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Metric Tensor and Riemannian Metric CHAPTER – 3 METRIC TENSOR AND RIEMANNIAN METRIC 3.1 THE METRIC TENSOR In rectangular cartesian coordinates, the distance between two neighbouring point are (x, y, z) and (,,)x+ dx y + dy z + dz is given by ds=+2 dx2 dy + 2 dz 2 . In n-dimensional space, Riemann defined the distance ds between two neighbouring points xi and xi+ dxi i ( = 1,2,... n ) by quadratic differential form 2 1 2 1 2 1 n +ds = g11 +()+ dx⋅⋅⋅ g12 dx dx g1n dx dx 2 1 2 2 2 n + g12 ()()dx dx g+22 dx ++⋅⋅⋅ g2n dx dx + . .. + n 1 n 2 n 2 + gn1 dx dx g+n2 dx dx⋅⋅⋅⋅ ++ gnn () dx 2 i j ds = gij dx dx (ij ,= 1,2,... n ) ...(1) using summation convention. i Where gij are the functions of the coordinates x such that g= gij 0≠ The quadratic differential form (1) is called the Riemannian Metric or Metric or line element for n- dimensional space and such n-dimensional space is called Riemannian space and denoted by Vn and gij is called Metric Tensor or Fundamental tensor.. The geometry based on Riemannian Metric is called the Riemannian Geometry. THEOREM 3.1 The Metric tensor gij is a covariant symmetry tensor of rank two. Proof: The metric is given by 2 i j ds = gij dx dx ...(1) 32 Tensors and Their Applications Let x i be the coordinates in X-coordinate system and x i be the coordinates in Y-coordinate 2 i j 2 i j system. Then metric ds = gij dx dx transforms to ds= gij dx dx . Since distance being scalar quantity. 2 i j i j So, ds = g=ijdx dx gij dx dx ...(2) The theorem will be proved in three steps. (i) To show that dxi is a contravariant vector. If x i = xi( x1 , x 2 ,... xn ) i i i i ∂x ∂x ∂x n dx = dx1 + dx2 ⋅⋅⋅++ dx ∂xi ∂x2 ∂xn i ∂x k dxi = dx ∂xk It is law of transformation of contravariant vector. So, dx i is contravariant vector.. (ii) To show that gij is a covariant tensor of rank two. Since i j ∂x k ∂x l dx=i dx and xd j = ∂x ∂xk ∂xl from equation (2) i j ∂x k ∂x l g= dxi dx j g dx dx ij ij ∂xk ∂xl i j ∂x ∂x k l g= dxi dx j g∂ x dx ij ij ∂x k ∂xl ∂x i ∂x j k l k l g dx dx = gij dx dx kl ∂xk ∂xl i j k l Since g=ijdx dx g kl(i, dx j are dx dummy indices). ∂x i ∂x j g− g dxk dx l kl ij k l = 0 ∂x ∂x ∂x i ∂x j k l or gkl− g ij = 0 as dx and dx are arbitrary.. ∂xk ∂xl ∂x j ∂x j g = g kl ij ∂xk ∂xl ∂x k ∂xl or g = g ij kl ∂x i ∂x j So, gij is covariant tensor of rank two. Metric Tensor and Riemannian Metric 33 (iii) To show that gij is symmetric. Then gij can be written as 1 1 g = ()g+ g + ()gij− g ji ij 2 ij ji 2 gij = ABij+ ij 1 where A = ()g+ij g ji = symmetric ij 2 1 B= ()g− g = Skew-symmetric ij 2 ij ji i j i j Now, g=ij dx dx ()+A ij B ij dx dx from (3) i j i j ()−gij= A ij dx dx Bij dx dx (4) i j Interchanging the dummy indices in Bij dx dx , we have i j i j Bij= dx dx Bji dx dx i j i j Bij= dx dx B−ij dx dx Since Bij is Skew-symmetric i.e., BBij−= ji i j i j Bij+dx dx Bij=0 dx dx i j 2Bij dx dx =0 i j ⇒ Bij dx dx =0 So, from (4), i j ()−gij A ij dx dx=0 i j ⇒ gij = Aij as dx, dx are arbitrary.. So, gij is symmetric since Aij is symmetric. Hence gij is a covariant symmetric tensor of rank two. This is called fundamental Covariant Tensor. i j THEOREM 3.2 To show that gij dx dx is an invariant. Proof: Let xi be coordinates of a point in X-coordinate system and x i be coordinates of a same point in Y-coordinate system. Since gij is a Covariant tensor of rank two. ∂x k ∂x1 Then, g = g ij kl ∂x i∂x j 34 Tensors and Their Applications ∂xk ∂xl ⇒ g− g = 0 ij kl ∂x i ∂x j ∂xk ∂xl g− g dxi dx j ij kl i j = 0 ∂x ∂x k l ∂x ∂x i j ()g= dxi dx j g dx dx ij kl ∂x i ∂x j k l ∂x i ∂x j = g dx dx kl ∂x i ∂x j i j k l g=ij dx dx gkl dx dx i j So, g ij dx dx is an ivariant. 3.2 CONJUGATE METRIC TENSOR: (CONTRAVARIANT TENSOR) ij The conjugate Metric Tensor to gij , which is written as g , is defined by Bij ij = (by Art.2.16, Chapter 2) g g where Bij is the cofactor of gij in the determinant g gij 0≠. = By theorem on page 26 kj k AAij= δi kj k So, g=ij g δi ij Note(i) Tensors gij and g are Metric Tensor or Fundamental Tensors. ij (ii)g ij is called first fundamental Tensor and g second fundamental Tensors. EXAMPLE 1 Find the Metric and component of first and second fundamental tensor is cylindrical coordinates. Solution Let (x1, x2, x3) be the Cartesian coordinates and (x1,, x 2 x 3) be the cylindrical coordinates of a point. The cylindrical coordinates are given by xrcos=θ , sinz=yθ = , z r So that x1=x,, x2 = y x 3= z = and x= 1 θ r,, x2 x 3 z ...(1) Let gij and gij be the metric tensors in Cartesian coordinates and cylindrical coordinates respectively. Metric Tensor and Riemannian Metric 35 The metric in Cartesian coordinate is given by ds 2 = dx+2 dy + 2 dz 2 1 2 2 2 3 2 ds 2 = ()()()dx+ dx + dx ...(2) 2 i j But ds = gij dx dx 1 2 1 2 1 3 2 1 = g11() dx+12 g dxdx + g13+ dx dx g21 dx dx 2 2 2 3 3 1 + g22() dx g+23 dx dx g+31 dx dx 3 2 3 3 g+32 dx dx g33+() dx ...(3) Comparing (2) and (3), we have g 11= g 22 =g 33 =1 and gggg12= 13 = 21 =23 gg = 31 = 32 =0 On transformation ∂xi ∂x j g= g , since g is Covariant Tensor of rank two. (i, j = 1, 2, 3) ij ij ∂xi ∂x j ij for i = j = 1. 2 2 ∂x1 ∂x2 ∂x3 g + g + g g11 = 11 1 22 1 33 1 ∂x ∂x ∂x since 12 gg=13. ⋅⋅⋅ == 32 g =0 2 2 2 ∂x ∂y ∂z g11 = g11 + g22 + g33 ∂r ∂r ∂r Since xrθ,= cos z = z sinyθ= , r ∂x ∂y ∂z cosθ= , sinθ= , 0= ∂r ∂r ∂r and g11= g 22 =g 33 =1. 2 2 g11 = cos sin+θ 0+θ g11 = 1 Put i = j = 2. 2 2 2 ∂x1 ∂x2 ∂x3 g + g + g g22 = 11 2 22 2 33 2 ∂x ∂x ∂x ∂x 2 ∂y 2 ∂z 2 g = g11 + g22 + g33 22 ∂θ ∂θ ∂ θ 36 Tensors and Their Applications Since g11= g 22 =g 33 =1 ∂x ∂y ∂z −r=sin θ , r cosθ= , 0= θ∂ θ∂ θ∂ 2 2 g22 = ( sin) ( cos−θ) r +θ 0+ r = rr2 sin2 2 cos 2 θ+θ 2 g22 = r Put i = j = 3. 2 ∂x1 ∂x 2 ∂x3 g + g + g g= 11 3 22 3 33 3 33 ∂x ∂x ∂x ∂x 2 ∂y ∂z = g11 + g22 + g 33 ∂z ∂z ∂z ∂x ∂y ∂z Since ,0= ,0= 1=. So, =1g. ∂z ∂z ∂z 33 = 2 = So, 11 g,1 g=22 r , 33 1g and gggg12= 13 = 21 =23 gg = 31 = 32 =0 (i) The metric in cylindrical coordinates 2 i j ds = gij dx dxij,= 1,2,3. 2 2 2 2 1 2 3 ds = g11++( dx ) g22 ( dx) g33 ( dx ) since 12 gg=13⋅⋅⋅ == 32 g =0 ds 2 = dr+2 rd2()θ +2d φ 2 (ii) The first fundamental tensor is g g g 1 0 0 11 12 13 = 2 gij = g21 g 22 g 23 0r 0 g31 g 32 g 33 0 0 1 1 0 0 2 since g = gij =0 r 0 0 0 1 g = r 2 Metric Tensor and Riemannian Metric 37 (iii) The cofactor of g are given by 2 = 2 B11= r , 22 B,1 B33= r and BB12= 21 =BB 13 =23 = BB 31 = 32 =0 ij Bij The second fundamental tensor or conjugate tensor is g = . g cofactor of g in g g11 = 11 g B r 2 g11 = 11 = =1 g r 2 B 1 g 22 = 12 = g r 2 B r 2 g 33 = 33 = =1 g r 2 and gggggg12= 13 = 21 =23 = 31 = 32 =0 1 0 0 1 Hence the second fundamental tensor in matrix form is 0 0 . r 2 0 0 1 EXAMPLE 2 Find the matrix and component of first and second fundamental tensors in spherical coordinates. Solution Let (x1 , x 2 , x 3 ) be the cartesian coordinates and (,,)x 1 x 2 x 3 be the spherical coordinates of a point.
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