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 +++ dzzdyydxx ),,( is given by ds 2 ++= dzdydx 222 .
In n-dimensional space, Riemann defined the distance ds between two neighbouring points xi and ii =+ nidxx ),...2,1( by quadratic differential form
2 21 21 1 n ds = 11 +++ ⋅⋅⋅ )( 12dxgdxg 1n dxdxgdx
12 22 2 n + g12 22 dxgdxdx )()( +++ ⋅⋅⋅ 2n dxdxg + ...... +
n 1 n 2 n 2 + n1 n2dxgdxdxg +++ ⋅⋅⋅⋅ nn dxgdx )( 2 ji ds = ij dxdxg = nji ),...2,1,( ...(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 ji ds = ij dxdxg ...(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 ji system. Then metric ds = gij dx dx transforms to = ij xdxdgds . Since distance being scalar quantity.
2 ji ji So, ds = g=ij ij xdxdgdxdx ...(2) The theorem will be proved in three steps. (i) To show that dxi is a contravariant vector. If x i = 21 xxxx ni ),...,(
i i i i ∂x ∂x ∂x n xd = dx1 + dx2 ++⋅⋅⋅ dx ∂xi ∂x2 ∂xn i ∂x k xd i = 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 =xd i dx and xd j = ∂x ∂xk ∂xl from equation (2) i j ∂x k ∂x l = dxdxg ji g dx dx ij ij ∂xk ∂xl i j ∂x ∂x lk = dxdxg ji g∂ dxx ij ij ∂x k ∂xl ∂x i ∂x j lk lk dxdxg = gij dxdx kl ∂xk ∂xl
ji lk Since g=ij kl(i, j aredxdxgdxdx dummy indices).
∂x i ∂x j −gg dxdx lk ijkl k l = 0 ∂x ∂x
∂x i ∂x j k l or −gg ijkl = 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 = gg )( ++ −gg jiij )( ij 2 jiij 2
gij = + BA ijij 1 where A = + gg jiij )( = symmetric ij 2 1 B= − gg )( = Skew-symmetric ij 2 jiij ji ji Now, =ij dxdxg + ijij )( dxdxBA from (3)
ji ji − = ijij )( dxdxAg ij dxdxB (4)
ji Interchanging the dummy indices in ij dxdxB , we have
ji ji ij= dxdxB ji dxdxB ji ji ij= dxdxB −ij dxdxB
Since Bij is Skew-symmetric i.e., −= BB jiij
ji ji Bij+ ij=0 dxdxBdxdx
ji ij2dxdxB = 0
ji ⇒ ij dxdxB =0 So, from (4), ji − ijij )( dxdxAg =0
ji ⇒ gij = Aij as , dxdx 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.
ji THEOREM 3.2 To show that ij dxdxg 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 ∂x ji 34 Tensors and Their Applications
∂xk ∂xl ⇒ −gg = 0 klij ∂x i ∂x j
∂xk ∂xl −gg dxdx ji klij i j = 0 ∂x ∂x
k l ∂x ∂x ji = dxdxg ji )( g dxdx ij kl ∂x i ∂x j k l ∂x i ∂x j = g dx dx kl ∂x i ∂x j
ji lk =ij dxdxg kl dxdxg
ji So, ij dxdxg 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 gg ij 0≠. = By theorem on page 26
kj k ij=AA δi
kj k So, =ij gg 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 ( ,, xxx 321 ) be the cylindrical coordinates of a point. The cylindrical coordinates are given by =θ ,cos rx =θ = ,sin zz ry So that x1 ,, 32 === zxyxx and =1 θ== ,, 32 zxxrx ...(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 = ++ dzdydx 222
1 23222 ds 2 = ++ dxdxdx )()()( ...(2)
2 ji But ds = ij dxdxg
21 21 31 12 = 11 )( 12 dxdxgdxg ++ 13+ dxg 21 dxdxgdx
22 32 13 + 22 dxg )( +23 dxdxg +31 dxdxg
23 33 +32 dxdxg 33+ dxg )( ...(3) Comparing (2) and (3), we have
ggg 332211 === 1 and == 211312 gggggg 323123 ==== 0 On transformation ∂xi ∂x j =gg , since g is Covariant Tensor of rank two. (i, j = 1, 2, 3) ijij ∂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 . == ⋅⋅⋅ 3213 == 0ggg 2 2 2 ∂x ∂y ∂z g11 = g11 + g22 + g33 ∂r ∂r ∂r Since θ,= cos rx = zz θ= ,sin ry ∂x ∂y ∂z θ=,cos θ=,sin 0= ∂r ∂r ∂r and ggg 332211 === 1.
22 g11 = +θ0sincos +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 ∂ ∂θθ 36 Tensors and Their Applications
Since ggg 332211 === 1 ∂x ∂y ∂z r= θ− ,sin r θ=,cos 0= θ∂ θ∂ θ∂
22 g22 = ( ) ( ) +θ0cossin +θ− rr
= 2 rr cossin 222 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 + g + g3322 ∂z ∂z ∂z ∂x ∂y ∂z Since ,0= ,0= 1=. So, =1g. ∂z ∂z ∂z 33 = 2 = So, 11 g,1 =22 rg , 33 1g
and == 211312 gggggg 323123 ==== 0 (i) The metric in cylindrical coordinates
2 ji ds = ij xdxdg =ji .3,2,1,
2 2 2 2 1 2 3 ds = 11( xdg ) 22 ( ) ++ 33 ( xdgxdg )
since 12 == ⋅⋅⋅ 3213 == 0ggg
ds 2 = 2 )( ++ ddrdr φθ 222 (ii) The first fundamental tensor is
ggg 001 131211 = 2 gij = ggg 232221 r 00 ggg 333231 100
001 2 since g = ij = rg 00 100 g = r 2 Metric Tensor and Riemannian Metric 37
(iii) The cofactor of g are given by 2 = 2 11= rB , 22 B ,1 33= rB and == 132112 BBBBBB 323123 ==== 0
ij Bij The second fundamental tensor or conjugate tensor is g = . g
in ofcofactor ofcofactor in gg 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 == 211312 gggggg 323123 ==== 0
001 1 Hence the second fundamental tensor in matrix form is 0 0 . r 2 100
EXAMPLE 2 Find the matrix and component of first and second fundamental tensors in spherical coordinates. Solution
Let xxx 321 ),,( be the cartesian coordinates and xxx 321 ),,( be the spherical coordinates of a point. The spherical coordinates are given by θ=rx φ,cossin θ=ry φ,sinsin θ= cosrz
So that x1 ,, 32 === zxyxx and 1 ,, xxrx 32 === φθ
Let gij and gij be the metric tensors in cartesian and spherical coordinates respectively.. The metric in cartesian coordinates is given by ds 2 = ++ dzdydx 222
2 2 2 ds 2 = ( 1 ) ( 2 ) ++ (dxdxdx 3 )
2 ji But ds = ij dxdxg ; ( =ji 3,2,1, ) 38 Tensors and Their Applications
======⇒ ggg 332211 === 1 and 132312 gggggg 323121 0 On transformation
∂xi ∂x j gij = gij ∂xi ∂x j
(since gij is covariant tensor of rank two) (where i, j = 1,2,3).
∂x ∂x11 ∂x ∂x 22 ∂x ∂x33 g = g + g + g ij 11 ∂x ∂x ji 22 ∂x ∂x 33 ∂x iji ∂x1 since i, j are dummy indices. Put i = j = 1
2 2 2 ∂x1 ∂x 2 ∂x3 g + g + g g= 11 1 22 1 33 1 11 ∂x ∂x ∂x
∂x 2 ∂y 2 ∂z 2 g= g11 + g22 + g33 11 ∂r ∂r ∂r Since θ =rx φ ,cossin =θ θry = φcos,sinsin rz
∂x ∂y ∂z = φθ ,cossin = φθ ,sinsin cosθ= ∂r ∂r ∂r
and ggg 332211 === 1. So, 22 2 g11 = ( cossin ) ( ) cossinsin +θφθ +φθ
g11 = 1 put i = j = 2
2 2 ∂x1 ∂x 2 ∂x3 g + g + g g= 11 2 22 2 33 2 22 ∂x ∂x ∂x
∂xz 2 ∂y 2 ∂ 2 g= g11 + g22 + g33 22 θ∂ θ∂ θ∂
since ggg 332211 === 1 ∂x ∂y ∂z r = φθ ,coscos r = φθ ,sincos =sinrθ− θ∂ θ∂ θ∂
2 22 g=22 (r coscos ) ( ) (+φθ rr +sinsincos φθ θ− )
2 g=22 r Metric Tensor and Riemannian Metric 39
Put i = j = 3
2 2 2 ∂x1 ∂x 2 ∂x3 g + g + g g= 11 3 22 3 33 3 33 ∂x ∂x ∂x
2 2 2 ∂xz ∂y ∂ g= g11 + g22 + g33 33 φ∂ φ∂ φ∂
since ggg 332211 === 1 ∂x ∂y ∂z r= φθ− ,sinsin r = φθ =0,cossin and φ∂ φ∂ φ∂
2 2 g=33 ( sinsin ) ( +) φθ− +φθ 0cossin rr
22 g=33 sinr θ So, we have = 2 22 11 g,1 =22 rg , 33 sinrg θ=
and == 211312 gggggg 323123 ==== 0 (i) The Metric in spherical coordinates is
2 ji ds= ij xdxdg ; =ji 3,2,1,
2 2 2 2 1 2 3 ds= 11( xdg ) 22 ( ) ++ 33( xdgxdg ) ds=2 222 sin +φθ θ222 + drdrdr (ii) The Metric tensor or first fundamental tensor is 001 ggg 131211 2 gij = ggg 232221 = r 00 22 ggg 333231 r sin00 θ and
001 2 24 g= ij rg 00 == r sin θ r sin00 22 θ = 2 22 == (iii) The cofactor of g are given by 11 ,1B 22= rB , 33 sinrB θ= and BB 2112
BB 1331 == 0 BB 3223 == B The second fundamental tensor or conjugate tensor is g ij = ij . g 40 Tensors and Their Applications
in ofcofactor ofcofactor in gg B g11 = = 1111 g g r sin 24 θ = r sin 24 θ g11 = 1 22 B22 r sin θ g 22 = = g r sin 24 θ 1 g=22 r 2 2 B33 r g=33 = g sinr24 θ 1 g=33 r sin 22 θ and = 1312 ggggg 323121 ==== 0 Hence the fundamental tensor in matrix form is
000 ggg 131211 1 0 0 g ij ggg 232221 2 = = r 333231 1 ggg 00 r sin 22 θ EXAMPLE 3 If the metric is given by
2 2 2 ds 2 = ( 1 ) 35 (dxdx 2 ) ++ 4 (dx3 ) − +1 46 dxdxdxdx 322
Evaluate (i) g and (ii) g ij .
Solution
2 ji The metric is ds = ij dxdxg ; =ji )3,2,1,(
1 2 21 31 12 2 + ++ ds = 11(dxg ) 12 gdxdxg 13 21 dxdxgdxdx 22 32 13 23 23 + g22 )( 23 dxdxgdx ++ 31 gdxdxg 32 ++ 33 dxgdxdx )(
Since gij is symmetric =⇒gg jiij
i.e., g12 = g ,21 23 , == gggg 311332 Metric Tensor and Riemannian Metric 41
2 21 22 23 21 So, ds = 11 dxg )( + 22 gdxg 33 ++ 2)()( 12 dxdxgdx
32 31 23 ++ 22 13 dxdxgdxdxg ...(1) Now, the given metric is
ds 2 = )(3)(5 2221 dxdxdx 123 +−46)(4 ++ dxdxdxdx 322 ...(2) Comparing (1) and (2) we have
g11 = ,5 22 = ,4,3 ggg 1233 362 =−gg 2112 === ⇒−
23 42 gg ⇒= 3223 0,2 ==== ggg 3113
ggg 131211 − 035
g= gij ggg 232221 == 233 =4−
ggg 333231 420
(ii) Let Bij be the cofactor of gij in g. Then 23 B11 = ofCofactor g == 8 11 42 05 B22 = ofCofactor g == 20 22 40
−35 B = ofCofactor =g =6 33 33 − 33 − 23 B = ofCofactor −= 12 == Bg 12 12 40 21
− 33 B = ofCofactor =−== 6 Bg 13 13 20 31 −35 B = ofCofactor g=23−==– 10 B32 23 20
Bij Sinceg ij = g We have
11 B11 8 22 33 3 2112 3113 3 3223 5 g === ;2 =g,5 =g , gg == ,3 gg −== , gg −== g 4 2 2 2 42 Tensors and Their Applications
Hence,
3 32 − 2 5 g ij = 53 − 2 3 5 3 −− 2 2 2 3.3 LENGTH OF A CURVE i Consider a continuous curve in a Riemannian Vn i.e., a curve such that the coordinate x of any current point on it are expressible as functions of some parameter, say t. The equation of such curve can be expressed as x=i i tx )(
The length ds of the arc between the points whose coordinate s are xi and + dxx ii given by
2 ji ds= ij dxdxg
If s be arc length of the curve between the points P1 and P2 on the curve which correspond to the two values t 1 and t 2 of the parameter t.
21 ji P2 t2 dx dx s = gds ij dt = ∫P ∫t 1 1 dt dt
NULL CURVE dx dx ji If g 0= along a curve. Then s = 0. Then the points P and P are at zero distance, despite ij dt dt 1 2 of the fact that they are not coincident. Such a curve is called minimal curve or null curve.
EXAMPLE 4 A curve is in spherical coordinate xi is given by
− 1 x1 = ,t x =sin 12 and 23 −12 = tx t Find length of arc 1 ≤ t ≤ 2. Solution In spherical coordinate, the metric is given by ds 2 = xdx 121 )()()( ++ 1222 dxxxdx )()sin( 2322 Metric Tensor and Riemannian Metric 43
− 1 given x1 = ,t x =sin 12 , 23 −12 = tx t
1 1 1 − 21 dx=1 dt, dx 2 = − dt, dx 2 = 23 −⋅ 21 dttt 2 ( ) 1 2 t 2 1 − t
dt 3 2t dx 2 = − , dx = dt tt 2 −1 t 2 −1
2 2 2 dt 21 t 2 = 2 tdt 2 −+ + t sinsin −1 dt ds () tt 2 −1 t t 2 −1
2 2 2 dt 4t 2 2 dt + + ()dt ds = 2 2 t −1 t −1
2 5t 2 2 dt ds = 2 t −1 t ds = 5 dt t2 −1 Now, the length of arc, t ≤≤ ,21 is
2 2 t2 2 t 5 t −1 ds = 5 dt = = units ∫ ∫ 2 15 t1 1 t −1 2 21 1
3.4 ASSOCIATED TENSOR A 21 ...iii r A tensor obtained by the process of inner product of any tensor 21 ... jjj s with either of the fundamental ij tensor gij or g is called associated tensor of given tensor..
e.g. Consider a tensor Aijk and form the following inner product
αi α αj k ααα Ag ijk =A jk ; ; Ag ikijk == AAgA ijijk
All these tensors are called Associated tensor of Aijk . Associated Vector ik k Consider a covariant vector Ai . Then = i AAg is called associated vector of Ai . Consider a j j j contravariant vector A . Then =jk is AAg calledk associated vector of A . 44 Tensors and Their Applications
3.5 MAGNITUDE OF VECTOR i The magnitude or length A of contravariant vector A . Then A is defined by
ji A = ij AAg
2 ji or A = ij AAg 2 j i Also, = j AAA as =ij AAg j i.e., square of the magnitude is equal to scalar product of the vector and its associate.
The magnitude or length A of covariant vector Ai . Then A is defined by
ij A = AAg ji
2 ij or A = AAg ji A vector of magnitude one is called Unit vector. A vector of magnitude zero is called zero vector or Null vector.
3.6� SCALAR� PRODUCT OF TWO VECTORS �� Let A and B be two vectors. Their scalar product is written as ⋅BA and defined by �� i ⋅BA = BA i
�� i ji j Also, ⋅BA = = iji BAgBA since = iji BgB
�� iji iji ⋅BA = i= BAgBA sinceji = BgB j Thus �� i ji 2 ⋅AA = A iji == AAAgA � ji i.e., A = = ij AAgA
Angle� between� two vectors Let A and B be two vectors. Then �� �� ⋅BA = θcosBA
�� ji ⋅BA ij BAg ⇒ θcos =�� = ji ji BA ij ij BBgAAg � � ji ji since A = ij AAg ; B = ij BBg This is required formula for θcos . Definition � � The inner product of two contravariant vectors A ( Ai )or and B ( Bi )or associated with a symmetric ji tensor gij is defined as ij BAg . It is denoted by �� ji BAg ),( = ij BAg Metric Tensor and Riemannian Metric 45
� � The necessary and sufficient condition that the two vectors and B at 0 be orthogonal THEOREM�� 3.3 A if =BAg 0),( � � Proof: Let θ be angle between the vectors A and B then �� �� ⋅BA = θcosBA �� or ⋅BA =θcosAB
ji =ij θcosBAg AB ji ij BAg ⇒ θcos= ...(1) AB � � π If A and B are orthogonal then =θ =0cos θ⇒ then from (1) 2 ji ij BAg = 0
�� �� ji ⇒ ( ,BAg ) = 0 since ( ,BAg ) = ij BAg ji Conversely if ij BAg 0= then from (1) π θcos= 0 =. θ⇒ � � 2 So, two vectors A & B are orthogonal. Proved. � � �� Note: (i) If A and B be unit vectors. Then = BA = 1. Then �� ji θcos= ⋅BA = ij BAg � � π π (ii) Two vectors A and B will be orthogonal if angle between them is i.e., =θ then 2 2 π θcos=cos =θ = 0 2
3.7 ANGLE BETWEEN TWO VECTORS
THEOREM 3.4 To show that the definition of the angle between two vectors is consistent with the requirement cos 2θ ≤ 1. OR To justify the definition of the angle between two vectors. OR To show that the angle between the contravariant vectors is real when the Riemannian Metric is positive definition. � � Proof: Let θ be the angle between unit vectors A and B then ji j ij ij i θcosij = BAg = j j BBABA i ij === BABAg i 46 Tensors and Their Applications
To show that θ is real i.e., |cosθ| ≤ 1. Consider the vector + mBlA ii when l and m are scalars. The square of the magnitude of
ii i jji + mBlA = ij (lAg ++ mBlAmB )()
2 ji ji ji 2 ji = ij AAlg ij mBlmAg lg ij +++ ij BBgmAB
= +2 θ+ cos2 mlml 2 Since ji 2 ji 2 ij AAg ==A ;1 ij BBBg == .1 and � ji � � 2 =ij BAg θ;cos as A & B are unit vector i.e., =11 ⇒.= AA Since square of magnitude of any vector .0≥ So, the square of the magnitude of mBlA ii .0≥+ or +2 θ+ cos2 mlml 2 ≥ 0 ( )2 mmml +θ+coscos 222 θ− ≥ 0 cos( +θ+222 θ− )cos1() ≥ 0 mml This inequality holds for the real values of l & m. if cos1 2 θ− ≥ 0 ⇒ cos2 θ ≤ 1 θcos1≤ Proved. THEOREM 3.5 The magnitude of two associated vectors are equal. i Proof: Let A and B be magnitudes of associate vectors A and Ai respectively. Then 2 ji A= ij AAg ...(1) and 2 ij B= AAg ji ...(2) From equation (1) 2 ji A= ij )( AAg 2 j A= j AA ...(3)
ji since =ij (AssociateAAg vector) From equation (2) ij B 2 = )( AAg ji
2 j B = AA j ...(4) Metric Tensor and Riemannian Metric 47
ij j since = i AAg from (3) and (4) A2 = B 2 ⇒ A = B i So, magnitude of Ai and A are equal.
3.8 ANGLE BETWEEN TWO COORDINATE CURVES i l Let a Vn referred to coordinate = nix )...,2,1(, . For a coordinate curve of parameter x , the coordinate xl alone varies. Thus the coordinate curve of parameter xl is defined as i xi = c , i∀ except = li ...(1) where i, sC are constants. Differentiating it, we get dx i = 0, i∀, except = li and dxl 0≠ Let Ai and B i be the tangent vectors to a coordinate curve of parameters x p and x q respectively.. Then Ai = = ( xdx pi )0...0,,0,...0 ...(2)
Bi = = ( xdx qi )0...0,,0,...0 ...(3) If θ is required angle then
ji ij BAg θcos = ji ji ij ij BBgAAg
qp pq BAg = pp qq g pp qq BBgAA
qp pq BAg = qp qqpp BAgg
g pq θcos = ...(4) gg qqpp
which is required formula for θ.
i j The angle wij between the coordinate curves of parameters x and x is given by
gij wijcos= gg jjii 48 Tensors and Their Applications
If these curves are orthogonal then π wijcos= cos= 0 2 ⇒ gij = 0
i j Hence the x coordinate curve and x coordinate curve are orthogonal if gij 0=.
3.9 HYPERSURFACE i i 1 1 The n equations x = x (u ) represent a subspace of Vn . If we eliminate the parameter u , we get (n –1) equations in x j, s which represent one dimensional curve. i i 1 2 Similarly the n equations x = x (u ,u ) represent two dimensional subspace of Vn. If we eliminating 1 2 i, the parameters u , u , we get n –2 equations in x s which represent two dimensional curve V n. This two dimensional curve define a subspace, denoted by V2 of Vn. i i 1 2 n–1 Then n equations x = x (u , u , ... u ) represent n – 1 dimensional subspace Vn–1 of Vn. If we eliminating the parameters u 1, u 2, ...un–1, we get only one equation in i, sx which represent n –1 dimensional curve in Vn. This particular curve is called hypersurface of V n. i Let φ be a scalar function of coordinates x i . Then xφ)( = constant determines a family of hypersurface of Vn.
3.10 ANGLE BETWEEN TWO COORDINATE HYPERSURFACE Let xφi )( = constant ...(1)
and xψi )( = constant ...(2) represents two families of hypersurfaces. Differentiating equation (1), we get
∂φ i dx = 0 ...(3) x∂i φ∂ φ∂ This shows that is orthogonal to dx i . Hence is normal to φ constant,=since dx i is x∂i x∂i tangential to hypersurface (1). ψ∂ Similarly is normal to the hypersurface (2). If ω is the angle between the hypersurface (1) x∂i and (2) then ω is also defined as the angle between their respective normals. Hence required angle ω is given by ∂ ∂ψφ gij ∂ ∂xx ji ωcos = ...(4) ∂ ∂ ∂ ∂ψψφφ gij gij ∂ ∂xx ji ∂ ∂xx ji Metric Tensor and Riemannian Metric 49
If we take φ =x p constant = ...(5) and ψ =xq constant= ...(6) The angle ω between (5) and (6) is given by
∂x p ∂xq gij ∂xi ∂x j ωcos = ∂x p ∂x p ∂xq ∂x q gij g ij ∂xi ∂x j ∂xi ∂x j
ij p q g i δδ j = ij p ijq q q i j gg i δδδδ j
g pq ωcos= ...(7) gg qqpp
i j The angle ωij between the coordinate hypersurfaces of parameters x and x is given by
gij ωcosij = ...(8) gg jjii
If the coordinate hypersurfaces of parameters xi and x j are orthogonal then π ω = ij 2
⇒ ωcosij = 0 from (8), we have g.ij 0= Hence the coordinate hypersurfaces of parameters xi and xj are orthogonal if g ij 0=.
3.11n-PLY ORTHOGONAL SYSTEM OF HYPERSURFACES
If in a Vn there are n families of hypersurfaces such that, at every point, each hypersurface is orthogonal to the 1− hypersurfacen of the other families which pass through that point, they are said to form as n-ply orthogonal system of hypersurfaces.
3.12 CONGRUENCE OF CURVES
A family of curves one of which passes through each point of Vn is called a congruence of curves.
3.13 ORTHOGONAL ENNUPLE
An orthogonal ennuple in a Riemannian Vn consists of n mutually orthogonal congruence of curves. 50 Tensors and Their Applications
ij THEOREM 3.6 To find the fundamental tensors gij and g in terms of the components of the unit tangent h = nhe ),...2,1( to an orthogonal ennuple. i = Proof: Consider n unit tangents h = nhe ),...2,1( to conguence h nhe ),...2,1( of an orthogonal ennuple in a Riemannian Vn . The subscript h followed by an upright bar simply distinguishes one congruence from other. It does not denote tensor suffix.
The contravariant and covariant components of e |h are denoted by e |h and e |ih respectively..
Suppose any two congruences of orthogonal ennuple are e |h and e |k so that
i j h hij eeg k|| = δk ...(1)
i ee h h || ik = δk from (1), i j hij eeg k|| = 0
i j and hij eeg h|| = 1 We define ee i ofcofactor |ih in determinant |ih eh| = e |ih Also, from the determinant property, we get n i ee i ∑ h || jh = δ j ...(2) h=1 Multiplying by e jk
n i ee j k jki ∑ h || jh g = j gδ h=1
n i ee k ik or ∑ h h|| = g ...(3) h=1
Again multiplying (2) by gik . n i gee i ∑ h || ikjh = j gδik h=1
g or jk = ∑ ee || jhkh ...(4) from (3) and (4)
n g ee ij = ∑ || jhih ...(5) h=1 Metric Tensor and Riemannian Metric 51
n ij i ee j g = ∑ h h|| ...(6) h=1 This is the required results.
Corollary: To find the magnitude of any vector u is zero if the projections of u on e |h are all zero. Proof: Let n i eC i u= ∑ hh | ...(7) h=1 Then
n n i C i h =CCee δ= eu |ik = hh || ik ∑∑ kh k h h== 11
i or Ck = eu |ik ...(8)
i i.e., Ck = projection of u on e |ik Using (8), equation (7) becomes
n i j eeu i u = ∑ kjh || h=1 Now,
u2 iuu = i eCeC = hhi ∑∑ || ikk from (7) h k
i = ∑ hkh eeCC || ik ,kh
h = ∑ CC δkkh ,kh
CC = ∑ hh h
n 2 C 2 u = ∑()h h=1
2 This implies that u = 0 iff = 0u C hiff 0=. Hence the magnitude of a vector u is zero iff all the projections of u (i.e. of u i) on n mutually
i orthogonal directions eh| are zero. 52 Tensors and Their Applications
Miscellaneous Examples 1. If p and q are orthogonal unit vectors, show that kjih − gg ijhkikhj )( =1qpqpgg Solution Since p and q are orthogonal unit vectors. Then j 22 iij qpg =0, qp == 1. Now, kjih kijh jikh − =gg ijhkikhj )( qpqpgg − ikhj qqppgg ijhk pqqpgg
jh ki kj ji = −hi ik qqgppg )()( hk ij pqgqpg )()( =p 2.q2 – 0.0 = 1 . 1 jh kh = 1 (since ghi hk qpgpp == 0&1 ) 2. If θ is the inclination of two vectors A and B show that
kijh − gg ijhkikhi )( BBAAgg 2 sinθ = kjjh ikhj BBAAgg
Solution If θ be the angle between the vectors A and B then
ij ij BAg θcos = ji ki ij ik BBgAAg But cos1sin =θ22 θ−
ji kh 2 ij hk BAgABg )()( sinθ = 1− jh ki hj ik BBgAAg )()(
kijh − gg ijhkikhj )( BBAAgg = kijh ikhj BBAAgg
3. If X ij are components of a symmetric covariant tensor and u, v are unit orthogonal to w and satisfying the relations i γ+α−)(ijij =0wugX j i δ+β−)(ijij =0wvgX j where α ≠ β prove that u and v are orthogonal and that Metric Tensor and Riemannian Metric 53
ji ij vuX =0 Solution
ji i Suppose Xij is a symmetric tensor. Since ,vu are orthogonal to w then i wu i =0 ...(1) i wv i =0 ...(2)
i given γ+α−)(ijij =0wugX j ...(3) i δ+β−)(ijij =0wvgX j ...(4) where α ≠ β. Multiply (3) & (4) by , uv jj respectively and using (1) and (2), we have ji α− ijij )( vugX =0 ...(5) ji β− ijij )( uvgX =0 ...(6)
Interchanging the suffixes i & j in the equation (6) and since , Xg ijij are symmetric, we get ji α− ijij )( vugX =0 ... (7) Subtract (6) & (7) we get ji α−β )( ij vug = 0 Since β α≠ and β .0≠α− Hence, ji ij vug =0 ...(8) So, u and v are orthogonal. Using (8) in equation (5) & (6), we get ii ij vuX =0 Proved. 4. Prove the invariance of the expression ... 21 dxdxdxg n for the element volume. Solution
Since gij is a symmetric tensor of rank two. Then ∂xk ∂xl g = g ij ∂x i ∂x j kl Taking determinant of both sides ∂x k ∂xl g = gkl ij ∂x i ∂x j ∂x Since =J (Jacobian) ∂x =gg g=klg& ij 54 Tensors and Their Applications
So, g = gJ 2 or g J = g Now, the transformation of coordinates from xl to x i , we get
∂x 21 n 21 n = ... xdxdxd ... dxdxdx ∂x = 21 ... xdxdxJd n
g 21 n 21 n ... xdxdxd ... dxdxdx = g
... 21 dxdxdxg n = ... 21 xdxdxdg n
So, the volume element dv...= 21 dxdxdxg n is invariant.
EXERCISES
kl 1. For the Metric tensor gij defined g and prove that it is a contravariant tensor. i j 2. Calculate the quantities g for a V3 whose fundamental form in coordinates u, v, w, is bdvadu ++ 222 fdvdwcdw +++ 222 hdudvgdwdu 3. Show that for an orthogonal coordinate system 1 1 1 g11 = , g22 = , g33 = g11 g22 g33
4. For a V2 in which g 11 ,, 2112 === GgFgE prove that
g = EG , 112 ggGgF −,, 2212 =−== gEggF 1 5. Prove that the number of independent components of the metric g cannot exceed +nn )1( . ij 2 i i i ij i ij ij j i i i j ij 6. If vectors u , v are defined by u = g uj, v = g v j show that ui = g u , u vi = uiv and u giju = uig uj 7. Define magnitude of a unit vector. prove that the relation of a vector and its associate vector is reciprocal.
8. If θ is the angle between the two vectors Ai and Bi at a point, prove that
kjih − gg ijhkikhi )( BBAAgg 2 sin θ = kjih jkhi BBAAgg 9. Show that the angle between two contravariant vectors is real when the Riemannian metric is positive definite. CHAPTER – 4
CHRISTOFFEL'S SYMBOLS AND COVARIANT DIFFERENTIATION
4.1 CHRISTOFFEL'S SYMBOLS The German Mathematician Elwin Bruno Christoffel defined symbols
1 ∂g ∂g kj ∂g ji ki + − [ ,kij ] = j i k , ( = ,...2,1,, nkji ) ...(1) 2 ∂x ∂x ∂x called Christoffel 3-index symbols of the first kind. k and = lk [], lijg ...(2) ji called Christoffel 3-index symbols of second kind, where g ji are the components of the metric Tensor or fundamental Tensor.
There are n distinct Christoffel symbols of each kind for each independent g ji . Since g ji is 1 symmetric tensor of rank two and has +nn )1( independent components. So, the number of 2 1 1 independent components of Christoffel’s symbols are ().⋅1 2 ()+=+ 1 nnnnn 2 2 k THEOREM 4.1 The Christoffel's symbols [ , kij ] and are symmetric with respect to the indices i ji and j. Proof: By Christoffel’s symbols of first kind
1 ∂g ∂g jk ∂gij ik + −∂ [ ,kij ] = j i k 2 ∂x ∂x ∂x Interchanging i and j, we get
1 ∂g jk ∂g ∂g ji + ik − [ ,kji ] = i j k 2 ∂x ∂x ∂x