3、 general curvilinear coordinates in Euclidean 3-D 3-1 coordinate system and general in Euclidean 3-D Suppose that general coordinates are(El,子，3)：this means that the position vectors x of a point is a known function of号'，ξ2andE 8x then the choice that is usually made for the base vectors is ; For consistency with the right-handedness of the,the coordinate must be numbered in such a way that OxOxOx >0 50253 8x g 8x o52 Ox aer Ox a a同 Ox Ox Ox O 52 ⊙30 53
Suppose that general coordinates are( ); this means that the position vectors x of a point is a known function of , and , then the choice that is usually made for the base vectors is 3、general curvilinear coordinates in Euclidean 3-D 1 2 3 ξ ,ξ ,ξ 1 ξ 2 ξ For consistency with the right-handedness of the εi , the coordinates must be numbered in such a way that 3 ξ = i x i ε ξ 1 2 3 0 x x x ξ ξ ξ 3-1 coordinate system and general in Euclidean 3-D 1 2 3 1 x 2 x 3 x 1 x 2 x 3 x 0 1 2 3 x x x
As an example,consider the cylindrical coordinate 52 =0 53 -Z WithX=xe in terms of the,Cartesian coordinates x and the Cartesian base vectors ewhere x=x'e+xe2+xe3 we have x=rcos0 x2=rsin 0 And so ax 81= o[(rcoso)e,+(rsine)e2 +zel=(cosO)e,+(sin O)e2 Or Or Ox 82 三 ol(rcosO)e,+(rsine)e2+zesl=(-rsinO)e,+(rcosO)e. ∂8 a0 Ox 83= of(rcosO)e,+(rsin)e+ze;=e, 8z 02 =(cose)e+(sine)e2 1 00 2 =(-rsine)e+(rcose)e2 马订 0 e3=e3 0
With in terms of the Cartesian coordinates and the Cartesian base vectors , where , we have i i X e = x i x i e x z x r x r = = = 3 2 1 sin cos 1 2 3 1 1 2 1 2 3 2 1 2 1 2 3 3 3 [( cos ) ( sin ) ] (cos ) (sin ) [( cos ) ( sin ) ] ( sin ) ( cos ) [( cos ) ( sin ) ] r r z r r r r z r r r r z z z + + = = = + + + = = = − + + + = = = X e e e ε e e X e e e ε e e X e e e ε e And so As an example , consider the cylindrical coordinate 1 2 3 r z = = = ξ ξ ξ 1 2 3 1 2 3 X e e = + + x x x e 1 1 2 2 1 2 3 3 (cos ) (sin ) ( sin ) ( cos ) r r = + = − + = ε e e ε e e ε e 2 1 0 0 0 r 0 0 0 1 ij g =
3-2 metric tensor and jacobian We have already seen thatis a tensor;it will now be shown why it is called the metric tensor he dentiontogether wth 8x Ox Ox give y Note that dx =x og'agl ∂ξ1 So that an element of arc lengthds satisfies (ds)2=dk·dk= oxodg a形lJ
We have already seen that is a tensor ; it will now be shown why it is called the metric tensor The definition , together with ,give Note that So that an element of arc length satisfies x x g ij i j = ξ ξ 2 ( ) i j i i j j ij x x ds dx dx d d g d d = = = ξ ξ ξ ξ ξ ξ i j ij ε = ε g x i dx d i = ξ ξ 3-2 metric tensor and jacobian g ij ds = i x i ε ξ
Ox Note that dx= d?is the same as o51 dx =(d The jacobian of the transformation relating Cartesian coordinates and curvilinear coordinates is determinant of the array x and the element of volume having the vectors 毫警5 As edges is dv Jde d'd=(x8)de de'd J-g
The jacobian of the transformation relating Cartesian coordinates and curvilinear coordinates is determinant of the array and the element of volume having the vectors As edges is 1 2 3 1 2 3 ( ).( ),( ) x x x d d d ξ ξ ξ ξ ξ ξ 1 2 3 1 2 3 1 2 3 dV Jd d d d d d = = ξ ξ ξ ( ) ε ε ε ξ ξ ξ i j x ξ Note that is the same as i i x dx d = ξ ξ ( )i i dx d = ξ ε J = g
3-3 Transformation rule for change of coordinates Suppose a new set general coordinates is introduced,with the understanding that the relations between and are known,at least in principle.The rule for changing to new tensor components is 7=T9ep2e,/Xe) =,=g(%g人86=5ty0a88- -7P9产aXg.E gru ex exon 泥P .03"m6a39 ag0”ak
3-3 Transformation rule for change of coordinates Suppose a new set general coordinates is introduced, with the understanding that the relations between and are known, at least in principle. The rule for changing to new tensor components is i ξ i ξ i ( )( )( ) .. .. ij i j pq r T T k k r p q = ε ε ε ε ε ε ( ), , i ij ij ij i j j j ij jk i k = = = = g g g and g g x ε ε ε ε ξ .. .. .. ( )( )( ) ( )( )( ) ij is jt pq ru k r p s q t u k l m n pq is jt ru r l s p m t q u n k x x x x x x T T g g g x x x x x x T g g g = = ξ ξ ξ ξ ξ ξ ξ ξ ξ ξ ξ ξ ξ ξ ξ ξ ξ ξ .. .. ( )( )( ) i j r ij pq k r p q k T T = ξ ξ ξ ξ ξ ξ
4、tensor calculus 4-1 gradient of a scalar of of ax Ox If f (is a scalar function,then ai ax ag! a utgrad-T-%‘:hence af 61 =f) af is the iconvariant component of Vf ag' Yf- 影 An alternative way to conclude that af i is a vector is to note thatcafor allrecll tsvectorand invoke the appropriate quotient law
If f ( ) is a scalar function, then 1 2 3 ξ , , ξ ξ 4-1 gradient of a scalar But grad ; hence An alternative way to conclude that is a vector is to note that is a scalar for all recall that is a vector ,and invoke the appropriate quotient law. j ( ) i j i j i j f f x f x x x = = e ξ ξ ξ j j f f f x = e ( ) i i f f = ε ξ i f j j f df d = j d 4、tensor calculus is the convariant component of i f ξ th i f i i f f = ε j d
4-2 Derivative of a vector christoffel'symbol;covariant derivative aF/ Consider the partial derivative of a vector F.with F we have OF ∂Fl ag! write 三 christoffel system of the second kind the tcontravriant component of the derivative with respect tog of the base vector.Note that 82x ag1 ogiati ∂s
4-2 Derivative of a vector ; christoffel` symbol; covariant derivative Consider the partial derivative of a vector F. with F = , we have write the contravriant component of the derivative with respect to of the base vector. Note that i i j j j i i F F = + F ε ε ξ ξ ξ j F ξ i k j ij k ε ε ξ 2 j i j j i i x = = ε ε ξ ξ ξ ξ i F i ε j th k christoffel system of the second kind
修喷 We can now write OF=( F+Fkrkae, (4-2-1) Introduce the notation OF 三 (4-2-2） agJ This means that called the convariant derivative of-is definded as the i contravariant component of the vector ag/comparing(4-2-1)and(4- 2-2)then gives us the formula Fi= +r ougtronefor -
k k ij ji = We can now write ( ) k i j j ki i i F F = + F ε ξ ξ Introduce the notation , i j j i F ε F ξ This means that ----- called the convariant derivative of --- is definded as the contravariant component of the vector comparing (4-2-1) and (4- 2-2) then gives us the formula , i F j i F th i j F , i i k i j kj j F F F = + ξ , ( ) j j i j j i d d F d = = F F ξ ξ ε ξ Although is not necessarily a tensor, is one , for i F j ξ i F , j (4-2-1) (4-2-2)
The covariant derivative of writing as F is defined as theicovariant component ofhence (4-2-3) Adirect calculation ofis more instructive:with ,we have OF Now wherce 11 3-50）-时 -E ag) And therefore 5 consequently aF And while this be the same as(4-2-3)it shows the explicit addition to i toprovid theri derivative of 5J
A direct calculation of is more instructive; with F= ,we have Now , whence The covariant derivative of writing as , is defined as the covariant component of ; hence Fi, j k Fi, j = gkiF , j Fi th i i F i ε , i ( ) k j i j j j i j k F F F = + F ε ε ε ξ ξ ξ k k i i ε = ε ( ) k k k l k i i ij l ij j j = − = − = − ε ε ε ε ε ε k k i j ij = − ε ε , i k i j k ij j F F F = − And therefore consequently And while this be the same as (4-2-3) it shows the explicit addition to needed to provide the covariant derivative of j F , F i j (4-2-3) i F i j F
Other notations are common for convariant derivatiives;they are,in approximate order of popularity F Rai Div Although Iis not a third-order tensor,the superscript can nevertheless be lowered by means of the operation Os and the resultant quantity,denoted by [i,p],is the Christoffel symbol of the first kind.The following relations are easily verified 0心1=k oEi oEk6E刀 [U,k]=ka形J ae1=9 8x 82x =[k,]+[Uk,]
Other notations are common for convariant derivatiives; they are, in approximate order of popularity | ; F F D F F i j i j j i j i Although is not a third-order tensor, the superscript can nevertheless be lowered by means of the operation k ij [ , ] k k i kp ij kp p j j i g ij p g = = = ε ε ε ε ξ ξ and the resultant quantity , denoted by , is the Christoffel symbol of the first kind. The following relations are easily verified [ij, p] 2 [ , ] j i k k j i k j i x x ij k = = = ε ε ε ε ξ ξ ξ ξ ξ ( ) [ , ] [ , ] j i k k k i j j i g ij ik j jk i k = = + = + ε ε ε ε ε ε ξ ξ ξ ξ