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{{main|Curvilinear coordinates}}
The stylish wardrobe of Maggie Gyllenhaal�s role in BBC Two�s hard-hitting political thriller, The Honourable Woman, has caught the attention of the eagle-eyed viewers.<br><br>


'''[[Curvilinear coordinates]]''' can be formulated in [[tensor calculus]], with important applications in [[physics]] and [[engineering]], particularly for describing transportation of physical quatities and deformation of matter in [[fluid mechanics]] and [[continuum mechanics]].
The eight-part series, set against the backdrop of the Israeli-Palestinian conflict, centres around Nessa Stein, played by Gyllenhaal. Stein is an Anglo-Israeli businesswoman recently ennobled in the House of Lords who devotes herself to philanthropic purposes across the Middle East, but hides a secret past from her time spent in Gaza eight years earlier.<br><br>Through the unravelling of her public and private life played out on an international, political stage, Stein parades in an increasingly impressive selection of outfits.<br>�Because the character of Nessa is so complicated and multi layered, we looked at all sorts of different people as reference. I suppose we started off by looking at other powerful and stylish women through history, Jackie Kennedy, Eva Peron, Margaret Thatcher, Cleopatra� Edward K Gibbon costume designer for the series told The Independent<br><br>
Maggie Gyllenhaal The Honourable Woma<br>
�And then we kind of threw all the reference away and started afresh. The way Maggie looked as Nessa was constantly evolving throughout the six month shoot.�
The series opens with Nessa clad in a Roland Mouret [https://www.Google.com/search?hl=en&gl=us&tbm=nws&q=power+dress power dress]. Her day to day look is a sartorial dream with [http://Www.Dailymail.Co.uk/home/search.html?sel=site&searchPhrase=tailored tailored] suits by the likes of Stella McCartney, Acne, Escada, Pringle and vintage Chane<br>
�Silk blouses and wide legged pants based on 1970s Yves Saint Laurent originals were created by Hilary Marschner� explains Gibb<br><br>


==Vector and tensor algebra in three-dimensional curvilinear coordinates==
Outerwear includes coats by Mulberry, vintage finds from Jil Sander and a 1980s Gieves and Hawkes men�s co<br>.
{{Einstein_summation_convention}}
Even curled up in her panic room at night she sports silk slips by haute couture Parisian lingerie designer Carine Gilson and London based lingerie label Bod<br>.
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.<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>
In pictures: Nessa Stein's wardrobe in The Honourable Woma<br>  
Shoes are by Acne, Christian Louboutin and [http://www.pcs-systems.co.uk/Images/celinebag.aspx Celine Bag Online]. With bags from Mulberry and John Lewis. �Nessa's wardrobe runs the full gamut from designer, through High Street, Charity shops and bespoke pieces� says Gib<br><br>


===Vectors in curvilinear coordinates===
�The clothing is always the way in [to the character]� Gyllenhaal told WWD. �I never played a character that didn�t care about what they were wearing.
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}}
The Honourable Woman continues tonight, BBC2 at 9pm.
:<math>
    \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>
  \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>
    \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>
  \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>
    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>
  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>
  \mathbf{v}\cdot\mathbf{b}^i = v^k~\mathbf{b}_k\cdot\mathbf{b}^i = v^k~\delta^i_k = v^i </math>
:<math>
  \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>
  \mathbf{v}\cdot\mathbf{b}_i = v^k~\mathbf{b}_k\cdot\mathbf{b}_i = g_{ki}~v^k </math>
:<math>
  \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>
  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>
  \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>
  \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>
  \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>
  \mathbf{b}_i\times\mathbf{b}_j = J~\varepsilon_{ijp}~\mathbf{b}^p = \sqrt{g}~\varepsilon_{ijp}~\mathbf{b}^p
</math>
Hence,
:<math>
  \mathcal{E}_{ijk} = J~\varepsilon_{ijk} = \sqrt{g}~\varepsilon_{ijk}
</math>
Similarly, we can show that
:<math>
  \mathcal{E}^{ijk} = \cfrac{1}{J}~\varepsilon^{ijk} = \cfrac{1}{\sqrt{g}}~\varepsilon^{ijk}
</math>
 
===Vector operations===
 
<ol>
<li>'''Identity map'''
The identity map '''I''' defined by <math>\mathsf{I}\cdot\mathbf{v} = \mathbf{v}</math> can be shown to be<ref name=Simmonds/>{{rp|page=39}}
:<math>
  \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></li>
 
<li>'''Scalar (dot) product'''
The scalar product of two vectors in curvilinear coordinates is<ref name=Simmonds/>{{rp|page=32}}
:<math>
  \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></li>
 
<li>'''Vector (cross) product'''
The [[cross product]] of two vectors is given by<ref name=Simmonds/>{{rp|pages=32–34}}
:<math>
  \mathbf{u}\times\mathbf{v} = \varepsilon_{ijk}~{u}_j~{v}_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>
  \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
</math>
where <math>\mathcal{E}_{ijk}</math> is the [[Curvilinear_coordinates#The_alternating_tensor|third-order alternating tensor]].
 
The [[cross product]] of two vectors is given by
:<math>
  \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>
  \mathbf{e}_p\times\mathbf{e}_q = \varepsilon_{ipq}~\mathbf{e}_i
</math>
and
:<math>
\begin{align}
  \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} \\[8pt]
& = \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
\end{align}
</math>
 
Hence,
 
:<math>
  (\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 back to the vector product and using the relations
:<math>
  \hat{u}_j = \frac{\partial x_j}{\partial q^m}~u^m ~;~~
  \hat{v}_k = \frac{\partial x_k}{\partial q^n}~v^n ~;~~
  \mathbf{e}_i = \frac{\partial x_i}{\partial q^s}~\mathbf{b}^s
</math>
 
gives us
 
:<math>
\begin{align}
  \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 \\[8pt]
  & = [(\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
\end{align}
</math></li>
</ol>
 
===Tensor operations===
<ol>
<li>'''[[Identity function|Identity map]]:'''
The 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>
  \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></li>
<li>'''Action of a second-order tensor on a vector:'''
The action <math>\mathbf{v} = \boldsymbol{S}\cdot\mathbf{u}</math> can be expressed in curvilinear coordinates as
:<math>
  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></li>
 
<li>'''[[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>
  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>
  \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></li>
 
<li>'''[[Determinant]] of a second-order tensor:'''
If <math>\boldsymbol{S}</math> is a second-order tensor, then the determinant is defined by the relation
:<math>
  \left[\boldsymbol{S}\cdot\mathbf{u}, \boldsymbol{S}\cdot\mathbf{v}, \boldsymbol{S}\cdot\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>
  \left[\mathbf{u},\mathbf{v},\mathbf{w}\right] := \mathbf{u}\cdot(\mathbf{v}\times\mathbf{w}).
</math></li>
</ol>
 
===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>
  \mathbf{b}_i = \boldsymbol{F}\cdot\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>
  \mathbf{b}_i\otimes\mathbf{e}_i = (\boldsymbol{F}\cdot\mathbf{e}_i)\otimes\mathbf{e}_i = \boldsymbol{F}\cdot(\mathbf{e}_i\otimes\mathbf{e}_i) = \boldsymbol{F}~.
</math>
From this relation we can show that
:<math>
  \mathbf{b}^i = \boldsymbol{F}^{-\rm{T}}\cdot\mathbf{e}^i ~;~~ g^{ij} = [\boldsymbol{F}^{-\rm{1}}\cdot\boldsymbol{F}^{-\rm{T}}]_{ij} ~;~~ g_{ij} = [g^{ij}]^{-1} = [\boldsymbol{F}^{\rm{T}}\cdot\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>
  \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>
  \left[\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3\right] = 1
</math>
we have
:<math>
  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>
  g := \det[g_{ij}]\,
</math>
Then
:<math>
  g = \det[\boldsymbol{F}^{\rm{T}}]\cdot\det[\boldsymbol{F}] = J\cdot J = J^2
</math>
Similarly, we can show that
:<math>
  \det[g^{ij}] = \cfrac{1}{J^2}
</math>
Therefore, using the fact that <math>[g^{ij}] = [g_{ij}]^{-1}</math>,
:<math>
  \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>
  \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>
  \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>
  \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>
  \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>
  \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>
  \mathbf{e}_i = \mathbf{b}_j~\cfrac{\partial q^j}{\partial x_i}
</math>
From these results we have
:<math>
  \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>
  \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 [[Albert Einstein]] saying<ref name=Lanczos>{{cite book | last=Einstein | first=A. | year=1915 | contribution=Contribution to the Theory of General Relativity | editor=Laczos, C. | title=The Einstein Decade | page=213 | isbn=0-521-38105-3 }}</ref>
<blockquote>
  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.
</blockquote>
Vector and tensor calculus in general curvilinear coordinates is used in tensor analysis on four-dimensional curvilinear [[manifold]]s in [[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 [[solid mechanics|mechanics]] of curved [[Plate theory|shells]],<ref name=Ciarlet/> in examining the [[invariant (mathematics)|invariance]] properties of [[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}}</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 | page=247 |arxiv = cond-mat/0607418 |bibcode = 2006NJPh....8..247L |doi = 10.1088/1367-2630/8/10/247 }}</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>
  \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''') = '''ψ'''<sup>−1</sup>('''x''').
 
The ''q<sup>i</sup>'' '''coordinate curves''' are defined by the one-parameter family of functions given by
:<math>
  \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>
  \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>
  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>
  [\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>
  q^i + \alpha~c^i = \psi^i(\mathbf{x} + \alpha~\mathbf{c})
</math>
then
:<math>
  [\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>
  [\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>
  [\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>
  [\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>
  \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>
  \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>
  \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>
  \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>
  \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>
  \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 symbol]]s of the second kind are defined as
:<math> \Gamma_{ij}^k = \Gamma_{ji}^k </math>
in which
 
:<math>\cfrac{\partial \mathbf{b}_i}{\partial q^j} = \Gamma_{ij}^k~\mathbf{b}_k
</math>
 
This implies that
:<math>
  \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>
    \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>
  \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>
  \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>
  \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>
  \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>
  \boldsymbol{\nabla}\boldsymbol{S} = \cfrac{\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>
  \boldsymbol{\nabla}\boldsymbol{S} = \cfrac{\partial}{\partial q^k}[S_{ij}~\mathbf{b}^i\otimes\mathbf{b}^j]\otimes\mathbf{b}^k
  = \left[\cfrac{\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>
  \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>
  \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>
  \hat{S}_{ij} = S_{ij}~\sqrt{g^{ii}~g^{jj}}
</math>
 
===Divergence===
 
====Vector field====
The [[divergence]] of a vector field (<math>\mathbf{v}</math>)is defined as
:<math>
  \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>
  \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>
  \boldsymbol{\nabla} \cdot \mathbf{v} = \frac{\partial v^i}{\partial q^i} + \Gamma_{\ell i}^i~v^\ell
</math>
Now,
:<math>
  \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>
  g^{mi}~\frac{\partial g_{\ell m}}{\partial q^i}  = g^{mi}~ \frac{\partial g_{i\ell}}{\partial q^m}
</math>
we have
:<math>
  \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>
  [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>
  g = \det([g_{ij}]) = \sum_i g_{ij}~A^{ij} \quad \Rightarrow \quad
  \frac{\partial g}{\partial g_{ij}} = A^{ij}
</math>
Hence,
:<math>
  [g^{ij}] = \cfrac{1}{g}~\frac{\partial g}{\partial g_{ij}}
</math>
Plugging this relation into the expression for the divergence gives
:<math>
  \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>
  \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]] of a second-order tensor field is defined using
:<math>
  (\boldsymbol{\nabla}\cdot\boldsymbol{S})\cdot\mathbf{a} = \boldsymbol{\nabla}\cdot(\boldsymbol{S}\cdot\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=[[Wikiversity]] | work=Introduction to Elasticity/Tensors | title=The divergence of a tensor field | accessdate=2010-11-26 }}</ref>
In curvilinear coordinates,
:<math>
  \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>
  \nabla^2 \varphi := \boldsymbol{\nabla} \cdot (\boldsymbol{\nabla} \varphi)
</math>
Using the alternative expression for the divergence of a vector field gives us
:<math>
  \nabla^2 \varphi =  \cfrac{1}{\sqrt{g}}~\frac{\partial }{\partial q^i}([\boldsymbol{\nabla} \varphi]^i~\sqrt{g})
</math>
Now
:<math>
  \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>
  \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>
  \boldsymbol{\nabla}\times\mathbf{v} = \mathcal{E}^{rst} v_{s|r}~ \mathbf{b}_t
</math>
where
:<math>
  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> \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> \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|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]].  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>
  \mathbf{x} = \sum_{i=1}^3 x_i~\mathbf{e}_i
</math>
 
Using the [[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>
  \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>
  \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>
  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 [[metric tensor|fundamental (or metric) tensor]] of the [[Euclidean space]] in curvilinear coordinates.
 
Note also that
:<math>
  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>
  \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> (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]] 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]]s we are interested in calculating
: <math> \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> \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]].  And from the definition of the Lamé coefficients,
: <math> {\partial \mathbf{x} \over \partial q^i} = \sum_{k} h_{ki}~ \mathbf{e}_{k} </math>
and thus
: <math>
\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>
\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]], the relevant calculation, with the parameterization of the surface in Cartesian coordinates is:
: <math>\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>
\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>
{\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>
  \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>\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>
        \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]] of a [[scalar (mathematics)|scalar]] or [[vector field]] as
:<math>
  \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>
  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]] of a vector field can then be written as
:<math>
  \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>
  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>
  \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>
  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>
  \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]] is given by
: <math>
    \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 coordinate]]s we have
:<math>
  (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>
  \{\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>
0 < r < \infty ~, ~~ 0 < \theta < 2\pi ~,~~ -\infty < z < \infty
</math>
 
Then the covariant and contravariant basis vectors are
:<math>
  \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>
    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>
  \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>
  \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>
  (\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>
  \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>
  \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>
  \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>
  \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>
  \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>
  \begin{align}
  \hat{S}_{11} & = S_{11} =: S_{rr} ~;~~\hat{S}_{12} = \cfrac{S_{12}}{r} =: S_{r\theta} ~;~~ \hat{S}_{13} & = S_{13} =: S_{rz} \\
    \hat{S}_{21} & = \cfrac{S_{11}}{r} =: S_{\theta r} ~;~~\hat{S}_{22} = \cfrac{S_{22}}{r^2} =: S_{\theta\theta} ~;~~ \hat{S}_{23} & = \cfrac{S_{23}}{r} =: S_{\theta z} \\
    \hat{S}_{31} & = S_{31} =: S_{zr} ~;~~\hat{S}_{32} = \cfrac{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>
\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>
\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_{\theta r}}{\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]]
* [[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]]
 
[[de:Krummlinige Koordinaten]]
[[fr:Système de coordonnées curvilignes]]
[[it:Coordinate curvilinee]]
[[ru:Криволинейная система координат]]
[[sl:Krivočrtni koordinatni sistem]]

Latest revision as of 17:15, 1 September 2014

The stylish wardrobe of Maggie Gyllenhaal�s role in BBC Two�s hard-hitting political thriller, The Honourable Woman, has caught the attention of the eagle-eyed viewers.

The eight-part series, set against the backdrop of the Israeli-Palestinian conflict, centres around Nessa Stein, played by Gyllenhaal. Stein is an Anglo-Israeli businesswoman recently ennobled in the House of Lords who devotes herself to philanthropic purposes across the Middle East, but hides a secret past from her time spent in Gaza eight years earlier.

Through the unravelling of her public and private life played out on an international, political stage, Stein parades in an increasingly impressive selection of outfits.
�Because the character of Nessa is so complicated and multi layered, we looked at all sorts of different people as reference. I suppose we started off by looking at other powerful and stylish women through history, Jackie Kennedy, Eva Peron, Margaret Thatcher, Cleopatra� Edward K Gibbon costume designer for the series told The Independent

Maggie Gyllenhaal The Honourable Woma
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