Curvilinear coordinates can be formulated intensor calculus, with important applications inphysics andengineering, particularly for describing transportation of physical quantities and deformation of matter influid mechanics andcontinuum mechanics.

Elementary vector and tensor algebra in curvilinear coordinates is used in some of the older scientific literature inmechanics andphysics 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 ofadmissible coordinates for${\displaystyle Z^{\acute{i}}}$. If${\displaystyle T}$ is linear the coordinate system${\displaystyle Z^i}$ will be called anaffine coordinate system, otherwise${\displaystyle Z^i}$ is called acurvilinear 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

$${\displaystyle \frac{\mathrm{\partial }{Z}^i}{\mathrm{\partial }{Z}^{\acute{i}}}\stackrel{\underset{\mathrm{d}\mathrm{e}\mathrm{f}}{}}{=}{J}_{\acute{i}}^i}$$ 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

The resultant matrix is called the Jacobian matrix.

Let${\displaystyle ({\mathbf{b}}_1,{\mathbf{b}}_2,{\mathbf{b}}_3)}$ be an arbitrary basis for three-dimensional Euclidean space. In general, the basis vectors areneither unit vectors nor mutually orthogonal. However, they are required to be linearly independent. Then a vector${\displaystyle \mathbf{v}}$ can be expressed as^[4]: 27$${\displaystyle \mathbf{v}=v^k{\mathbf{b}}_k}$$The components${\displaystyle v^k}$ are thecontravariant components of the vector${\displaystyle \mathbf{v}}$.

Thereciprocal basis${\displaystyle ({\mathbf{b}}^1,{\mathbf{b}}^2,{\mathbf{b}}^3)}$ is defined by the relation^{[4]: 28–29}$${\displaystyle {\mathbf{b}}^i\cdot {\mathbf{b}}_j={\delta }_j^i}$$where${\displaystyle {\delta }_j^i}$ is theKronecker delta.

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

A second-order tensor can be expressed as$${\displaystyle \mathit{S}=S^{ij}{\mathbf{b}}_i\otimes {\mathbf{b}}_j=S_j^i{\mathbf{b}}_i\otimes {\mathbf{b}}^j=S_i^j{\mathbf{b}}^i\otimes {\mathbf{b}}_j=S_{ij}{\mathbf{b}}^i\otimes {\mathbf{b}}^j}$$The components${\displaystyle S^{ij}}$ are called thecontravariant components,${\displaystyle S_j^i}$ themixed right-covariant components,${\displaystyle S_i^j}$ themixed left-covariant components, and${\displaystyle S_{ij}}$ thecovariant components of the second-order tensor.

The quantities${\displaystyle g_{ij}}$,${\displaystyle g^{ij}}$ 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-orderalternating tensor is defined as$${\displaystyle \mathcal{E}={\epsilon }_{ijk}{\mathbf{e}}^i\otimes {\mathbf{e}}^j\otimes {\mathbf{e}}^k}$$In a general curvilinear basis the same tensor may be expressed as$${\displaystyle \mathcal{E}={\mathcal{E}}_{ijk}{\mathbf{b}}^i\otimes {\mathbf{b}}^j\otimes {\mathbf{b}}^k={\mathcal{E}}^{ijk}{\mathbf{b}}_i\otimes {\mathbf{b}}_j\otimes {\mathbf{b}}_k}$$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,$${\displaystyle {\mathbf{b}}_i\times {\mathbf{b}}_j=J{\epsilon }_{ijp}{\mathbf{b}}^p=\sqrt{g}{\epsilon }_{ijp}{\mathbf{b}}^p}$$Hence,$${\displaystyle {\mathcal{E}}_{ijk}=J{\epsilon }_{ijk}=\sqrt{g}{\epsilon }_{ijk}}$$Similarly, we can show that$${\displaystyle {\mathcal{E}}^{ijk}=\frac{1}{J}{\epsilon }^{ijk}=\frac{1}{\sqrt{g}}{\epsilon }^{ijk}}$$

The identity map${\displaystyle \mathbf{I}}$ 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

$${\displaystyle \mathbf{u}\cdot \mathbf{v}=u^i{v}_i=u_i{v}^i=g_{ij}{u}^i{v}^j=g^{ij}{u}_i{v}_j}$$

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

$${\displaystyle \mathbf{u}\times \mathbf{v}={\epsilon }_{ijk}{u}_j{v}_k{\mathbf{e}}_i}$$

where${\displaystyle {\epsilon }_{ijk}}$ is thepermutation symbol and${\displaystyle {\mathbf{e}}_i}$ is a Cartesianbasis 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 thethird-order alternating tensor. Thecross product of two vectors is given by:

$${\displaystyle \mathbf{u}\times \mathbf{v}={\epsilon }_{ijk}{\hat{u}}_j{\hat{v}}_k{\mathbf{e}}_i}$$

where${\displaystyle {\epsilon }_{ijk}}$ is thepermutation symbol and${\displaystyle {\mathbf{e}}_i}$ is a Cartesian basis vector. Therefore,

$${\displaystyle {\mathbf{e}}_p\times {\mathbf{e}}_q={\epsilon }_{ipq}{\mathbf{e}}_i}$$

and

$${\displaystyle {\mathbf{b}}_m\times {\mathbf{b}}_n=\frac{\mathrm{\partial }\mathbf{x}}{\mathrm{\partial }{q}^m}\times \frac{\mathrm{\partial }\mathbf{x}}{\mathrm{\partial }{q}^n}=\frac{\mathrm{\partial }(x_p{\mathbf{e}}_p)}{\mathrm{\partial }{q}^m}\times \frac{\mathrm{\partial }(x_q{\mathbf{e}}_q)}{\mathrm{\partial }{q}^n}=\frac{\mathrm{\partial }{x}_p}{\mathrm{\partial }{q}^m}\frac{\mathrm{\partial }{x}_q}{\mathrm{\partial }{q}^n}{\mathbf{e}}_p\times {\mathbf{e}}_q={\epsilon }_{ipq}\frac{\mathrm{\partial }{x}_p}{\mathrm{\partial }{q}^m}\frac{\mathrm{\partial }{x}_q}{\mathrm{\partial }{q}^n}{\mathbf{e}}_i.}$$

Hence,

$${\displaystyle ({\mathbf{b}}_m\times {\mathbf{b}}_n)\cdot {\mathbf{b}}_s={\epsilon }_{ipq}\frac{\mathrm{\partial }{x}_p}{\mathrm{\partial }{q}^m}\frac{\mathrm{\partial }{x}_q}{\mathrm{\partial }{q}^n}\frac{\mathrm{\partial }{x}_i}{\mathrm{\partial }{q}^s}}$$

Returning to the vector product and using the relations:

$${\displaystyle {\hat{u}}_j=\frac{\mathrm{\partial }{x}_j}{\mathrm{\partial }{q}^m}{u}^m,{\hat{v}}_k=\frac{\mathrm{\partial }{x}_k}{\mathrm{\partial }{q}^n}{v}^n,{\mathbf{e}}_i=\frac{\mathrm{\partial }{x}_i}{\mathrm{\partial }{q}^s}{\mathbf{b}}^s,}$$

gives us:

$${\displaystyle \mathbf{u}\times \mathbf{v}={\epsilon }_{ijk}{\hat{u}}_j{\hat{v}}_k{\mathbf{e}}_i={\epsilon }_{ijk}\frac{\mathrm{\partial }{x}_j}{\mathrm{\partial }{q}^m}\frac{\mathrm{\partial }{x}_k}{\mathrm{\partial }{q}^n}\frac{\mathrm{\partial }{x}_i}{\mathrm{\partial }{q}^s}{u}^m{v}^n{\mathbf{b}}^s=[({\mathbf{b}}_m\times {\mathbf{b}}_n)\cdot {\mathbf{b}}_s]u^m{v}^n{\mathbf{b}}^s={\mathcal{E}}_{smn}{u}^m{v}^n{\mathbf{b}}^s}$$

Theidentity 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${\displaystyle \mathbf{v}=\mathit{S}\mathbf{u}}$ can be expressed in curvilinear coordinates as

$${\displaystyle v^i{\mathbf{b}}_i=S^{ij}{u}_j{\mathbf{b}}_i=S_j^i{u}^j{\mathbf{b}}_i;v_i{\mathbf{b}}^i=S_{ij}{u}^j{\mathbf{b}}^i=S_i^j{u}_j{\mathbf{b}}^i}$$

Theinner product of two second-order tensors${\displaystyle \mathit{U}=\mathit{S}\cdot \mathit{T}}$ 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,

$${\displaystyle \mathit{U}=S^{ij}{T}_{.n}^m{g}_{jm}{\mathbf{b}}_i\otimes {\mathbf{b}}^n=S_{.m}^i{T}_{.n}^m{\mathbf{b}}_i\otimes {\mathbf{b}}^n=S^{ij}{T}_{jn}{\mathbf{b}}_i\otimes {\mathbf{b}}^n}$$

If${\displaystyle \mathit{S}}$ is a second-order tensor, then thedeterminant is defined by the relation

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

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${\displaystyle ({\mathbf{e}}_1,{\mathbf{e}}_2,{\mathbf{e}}_3)}$ be the usual Cartesian basis vectors for the Euclidean space of interest and let$${\displaystyle {\mathbf{b}}_i=\mathit{F}{\mathbf{e}}_i}$$where${\displaystyle \mathit{F}}$ is a second-order transformation tensor that maps${\displaystyle {\mathbf{e}}_i}$ to${\displaystyle {\mathbf{b}}_i}$. Then,$${\displaystyle {\mathbf{b}}_i\otimes {\mathbf{e}}_i=(\mathit{F}{\mathbf{e}}_i)\otimes {\mathbf{e}}_i=\mathit{F}({\mathbf{e}}_i\otimes {\mathbf{e}}_i)=\mathit{F}.}$$From this relation we can show that$${\displaystyle {\mathbf{b}}^i={\mathit{F}}^{-\mathrm{T}}{\mathbf{e}}^i;g^{ij}=[{\mathit{F}}^{-1}{\mathit{F}}^{-\mathrm{T}}{]}_{ij};g_{ij}=[g^{ij}{]}^{-1}=[{\mathit{F}}^{\mathrm{T}}\mathit{F}{]}_{ij}}$$Let${\displaystyle J:=det\mathit{F}}$ be the Jacobian of the transformation. Then, from the definition of the determinant,$${\displaystyle \left[{\mathbf{b}}_1,{\mathbf{b}}_2,{\mathbf{b}}_3\right]=det\mathit{F}\left[{\mathbf{e}}_1,{\mathbf{e}}_2,{\mathbf{e}}_3\right].}$$Since$${\displaystyle \left[{\mathbf{e}}_1,{\mathbf{e}}_2,{\mathbf{e}}_3\right]=1}$$we have$${\displaystyle J=det\mathit{F}=\left[{\mathbf{b}}_1,{\mathbf{b}}_2,{\mathbf{b}}_3\right]={\mathbf{b}}_1\cdot ({\mathbf{b}}_2\times {\mathbf{b}}_3)}$$A number of interesting results can be derived using the above relations.

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

Another interesting relation is derived below. Recall that$${\displaystyle {\mathbf{b}}^i\cdot {\mathbf{b}}_j={\delta }_j^i\Rightarrow {\mathbf{b}}^1\cdot {\mathbf{b}}_1=1,{\mathbf{b}}^1\cdot {\mathbf{b}}_2={\mathbf{b}}^1\cdot {\mathbf{b}}_3=0\Rightarrow {\mathbf{b}}^1=A({\mathbf{b}}_2\times {\mathbf{b}}_3)}$$where${\displaystyle A}$ is a, yet undetermined, constant. Then$${\displaystyle {\mathbf{b}}^1\cdot {\mathbf{b}}_1=A{\mathbf{b}}_1\cdot ({\mathbf{b}}_2\times {\mathbf{b}}_3)=AJ=1\Rightarrow A=\frac{1}{J}}$$This observation leads to the relations$${\displaystyle {\mathbf{b}}^1=\frac{1}{J}({\mathbf{b}}_2\times {\mathbf{b}}_3);{\mathbf{b}}^2=\frac{1}{J}({\mathbf{b}}_3\times {\mathbf{b}}_1);{\mathbf{b}}^3=\frac{1}{J}({\mathbf{b}}_1\times {\mathbf{b}}_2)}$$In index notation,$${\displaystyle {\epsilon }_{ijk}{\mathbf{b}}^k=\frac{1}{J}({\mathbf{b}}_i\times {\mathbf{b}}_j)=\frac{1}{\sqrt{g}}({\mathbf{b}}_i\times {\mathbf{b}}_j)}$$where${\displaystyle {\epsilon }_{ijk}}$ is the usualpermutation symbol.

We have not identified an explicit expression for the transformation tensor${\displaystyle \mathit{F}}$ 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$${\displaystyle {\mathbf{b}}_i=\frac{\mathrm{\partial }\mathbf{x}}{\mathrm{\partial }{q}^i}=\frac{\mathrm{\partial }\mathbf{x}}{\mathrm{\partial }{x}_j}\frac{\mathrm{\partial }{x}_j}{\mathrm{\partial }{q}^i}={\mathbf{e}}_j\frac{\mathrm{\partial }{x}_j}{\mathrm{\partial }{q}^i}}$$Similarly,$${\displaystyle {\mathbf{e}}_i={\mathbf{b}}_j\frac{\mathrm{\partial }{q}^j}{\mathrm{\partial }{x}_i}}$$From these results we have$${\displaystyle {\mathbf{e}}^k\cdot {\mathbf{b}}_i=\frac{\mathrm{\partial }{x}_k}{\mathrm{\partial }{q}^i}\Rightarrow \frac{\mathrm{\partial }{x}_k}{\mathrm{\partial }{q}^i}{\mathbf{b}}^i={\mathbf{e}}^k\cdot ({\mathbf{b}}_i\otimes {\mathbf{b}}^i)={\mathbf{e}}^k}$$and$${\displaystyle {\mathbf{b}}^k=\frac{\mathrm{\partial }{q}^k}{\mathrm{\partial }{x}_i}{\mathbf{e}}^i}$$

Simmonds,^[4] in his book ontensor analysis, quotesAlbert 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 curvilinearmanifolds ingeneral relativity,^[8] in themechanics of curvedshells,^[6] in examining theinvariance properties ofMaxwell's equations which has been of interest inmetamaterials^[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${\displaystyle (q^1,q^2,q^3)}$.

Thecoordinate curve${\displaystyle q^1}$ represents a curve on which${\displaystyle q^2}$ and${\displaystyle q^3}$ are constant. Let${\displaystyle \mathbf{x}}$ be theposition 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$${\displaystyle \mathbf{x}=\mathit{\phi }(q^1,q^2,q^3);q^i={\psi }^i(\mathbf{x})=[{\mathit{\phi }}^{-1}(\mathbf{x}){]}^i}$$The fields${\displaystyle {\psi }^i(\mathbf{x})}$ are called thecurvilinear coordinate functions of thecurvilinear coordinate system${\displaystyle \mathit{\psi }(\mathbf{x})={\mathit{\phi }}^{-1}(\mathbf{x})}$.

The${\displaystyle q^i}$coordinate curves are defined by the one-parameter family of functions given by$${\displaystyle {\mathbf{x}}_i(\alpha )=\mathit{\phi }(\alpha ,q^j,q^k),i\ne j\ne k}$$with${\displaystyle q^j}$,${\displaystyle q^k}$ fixed.

Thetangent vector to the curve${\displaystyle {\mathbf{x}}_i}$ at the point${\displaystyle {\mathbf{x}}_i(\alpha )}$ (or to the coordinate curve${\displaystyle q_i}$ at the point${\displaystyle \mathbf{x}}$) is$${\displaystyle \frac{\mathrm{d}{\mathbf{x}}_{\mathrm{i}}}{\mathrm{d}\alpha }\equiv \frac{\mathrm{\partial }\mathbf{x}}{\mathrm{\partial }{q}^i}}$$

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

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

If${\displaystyle {\mathbf{b}}_i}$ is the covariant (or natural) basis at a point, and if${\displaystyle {\mathbf{b}}^i}$ is the contravariant (or reciprocal) basis at that point, then$${\displaystyle [\mathbf{\nabla }f(\mathbf{x})]\cdot \mathbf{c}=\frac{\mathrm{\partial }f}{\mathrm{\partial }{q}^i}{c}^i=\left(\frac{\mathrm{\partial }f}{\mathrm{\partial }{q}^i}{\mathbf{b}}^i\right)\left(c^i{\mathbf{b}}_i\right)\Rightarrow \mathbf{\nabla }f(\mathbf{x})=\frac{\mathrm{\partial }f}{\mathrm{\partial }{q}^i}{\mathbf{b}}^i}$$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${\displaystyle \mathbf{f}(\mathbf{x})}$. The gradient is given by$${\displaystyle [\mathbf{\nabla }\mathbf{f}(\mathbf{x})]\cdot \mathbf{c}=\frac{\mathrm{\partial }\mathbf{f}}{\mathrm{\partial }{q}^i}{c}^i}$$If we consider the gradient of the position vector field${\displaystyle \mathbf{r}(\mathbf{x})=\mathbf{x}}$, then we can show that$${\displaystyle \mathbf{c}=\frac{\mathrm{\partial }\mathbf{x}}{\mathrm{\partial }{q}^i}{c}^i={\mathbf{b}}_i(\mathbf{x})c^i;{\mathbf{b}}_i(\mathbf{x}):=\frac{\mathrm{\partial }\mathbf{x}}{\mathrm{\partial }{q}^i}}$$The vector field${\displaystyle {\mathbf{b}}_i}$ is tangent to the${\displaystyle q^i}$ coordinate curve and forms anatural basis at each point on the curve. This basis, as discussed at the beginning of this article, is also called thecovariant curvilinear basis. We can also define areciprocal basis, orcontravariant curvilinear basis,${\displaystyle {\mathbf{b}}^i}$. 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${\displaystyle \mathbf{x}}$.

Since${\displaystyle \mathbf{c}}$ is arbitrary, we can write$${\displaystyle \mathbf{\nabla }\mathbf{f}(\mathbf{x})=\frac{\mathrm{\partial }\mathbf{f}}{\mathrm{\partial }{q}^i}\otimes {\mathbf{b}}^i}$$

Note that the contravariant basis vector${\displaystyle {\mathbf{b}}^i}$ is perpendicular to the surface of constant${\displaystyle {\psi }^i}$ and is given by$${\displaystyle {\mathbf{b}}^i=\mathbf{\nabla }{\psi }^i}$$

TheChristoffel symbols of the first kind are defined as$${\displaystyle {\mathbf{b}}_{i,j}=\frac{\mathrm{\partial }{\mathbf{b}}_i}{\mathrm{\partial }{q}^j}:={\mathrm{\Gamma }}_{ijk}{\mathbf{b}}^k\Rightarrow {\mathbf{b}}_{i,j}\cdot {\mathbf{b}}_l={\mathrm{\Gamma }}_{ijl}}$$To express${\displaystyle {\mathrm{\Gamma }}_{ijk}}$ in terms of${\displaystyle g_{ij}}$ we note thatSince${\displaystyle {\mathbf{b}}_{i,j}={\mathbf{b}}_{j,i}}$ we have${\displaystyle {\mathrm{\Gamma }}_{ijk}={\mathrm{\Gamma }}_{jik}}$. Using these to rearrange the above relations gives$${\displaystyle {\mathrm{\Gamma }}_{ijk}=\frac{1}{2}(g_{ik,j}+g_{jk,i}-g_{ij,k})=\frac{1}{2}[({\mathbf{b}}_i\cdot {\mathbf{b}}_k{)}_{,j}+({\mathbf{b}}_j\cdot {\mathbf{b}}_k{)}_{,i}-({\mathbf{b}}_i\cdot {\mathbf{b}}_j{)}_{,k}]}$$

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

$${\displaystyle \frac{\mathrm{\partial }{\mathbf{b}}_i}{\mathrm{\partial }{q}^j}={\mathrm{\Gamma }}_{ij}^k{\mathbf{b}}_k}$$

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

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

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

The vector field${\displaystyle \mathbf{v}}$ 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 (nosummation)$${\displaystyle {\hat{\mathbf{b}}}^i=\frac{{\mathbf{b}}^i}{\sqrt{g^{ii}}}}$$is the normalized contravariant basis vector.

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

If we consider the expression for the tensor in terms of a contravariant basis, then$${\displaystyle \mathbf{\nabla }\mathit{S}=\frac{\mathrm{\partial }}{\mathrm{\partial }{q}^k}[S_{ij}{\mathbf{b}}^i\otimes {\mathbf{b}}^j]\otimes {\mathbf{b}}^k=\left[\frac{\mathrm{\partial }{S}_{ij}}{\mathrm{\partial }{q}^k}-{\mathrm{\Gamma }}_{ki}^l{S}_{lj}-{\mathrm{\Gamma }}_{kj}^l{S}_{il}\right]{\mathbf{b}}^i\otimes {\mathbf{b}}^j\otimes {\mathbf{b}}^k}$$We may also write