投影片 1 - National Cheng Kung University
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3、general curvilinear coordinates in Euclidean 3-D
3-1 coordinate system and general in Euclidean 3-D
Suppose that general coordinates are( ξ1 ,ξ 2 ,ξ 3 ); this means that the
1
2
position vectors x of a point is a known function of ξ , ξ and ξ3 ,
x
then the choice that is usually made for the base vectors is εi =
i
ξ
For consistency with the right-handedness of the εi , the coordinates
x
x
x
must be numbered in such a way that
0
1
2
3
ξ
ξ
ξ
3
x
3
x
2
x
1
2
1
x
3
x
1
x
2
x
x
x
0
1
2
3
ξ1 r
As an example , consider the cylindrical coordinate
ξ2
ξ3 z
i
With X x eiin terms of the Cartesian coordinates xi and the Cartesian base
1
2
3
vectors e i, where X x e1 x e2 x e3 , we have
x1 r cos
x 2 r sin
x3 z
And so
X [(r cos )e1 (r sin )e 2 ze3 ]
(cos )e1 (sin )e2
r
r
X [(r cos )e1 (r sin )e2 ze3 ]
ε2
(r sin )e1 ( r cos )e 2
X [(r cos )e1 (r sin )e 2 ze3 ]
ε3
e3
z
z
ε1 (cos )e1 (sin )e2
1 0 0
ε 2 (r sin )e1 (r cos )e2
g 0 r 2 0
ij
ε3 e3
0 0 1
ε1
3-2 metric tensor and jacobian
g
We have already seen that ij is a tensor ; it will now be shown why it is
called the metric tensor
x x
x
ε
ε
g
g
The definition i j
ij, together with εi = i ,give ij
i
j
ξ
ξ
ξ
x
dξi
Note that dx
ξi
So that an element of arc length ds satisfies
x x i j
j
(ds) dx dx
dξ dξ g dξi dξ
ij
ξi ξ j
2
Note that dx
x
dξi is the same as
ξi
dx (dξi )ε
i
The jacobian of the transformation relating Cartesian coordinates and
curvilinear coordinates is determinant of the array xi ξ j and the
element of volume having the vectors
(
x 1 x 2 x 3
dξ ).( 2 dξ ), ( 3 dξ )
1
ξ
ξ
ξ
As edges is
dV Jdξ1dξ2dξ3 (ε1 ε2 ε3 )dξ1dξ2dξ3
J
g
3-3 Transformation rule for change of coordinates
i
Suppose a new set general coordinates ξ is introduced, with the
i
understanding that the relations between ξi and are known, at least in
principle. The rule for changing to new tensor components is
ij
i
j
pq
T ..k T (ε ε )(ε ε )(ε r ε k )
..r
p
q
ε g ε j g ( x
i
ij
ij
ij
ij
ξ
i
),
g
ε
ε
,
and
g
g
j
ij
jk
j
i
k
x x
jt x x
pq is x x
(g
)( g
)( g ru
)
..r
p
s
q
t
u
k
ξ
T ..k T
ξ
ξ
ξ
ξ
ξ
l
m
x
x
ξ
x
x
ξ
x x ξ n
is
jt
pq
ru
T (g
)( g
)( g
)
..r
l
s
p
m
t
q
u
n
ξ ξ ξ
ξ ξ ξ k
ξ ξ ξ
i
j
ξ
ξ r
ij
pq ξ
T ..k T
(
)(
)(
)
..r
p
q
k
ξ
ξ
ξ
4、tensor calculus
4-1 gradient of a scalar
If f ξ , ξ , ξ
1
2
3
f
is a scalar function, then
f
f
f
But grad
f
x
ξi
f x
j
j
i
x ξ
(
f
x
j
e )
j
x
ξi
j e j ; hence
f
ξi
(f ) ε
is the i th convariant component of
ξi
f
f
εi
i
f
i
f
An alternative way to conclude that
i is a vector is to note
f
j
d is a scalar for all d j recall that d j is a vector ,and
that df
j
invoke the appropriate quotient law.
4-2 Derivative of a vector ; christoffel` symbol; covariant derivative
Consider the partial derivative
F
ξ
i
j of a vector F. with F = F ε ,
i
we have
F
ξ
ε
write
ξ
j
ε
i
i
ε F
j i
j
F i
ξ
ξ
i k ε
ij k
j
christoffel system of the second kind
the k th contravriant component of the derivative with respect to j of the base
vector. Note that
ε
ξ
i
j
2 x
ξi ξ
ε
j
j
ξi
k k
ij
ji
We can now write
F
ξ
(
j
F i
ξ
j
F k i )ε
(4-2-1)
ki i
Introduce the notation
F
ξ
j
Fi ε
(4-2-2)
,j i
i
F
This means that , j ----- called the convariant derivative of F i --- is definded as
F
th
the i contravariant component of the vector
j comparing (4-2-1) and (4
2-2) then gives us the formula
i
F
Fi
F k i
,j
kj
j
ξ
F i
Although
i
ξ
j is not necessarily a tensor,F, j is one , for
dF
F
ξ
j
j
j
dξ ( F i dξ )ε
,j
i
th
The covariant derivative of Fi writing as Fi , j, is defined as the i covariant
component of F j ; hence
Fi , j gki F, kj
(4-2-3)
A direct calculation of Fi, j is more instructive; with F= F εi ,we have
i
F
ε k
i
i
F ε (
ε F
)
i, j
k j
j
j
ξ
ξ
ξ
F
j
k
k
Now ε εi i , whence
k
ε
i
And therefore
consequently
ε
k
ε i ε k (l ε ) k
ij l
ij
j
j
ε
ε k
F
i, j
j
k εi
ij
F
i F k
k ij
j
F
i
And while this be the same as (4-2-3) it shows the explicit addition to
j
needed to provide the covariant derivative of F
i
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 ijkis not a third-order tensor, the superscript can nevertheless be
lowered by means of the operation ε
g
k [ij , p] g
kp ij
kp
εk
ξ
i ε
j
p
ε
ξ
i
j
and the resultant quantity , denoted by [ij, p] , is the Christoffel symbol of the
first kind. The following relations are easily verified
ε
ε
x x
j
[ij, k ] ε i ε
k
k
j
ξi
ξ k ξi ξ j
ξ
2
g
ε
ε
ij
j
i ε ε
(ε ε )
[ik , j ] [ jk , i]
i
j
j
i
k
k
k
k
ξ
ξ
ξ
ξ
g
g
g
1
jk
ij
p
ik
g [ij, k ] {
}
kp ij
2 j
i
k
[Prove] :
g
g
g
1
ij
jk
{ ik
}
2 j
i k
1
{[kj , i ] [ij, k ] [ ki, j ] [ ji, k ] [ jk , i] [ik , j ]}
2
1
{[ij , k ] [ ji, k ]}
2
[ij , k ]
(4-2-4)
4-3 covariant derivatives of Nth –order Tensors
ij
Let us work out the formula for the covariant derivatives of A..k. write
ij
ε ε εk
..k i j
A A
By definition
A
ij
A
ε ε εk
..k , p i j
p
ξ
ij
( A ε ε εk )
..k i j
p
ξ
This leads directly to the formula
ij
..k , p
A
ij
..k Arj i Air j Aij r
..k rp
..k rp
..r pk
p
A
4-4 divergence of a vector
A useful formula for
i
p i
i
F
F divF F (
)
F
i
,i
ip
will be developed for general coordinate systems. We have
i
ip
g is [ip, s ]
g
g
1 is gis
ps
ip
g [
]
p
i
s
2
ξ
ξ
ξ
1 is gis
g
p
2
ξ
But, by determinant theory
g
is
is
gg
p
p
g
ξ
( g )
Hence i 1 g 1
ip 2 g
p
g ξ p
ξ
And therefore
1
F
(F i g )
g ξi
ξ
4-5 Riemann-Christoffel Tensor
Since the order of differentiation in repeated partial differentiation of Cartesian
tensor is irrelevant, it follows that the indices in repeated covariant differentiation of
general tensors in Euclidean 3-D may also be interchanged at will. Thus, identities
like
,ij , ji
and
f k ,ij f k , ji
(4-5-1)
Eq (4-5-1) is easily verified directly, since
,ij
2
ξ i ξ
j
p
, p ij
However, the assertion of (4-5-1) in Euclidean 3-D leads to some nontrivial
information. By direct calculation it can be shown that
p
f k ,ij f k , ji R.kij
fp
p
p
kj
p
ki p r p r
R
.kij
ri kj
rj ki
j
ξi
ξ
With help of (4-2-4) it can be shown that R pkij , the Riemann-christoffel tensor,
is given by
2 g
2 g
2 g
2 g
1
pj
pi
kj
ki
R
[
] g [ r m r m ]
pkij 2
rm pj ki
pi kj
ξ k ξi ξ p ξ j ξ k ξ j ξ p ξi
But since the left-hand side of vanishes for all vectors
f k , it follows that
Rpkij 0
(4-5-2)
Although (4-5-2) represents 81equations, most of them are either identities
or redundant, since R pkij R pkji Rkpij Rijpk . Only six distinct
nontrivial conditions are specified by (4-5-2), and they may be written as
Rpkij R1223 R1231 R2323 R2331 R3131 0
(4-5-3)
[Note]
p
p
kj
ki p r p r ) f
f
f
(
k ,ij
k , ji
rj kj
rj ki p
j
ξi
ξ
[prove]
f
k ,ij
f
k ,i
p
p
f
kj pi
ij
ξ i
f
f
f
p
p
p
( k f ) (
f ) r ( k f )
ki p
kj
kp r
ij
kp
ξ i ξ i
ξi
ξ r
f
k , ji
f
(
k p f )p (
kj p
kj
j
ξ i ξ
f
ξ
p
p f
r f ) ( k f )
pj r
ji
kp
j
p
ξ
p
f
f
p
p
p
p
p
p f k
p
k
ki
f
f
r f
f
k ,ij
ki
kj
kj pi r
ij
ij kp
j
j p
j
p
ξi
ξ ξ i ξ
ξ
ξ
2 f
p
f
f
k, j
p
p
p
p
p
p
p
k
f
r f f
k , ji
kj
ki
kj pj r
ji
ji kr
j
j
ξi
ξi
ξ i ξ
ξ
2 f
p
p
jk
kj
p
p
f
f
r f
f pjk r f
f
k ,ij
k , ji
kj pi r
p
jk
pj r
p
j
i
ξ
ξ
p
p
kj ki
p
p
(
r r ) f
ri kj
rj ki p
j
ξi
ξ
Since Rpkij is antisymmetrical in i and j as well as in p and k, no information is
lost if (4-5-2) is multiplied by ε spk εtij . Consequently , a set of six equations
equivalent to (4-5-3) is given neatly by
S
st
0
Where S st is the symmetrical, second-order tensor
S st
1 spk tij
ε
ε
R
pkij
4
p
The tensor Sij is related simply to the Ricci tensor R R
ij
.ijp
p
R S g S
ij
ij
ij p
So that (4-5-3) is also equivalent to the assertion Rij 0
4-6 Integral Relations
The familiar divergence theorem relating integrals over a volume V and its
boundary surface S can obvious be written in tensor notation as
i
i
f, i dv f Ni ds
v
s
Where Ni is the unit outward normal vector to S . Similar stokes,theorem
for integrals over a surface S and its boundary line C is just
ijk
k t ds
ε
f
N
ds
f
c
k, j i
k
s
Where tk is the unit tangent vector to C , and the usual handedness rules
apply for direction of Ni and ti
廣義相對論
T PU U Pg
1
R g R 8GT
2
R R
ij
ij
Ri
jkl
ij
Rg R
ij
xk
i
jk
x
i
j
i
k
jk
i
i
g
jk ,
jk
i
l
jl
1 g j gk g jk
jk , k j
2 x
x
x
ds 2 g dxi dx
ij
j
A
j,kl
A
j,lk
Ri A
jkl i
R
ijkl
R
klij
Rij kl
R
R
jikl
ijkl
R
R
ijlk
ijkl
i
i
R R Ri 0
jkl
klj
ljk
Rijkl 0
2
2
1
nn 1nn 1 1 nn 1n 2n 3 N n n 1
2
4!
12
xi
A
i
Ai
A
j
,j
j
x
A
i, j
A
i A
ij
j