张量分析第五章5.3

;
i ∂ j r i = −Γ kj r k
（5.3-9）
Γ ijk =
1 kr g ( g jr ,i + g ri , j − gij ,r ) 2
2． 3．
Γ ijk = ( g jk ,i + g ki , j − gij ,k )
k Γ ik = ∂ i (ln g ) =
1 2
;
（5.3-10） （5.3-11）
（5.3-13）
1 gii , j g kk 2
(i不求和)
将Γ
ijk
, Γ ijk
Γ i jk ：
当i , j , k =1,2,3时的值用h1, h2, h3表示时有：
Γ123 = Γ 231 = Γ 312 = Γ132 = Γ 321 = Γ 213 = 0 Γ121 = Γ 211 = −Γ112 = h1∂ 2 h1 Γ131 = Γ 311 = −Γ113 = h1∂ 3 h1 Γ 212 = Γ122 = −Γ 221 = h2 ∂ 1 h2 Γ 232 = Γ 322 = −Γ 223 = h2 ∂ 3 h2 Γ 313 = Γ133 = −Γ 331 = h3 ∂ 1 h3 Γ 323 = Γ 233 = −Γ 332 = h3 ∂ 2 h3 Γ111 = h1∂ 1 h1 Γ 222 = h2 ∂ 2 h2 Γ 333 = h3 ∂ 3 h3
h =1 , h = r , h =1 ∵ 由（5.3-14）、（5.3-15）式得，除 ∂1h 2 的偏导数为1外，其 余的偏导数均为零。
1 2 3
∴
Γ 212 = Γ 122 = −Γ 221 = h 2 = r Γ i j k = 0 ; i , j , k 的其它取值
1 Γ 22 = − r 1 2 2 Γ 12 = Γ 21 = r Γ ikj = 0 ; i , j , k 的其它取值
（5.3-14）
Γ ikj ：
3 1 2 2 1 3 Γ 12 = Γ 23 = Γ 31 = Γ 13 = Γ 32 = Γ 21 = 0 2 Γ 11 = −h1 (h2 ) −2 ∂ 2 h1 3 Γ 11 = −h1 (h3 ) −2 ∂ 3 h1 1 Γ 22 = −h 2 (h1 ) −2 ∂1h 2 3 Γ 22 = −h 2 (h3 ) −2 ∂ 3h 2
∂ k [ r1 ⋅ ( r2 × r3 )] = ∂ k g
r1 ⋅ ( r2 Fra Baidu bibliotek r3 ) = g
∂ k [r1 ⋅ (r2 × r3 )] = (∂ k r1 ) ⋅ (r2 × r3 ) + r1 ⋅ [(∂ k r2 ) × r3 ] + r1 ⋅ [r2 × (∂ k r3 )]
i = (Γ 1ik ri ) ⋅ (r2 × r3 ) + r1 ⋅ [(Γ 2 k ri ) × r3 ] + r1 ⋅ [r2 × (Γ 3i k ri )]
Γ ijk = 0
; 1 2 ; ; ; 1 2 ; ;
i≠ j≠k
Γ ijk = − gii ,k Γ iji = Γ jii = Γ ijk = 0 Γ iik = − gii ,k g kk
i Γ ij =
i≠k
(i不求和)
（5.3-12）
1 gii , j 2
(i不求和) i≠ j≠k i≠k (i, k 不求和)
k ij rk ijr
ijk
r ij
rk
证：
1．∵ ∴ 2．∵ ∴ 例9： 证明：
(∂ j ri ) ⋅ r k = (Γ ijs rs ) ⋅ r k = Γ ijsδ sk = Γ ijk
(∂ j ri ) ⋅ r k = (Γ ijr r r ) ⋅ r k = Γ ijr r r ⋅ r k = Γ ijr g rk
= Γ 1ik ri ⋅ (r2 × r3 ) + Γ 2i k r1 ⋅ (ri × r3 ) + Γ 3i k r1 ⋅ (r2 × ri ) = Γ 11k r1 ⋅ (r2 × r3 ) + Γ 12k r2 ⋅ (r2 × r3 ) + Γ 13k r3 ⋅ (r2 × r3 )
1 3 +Γ 2 k r1 ⋅ (r1 × r3 ) + Γ 22k r1 ⋅ (r2 × r3 ) + Γ 2 k r1 ⋅ (r3 × r3 ) 1 +Γ 3k r1 ⋅ (r2 × r1 ) + Γ 32k r1 ⋅ (r2 × r2 ) + Γ 33k r1 ⋅ (r2 × r3 )
1 2
g jk ,i = Γ ijk + Γ ikj
;
g jk ,i + g ki , j − gij , k = 2Γ ijk
∴ ∵
Γ ijk = ( g jk ,i + g ki, j − g ij ,k )
Γ ijk = g kr Γ ijr
∴ 3．∵
Γ ijk =
1 kr g ( g jr ,i + g ri , j − g ij ,r ) 2
i i i i
i 1Li r
i1
ir
∂ ri ∂x
j
;
∂r i ∂x j
为书写简明，记：
∂ ri
j
（5.3-2） 因为 ∂ j ri 是矢量，且 ∂ j ri 可以在协变基底上线性表示，因此 有：
∂x
= ∂ j ri
;
∂r i = ∂ jri ∂x j
（5.3-3） 式中 (∂ j ri ) ⋅ r k 是矢量 ∂ j ri 在协变基矢量rk上的线性表示系数（ ∂ 或称为 rk 上的坐标）。同理， j ri 也可以在逆变基底上线性 表示为： ∂ r = (∂ r ) ⋅ r r （5.3-4） 定义： k Γ = = (∂ r ) ⋅ r ; Γ = { i j , k }= (∂ r ) ⋅ r （5.3-5） i j
例11： 试求球坐标：
x1 = r sin θ cos ϕ ; x2 = r sin θ sin ϕ ; x3 = r cos θ
（式中 解： ∵
x 1 = r , x 2 = θ , x 3 = ϕ ）的 Γ ikj 。
（5.3-8）
证： ∵ ∴ 又 ∵
∂ j (δ ik ) = 0 ∂ j (ri ⋅ r k ) = 0
; ;
δ ik = r i ⋅ r k
(∂ j ri ) ⋅ r k + ri ⋅ (∂ j r k ) = 0
ri ⋅ (∂ j r k ) = −(∂ j ri ) ⋅ r k = −Γ ijs rs ⋅ r k = −Γ ijk
（5.3-15）
1 Γ 11 = ( h1 ) −1 ∂1h1 2 Γ 22 = ( h2 ) −1 ∂ 2 h2 3 Γ 33 = (h3 ) −1 ∂ 3 h3
例10： 试求柱坐标： （式中 解：
x1 = r cos θ
；
x2 = r sin θ
；
x3 = z
x 1 = r , x 2 = θ , x 3 = z ）的 Γ i j k , Γ ikj 。
ri ⋅ (∂ j r k ) = −Γ ijk = −Γ ljk ri ⋅ r l = ri ⋅ (−Γ ljk r l )
∴ Christoffel符号基本性质： 符号基本性质： 符号基本性质 ; Γ =Γ 1． Γ = Γ
ijk jik k ij k ji
∂ j r k = −Γ ljk r l
5.3 协变（逆变）基底矢量导数 协变（逆变）
在曲线坐标系中，在位置矢量 x =xi ii 处的自然协变和逆变 基底矢量是 x 的矢量函数： r = r ( x) ; r = r ( x) （5.3-1） 当位置矢量 x 处的张量在曲线坐标系的自然基底上表示时 （如 A = A r L r ），对张量的分析涉及到自然基底的导数 。即需要确定：
; ;
( i = 1, 2,3) (i ≠ j )
因此有：
g ij = gij = 0 ; ( i ≠ j ) g11 = ( g 11 ) −1 = (h1 ) 2 g 22 = ( g 22 ) −1 = (h2 ) 2 g33 = ( g 33 ) −1 = (h3 ) 2
由（5.3-10）式得：
（5.3-15）
1 Γ 33 = −h3 (h1 ) −2 ∂1h3 2 Γ 33 = −h3 (h 2 ) −2 ∂ 2 h3 1 1 Γ 12 = Γ 21 = (h1 ) −1 ∂ 2 h1 1 1 Γ 13 = Γ 31 = (h1 ) −1 ∂ 3h1 2 2 Γ 21 = Γ 12 = (h 2 ) −1 ∂1h 2 2 2 Γ 23 = Γ 32 = (h 2 ) −1 ∂ 3h 2 3 3 Γ 31 = Γ 13 = (h3 ) −1 ∂1h3 3 3 Γ 32 = Γ 23 = (h3 ) −1 ∂ 2 h3
由（5.3-6）式：
∂ j ri = Γ ijk rk ; ( i, j = 1, 2,3)
得：
∂ r1 ∂ r1 1 2 3 = = Γ 11r1 + Γ 11r2 + Γ 11r3 = o ∂ x1 ∂r 1 ∂ r1 ∂ r1 1 2 3 = = Γ 12 r1 + Γ 12 r2 + Γ 21r3 = r2 ∂ x 2 ∂θ r ∂ r1 ∂ r1 1 2 3 = = Γ 13 r1 + Γ 13r2 + Γ 13 r3 = o ∂x3 ∂z
Γ ijk = (∂ j ri ) ⋅ r k = (∂ i r j ) ⋅ r k = Γ k ji
gij ,k = ∂ k (ri ⋅ rj ) = (∂ k ri ) ⋅ r j + ri ⋅ (∂ k r j ) = Γ ikj + Γ jki = Γ kij + Γ kji
; g ki , j = Γ jki + Γ jik
= Γ 11k g + Γ 22k g + Γ 33k g
i = Γ ik g
∴
i Γ ik =
1 g
∂ k g = ∂ k (ln g )
对一般曲线坐标系（5.3-6）和（5.3-8）式给出了协变基矢 量和逆变基矢量的曲线坐标偏导数 。当曲线坐标系是正交 曲线坐标系时。由于：
ri = r i ri ⋅ r j = 0
Γ ijk = Γ ijr g rk
(∂ j ri ) ⋅ rk = (Γ ijr rr ) ⋅ rk = Γ ijr g rk
(∂ j ri ) ⋅ rk = (Γ ijr r r ) ⋅ rk = Γ ijr r r ⋅ rk = Γ ijk
Γ ijk = Γ ijr g rk
证毕。
i ∂ j r i = −Γ kj r k
1 ∂i g g
∂x i ∂x ∂2 x ∂2 x ∂ = = i j = i j i ∂x ∂x ∂x ∂x ∂x
;
证： 1．∵ ∴ 2． 同理有：
∂x ∂ ri = i ∂ j ri = ∂x ∂x j
∂x j ∂x
= ∂i rj
Γ ijk = (∂ j ri ) ⋅ rk = (∂ i r j ) ⋅ rk = Γ jik
k j i j i k
k ij k j i i jk j i k
∂ j ri = (∂ j ri ) ⋅ r k rk
Γ ikj , Γ i j k 式中
称为第一类和第二类Christoffel符号。由Chris∂ toffel符号， j ri 矢量可表示为： ∂ j ri = Γ ijk rk = Γ ijk r k （5.3-6） 两个基本关系： 两个基本关系： （5.3-7a） 1． Γ = Γ g （5.3-7b） 2． Γ = Γ g
∂ r2 ∂ r2 1 1 2 3 = = Γ 21r1 + Γ 21r2 + Γ 21r3 = r2 ∂ x 1 ∂r r ∂ r2 ∂ r2 1 2 3 = = Γ 22 r1 + Γ 22 r2 + Γ 22 r3 = −rr2 2 ∂x ∂θ ∂ r2 ∂ r2 1 2 3 = = Γ 23 r1 + Γ 23r2 + Γ 23r3 = o 3 ∂x ∂z ∂ r3 ∂ r3 1 2 3 = = Γ 31r1 + Γ 31r2 + Γ 31r3 = o 1 ∂x ∂r ∂ r3 ∂ r3 1 2 3 = = Γ 32 r1 + Γ 32 r2 + Γ 32 r3 = o 2 ∂x ∂θ ∂ r3 ∂ r3 1 2 3 = = Γ 33 r1 + Γ 33r2 + Γ 33r3 = o 3 ∂x ∂z