i Chapter II: General Coordinate Transformations
Before beginning this chapter, please note the Cartesian coordinate system belowand the definitions of the angles θ and φ in the spherical coordinate system. In the spherical coordinate system, (r,θ,φ) we shall use:
x = r sinθcosφ y = r sinθsinφ z = r cosΘ
and in the cylindrical coordinate system (ρ,φ,z):
x = ρ cosφ y = ρ sinφ ρ2 = x2 + y2 .
Figure 1. The Cartesian coordinate system for 3- dimensional Euclidian space. II-1
Chapter II: General Coordinate Transformations
Consider two coordinate systems in 3-dimensional Euclidian space:
1. a Cartesian system where a point is specified by (x1, x2, x3) (x,y,z)
2. a general "qi" system where a point is specified by (q1, q2, q3).
Each point (x1, x2, x3) corresponds to a unique set of real numbers (q1, q2, q3). Further, each xi is a function of the qj, fi (q1,q2,q3), and each qj = hj (x1,x2,x3) where all first partial derivatives of fi and hj exist. Using only the chain rule for differentiation, the following equations can be obtained:
f i f i f i dx i dq 1 dq 2 dq 3 q 1 q 2 q 3 i x j j dq q j also k q k x i x i q k using chain rule
where we have used the notation x i (q1,q2,q3) f i (q1,q2,q3) and q j(x1,x2,x3) hj(x1,x2,x3) .
Note that the differential, dxi, "transforms with" [ xi/qj ] and the partial derivative "transforms with" [ qj/xi]. So, using the summation notation (repeated indices ==> summation):
i i x j i j dx dq A .j dq q j j q j Bi . x i x i q j q j
Note the placement of the indices. II-2
i j j i i i In general, x /q ! q /x . This means that dx and /x transform differently under the coordinate transformation. The two transformations have been named contravariant and covariant, respectively.
i i x A .j contravariant transformation matrix q j . j j q Bi. covariant transformation matrix x i
Example: Transformation from Cartesian to spherical coordinates:
x = rsinθcosφ x1 r = [x² + y² + z²]½ q1 y = rsinθsinφ x2 θ = cos-1(z/r) q2 z = rcosθ x3 φ = tan-1(y/x) q3
The differentials are given by:
dx = sinθcosφ dr + rcosθcosφ dθ + -rsinθsinφ dφ dy = sinθsinφ dr + rcosθsinφ dθ + rsinθcosφ dφ dz = cos θ dr + -rsinθ dθ + 0 · dφ
Thus,
dx dx 1 sincos rcoscos rsinsin dr dy dx 2 sinsin rcossin rsincos d dz dx 3 cos rsin 0 d
dr dq 1 d dq 2 d dq 3 II-3
1 3 In the above equation A 1 = sinθcosφ and A 2 = -rsinθ, etc. The elements of the contravariant transformation matrix are obtained from the expression for the differentials of dx, dy and dz.
One can also show that:
dr sinθϕ cos sin θϕ sin cos θ dx θ =− θϕ θϕ θ ⋅ d (coscos)/rrr (cossin)/ sin/ dy d − sinϕθϕθ / (rr sin ) cos / ( sin ) 0 dz And using the chain rule one finds:
k k Mq j &1 k j dq ' dx ' [A ] .j dx Mx j
m M Mx M &1 m M ' ' [B ]n. Mq n Mq n Mx m Mx m
Note that My/Mθ = rcosθsinφ is not simply related to Mθ/My = (cosθsinφ)/r. One can see also from the above -1 T -1 T equations that A = B and B = A . In the above expressions different summation indices were used for each of the sums. This is a good habit to adopt. Try to do this in all your derivations.
j j Sometimes you will find the notation: Ai for Bi . We won't use this notation since it is i j confusing. Keep in mind that A j … Ai . If you don't understand this comment look again carefully at the indices on the above equations.
Exercise: Evaluate the A (contravariant) matrix for the transformation (from Cartesian to spherical coordinates) if r = 3, θ = π/4 and φ = π/2.
-1 Exercise: Find the A matrix for the same values of r, θ, and φ in the above exercise. II-4 Rules for keeping track of indices:
1. Always use different index symbols for different summations; 2. Indices on the left side of the equation match with the "left-most" index on the right hand side; 3. "Superscripts" found in the "denominator" (such as the k in /xk) function as subscripts in the expression; i m 4. A j and Bk have four "positions", two "upper" and two "lower" (see below); i m 5. A "left" position on A j or Bk denotes a row index; the "right" position denotes column index; 6. Indices are always "matched" in an equation (look for the pairs);
left index denotes a matrix "row" i m j k right index denotes a matrix "column"
Contravariant and covariant vectors:
A triplet of quantities in the Cartesian system, (V1,V2,V3), [such as (dx1,dx2,dx3) ] which transforms into its 1 2 3 i 1 2 3 i counterpart, (V' ,V' ,V' ), in the q system [such as (dq ,dq ,dq )] via the A j matrix is said to be a set of contravariant vector components. Contravariant vector components are denoted by superscripts.
i i j V A .j V ' contravariant vector transformation
A triplet of quantities in the Cartesian system, (V1,V2,V3), [such as ( / x, / y, / z) ] which transforms into its i 1 2 3 m counterpart, (V'1,V'2,V'3), in the q system [such as ( / q , / q , / q )] via the Bk matrix is said to be a set of covariant vector components. Covariant vector components are denoted by subscripts.
n Vk Bk. V 'n covariant vector transformation II-5
The differentials, dxi, transform like contravariant vector components under a generalized (linear) coordinate transformation. The "linear" means that the transformation is linear in the differentials. The partial derivatives, /xk, transform like covariant vector components under a generalized (linear) coordinate transformation. In fact, any triplet which transforms "like the differentials" under a generalized linear coordinate transformation is said to be a contravariant vector. A triplet which transforms "like the partial derivatives" is said to be a covariant vector under the generalized linear coordinate transformation.
Later we shall find out how to find the proper contravariant and covariant components of a physical vector. Two "physical" vectors are always simple to deal with: dr and f(x,y,z).
------Exercise: Find the contravariant components of the vector dr in the Cartesian coordinate system.
Exercise: Find the contravariant components of the vector dr in the spherical coordinate system.
Exercise: Find the covariant components of the vector ^x i (/xi) in the spherical coordinate system.
Exercise: Find the covariant components of the vector ^x i (/xi) in the cylindrical coordinate system. II-6 Kronecker Delta Symbol and Other Notation
j ! The Kronecker delta symbol is į i = { 0 if i j and 1 if i=j}.
Generalized Kronecker delta: + * 0 if any two superscripts (or subscripts) are equal * or if {i,j,k,...m} ! {i',j',k',...m'} .. . m i j k, * į i' j' k', ... m' = +1 if subscripts are even permutation of superscripts * -1 if subscripts are odd permutation of superscripts .
123 Examples: į 132 = -1 123 į 312 = +1 122 į 132 = 0
i ij Numerically, į j = įij = į . But in an expression, these can carry different information.
PERMUTATION SYMBOLS (COMPLETELY ANTISYMMETRIC "TENSORS")
+ ijk..m ijk...m * İ į 123...n = 0 if any two superscripts are equal * or if superscripts are not same set as 1,2,3...n * +1 if the superscripts are an even permutation of 1,2,3,...n * -1 if the superscripts are an odd permutation of 1,2,3,...n . + 123...n * İijk...m įijk...m = 0 if any two subscripts are equal * or if subscripts are not same set as 1,2,3...n * +1 if the subscripts are an even permutation of 1,2,3,...n * -1 if the subscripts are an odd permutation of 1,2,3,...n .
The above symbols are special cases of the generalized Kronecker delta and are often used with expressions involving permutations. The above symbols are "almost" tensors as we shall see in the chapter on tensors.
Examples: İ123 = İ312 = İ231 =+1 İ213 = İ132 = İ321 =-1 İ112 = İ223 = İ323 = 0, etc.
İ123 = İ312 = İ231 =+1 İ213 = İ132 = İ321 =-1 İ313 = İ322 = İ211 = 0, etc.
Example: İ1234 = +1
İ0123 = 0 II-7 The Metric Tensor
i The metric tensor, gjk, for the q system is defined as follows:
j k ds² gjk dq dq . j,k .summation implied g metric tensor
i where ds² = (dx)² + (dy)² + (dz)² = dx dxi (summation over i implied)
What is gjk ?
i ds² = dr·dr =dxdxi = [xi/qj]dqj · [xi/qk]dqk Summation over i
= [xi/qj] [xi/qk] dqj dqk Summation over i
j k = gjk dq dq
i j i k Thus gjk = [ x / q ] [ x / q ] (Summation over i)
The metric tensor is symmetric since
i k i j i j i k [ x / q ] [ x / q ] = [ x / q ] [ x / q ] = gkj = gjk. Summation over i
THE METRIC TENSOR IN CARTESIAN SYSTEMS:
i -1 i k i k k dx = [R ] k dx' =[x / x' ] dx'
-1 i -1 i Thus gjk = i [R ] j [R ] k
-1 T i -1 i = i [R ] j [R ] k
i -1 i = i Rj [R ] k
-1 = i [RR ]jk = 1jk and gcartesian = 1 II-8
The Jacobian:
i j i Jacobian det [ q / x ]= det [Bj ] = det(B) where [qi/xj] implies a matrix whose i,j component is qi/xj.
T The metric tensor, g = A A
proof:
i j i k gjk = i [ x / q ] [ x / q ] i i = i A j A k T i i = i [A ]j A k T = [A A]jk.
Exercise: Find g for the spherical coordinate system.
...... gspherical ...... II-9
Exercise: Find the metric tensor for a qi system if
ds² = 4dq1dq1 - q2 dq1dq2 + π dq2dq3 + 7q1 dq2dq2 - dq3dq3.
Exercise: Find the metric tensor for the cylindrical coordinate system:
x=ρ cosφρ = [x²+y²]½ y= ρ sinφφ = tan-1[y/x]. z=z II-10 The unit vectors, ^qi in the qi system
The equations qi = qi(x1,x2,x3) = ci where i = 1,2,3 each represent a surface of constant qi in the three dimensional space. Below is a schematic of three such surfaces which intersect at a point. This point can be labelled by its (x,y,z) value, or by (q1,q2,q3) = (c1,c2,c3).
Unit vectors perpendicular to a surface of constant qi can be obtained from the qi:
qi Rqi 1 i k q i k x i q k Rx q no sum on i i 1 / 2 Rq 2 qi j Rx j
i i j i i i Note that q^ = Bj x^ / [β ] (with no sum on i !) where β | q |.
Later we shall put the βi into another more convenient form.
Exercise: Find ^r, ^θ, and φ^ in the spherical coordinate system. II-11
answers: ^r = sinθcosφ^x + sinθsinφ^y + cosθ^z ^θ = cosθcosφ^x + cosθsinφ^y - sinθ^z φ^ = -sinφ ^x + cosφ ^y
The Inverse Metric Tensor
-1 T Claim: g = B B. proof:
1. We note from page II-2 that expression a) below holds; from page II-3 we find expression b):
k j k a) [ q / x ] = Bj .
k j -1 k b) [ q / x ] = [A ] j.
T k -1 k T -1 Thus B j = [A ] j and B = A .
-1 T -1 -1 T -1 T 2. But g = [A A] = A [A ] = B B
T -1 -1 T T T where we have used [A ] = [A ] = [B ] = B. (to be shown later).
-1 jk T j k Note that [g ] = n [B ] n Bn . II-12
-1 i j The components of the inverse metric tensor, g , are denoted by g -- with superscripts! Later we shall -1 see that the superscripts on g and the subscripts on g have definite implications. Normally, one does not use a special symbol for the inverse metric tensor -- when written with components.
gij metric tensor components . g ij inverse metric tensor components
ii Claim: βi |qi| = g (no sum on i). proof:
ii 1. g (no sum on i) is the diagonal (ii) component of the inverse metric tensor.
ii T ii -1 i i g = [B B] = [A ] k Bk sum on k, no sum on i.
i i = Bk Bk sum on k only.
= [qi/xk] [qi/xk] sum on k only.
= [ (qi/xk) ^xk] · [ (qi/xn) ^xn] = qi · qi ( no sum on i).
= (qi/xk) (qi/xn) ^xk·^xn = (qi/xk) (qi/xn) δkn
ii i i i 2. Thus g = q ·q = |q |² no sum on i.
i ii 3. So, βi = |q | = g (no sum on i). II-13
T i 1 i i B k A k B k T m 1 m m A n B n A n
Use of g to convert from contravariant to covariant components
dx [g ] dx j and dq g dq m Claim: i x ij k km
proof:
j j n 1. let dx = A n dq ; multiply on the left by [gx] ij [gcartesian]ij = 1ij and sum over j:
j j n j n n' [gx]ijdx = [gx]ij A n dq = [gx]ij A n δ n' dq
j -1 n n' -1 i = [gx]ij A n [g g] n' dq where 1 is replaced by g g and g is for q system.
j nm n' -1 = [gx]ij A n g gmn' dq (recall that superscripts ==> g )
-1 T 2. Now use g = B B:
j j T nm n' [gx]ijdx = [gx]ij A n [B B] gmn' dq
T jm n' = δij [AB B] gmn' dq
T m n' -1 m n' = [AB B]i gmn' dq = [AA B]i gmn' dq
m n' = Bi gmn' dq
m n' = Bi [gmn' dq ]
j n' m 3. Thus [gx]ijdx dxi and [gmn' dq ] dqm transform like covariant vectors and dxi = Bi dqm . QED
j n' Note that we have defined the dx i and dq m in terms of dx and dq . II-14
The previous proof can be used to show that for any vi in the Cartesian system, the proper covariant vector j k components can be formed from vi = [gx]ij v -- and in the q system,
n' v'm = gmn' v' . ------n nk Exercise: Show that v' = g v'k.
------
ij Sij claim: g is diagonal iff qq (diagonal g <==> orthogonal coordinate system)
proof:
ij 1. case I: show that ^qi·^qj = δ ==> g is diagonal.
i j i ii j jj ij a) assume ^q·^q = [ q / g ]·[ q / g ] no sum on i or j = δ b) thus, ii jj ij ij ii i j g g δ no sum on i or j = δ g = q · q (no sum on i) = [qi/xk] ^xk ·[qj/xn] ^xn
i j kn = Bk Bn δ Ti j kn = B k Bn δ T ij = [B B] -1 ij = [g ]
ij ii -1 ij -1 c) so δ g no sum on i = [g ] and g is diagonal. -1 d) if g is diagonal, g is also.
ij 2. case II: assume g is diagonal and show ^qi·^qj = δ .
-1 a) g is diagonal if g is. b) but, -1 ij T ij i j kn i j [g ] = [B B] = Bk Bn δ = q · q (all the steps in case I are reversible) c) therefore (since steps in case I are reversible):
-1 ij ij ii i j ij [g ] = δ g no sum on i ==> q · q = δ C , where C = constant. II-15
Claim: If A is orthogonal g = 1. proof:
T -1 1. A = A since A is orthogonal. T -1 2. thus, A A = g = A A = 1.
j k The u and ui basis vectors for the q system
ii i i k Claim: g ^q no sum on i u transforms into ^x via A. proof:
ii i i k i 1. g ^q no sum on i = Bk ^x (see box on page II-10 and β on page II-12).
i i k T i k -1 i k n 2. so, u = Bk ^x = [B ] k ^x = [A ] k ^x ; multiply both sides by A i and sum over i:
n i n -1 i k -1 n k A i u = A i [A ] k ^x = [AA ] k ^x .
n k n = δ k ^x = ^x . and,
n i xu n A i
Since the ui and ^xn transform "according to" A we can (using same steps as on page II-13) show that:
i uj = gji u gives the corresponding basis "vectors" which transform according to B:
j ^xk = Bk uj.
i k The uj and u vectors serve as basis vectors -- playing the "role" of ^x and x^n in the "general" coordinate system. We shall show on the next page that these generalized basis vectors obey an orthogonality condition and their dot k products can be determined from the metric tensor, g. The geometrical interpretation of the u and un vectors can be seen from a simple example, after we derive the special conditions on the next page. II-16
Orthogonality and dot products for u i u j gij "basis vectors". u i · u j g ij i i u ·u j j derivations:
-1 k -1 n 1. u i· uj = [B ]i ^xk · [B ]j ^xn
T k T n = [A ]i [A ]j ^xk·x^n
T k n = [A ]i A j δkn
T k k = [A ]i A j sum over k
T = [A A]ij = gij.
i j -1 i k -1 j n 2. u · u = [A ] k x^ · [A ] n x^
T i j k n = [B ] k Bn ^x · x^
T i j kn = [B ] k Bn δ
T ij = [B B]
ij = g .
i -1 i k -1 n 3. u · uj = [A ] kx^ · [B ]j x^n
-1 i T n k = [A ] k [A ]j x^ ·x^n
-1 i T n k = [A ] k [A ]j δ n
-1 i k = [A ] k A j
-1 i = [A A] j
i = δ j. II-17
Ordinary ( f i ) , contra- and co-variant components of F:
1 2 3 F = f 1 q^ +f 2 q^ +f 3 q^
i F = F i u
j F = F u j
k k F i = F u i and F = F u
^ [ When bold face is not available, use and ----- (bars) over symbols to denote vectors.]
Derivations:
1 2 3 i 1. F = f1q^ + f2 q^ + f3 q^ = ifiq^ where fi ordinary components of F
i ii i 2. Substitute u = g q^ no sum on i in above.
i i ii F = i fiq^ = i fi [ u / g ]
ii i = i [fi/ g ] u
ii i 3. F·uj = i [fi/ g ] u ·uj ii i = i [fi/ g ] δ j
jj =[fj/ g ]. no sum on j
On the following page we shall show that F·uj = F'j, and transforms like the covariant component of F. (Note that in the qi system, the components are distinguished from those in the Cartesian system by primes.) Once k k k n this is accomplished, we can find the contravariant component of F, F' , from F' = g F'n. II-18 Claim: F·u j F 'j covariant component of`F in q i system proof:
i n nn n 1. F =Fi ^x =fn^q =(fn/ g )u ; dot with ^xj.
i nn n n 2. Fi ^x ·^xj =(fn/ g ) u · ^xj transform u (see page II-15)
nn -1 n k Fj =(fn/ g ) [A ] k^x ^xj
T n k = F·un [B ] k δ j (see p. II-17 )
n F·^xj = Bj F·un
4. So F·un and F·^xj are covariant components of F (they transform via B).
......
Claim: F·u k F'k contravariant component proof:
k n k nk 1. First note that F·u =[F'n u ]·u =F'n g (see page II-16)
n kj 2. Consider Fk = Bk F'n and multiply by [gx] and sum over k
kj kj n n' 3Fk[gx] =[gx] Bk δn F'n'
kj n -1 n' = δ Bk [gg ]n F'n'
kj n mn' = δ Bk gnm g F'n' kj n T i i mn' T -1 = δ Bk i[A ]n A m g F'n' use A = B -1 ji i mn' j mn' = i[BB ] A m [g F'n']=A m [g F'n'].
mn' kj j so g F'n' and Fk[gx] transform via A m and are contravariant components of F. II-19
....
k j F F'k u F ' u j
Derivation:
n nn n n n 1. F =fn^q = (fn/ g ) u = (F·un)u = F'nu ----as shown on p. II-18.
m n m n m 2. = gnm F' u =F' [gmn u ]= F' um .
j F·F F'j F'
Derivation:
i k k i i 1. F·F =F'i u · F' uk = F'i F' δ k = F'i F' .
......
i k Fi F F'k F'
Derivation:
i m i k 1. F · F = FiF = Bi F'm A k F'
-1 m i k -1 m k = [A ] i A k F'm F' = [A A] k F'm F'
m k = δ k F'm F'
k = F'k F' II-20 Exercises:
1 2 3 1. Assume that in the s,t,w ( q ,q ,q ) a physical quantity is given by F = 6u1 + 4u3. Find the contravariant components of F in the s,t,w system (leave answer in terms of s,t,w).
2. Given that for the s,t,w system g is as shown, 625 find the covariant components of F in the stw system. g 232 Leave answer in terms of s,t,w. 5210
i j 3. Find FiF (note unprimed components ==> x system).
4. Find u1·u3. II-21 Covariant Form When expressions are written totally in terms of covariant and contravariant vector components with proper i i summation over all indices (such as FiF ) they can be easily evaluated in either (x,y,z or the q ) system. Further, one can set the expression for the quantity in, say, the xyz system equal to the expression in the expression in terms of the qi. This is a powerful tool. It means you don't have to use A (or B). This is used extensively in special relativity and in general relativity. It can also be of use in any problem where one wants the answer, say, in the xyz coordinate system, but finds it easier to use a coordinate system which follows fluid flows, or isobars, etc. A Physical Picture of the Coordinate Systems
1. assume that s q1 = 3x·cosα + 3y·sinα t q2 = 3x·cosβ + 3y·sinβ w q3 = 5z
2. Let α = π/6 and β = π/3 to simplify the problem.
s = 3x(3/2) + 3y/2 t = 3x(1/2) + 3y(3/2) w= 5z
-1 T -1 T You can determine A = B and g = B B easily:
4 3 3 3 2 0 3 3/2 0 99 0 9 9 2 2 3 4 A 1 3 ; g 1 B T B 3 ; g 2 0 3/2 3 0 9 90 9 9 2 2 1 005 0025 00 25
11 22 3. Note that ^s · ^t = u1/g ·u2/g = (1/9)·9·3/2 = 3/2 =/ 0.
1 2 3 m also, u = 3 ^s;u = 3 ^t; and u = 5 w^ . u1 = g1mu II-22
i ii i uq g no sum on i Note easiest way to find covariant and uu j contravariant components of F: k gkj j 1) "pick off" F'j as the coefficients of u in the expression for F.
k k j 2) find F' from g F'j.
Exercises:
1. Find the ui ( i = 1,2,3) in the spherical coordinate system in terms of r,θ. and φ.
2. Find covariant and contravariant components of the position vector, r, and dr, in the spherical coordinate system. Leave answer in terms of r,θ,φ and dr, dθ, dφ. Hint for dr: recall how dqi are related to the definition of contravariant components. II-23 Use of A and B to find F·x^i
k The components of F in the xyz system can be obtained from the F' using A, or from the F'j using B:
! 1 FxS '1 " 2 !F 1 "FyS 2 A"F'2 2 " 2 " '3 2 S #F 3 #"Fz32
Example: Find F·^xi if F = 2^s + 4^t in the example on page II-21.
1 2 -1 T First, find the F'j: F = 2/ 9 u + 4/ 9 u so F'1 = 2/ 9 and F'2 = 4/ 9. Then use B = [A ] to relate the covariant components in the xyz and stw system:
! 33 3 1 " 02 ! 1 2 2 !2 1 FxS " 2 3 ! 32 1 " 2 3 33 " 2 " S 2 " 2"4 2 " 2 Fy " 02 3 "1232 " 2 2 2 " 2 0 "FzS 2 " 0052" 0 2 # 3 # 3 " 2# 3 #" 32
Exercise: In the above example, find F'k. II-24
The Determinant in Summation Notation
* * detM = m11 m12 m13 * * m21 m22 m23 * * m31 m32 m33
(1+1) (2+1) (3+1) = (-1) m11(m22·m33 - m23·m32) + (-1) m21(m12·m33 - m13·m32) + (-1) m31(m12·m23 - m13·m22)
= i j k m 1 i m 2 j m 3 k · {"sign factor"} --Note six terms for which i,j, and k are all different.
sign factor = +1 for ij k even permutation of 123 -1 for ij k odd permutation of 123 0 otherwise
= i j k
Thus the εijk symbol can be used to form a concise summation notation for detM.
ijk det M m1i m2j m3k
Exercise: Write the determinant of g in summation notation.
* -1* ijk Exercise: Write g in summation notation. Hint, use εijk rather than ε .
* * -1 i Exercise: Write [ A A ] k in summation notation. II-25
imn m n n m ijk j k j k
kji nm Exercise: Write out the terms in εknmε g gji and try to find the simplest final result.
Exercise: There is a "similar" expression (to that in the box on page II-24) for the determinant of a 4x4 matrix, using εijkm. Using what you know about determinants write down what you think it should be.
------
An interesting "expression" for εijk is:
* i j k * δ 1 δ 1 δ 1 ijk * i j k * ε = δ 2 δ 2 δ 2 * i j k * δ 3 δ 3 δ 3 .
i j k i j k i j k i j k i j k i j k =[ δ 1δ 2δ 3 + δ 2δ 3δ 1 + δ 3δ 1δ 2 ] - [ δ 1δ 3δ 2 + δ 2δ 1δ 3 + δ 3δ 2δ 1 ] || ||
3 even perm. 3 odd perm.
ijk A word of warning: in Cartesian systems one can use ε and εijk in an "equivalent" way. But in the general coordinate system care must be taken. In strict covariant notation, these symbols are usually accompanied by either |g| or 1/|g|. In Cartesian systems, since |g|= 1, one cannot see the distinction. We shall discuss this later. II-26 Vector Operators in Covariant Form
1. The gradient, φ
φ = ^xk φ/xk k j m m = A j u Bk φ/ q k -1 m m j = A j [A ] k φ/ q u -1 m k m j = [A ] k A j φ/ q u -1 m j m = [A A] j u φ/ q m j m = δ j u φ/ q = uj φ/qj
j u 'j . wherewedefine 'j q j
Exercise: Find φ(r,θ,φ) in spherical coordinates, as a function of r, θ, φ and ^r, ^θ, and φ^ only.
Exercise: Find φ(ρ,θ,z) in cylindrical coordinates, as a function of ρ, θ, z and ^ρ, ^θ, and ^z only. II-27
2. The Divergence, ·F.
k n n k 1. ·F = ^x k ·^xnF = kF ^x ·^xn n k k = k F δ n = k F i k j =(Bk 'i) (A j F' ) Note that 'i acts on all that follows.
i k j i j k = Bk A j 'i F' + Bk F' 'i A j
-1 i k j j i k = [A ] k A j 'i F' + F' { Bk 'i A j}
-1 i j j i k = [A A] j 'i F' + F' { Bk 'i A j}
i j j = δ j 'i F' + F' {...... }
i j = 'i F' + F' {...... }
2. We now derive three relationships which will be used to evaluate {...... }.
i a) First note that (see page II-12 for g in terms of derivatives)
i j i m m j n i n ½g gi / q = ½[ m ( q / x )( q / x )] / q [ n ( x /q )( x / q )] sum over i and implied!
i m m n i j n n i n j =½ m,n ( q / x )( q / x ) [ ( ²x /( q q ))( x / q ) + ( x / q )( ²x /( q q )) ] rearrange terms:
i m n i j n m m n j n i i m =½ m,n [ ( q / x ) ( ²x /( q q )) ( x / q )( q / x ) + ( q / x )( ²x /( q q )) ( x / q )( q / x )]
i m n i j n m m n j n m =½ m,n [ ( q / x ) ( ²x /( q q )) ( x / x ) + ( q / x )( ²x /( q q )) ( x / x )]
i m n i j n m n j n n =½ m,n [ ( q / x ) ( ²x /( q q )) δm + ( q / x )( ²x /( q q )) δm ] -- use δm to do n sum
i m m m m j =½ m [ ( q / x ) 'i ( x / qj) + ( q / x ) '( x / q ) ] i, dummy indices ==> add terms i m m = m [ ( q / x ) 'i ( x / qj)
i m i k = Bm 'iA j = Bk 'iA j so, i i k ½ g 'j gi Bk 'i A j
i and {...... } = ½g 'j gi . II-28
cof ik b) Next we consider detg g = gik [g ] sum on k, no sum on i: i can be 1,2,or 3.
T cof ik cof ki Since gik is symmetric and detg = detg = g, [g ] = [g ] . Thus,
cof ik g/ gik = [g ] c) Also, we know that,
-1 ik cof T ik [g ] = [g ] /g, note: g detg
ik cof ki g = [g ] /g,
= [1/g] g/ gki = [1/g] g/ gik using result above for cofactor.
= [ln g] / gik
3. Finally, using the results in 2 a), b), and c):
j j i F' {...... } = F' [½ g 'j gi ] note sums on i, , j
j j = F' [½ ( [ln g]/ gi) ( gi/ q ) ]
= F'j [(lng)/qj]
= F'j [(1/g) (g/qj)]
4. So,
i j j i · F = [ g/ g] 'i F' + F' [(1/ g) g/ q ] = (1/ g)[ 'i g F' ]
and,
1 ·F ' [gF'i ] g i
II-29
Exercise: Find ·F in the spherical coordinate system, using the generalized formula. Put answer in terms of r,θ,
φ and ^r, ^θ, φ^ . Use F = fr ^r + fθ ^θ + fφ φ^.
Exercise: Repeat the above exercise for the cylindrical coordinate system.
1 2 3 Exercise: Given g = st(sinw) in the s,t,w ( q ,q ,q ) system, find ·F for F = [st/w²]u1 + [(sinw)/t]u2 -wu3. II-30
3. The Laplacian, · φ = ²θ.
· φ = · [φ] use form for divergence and gradient:
j k j = ·[u 'jφ]= ·[ uk u ·(u 'jφ) ]
k j =[1/g] 'k [ g u ·u 'jφ] note contravariant components of φ
kj =[1/g] 'k [ gg 'jφ]
1 ' [ggkj ' ] g k j
Exercise: Find ²[r²esinθ] by working totally in the r,θ,φ system. II-31 4. The Curl, x F
ijk 1. Before deriving the general expression for the curl, we need to consider how ε "transforms" under the generalized coordinate transformation.
Claim:
ε ijk ε'mn A i A j A k m n gx g where gx xyz system metric tensor
Derivation:
Note that gx metric tensor in the Cartesian system and gx |gx| = 1. Consider the following:
ijk i j k mn ε |A|= A A m A n ε' =+(-)|A| if ijk is cyclic (non-cyclic) = 0 if ijk is not a permutation of 123.
where ε'mn is numerically equal to εmn
ijk i j k mn -1 and ε = A A m A n ε' |A|
i j k mn = A A m A n ε' / |A|
i j k mn T 1/2 = A A m A n ε' / [|AA |]
ijk i j k mn ε / gx = A A m A n ε' / g
where one notes that in all cases ( e.g., when |A| <0 ):
g |A|
2. To derive the form for the curl, start in the Cartesian system:
xF = ^x[ 2F3 - 3F2] + ^y[ 3F1 - 1F3] + ^z[ 1F2 - 2F1]
123 132 231 213 312 321 = ε ^x1 2F3 + ε ^x1 3F2 + ε ^x2 3F1 +ε ^x2 1F3 + ε ^x3 1F2 + ε ^x3 2F1
ijk = ε ^xi jFk
ijk =[1/gx]ε ^xi jFk where gx = 1.
That is, the curl can be written like a determinant. II-32
In fact, any cross product in the Cartesian system can be written like a determinant:
FGP ijk xFG ij k lm n lmn xF G Cartesian systems only
3. Now we transform the "determinant" form of the curl:
ijk ijk x F = ε ^xi j Fk =[1/gx] ε ^xi j Fk
i j k mn r s t =[1/g]A A m A n ε' Bi ur Bj 's Bk F't
r i s j t k mn t =[1/g] Bi A Bj A m Bk A n ε' ur 's F't + {... 'sBk ....}
-1T r i -1T s j -1T t k mn =[1/g] [A ]i A [A ]j A m [A ]k A n ε' ur 's F't + {.....}
-1 r i -1 s j -1 t k mn =[1/g] [A ] i A [A ] jA m [A ] kA n ε' ur 's F't + {...}
-1 r -1 s -1 t mn =[1/g] [A A] [A A] m [A A] n ε' ur 's F't + {....}
r s t mn =[1/g] δ δ mδ n ε' ur 's F't + {....}
mn =[1/g] ε' u 'm F'n + {....} where we shall next show that the second term, {...... } = 0.
r i s j k mn t 4. {....} =[1/g] Bi A Bj A m A n ε' ur F't 's Bk
r s k mn t =[1/g] δ δ m A n ε' ur F't 's Bk
r s mn k t =[1/g] δ δ m ur F't ε' A n 's Bk
mn k t =[1/g] u F't ε' A n 'm Bk mn k n t k =[1/g] u F't ε' ( x / q ) 'm ( q / x ) II-33
mn m t k k n t k m k n {...} =[1/g] u F't ε' [ / q ( q / x )( x / q ) - ( q / x ) / q ( x / q )]
mn m t n t m n k =[1/g] u F't ε' [ / q ( q / q ) - Bk / q ( / q )x ]
mn m t t k =[1/g] u F't ε' [ / q (δ n ) - Bk 'm 'n x ]
mn t k =[1/g] u F't ε' [ 0 - Bk 'm 'n x ]
t mn k = -[1/ g] F't Bk ε' u 'm 'n x
t nm k mn nm =[1/g] F't Bk ε' u 'm 'n x since ε' = - ε'
t nm k =[1/g] F'n Bk ε' u 'n 'm x since 'm 'n = 'n 'm
t mn k =[1/g] F'n Bk ε' u 'm 'n x since m and n are dummy indices.
= 0 (note: we showed {...} = - {...} so {...} = 0)
Thus,
ijk xF [1/ g] ' ui ' j F ' k (general q i system )
Note:
ijk ε xixjFk = 0 ==> (rxr)·F = 0
ijk ε i jFk = 0 ==> ( x )·F = 0 if the operators do not act on a function following F.
ijk ε Fixjxk = 0 ==> F·(rxr) = 0
ijk In general, in an expression involving ε or εijk: if the remainder of the expression is symmetric under the interchange of any two of the indices ijk then the total expression is zero. No conclusion can be made if the remainder of the expression is antisymmetric under the interchange of two indices.
II-34
Writing the curl as a determinant can sometimes make a calculation easier:
u1 u2 u3 / / L xF' [1//g] 0 L' L' L' 0 0 1 2 3 0 0 0 0 F '1 F '2 F '3 0
Transforming the general expression for the curl using the metric tensor:
1. First, we note that:
ijk -1 iR jm kn Ri mj nk ε' |g | = g g g ε'Rmn = g g g ε'Rmn
where ε'Rmn is equal numerically to εRmn
ijk Ri mj nk -1 ε' =g g g /|g | ε'Rmn
2. Thus starting with the general form given on page II-33,
ijk ijk L x F =[1//g]ε' ui L'j F'k use the form for ε' given above,
Ri mj nk -1 =[1//g] g g g /|g | ε'Rmn ui L'j F'k
-1 Ri mj nk =[1//g] /|g | ε'Rmn g g g ui L'j F'k
R m nk mj nk =[1//g] g ε'Rmn u [ L' g F'k - F'kg L'jg ] -1| since |g = 1/g R m n R mj nk =[1//g] g ε'Rmn u L' F' - {[1//g] g F'k ε'Rmn u g L'jg ]}
R m n … /g ε'Rmn u L' F'
m mi note that L' = g L'i … L'm R m n and L x F … /g ε'Rmn u L' F' II-35
In general,
m n ijk F x G = g ε'mn u F' G' = [1/ g]ε' ui F'j G'k
m ijk ijk m m W·(F x G) = u W'm · [1/ g]ε' ui F'j G'k = [1/ g]ε' Wi F'j G'k (recall u ·ui = δ i)
Exercise care with such formula when any vector is replaced by . Some other expressions which might be useful:
g g g i mj nk ijk ' ' mn |g | . B B mB n i j k ' ijk |B | mn . i j k A A A ijk m n 'mn |A |
Exercise:
-1 In the stw coordinate system where g and g are given by and F = s²tw^s -t²sw^t+ ln(stw)w^ find u1·xF. 701 12 0 3 1 g 030; g 1 0270 81 10 4 3021