21 l 22 l 23 l 31 l 32 l 33 where l j s the cosne of the angle between ˆx and ν j (ths matrx s not symmetrc). c) Show that n the present example the matrx of l j values has the property descrbed n Problem 1.1ab, namely that pre-multplcaton or post-multplcaton of l by ts transpose gves the dentty matrx, and hence that ( l 1) j = l j. d) Show n ths example that when the prncpal axes are taken as coordnates, the components of τ become σ 1 0 0 τ = 0 σ 2 0. 0 0 σ 3 [Note then that f the prncpal axes are taken as the coordnates OX 1 X 2 X 3, the equaton of the stress quadrc s smply σ 1 X1 2 + σ 2X2 2 + σ 3X3 2 = constant.] 1.4 Suppose that at a pont nsde a flud wth volume V the stress feld has the property that the normal stress across any surface through the pont s a constant, P

Problems 27 (ndependent of the orentaton of the surface). Show then that the correspondng stress tensor at a pont n a flud s sotropc, and has components τ j = Pδ j. 1.5 Show that e kk = u l,l = u = ncrease n volume orgnal volume for a small volume element of materal that s deformed. [Hnt: to prove the result, consder a small cube wth faces parallel to the prncpal axes of stran (the symmetry axes of the stran tensor). Alternatvely, use the physcal defnton of dvergence (flux out of a volume, per unt volume). The result tself s the reason that e kk and uare sometmes referred to as volumetrc stran. ] 1.6 The bulk modulus κ of an sotropc materal s defned as P κ = volumetrc stran when the materal s compressed by an all-round pressure P. a) Show that κ = λ + 2 3 µ where λ and µ are the Lamé modul. b) κ s sometmes gven other names, such the compressblty or the ncompressblty. Whch of these two names makes more sense? 1.7 Use the alternatng tensor ε jk dscussed n (1.6) and (1.7) to show for vectors a, b, and c, that Usng ( 2 u) = 2 (u ), show also that (a b) c = (a c)b (b c)a. 2 u = ( u) ( u). [Ths last result, obtaned here by usng cartesan components, s essentally a defnton of 2 u, that can be appled to non-cartesan components.] 1.8 Usng u, j to denote u, show that ε jk ε jkm u m,l δx k = (u, j u j, )δx j and hence u j that (u, j u j, )δx j = (curl u δx). 1.9 Show that Posson s rato (ν) s 1 2 for a materal that s ncompressble. 1.10 Young s modulus (E)sdefned as the rato of stretchng stress to stretchng stran n the example () of Secton 1.3 (see also (1.39)). Show that µ(3λ + 2µ) E = λ + µ. 1.11 For an sotropc elastc materal, the stress s gven n terms of strans by (1.37). Show that stran s then gven n terms of the stresses by λ 2µe j = 3λ + 2µ τ kkδ j + τ j.