View source for Tensors in curvilinear coordinates

[[Curvilinear coordinates]] can be formulated in [[Tensor calculus|tensor calculus]], with important applications in [[Physics:Physics|physics]] and [[Engineering:Engineering|engineering]], particularly for describing transportation of physical quantities and deformation of matter in [[Physics:Fluid mechanics|fluid mechanics]] and [[Physics:Continuum mechanics|continuum mechanics]].
==Vector and tensor algebra in three-dimensional curvilinear coordinates==
{{Einstein_summation_convention}}
Elementary vector and tensor algebra in curvilinear coordinates is used in some of the older scientific literature in [[Physics:Mechanics|mechanics]] and [[Physics:Physics|physics]] and can be indispensable to understanding work from the early and mid 1900s, for example the text by Green and Zerna.<ref name=Green>{{cite book | last1=Green | first1=A. E. | last2=Zerna | first2=W. | year=1968 | title=Theoretical Elasticity | publisher=Oxford University Press | isbn=0-19-853486-8 }}</ref>  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,<ref name=Ogden00/> Naghdi,<ref name=Naghdi>{{cite book | first1=P. M. | last1=Naghdi | year=1972 | contribution=Theory of shells and plates | editor=S. Flügge | title=Handbook of Physics | volume=VIa/2 | pages=425–640}}</ref> Simmonds,<ref name=Simmonds>{{cite book | last=Simmonds | first=J. G. | year=1994 | title=A brief on tensor analysis | publisher=Springer | isbn=0-387-90639-8}}</ref> Green and Zerna,<ref name=Green/>  Basar and Weichert,<ref name=Basar>{{cite book | last1=Basar | first1=Y. | last2=Weichert | first2=D. | year=2000 | title=Numerical continuum mechanics of solids: fundamental concepts and perspectives | publisher=Springer}}</ref> and Ciarlet.<ref name=Ciarlet>{{cite book | last=Ciarlet | first=P. G. | year=2000 | title=Theory of Shells | volume=1 | publisher=Elsevier Science }}</ref>

=== Coordinate transformations ===
Consider two coordinate systems with coordinate variables <math>(Z^1 , Z^2 ,Z^3 )</math> and <math>(Z^\acute{1} , Z^\acute{2} ,Z^\acute{3} )</math>,  which we shall represent in short as just <math>Z^i </math> and <math>Z^\acute{i}</math> respectively and always assume our index <math>i</math> runs from 1 through 3. We shall assume that  these coordinates systems are embedded in the three-dimensional euclidean space. Coordinates  <math>Z^i </math> and <math>Z^\acute{i}</math> 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  <math>Z^i </math> and <math>Z^\acute{i}</math> are functions of each other

<math display="block">Z^i= f^{i}(Z^\acute{1} , Z^\acute{2} ,Z^\acute{3} ) </math> for <math>i=1,2,3 </math>

which can be written as

<math display="block">Z^i= Z^{i}(Z^\acute{1} , Z^\acute{2} ,Z^\acute{3} )= Z^{i}(Z^\acute{i}) </math>  for <math>\acute{i},i=1,2,3 </math>

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

<math display="block">Z= T(\acute{z}) </math>

Similarly we can represent <math>Z^\acute{i}</math> as a function of <math>Z^i </math> as follows:

<math display="block">Z^\acute{i}= g^{\acute{i}}(Z^{1}, Z^{2} ,Z^{3} )  </math> for <math>\acute{i}=1,2,3 </math>

similarly we can write the free equations more compactly as

<math display="block">Z^\acute{i}= Z^\acute{i}(Z^{1} , Z^{2} ,Z^{3} )= Z^\acute{i}(Z^{i})  </math> for <math>\acute{i},i=1,2,3 </math>

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

<math display="block">\acute{z}= S(z) </math>

If the transformation <math>T </math> is bijective then we call the image of the transformation,namely <math>Z^i </math>, a set of '''admissible coordinates for''' <math>Z^\acute{i}</math>. If <math>T </math> is linear the coordinate system <math>Z^i </math> will be called an '''affine coordinate system''' ,otherwise <math>Z^i </math> is called a '''curvilinear coordinate system'''

==== The Jacobian ====
As we now see that the Coordinates  <math>Z^i </math> and <math>Z^\acute{i}</math> are functions of each other, we can take the derivative of the coordinate variable <math>Z^i </math> with respect to the coordinate variable <math>Z^\acute{i}</math>

consider

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

<math display="block">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}  </math>

The resultant matrix is called the Jacobian matrix.

===Vectors in curvilinear coordinates===
Let ('''b'''<sub>1</sub>, '''b'''<sub>2</sub>, '''b'''<sub>3</sub>) 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<ref name=Simmonds/>{{rp|page=27}}
<math display="block">
    \mathbf{v} = v^k\,\mathbf{b}_k
 </math>
The components ''v<sup>k</sup>'' are the '''contravariant''' components of the vector '''v'''.

The '''reciprocal basis''' ('''b'''<sup>1</sup>, '''b'''<sup>2</sup>, '''b'''<sup>3</sup>) is defined by the relation <ref name=Simmonds/>{{rp|pages=28–29}}
<math display="block">
   \mathbf{b}^i\cdot\mathbf{b}_j = \delta^i_j
 </math>
where ''δ<sup>i</sup> <sub>j</sub>'' is the [[Kronecker delta]].

The vector '''v''' can also be expressed in terms of the reciprocal basis:
<math display="block">
    \mathbf{v} = v_k~\mathbf{b}^k
 </math>
The components ''v<sub>k</sub>'' are the '''covariant''' components of the vector <math>\mathbf{v}</math>.

===Second-order tensors in curvilinear coordinates===
A second-order tensor can be expressed as
<math display="block">
   \boldsymbol{S} = S^{ij}~\mathbf{b}_i\otimes\mathbf{b}_j = S^{i}_{~j}~\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
 </math>
The components ''S<sup>ij</sup>'' are called the '''contravariant''' components, ''S<sup>i</sup> <sub>j</sub>'' the '''mixed right-covariant''' components, ''S<sub>i</sub> <sup>j</sup>'' the '''mixed left-covariant''' components, and ''S<sub>ij</sub>'' the '''covariant''' components of the second-order tensor.

====Metric tensor and relations between components====
The quantities ''g<sub>ij</sub>'', ''g<sup>ij</sup>'' are defined as<ref name=Simmonds/>{{rp|page=39}}

<math display="block">
    g_{ij} = \mathbf{b}_i \cdot \mathbf{b}_j = g_{ji} ~;~~ g^{ij} = \mathbf{b}^i \cdot \mathbf{b}^j = g^{ji}
 </math>
From the above equations we have
<math display="block">
   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
 </math>

The components of a vector are related by<ref name=Simmonds/>{{rp|pages=30–32}}
<math display="block">
   \mathbf{v}\cdot\mathbf{b}^i = v^k~\mathbf{b}_k\cdot\mathbf{b}^i = v^k~\delta^i_k = v^i </math>
<math display="block">
   \mathbf{v}\cdot\mathbf{b}_i = v_k~\mathbf{b}^k\cdot\mathbf{b}_i = v_k~\delta_i^k = v_i </math>
Also,
<math display="block">
   \mathbf{v}\cdot\mathbf{b}_i = v^k~\mathbf{b}_k\cdot\mathbf{b}_i = g_{ki}~v^k </math>
<math display="block">
   \mathbf{v}\cdot\mathbf{b}^i = v_k~\mathbf{b}^k\cdot\mathbf{b}^i = g^{ki}~v_k
 </math>

The components of the second-order tensor are related by
<math display="block">
   S^{ij} = g^{ik}~S_k^{~j} = g^{jk}~S^i_{~k} = g^{ik}~g^{jl}~S_{kl}
 </math>

===The alternating tensor===
In an orthonormal right-handed basis, the third-order [[Levi-Civita symbol|alternating tensor]] is defined as
<math display="block">
  \boldsymbol{\mathcal{E}} = \varepsilon_{ijk}~\mathbf{e}^i\otimes\mathbf{e}^j\otimes\mathbf{e}^k
</math>
In a general curvilinear basis the same tensor may be expressed as
<math display="block">
  \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
</math>
It can be shown that
<math display="block">
  \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]
</math>
Now,
<math display="block">
  \mathbf{b}_i\times\mathbf{b}_j = J~\varepsilon_{ijp}~\mathbf{b}^p = \sqrt{g}~\varepsilon_{ijp}~\mathbf{b}^p
</math>
Hence,
<math display="block">
  \mathcal{E}_{ijk} = J~\varepsilon_{ijk} = \sqrt{g}~\varepsilon_{ijk}
</math>
Similarly, we can show that
<math display="block">
  \mathcal{E}^{ijk} = \cfrac{1}{J}~\varepsilon^{ijk} = \cfrac{1}{\sqrt{g}}~\varepsilon^{ijk}
</math>

===Vector operations===

====Identity map====
The identity map '''I''' defined by <math>\mathbf{I}\cdot\mathbf{v} = \mathbf{v}</math> can be shown to be:<ref name=Simmonds/>{{rp|page=39}}

<math display="block">\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</math>

====Scalar (dot) product====
The scalar product of two vectors in curvilinear coordinates is<ref name=Simmonds/>{{rp|page=32}}

<math display="block">\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</math>

====Vector (cross) product====
The [[Cross product|cross product]] of two vectors is given by:<ref name=Simmonds/>{{rp|pages=32–34}}

<math display="block">\mathbf{u}\times\mathbf{v} = \varepsilon_{ijk}u_jv_k\mathbf{e}_i</math>

where ε<sub>''ijk''</sub> is the permutation symbol and '''e'''<sub>''i''</sub> is a Cartesian basis vector.  In curvilinear coordinates, the equivalent expression is:

<math display="block">\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^mv^n\mathbf{b}^s</math>

where <math>\mathcal{E}_{ijk}</math> is the [[Curvilinear coordinates#The alternating tensor|third-order alternating tensor]]. The [[Cross product|cross product]] of two vectors is given by:

<math display="block">\mathbf{u}\times\mathbf{v} = \varepsilon_{ijk}\hat{u}_j\hat{v}_k\mathbf{e}_i</math>

where ε<sub>''ijk''</sub> is the permutation symbol and <math>\mathbf{e}_i</math> is a Cartesian basis vector. Therefore,

<math display="block">\mathbf{e}_p\times\mathbf{e}_q = \varepsilon_{ipq}\mathbf{e}_i</math>

and

<math display="block">\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.</math>

Hence,

<math display="block">(\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}</math>

Returning to the vector product and using the relations:

<math display="block">\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,</math>

gives us:

<math display="block">\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</math>

===Tensor operations===

====Identity map====
The [[Identity function|identity map]] <math>\mathsf{I}</math> defined by <math>\mathsf{I}\cdot\mathbf{v} = \mathbf{v}</math> can be shown to be<ref name=Simmonds/>{{rp|page=39}}

<math display="block">\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</math>

====Action of a second-order tensor on a vector====
The action <math>\mathbf{v} = \boldsymbol{S}\mathbf{u}</math> can be expressed in curvilinear coordinates as

<math display="block">v^i\mathbf{b}_i = S^{ij}u_j\mathbf{b}_i = S^i_{j}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</math>

====Inner product of two second-order tensors====
The inner product of two second-order tensors <math>\boldsymbol{U} = \boldsymbol{S}\cdot\boldsymbol{T}</math> can be expressed in curvilinear coordinates as

<math display="block">U_{ij}\mathbf{b}^i\otimes\mathbf{b}^j = S_{ik}T^k_{.j} \mathbf{b}^i\otimes\mathbf{b}^j= S_i^{.k}T_{kj}\mathbf{b}^i\otimes\mathbf{b}^j</math>

Alternatively,

<math display="block">\boldsymbol{U} = S^{ij}T^m_{.n}g_{jm}\mathbf{b}_i\otimes\mathbf{b}^n = S^i_{.m}T^m_{.n}\mathbf{b}_i\otimes\mathbf{b}^n= S^{ij}T_{jn} \mathbf{b}_i \otimes\mathbf{b}^n</math>

====Determinant of a second-order tensor====
If <math>\boldsymbol{S}</math> is a second-order tensor, then the [[Determinant|determinant]] is defined by the relation

<math display="block">\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]</math>

where <math>\mathbf{u}, \mathbf{v}, \mathbf{w}</math> are arbitrary vectors and

<math display="block">\left[\mathbf{u},\mathbf{v},\mathbf{w}\right] := \mathbf{u}\cdot(\mathbf{v}\times\mathbf{w}).</math>

===Relations between curvilinear and Cartesian basis vectors===
Let ('''e'''<sub>1</sub>, '''e'''<sub>2</sub>, '''e'''<sub>3</sub>) be the usual Cartesian basis vectors for the Euclidean space of interest and let
<math display="block">
  \mathbf{b}_i = \boldsymbol{F}\mathbf{e}_i
</math>
where '''''F'''<sub>i</sub>'' is a second-order transformation tensor that maps '''e'''<sub>''i''</sub> to '''b'''<sub>''i''</sub>.  Then,
<math display="block">
  \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}~.
</math>
From this relation we can show that
<math display="block">
   \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}
 </math>
Let <math>J := \det\boldsymbol{F}</math> be the Jacobian of the transformation.  Then, from the definition of the determinant,
<math display="block">
  \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] ~.
</math>
Since
<math display="block">
  \left[\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3\right] = 1 
</math>
we have
<math display="block">
  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)
</math>
A number of interesting results can be derived using the above relations.

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

Another interesting relation is derived below.  Recall that
<math display="block">
  \mathbf{b}^i\cdot\mathbf{b}_j = \delta^i_j \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)
</math>
where ''A'' is a, yet undetermined, constant.  Then
<math display="block">
  \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}
</math>
This observation leads to the relations
<math display="block">
  \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)
</math>
In index notation,
<math display="block">
  \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)
</math>
where <math>\varepsilon_{ijk}</math> 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
<math display="block">
   \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} 
 </math>
Similarly,
<math display="block">
  \mathbf{e}_i = \mathbf{b}_j~\cfrac{\partial q^j}{\partial x_i}
 </math>
From these results we have
<math display="block">
  \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
</math>
and
<math display="block">
  \mathbf{b}^k = \frac{\partial q^k}{\partial x_i}~\mathbf{e}^i
</math>

==Vector and tensor calculus in three-dimensional curvilinear coordinates==
{{Einstein_summation_convention}}
Simmonds,<ref name=Simmonds/> in his book on tensor analysis, quotes [[Biography:Albert Einstein|Albert Einstein]] saying<ref name=Lanczos>{{cite book | last=Einstein | first=A. | year=1915 | contribution=Contribution to the Theory of General Relativity | editor-last=Laczos |editor-first=C. | title=The Einstein Decade | page=213}}</ref>

{{quote|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 [[Manifold|manifold]]s in [[Physics:General relativity|general relativity]],<ref name=Misner>{{cite book | last1=Misner | first1=C. W. | last2=Thorne | first2=K. S. | last3=Wheeler | first3=J. A. | year=1973 | title=Gravitation | publisher=W. H. Freeman and Co. | isbn=0-7167-0344-0}}</ref> in the [[Physics:Solid mechanics|mechanics]] of curved [[Physics:Plate theory|shells]],<ref name=Ciarlet/> in examining the [[Invariant (mathematics)|invariance]] properties of [[Physics:Maxwell's equations|Maxwell's equations]] which has been of interest in metamaterials<ref name=Greenleaf>{{cite journal | doi=10.1088/0967-3334/24/2/353 | last1=Greenleaf | first1=A. | last2=Lassas | first2=M. | last3=Uhlmann | first3=G. | year=2003 | title=Anisotropic conductivities that cannot be detected by EIT | journal=Physiological Measurement | volume=24 | issue=2 | pages=413–419 | pmid=12812426| s2cid=250813768 }}</ref><ref name=Leonhardt>{{cite journal | last1=Leonhardt | first1=U. | last2=Philbin | first2=T. G. | year=2006 | title=General relativity in electrical engineering | journal=New Journal of Physics | volume=8 | issue=10 | page=247 |arxiv = cond-mat/0607418 |bibcode = 2006NJPh....8..247L |doi = 10.1088/1367-2630/8/10/247 | s2cid=12100599 }}</ref> 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,<ref name=Ogden00/> Simmonds,<ref name=Simmonds /> Green and Zerna,<ref name=Green/>  Basar and Weichert,<ref name=Basar/> and Ciarlet.<ref name=Ciarlet/>

===Basic definitions===
Let the position of a point in space be characterized by three coordinate variables <math>(q^1, q^2, q^3)</math>.

The coordinate curve ''q''<sup>1</sup> represents a curve on which ''q''<sup>2</sup>, ''q''<sup>3</sup> 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 <ref name=Ogden00>{{cite book | last=Ogden | first=R. W. | year=2000 | title=Nonlinear elastic deformations | publisher=Dover}}</ref>{{rp|page=55}}
<math display="block">
   \mathbf{x} = \boldsymbol{\varphi}(q^1, q^2, q^3) ~;~~ q^i = \psi^i(\mathbf{x}) = [\boldsymbol{\varphi}^{-1}(\mathbf{x})]^i
 </math>
The fields ψ<sup>''i''</sup>('''x''') are called the '''curvilinear coordinate functions''' of the '''curvilinear coordinate system''' '''ψ'''('''x''') = '''φ'''<sup>−1</sup>('''x''').

The ''q<sup>i</sup>'' '''coordinate curves''' are defined by the one-parameter family of functions given by
<math display="block">
   \mathbf{x}_i(\alpha) = \boldsymbol{\varphi}(\alpha, q^j, q^k) ~,~~ i\ne j \ne k
 </math>
with ''q<sup>j</sup>, q<sup>k</sup>'' fixed.

===Tangent vector to coordinate curves===
The '''tangent vector''' to the curve '''x'''<sub>''i''</sub> at the point '''x'''<sub>''i''</sub>(α) (or to the coordinate curve ''q<sub>i</sub>'' at the point '''x''') is
<math display="block">
   \cfrac{\rm{d}\mathbf{x}_i}{\rm{d}\alpha} \equiv \cfrac{\partial\mathbf{x}}{\partial q^i}
 </math>

===Gradient===

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

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

If '''b'''<sub>''i''</sub> is the covariant (or natural) basis at a point, and if '''b'''<sup>''i''</sup> is the contravariant (or reciprocal) basis at that point, then
<math display="block">
   [\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
 </math>
A brief rationale for this choice of basis is given in the next section.

====Vector field====
A similar process can be used to arrive at the gradient of a vector field '''f'''('''x''').  The gradient is given by
<math display="block">
   [\boldsymbol{\nabla}\mathbf{f}(\mathbf{x})]\cdot\mathbf{c} = \cfrac{\partial \mathbf{f}}{\partial q^i}~c^i
 </math>
If we consider the gradient of the position vector field '''r'''('''x''') = '''x''', then we can show that
<math display="block">
   \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}
 </math>
The vector field '''b'''<sub>''i''</sub> is tangent to the ''q<sup>i</sup>'' 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, '''b'''<sup>''i''</sup>.  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
<math display="block">
  \boldsymbol{\nabla}\mathbf{f}(\mathbf{x}) = \cfrac{\partial \mathbf{f}}{\partial q^i}\otimes\mathbf{b}^i
 </math>

Note that the contravariant basis vector '''b'''<sup>''i''</sup> is perpendicular to the surface of constant ψ<sup>''i''</sup> and is given by
<math display="block">
   \mathbf{b}^i = \boldsymbol{\nabla}\psi^i
 </math>

====Christoffel symbols of the first kind====
The [[Christoffel symbols]] of the first kind are defined as
<math display="block">
   \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}
</math>
To express Γ<sub>''ijk''</sub> in terms of ''g<sub>ij</sub>'' we note that
<math display="block">
   \begin{align}
     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{align}
</math>
Since '''b'''<sub>''i,j''</sub> = '''b'''<sub>''j,i''</sub> we have Γ<sub>''ijk''</sub> = Γ<sub>''jik''</sub>.  Using these to rearrange the above relations gives
<math display="block">
   \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}]
</math>

====Christoffel symbols of the second kind====
The Christoffel symbols of the second kind are defined as
<math display="block"> \Gamma_{ij}^k = \Gamma_{ji}^k </math>
in which

<math display="block">\cfrac{\partial \mathbf{b}_i}{\partial q^j} = \Gamma_{ij}^k~\mathbf{b}_k
 </math>

This implies that
<math display="block">
   \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}
 </math>
Other relations that follow are
<math display="block">
     \cfrac{\partial \mathbf{b}^i}{\partial q^j} = -\Gamma^i_{jk}~\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
 </math>

Another particularly useful relation, which shows that the Christoffel symbol depends only on the metric tensor and its derivatives, is
<math display="block">
   \Gamma^k_{ij} = \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)
 </math>

====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.
<math display="block">
   \begin{align}
   \boldsymbol{\nabla}\mathbf{v} & = \left[\cfrac{\partial v^i}{\partial q^k} + \Gamma^i_{lk}~v^l\right]~\mathbf{b}_i\otimes\mathbf{b}^k \\[8pt]
    & = \left[\cfrac{\partial v_i}{\partial q^k} - \Gamma^l_{ki}~v_l\right]~\mathbf{b}^i\otimes\mathbf{b}^k
   \end{align}
 </math>

====Representing a physical vector field====
The vector field '''v''' can be represented as
<math display="block">
   \mathbf{v} = v_i~\mathbf{b}^i = \hat{v}_i~\hat{\mathbf{b}}^i 
 </math>
where <math>v_i</math> are the covariant components of the field, <math>\hat{v}_i</math> are the physical components, and (no [[Einstein notation|summation]])
<math display="block">
   \hat{\mathbf{b}}^i = \cfrac{\mathbf{b}^i}{\sqrt{g^{ii}}}
 </math>
is the normalized contravariant basis vector.

===Second-order tensor field===
The gradient of a second order tensor field can similarly be expressed as
<math display="block">
   \boldsymbol{\nabla}\boldsymbol{S} = \frac{\partial \boldsymbol{S}}{\partial q^i}\otimes\mathbf{b}^i
 </math>

====Explicit expressions for the gradient====
If we consider the expression for the tensor in terms of a contravariant basis, then
<math display="block">
   \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^l_{ki}~S_{lj} - \Gamma^l_{kj} ~S_{il}\right] ~\mathbf{b}^i \otimes\mathbf{b}^j \otimes \mathbf{b}^k
 </math>
We may also write
<math display="block">
  \begin{align}
   \boldsymbol{\nabla}\boldsymbol{S} & = \left[\cfrac{\partial S^{ij}}{\partial q^k} + \Gamma^i_{kl} ~ S^{lj} + \Gamma^j_{kl} ~ S^{il}\right]~ \mathbf{b}_i\otimes\mathbf{b}_j\otimes\mathbf{b}^k \\[8pt]
   & = \left[\cfrac{\partial S^i_{~j}}{\partial q^k} + \Gamma^i_{kl}~S^l_{~j} - \Gamma^l_{kj}~S^i_{~l}\right]~\mathbf{b}_i\otimes\mathbf{b}^j\otimes\mathbf{b}^k \\[8pt]
   & = \left[\cfrac{\partial S_i^{~j}}{\partial q^k} - \Gamma^l_{ik}~S_l^{~j} + \Gamma^j_{kl}~S_i^{~l}\right]~\mathbf{b}^i\otimes\mathbf{b}_j\otimes\mathbf{b}^k
  \end{align}
 </math>

====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.,
<math display="block">
   \boldsymbol{S} = S_{ij}~\mathbf{b}^i\otimes\mathbf{b}^j = \hat{S}_{ij}~\hat{\mathbf{b}}^i\otimes\hat{\mathbf{b}}^j
 </math>
where the hatted basis vectors have been normalized. This implies that (again no summation)

<math display="block"> \hat{S}_{ij} = S_{ij} ~ \sqrt{g^{ii}~g^{jj}} </math>

===Divergence===

====Vector field====
The [[Divergence|divergence]] of a vector field (<math>\mathbf{v}</math>)is defined as
<math display="block">
   \operatorname{div}~\mathbf{v} = \boldsymbol{\nabla}\cdot\mathbf{v} = \text{tr}(\boldsymbol{\nabla}\mathbf{v})
 </math>
In terms of components with respect to a curvilinear basis
<math display="block">
   \boldsymbol{\nabla}\cdot\mathbf{v} = \cfrac{\partial v^i}{\partial q^i} + \Gamma^i_{\ell i}~v^\ell
    = \left[\cfrac{\partial v_i}{\partial q^j} - \Gamma^\ell_{ji}~v_\ell\right]~g^{ij}
 </math>

An alternative equation for the divergence of a vector field is frequently used.  To derive this relation recall that
<math display="block">
  \boldsymbol{\nabla} \cdot \mathbf{v} = \frac{\partial v^i}{\partial q^i} + \Gamma_{\ell i}^i~v^\ell
</math>
Now,
<math display="block">
  \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]
</math>
Noting that, due to the symmetry of <math>\boldsymbol{g}</math>,
<math display="block">
  g^{mi}~\frac{\partial g_{\ell m}}{\partial q^i}  = g^{mi}~ \frac{\partial g_{i\ell}}{\partial q^m}
</math>
we have
<math display="block">
  \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
</math>
Recall that if [''g<sub>ij</sub>''] is the matrix whose components are ''g<sub>ij</sub>'', then the inverse of the matrix is <math>[g_{ij}]^{-1} = [g^{ij}]</math>.  The inverse of the matrix is given by
<math display="block">
   [g^{ij}] = [g_{ij}]^{-1} = \cfrac{A^{ij}}{g} ~;~~ g := \det([g_{ij}]) = \det\boldsymbol{g}
</math>
where ''A<sup>ij</sup>'' are the Cofactor matrix of the components ''g<sub>ij</sub>''.  From matrix algebra we have
<math display="block">
  g = \det([g_{ij}]) = \sum_i g_{ij}~A^{ij} \quad \Rightarrow \quad 
  \frac{\partial g}{\partial g_{ij}} = A^{ij}
</math>
Hence,
<math display="block">
   [g^{ij}] = \cfrac{1}{g}~\frac{\partial g}{\partial g_{ij}}
</math>
Plugging this relation into the expression for the divergence gives
<math display="block">
  \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
</math>
A little manipulation leads to the more compact form
<math display="block">
  \boldsymbol{\nabla} \cdot \mathbf{v} = \cfrac{1}{\sqrt{g}}~\frac{\partial }{\partial q^i}(v^i~\sqrt{g})
</math>

====Second-order tensor field====

The [[Divergence|divergence]] of a second-order tensor field is defined using
<math display="block">
   (\boldsymbol{\nabla}\cdot\boldsymbol{S})\cdot\mathbf{a} = \boldsymbol{\nabla}\cdot(\boldsymbol{S}\mathbf{a})
 </math>
where '''a''' is an arbitrary constant vector.
<ref>{{cite web | url=http://en.wikiversity.org/wiki/Introduction_to_Elasticity/Tensors#The_divergence_of_a_tensor_field | publisher=[[Social:Wikiversity|Wikiversity]] | work=Introduction to Elasticity/Tensors | title=The divergence of a tensor field | accessdate=2010-11-26 }}</ref>
In curvilinear coordinates,
<math display="block">
  \begin{align}
   \boldsymbol{\nabla}\cdot\boldsymbol{S} & = \left[\cfrac{\partial S_{ij}}{\partial q^k} - \Gamma^l_{ki}~S_{lj} - \Gamma^l_{kj}~S_{il}\right]~g^{ik}~\mathbf{b}^j \\[8pt]
   & = \left[\cfrac{\partial S^{ij}}{\partial q^i} + \Gamma^i_{il}~S^{lj} + \Gamma^j_{il}~S^{il}\right]~\mathbf{b}_j \\[8pt]
   & = \left[\cfrac{\partial S^i_{~j}}{\partial q^i} + \Gamma^i_{il}~S^l_{~j} - \Gamma^l_{ij}~S^i_{~l}\right]~\mathbf{b}^j \\[8pt]
   & = \left[\cfrac{\partial S_i^{~j}}{\partial q^k} - \Gamma^l_{ik}~S_l^{~j} + \Gamma^j_{kl}~S_i^{~l}\right]~g^{ik}~\mathbf{b}_j
  \end{align}
 </math>

===Laplacian ===

====Scalar field====
The Laplacian of a scalar field φ('''x''') is defined as
<math display="block">
  \nabla^2 \varphi := \boldsymbol{\nabla} \cdot (\boldsymbol{\nabla} \varphi)
</math>
Using the alternative expression for the divergence of a vector field gives us
<math display="block">
  \nabla^2 \varphi =  \cfrac{1}{\sqrt{g}}~\frac{\partial }{\partial q^i}([\boldsymbol{\nabla} \varphi]^i~\sqrt{g})
</math>
Now
<math display="block">
  \boldsymbol{\nabla} \varphi = \frac{\partial \varphi}{\partial q^l}~\mathbf{b}^l = g^{li}~\frac{\partial \varphi}{\partial q^l}~\mathbf{b}_i
  \quad \Rightarrow \quad
  [\boldsymbol{\nabla} \varphi]^i = g^{li}~\frac{\partial \varphi}{\partial q^l}
</math>
Therefore,
<math display="block">
  \nabla^2 \varphi =  \cfrac{1}{\sqrt{g}}~\frac{\partial }{\partial q^i}\left(g^{li}~\frac{\partial \varphi}{\partial q^l}
~\sqrt{g}\right)
</math>

===Curl of a vector field===
The curl of a vector field '''v''' in covariant curvilinear coordinates can be written as
<math display="block">
   \boldsymbol{\nabla}\times\mathbf{v} = \mathcal{E}^{rst} v_{s|r}~ \mathbf{b}_t
 </math>
where
<math display="block">
   v_{s|r} = v_{s,r} - \Gamma^i_{sr}~v_i
 </math>

==Orthogonal curvilinear coordinates==
Assume, for the purposes of this section, that the curvilinear coordinate system is orthogonal, i.e.,
<math display="block"> \mathbf{b}_i\cdot\mathbf{b}_j = 
   \begin{cases} g_{ii} & \text{if } i = j \\
                0 & \text{if } i \ne j,
   \end{cases}
 </math>
or equivalently,
<math display="block"> \mathbf{b}^i\cdot\mathbf{b}^j = 
   \begin{cases} g^{ii} & \text{if } i = j \\
                0 & \text{if } i \ne j,
   \end{cases}
 </math>
where <math>g^{ii} = g_{ii}^{-1}</math>. As before, <math>\mathbf{b}_i, \mathbf{b}_j</math> are covariant basis vectors and '''b'''<sup>''i''</sup>, '''b'''<sup>''j''</sup> are contravariant basis vectors.  Also, let ('''e'''<sup>1</sup>, '''e'''<sup>2</sup>, '''e'''<sup>3</sup>) be a background, fixed, [[Cartesian coordinate system|Cartesian]] basis.  A list of orthogonal curvilinear coordinates is given below.

===Metric tensor in orthogonal curvilinear coordinates===
{{Main article|Physics:Metric tensor}}

Let '''r'''('''x''') be the position vector of the point '''x''' with respect to the origin of the coordinate system.  The notation can be simplified by noting that '''x''' = '''r'''('''x''').  At each point we can construct a small line element d'''x'''.  The square of the length of the line element is the scalar product d'''x''' • d'''x''' and is called the [[Metric (mathematics)|metric]] of the [[Space|space]].  Recall that the space of interest is assumed to be [[Euclidean space|Euclidean]] when we talk of curvilinear coordinates. Let us express the position vector in terms of the background, fixed, Cartesian basis, i.e.,
<math display="block">
   \mathbf{x} = \sum_{i=1}^3 x_i~\mathbf{e}_i
 </math>

Using the [[Chain rule|chain rule]], we can then express d'''x''' in terms of three-dimensional orthogonal curvilinear coordinates (''q''<sup>1</sup>, ''q''<sup>2</sup>, ''q''<sup>3</sup>) as
<math display="block">
   \mathrm{d}\mathbf{x} = \sum_{i=1}^3 \sum_{j=1}^3 \left(\cfrac{\partial x_i}{\partial q^j}~\mathbf{e}_i\right)\mathrm{d}q^j
 </math>
Therefore, the metric is given by
<math display="block"> 
  \mathrm{d}\mathbf{x}\cdot\mathrm{d}\mathbf{x} = \sum_{i=1}^3 \sum_{j=1}^3 \sum_{k=1}^3 \cfrac{\partial x_i}{\partial q^j}~\cfrac{\partial x_i}{\partial q^k}~\mathrm{d}q^j~\mathrm{d}q^k
 </math>

The symmetric quantity
<math display="block">
  g_{ij}(q^i,q^j) = \sum_{k=1}^3 \cfrac{\partial x_k}{\partial q^i}~\cfrac{\partial x_k}{\partial q^j} = \mathbf{b}_i\cdot\mathbf{b}_j
</math>
is called the [[Physics:Metric tensor|fundamental (or metric) tensor]] of the [[Euclidean space]] in curvilinear coordinates.

Note also that
<math display="block">
   g_{ij} = \cfrac{\partial\mathbf{x}}{\partial q^i}\cdot\cfrac{\partial\mathbf{x}}{\partial q^j}
          = \left(\sum_{k} h_{ki}~\mathbf{e}_k\right)\cdot\left(\sum_{m} h_{mj}~\mathbf{e}_m\right)
          = \sum_{k} h_{ki}~h_{kj}
 </math>
where ''h<sub>ij</sub>'' are the Lamé coefficients.

If we define the scale factors, ''h<sub>i</sub>'',  using
<math display="block">
   \mathbf{b}_i\cdot\mathbf{b}_i = g_{ii} = \sum_{k} h_{ki}^2 =: h_i^2
   \quad \Rightarrow \quad \left|\cfrac{\partial\mathbf{x}}{\partial q^i}\right| = \left|\mathbf{b}_i\right| = \sqrt{g_{ii}} = h_i
 </math>
we get a relation between the fundamental tensor and the Lamé coefficients.

====Example: Polar coordinates====
If we consider polar coordinates for '''R'''<sup>2</sup>, note that
<math display="block"> (x, y)=(r \cos \theta, r \sin \theta) </math>
(r, θ) are the curvilinear coordinates, and the Jacobian determinant of the transformation (''r'',θ) → (''r'' cos θ, ''r'' sin θ) is ''r''.

The orthogonal basis vectors are '''''b'''''<sub>''r''</sub> = (cos θ, sin θ), '''''b'''''<sub>θ</sub> = (−''r'' sin θ, ''r'' cos θ).  The normalized basis vectors are '''''e'''''<sub>''r''</sub> = (cos θ, sin θ), '''''e'''''<sub>θ</sub> = (−sin θ, cos θ) and the scale factors are ''h''<sub>''r''</sub> = 1 and ''h''<sub>θ</sub>= ''r''. The fundamental tensor is ''g''<sub>11</sub> =1, ''g''<sub>22</sub> =''r''<sup>2</sup>, ''g''<sub>12</sub> = ''g''<sub>21</sub> =0.

===Line and surface integrals===
If we wish to use curvilinear coordinates for [[Vector calculus|vector calculus]] calculations, adjustments need to be made in the calculation of line, surface and volume integrals.  For simplicity, we again restrict the discussion to three dimensions and orthogonal curvilinear coordinates.  However, the same arguments apply for <math>n</math>-dimensional problems though there are some additional terms in the expressions when the coordinate system is not orthogonal.

====Line integrals====
Normally in the calculation of [[Line integral|line integral]]s we are interested in calculating
<math display="block"> \int_C f \,ds = \int_a^b f(\mathbf{x}(t))\left|{\partial \mathbf{x} \over \partial t}\right|\; dt</math>
where '''''x'''''(''t'') parametrizes C in Cartesian coordinates.
In curvilinear coordinates, the term

<math display="block"> \left|{\partial \mathbf{x} \over \partial t}\right| = \left| \sum_{i=1}^3 {\partial \mathbf{x} \over \partial q^i}{\partial q^i \over \partial t}\right|</math>

by the [[Chain rule|chain rule]].  And from the definition of the Lamé coefficients,

<math display="block"> {\partial \mathbf{x} \over \partial q^i} = \sum_{k} h_{ki}~ \mathbf{e}_{k} </math>

and thus

<math display="block">
\begin{align}
\left|{\partial \mathbf{x} \over \partial t}\right| & = \left| \sum_k\left(\sum_i h_{ki}~\cfrac{\partial q^i}{\partial t}\right)\mathbf{e}_k\right| \\[8pt]
& = \sqrt{\sum_i\sum_j\sum_k h_{ki}~h_{kj}\cfrac{\partial q^i}{\partial t}\cfrac{\partial q^j}{\partial t}} = \sqrt{\sum_i\sum_j g_{ij}~\cfrac{\partial q^i}{\partial t}\cfrac{\partial q^j}{\partial t}}
\end{align}
</math>

Now, since <math>g_{ij} = 0</math> when <math> i \ne j </math>, we have
<math display="block"> 
\left|{\partial \mathbf{x} \over \partial t}\right| =  \sqrt{\sum_i g_{ii}~\left(\cfrac{\partial q^i}{\partial t}\right)^2} = \sqrt{\sum_i h_{i}^2~\left(\cfrac{\partial q^i}{\partial t}\right)^2} 
</math>
and we can proceed normally.

====Surface integrals====
Likewise, if we are interested in a [[Surface integral|surface integral]], the relevant calculation, with the parameterization of the surface in Cartesian coordinates is:
<math display="block">\int_S f \,dS = \iint_T f(\mathbf{x}(s, t)) \left|{\partial \mathbf{x} \over \partial s}\times {\partial \mathbf{x} \over \partial t}\right| \, ds \, dt</math>
Again, in curvilinear coordinates, we have
<math display="block">
\left|{\partial \mathbf{x} \over \partial s}\times {\partial \mathbf{x} \over \partial t}\right| = \left|\left(\sum_i {\partial \mathbf{x} \over \partial q^i}{\partial q^i \over \partial s}\right) \times \left(\sum_j {\partial \mathbf{x} \over \partial q^j}{\partial q^j \over \partial t}\right)\right|
</math>
and we make use of the definition of curvilinear coordinates again to yield
<math display="block">
{\partial \mathbf{x} \over \partial q^i}{\partial q^i \over \partial s} = \sum_k \left(\sum_{i=1}^3 h_{ki}~{\partial q^i \over \partial s}\right) \mathbf{e}_k ~;~~ 
{\partial \mathbf{x} \over \partial q^j}{\partial q^j \over \partial t} = \sum_m \left(\sum_{j=1}^3 h_{mj}~{\partial q^j \over \partial t}\right) \mathbf{e}_{m}
</math>

Therefore,
<math display="block">
  \begin{align}
\left|{\partial \mathbf{x} \over \partial s}\times {\partial \mathbf{x} \over \partial t}\right|
  & = \left|
     \sum_k \sum_m \left(\sum_{i=1}^3 h_{ki}~{\partial q^i \over \partial s}\right)\left(\sum_{j=1}^3 h_{mj}~{\partial q^j \over \partial t}\right) \mathbf{e}_k\times\mathbf{e}_m
    \right| \\[8pt]
  & = \left|\sum_p \sum_k \sum_m \mathcal{E}_{kmp}\left(\sum_{i=1}^3 h_{ki}~{\partial q^i \over \partial s}\right)\left(\sum_{j=1}^3 h_{mj}~{\partial q^j \over \partial t}\right) \mathbf{e}_p \right|
  \end{align}
  </math>
where <math>\mathcal{E}</math> is the permutation symbol.

In determinant form,  the cross product in terms of curvilinear coordinates will be:
<math display="block">\begin{vmatrix} 
\mathbf{e}_1                    & \mathbf{e}_2                    & \mathbf{e}_3 \\
 && \\
\sum_i h_{1i} {\partial q^i \over \partial s} & \sum_i h_{2i} {\partial q^i \over \partial s} & \sum_i h_{3i} {\partial q^i \over \partial s} \\
&& \\
\sum_j h_{1j} {\partial q^j \over \partial t} & \sum_j h_{2j} {\partial q^j \over \partial t} & \sum_j h_{3j} {\partial q^j \over \partial t} \end{vmatrix}</math>

===Grad, curl, div, Laplacian===

In [[Orthogonality|orthogonal]] curvilinear coordinates of 3 dimensions, where
<math display="block">
        \mathbf{b}^i = \sum_k g^{ik}~\mathbf{b}_k ~;~~ g^{ii} = \cfrac{1}{g_{ii}} = \cfrac{1}{h_i^2}
</math>
one can express the [[Gradient|gradient]] of a [[Scalar (mathematics)|scalar]] or [[Vector field|vector field]] as
<math display="block">
   \nabla\varphi = \sum_{i} {\partial\varphi \over \partial q^i}~ \mathbf{b}^i = \sum_{i} \sum_j {\partial\varphi \over \partial q^i}~ g^{ij}~\mathbf{b}_j = \sum_i \cfrac{1}{h_i^2}~{\partial f \over \partial q^i}~\mathbf{b}_i ~;~~
   \nabla\mathbf{v} = \sum_i \cfrac{1}{h_i^2}~{\partial \mathbf{v} \over \partial q^i}\otimes\mathbf{b}_i
 </math>
For an orthogonal basis
<math display="block">
   g = g_{11}~g_{22}~g_{33} = h_1^2~h_2^2~h_3^2 \quad \Rightarrow \quad \sqrt{g} = h_1 h_2 h_3
</math>
The [[Divergence|divergence]] of a vector field can then be written as
<math display="block">
  \boldsymbol{\nabla} \cdot \mathbf{v} = \cfrac{1}{h_1 h_2 h_3}~\frac{\partial }{\partial q^i}(h_1 h_2 h_3~v^i)
</math>
Also,
<math display="block">
  v^i = g^{ik}~v_k \quad \Rightarrow v^1 = g^{11}~v_1 = \cfrac{v_1}{h_1^2} ~;~~ v^2 = g^{22}~v_2  = \cfrac{v_2}{h_2^2}~;~~ v^3 = g^{33}~v_3 = \cfrac{v_3}{h_3^2}
</math>
Therefore,
<math display="block">
  \boldsymbol{\nabla} \cdot \mathbf{v} = \cfrac{1}{h_1 h_2 h_3}~\sum_i \frac{\partial }{\partial q^i}\left(\cfrac{h_1 h_2 h_3}{h_i^2}~v_i\right)
</math>
We can get an expression for the Laplacian in a similar manner by noting that
<math display="block">
  g^{li}~\frac{\partial \varphi}{\partial q^l} = 
   \left\{ g^{11}~\frac{\partial \varphi}{\partial q^1}, g^{22}~\frac{\partial \varphi}{\partial q^2}, 
    g^{33}~\frac{\partial \varphi}{\partial q^3} \right\} = 
   \left\{ \cfrac{1}{h_1^2}~\frac{\partial \varphi}{\partial q^1}, \cfrac{1}{h_2^2}~\frac{\partial \varphi}{\partial q^2}, 
    \cfrac{1}{h_3^2}~\frac{\partial \varphi}{\partial q^3} \right\} 
</math>
Then we have
<math display="block">
  \nabla^2 \varphi =  \cfrac{1}{h_1 h_2 h_3}~\sum_i\frac{\partial }{\partial q^i}\left(\cfrac{h_1 h_2 h_3}{h_i^2}~\frac{\partial \varphi}{\partial q^i}\right)
</math>
The expressions for the gradient, divergence, and Laplacian can be directly extended to ''n''-dimensions.

The [[Curl (mathematics)|curl]] of a [[Vector field|vector field]] is given by
<math display="block"> 
     \nabla\times\mathbf{v} = \frac{1}{h_1h_2h_3} \sum_{i=1}^n  \mathbf{e}_i
\sum_{jk} \varepsilon_{ijk} h_i \frac{\partial (h_k v_k)}{\partial q^j}
</math>
where ε<sub>''ijk''</sub> is the [[Levi-Civita symbol]].

==Example: Cylindrical polar coordinates==
For cylindrical coordinates we have
<math display="block">
   (x_1, x_2, x_3) = \mathbf{x} = \boldsymbol{\varphi}(q^1, q^2, q^3) = \boldsymbol{\varphi}(r, \theta, z)
     = \{r\cos\theta, r\sin\theta, z\}
 </math>
and
<math display="block">
   \{\psi^1(\mathbf{x}), \psi^2(\mathbf{x}), \psi^3(\mathbf{x})\} = (q^1, q^2, q^3) \equiv (r, \theta, z) 
   = \{ \sqrt{x_1^2+x_2^2}, \tan^{-1}(x_2/x_1), x_3\}
 </math>
where
<math display="block">
0 < r < \infty ~, ~~ 0 < \theta < 2\pi ~,~~ -\infty < z < \infty
 </math>

Then the covariant and contravariant basis vectors are
<math display="block">
  \begin{align}
  \mathbf{b}_1 & = \mathbf{e}_r = \mathbf{b}^1 \\
  \mathbf{b}_2 & = r~\mathbf{e}_\theta = r^2~\mathbf{b}^2 \\
  \mathbf{b}_3 & = \mathbf{e}_z = \mathbf{b}^3
  \end{align}
 </math>
where <math>\mathbf{e}_r, \mathbf{e}_\theta, \mathbf{e}_z</math> are the unit vectors in the <math>r, \theta, z</math> directions.

Note that the components of the metric tensor are such that
<math display="block">
    g^{ij} = g_{ij} = 0 (i \ne j) ~;~~ \sqrt{g^{11}} = 1,~\sqrt{g^{22}} = \cfrac{1}{r},~\sqrt{g^{33}}=1
 </math>
which shows that the basis is orthogonal.

The non-zero components of the Christoffel symbol of the second kind are
<math display="block">
   \Gamma_{12}^2 = \Gamma_{21}^2 = \cfrac{1}{r} ~;~~ \Gamma_{22}^1 = -r
 </math>

===Representing a physical vector field===
The normalized contravariant basis vectors in cylindrical polar coordinates are
<math display="block">
   \hat{\mathbf{b}}^1 = \mathbf{e}_r ~;~~\hat{\mathbf{b}}^2 = \mathbf{e}_\theta ~;~~\hat{\mathbf{b}}^3 = \mathbf{e}_z
 </math>
and the physical components of a vector '''v''' are
<math display="block">
   (\hat{v}_1, \hat{v}_2, \hat{v}_3) = (v_1, v_2/r, v_3) =: (v_r, v_\theta, v_z)
 </math>

===Gradient of a scalar field===
The gradient of a scalar field, ''f''('''x'''), in cylindrical coordinates can now be computed from the general expression in curvilinear coordinates and has the form
<math display="block">
   \boldsymbol{\nabla}f = \cfrac{\partial f}{\partial r}~\mathbf{e}_r + \cfrac{1}{r}~\cfrac{\partial f}{\partial \theta}~\mathbf{e}_\theta +  \cfrac{\partial f}{\partial z}~\mathbf{e}_z
 </math>

===Gradient of a vector field===
Similarly, the gradient of a vector field, '''v'''('''x'''), in cylindrical coordinates can be shown to be
<math display="block">
   \begin{align}
   \boldsymbol{\nabla}\mathbf{v} & = \cfrac{\partial v_r}{\partial r}~\mathbf{e}_r\otimes\mathbf{e}_r + 
     \cfrac{1}{r}\left(\cfrac{\partial v_r}{\partial \theta} - v_\theta\right)~\mathbf{e}_r\otimes\mathbf{e}_\theta + \cfrac{\partial v_r}{\partial z}~\mathbf{e}_r\otimes\mathbf{e}_z \\[8pt]
 & + \cfrac{\partial v_\theta}{\partial r}~\mathbf{e}_\theta\otimes\mathbf{e}_r + 
     \cfrac{1}{r}\left(\cfrac{\partial v_\theta}{\partial \theta} + v_r \right)~\mathbf{e}_\theta\otimes\mathbf{e}_\theta + \cfrac{\partial v_\theta}{\partial z}~\mathbf{e}_\theta\otimes\mathbf{e}_z \\[8pt]
 & + \cfrac{\partial v_z}{\partial r}~\mathbf{e}_z\otimes\mathbf{e}_r + 
     \cfrac{1}{r}\cfrac{\partial v_z}{\partial \theta}~\mathbf{e}_z\otimes\mathbf{e}_\theta + \cfrac{\partial v_z}{\partial z}~\mathbf{e}_z\otimes\mathbf{e}_z
   \end{align} 
 </math>

===Divergence of a vector field===
Using the equation for the divergence of a vector field in curvilinear coordinates, the divergence in cylindrical coordinates can be shown to be
<math display="block">
   \begin{align}
   \boldsymbol{\nabla}\cdot\mathbf{v} & = \cfrac{\partial v_r}{\partial r} + 
     \cfrac{1}{r}\left(\cfrac{\partial v_\theta}{\partial \theta} + v_r \right)
 + \cfrac{\partial v_z}{\partial z}
   \end{align} 
 </math>

===Laplacian of a scalar field===
The Laplacian is more easily computed by noting that <math>\boldsymbol{\nabla}^2 f = \boldsymbol{\nabla}\cdot\boldsymbol{\nabla}f</math>.  In cylindrical polar coordinates
<math display="block">
  \mathbf{v} = \boldsymbol{\nabla}f = \left[v_r~~ v_\theta~~ v_z\right] = \left[\cfrac{\partial f}{\partial r}~~  \cfrac{1}{r}\cfrac{\partial f}{\partial \theta}~~ \cfrac{\partial f}{\partial z} \right]
</math>
Hence,
<math display="block">
  \boldsymbol{\nabla}\cdot\mathbf{v} = \boldsymbol{\nabla}^2 f = \cfrac{\partial^2 f}{\partial r^2} + 
     \cfrac{1}{r}\left(\cfrac{1}{r}\cfrac{\partial^2f}{\partial \theta^2} + \cfrac{\partial f}{\partial r} \right)
 + \cfrac{\partial^2 f}{\partial z^2}
   = \cfrac{1}{r}\left[\cfrac{\partial}{\partial r}\left(r\cfrac{\partial f}{\partial r}\right)\right] + \cfrac{1}{r^2}\cfrac{\partial^2f}{\partial \theta^2} + \cfrac{\partial^2 f}{\partial z^2}
 </math>

===Representing a physical second-order tensor field===
The physical components of a second-order tensor field are those obtained when the tensor is expressed in terms of a normalized contravariant basis.  In cylindrical polar coordinates these components are:

<math display="block">\begin{align}
\hat{S}_{11} &= S_{11} =: S_{rr}, & \hat{S}_{12} &= \frac{S_{12}}{r} =: S_{r\theta}, & \hat{S}_{13} &= S_{13} =: S_{rz} \\[6pt]
\hat{S}_{21} &= \frac{S_{21}}{r} =: S_{\theta r}, & \hat{S}_{22} &= \frac{S_{22}}{r^2} =: S_{\theta\theta}, & \hat{S}_{23} &= \frac{S_{23}}{r} =: S_{\theta z} \\[6pt]
\hat{S}_{31} &= S_{31} =: S_{zr}, & \hat{S}_{32} &= \frac{S_{32}}{r} =: S_{z\theta}, & \hat{S}_{33} &= S_{33} =: S_{zz}
\end{align}</math>

===Gradient of a second-order tensor field===
Using the above definitions we can show that the gradient of a second-order tensor field in cylindrical polar coordinates can be expressed as
<math display="block">
\begin{align}
\boldsymbol{\nabla} \boldsymbol{S} & = \frac{\partial S_{rr}}{\partial r}~\mathbf{e}_r\otimes\mathbf{e}_r\otimes\mathbf{e}_r +
\cfrac{1}{r}\left[\frac{\partial S_{rr}}{\partial \theta} - (S_{\theta r}+S_{r\theta})\right]~\mathbf{e}_r\otimes\mathbf{e}_r\otimes\mathbf{e}_\theta +
\frac{\partial S_{rr}}{\partial z}~\mathbf{e}_r\otimes\mathbf{e}_r\otimes\mathbf{e}_z \\[8pt]
 & + \frac{\partial S_{r\theta}}{\partial r}~\mathbf{e}_r\otimes\mathbf{e}_\theta\otimes\mathbf{e}_r +
\cfrac{1}{r}\left[\frac{\partial S_{r\theta}}{\partial \theta} + (S_{rr}-S_{\theta\theta})\right]~\mathbf{e}_r\otimes\mathbf{e}_\theta\otimes\mathbf{e}_\theta +
\frac{\partial S_{r\theta}}{\partial z}~\mathbf{e}_r\otimes\mathbf{e}_\theta\otimes\mathbf{e}_z \\[8pt]
 & + \frac{\partial S_{rz}}{\partial r}~\mathbf{e}_r\otimes\mathbf{e}_z\otimes\mathbf{e}_r +
\cfrac{1}{r}\left[\frac{\partial S_{rz}}{\partial \theta} -S_{\theta z}\right]~\mathbf{e}_r\otimes\mathbf{e}_z\otimes\mathbf{e}_\theta +
\frac{\partial S_{rz}}{\partial z}~\mathbf{e}_r\otimes\mathbf{e}_z\otimes\mathbf{e}_z \\[8pt]
 & + \frac{\partial S_{\theta r}}{\partial r}~\mathbf{e}_\theta\otimes\mathbf{e}_r\otimes\mathbf{e}_r +
\cfrac{1}{r}\left[\frac{\partial S_{\theta r}}{\partial \theta} + (S_{rr}-S_{\theta\theta})\right]~\mathbf{e}_\theta\otimes\mathbf{e}_r\otimes\mathbf{e}_\theta +
\frac{\partial S_{\theta r}}{\partial z}~\mathbf{e}_\theta\otimes\mathbf{e}_r\otimes\mathbf{e}_z \\[8pt]
 & + \frac{\partial S_{\theta\theta}}{\partial r}~\mathbf{e}_\theta\otimes\mathbf{e}_\theta\otimes\mathbf{e}_r +
\cfrac{1}{r}\left[\frac{\partial S_{\theta\theta}}{\partial \theta} + (S_{r\theta}+S_{\theta r})\right]~\mathbf{e}_\theta\otimes\mathbf{e}_\theta\otimes\mathbf{e}_\theta +
\frac{\partial S_{\theta\theta}}{\partial z}~\mathbf{e}_\theta\otimes\mathbf{e}_\theta\otimes\mathbf{e}_z \\[8pt]
 & + \frac{\partial S_{\theta z}}{\partial r}~\mathbf{e}_\theta\otimes\mathbf{e}_z\otimes\mathbf{e}_r +
\cfrac{1}{r}\left[\frac{\partial S_{\theta z}}{\partial \theta} + S_{rz}\right]~\mathbf{e}_\theta\otimes\mathbf{e}_z\otimes\mathbf{e}_\theta +
\frac{\partial S_{\theta z}}{\partial z}~\mathbf{e}_\theta\otimes\mathbf{e}_z\otimes\mathbf{e}_z \\[8pt]
 & + \frac{\partial S_{zr}}{\partial r}~\mathbf{e}_z\otimes\mathbf{e}_r\otimes\mathbf{e}_r +
\cfrac{1}{r}\left[\frac{\partial S_{zr}}{\partial \theta} - S_{z\theta}\right]~\mathbf{e}_z\otimes\mathbf{e}_r\otimes\mathbf{e}_\theta +
\frac{\partial S_{zr}}{\partial z}~\mathbf{e}_z\otimes\mathbf{e}_r\otimes\mathbf{e}_z \\[8pt]
 & + \frac{\partial S_{z\theta}}{\partial r}~\mathbf{e}_z\otimes\mathbf{e}_\theta\otimes\mathbf{e}_r +
\cfrac{1}{r}\left[\frac{\partial S_{z\theta}}{\partial \theta} + S_{zr}\right]~\mathbf{e}_z\otimes\mathbf{e}_\theta\otimes\mathbf{e}_\theta +
\frac{\partial S_{z\theta}}{\partial z}~\mathbf{e}_z\otimes\mathbf{e}_\theta\otimes\mathbf{e}_z \\[8pt]
 & + \frac{\partial S_{zz}}{\partial r}~\mathbf{e}_z\otimes\mathbf{e}_z\otimes\mathbf{e}_r +
\cfrac{1}{r}~\frac{\partial S_{zz}}{\partial \theta}~\mathbf{e}_z\otimes\mathbf{e}_z\otimes\mathbf{e}_\theta +
\frac{\partial S_{zz}}{\partial z}~\mathbf{e}_z\otimes\mathbf{e}_z\otimes\mathbf{e}_z
\end{align}
</math>

===Divergence of a second-order tensor field===
The divergence of a second-order tensor field in cylindrical polar coordinates can be obtained from the expression for the gradient by collecting terms where the scalar product of the two outer vectors in the dyadic products is nonzero. Therefore,
<math display="block">
\begin{align}
\boldsymbol{\nabla}\cdot \boldsymbol{S} & = \frac{\partial S_{rr}}{\partial r}~\mathbf{e}_r 
   + \frac{\partial S_{r\theta}}{\partial r}~\mathbf{e}_\theta
   + \frac{\partial S_{rz}}{\partial r}~\mathbf{e}_z  \\[8pt]
 &  + \cfrac{1}{r}\left[\frac{\partial S_{r \theta}}{\partial \theta} + (S_{rr}-S_{\theta\theta})\right]~\mathbf{e}_r  +
\cfrac{1}{r}\left[\frac{\partial S_{\theta\theta}}{\partial \theta} + (S_{r\theta}+S_{\theta r})\right]~\mathbf{e}_\theta   +\cfrac{1}{r}\left[\frac{\partial S_{\theta z}}{\partial \theta} + S_{rz}\right]~\mathbf{e}_z  \\[8pt]
 &  +
\frac{\partial S_{zr}}{\partial z}~\mathbf{e}_r +
\frac{\partial S_{z\theta}}{\partial z}~\mathbf{e}_\theta +
\frac{\partial S_{zz}}{\partial z}~\mathbf{e}_z
\end{align}
</math>

==See also==
* Covariance and contravariance
* Basic introduction to the mathematics of curved spacetime
* [[Orthogonal coordinates]]
* [[Frenet–Serret formulas]]
* [[Covariant derivative]]
* [[Physics:Tensor derivative (continuum mechanics)|Tensor derivative (continuum mechanics)]]
* [[Curvilinear perspective]]
* [[Del in cylindrical and spherical coordinates]]

==References==
;Notes
{{reflist|2}}

;Further reading
{{refbegin}}
*{{Cite book| first=M. R. | last=Spiegel | title=Vector Analysis | publisher=Schaum's Outline Series | location=New York | year=1959| isbn=0-07-084378-3 }}
*{{Cite book| last=Arfken | first=George | title=Mathematical Methods for Physicists | publisher=Academic Press | year=1995| isbn=0-12-059877-9}}
{{refend}}

==External links==
* [http://planetmath.org/?method=l2h&from=collab&id=83&op=getobj Derivation of Unit Vectors in Curvilinear Coordinates]
* [http://mathworld.wolfram.com/CurvilinearCoordinates.html MathWorld's page on Curvilinear Coordinates]
* [http://www.mech.utah.edu/~brannon/public/curvilinear.pdf Prof. R. Brannon's E-Book on Curvilinear Coordinates]

{{Orthogonal coordinate systems}}

{{tensors}}

{{DEFAULTSORT:Curvilinear Coordinates}}
[[Category:Coordinate systems]]
[[Category:Metric tensors|*3]]

{{Sourceattribution|Tensors in curvilinear coordinates}}