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C:\Book\Booktex\C1s3.DVI 65 1.3 SPECIAL TENSORS § Knowing how tensors are defined and recognizing a tensor when it pops up in front of you are two different things. Some quantities, which are tensors, frequently arise in applied problems and you should learn to recognize these special tensors when they occur. In this section some important tensor quantities are defined. We also consider how these special tensors can in turn be used to define other tensors. Metric Tensor Define yi,i=1,...,N as independent coordinates in an N dimensional orthogonal Cartesian coordinate system. The distance squared between two points yi and yi + dyi,i=1,...,N is defined by the expression ds2 = dymdym =(dy1)2 +(dy2)2 + +(dyN )2. (1.3.1) ··· Assume that the coordinates yi are related to a set of independent generalized coordinates xi,i=1,...,N by a set of transformation equations yi = yi(x1,x2,...,xN ),i=1,...,N. (1.3.2) To emphasize that each yi depends upon the x coordinates we sometimes use the notation yi = yi(x), for i =1,...,N. The differential of each coordinate can be written as ∂ym dym = dxj ,m=1,...,N, (1.3.3) ∂xj and consequently in the x-generalized coordinates the distance squared, found from the equation (1.3.1), becomes a quadratic form. Substituting equation (1.3.3) into equation (1.3.1) we find ∂ym ∂ym ds2 = dxidxj = g dxidxj (1.3.4) ∂xi ∂xj ij where ∂ym ∂ym g = ,i,j=1,...,N (1.3.5) ij ∂xi ∂xj i are called the metrices of the space defined by the coordinates x ,i=1,...,N. Here the gij are functions of the x coordinates and is sometimes written as gij = gij (x). Further, the metrices gij are symmetric in the indices i and j so that gij = gji for all values of i and j over the range of the indices. If we transform to another coordinate system, say xi,i=1,...,N, then the element of arc length squared is expressed in terms 2 i j of the barred coordinates and ds = gij dx dx , where gij = gij (x) is a function of the barred coordinates. The following example demonstrates that these metrices are second order covariant tensors. 66 EXAMPLE 1.3-1. Show the metric components gij are covariant tensors of the second order. Solution: In a coordinate system xi,i=1,...,N the element of arc length squared is 2 i j ds = gij dx dx (1.3.6) while in a coordinate system xi,i=1,...,N the element of arc length squared is represented in the form 2 m n ds = gmndx dx . (1.3.7) The element of arc length squared is to be an invariant and so we require that m n i j gmndx dx = gij dx dx (1.3.8) Here it is assumed that there exists a coordinate transformation of the form defined by equation (1.2.30) together with an inverse transformation, as in equation (1.2.32), which relates the barred and unbarred coordinates. In general, if xi = xi(x), then for i =1,...,N we have ∂xi ∂xj dxi = dxm and dxj = dxn (1.3.9) ∂xm ∂xn Substituting these differentials in equation (1.3.8) gives us the result i j i j m n ∂x ∂x m n ∂x ∂x m n g dx dx = gij dx dx or g gij dx dx =0 mn ∂xm ∂xn mn − ∂xm ∂xn ∂xi ∂xj For arbitrary changes in dxm this equation implies that g = g and consequently g transforms mn ij ∂xm ∂xn ij as a second order absolute covariant tensor. EXAMPLE 1.3-2. (Curvilinear coordinates) Consider a set of general transformation equations from rectangular coordinates (x, y, z) to curvilinear coordinates (u, v, w). These transformation equations and the corresponding inverse transformations are represented x = x(u, v, w) u = u(x, y, z) y = y(u, v, w) v = v(x, y, z) (1.3.10) z = z(u, v, w). w = w(x, y, z) Here y1 = x, y2 = y, y3 = z and x1 = u, x2 = v, x3 = w are the Cartesian and generalized coordinates and N =3. The intersection of the coordinate surfaces u = c1,v = c2 and w = c3 define coordinate curves of the curvilinear coordinate system. The substitution of the given transformation equations (1.3.10) into the position vector ~r = x e1 + y e2 + z e3 produces the position vector which is a function of the generalized coordinates and b b b ~r = ~r(u, v, w)=x(u, v, w) e1 + y(u, v, w) e2 + z(u, v, w) e3 b b b 67 ∂~r ∂~r ∂~r and consequently d~r = du + dv + dw, where ∂u ∂v ∂w ∂~r ∂x ∂y ∂z E~ = = e + e + e 1 ∂u ∂u 1 ∂u 2 ∂u 3 ∂~r ∂x ∂y ∂z E~ = = e + e + e (1.3.11) 2 ∂v ∂v b1 ∂v b2 ∂v b3 ∂~r ∂x ∂y ∂z E~ = = e + e + e . 3 ∂w ∂w b 1 ∂wb 2 ∂wb 3 are tangent vectors to the coordinate curves. The elementb ofb arc lengthb in the curvilinear coordinates is ∂~r ∂~r ∂~r ∂~r ∂~r ∂~r ds2 = d~r d~r = dudu + dudv + dudw · ∂u · ∂u ∂u · ∂v ∂u · ∂w ∂~r ∂~r ∂~r ∂~r ∂~r ∂~r + dvdu + dvdv + dvdw (1.3.12) ∂v · ∂u ∂v · ∂v ∂v · ∂w ∂~r ∂~r ∂~r ∂~r ∂~r ∂~r + dwdu + dwdv + dwdw. ∂w · ∂u ∂w · ∂v ∂w · ∂w Utilizing the summation convention, the above can be expressed in the index notation. Define the quantities ∂~r ∂~r ∂~r ∂~r ∂~r ∂~r g = g = g = 11 ∂u · ∂u 12 ∂u · ∂v 13 ∂u · ∂w ∂~r ∂~r ∂~r ∂~r ∂~r ∂~r g = g = g = 21 ∂v · ∂u 22 ∂v · ∂v 23 ∂v · ∂w ∂~r ∂~r ∂~r ∂~r ∂~r ∂~r g = g = g = 31 ∂w · ∂u 32 ∂w · ∂v 33 ∂w · ∂w and let x1 = u, x2 = v, x3 = w. Then the above element of arc length can be expressed as 2 i j i j ds = E~i E~j dx dx = gij dx dx ,i,j=1, 2, 3 · where ∂~r ∂~r ∂ym ∂ym gij = E~i E~j = = ,i,jfree indices (1.3.13) · ∂xi · ∂xj ∂xi ∂xj are called the metric components of the curvilinear coordinate system. The metric components may be thought of as the elements of a symmetric matrix, since gij = gji. In the rectangular coordinate system x, y, z, the element of arc length squared is ds2 = dx2 + dy2 + dz2. In this space the metric components are 100 gij = 010 . 0 0011 @ A 68 EXAMPLE 1.3-3. (Cylindrical coordinates (r, θ, z)) The transformation equations from rectangular coordinates to cylindrical coordinates can be expressed as x = r cos θ, y = r sin θ, z = z. Here y1 = x, y2 = y, y3 = z and x1 = r, x2 = θ, x3 = z, and the position vector can be expressed ~r = ~r(r, θ, z)=r cos θ e1 + r sin θ e2 + z e3. The derivatives of this position vector are calculated and we find b b b ∂~r ∂~r ∂~r E~ = =cosθ e +sinθ e , E~ = = r sin θ e + r cos θ e , E~ = = e . 1 ∂r 1 2 2 ∂θ − 1 2 3 ∂z 3 From the results in equationb (1.3.13),b the metric components ofb this spaceb are b 100 2 gij = 0 r 0 . 0 0011 @ A We note that since gij =0wheni = j, the coordinate system is orthogonal. 6 Given a set of transformations of the form found in equation (1.3.10), one can readily determine the metric components associated with the generalized coordinates. For future reference we list several differ- ent coordinate systems together with their metric components. Each of the listed coordinate systems are orthogonal and so gij =0fori = j. The metric components of these orthogonal systems have the form 6 2 h1 00 2 gij = 0 h2 0 0 2 1 00h3 @ A and the element of arc length squared is 2 2 1 2 2 2 2 2 3 2 ds = h1(dx ) + h2(dx ) + h3(dx ) . 1. Cartesian coordinates (x, y, z) x = x h1 =1 y = y h2 =1 z = z h3 =1 The coordinate curves are formed by the intersection of the coordinate surfaces x =Constant, y =Constant and z =Constant. 69 Figure 1.3-1. Cylindrical coordinates. 2. Cylindrical coordinates (r, θ, z) x = r cos θ r 0 h =1 ≥ 1 y = r sin θ 0 θ 2π h = r ≤ ≤ 2 z = z <z< h =1 −∞ ∞ 3 The coordinate curves, illustrated in the figure 1.3-1, are formed by the intersection of the coordinate surfaces x2 + y2 = r2, Cylinders y/x =tanθ Planes z = Constant Planes. 3. Spherical coordinates (ρ, θ, φ) x = ρ sin θ cos φ ρ 0 h =1 ≥ 1 y = ρ sin θ sin φ 0 θ π h = ρ ≤ ≤ 2 z = ρ cos θ 0 φ 2π h = ρ sin θ ≤ ≤ 3 The coordinate curves, illustrated in the figure 1.3-2, are formed by the intersection of the coordinate surfaces x2 + y2 + z2 = ρ2 Spheres x2 + y2 =tan2 θz2 Cones y = x tan φ Planes. 4. Parabolic cylindrical coordinates (ξ,η,z) x = ξη <ξ< h = ξ2 + η2 −∞ ∞ 1 1 2 2 2 2 y = (ξ η ) <z< h2 = pξ + η 2 − −∞ ∞ z = z η 0 h =1p ≥ 3 70 Figure 1.3-2. Spherical coordinates. The coordinate curves, illustrated in the figure 1.3-3, are formed by the intersection of the coordinate surfaces ξ2 x2 = 2ξ2(y ) Parabolic cylinders − − 2 η2 x2 =2η2(y + ) Parabolic cylinders 2 z = Constant Planes.
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