矢量分析 各种坐标系下梯度算符、拉普拉斯算符形式

(27)
2
h
2
dc dc − f (c , c , c )h h dc dc
1
2
31Biblioteka 231 2
1
∂ (h2 h3 f 1 ) 1 2 3 ∂ (h3 h1 f 2 ) 2 1 3 ∂ (h1 h2 f 3 ) 3 1 2 dc dc dc + dc dc dc + dc dc dc ∂c1 ∂c2 ∂c3
• Scales. We use hp as the scales in the p direction in general coordinates. (see below)
Scales
• Definition of unit vectors for general coordinates. The directions of unit vectors equal the direction of gradient of corresponding coordinate; The length of unit vectors should be 1. • Thus we introduce Scales denoted as hp :
∇f ≡
p
(22)
The scales in orthogonal coordinates can be calculated use the method in the former section. • Examples. – In Spherical coordinates we have eq.(20) Thus the Gradient Operation in Spherical coordinates is: ∇f =
(8)
= np i np i
chain rule
q
=
eq. (7)
q
=
q
q ∂x np i ni ∂cq hq
change the order of summation
=
q
∂x δpq ∂cq hq 1 ∂x hp ∂cp
eq. (2)
=
summation with δ f unction
• Through the deduction, eq.(8), we have: ∂x p ˆ = hp e ∂cp (9)
3
• Einstein summation convention ONLY for subscripts. Superscripts is free. e.g. ai bi = ap bp =ap · bp , p = 1, 2, 3.
i=1
ai · bi , however,
• For convenience, in Cartesian coordinate, we also use xi as the coordinates of x, namely, x = (x, y, z ) = (x1 , x2 , x3 ). • Similarly as above, we also will use x ˆi denote unit vectors in Cartesian coordinate, namely, x ˆ1 = e ˆ x, x ˆ2 = e ˆ ˆ3 = e ˆ y, x z . Thus we have: x ˆi = ∂x ∂xi (1)
p is the general coordinates unit vector along ˆ the Spherical coordinates unit vector along the r direction, e
p direction. p = 1, 2, 3 in a 3-D problem. • cp is the coordinate of a point in general coordinates. e.g. coordinate of P in Cartesian is (x, y, z ), in spherical Coordinate is (r, θ, φ), in general orthogonal coordinates is (c1 , c2 , c3 ).
Thus, (h1 )2 = ( ∂x 2 ∂y ∂z ) + ( 1 )2 + ( 1 )2 1 ∂c ∂c ∂c (18)
• Take spherical coordinates as an example. We have: x = r sin θ cos φ y = r sin θ sin φ (19) z = r cos θ c1 = r After calculation, one can easily find that: hr = 1 hθ = r hφ = r sin θ (20) c2 = θ c3 = φ
∆V ∆V →0
∆V
(25)
use Gauss Theorem: ∇ · f dV =
∆V ∂ ∆V
f · dS
(26)
• Calculate the closed surface integral. Let surface units pointing outside in the figure ˆ 1 + on the right, decompose f into f = f 1 e ˆ ˆ 2 + f 3e 3 , we have: f 2e f · dS = f 1 (c1 + dc1 , c2 , c3 )h2 h3 dc2 dc3 − f 1 (c1 , c2 , c3 )h2 h3 dc2 dc3
p = hp · ∇cp ˆ e
(4)
• Use Cartesian coordinates solve the gradient. (∇f =
∂ ∂ ∂ e ˆ e ˆ e ˆ x+ y + z ): ∂x ∂y ∂z ∂cp x ˆi ∂xi (5)
p = hp ˆ e
p in Cartesian coordinates: ˆ • Decompose e p = np x ˆ e i ˆi
q
(
∂x q dc ) · ∂cq
(
p
∂x p dc ) ∂cp
(chain rule)
(10)
Use eq.(9) in the above equation: (dl)2 =
q q dcq ) · ˆ (hq e p p dcp ) ˆ (hp e
(11)
Remember we have the orthogonal condition, eq.(2), we have : (dl)2 =
Gradient
• Consider a scalar function f (x, y, z ). Use chain rule on the gradient: ∇f =
p
∂f ∇cp ∂cp
(21)
And we have eq.(4), so the gradient in general coordinates is: 1 ∂f p ˆ e hp ∂cp
(6)
• Compare equation (5) and (6) we have:
np ∂cp i = p ∂xi h
(7)
• Pure mathematic deduction:
p = ˆ e
np ˆi ix ∂x ∂cq ∂cq ∂xi ∂x nq i ∂cq hq
= np i
∂x ∂xi
eq. (1) & (6)
p
1 ∂f p ∂f 1 ∂f 1 ∂f ˆ = e e ˆ e ˆ e ˆ r+ θ+ φ p p h ∂c ∂r r ∂θ r sin θ ∂φ
(23)
Namely, in spherical coordinates: ∂ 1 ∂ 1 ∂ e ˆ e ˆ e ˆ r+ θ+ φ ∂r r ∂θ r sin θ ∂φ
p
(hp dcp )2 = (h1 dc1 )2 + (h2 dc2 )2 + (h3 dc3 )2
(12)
• Thus we come to a very very important equation, eq.(12). – if the path is along one coordinates axis, its length is: dl1 = h1 dc1 (13)
• The Divergence in general coordinates. – combine with eq.(15), we have: 1 h1 h2 h3 ∂ (h2 h3 f 1 ) ∂ (h3 h1 f 2 ) ∂ (h1 h2 f 3 ) + + ∂c1 ∂c2 ∂c3
Denotation and Convention
We will use the following denotation in the discussion. • e with a hat for unit vector. e.g. e ˆ ˆ x is the Cartesian coordinates unit vector along the x direction, e r is
dc2 =0 dc3 =0
– which means the area of surface element is: dS = dl1 · dl2 = h1 h2 dc1 dc2 (14)
– and the volume of volume element is: dV = dl1 · dl2 · dl3 = h1 h2 h3 dc1 dc2 dc3 (15)
Orthogonal Coordinates