547curvilinear

1 Curvilinear Coordinates 1.1 Basis vectors The usual basis vectors are denoted by i,j,kand are as the following picture describes. 6 ¡¡ª-k ji The vectors, i,j,k,arefixed. If vis a vector, there are unique scalars called components such that v=v1i+v2j+v3k .T h i si sw h a tw em e a nw h e nw es a y i,j,kis a basis. Now suppose e1,e2,e3are three vectors which satisfy e1×e2·e36=0. Recall this means the volume of the box spanned by the three vectors is not zero. £ £ ££°»»: @@R@@»» @@»»££££»» ££££ e1e3 e2 Suppose e1,e2,e3are as just described. Does it follow that they form a basis? We show this is the case. Thus there are unique scalars, v1,v2,andv3such that v=viei. This is the content of the following theorem. Theorem 1 Ife1,e2,e3are three vectors, then they form a basis if and only if e1×e2·e36=0. Proof: Suppose first the above condition holds. Let i1≡i,i2≡j,i3≡kand suppose v=ujij.Therefore, u1,u2,u3are the components of the vector vwith respect to the usual basis vectors. Also let ei=aj iij thus writing each eiin terms of the vectors ij.Then from the de finition of the box product in terms of the usual basis vectors, we see 06=e1×e2·e3=d e t³ aj i´ ≡det a1 1a21a31 a1 2a22a32 a13a23a33 =d e t a1 1a12a13 a2 1a22a23 a31a32a33  (1) and we want to show there exists a unique solution to v=ujij=viei=aj iviij. (2) In other words, we want a unique solution to the system of equations uj=aj ivi,j=1,2,3. 1 This is in the form u=Av (3) where Ais matrix which, by 1, has non zero determinant. If the box product is equal to zero, then the system which needs to be solved is of the form 3 where det ( A)=0 . Therefore, Ahas an eigenvector, v=(v1,v2,v3)Twhich corresponds to the eigen value λ=0.Therefore, 2 shows 0 has more than one set of components with respect to the vectors e1,e2,e3and this set of vectors is therefore not a basis. This gives a simple geometric condition which determines whether a list of three vectors forms a basis in R3.One simply takes the box product. If the box product is not equal to zero, then the vectors form a basis. If not, the list of three vectors does not form a basis. This condition generalizes to Rnas follows. If ei=aj iij,then {ei}n i=1forms a basis if and only if det³ aj i´ 6=0. These vectors may or may not be orthonormal. In any case, it is convenient to de fine something called the dual basis. Definition 2 Let{ei}n i=1form a basis for Rn.Then© eiªn i=1is called the dual basis if ei·ej=δi j≡½1ifi=j 0ifi6=j. (4) Theorem 3 If{ei}n i=1is a basis then© eiªn i=1is also a basis provided 4 holds. Proof: Suppose v=viei. (5) Then taking the dot product of both sides of 5 with ej,yields vj=v·ej. (6) Thus there is at most one choice of scalars vjsuch that v=vjejand it is given by 6. ¡ v−v·ejej¢ ·ek=0 and so, since {ei}n i=1is a basis, ¡ v−v·ejej¢ ·w=0 for all vectors, w.It follows v−v·ejej=0and this shows© eiªn i=1is a basis. In the above argument we obtained formulas for the components of a vector v,vi,with respect to the dual basis, found to be vj=v·ej.I nt h es a m ew a y ,w em a y find the components of a vector with respect to the basis {ei}n i=1. Letvbe any vector and let v=vjej. (7) Then using 4 and taking the dot product of both sides of 7 with eiwe see vi=ei·v. Does there exist a dual basis and is it uniquely determined? Theorem 4 If{ei}n i=1is a basis for Rn,then there exists a unique dual basis,© ejªn j=1satisfying ej·ei=δj i. 2 Proof: First we show the dual basis is unique. Suppose© fjªn j=1is another set of vectors which satis fiesfj·ei=δj i. Then fj=fj·eiei=δj iei=ej. Note that from the de finition, the dual basis to {ij}n j=1is just ij=ij.Letting ei=aj iij where the vectors, {ij}n j=1are the standard basis vectors, it follows that since the eiform a basis that the matrix whose ij th entry is aj iis an invertible matrix. Letting ek=bk rir, we need to choose bk rsuch that bk rir·aj iij=bk raj iδr j=bk jaj i=δk i. But this is nothing more than the matrix equation for Bwhich is of the form AB=I where A=³ aj i´ and has an inverse. There exists a unique solution to this equation given by B=A−1and this proves the existence of the dual basis. Summarizing what has been shown so far, we know that {ei}n i=1is a basis for Rnif and only if when ei=aj iij, det³ aj i´ 6=0. (8) If{ei}n i=1is a basis, then there exists a unique dual basis,© ejªn j=1satisfying ej·ei=δj i, (9) and that if vis any vector, v=vjej,v=vjej. (10) The components of vwhich have the index on the top are called the contravariant components of the vector while the components which have the index on the bottom ar e called the covariant components. In general vi6=vj!W e also have formulae for these components in terms of the dot product. vj=v·ej,vj=v·ej. (11) We de finegij≡ei·ejandgij≡ei·ej.The next theorem describes the process of raising or lowering an index. Theorem 5 The following hold. gijej=ei,gijej=ei, (12) gijvj=vi,gijvj=vi, (13) gijgjk=δi k, (14) det (gij)>0,det¡ gij¢ >0. (15) 3 Proof: First, ei=ei·ejej=gijej by 10 and 11. Similarly, by 10 and 11, ei=ei·ejej=gijej. This veri fies 12. To verify 13, vi=ei·v=gijej·v=gijvj. The proof of the remaining formula in 13 is similar. To verify 14, gijgjk=ei·ejej·ek=¡¡ ei·ej¢ ej¢ ·ek=ei·ek=δi k. This shows the two determinants in 15 are non zero because the two matrices are inverses of each other. It only remains to verify that one of these is greater than zero. Letting ei=aj iij=bi jij,we see that since ij=ij,aj i=bi j. Therefore, ei·ej=ar iir·bj kik=ar ibj kδk r=ak ibj k=ak iakj. It follows that for Gthe matrix whose ij th entry is ei·ej,G=AATwhere the ik th entry of Aisak i.Therefore, det (G)=d e t ( A)d e t¡ AT¢ =d e t ( A)2>0.It follows from 14 that if His the matrix whose ij th entry is gij,then GH=Iand so H=G−1and det (G)d e t¡ G−1¢ =d e t¡ gij¢ det (G)=1 . Therefore, det¡ G−1¢ >0 also. This proves the theorem. Definition 6 The matrix (gij)=Gis called the metric tensor. 1.2 Exercises 1. Let e1=i+j,e2=i−j,e3=j+k.Find e1,e2,e3,(gij),¡ gij¢ .Ifv=i+2j+k,findviandvj,the con- travariant and covariant components of the vector. 2. Let e1=2i+j,e2=i−2j,e3=k.Finde1,e2,e3,(gij),¡ gij¢ .Ifv=2i−2j+k,findviandvj,the contravari- ant and covariant components of the vector. 3. Suppose e1,e2,e3have the property that ei·ej= 0 whenever i6=j.Show the same is true of the dual basis and that in fact, eiis a multiple of ei. 4. Let e1,···,e3be a basis for Rnand let v=viei=viei,w=wjej=wjejbe two vectors. Show v·w=gijviwj=gijviwj. 5. Show if {ei}3 i=1is a basis in R3 e1=e2×e3 e2×e3·e1,e2=e1×e2 e1×e3·e2,e3=e1×e2 e1×e2·e3. 4 6. Let {ei}n i=1be a basis and de fine e∗ i≡ei |ei|,e∗i≡ei|ei|. Show e∗i·e∗ j=δi j. 7. Ifvis a vector, v∗ iandv∗i,are de fined by v≡v∗ ie∗i≡v∗ie∗ i. These are called the physical components of v.Show v∗ i=vi |ei|,v∗i=vi|ei|( No summation on i). 1.3 Curvilinear Coordinates With the algebraic preparation of the last section, we are ready to consider curvilinear coordinates. Let D⊆Rnbe an open set and let M:D→Rnsatisfy MisC2, (16) Mis one to one. (17) Letting x∈D,we can write M(x)=Mk(x)ik where, as usual, ikare the standard basis vectors for Rn,ikbeing the vector in Rnwhich has a one in the kth coordinate and a 0 in every other spot. For a fixedx∈D,we can consider the curves, t→M(x+tik) fort∈I,some open interval containing 0 .Then for the point x,we let ek≡∂M ∂xk(x)≡d dt(M(x+tik))|t=0. We will denote this vector as ek(x) to emphasize its dependence on x.The following picture illustrates the situation inR3. ¡¡¡µ ¢¢¸ HHHHYe1 e2 e3 t→M(x1 0,x20,t)t→M(t, x2 0,x30)t→M(x1 0,t ,x30) We want {ek}n k=1to be a basis. Thus we need detµ∂Mi ∂xk¶ 6=0. (18) 5 Let yi=Mi(x)i=1,···,n (19) so that the yiare the usual coordinates with respect to the usual basis vectors {ik}n k=1of the point M(x).Letting x≡¡ x1,···,xn¢ ,it follows from the inverse function theorem of advanced calculus that M(D)i so p e n ,a n dt h a t 18, 16, and 17 imply the equations 19 de fine each xias aC2function of y≡¡ y1,···,yn¢T.Thus, abusing notation slightly, the equations 19 are equivalent to xi=xi(y),i=1,···,n where xiis aC2function. Thus ∇xk(y)=∂xk(y) ∂yjij. Then ∇xk(y)·ej=∂xk ∂ysis·∂yr ∂xjir=∂xk ∂ys∂ys ∂xj=δk j by the chain rule. Therefore, the dual basis is given by ek(x)=∇xk(y). (20) Notice that it might be hard or even impossible to solve algebraically for xiin terms of the yj.Thus the straight forward approach to finding ekby 20 might be impossible. Also, this approach leads to an expression in terms of the ycoordinates rather than the desired xcoordinates. Therefore, it is expedient to use another method to obtain these vectors. The vectors, ek(x) may always be found by using formula 12 and the result is in terms of the curvilinear coordinates, x. We illustrate in the following example. Example 7 D≡(0,∞)×(0,π)×(0,2π)and  y1 y2 y3 = x1sin¡ x2¢ cos¡ x3¢ x1sin¡ x2¢ sin¡ x3¢ x1cos¡ x2¢  (We usually write this as  x y z = ρsin (φ)c o s(θ) ρsin (φ)s i n(θ) ρcos (φ)  where (ρ,φ,θ)are the spherical coordinates. We are calling them x1,x2,andx3to preserve the notation just dis- cussed.) Thus e1(x)=s i n¡ x2¢ cos¡ x3¢ i1+s i n¡ x2¢ sin¡ x3¢ i2+c o s¡ x2¢ i3, e2(x)=x1cos¡ x2¢ cos¡ x3¢ i1 +x1cos¡ x2¢ sin¡ x3¢ i2−x1sin¡ x2¢ i3, e3(x)=−x1sin¡ x2¢ sin¡ x3¢ i1+x1sin¡ x2¢ cos¡ x3¢ i2+0i3. 6 It follows the metric tensor is G= 10 0 0¡ x1¢20 00¡ x1¢2sin2¡ x2¢ =(gij)=(ei·ej). (21) Therefore, by Theorem 5 G−1=¡ gij¢ =¡ ei,ej¢ = 10 0 0¡ x1¢−20 00¡ x1¢−2sin−2¡ x2¢ . To obtain the dual basis, use Theorem 5 to write e1=g1jej=e1 e2=g2jej=¡ x1¢−2e2 e3=g3jej=¡ x1¢−2sin−2¡ x2¢ e3. It is natural to ask if we can get a transformation Msuch that ∂M ∂x1=i=i1,∂M ∂x2=j=i2,∂M ∂x3=k=i3. (22) T h ea n s w e ri st h a tw ec a n .L e t M¡ x1,x2,x3¢ ≡x1i+x2j+x3k. Then 22 holds for this transformation. 1.4 Exercises 1. Let y1 y2 y3 = x1+2x2 x2+x3 x1−2x2  where the yiare the rectangular coordinates of the point. Find ei,ei,i=1,2,3,and find (gij)(x)a n d¡ gij(x)¢ . 2. Let y=y(x,t)w h e r e tsigni fies time and x∈U⊆RmforUan open set, while y∈Rnand suppose xis a function of t.Physically, this corresponds to an object moving over a surface in Rnwhich may be changing as a function of t.The point y=y(x(t),t)i st h ep o i n ti n Rncorresponding to t.F o re x a m p l e ,c o n s i d e rt h e pendulum D D D D DD•ml θ in which n=2,lisfixed and y1=lsinθ,y2=l−lcosθ.Thus, in this simple example, m=1.Iflwere changing in a known way with respect to t,then this would be of the form y=y(x,t).The kinetic energy is 7 defined as T≡1 2m˙y·˙y (∗) where the dot on the top signi fies differentiation with respect to t.Show ∂T ∂˙xk=m˙y·∂y ∂xk. Hint: First show ˙y=∂y ∂xj˙xj+∂y ∂t(∗∗) and so ∂˙y ∂˙xj=∂y ∂xj. 3.↑Show d dtµ∂T ∂˙xk¶ =m¨y·∂y ∂xk+m˙y·∂2y ∂xk∂xr˙xr+m˙y·∂2y ∂t∂xk. 4.↑Show ∂T ∂xk=m˙y·µ∂2y ∂xr∂xk˙xr+∂2y ∂t∂xk¶ . Hint: Use∗and∗∗. 5.↑Now show from Newton’s second law ( mass times acceleration equals force ) that for Fthe force, d dtµ∂T ∂˙xk¶ −∂T ∂xk=m¨y·∂y ∂xk=F·∂y ∂xk. (∗∗∗) 6.↑In the example of the simple pendulum above, y=µlsinθ l−lcosθ¶ =lsinθi+(l−lcosθ)j. Use∗∗∗ tofind a differential equation which describes the vibrations of the pendulum in terms of θ.First write the kinetic energy and then consider the force acting on the mass which is −mgj. 7. The above problem is fairly easy to do without th e formalism developed. Now consider the case where x=(ρ,θ,φ),spherical coordinates, and write di fferential equations for ρ,θ,andφto describe the motion of an object in terms of these coordinates given a force, F. 8. Suppose the pendulum is not assumed to vibrate in a p lane. Let it be suspended at the origin and consider spherical coordinates. Find di fferential equations for θandφ. 8 9. If there are many masses, mα,α=1,···,R ,the kinetic energy is the sum of the k inetic energies of the individual masses. Thus, T≡1 2RX α=1mα|˙yα|2. Generalize the above problems to show that, assuming yα=yα(x,t), d dtµ∂T ∂˙xk¶ −∂T ∂xk=RX α=1Fα·∂yα ∂xk where Fαis the force acting on mα. 10. Discuss the equivalence of these formulae with Newt on’s second law, force equals mass times acceleration. What is gained from the above so called Lagrangian formalism? 11. The double pendulum has two masses instead of only one. D D D D DD•m1l1 θ D D D D DD•m2l2 φ Write differential equations for θandφto describe the motion of the double pendulum. 1.5 Transformation of coordinates. The scalars© xiª are called cuvilinear coordinates. Note they can be used to identify a point in Rnandx=¡ x1,···,xn¢ is a point in Rn.The basis vectors associated with this particula r set of curvilinear coordinates at a point identi fied byxare denoted by ei(x) and the dual basis vectors at this point are denoted by ej(x). What if other curvilinear coordinates are used? How do we write ek(x) in terms of the vectors, ej(z)w h e r e zis some other type of curvilinear coordinates? We consider this topic next. Consider the following picture in which Uis an open set in Rn,D ,andbDare open sets in Rn,andM,NareC2 mappings which are one to one from DandbDrespectively. We will suppose that a point in Uis identi fied by the curvilinear coordinates xinDandzinbD. - ¾U D bDMN (x1,x2,x3)( z1,z2,z3) Thus M(x)=N(z).Now by the chain rule, ei(z)≡∂N ∂zi=∂M ∂xj∂xj ∂zi=∂xj ∂ziej(x). (23) 9 We de fine the covariant and contravariant coordinates for the various curvilinear coordinates in the obvious way. Thus, v=vi(x)ei(x)=vi(x)ei(x)=vj(z)ej(z)=vj(z)ej(z). Then we have the following theorem about transforming the vectors and coordinates. Theorem 8 The following transformation rules hold for pairs of curvilinear coordinates. vi(z)=∂xj ∂zivj(x),vi(z)=∂zi ∂xjvj(x), (24) ei(z)=∂xj ∂ziej(x),ei(z)=∂zi ∂xjej(x), (25) gij(z)=∂xr ∂zi∂xs ∂zjgrs(x),gij(z)=∂zi ∂xr∂zj ∂xsgrs(x). (26) Proof: We already have shown the first part of 25 in 23. Then, from 23, ei(z)= ei(z)·ej(x)ej(x)=ei(z)·∂zk ∂xjek(z)ej(x) =δi k∂zk ∂xjej(x)=∂zi ∂xjej(x) and this proves the second part of 25. Now to show 24, vi(z)=v·ei(z)=v·∂xj ∂ziej(x)=∂xj ∂zivj(x) and vi(z)=v·ei(z)=v·∂zi ∂xjej(x)=∂zi ∂xjvj(x). To verify 26, gij(z)=ei(z)·ej(z)=er(x)∂xr ∂zi·es(x)∂xs ∂zj=grs(x)∂xr ∂zi∂xs ∂zj. This proves the theorem. We will denote by ythe curvilinear coordinates with the property that ek(y)=ik=ek(y). 1.6 Differentiation and Christo ffel Symbols LetF:U→Rnbe differentiable. We call Fav e c t o r field and it is used to model force, velocity, acceleration, or any other vector quantity which may change from point to point in U.Then ∂F(x) ∂xj is a vector and so there exist scalars, Fi ,j(x)a n d Fi,j(x)s u c ht h a t ∂F(x) ∂xj=Fi ,j(x)ei(x)=Fi,j(x)ej(x). (27) We will see how these scalars transform when the coordinates are changed. 10 Theorem 9 Ifxandzare curvilinear coordinates, Fr ,s(x)=Fi ,j(z)∂xr ∂zi∂zj ∂xs,Fr,s(x)∂xr ∂zi∂xs ∂zj=Fi,j(z). (28) Proof: Fr ,s(x)er(x)≡∂F(x) ∂xs=∂F(z) ∂zj∂zj ∂xs≡ Fi ,j(z)ei(z)∂zj ∂xs=Fi ,j(z)∂xr ∂zi∂zj ∂xser(x) which shows the first formula of 27. To show the other formula, Fi,j(z)ei(z)≡∂F(z) ∂zj=∂F(x) ∂xs∂xs ∂zj≡ Fr,s(x)er(x)∂xs ∂zj=Fr,s(x)∂xr ∂zi∂xs ∂zjei(z), and this shows the second formula for transforming these scalars. NowF(x)=Fi(x)ei(x) and so by the product rule, ∂F ∂xj=∂Fi ∂xjei(x)+Fi(x)∂ei(x) ∂xj. Now∂ei(x) ∂xjis a vector and so there exist scalars,©k ijª such that ∂ei(x) ∂xj=½k ij¾ ek(x). Therefore, ∂F ∂xj=∂Fk ∂xjek(x)+Fi(x)½k ij¾ ek(x) which shows Fk ,j(x)=∂Fk ∂xj+Fi(x)½k ij¾ . This is sometimes called the covariant derivative. These scalars are called the Christo ffel symbols of the second kind. The next theorem is devoted to properties of these Christo ffel symbols. Before stating the theo rem, we recall that the mapping, Mwhich de fines the curvilinear coordinates is C2.The reason for this is that we want to be able to as sert the mixed partial derivatives are equal. Theorem 10 The Christo ffel symbols of the second kind satisfy the following ∂ei(x) ∂xj=½k ij¾ ek(x), (29) ∂ei(x) ∂xj=−½i kj¾ ek(x), (30) ½k ij¾ =½k ji¾ , (31) ½m ik¾ =gjm 2·∂gij ∂xk+∂gkj ∂xi−∂gik ∂xj¸ . (32) 11 Proof: Formula 29 is the de finition of the Christo ffel symbols. We verify 30 next. To do so, note ei(x)·ek(x)=δi k. Then from the product rule, ∂ei(x) ∂xj·ek(x)+ei(x)·∂ek(x) ∂xj=0. Now from the de finition, ∂ei(x) ∂xj·ek(x)=−ei(x)·½r kj¾ er(x)=−½i kj¾ . But also, ∂ei(x) ∂xj=∂ei(x) ∂xj·ek(x)ek(x)=−½i kj¾ ek(x). This veri fies 30. Letting∂M(x) ∂xj=ej(x),it follows from equality of mixed partial derivatives, ½k ij¾ ek(x)=∂ei ∂xj≡∂2M ∂xj∂xi=∂2M ∂xi∂xj=∂ej ∂xi=½k ji¾ ek(x), which shows 31. It remains to show 32. ∂gij ∂xk=∂ei ∂xk·ej+ei·∂ej ∂xk=½r ik¾ er·ej+ei·er½r jk¾ . Therefore, ∂gij ∂xk=½r ik¾ grj+½r jk¾ gri. (33) Switching iandkwhile remembering 31 yields ∂gkj ∂xi=½r ik¾ grj+½r ji¾ grk. (34) Now switching jandkin 33, we obtain, ∂gik ∂xj=½r ij¾ grk+½r jk¾ gri. (35) Adding 33 to 34 and subtracting 35 yields ∂gij ∂xk+∂gkj ∂xi−∂gik ∂xj=2½r ik¾ grj. Now multiplying both sides by gjmand using the fact shown earlier in Theorem 5 that grjgjm=δm r, we obtain 2½m ik¾ =gjmµ∂gij ∂xk+∂gkj ∂xi−∂gik ∂xj¶ which proves 32. This is a very interesting formula because it shows the Christo ffel symbols are completely determined by the metric tensor and its derivatives. 12 1.7 Gradients and divergence In this section we express the gradient and the divergence of a vector field in general curvilinear coordinates. As before, ywill denote the standard coordinates with respect to the usual basis vectors. Thus N(y)≡ykik,ek(y)=ik=ek(y). Letφ:U→Rbe a differentiable scalar function, sometimes called a “scalar field” in this subject. We write φ(x) to denote the value of φat the point whose coordinates are x.In general, we follow this practice for any field, vector or scalar. Thus F(x) is the value of a vector fie l da tt h ep o i n to f Udetermined by the coordinates x.If we are using the standard coordinates, we know what we mean by the gradient of φ. It is given by the following formula. ∇φ(y)=∂φ(y) ∂ykek(y). Therefore, using the chain rule, if the coordinates of the point of Uare given as x, ∇φ(x)=∇φ(y) =∂φ(x) ∂xr∂xr ∂yk∂yk ∂xses(x)=∂φ(x) ∂xrδr ses(x)=∂φ(x) ∂xrer(x). This shows the covariant components of ∇φ(x)a r e (∇φ(x))r=∂φ(x) ∂xr. (36) Tofind the contravariant components, we “raise the index” in the usual way. Thus (∇φ(x))r=grk(x)(∇φ(x))k=grk(x)∂φ(x) ∂xk. (37) What about the divergence of a vector field? The divergence of a vector field,Fdefined on Uis a scalar field, div (F) which we know from calculus to be ∂Fk ∂yk(y)=Fk ,k(y) in terms of the usual coordinates y.The reason the above equation holds in this case is that ek(y)i sac o n s t a n ta n d so the Christo ffel symbols are zero. We want an expression for the d ivergence in an arbitrary coordinate system. From Theorem 9, Fi ,j(y)=Fr ,s(x)∂xs ∂yj∂yi ∂xr =µ∂Fr(x) ∂xs+Fk(x)½r ks¾ (x)¶∂xs ∂yj∂yi ∂xr. Letting j=iyields div (F)=µ∂Fr(x) ∂xs+Fk(x)½r ks¾ (x)¶∂xs ∂yi∂yi ∂xr =µ∂Fr(x) ∂xs+Fk(x)½r ks¾ (x)¶ δs r =µ∂Fr(x) ∂xr+Fk(x)½r kr¾ (x)¶ . (38) 13 We will simplify©r krª using the description of it in Theorem 10. Thus, from this theorem, ½r rk¾ =gjr 2·∂grj ∂xk+∂gkj ∂xr−∂grk ∂xj¸ Now considergjr 2times the last two terms in [ ·].Relabeling the indices randjin the second term implies gjr 2∂gkj ∂xr−gjr 2∂grk ∂xj=gjr 2∂gkj ∂xr−grj 2∂gjk ∂xr=0. Therefore, ½r rk¾ =gjr 2∂grj ∂xk. (39) Now recall g≡det (gij)=d e t ( G)>0 from Theorem 5. Also from the formula for the inverse of a matrix and this theorem, gjr=Arj(detG)−1=Ajr(detG)−1 where Arjis the rjth cofactor of the matrix ( gij).Also recall that g=nX r=1grjArjno sum on j. Therefore, gis a function of the variables {grj}and ∂g ∂grj=Arj. From 39, ½r rk¾ =gjr 2∂grj ∂xk=1 2g∂grj ∂xkAjr=1 2g∂g ∂grj∂grj ∂xk=1 2g∂g ∂xk and so from 38, div (F)=∂Fk(x) ∂xk+ +Fk(x)1 2g(x)∂g(x) ∂xk=1p g(x)∂ ∂xi³ Fi(x)p g(x)´ . (40) This is our formula for the divergence of a vector field in general curvilinear coordinates. The Laplacian of a scalar field is nothing more than the divergence of the gradient. In symbols, ∆φ≡∇ ·∇φ From 40 and 37 it follows ∆φ(x)=1p g(x)∂ ∂xiµ gik(x)∂φ(x) ∂xkp g(x)¶ . (41) We summarize the conclusions of this section in the following theorem. 14 Theorem 11 The following formulas hold for the gradient, divergence and Laplacian in general curvilinear coordi- nates. (∇φ(x))r=∂φ(x) ∂xr, (42) (∇φ(x))r=grk(x)∂φ(x) ∂xk, (43) div (F)=1p g(x)∂ ∂xi³ Fi(x)p g(x)´ , (44) ∆φ(x)=1p g(x)∂ ∂xiµ gik(x)∂φ(x) ∂xkp g(x)¶ . (45) 1.8 Exercises 1. Let y1=x1+2x2,y2=x2+3x3,y3=x1+x3.Let F(x)=x1e1(x)+x2e2(x)+¡ x3¢2e(x). Find div ( F)(x). 2. For the coordinates of the preceding problem, and φas c a l a r field, find (∇φ(x))3 in terms of the partial derivatives of φtaken with respect to the variables xi. 3. Let y1=7x1+2x2,y2=x2+3x3,y3=x1+x3.Letφbe a scalar field. Find ∇2φ(x). 4. Derive ∇2uin cylindrical coordinates, r,θ,z,where uis a scalar field on R3. x=rcosθ,y=rsinθ,z=z. 5.↑Find all solutions to ∇2u= 0 which depend only on rwhere r≡p x2+y2. 6. Let ube a scalar field on R3.Find all solutions to ∇2u= 0 which depend only on ρ≡p x2+y2+z2. 7. The temperature, u,in a solid satis fies∇2u= 0 after a long time. Suppose in a long pipe of inner radius 9 and outer radius 10 the exterior surface is held at 100◦while the inner surface is held at 200◦find the temperature in the solid part of the pipe. 8. Show ½l ij¾ =∂ei ∂xj·el. Find the Christo ffel symbols of the second kind for spherical coordinates in which x1=φ,x2=θ,a n d x3=ρ. Do the same for cylindrical coordinates letting x1=r,x2=θ,x3=z. 15 9. Show velocity can be expressed as v=vi(x)ei(x),where vi(x)=∂ri ∂xjdxj dt−rp(x)½p ik¾dxk dt andri(x) are the covariant components of the displacement vector, r=ri(x)ei(x). 10.↑Using problem 8 and 9, show the covariant components of velocity in spherical coordinates are v1=ρ2dφ dt,v2=ρ2sin2(φ)dθ dt,v3=dρ dt. Hint: First observe that if ris the position vector from the origin, then r=ρe3sor1=0= r2,a n d r3=ρ. Now use 9. 1.9 Curl and cross products In this section we consider the curl and cross pr oduct in general curvilinear coordinates in R3.W e w i l l a l w a y s a s s u m e that for xa set of curvilinear coordinates, detµ∂yi ∂xj¶ >0 (46) Where the yiare the usual coordinates in which ek(y)=ik. Theorem 12 Let 46 hold. Then detµ∂yi ∂xj¶ =p g(x) (47) and detµ∂xi ∂yj¶ =1p g(x). (48) Proof: ei(x)=∂yk ∂xiik and so gij(x)=∂yk ∂xiik·∂yl ∂xjil=∂yk ∂xi∂yk ∂xj. Therefore, g=d e t( gij(x)) =³ det³ ∂yk ∂xi´´2 .By 46,√g=d e t³ ∂yk ∂xi´ as claimed. Now ∂yk ∂xi∂xi ∂yr=δk r and so detµ∂xi ∂yr¶ =1p g(x). 16 This proves the theorem. To get the curl and cross product in curvilinear coordinates, let ²ijkbe the usual permutation symbol. Thus, ²123=1 and when any two indices in ²ijkare switched, the sign changes. Thus ²132=−1,²312=1,etc. Now de fine εijk(x)≡²ijk 1p g(x). Then for xandzsatisfying 46, εijk(x)∂zr ∂xi∂zs ∂xj∂zt ∂xk=²ijkdetµ∂xp ∂yq¶∂zr ∂xi∂zs ∂xj∂zt ∂xk =²rstdetµ∂xp ∂yq¶ detµ∂zi ∂xk¶ =²rstdet (MN) where Nis the matrix whose pq th entry is∂xp ∂yqandMis the matrix whose ik th entry is∂zi ∂xk. Therefore, from the definition of matrix multiplication and the chain rule, this equals =²rstdetµ∂zi ∂yp¶ ≡εrst(z) from the above discussion. Nowεijk(y)=²ijkand for a vector field,F, curl (F)≡εijk(y)Fk,j(y)ei(y). Therefore, since we know how everything transforms assuming 46, it is routine to write this in terms of x. curl (F)=εrst(x)∂yi ∂xr∂yj ∂xs∂yk ∂xtFp,q(x)∂xp ∂yk∂xq ∂yjem(x)∂xm ∂yi =εrst(x)δm rδqsδp tFp,q(x)em(x)=εmqp(x)Fp,q(x)em(x). (49) More simpli fication is possible. Recalling the de finition of Fp,q(x), ∂F ∂xq≡Fp,q(x)ep(x)=∂ ∂xq[Fp(x)ep(x)] =∂Fp(x) ∂xqep(x)+Fp(x)∂ep ∂xq=∂Fp(x) ∂xqep(x)−Fr(x)½r pq¾ ep(x) by Theorem 10. Therefore, Fp,q(x)=∂Fp(x) ∂xq−Fr(x)½r pq¾ 17 and so curl (F)=εmqp(x)∂Fp(x) ∂xqem(x)−εmqp(x)Fr(x)½r pq¾ em(x). However, because©r pqª =©r qpª ,the second term in this expression equals 0. To see this, εmqp(x)½r pq¾ =εmpq(x)½r qp¾ =−εmqp(x)½r pq¾ . Therefore, by 49, curl(F)=εmqp(x)∂Fp(x) ∂xqem(x). (50) What about the cross product of two vector fields? Let FandGbe two vector fie l d s . T h e ni nt e r m so fs t a n d a r d coordinates, y, F×G=εijk(y)Fj(y)Gk(y)ei(y) =εrst(x)∂yi ∂xr∂yj ∂xs∂yk ∂xtFp(x)∂xp ∂yjGq(x)∂xq ∂ykel(x)∂xl ∂yi =εrst(x)δp sδq tδl rFp(x)Gq(x)el(x)=εlpq(x)Fp(x)Gq(x)el(x). (51) We summarize these results in the following theorem. Theorem 13 Suppose xis a system of curvilinear coordinates in R3such that detµ∂yi ∂xj¶ >0. Let εijk(x)≡²ijk 1p g(x). Then the following formulas for curl and cross product hold in this system of coordinates. curl (F)=εmqp(x)∂Fp(x) ∂xqem(x), and F×G=εlpq(x)Fp(x)Gq(x)el(x). 18