Compatibility (mechanics)

Compatibility conditions for finite strains

For solids in which the deformations are not required to be small, the compatibility conditions take the form $${\displaystyle \mathbf{\nabla }\times \mathit{F}=\mathbf{0}}$$ where ${\displaystyle \mathit{F}}$ is the deformation gradient. In terms of components with respect to a Cartesian coordinate system we can write these compatibility relations as $${\displaystyle e_{ABC}\frac{\mathrm{\partial }{F}_{iB}}{\mathrm{\partial }{X}_A}=0}$$ This condition is necessary if the deformation is to be continuous and derived from the mapping ${\displaystyle \mathbf{x}=\mathit{\chi }(\mathbf{X},t)}$ (see Finite strain theory). The same condition is also sufficient to ensure compatibility in a simply connected body.

Compatibility condition for the right Cauchy-Green deformation tensor

The compatibility condition for the right Cauchy-Green deformation tensor can be expressed as $${\displaystyle R_{\alpha \beta \rho }^{\gamma }:=\frac{\mathrm{\partial }}{\mathrm{\partial }{X}^{\rho }}[{\mathrm{\Gamma }}_{\alpha \beta }^{\gamma }]-\frac{\mathrm{\partial }}{\mathrm{\partial }{X}^{\beta }}[{\mathrm{\Gamma }}_{\alpha \rho }^{\gamma }]+{\mathrm{\Gamma }}_{\mu \rho }^{\gamma }{\mathrm{\Gamma }}_{\alpha \beta }^{\mu }-{\mathrm{\Gamma }}_{\mu \beta }^{\gamma }{\mathrm{\Gamma }}_{\alpha \rho }^{\mu }=0}$$ where ${\displaystyle {\mathrm{\Gamma }}_{ij}^k}$ is the Christoffel symbol of the second kind. The quantity ${\displaystyle R_{ijk}^m}$ represents the mixed components of the Riemann-Christoffel curvature tensor.

The general compatibility problem

The problem of compatibility in continuum mechanics involves the determination of allowable single-valued continuous fields on simply connected bodies. More precisely, the problem may be stated in the following manner.

Consider the deformation of a body shown in Figure 1. If we express all vectors in terms of the reference coordinate system ${\displaystyle \{({\mathbf{E}}_1,{\mathbf{E}}_2,{\mathbf{E}}_3),O\}}$, the displacement of a point in the body is given by $${\displaystyle \mathbf{u}=\mathbf{x}-\mathbf{X};u_i=x_i-X_i}$$ Also $${\displaystyle \mathbf{\nabla }\mathbf{u}=\frac{\mathrm{\partial }\mathbf{u}}{\mathrm{\partial }\mathbf{X}};\mathbf{\nabla }\mathbf{x}=\frac{\mathrm{\partial }\mathbf{x}}{\mathrm{\partial }\mathbf{X}}}$$

What conditions on a given second-order tensor field ${\displaystyle \mathit{A}(\mathbf{X})}$ on a body are necessary and sufficient so that there exists a unique vector field ${\displaystyle \mathbf{v}(\mathbf{X})}$ that satisfies $${\displaystyle \mathbf{\nabla }\mathbf{v}=\mathit{A}\equiv v_{i,j}=A_{ij}}$$

Necessary conditions

For the necessary conditions we assume that the field ${\displaystyle \mathbf{v}}$ exists and satisfies ${\displaystyle v_{i,j}=A_{ij}}$. Then $${\displaystyle v_{i,jk}=A_{ij,k};v_{i,kj}=A_{ik,j}}$$ Since changing the order of differentiation does not affect the result we have $${\displaystyle v_{i,jk}=v_{i,kj}}$$ Hence $${\displaystyle A_{ij,k}=A_{ik,j}}$$ From the well known identity for the curl of a tensor we get the necessary condition $${\displaystyle \mathbf{\nabla }\times \mathit{A}=\mathbf{0}}$$

Sufficient conditions

To prove that this condition is sufficient to guarantee existence of a compatible second-order tensor field, we start with the assumption that a field ${\displaystyle \mathit{A}}$ exists such that ${\displaystyle \mathbf{\nabla }\times \mathit{A}=\mathbf{0}}$. We will integrate this field to find the vector field ${\displaystyle \mathbf{v}}$ along a line between points ${\displaystyle A}$ and ${\displaystyle B}$ (see Figure 2), i.e., $${\displaystyle \begin{array}{cc}\mathbf{v}({\mathbf{X}}_B)-\mathbf{v}({\mathbf{X}}_A)& ={\int }_{{\mathbf{X}}_A}^{{\mathbf{X}}_B}\mathbf{\nabla }\mathbf{v}\cdot d\mathbf{X}\\ & ={\int }_{{\mathbf{X}}_A}^{{\mathbf{X}}_B}\mathit{A}(\mathbf{X})\cdot d\mathbf{X}\end{array}}$$ If the vector field ${\displaystyle \mathbf{v}}$ is to be single-valued then the value of the integral should be independent of the path taken to go from ${\displaystyle A}$ to ${\displaystyle B}$.

From Stokes' theorem, the integral of a second order tensor along a closed path is given by $${\displaystyle {\oint }_{\mathrm{\partial }\mathrm{\Omega }}\mathit{A}\cdot d\mathbf{s}={\int }_{\mathrm{\Omega }}\mathbf{n}\cdot (\mathbf{\nabla }\times \mathit{A})da}$$ Using the assumption that the curl of ${\displaystyle \mathit{A}}$ is zero, we get $${\displaystyle \begin{array}{cc}& {\oint }_{\mathrm{\partial }\mathrm{\Omega }}\mathit{A}\cdot d\mathbf{s}=0\\ ⟹& {\int }_{AB}\mathit{A}\cdot d\mathbf{X}+{\int }_{BA}\mathit{A}\cdot d\mathbf{X}=0\end{array}}$$ Hence the integral is path independent and the compatibility condition is sufficient to ensure a unique ${\displaystyle \mathbf{v}}$ field, provided that the body is simply connected.

Compatibility of the deformation gradient

The compatibility condition for the deformation gradient is obtained directly from the above proof by observing that $${\displaystyle \mathit{F}=\frac{\mathrm{\partial }\mathbf{x}}{\mathrm{\partial }\mathbf{X}}=\mathbf{\nabla }\mathbf{x}}$$ Then the necessary and sufficient conditions for the existence of a compatible ${\displaystyle \mathit{F}}$ field over a simply connected body are $${\displaystyle \mathbf{\nabla }\times \mathit{F}=\mathbf{0}}$$

Compatibility of infinitesimal strains

The compatibility problem for small strains can be stated as follows.

Given a symmetric second order tensor field ${\displaystyle \mathit{ϵ}}$ when is it possible to construct a vector field ${\displaystyle \mathbf{u}}$ such that $${\displaystyle \mathit{ϵ}=\frac{1}{2}[\mathbf{\nabla }\mathbf{u}+(\mathbf{\nabla }\mathbf{u}{)}^T]}$$

Necessary conditions

Suppose that there exists ${\displaystyle \mathbf{u}}$ such that the expression for ${\displaystyle \mathit{ϵ}}$ holds. Now $${\displaystyle \mathbf{\nabla }\mathbf{u}=\mathit{ϵ}+\mathit{\omega }}$$ where $${\displaystyle \mathit{\omega }:=\frac{1}{2}[\mathbf{\nabla }\mathbf{u}-(\mathbf{\nabla }\mathbf{u}{)}^T]}$$ Therefore, in index notation, $${\displaystyle \begin{array}{cc}\mathbf{\nabla }\mathit{\omega }\equiv {\omega }_{ij,k}& =\frac{1}{2}(u_{i,jk}-u_{j,ik})\\ & =\frac{1}{2}(u_{i,jk}+u_{k,ji}-u_{j,ik}-u_{k,ji})\\ & ={\epsilon }_{ik,j}-{\epsilon }_{jk,i}\end{array}}$$ If ${\displaystyle \mathit{\omega }}$ is continuously differentiable we have ${\displaystyle {\omega }_{ij,kl}={\omega }_{ij,lk}}$. Hence, $${\displaystyle {\epsilon }_{ik,jl}-{\epsilon }_{jk,il}-{\epsilon }_{il,jk}+{\epsilon }_{jl,ik}=0}$$ In direct tensor notation $${\displaystyle \mathbf{\nabla }\times (\mathbf{\nabla }\times \mathit{ϵ}{)}^T=\mathbf{0}}$$ The above are necessary conditions. If ${\displaystyle \mathbf{w}}$ is the infinitesimal rotation vector then ${\displaystyle \mathbf{\nabla }\times \mathit{ϵ}=\mathbf{\nabla }\mathbf{w}+\mathbf{\nabla }{\mathbf{w}}^T}$. Hence the necessary condition may also be written as ${\displaystyle \mathbf{\nabla }\times (\mathbf{\nabla }\mathbf{w}+\mathbf{\nabla }{\mathbf{w}}^T{)}^T=\mathbf{0}}$.

Sufficient conditions

Let us now assume that the condition ${\displaystyle \mathbf{\nabla }\times (\mathbf{\nabla }\times \mathit{ϵ}{)}^T=\mathbf{0}}$ is satisfied in a portion of a body. Is this condition sufficient to guarantee the existence of a continuous, single-valued displacement field ${\displaystyle \mathbf{u}}$?

The first step in the process is to show that this condition implies that the infinitesimal rotation tensor ${\displaystyle \mathit{\omega }}$ is uniquely defined. To do that we integrate ${\displaystyle \mathbf{\nabla }\mathbf{w}}$ along the path ${\displaystyle {\mathbf{X}}_A}$ to ${\displaystyle {\mathbf{X}}_B}$, i.e., $${\displaystyle \mathbf{w}({\mathbf{X}}_B)-\mathbf{w}({\mathbf{X}}_A)={\int }_{{\mathbf{X}}_A}^{{\mathbf{X}}_B}\mathbf{\nabla }\mathbf{w}\cdot d\mathbf{X}={\int }_{{\mathbf{X}}_A}^{{\mathbf{X}}_B}(\mathbf{\nabla }\times \mathit{ϵ})\cdot d\mathbf{X}}$$ Note that we need to know a reference ${\displaystyle \mathbf{w}({\mathbf{X}}_A)}$ to fix the rigid body rotation. The field ${\displaystyle \mathbf{w}(\mathbf{X})}$ is uniquely determined only if the contour integral along a closed contour between ${\displaystyle {\mathbf{X}}_A}$ and ${\displaystyle {\mathbf{X}}_b}$ is zero, i.e., $${\displaystyle {\oint }_{{\mathbf{X}}_A}^{{\mathbf{X}}_B}(\mathbf{\nabla }\times \mathit{ϵ})\cdot d\mathbf{X}=\mathbf{0}}$$ But from Stokes' theorem for a simply-connected body and the necessary condition for compatibility $${\displaystyle {\oint }_{{\mathbf{X}}_A}^{{\mathbf{X}}_B}(\mathbf{\nabla }\times \mathit{ϵ})\cdot d\mathbf{X}={\int }_{{\mathrm{\Omega }}_{AB}}\mathbf{n}\cdot (\mathbf{\nabla }\times \mathbf{\nabla }\times \mathit{ϵ})da=\mathbf{0}}$$ Therefore, the field ${\displaystyle \mathbf{w}}$ is uniquely defined which implies that the infinitesimal rotation tensor ${\displaystyle \mathit{\omega }}$ is also uniquely defined, provided the body is simply connected.

In the next step of the process we will consider the uniqueness of the displacement field ${\displaystyle \mathbf{u}}$. As before we integrate the displacement gradient $${\displaystyle \mathbf{u}({\mathbf{X}}_B)-\mathbf{u}({\mathbf{X}}_A)={\int }_{{\mathbf{X}}_A}^{{\mathbf{X}}_B}\mathbf{\nabla }\mathbf{u}\cdot d\mathbf{X}={\int }_{{\mathbf{X}}_A}^{{\mathbf{X}}_B}(\mathit{ϵ}+\mathit{\omega })\cdot d\mathbf{X}}$$ From Stokes' theorem and using the relations ${\displaystyle \mathbf{\nabla }\times \mathit{ϵ}=\mathbf{\nabla }\mathbf{w}=-\mathbf{\nabla }\times \omega }$ we have $${\displaystyle {\oint }_{{\mathbf{X}}_A}^{{\mathbf{X}}_B}(\mathit{ϵ}+\mathit{\omega })\cdot d\mathbf{X}={\int }_{{\mathrm{\Omega }}_{AB}}\mathbf{n}\cdot (\mathbf{\nabla }\times \mathit{ϵ}+\mathbf{\nabla }\times \mathit{\omega })da=\mathbf{0}}$$ Hence the displacement field ${\displaystyle \mathbf{u}}$ is also determined uniquely. Hence the compatibility conditions are sufficient to guarantee the existence of a unique displacement field ${\displaystyle \mathbf{u}}$ in a simply-connected body.

Compatibility for Right Cauchy-Green Deformation field

The compatibility problem for the Right Cauchy-Green deformation field can be posed as follows.

Problem: Let ${\displaystyle \mathit{C}(\mathbf{X})}$ be a positive definite symmetric tensor field defined on the reference configuration. Under what conditions on ${\displaystyle \mathit{C}}$ does there exist a deformed configuration marked by the position field ${\displaystyle \mathbf{x}(\mathbf{X})}$ such that $${\displaystyle (1){\left(\frac{\mathrm{\partial }\mathbf{x}}{\mathrm{\partial }\mathbf{X}}\right)}^T\left(\frac{\mathrm{\partial }\mathbf{x}}{\mathrm{\partial }\mathbf{X}}\right)=\mathit{C}}$$

Necessary conditions

Suppose that a field ${\displaystyle \mathbf{x}(\mathbf{X})}$ exists that satisfies condition (1). In terms of components with respect to a rectangular Cartesian basis $${\displaystyle \frac{\mathrm{\partial }{x}^i}{\mathrm{\partial }{X}^{\alpha }}\frac{\mathrm{\partial }{x}^i}{\mathrm{\partial }{X}^{\beta }}=C_{\alpha \beta }}$$ From finite strain theory we know that ${\displaystyle C_{\alpha \beta }=g_{\alpha \beta }}$. Hence we can write $${\displaystyle {\delta }_{ij}\frac{\mathrm{\partial }{x}^i}{\mathrm{\partial }{X}^{\alpha }}\frac{\mathrm{\partial }{x}^j}{\mathrm{\partial }{X}^{\beta }}=g_{\alpha \beta }}$$ For two symmetric second-order tensor field that are mapped one-to-one we also have the relation $${\displaystyle G_{ij}=\frac{\mathrm{\partial }{X}^{\alpha }}{\mathrm{\partial }{x}^i}\frac{\mathrm{\partial }{X}^{\beta }}{\mathrm{\partial }{x}^j}{g}_{\alpha \beta }}$$ From the relation between of ${\displaystyle G_{ij}}$ and ${\displaystyle g_{\alpha \beta }}$ that ${\displaystyle {\delta }_{ij}=G_{ij}}$, we have $${\displaystyle {}_{(x)}{\mathrm{\Gamma }}_{ij}^k=0}$$ Then, from the relation $${\displaystyle \frac{{\mathrm{\partial }}^2{x}^m}{\mathrm{\partial }{X}^{\alpha }\mathrm{\partial }{X}^{\beta }}=\frac{\mathrm{\partial }{x}^m}{\mathrm{\partial }{X}^{\mu }}{}_{(X)}{\mathrm{\Gamma }}_{\alpha \beta }^{\mu }-\frac{\mathrm{\partial }{x}^i}{\mathrm{\partial }{X}^{\alpha }}\frac{\mathrm{\partial }{x}^j}{\mathrm{\partial }{X}^{\beta }}{}_{(x)}{\mathrm{\Gamma }}_{ij}^m}$$ we have $${\displaystyle \frac{\mathrm{\partial }{F}_{\alpha }^m}{\mathrm{\partial }{X}^{\beta }}=F_{\mu }^m{}_{(X)}{\mathrm{\Gamma }}_{\alpha \beta }^{\mu };F_{\alpha }^i:=\frac{\mathrm{\partial }{x}^i}{\mathrm{\partial }{X}^{\alpha }}}$$ From finite strain theory we also have $${\displaystyle \begin{array}{cc}{}_{(X)}{\mathrm{\Gamma }}_{\alpha \beta \gamma }& =\frac{1}{2}\left(\frac{\mathrm{\partial }{g}_{\alpha \gamma }}{\mathrm{\partial }{X}^{\beta }}+\frac{\mathrm{\partial }{g}_{\beta \gamma }}{\mathrm{\partial }{X}^{\alpha }}-\frac{\mathrm{\partial }{g}_{\alpha \beta }}{\mathrm{\partial }{X}^{\gamma }}\right);\\ {}_{(X)}{\mathrm{\Gamma }}_{\alpha \beta }^{\nu }& =g^{\nu \gamma }{}_{(X)}{\mathrm{\Gamma }}_{\alpha \beta \gamma };\\ g_{\alpha \beta }& =C_{\alpha \beta };g^{\alpha \beta }=C^{\alpha \beta }\end{array}}$$ Therefore, $${\displaystyle {}_{(X)}{\mathrm{\Gamma }}_{\alpha \beta }^{\mu }=\frac{C^{\mu \gamma }}{2}\left(\frac{\mathrm{\partial }{C}_{\alpha \gamma }}{\mathrm{\partial }{X}^{\beta }}+\frac{\mathrm{\partial }{C}_{\beta \gamma }}{\mathrm{\partial }{X}^{\alpha }}-\frac{\mathrm{\partial }{C}_{\alpha \beta }}{\mathrm{\partial }{X}^{\gamma }}\right)}$$ and we have $${\displaystyle \frac{\mathrm{\partial }{F}_{\alpha }^m}{\mathrm{\partial }{X}^{\beta }}=F_{\mu }^m\frac{C^{\mu \gamma }}{2}\left(\frac{\mathrm{\partial }{C}_{\alpha \gamma }}{\mathrm{\partial }{X}^{\beta }}+\frac{\mathrm{\partial }{C}_{\beta \gamma }}{\mathrm{\partial }{X}^{\alpha }}-\frac{\mathrm{\partial }{C}_{\alpha \beta }}{\mathrm{\partial }{X}^{\gamma }}\right)}$$ Again, using the commutative nature of the order of differentiation, we have $${\displaystyle \begin{array}{cc}& \frac{{\mathrm{\partial }}^2{F}_{\alpha }^m}{\mathrm{\partial }{X}^{\beta }\mathrm{\partial }{X}^{\rho }}=\frac{{\mathrm{\partial }}^2{F}_{\alpha }^m}{\mathrm{\partial }{X}^{\rho }\mathrm{\partial }{X}^{\beta }}\\ ⟹& \frac{\mathrm{\partial }{F}_{\mu }^m}{\mathrm{\partial }{X}^{\rho }}{}_{(X)}{\mathrm{\Gamma }}_{\alpha \beta }^{\mu }+F_{\mu }^m\frac{\mathrm{\partial }}{\mathrm{\partial }{X}^{\rho }}\left[{}_{(X)}{\mathrm{\Gamma }}_{\alpha \beta }^{\mu }\right]=\frac{\mathrm{\partial }{F}_{\mu }^m}{\mathrm{\partial }{X}^{\beta }}{}_{(X)}{\mathrm{\Gamma }}_{\alpha \rho }^{\mu }+F_{\mu }^m\frac{\mathrm{\partial }}{\mathrm{\partial }{X}^{\beta }}\left[{}_{(X)}{\mathrm{\Gamma }}_{\alpha \rho }^{\mu }\right]\end{array}}$$ or $${\displaystyle F_{\gamma }^m{}_{(X)}{\mathrm{\Gamma }}_{\mu \rho }^{\gamma }{}_{(X)}{\mathrm{\Gamma }}_{\alpha \beta }^{\mu }+F_{\mu }^m\frac{\mathrm{\partial }}{\mathrm{\partial }{X}^{\rho }}\left[{}_{(X)}{\mathrm{\Gamma }}_{\alpha \beta }^{\mu }\right]=F_{\gamma }^m{}_{(X)}{\mathrm{\Gamma }}_{\mu \beta }^{\gamma }{}_{(X)}{\mathrm{\Gamma }}_{\alpha \rho }^{\mu }+F_{\mu }^m\frac{\mathrm{\partial }}{\mathrm{\partial }{X}^{\beta }}\left[{}_{(X)}{\mathrm{\Gamma }}_{\alpha \rho }^{\mu }\right]}$$ After collecting terms we get $${\displaystyle F_{\gamma }^m\left({}_{(X)}{\mathrm{\Gamma }}_{\mu \rho }^{\gamma }{}_{(X)}{\mathrm{\Gamma }}_{\alpha \beta }^{\mu }+\frac{\mathrm{\partial }}{\mathrm{\partial }{X}^{\rho }}[{}_{(X)}{\mathrm{\Gamma }}_{\alpha \beta }^{\gamma }]-{}_{(X)}{\mathrm{\Gamma }}_{\mu \beta }^{\gamma }{}_{(X)}{\mathrm{\Gamma }}_{\alpha \rho }^{\mu }-\frac{\mathrm{\partial }}{\mathrm{\partial }{X}^{\beta }}[{}_{(X)}{\mathrm{\Gamma }}_{\alpha \rho }^{\gamma }]\right)=0}$$ From the definition of ${\displaystyle F_{\gamma }^m}$ we observe that it is invertible and hence cannot be zero. Therefore, $${\displaystyle R_{\alpha \beta \rho }^{\gamma }:=\frac{\mathrm{\partial }}{\mathrm{\partial }{X}^{\rho }}[{}_{(X)}{\mathrm{\Gamma }}_{\alpha \beta }^{\gamma }]-\frac{\mathrm{\partial }}{\mathrm{\partial }{X}^{\beta }}[{}_{(X)}{\mathrm{\Gamma }}_{\alpha \rho }^{\gamma }]+{}_{(X)}{\mathrm{\Gamma }}_{\mu \rho }^{\gamma }{}_{(X)}{\mathrm{\Gamma }}_{\alpha \beta }^{\mu }-{}_{(X)}{\mathrm{\Gamma }}_{\mu \beta }^{\gamma }{}_{(X)}{\mathrm{\Gamma }}_{\alpha \rho }^{\mu }=0}$$ We can show these are the mixed components of the Riemann-Christoffel curvature tensor. Therefore, the necessary conditions for ${\displaystyle \mathit{C}}$-compatibility are that the Riemann-Christoffel curvature of the deformation is zero.

Sufficient conditions

The proof of sufficiency is a bit more involved. We start with the assumption that $${\displaystyle R_{\alpha \beta \rho }^{\gamma }=0;g_{\alpha \beta }=C_{\alpha \beta }}$$ We have to show that there exist ${\displaystyle \mathbf{x}}$ and ${\displaystyle \mathbf{X}}$ such that $${\displaystyle \frac{\mathrm{\partial }{x}^i}{\mathrm{\partial }{X}^{\alpha }}\frac{\mathrm{\partial }{x}^i}{\mathrm{\partial }{X}^{\beta }}=C_{\alpha \beta }}$$ From a theorem by T.Y.Thomas we know that the system of equations $${\displaystyle \frac{\mathrm{\partial }{F}_{\alpha }^i}{\mathrm{\partial }{X}^{\beta }}=F_{\gamma }^i{}_{(X)}{\mathrm{\Gamma }}_{\alpha \beta }^{\gamma }}$$ has unique solutions ${\displaystyle F_{\alpha }^i}$ over simply connected domains if $${\displaystyle {}_{(X)}{\mathrm{\Gamma }}_{\alpha \beta }^{\gamma }{=}_{(X)}{\mathrm{\Gamma }}_{\beta \alpha }^{\gamma };R_{\alpha \beta \rho }^{\gamma }=0}$$ The first of these is true from the defining of ${\displaystyle {\mathrm{\Gamma }}_{jk}^i}$ and the second is assumed. Hence the assumed condition gives us a unique ${\displaystyle F_{\alpha }^i}$ that is ${\displaystyle C^2}$ continuous.

Next consider the system of equations $${\displaystyle \frac{\mathrm{\partial }{x}^i}{\mathrm{\partial }{X}^{\alpha }}=F_{\alpha }^i}$$ Since ${\displaystyle F_{\alpha }^i}$ is ${\displaystyle C^2}$ and the body is simply connected there exists some solution ${\displaystyle x^i(X^{\alpha })}$ to the above equations. We can show that the ${\displaystyle x^i}$ also satisfy the property that $${\displaystyle det\left|\frac{\mathrm{\partial }{x}^i}{\mathrm{\partial }{X}^{\alpha }}\right|\ne 0}$$ We can also show that the relation $${\displaystyle \frac{\mathrm{\partial }{x}^i}{\mathrm{\partial }{X}^{\alpha }}{g}^{\alpha \beta }\frac{\mathrm{\partial }{x}^j}{\mathrm{\partial }{X}^{\beta }}={\delta }^{ij}}$$ implies that $${\displaystyle g_{\alpha \beta }=C_{\alpha \beta }=\frac{\mathrm{\partial }{x}^k}{\mathrm{\partial }{X}^{\alpha }}\frac{\mathrm{\partial }{x}^k}{\mathrm{\partial }{X}^{\beta }}}$$ If we associate these quantities with tensor fields we can show that ${\displaystyle \frac{\mathrm{\partial }\mathbf{x}}{\mathrm{\partial }\mathbf{X}}}$ is invertible and the constructed tensor field satisfies the expression for ${\displaystyle \mathit{C}}$.