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4.9 Metric tensor for general coordinate system

Consider a general coordinate system (ψ,πœƒ,ΞΆ). I define the metric tensor as the transformation matrix between the covariant basis vectors and the contravariant ones. Equations (89) and (90) express the relation between the two sets of basis vectors using cross product. Next, let us express the relation in matrix from. To obtain the metric matrix, we write the contrariant basis vectors in terms of the covariant ones, such as

1 2 3 βˆ‡Οˆ = a π’₯ βˆ‡πœƒΓ— βˆ‡ ΞΆ + a π’₯ βˆ‡ΞΆ Γ— βˆ‡Οˆ + a π’₯βˆ‡ ψ Γ— βˆ‡πœƒ.
(117)

Taking the scalar product respectively with βˆ‡Οˆ, βˆ‡πœƒ, and βˆ‡ΞΆ, Eq. (117) is written as

a1 = |βˆ‡Οˆ |2,
(118)

2 a = βˆ‡ Οˆβ‹…βˆ‡ πœƒ,
(119)

a3 = βˆ‡ Οˆβ‹…βˆ‡ ΞΆ.
(120)

Similarly, we write

βˆ‡ πœƒ = b1π’₯βˆ‡ πœƒΓ— βˆ‡ ΞΆ + b2π’₯βˆ‡ ΞΆ Γ— βˆ‡ ψ + b3π’₯ βˆ‡Οˆ Γ— βˆ‡ πœƒ,
(121)

Taking the scalar product with βˆ‡Οˆ, βˆ‡πœƒ, and βˆ‡ΞΆ, respectively, the above becomes

1 b = βˆ‡ πœƒβ‹…βˆ‡ ψ
(122)

b2 = |βˆ‡πœƒ|2,
(123)

b3 = βˆ‡ πœƒβ‹…βˆ‡ ΞΆ.
(124)

The same situation applies for the βˆ‡ΞΆ basis vector,

1 2 3 βˆ‡ ΞΆ = c βˆ‡πœƒ Γ— βˆ‡ΞΆπ’₯ + c βˆ‡ΞΆ Γ— βˆ‡Οˆ π’₯ + cβˆ‡ ψ Γ— βˆ‡πœƒπ’₯ ,
(125)

Taking the scalar product with βˆ‡Οˆ, βˆ‡πœƒ, and βˆ‡ΞΆ, respectively, the above equation becomes

c1 = βˆ‡ ΞΆ β‹…βˆ‡ ψ
(126)

c2 = βˆ‡ ΞΆ β‹…βˆ‡ πœƒ
(127)

c3 = |βˆ‡ ΞΆ|2
(128)

Summarizing the above results in matrix form, we obtain

( ) ( 2 ) ( ) βˆ‡ ψ |βˆ‡ ψ| βˆ‡ Οˆβ‹…βˆ‡2πœƒ βˆ‡ ψ β‹…βˆ‡ΞΆ βˆ‡πœƒ Γ—βˆ‡ ΞΆπ’₯ ( βˆ‡πœƒ ) = ( βˆ‡πœƒ β‹…βˆ‡Οˆ |βˆ‡πœƒ| βˆ‡πœƒ β‹…βˆ‡2ΞΆ) ( βˆ‡ ΞΆ Γ— βˆ‡ ψπ’₯ ) βˆ‡ΞΆ βˆ‡ΞΆ β‹…βˆ‡Οˆ βˆ‡ ΞΆ β‹…βˆ‡ πœƒ |βˆ‡ ΞΆ| βˆ‡ ψ Γ— βˆ‡πœƒπ’₯
(129)

Similarly, to convert contravariant basis vector to covariant one, we write

βˆ‡πœƒ Γ— βˆ‡ΞΆπ’₯ = d1βˆ‡ ψ+ d2βˆ‡ πœƒ+ d3βˆ‡ ΞΆ
(130)

Taking the scalar product respectively with βˆ‡πœƒ Γ—βˆ‡ΞΆπ’₯ , βˆ‡ΞΆ Γ—βˆ‡Οˆπ’₯ , and βˆ‡Οˆ Γ—βˆ‡πœƒπ’₯ , the above equation becomes

2 2 d1 = |βˆ‡ πœƒΓ— βˆ‡ ΞΆ|π’₯
(131)

d2 = (βˆ‡ πœƒΓ— βˆ‡ ΞΆπ’₯)β‹…(βˆ‡ ΞΆ Γ— βˆ‡ ψπ’₯)
(132)

d3 = (βˆ‡ πœƒΓ— βˆ‡ ΞΆπ’₯)β‹…(βˆ‡ ΟˆΓ— βˆ‡ πœƒπ’₯)
(133)

For the second contravariant basis vector

βˆ‡Ο• Γ— βˆ‡Οˆ π’₯ = e1βˆ‡ ψ + e2βˆ‡ πœƒ+ e3βˆ‡ ΞΆ
(134)

e1 = (βˆ‡ ΞΆ Γ— βˆ‡ ψ)β‹…(βˆ‡ πœƒΓ— βˆ‡ ΞΆ)π’₯ 2
(135)

e2 = (βˆ‡ ΞΆ Γ— βˆ‡ ψ)β‹…(βˆ‡ ΞΆ Γ— βˆ‡Οˆ )π’₯ 2
(136)

2 e3 = (βˆ‡ ΞΆ Γ— βˆ‡ ψ)β‹…(βˆ‡ ΟˆΓ— βˆ‡ πœƒ)π’₯
(137)

For the third contravariant basis vector

βˆ‡Οˆ Γ— βˆ‡πœƒπ’₯ = f1βˆ‡ ψ + f2βˆ‡ πœƒ+ f3βˆ‡ ΞΆ
(138)

f1 = (βˆ‡Οˆ Γ— βˆ‡πœƒ)β‹…(βˆ‡ πœƒΓ— βˆ‡ ΞΆ)π’₯ 2
(139)

f = (βˆ‡ ψ Γ— βˆ‡πœƒ)β‹…(βˆ‡ ΞΆ Γ— βˆ‡Οˆ )π’₯ 2 2
(140)

2 f3 = (βˆ‡ ψ Γ— βˆ‡πœƒ)β‹…(βˆ‡ ΟˆΓ— βˆ‡ πœƒ)π’₯
(141)

Summarizing these results, we obtain

( ) ( ) ( βˆ‡πœƒ Γ— βˆ‡ΞΆπ’₯ ) ( βˆ‡Οˆ ) βˆ‡ΞΆ Γ— βˆ‡Οˆπ’₯ = M βˆ‡ πœƒ , βˆ‡Οˆ Γ— βˆ‡πœƒπ’₯ βˆ‡ ΞΆ
(142)

where

M = ( ) |βˆ‡πœƒ Γ— βˆ‡ΞΆ|2π’₯2 (βˆ‡πœƒ Γ— βˆ‡ΞΆ)β‹…(βˆ‡ ΞΆ Γ— βˆ‡Οˆ )π’₯ 2 (βˆ‡πœƒ Γ—βˆ‡ ΞΆ)β‹…(βˆ‡Οˆ Γ— βˆ‡ πœƒ)π’₯ 2 ( (βˆ‡ΞΆ Γ— βˆ‡Οˆ )β‹…(βˆ‡πœƒ Γ—βˆ‡ ΞΆ)π’₯2 (βˆ‡ ΞΆ Γ— βˆ‡ ψ)β‹…(βˆ‡ ΞΆ Γ— βˆ‡ ψ)π’₯2 (βˆ‡ ΞΆ Γ— βˆ‡ ψ)β‹…(βˆ‡ ΟˆΓ— βˆ‡ πœƒ)π’₯2) (βˆ‡ ΟˆΓ— βˆ‡ πœƒ)β‹…(βˆ‡πœƒΓ— βˆ‡ ΞΆ)π’₯2 (βˆ‡ ψ Γ— βˆ‡πœƒ)β‹…(βˆ‡ ΞΆ Γ— βˆ‡ ψ)π’₯2 (βˆ‡ ψ Γ— βˆ‡πœƒ)β‹…(βˆ‡ ΟˆΓ— βˆ‡ πœƒ)π’₯2,

This matrix and the matrix in Eqs. (129) should be the inverse of each other. It is ready to prove this by directly calculating the product of the two matrix.

Special case: metric tensor for (ψ,πœƒ,Ο•) coordinate system

Suppose that (ψ,πœƒ,Ο•) are arbitrary general coordinates except that Ο• is the usual toroidal angle in cylindrical coordinates. Then βˆ‡Ο• = 1βˆ•RΛ† Ο• is perpendicular to both βˆ‡Οˆ and βˆ‡πœƒ. Using this, Eq. (129) is simplified to

( ) ( 2 ) ( ) ( βˆ‡Οˆ ) ( |βˆ‡ ψ| βˆ‡ ψ β‹…βˆ‡2πœƒ 0 ) ( βˆ‡ πœƒΓ— βˆ‡ Ο•π’₯ ) βˆ‡πœƒ = βˆ‡ Οˆβ‹…βˆ‡ πœƒ |βˆ‡ πœƒ| 0 2 βˆ‡ ϕ× βˆ‡ ψπ’₯ βˆ‡Ο• 0 0 1βˆ•R βˆ‡ ΟˆΓ— βˆ‡ πœƒπ’₯
(143)

Similarly, Eq. (142) is simplified to

( ) ( ) ( ) βˆ‡πœƒΓ— βˆ‡ Ο•π’₯ |βˆ‡πœƒ|2π’₯2βˆ•R2 βˆ’ βˆ‡ πœƒβ‹…βˆ‡ ψπ’₯ 2βˆ•R2 0 βˆ‡ ψ ( βˆ‡ ϕ× βˆ‡ ψπ’₯ ) = ( βˆ’ βˆ‡Οˆ β‹…βˆ‡ πœƒπ’₯2βˆ•R2 |βˆ‡ ψ|2π’₯ 2βˆ•R2 0 ) ( βˆ‡ πœƒ) βˆ‡ ψ Γ— βˆ‡πœƒπ’₯ 0 0 R2 βˆ‡ Ο•
(144)

[Note that the matrix in Eqs. (143) and (144) should be the inverse of each other. The product of the two matrix,

( |βˆ‡ πœƒ|2π’₯ 2βˆ•R2 βˆ’ βˆ‡ πœƒβ‹…βˆ‡Οˆ π’₯2βˆ•R2 0 ) ( 2 ) ( 2 2 π’₯2- 2 ) ( |βˆ‡ ψ| βˆ‡ Οˆβ‹…βˆ‡2πœƒ 0 ) βˆ’ βˆ‡ ψ β‹…βˆ‡πœƒπ’₯ βˆ•R R2|βˆ‡Οˆ | 02 βˆ‡Οˆ β‹…βˆ‡πœƒ |βˆ‡πœƒ| 0 2 , 0 0 R 0 0 1βˆ•R
(145)

can be calculated to give

( ) A 0 0 ( 0 A 0 ) 0 0 1,

where

A = |βˆ‡πœƒ|2|βˆ‡Οˆ|2π’₯2βˆ•R2 βˆ’ (βˆ‡πœƒ β‹…βˆ‡Οˆ)2π’₯2.

By using the definition of the Jacobian in Eq. (79), it is easy to verify that A = 1, i.e.,

2 |βˆ‡πœƒ|2|βˆ‡ ψ|2 βˆ’ (βˆ‡πœƒ β‹…βˆ‡ ψ)2 = R2, π’₯
(146)

]

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