Curvilinear coordinates can be formulated in tensor calculus, with important applications in physics and engineering, particularly for describing transportation of physical quantities and deformation of matter in fluid mechanics and continuum mechanics.

Vector and tensor algebra in three-dimensional curvilinear coordinates

Elementary vector and tensor algebra in curvilinear coordinates is used in some of the older scientific literature in mechanics and physics and can be indispensable to understanding work from the early and mid 1900s, for example the text by Green and Zerna.^[1] Some useful relations in the algebra of vectors and second-order tensors in curvilinear coordinates are given in this section. The notation and contents are primarily from Ogden,^[2] Naghdi,^[3] Simmonds,^[4] Green and Zerna,^[1] Basar and Weichert,^[5] and Ciarlet.^[6]

Coordinate transformations

Consider two coordinate systems with coordinate variables ${\displaystyle (Z^{1},Z^{2},Z^{3})}$ and ${\displaystyle (Z^{\acute {1)),Z^{\acute {2)),Z^{\acute {3)))}$, which we shall represent in short as just ${\displaystyle Z^{i))$ and ${\displaystyle Z^{\acute {i))}$ respectively and always assume our index ${\displaystyle i}$ runs from 1 through 3. We shall assume that these coordinates systems are embedded in the three-dimensional euclidean space. Coordinates ${\displaystyle Z^{i))$ and ${\displaystyle Z^{\acute {i))}$ may be used to explain each other, because as we move along the coordinate line in one coordinate system we can use the other to describe our position. In this way Coordinates ${\displaystyle Z^{i))$ and ${\displaystyle Z^{\acute {i))}$ are functions of each other

$${\displaystyle Z^{i}=f^{i}(Z^{\acute {1)),Z^{\acute {2)),Z^{\acute {3)))}$$

${\displaystyle i=1,2,3}$

which can be written as

$${\displaystyle Z^{i}=Z^{i}(Z^{\acute {1)),Z^{\acute {2)),Z^{\acute {3)))=Z^{i}(Z^{\acute {i)))}$$

${\displaystyle {\acute {i)),i=1,2,3}$

These three equations together are also called a coordinate transformation from ${\displaystyle Z^{\acute {i))}$ to ${\displaystyle Z^{i))$.Let us denote this transformation by ${\displaystyle T}$. We will therefore represent the transformation from the coordinate system with coordinate variables ${\displaystyle Z^{\acute {i))}$ to the coordinate system with coordinates ${\displaystyle Z^{i))$ as:

$${\displaystyle Z=T({\acute {z)))}$$

Similarly we can represent ${\displaystyle Z^{\acute {i))}$ as a function of ${\displaystyle Z^{i))$ as follows:

$${\displaystyle Z^{\acute {i))=g^{\acute {i))(Z^{1},Z^{2},Z^{3})}$$

${\displaystyle {\acute {i))=1,2,3}$

similarly we can write the free equations more compactly as

$${\displaystyle Z^{\acute {i))=Z^{\acute {i))(Z^{1},Z^{2},Z^{3})=Z^{\acute {i))(Z^{i})}$$

${\displaystyle {\acute {i)),i=1,2,3}$

These three equations together are also called a coordinate transformation from ${\displaystyle Z^{i))$ to ${\displaystyle Z^{\acute {i))}$. Let us denote this transformation by ${\displaystyle S}$. We will represent the transformation from the coordinate system with coordinate variables ${\displaystyle Z^{i))$ to the coordinate system with coordinates ${\displaystyle Z^{\acute {i))}$ as:

$${\displaystyle {\acute {z))=S(z)}$$

If the transformation ${\displaystyle T}$ is bijective then we call the image of the transformation,namely ${\displaystyle Z^{i))$, a set of admissible coordinates for ${\displaystyle Z^{\acute {i))}$. If ${\displaystyle T}$ is linear the coordinate system ${\displaystyle Z^{i))$ will be called an affine coordinate system ,otherwise ${\displaystyle Z^{i))$ is called a curvilinear coordinate system

The Jacobian

As we now see that the Coordinates ${\displaystyle Z^{i))$ and ${\displaystyle Z^{\acute {i))}$ are functions of each other, we can take the derivative of the coordinate variable ${\displaystyle Z^{i))$ with respect to the coordinate variable ${\displaystyle Z^{\acute {i))}$

consider

$${\displaystyle {\frac {\partial {Z^{i))}{\partial {Z^{\acute {i))))}\;{\overset {\underset {\mathrm {def} }{)){=))\;J_{\acute {i))^{i))$$

${\displaystyle {\acute {i)),i=1,2,3}$

${\displaystyle J}$

${\displaystyle J_{\acute {i))^{i))$

${\displaystyle i}$

${\displaystyle {\acute {i))}$

$${\displaystyle J={\begin{pmatrix}J_{\acute {1))^{1}&J_{\acute {2))^{1}&J_{\acute {3))^{1}\\J_{\acute {1))^{2}&J_{\acute {2))^{2}&J_{\acute {3))^{2}\\J_{\acute {1))^{3}&J_{\acute {2))^{3}&J_{\acute {3))^{3}\end{pmatrix))={\begin{pmatrix}{\partial {Z^{1)) \over \partial {Z^{\acute {1))))&{\partial {Z^{1)) \over \partial {Z^{\acute {2))))&{\partial {Z^{1)) \over \partial {Z^{\acute {3))))\\{\partial {Z^{2)) \over \partial {Z^{\acute {1))))&{\partial {Z^{2)) \over \partial {Z^{\acute {2))))&{\partial {Z^{2)) \over \partial {Z^{\acute {3))))\\{\partial {Z^{3)) \over \partial {Z^{\acute {1))))&{\partial {Z^{3)) \over \partial {Z^{\acute {2))))&{\partial {Z^{3)) \over \partial {Z^{\acute {3))))\end{pmatrix))}$$

The resultant matrix is called the Jacobian matrix.

Vectors in curvilinear coordinates

Let (b₁, b₂, b₃) be an arbitrary basis for three-dimensional Euclidean space. In general, the basis vectors are neither unit vectors nor mutually orthogonal. However, they are required to be linearly independent. Then a vector v can be expressed as^[4]: 27

$${\displaystyle \mathbf {v} =v^{k}\,\mathbf {b} _{k))$$

The reciprocal basis (b¹, b², b³) is defined by the relation ^{[4]: 28–29}

$${\displaystyle \mathbf {b} ^{i}\cdot \mathbf {b} _{j}=\delta _{j}^{i))$$

The vector v can also be expressed in terms of the reciprocal basis:

$${\displaystyle \mathbf {v} =v_{k}~\mathbf {b} ^{k))$$

${\displaystyle \mathbf {v} }$

Second-order tensors in curvilinear coordinates

A second-order tensor can be expressed as

$${\displaystyle {\boldsymbol {S))=S^{ij}~\mathbf {b} _{i}\otimes \mathbf {b} _{j}=S_{~j}^{i}~\mathbf {b} _{i}\otimes \mathbf {b} ^{j}=S_{i}^{~j}~\mathbf {b} ^{i}\otimes \mathbf {b} _{j}=S_{ij}~\mathbf {b} ^{i}\otimes \mathbf {b} ^{j))$$

Metric tensor and relations between components

The quantities g_ij, g^ij are defined as^[4]: 39

$${\displaystyle g_{ij}=\mathbf {b} _{i}\cdot \mathbf {b} _{j}=g_{ji}~;~~g^{ij}=\mathbf {b} ^{i}\cdot \mathbf {b} ^{j}=g^{ji))$$

$${\displaystyle v^{i}=g^{ik}~v_{k}~;~~v_{i}=g_{ik}~v^{k}~;~~\mathbf {b} ^{i}=g^{ij}~\mathbf {b} _{j}~;~~\mathbf {b} _{i}=g_{ij}~\mathbf {b} ^{j))$$

The components of a vector are related by^{[4]: 30–32}

$${\displaystyle \mathbf {v} \cdot \mathbf {b} ^{i}=v^{k}~\mathbf {b} _{k}\cdot \mathbf {b} ^{i}=v^{k}~\delta _{k}^{i}=v^{i))$$

$${\displaystyle \mathbf {v} \cdot \mathbf {b} _{i}=v_{k}~\mathbf {b} ^{k}\cdot \mathbf {b} _{i}=v_{k}~\delta _{i}^{k}=v_{i))$$

$${\displaystyle \mathbf {v} \cdot \mathbf {b} _{i}=v^{k}~\mathbf {b} _{k}\cdot \mathbf {b} _{i}=g_{ki}~v^{k))$$

$${\displaystyle \mathbf {v} \cdot \mathbf {b} ^{i}=v_{k}~\mathbf {b} ^{k}\cdot \mathbf {b} ^{i}=g^{ki}~v_{k))$$

The components of the second-order tensor are related by

$${\displaystyle S^{ij}=g^{ik}~S_{k}^{~j}=g^{jk}~S_{~k}^{i}=g^{ik}~g^{jl}~S_{kl))$$

The alternating tensor

In an orthonormal right-handed basis, the third-order alternating tensor is defined as

$${\displaystyle {\boldsymbol {\mathcal {E))}=\varepsilon _{ijk}~\mathbf {e} ^{i}\otimes \mathbf {e} ^{j}\otimes \mathbf {e} ^{k))$$

$${\displaystyle {\boldsymbol {\mathcal {E))}={\mathcal {E))_{ijk}~\mathbf {b} ^{i}\otimes \mathbf {b} ^{j}\otimes \mathbf {b} ^{k}={\mathcal {E))^{ijk}~\mathbf {b} _{i}\otimes \mathbf {b} _{j}\otimes \mathbf {b} _{k))$$

$${\displaystyle {\mathcal {E))_{ijk}=\left[\mathbf {b} _{i},\mathbf {b} _{j},\mathbf {b} _{k}\right]=(\mathbf {b} _{i}\times \mathbf {b} _{j})\cdot \mathbf {b} _{k}~;~~{\mathcal {E))^{ijk}=\left[\mathbf {b} ^{i},\mathbf {b} ^{j},\mathbf {b} ^{k}\right]}$$

$${\displaystyle \mathbf {b} _{i}\times \mathbf {b} _{j}=J~\varepsilon _{ijp}~\mathbf {b} ^{p}={\sqrt {g))~\varepsilon _{ijp}~\mathbf {b} ^{p))$$

$${\displaystyle {\mathcal {E))_{ijk}=J~\varepsilon _{ijk}={\sqrt {g))~\varepsilon _{ijk))$$

$${\displaystyle {\mathcal {E))^{ijk}={\cfrac {1}{J))~\varepsilon ^{ijk}={\cfrac {1}{\sqrt {g))}~\varepsilon ^{ijk))$$

Vector operations

Identity map

The identity map I defined by ${\displaystyle \mathbf {I} \cdot \mathbf {v} =\mathbf {v} }$ can be shown to be:^[4]: 39

$${\displaystyle \mathbf {I} =g^{ij}\mathbf {b} _{i}\otimes \mathbf {b} _{j}=g_{ij}\mathbf {b} ^{i}\otimes \mathbf {b} ^{j}=\mathbf {b} _{i}\otimes \mathbf {b} ^{i}=\mathbf {b} ^{i}\otimes \mathbf {b} _{i))$$

Scalar (dot) product

The scalar product of two vectors in curvilinear coordinates is^[4]: 32

$${\displaystyle \mathbf {u} \cdot \mathbf {v} =u^{i}v_{i}=u_{i}v^{i}=g_{ij}u^{i}v^{j}=g^{ij}u_{i}v_{j))$$

Vector (cross) product

The cross product of two vectors is given by:^{[4]: 32–34}

$${\displaystyle \mathbf {u} \times \mathbf {v} =\varepsilon _{ijk}u_{j}v_{k}\mathbf {e} _{i))$$

where ε_ijk is the permutation symbol and e_i is a Cartesian basis vector. In curvilinear coordinates, the equivalent expression is:

$${\displaystyle \mathbf {u} \times \mathbf {v} =[(\mathbf {b} _{m}\times \mathbf {b} _{n})\cdot \mathbf {b} _{s}]u^{m}v^{n}\mathbf {b} ^{s}={\mathcal {E))_{smn}u^{m}v^{n}\mathbf {b} ^{s))$$

where ${\displaystyle {\mathcal {E))_{ijk))$ is the third-order alternating tensor. The cross product of two vectors is given by:

$${\displaystyle \mathbf {u} \times \mathbf {v} =\varepsilon _{ijk}{\hat {u))_{j}{\hat {v))_{k}\mathbf {e} _{i))$$

where ε_ijk is the permutation symbol and ${\displaystyle \mathbf {e} _{i))$ is a Cartesian basis vector. Therefore,

$${\displaystyle \mathbf {e} _{p}\times \mathbf {e} _{q}=\varepsilon _{ipq}\mathbf {e} _{i))$$

and

$${\displaystyle \mathbf {b} _{m}\times \mathbf {b} _{n}={\frac {\partial \mathbf {x} }{\partial q^{m))}\times {\frac {\partial \mathbf {x} }{\partial q^{n))}={\frac {\partial (x_{p}\mathbf {e} _{p})}{\partial q^{m))}\times {\frac {\partial (x_{q}\mathbf {e} _{q})}{\partial q^{n))}={\frac {\partial x_{p)){\partial q^{m))}{\frac {\partial x_{q)){\partial q^{n))}\mathbf {e} _{p}\times \mathbf {e} _{q}=\varepsilon _{ipq}{\frac {\partial x_{p)){\partial q^{m))}{\frac {\partial x_{q)){\partial q^{n))}\mathbf {e} _{i}.}$$

Hence,

$${\displaystyle (\mathbf {b} _{m}\times \mathbf {b} _{n})\cdot \mathbf {b} _{s}=\varepsilon _{ipq}{\frac {\partial x_{p)){\partial q^{m))}{\frac {\partial x_{q)){\partial q^{n))}{\frac {\partial x_{i)){\partial q^{s))))$$

Returning to the vector product and using the relations:

$${\displaystyle {\hat {u))_{j}={\frac {\partial x_{j)){\partial q^{m))}u^{m},\quad {\hat {v))_{k}={\frac {\partial x_{k)){\partial q^{n))}v^{n},\quad \mathbf {e} _{i}={\frac {\partial x_{i)){\partial q^{s))}\mathbf {b} ^{s},}$$

gives us:

$${\displaystyle \mathbf {u} \times \mathbf {v} =\varepsilon _{ijk}{\hat {u))_{j}{\hat {v))_{k}\mathbf {e} _{i}=\varepsilon _{ijk}{\frac {\partial x_{j)){\partial q^{m))}{\frac {\partial x_{k)){\partial q^{n))}{\frac {\partial x_{i)){\partial q^{s))}u^{m}v^{n}\mathbf {b} ^{s}=[(\mathbf {b} _{m}\times \mathbf {b} _{n})\cdot \mathbf {b} _{s}]u^{m}v^{n}\mathbf {b} ^{s}={\mathcal {E))_{smn}u^{m}v^{n}\mathbf {b} ^{s))$$

Tensor operations

Identity map

The identity map ${\displaystyle {\mathsf {I))}$ defined by ${\displaystyle {\mathsf {I))\cdot \mathbf {v} =\mathbf {v} }$ can be shown to be^[4]: 39

$${\displaystyle {\mathsf {I))=g^{ij}\mathbf {b} _{i}\otimes \mathbf {b} _{j}=g_{ij}\mathbf {b} ^{i}\otimes \mathbf {b} ^{j}=\mathbf {b} _{i}\otimes \mathbf {b} ^{i}=\mathbf {b} ^{i}\otimes \mathbf {b} _{i))$$

Action of a second-order tensor on a vector

The action ${\displaystyle \mathbf {v} ={\boldsymbol {S))\mathbf {u} }$ can be expressed in curvilinear coordinates as

$${\displaystyle v^{i}\mathbf {b} _{i}=S^{ij}u_{j}\mathbf {b} _{i}=S_{j}^{i}u^{j}\mathbf {b} _{i};\qquad v_{i}\mathbf {b} ^{i}=S_{ij}u^{i}\mathbf {b} ^{i}=S_{i}^{j}u_{j}\mathbf {b} ^{i))$$

Inner product of two second-order tensors

The inner product of two second-order tensors ${\displaystyle {\boldsymbol {U))={\boldsymbol {S))\cdot {\boldsymbol {T))}$ can be expressed in curvilinear coordinates as

$${\displaystyle U_{ij}\mathbf {b} ^{i}\otimes \mathbf {b} ^{j}=S_{ik}T_{.j}^{k}\mathbf {b} ^{i}\otimes \mathbf {b} ^{j}=S_{i}^{.k}T_{kj}\mathbf {b} ^{i}\otimes \mathbf {b} ^{j))$$

Alternatively,

$${\displaystyle {\boldsymbol {U))=S^{ij}T_{.n}^{m}g_{jm}\mathbf {b} _{i}\otimes \mathbf {b} ^{n}=S_{.m}^{i}T_{.n}^{m}\mathbf {b} _{i}\otimes \mathbf {b} ^{n}=S^{ij}T_{jn}\mathbf {b} _{i}\otimes \mathbf {b} ^{n))$$

Determinant of a second-order tensor

If ${\displaystyle {\boldsymbol {S))}$ is a second-order tensor, then the determinant is defined by the relation

$${\displaystyle \left[{\boldsymbol {S))\mathbf {u} ,{\boldsymbol {S))\mathbf {v} ,{\boldsymbol {S))\mathbf {w} \right]=\det {\boldsymbol {S))\left[\mathbf {u} ,\mathbf {v} ,\mathbf {w} \right]}$$

where ${\displaystyle \mathbf {u} ,\mathbf {v} ,\mathbf {w} }$ are arbitrary vectors and

$${\displaystyle \left[\mathbf {u} ,\mathbf {v} ,\mathbf {w} \right]:=\mathbf {u} \cdot (\mathbf {v} \times \mathbf {w} ).}$$

Relations between curvilinear and Cartesian basis vectors

Let (e₁, e₂, e₃) be the usual Cartesian basis vectors for the Euclidean space of interest and let

$${\displaystyle \mathbf {b} _{i}={\boldsymbol {F))\mathbf {e} _{i))$$

$${\displaystyle \mathbf {b} _{i}\otimes \mathbf {e} _{i}=({\boldsymbol {F))\mathbf {e} _{i})\otimes \mathbf {e} _{i}={\boldsymbol {F))(\mathbf {e} _{i}\otimes \mathbf {e} _{i})={\boldsymbol {F))~.}$$

$${\displaystyle \mathbf {b} ^{i}={\boldsymbol {F))^{-{\rm {T))}\mathbf {e} ^{i}~;~~g^{ij}=[{\boldsymbol {F))^{-{\rm {1))}{\boldsymbol {F))^{-{\rm {T))}]_{ij}~;~~g_{ij}=[g^{ij}]^{-1}=[{\boldsymbol {F))^{\rm {T)){\boldsymbol {F))]_{ij))$$

${\displaystyle J:=\det {\boldsymbol {F))}$

$${\displaystyle \left[\mathbf {b} _{1},\mathbf {b} _{2},\mathbf {b} _{3}\right]=\det {\boldsymbol {F))\left[\mathbf {e} _{1},\mathbf {e} _{2},\mathbf {e} _{3}\right]~.}$$

$${\displaystyle \left[\mathbf {e} _{1},\mathbf {e} _{2},\mathbf {e} _{3}\right]=1}$$

$${\displaystyle J=\det {\boldsymbol {F))=\left[\mathbf {b} _{1},\mathbf {b} _{2},\mathbf {b} _{3}\right]=\mathbf {b} _{1}\cdot (\mathbf {b} _{2}\times \mathbf {b} _{3})}$$

First, consider

$${\displaystyle g:=\det[g_{ij}]}$$

$${\displaystyle g=\det[{\boldsymbol {F))^{\rm {T))]\cdot \det[{\boldsymbol {F))]=J\cdot J=J^{2))$$

$${\displaystyle \det[g^{ij}]={\cfrac {1}{J^{2))))$$

${\displaystyle [g^{ij}]=[g_{ij}]^{-1))$

$${\displaystyle {\cfrac {\partial g}{\partial g_{ij))}=2~J~{\cfrac {\partial J}{\partial g_{ij))}=g~g^{ij))$$

Another interesting relation is derived below. Recall that

$${\displaystyle \mathbf {b} ^{i}\cdot \mathbf {b} _{j}=\delta _{j}^{i}\quad \Rightarrow \quad \mathbf {b} ^{1}\cdot \mathbf {b} _{1}=1,~\mathbf {b} ^{1}\cdot \mathbf {b} _{2}=\mathbf {b} ^{1}\cdot \mathbf {b} _{3}=0\quad \Rightarrow \quad \mathbf {b} ^{1}=A~(\mathbf {b} _{2}\times \mathbf {b} _{3})}$$

$${\displaystyle \mathbf {b} ^{1}\cdot \mathbf {b} _{1}=A~\mathbf {b} _{1}\cdot (\mathbf {b} _{2}\times \mathbf {b} _{3})=AJ=1\quad \Rightarrow \quad A={\cfrac {1}{J))}$$

$${\displaystyle \mathbf {b} ^{1}={\cfrac {1}{J))(\mathbf {b} _{2}\times \mathbf {b} _{3})~;~~\mathbf {b} ^{2}={\cfrac {1}{J))(\mathbf {b} _{3}\times \mathbf {b} _{1})~;~~\mathbf {b} ^{3}={\cfrac {1}{J))(\mathbf {b} _{1}\times \mathbf {b} _{2})}$$

$${\displaystyle \varepsilon _{ijk}~\mathbf {b} ^{k}={\cfrac {1}{J))(\mathbf {b} _{i}\times \mathbf {b} _{j})={\cfrac {1}{\sqrt {g))}(\mathbf {b} _{i}\times \mathbf {b} _{j})}$$

${\displaystyle \varepsilon _{ijk))$

We have not identified an explicit expression for the transformation tensor F because an alternative form of the mapping between curvilinear and Cartesian bases is more useful. Assuming a sufficient degree of smoothness in the mapping (and a bit of abuse of notation), we have

$${\displaystyle \mathbf {b} _{i}={\cfrac {\partial \mathbf {x} }{\partial q^{i))}={\cfrac {\partial \mathbf {x} }{\partial x_{j))}~{\cfrac {\partial x_{j)){\partial q^{i))}=\mathbf {e} _{j}~{\cfrac {\partial x_{j)){\partial q^{i))))$$

$${\displaystyle \mathbf {e} _{i}=\mathbf {b} _{j}~{\cfrac {\partial q^{j)){\partial x_{i))))$$

$${\displaystyle \mathbf {e} ^{k}\cdot \mathbf {b} _{i}={\frac {\partial x_{k)){\partial q^{i))}\quad \Rightarrow \quad {\frac {\partial x_{k)){\partial q^{i))}~\mathbf {b} ^{i}=\mathbf {e} ^{k}\cdot (\mathbf {b} _{i}\otimes \mathbf {b} ^{i})=\mathbf {e} ^{k))$$

$${\displaystyle \mathbf {b} ^{k}={\frac {\partial q^{k)){\partial x_{i))}~\mathbf {e} ^{i))$$

Vector and tensor calculus in three-dimensional curvilinear coordinates

Simmonds,^[4] in his book on tensor analysis, quotes Albert Einstein saying^[7]

The magic of this theory will hardly fail to impose itself on anybody who has truly understood it; it represents a genuine triumph of the method of absolute differential calculus, founded by Gauss, Riemann, Ricci, and Levi-Civita.

Vector and tensor calculus in general curvilinear coordinates is used in tensor analysis on four-dimensional curvilinear manifolds in general relativity,^[8] in the mechanics of curved shells,^[6] in examining the invariance properties of Maxwell's equations which has been of interest in metamaterials^[9][10] and in many other fields.

Some useful relations in the calculus of vectors and second-order tensors in curvilinear coordinates are given in this section. The notation and contents are primarily from Ogden,^[2] Simmonds,^[4] Green and Zerna,^[1] Basar and Weichert,^[5] and Ciarlet.^[6]

Basic definitions

Let the position of a point in space be characterized by three coordinate variables ${\displaystyle (q^{1},q^{2},q^{3})}$.

The coordinate curve q¹ represents a curve on which q², q³ are constant. Let x be the position vector of the point relative to some origin. Then, assuming that such a mapping and its inverse exist and are continuous, we can write ^[2]: 55

$${\displaystyle \mathbf {x} ={\boldsymbol {\varphi ))(q^{1},q^{2},q^{3})~;~~q^{i}=\psi ^{i}(\mathbf {x} )=[{\boldsymbol {\varphi ))^{-1}(\mathbf {x} )]^{i))$$

The qⁱ coordinate curves are defined by the one-parameter family of functions given by

$${\displaystyle \mathbf {x} _{i}(\alpha )={\boldsymbol {\varphi ))(\alpha ,q^{j},q^{k})~,~~i\neq j\neq k}$$

Tangent vector to coordinate curves

The tangent vector to the curve x_i at the point x_i(α) (or to the coordinate curve q_i at the point x) is

$${\displaystyle {\cfrac {\rm ((d}\mathbf {x} _{i))}{\rm ((d}\alpha ))}\equiv {\cfrac {\partial \mathbf {x} }{\partial q^{i))))$$

Gradient

Scalar field

Let f(x) be a scalar field in space. Then

$${\displaystyle f(\mathbf {x} )=f[{\boldsymbol {\varphi ))(q^{1},q^{2},q^{3})]=f_{\varphi }(q^{1},q^{2},q^{3})}$$

$${\displaystyle [{\boldsymbol {\nabla ))f(\mathbf {x} )]\cdot \mathbf {c} ={\cfrac {\rm {d)){\rm ((d}\alpha ))}f(\mathbf {x} +\alpha \mathbf {c} ){\biggr |}_{\alpha =0))$$

$${\displaystyle q^{i}+\alpha ~c^{i}=\psi ^{i}(\mathbf {x} +\alpha ~\mathbf {c} )}$$

$${\displaystyle [{\boldsymbol {\nabla ))f(\mathbf {x} )]\cdot \mathbf {c} ={\cfrac {\rm {d)){\rm ((d}\alpha ))}f_{\varphi }(q^{1}+\alpha ~c^{1},q^{2}+\alpha ~c^{2},q^{3}+\alpha ~c^{3}){\biggr |}_{\alpha =0}={\cfrac {\partial f_{\varphi )){\partial q^{i))}~c^{i}={\cfrac {\partial f}{\partial q^{i))}~c^{i))$$

If we set ${\displaystyle f(\mathbf {x} )=\psi ^{i}(\mathbf {x} )}$, then since ${\displaystyle q^{i}=\psi ^{i}(\mathbf {x} )}$, we have

$${\displaystyle [{\boldsymbol {\nabla ))\psi ^{i}(\mathbf {x} )]\cdot \mathbf {c} ={\cfrac {\partial \psi ^{i)){\partial q^{j))}~c^{j}=c^{i))$$

If b_i is the covariant (or natural) basis at a point, and if bⁱ is the contravariant (or reciprocal) basis at that point, then

$${\displaystyle [{\boldsymbol {\nabla ))f(\mathbf {x} )]\cdot \mathbf {c} ={\cfrac {\partial f}{\partial q^{i))}~c^{i}=\left({\cfrac {\partial f}{\partial q^{i))}~\mathbf {b} ^{i}\right)\left(c^{i}~\mathbf {b} _{i}\right)\quad \Rightarrow \quad {\boldsymbol {\nabla ))f(\mathbf {x} )={\cfrac {\partial f}{\partial q^{i))}~\mathbf {b} ^{i))$$

Vector field

A similar process can be used to arrive at the gradient of a vector field f(x). The gradient is given by

$${\displaystyle [{\boldsymbol {\nabla ))\mathbf {f} (\mathbf {x} )]\cdot \mathbf {c} ={\cfrac {\partial \mathbf {f} }{\partial q^{i))}~c^{i))$$

$${\displaystyle \mathbf {c} ={\cfrac {\partial \mathbf {x} }{\partial q^{i))}~c^{i}=\mathbf {b} _{i}(\mathbf {x} )~c^{i}~;~~\mathbf {b} _{i}(\mathbf {x} ):={\cfrac {\partial \mathbf {x} }{\partial q^{i))))$$

Since c is arbitrary, we can write

$${\displaystyle {\boldsymbol {\nabla ))\mathbf {f} (\mathbf {x} )={\cfrac {\partial \mathbf {f} }{\partial q^{i))}\otimes \mathbf {b} ^{i))$$

Note that the contravariant basis vector bⁱ is perpendicular to the surface of constant ψⁱ and is given by

$${\displaystyle \mathbf {b} ^{i}={\boldsymbol {\nabla ))\psi ^{i))$$

Christoffel symbols of the first kind

The Christoffel symbols of the first kind are defined as

$${\displaystyle \mathbf {b} _{i,j}={\frac {\partial \mathbf {b} _{i)){\partial q^{j))}:=\Gamma _{ijk}~\mathbf {b} ^{k}\quad \Rightarrow \quad \mathbf {b} _{i,j}\cdot \mathbf {b} _{l}=\Gamma _{ijl))$$

$${\displaystyle {\begin{aligned}g_{ij,k}&=(\mathbf {b} _{i}\cdot \mathbf {b} _{j})_{,k}=\mathbf {b} _{i,k}\cdot \mathbf {b} _{j}+\mathbf {b} _{i}\cdot \mathbf {b} _{j,k}=\Gamma _{ikj}+\Gamma _{jki}\\g_{ik,j}&=(\mathbf {b} _{i}\cdot \mathbf {b} _{k})_{,j}=\mathbf {b} _{i,j}\cdot \mathbf {b} _{k}+\mathbf {b} _{i}\cdot \mathbf {b} _{k,j}=\Gamma _{ijk}+\Gamma _{kji}\\g_{jk,i}&=(\mathbf {b} _{j}\cdot \mathbf {b} _{k})_{,i}=\mathbf {b} _{j,i}\cdot \mathbf {b} _{k}+\mathbf {b} _{j}\cdot \mathbf {b} _{k,i}=\Gamma _{jik}+\Gamma _{kij}\end{aligned))}$$

$${\displaystyle \Gamma _{ijk}={\frac {1}{2))(g_{ik,j}+g_{jk,i}-g_{ij,k})={\frac {1}{2))[(\mathbf {b} _{i}\cdot \mathbf {b} _{k})_{,j}+(\mathbf {b} _{j}\cdot \mathbf {b} _{k})_{,i}-(\mathbf {b} _{i}\cdot \mathbf {b} _{j})_{,k}]}$$

Christoffel symbols of the second kind

The Christoffel symbols of the second kind are defined as

$${\displaystyle \Gamma _{ij}^{k}=\Gamma _{ji}^{k))$$

$${\displaystyle {\cfrac {\partial \mathbf {b} _{i)){\partial q^{j))}=\Gamma _{ij}^{k}~\mathbf {b} _{k))$$

This implies that

$${\displaystyle \Gamma _{ij}^{k}={\cfrac {\partial \mathbf {b} _{i)){\partial q^{j))}\cdot \mathbf {b} ^{k}=-\mathbf {b} _{i}\cdot {\cfrac {\partial \mathbf {b} ^{k)){\partial q^{j))))$$

$${\displaystyle {\cfrac {\partial \mathbf {b} ^{i)){\partial q^{j))}=-\Gamma _{jk}^{i}~\mathbf {b} ^{k}~;~~{\boldsymbol {\nabla ))\mathbf {b} _{i}=\Gamma _{ij}^{k}~\mathbf {b} _{k}\otimes \mathbf {b} ^{j}~;~~{\boldsymbol {\nabla ))\mathbf {b} ^{i}=-\Gamma _{jk}^{i}~\mathbf {b} ^{k}\otimes \mathbf {b} ^{j))$$

Another particularly useful relation, which shows that the Christoffel symbol depends only on the metric tensor and its derivatives, is

$${\displaystyle \Gamma _{ij}^{k}={\frac {g^{km)){2))\left({\frac {\partial g_{mi)){\partial q^{j))}+{\frac {\partial g_{mj)){\partial q^{i))}-{\frac {\partial g_{ij)){\partial q^{m))}\right)}$$

Explicit expression for the gradient of a vector field

The following expressions for the gradient of a vector field in curvilinear coordinates are quite useful.

$${\displaystyle {\begin{aligned}{\boldsymbol {\nabla ))\mathbf {v} &=\left[{\cfrac {\partial v^{i)){\partial q^{k))}+\Gamma _{lk}^{i}~v^{l}\right]~\mathbf {b} _{i}\otimes \mathbf {b} ^{k}\\[8pt]&=\left[{\cfrac {\partial v_{i)){\partial q^{k))}-\Gamma _{ki}^{l}~v_{l}\right]~\mathbf {b} ^{i}\otimes \mathbf {b} ^{k}\end{aligned))}$$

Representing a physical vector field

The vector field v can be represented as

$${\displaystyle \mathbf {v} =v_{i}~\mathbf {b} ^{i}={\hat {v))_{i}~{\hat {\mathbf {b} ))^{i))$$

${\displaystyle v_{i))$

${\displaystyle {\hat {v))_{i))$

$${\displaystyle {\hat {\mathbf {b} ))^{i}={\cfrac {\mathbf {b} ^{i)){\sqrt {g^{ii))))}$$

Second-order tensor field

The gradient of a second order tensor field can similarly be expressed as

$${\displaystyle {\boldsymbol {\nabla )){\boldsymbol {S))={\frac {\partial {\boldsymbol {S))}{\partial q^{i))}\otimes \mathbf {b} ^{i))$$

Explicit expressions for the gradient

If we consider the expression for the tensor in terms of a contravariant basis, then

$${\displaystyle {\boldsymbol {\nabla )){\boldsymbol {S))={\frac {\partial }{\partial q^{k))}[S_{ij}~\mathbf {b} ^{i}\otimes \mathbf {b} ^{j}]\otimes \mathbf {b} ^{k}=\left[{\frac {\partial S_{ij)){\partial q^{k))}-\Gamma _{ki}^{l}~S_{lj}-\Gamma _{kj}^{l}~S_{il}\right]~\mathbf {b} ^{i}\otimes \mathbf {b} ^{j}\otimes \mathbf {b} ^{k))$$

$${\displaystyle {\begin{aligned}{\boldsymbol {\nabla )){\boldsymbol {S))&=\left[{\cfrac {\partial S^{ij)){\partial q^{k))}+\Gamma _{kl}^{i}~S^{lj}+\Gamma _{kl}^{j}~S^{il}\right]~\mathbf {b} _{i}\otimes \mathbf {b} _{j}\otimes \mathbf {b} ^{k}\\[8pt]&=\left[{\cfrac {\partial S_{~j}^{i)){\partial q^{k))}+\Gamma _{kl}^{i}~S_{~j}^{l}-\Gamma _{kj}^{l}~S_{~l}^{i}\right]~\mathbf {b} _{i}\otimes \mathbf {b} ^{j}\otimes \mathbf {b} ^{k}\\[8pt]&=\left[{\cfrac {\partial S_{i}^{~j)){\partial q^{k))}-\Gamma _{ik}^{l}~S_{l}^{~j}+\Gamma _{kl}^{j}~S_{i}^{~l}\right]~\mathbf {b} ^{i}\otimes \mathbf {b} _{j}\otimes \mathbf {b} ^{k}\end{aligned))}$$

Representing a physical second-order tensor field

The physical components of a second-order tensor field can be obtained by using a normalized contravariant basis, i.e.,

$${\displaystyle {\boldsymbol {S))=S_{ij}~\mathbf {b} ^{i}\otimes \mathbf {b} ^{j}={\hat {S))_{ij}~{\hat {\mathbf {b} ))^{i}\otimes {\hat {\mathbf {b} ))^{j))$$

$${\displaystyle {\hat {S))_{ij}=S_{ij}~{\sqrt {g^{ii}~g^{jj))))$$

Divergence

Vector field

The divergence of a vector field (${\displaystyle \mathbf {v} }$)is defined as

$${\displaystyle \operatorname {div} ~\mathbf {v} ={\boldsymbol {\nabla ))\cdot \mathbf {v} ={\text{tr))({\boldsymbol {\nabla ))\mathbf {v} )}$$

$${\displaystyle {\boldsymbol {\nabla ))\cdot \mathbf {v} ={\cfrac {\partial v^{i)){\partial q^{i))}+\Gamma _{\ell i}^{i}~v^{\ell }=\left[{\cfrac {\partial v_{i)){\partial q^{j))}-\Gamma _{ji}^{\ell }~v_{\ell }\right]~g^{ij))$$

An alternative equation for the divergence of a vector field is frequently used. To derive this relation recall that

$${\displaystyle {\boldsymbol {\nabla ))\cdot \mathbf {v} ={\frac {\partial v^{i)){\partial q^{i))}+\Gamma _{\ell i}^{i}~v^{\ell ))$$

$${\displaystyle \Gamma _{\ell i}^{i}=\Gamma _{i\ell }^{i}={\cfrac {g^{mi)){2))\left[{\frac {\partial g_{im)){\partial q^{\ell ))}+{\frac {\partial g_{\ell m)){\partial q^{i))}-{\frac {\partial g_{il)){\partial q^{m))}\right]}$$

${\displaystyle {\boldsymbol {g))}$

$${\displaystyle g^{mi}~{\frac {\partial g_{\ell m)){\partial q^{i))}=g^{mi}~{\frac {\partial g_{i\ell )){\partial q^{m))))$$

$${\displaystyle {\boldsymbol {\nabla ))\cdot \mathbf {v} ={\frac {\partial v^{i)){\partial q^{i))}+{\cfrac {g^{mi)){2))~{\frac {\partial g_{im)){\partial q^{\ell ))}~v^{\ell ))$$

${\displaystyle [g_{ij}]^{-1}=[g^{ij}]}$

$${\displaystyle [g^{ij}]=[g_{ij}]^{-1}={\cfrac {A^{ij)){g))~;~~g:=\det([g_{ij}])=\det {\boldsymbol {g))}$$

$${\displaystyle g=\det([g_{ij}])=\sum _{i}g_{ij}~A^{ij}\quad \Rightarrow \quad {\frac {\partial g}{\partial g_{ij))}=A^{ij))$$

$${\displaystyle [g^{ij}]={\cfrac {1}{g))~{\frac {\partial g}{\partial g_{ij))))$$