Tensors in curvilinear coordinates

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.

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]

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

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

which can be written as

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

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:

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

and we can write the free equations more compactly as

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

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.

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

$${\displaystyle {\frac {\partial {Z^{i}}}{\partial {Z^{\acute {i}}}}}\;{\overset {\underset {\mathrm {def} }{}}{=}}\;J_{\acute {i}}^{i}}$$ for ${\displaystyle {\acute {i}},i=1,2,3}$, these derivatives can be arranged in a matrix, say ${\displaystyle J}$, in which ${\displaystyle J_{\acute {i}}^{i}}$ is the element in the ${\displaystyle i}$-th row and ${\displaystyle {\acute {i}}}$-th column

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

The resultant matrix is called the Jacobian matrix.

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 components v^k are the contravariant components of the vector v.

The reciprocal basis (b¹, b², b³) is defined by the relation ^{[4]: 28–29} $${\displaystyle \mathbf {b} ^{i}\cdot \mathbf {b} _{j}=\delta _{j}^{i}}$$ where δⁱ _j is the Kronecker delta.

The vector v can also be expressed in terms of the reciprocal basis: $${\displaystyle \mathbf {v} =v_{k}~\mathbf {b} ^{k}}$$ The components v_k are the covariant components of the vector ${\displaystyle \mathbf {v} }$.

A second-order tensor can be expressed as $${\displaystyle {\boldsymbol {S}}=S^{ij}~\mathbf {b} _{i}\otimes \mathbf {b} _{j}=S_{~j}^{i}~\mathbf {b} _{i}\otimes \mathbf {b} ^{j}=S_{i}^{~j}~\mathbf {b} ^{i}\otimes \mathbf {b} _{j}=S_{ij}~\mathbf {b} ^{i}\otimes \mathbf {b} ^{j}}$$ The components S^ij are called the contravariant components, Sⁱ _j the mixed right-covariant components, S_i ^j the mixed left-covariant components, and S_ij the covariant components of the second-order tensor.

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

$${\displaystyle g_{ij}=\mathbf {b} _{i}\cdot \mathbf {b} _{j}=g_{ji}~;~~g^{ij}=\mathbf {b} ^{i}\cdot \mathbf {b} ^{j}=g^{ji}}$$ From the above equations we have $${\displaystyle v^{i}=g^{ik}~v_{k}~;~~v_{i}=g_{ik}~v^{k}~;~~\mathbf {b} ^{i}=g^{ij}~\mathbf {b} _{j}~;~~\mathbf {b} _{i}=g_{ij}~\mathbf {b} ^{j}}$$

The components of a vector are related by^{[4]: 30–32} $${\displaystyle \mathbf {v} \cdot \mathbf {b} ^{i}=v^{k}~\mathbf {b} _{k}\cdot \mathbf {b} ^{i}=v^{k}~\delta _{k}^{i}=v^{i}}$$ $${\displaystyle \mathbf {v} \cdot \mathbf {b} _{i}=v_{k}~\mathbf {b} ^{k}\cdot \mathbf {b} _{i}=v_{k}~\delta _{i}^{k}=v_{i}}$$ Also, $${\displaystyle \mathbf {v} \cdot \mathbf {b} _{i}=v^{k}~\mathbf {b} _{k}\cdot \mathbf {b} _{i}=g_{ki}~v^{k}}$$ $${\displaystyle \mathbf {v} \cdot \mathbf {b} ^{i}=v_{k}~\mathbf {b} ^{k}\cdot \mathbf {b} ^{i}=g^{ki}~v_{k}}$$

The components of the second-order tensor are related by $${\displaystyle S^{ij}=g^{ik}~S_{k}^{~j}=g^{jk}~S_{~k}^{i}=g^{ik}~g^{jl}~S_{kl}}$$

In an orthonormal right-handed basis, the third-order alternating tensor is defined as $${\displaystyle {\boldsymbol {\mathcal {E}}}=\varepsilon _{ijk}~\mathbf {e} ^{i}\otimes \mathbf {e} ^{j}\otimes \mathbf {e} ^{k}}$$ In a general curvilinear basis the same tensor may be expressed as $${\displaystyle {\boldsymbol {\mathcal {E}}}={\mathcal {E}}_{ijk}~\mathbf {b} ^{i}\otimes \mathbf {b} ^{j}\otimes \mathbf {b} ^{k}={\mathcal {E}}^{ijk}~\mathbf {b} _{i}\otimes \mathbf {b} _{j}\otimes \mathbf {b} _{k}}$$ It can be shown that $${\displaystyle {\mathcal {E}}_{ijk}=\left[\mathbf {b} _{i},\mathbf {b} _{j},\mathbf {b} _{k}\right]=(\mathbf {b} _{i}\times \mathbf {b} _{j})\cdot \mathbf {b} _{k}~;~~{\mathcal {E}}^{ijk}=\left[\mathbf {b} ^{i},\mathbf {b} ^{j},\mathbf {b} ^{k}\right]}$$ Now, $${\displaystyle \mathbf {b} _{i}\times \mathbf {b} _{j}=J~\varepsilon _{ijp}~\mathbf {b} ^{p}={\sqrt {g}}~\varepsilon _{ijp}~\mathbf {b} ^{p}}$$ Hence, $${\displaystyle {\mathcal {E}}_{ijk}=J~\varepsilon _{ijk}={\sqrt {g}}~\varepsilon _{ijk}}$$ Similarly, we can show that $${\displaystyle {\mathcal {E}}^{ijk}={\cfrac {1}{J}}~\varepsilon ^{ijk}={\cfrac {1}{\sqrt {g}}}~\varepsilon ^{ijk}}$$

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

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

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

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

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:

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

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

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

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

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

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,

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

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

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

Let (e₁, e₂, e₃) be the usual Cartesian basis vectors for the Euclidean space of interest and let $${\displaystyle \mathbf {b} _{i}={\boldsymbol {F}}\mathbf {e} _{i}}$$ where F_i is a second-order transformation tensor that maps e_i to b_i. Then, $${\displaystyle \mathbf {b} _{i}\otimes \mathbf {e} _{i}=({\boldsymbol {F}}\mathbf {e} _{i})\otimes \mathbf {e} _{i}={\boldsymbol {F}}(\mathbf {e} _{i}\otimes \mathbf {e} _{i})={\boldsymbol {F}}~.}$$ From this relation we can show that $${\displaystyle \mathbf {b} ^{i}={\boldsymbol {F}}^{-{\rm {T}}}\mathbf {e} ^{i}~;~~g^{ij}=[{\boldsymbol {F}}^{-{\rm {1}}}{\boldsymbol {F}}^{-{\rm {T}}}]_{ij}~;~~g_{ij}=[g^{ij}]^{-1}=[{\boldsymbol {F}}^{\rm {T}}{\boldsymbol {F}}]_{ij}}$$ Let ${\displaystyle J:=\det {\boldsymbol {F}}}$ be the Jacobian of the transformation. Then, from the definition of the determinant, $${\displaystyle \left[\mathbf {b} _{1},\mathbf {b} _{2},\mathbf {b} _{3}\right]=\det {\boldsymbol {F}}\left[\mathbf {e} _{1},\mathbf {e} _{2},\mathbf {e} _{3}\right]~.}$$ Since $${\displaystyle \left[\mathbf {e} _{1},\mathbf {e} _{2},\mathbf {e} _{3}\right]=1}$$ we have $${\displaystyle J=\det {\boldsymbol {F}}=\left[\mathbf {b} _{1},\mathbf {b} _{2},\mathbf {b} _{3}\right]=\mathbf {b} _{1}\cdot (\mathbf {b} _{2}\times \mathbf {b} _{3})}$$ A number of interesting results can be derived using the above relations.

First, consider $${\displaystyle g:=\det[g_{ij}]}$$ Then $${\displaystyle g=\det[{\boldsymbol {F}}^{\rm {T}}]\cdot \det[{\boldsymbol {F}}]=J\cdot J=J^{2}}$$ Similarly, we can show that $${\displaystyle \det[g^{ij}]={\cfrac {1}{J^{2}}}}$$ Therefore, using the fact that ${\displaystyle [g^{ij}]=[g_{ij}]^{-1}}$, $${\displaystyle {\cfrac {\partial g}{\partial g_{ij}}}=2~J~{\cfrac {\partial J}{\partial g_{ij}}}=g~g^{ij}}$$

Another interesting relation is derived below. Recall that $${\displaystyle \mathbf {b} ^{i}\cdot \mathbf {b} _{j}=\delta _{j}^{i}\quad \Rightarrow \quad \mathbf {b} ^{1}\cdot \mathbf {b} _{1}=1,~\mathbf {b} ^{1}\cdot \mathbf {b} _{2}=\mathbf {b} ^{1}\cdot \mathbf {b} _{3}=0\quad \Rightarrow \quad \mathbf {b} ^{1}=A~(\mathbf {b} _{2}\times \mathbf {b} _{3})}$$ where A is a, yet undetermined, constant. Then $${\displaystyle \mathbf {b} ^{1}\cdot \mathbf {b} _{1}=A~\mathbf {b} _{1}\cdot (\mathbf {b} _{2}\times \mathbf {b} _{3})=AJ=1\quad \Rightarrow \quad A={\cfrac {1}{J}}}$$ This observation leads to the relations $${\displaystyle \mathbf {b} ^{1}={\cfrac {1}{J}}(\mathbf {b} _{2}\times \mathbf {b} _{3})~;~~\mathbf {b} ^{2}={\cfrac {1}{J}}(\mathbf {b} _{3}\times \mathbf {b} _{1})~;~~\mathbf {b} ^{3}={\cfrac {1}{J}}(\mathbf {b} _{1}\times \mathbf {b} _{2})}$$ In index notation, $${\displaystyle \varepsilon _{ijk}~\mathbf {b} ^{k}={\cfrac {1}{J}}(\mathbf {b} _{i}\times \mathbf {b} _{j})={\cfrac {1}{\sqrt {g}}}(\mathbf {b} _{i}\times \mathbf {b} _{j})}$$ where ${\displaystyle \varepsilon _{ijk}}$ is the usual permutation symbol.

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 $${\displaystyle \mathbf {b} _{i}={\cfrac {\partial \mathbf {x} }{\partial q^{i}}}={\cfrac {\partial \mathbf {x} }{\partial x_{j}}}~{\cfrac {\partial x_{j}}{\partial q^{i}}}=\mathbf {e} _{j}~{\cfrac {\partial x_{j}}{\partial q^{i}}}}$$ Similarly, $${\displaystyle \mathbf {e} _{i}=\mathbf {b} _{j}~{\cfrac {\partial q^{j}}{\partial x_{i}}}}$$ From these results we have $${\displaystyle \mathbf {e} ^{k}\cdot \mathbf {b} _{i}={\frac {\partial x_{k}}{\partial q^{i}}}\quad \Rightarrow \quad {\frac {\partial x_{k}}{\partial q^{i}}}~\mathbf {b} ^{i}=\mathbf {e} ^{k}\cdot (\mathbf {b} _{i}\otimes \mathbf {b} ^{i})=\mathbf {e} ^{k}}$$ and $${\displaystyle \mathbf {b} ^{k}={\frac {\partial q^{k}}{\partial x_{i}}}~\mathbf {e} ^{i}}$$

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]

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 fields ψⁱ(x) are called the curvilinear coordinate functions of the curvilinear coordinate system ψ(x) = φ⁻¹(x).

The qⁱ coordinate curves are defined by the one-parameter family of functions given by $${\displaystyle \mathbf {x} _{i}(\alpha )={\boldsymbol {\varphi }}(\alpha ,q^{j},q^{k})~,~~i\neq j\neq k}$$ with q^j, q^k fixed.

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

Let f(x) be a scalar field in space. Then $${\displaystyle f(\mathbf {x} )=f[{\boldsymbol {\varphi }}(q^{1},q^{2},q^{3})]=f_{\varphi }(q^{1},q^{2},q^{3})}$$ The gradient of the field f is defined by $${\displaystyle [{\boldsymbol {\nabla }}f(\mathbf {x} )]\cdot \mathbf {c} ={\cfrac {\rm {d}}{\rm {{d}\alpha }}}f(\mathbf {x} +\alpha \mathbf {c} ){\biggr |}_{\alpha =0}}$$ where c is an arbitrary constant vector. If we define the components cⁱ of c are such that $${\displaystyle q^{i}+\alpha ~c^{i}=\psi ^{i}(\mathbf {x} +\alpha ~\mathbf {c} )}$$ then $${\displaystyle [{\boldsymbol {\nabla }}f(\mathbf {x} )]\cdot \mathbf {c} ={\cfrac {\rm {d}}{\rm {{d}\alpha }}}f_{\varphi }(q^{1}+\alpha ~c^{1},q^{2}+\alpha ~c^{2},q^{3}+\alpha ~c^{3}){\biggr |}_{\alpha =0}={\cfrac {\partial f_{\varphi }}{\partial q^{i}}}~c^{i}={\cfrac {\partial f}{\partial q^{i}}}~c^{i}}$$

If we set ${\displaystyle f(\mathbf {x} )=\psi ^{i}(\mathbf {x} )}$, then since ${\displaystyle q^{i}=\psi ^{i}(\mathbf {x} )}$, we have $${\displaystyle [{\boldsymbol {\nabla }}\psi ^{i}(\mathbf {x} )]\cdot \mathbf {c} ={\cfrac {\partial \psi ^{i}}{\partial q^{j}}}~c^{j}=c^{i}}$$ which provides a means of extracting the contravariant component of a vector c.

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 $${\displaystyle [{\boldsymbol {\nabla }}f(\mathbf {x} )]\cdot \mathbf {c} ={\cfrac {\partial f}{\partial q^{i}}}~c^{i}=\left({\cfrac {\partial f}{\partial q^{i}}}~\mathbf {b} ^{i}\right)\left(c^{i}~\mathbf {b} _{i}\right)\quad \Rightarrow \quad {\boldsymbol {\nabla }}f(\mathbf {x} )={\cfrac {\partial f}{\partial q^{i}}}~\mathbf {b} ^{i}}$$ A brief rationale for this choice of basis is given in the next section.

A similar process can be used to arrive at the gradient of a vector field f(x). The gradient is given by $${\displaystyle [{\boldsymbol {\nabla }}\mathbf {f} (\mathbf {x} )]\cdot \mathbf {c} ={\cfrac {\partial \mathbf {f} }{\partial q^{i}}}~c^{i}}$$ If we consider the gradient of the position vector field r(x) = x, then we can show that $${\displaystyle \mathbf {c} ={\cfrac {\partial \mathbf {x} }{\partial q^{i}}}~c^{i}=\mathbf {b} _{i}(\mathbf {x} )~c^{i}~;~~\mathbf {b} _{i}(\mathbf {x} ):={\cfrac {\partial \mathbf {x} }{\partial q^{i}}}}$$ The vector field b_i is tangent to the qⁱ coordinate curve and forms a natural basis at each point on the curve. This basis, as discussed at the beginning of this article, is also called the covariant curvilinear basis. We can also define a reciprocal basis, or contravariant curvilinear basis, bⁱ. All the algebraic relations between the basis vectors, as discussed in the section on tensor algebra, apply for the natural basis and its reciprocal at each point x.

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

The Christoffel symbols of the first kind are defined as $${\displaystyle \mathbf {b} _{i,j}={\frac {\partial \mathbf {b} _{i}}{\partial q^{j}}}:=\Gamma _{ijk}~\mathbf {b} ^{k}\quad \Rightarrow \quad \mathbf {b} _{i,j}\cdot \mathbf {b} _{l}=\Gamma _{ijl}}$$ To express Γ_ijk in terms of g_ij we note that $${\displaystyle {\begin{aligned}g_{ij,k}&=(\mathbf {b} _{i}\cdot \mathbf {b} _{j})_{,k}=\mathbf {b} _{i,k}\cdot \mathbf {b} _{j}+\mathbf {b} _{i}\cdot \mathbf {b} _{j,k}=\Gamma _{ikj}+\Gamma _{jki}\\g_{ik,j}&=(\mathbf {b} _{i}\cdot \mathbf {b} _{k})_{,j}=\mathbf {b} _{i,j}\cdot \mathbf {b} _{k}+\mathbf {b} _{i}\cdot \mathbf {b} _{k,j}=\Gamma _{ijk}+\Gamma _{kji}\\g_{jk,i}&=(\mathbf {b} _{j}\cdot \mathbf {b} _{k})_{,i}=\mathbf {b} _{j,i}\cdot \mathbf {b} _{k}+\mathbf {b} _{j}\cdot \mathbf {b} _{k,i}=\Gamma _{jik}+\Gamma _{kij}\end{aligned}}}$$ Since b_i,j = b_j,i we have Γ_ijk = Γ_jik. Using these to rearrange the above relations gives $${\displaystyle \Gamma _{ijk}={\frac {1}{2}}(g_{ik,j}+g_{jk,i}-g_{ij,k})={\frac {1}{2}}[(\mathbf {b} _{i}\cdot \mathbf {b} _{k})_{,j}+(\mathbf {b} _{j}\cdot \mathbf {b} _{k})_{,i}-(\mathbf {b} _{i}\cdot \mathbf {b} _{j})_{,k}]}$$

The Christoffel symbols of the second kind are defined as $${\displaystyle \Gamma _{ij}^{k}=\Gamma _{ji}^{k}}$$ in which

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

This implies that $${\displaystyle \Gamma _{ij}^{k}={\cfrac {\partial \mathbf {b} _{i}}{\partial q^{j}}}\cdot \mathbf {b} ^{k}=-\mathbf {b} _{i}\cdot {\cfrac {\partial \mathbf {b} ^{k}}{\partial q^{j}}}}$$ Other relations that follow are $${\displaystyle {\cfrac {\partial \mathbf {b} ^{i}}{\partial q^{j}}}=-\Gamma _{jk}^{i}~\mathbf {b} ^{k}~;~~{\boldsymbol {\nabla }}\mathbf {b} _{i}=\Gamma _{ij}^{k}~\mathbf {b} _{k}\otimes \mathbf {b} ^{j}~;~~{\boldsymbol {\nabla }}\mathbf {b} ^{i}=-\Gamma _{jk}^{i}~\mathbf {b} ^{k}\otimes \mathbf {b} ^{j}}$$

Another particularly useful relation, which shows that the Christoffel symbol depends only on the metric tensor and its derivatives, is $${\displaystyle \Gamma _{ij}^{k}={\frac {g^{km}}{2}}\left({\frac {\partial g_{mi}}{\partial q^{j}}}+{\frac {\partial g_{mj}}{\partial q^{i}}}-{\frac {\partial g_{ij}}{\partial q^{m}}}\right)}$$

The following expressions for the gradient of a vector field in curvilinear coordinates are quite useful. $${\displaystyle {\begin{aligned}{\boldsymbol {\nabla }}\mathbf {v} &=\left[{\cfrac {\partial v^{i}}{\partial q^{k}}}+\Gamma _{lk}^{i}~v^{l}\right]~\mathbf {b} _{i}\otimes \mathbf {b} ^{k}\\[8pt]&=\left[{\cfrac {\partial v_{i}}{\partial q^{k}}}-\Gamma _{ki}^{l}~v_{l}\right]~\mathbf {b} ^{i}\otimes \mathbf {b} ^{k}\end{aligned}}}$$

The vector field v can be represented as $${\displaystyle \mathbf {v} =v_{i}~\mathbf {b} ^{i}={\hat {v}}_{i}~{\hat {\mathbf {b} }}^{i}}$$ where ${\displaystyle v_{i}}$ are the covariant components of the field, ${\displaystyle {\hat {v}}_{i}}$ are the physical components, and (no summation) $${\displaystyle {\hat {\mathbf {b} }}^{i}={\cfrac {\mathbf {b} ^{i}}{\sqrt {g^{ii}}}}}$$ is the normalized contravariant basis vector.

The gradient of a second order tensor field can similarly be expressed as $${\displaystyle {\boldsymbol {\nabla }}{\boldsymbol {S}}={\frac {\partial {\boldsymbol {S}}}{\partial q^{i}}}\otimes \mathbf {b} ^{i}}$$

If we consider the expression for the tensor in terms of a contravariant basis, then $${\displaystyle {\boldsymbol {\nabla }}{\boldsymbol {S}}={\frac {\partial }{\partial q^{k}}}[S_{ij}~\mathbf {b} ^{i}\otimes \mathbf {b} ^{j}]\otimes \mathbf {b} ^{k}=\left[{\frac {\partial S_{ij}}{\partial q^{k}}}-\Gamma _{ki}^{l}~S_{lj}-\Gamma _{kj}^{l}~S_{il}\right]~\mathbf {b} ^{i}\otimes \mathbf {b} ^{j}\otimes \mathbf {b} ^{k}}$$ We may also write $${\displaystyle {\begin{aligned}{\boldsymbol {\nabla }}{\boldsymbol {S}}&=\left[{\cfrac {\partial S^{ij}}{\partial q^{k}}}+\Gamma _{kl}^{i}~S^{lj}+\Gamma _{kl}^{j}~S^{il}\right]~\mathbf {b} _{i}\otimes \mathbf {b} _{j}\otimes \mathbf {b} ^{k}\\[8pt]&=\left[{\cfrac {\partial S_{~j}^{i}}{\partial q^{k}}}+\Gamma _{kl}^{i}~S_{~j}^{l}-\Gamma _{kj}^{l}~S_{~l}^{i}\right]~\mathbf {b} _{i}\otimes \mathbf {b} ^{j}\otimes \mathbf {b} ^{k}\\[8pt]&=\left[{\cfrac {\partial S_{i}^{~j}}{\partial q^{k}}}-\Gamma _{ik}^{l}~S_{l}^{~j}+\Gamma _{kl}^{j}~S_{i}^{~l}\right]~\mathbf {b} ^{i}\otimes \mathbf {b} _{j}\otimes \mathbf {b} ^{k}\end{aligned}}}$$

The physical components of a second-order tensor field can be obtained by using a normalized contravariant basis, i.e., $${\displaystyle {\boldsymbol {S}}=S_{ij}~\mathbf {b} ^{i}\otimes \mathbf {b} ^{j}={\hat {S}}_{ij}~{\hat {\mathbf {b} }}^{i}\otimes {\hat {\mathbf {b} }}^{j}}$$ where the hatted basis vectors have been normalized. This implies that (again no summation)

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

The divergence of a vector field (${\displaystyle \mathbf {v} }$)is defined as $${\displaystyle \operatorname {div} ~\mathbf {v} ={\boldsymbol {\nabla }}\cdot \mathbf {v} ={\text{tr}}({\boldsymbol {\nabla }}\mathbf {v} )}$$ In terms of components with respect to a curvilinear basis $${\displaystyle {\boldsymbol {\nabla }}\cdot \mathbf {v} ={\cfrac {\partial v^{i}}{\partial q^{i}}}+\Gamma _{\ell i}^{i}~v^{\ell }=\left[{\cfrac {\partial v_{i}}{\partial q^{j}}}-\Gamma _{ji}^{\ell }~v_{\ell }\right]~g^{ij}}$$

An alternative equation for the divergence of a vector field is frequently used. To derive this relation recall that $${\displaystyle {\boldsymbol {\nabla }}\cdot \mathbf {v} ={\frac {\partial v^{i}}{\partial q^{i}}}+\Gamma _{\ell i}^{i}~v^{\ell }}$$ Now, $${\displaystyle \Gamma _{\ell i}^{i}=\Gamma _{i\ell }^{i}={\cfrac {g^{mi}}{2}}\left[{\frac {\partial g_{im}}{\partial q^{\ell }}}+{\frac {\partial g_{\ell m}}{\partial q^{i}}}-{\frac {\partial g_{il}}{\partial q^{m}}}\right]}$$ Noting that, due to the symmetry of ${\displaystyle {\boldsymbol {g}}}$, $${\displaystyle g^{mi}~{\frac {\partial g_{\ell m}}{\partial q^{i}}}=g^{mi}~{\frac {\partial g_{i\ell }}{\partial q^{m}}}}$$ we have $${\displaystyle {\boldsymbol {\nabla }}\cdot \mathbf {v} ={\frac {\partial v^{i}}{\partial q^{i}}}+{\cfrac {g^{mi}}{2}}~{\frac {\partial g_{im}}{\partial q^{\ell }}}~v^{\ell }}$$ Recall that if [g_ij] is the matrix whose components are g_ij, then the inverse of the matrix is ${\displaystyle [g_{ij}]^{-1}=[g^{ij}]}$. The inverse of the matrix is given by $${\displaystyle [g^{ij}]=[g_{ij}]^{-1}={\cfrac {A^{ij}}{g}}~;~~g:=\det([g_{ij}])=\det {\boldsymbol {g}}}$$ where A^ij are the Cofactor matrix of the components g_ij. From matrix algebra we have $${\displaystyle g=\det([g_{ij}])=\sum _{i}g_{ij}~A^{ij}\quad \Rightarrow \quad {\frac {\partial g}{\partial g_{ij}}}=A^{ij}}$$ Hence, $${\displaystyle [g^{ij}]={\cfrac {1}{g}}~{\frac {\partial g}{\partial g_{ij}}}}$$ Plugging this relation into the expression for the divergence gives $${\displaystyle {\boldsymbol {\nabla }}\cdot \mathbf {v} ={\frac {\partial v^{i}}{\partial q^{i}}}+{\cfrac {1}{2g}}~{\frac {\partial g}{\partial g_{mi}}}~{\frac {\partial g_{im}}{\partial q^{\ell }}}~v^{\ell }={\frac {\partial v^{i}}{\partial q^{i}}}+{\cfrac {1}{2g}}~{\frac {\partial g}{\partial q^{\ell }}}~v^{\ell }}$$ A little manipulation leads to the more compact form $${\displaystyle {\boldsymbol {\nabla }}\cdot \mathbf {v} ={\cfrac {1}{\sqrt {g}}}~{\frac {\partial }{\partial q^{i}}}(v^{i}~{\sqrt {g}})}$$