Deformation gradient ($F_{iI}=\partial x_i/\partial X_I$)

The Jacobian matrix of a deformation map carries an infinitesimal reference line element to its current image. Its determinant $J=\det F$ is the local volume ratio. An orientation-preserving deformation has $J>0$, and incompressibility imposes $J=1$.

Two constant deformation gradients are rank-one connected when their difference is an outer product $a\otimes N$. For a nonzero jump its matrix rank is one; equality $A=B$ is the trivial compatible case. The vector $N$ specifies a reference interface normal, and $a$ is the jump amplitude. The Hadamard jump condition identifies this algebraic condition with continuity of an affine deformation across a planar interface. The normal can be normalized by rescaling $a$ without changing the product.

The unrotated variants differ by $2\delta e_3\otimes e_1$, giving normal $e_1$. The rotation about $e_2$ with $\cos \theta =(1-{\delta }^2)/(1+{\delta }^2)$ and $\sin \theta =2\delta /(1+{\delta }^2)$ gives $U_{+}-\overline{R}{U}_{-}=[2\delta /(1+{\delta }^2)](e_1+\delta e_3)\otimes e_3$, giving the second normal $e_3$. If these are the only two compatible rotations, the only gradients in $SO(3)U_{+}$ compatible with $\overline{R}{U}_{-}$ are $U_{+}$ and $\overline{R}{U}_{+}$. Left multiplication by a rotation preserves rank, so the second classification follows directly from the first.

Suppose an affine deformation on each side of a plane with unit normal $N$ is continuous across it. Tangential derivatives agree, so $(A-B)t=0$ for every $t\perp N$. Decomposition of an arbitrary vector into its tangent component and $(x\cdot N)N$ then gives $(A-B)x=((A-B)N)(N\cdot x)$, proving the rank-one form. Conversely that form makes the two affine maps agree on the plane when their constants agree there.

Orientation preservation on both sides also requires $1-N\cdot A^{-1}a>0$: the matrix determinant lemma gives $\det B=\det A(1-N\cdot A^{-1}a)$. This inequality is additional geometric information supplied by the determinant signs, not by continuity alone.

For a twice continuously differentiable deformation map, each row of its cofactor matrix has zero divergence in reference coordinates. In three dimensions write $(\mathrm{cof}Dy{)}_{iA}=\frac{1}{2}{\epsilon }_{ijk}{\epsilon }_{ABC}{y}_{j,B}{y}_{k,C}$. Applying ${\partial }_A$ gives terms in which a symmetric pair of second partial derivatives contracts with an antisymmetric Levi-Civita symbol, so each term vanishes. This identity is the cancellation that makes the Piola-Kirchhoff stress tensor transform force balance correctly between configurations.

An elastic deformation map whose deformation gradient is independent of position at each time. Its spatial material acceleration is $\mathbf{a}=\ddot{F}{F}^{-1}(\mathbf{x}-\mathbf{c})+\ddot{\mathbf{c}}$. Incompressibility requires $\det F=1$.

The factor describing local stress-free growth in the morphoelasticity convention $\mathbf{F}=\mathbf{AG}$. Its determinant is the local grown-volume ratio. For $\mathbf{G}=\mathrm{diag}(1,1,\gamma )$, growth changes only the preferred third-direction length, by factor $\gamma$.

The ratio of current to reference lengths of an infinitesimal material line element. Along an elastic rod, it is $\alpha =ds/dS=|d\mathbf{r}/dS|>0$. A principal stretch is a stretch ratio along a principal direction of the stretch tensor.

For an orientation-preserving deformation gradient $F$, oriented area transforms by its cofactor matrix: $(Fa)\times (Fb)=(\det F)F^{-T}(a\times b)$. Taking the dot product with $Fc$ proves this from the determinant identity for the scalar triple product. Applying it to tangent vectors of a surface proves Nanson's formula. Equating current and reference tractions then gives the material-first nominal stress tensor $P=J{F}^{-1}\sigma$ for symmetric Cauchy stress tensor $\sigma$.

The factor $A$ can encode an internal relaxed configuration, while $F_{\ast }=F{A}^{-1}$ determines the instantaneous elastic energy. Differentiating $F_{\ast }$ at fixed $F$ or fixed $A$ produces the corresponding nominal stress tensor and thermodynamic force conjugate to an internal variable. This is a kinematic decomposition, not an assertion that each factor separately derives from a global deformation map.

For an orientation-preserving invertible deformation gradient, the polar decomposition of an invertible real matrix separates a rotation $R$ from the right stretch tensor $U$ or the left stretch tensor $V$. The rotation is shared by the two factorizations, with $V=RU{R}^T$.

The right Cauchy-Green tensor measures stretch on the reference configuration. Its eigenvalues are the same squared principal stretches as those of the left Cauchy-Green deformation tensor, but its eigenvectors are reference directions.

The left Cauchy-Green tensor measures stretch on the current configuration. It is a positive-definite symmetric matrix for an invertible deformation gradient, with eigenvalues equal to the squares of the principal stretches.

The first principal invariant of the left Cauchy-Green deformation tensor is the sum of the squares of the principal stretches. Since $B=F{F}^T$, it also equals ${\sum }_{i,I}{F}_{iI}^2$. Its derivative with respect to the deformation gradient is $2F$. For a strain energy density $W=w(I_1)$ with incompressibility, the Cauchy stress tensor is $T=2w^{\prime }(I_1)B-pI$.