The
deformation gradient tensor is a quantity related to both the reference and current configuration, and expresses motion locally around a point. Two types of deformation gradient tensor may be defined. The
material deformation gradient tensor \mathbf F(\mathbf X,t) = F_{jK} \mathbf e_j\otimes\mathbf I_K is a
second-order tensor that represents the gradient of the
smooth and invertible mapping function \chi(\mathbf X,t)\,\!, which describes the
motion of a continuum. In particular, the continuity of the mapping function \chi(\mathbf X,t)\,\! implies that
cracks and voids do not open or close during the deformation. The material deformation gradient tensor characterizes the local deformation at a material point with position vector \mathbf X\,\!, i.e., deformation at neighbouring points, by transforming (
linear transformation) a material
line element emanating from that point from the reference configuration to the current or deformed configuration. Thus we have, \begin{align} d\mathbf{x} &= \frac {\partial \mathbf{x}} {\partial \mathbf {X}}\,d\mathbf{X} \qquad &\text{or}& \qquad dx_j =\frac{\partial x_j}{\partial X_K}\,dX_K \\ &= \nabla \chi(\mathbf X,t) \,d\mathbf{X} \qquad &\text{or}& \qquad dx_j =F_{jK}\,dX_K \,. \\ & = \mathbf F(\mathbf X,t) \,d\mathbf{X} \end{align} Assuming that \chi(\mathbf X,t)\,\! has
a smooth inverse, \mathbf F has the inverse \mathbf H = \mathbf F^{-1} = \frac {\partial \mathbf{X}} {\partial \mathbf {x}}\,\!, which is the
spatial deformation gradient tensor. \mathbf F being invertible is equivalent to \text{det}\mathbf F \neq 0, which corresponds to the notion that the material cannot be infinitely compressed.
Relative displacement vector Consider a
particle or material point P with position vector \mathbf X = X_I \mathbf I_I in the undeformed configuration (Figure 2). After a displacement of the body, the new position of the particle indicated by p in the new configuration is given by the vector position \mathbf{x} = x_i \mathbf e_i\,\!. The coordinate systems for the undeformed and deformed configuration can be superimposed for convenience. Consider now a material point Q neighboring P\,\!, with position vector \mathbf{X}+ \Delta \mathbf{X} = (X_I+\Delta X_I) \mathbf I_I\,\!. In the deformed configuration this particle has a new position q given by the position vector \mathbf{x}+ \Delta \mathbf{x}\,\!. Assuming that the line segments \Delta X and \Delta \mathbf x joining the particles P and Q in both the undeformed and deformed configuration, respectively, to be very small, then we can express them as d\mathbf X and d\mathbf x\,\!. Thus from Figure 2 we have \begin{align} \mathbf{x} &= \mathbf{X} +\mathbf{u}(\mathbf{X}), \\ \mathbf{x}+ d\mathbf{x} &= \mathbf{X}+d\mathbf{X}+\mathbf{u}(\mathbf{X}+d\mathbf{X}), \\ \text{and}\text{ therefore}& \\ d\mathbf{x} &= \mathbf{X}-\mathbf{x}+d\mathbf{X}+ \mathbf{u}(\mathbf{X}+d\mathbf{X}) \\ &= d\mathbf{X}+\mathbf{u}(\mathbf{X}+d\mathbf{X})- \mathbf{u}(\mathbf{X}) \\ &= d\mathbf{X}+d\mathbf{u} \\ \end{align}, where \mathbf {du} is the
relative displacement vector, which represents the relative displacement of Q with respect to P in the deformed configuration.
Taylor approximation For an infinitesimal element d\mathbf X\,\!, and assuming continuity on the displacement field, it is possible to use a
Taylor series expansion around point P\,\!, neglecting higher-order terms, to approximate the components of the relative displacement vector for the neighboring particle Q as \begin{align} \mathbf{u}(\mathbf{X}+d\mathbf{X}) &= \mathbf{u}(\mathbf{X})+d\mathbf{u} \quad & \text{or} & \quad u_i^* = u_i+du_i \\ &\approx \mathbf{u}(\mathbf{X})+\nabla_{\mathbf X}\mathbf u\cdot d\mathbf X \quad & \text{or} & \quad u_i^* \approx u_i + \frac{\partial u_i}{\partial X_J}dX_J \,. \end{align} Thus, the previous equation d\mathbf x = d\mathbf{X} + d\mathbf{u} can be written as \begin{align} d\mathbf x&=d\mathbf X+d\mathbf u \\ &=d\mathbf X+\nabla_{\mathbf X}\mathbf u\cdot d\mathbf X\\ &=\left(\mathbf I + \nabla_{\mathbf X}\mathbf u\right)d\mathbf X\\ &=\mathbf F d\mathbf X \end{align}
Time-derivative of the deformation gradient Calculations that involve the time-dependent deformation of a body often require a
time derivative of the deformation gradient to be calculated. A geometrically consistent definition of such a derivative requires an excursion into
differential geometry but we avoid those issues in this article. The time derivative of \mathbf{F} is \dot{\mathbf{F}} = \frac{\partial \mathbf{F}}{\partial t} = \frac{\partial}{\partial t} \left[\frac{\partial \mathbf{x}(\mathbf{X}, t)}{\partial \mathbf{X}}\right] = \frac{\partial}{\partial \mathbf{X}}\left[\frac{\partial \mathbf{x}(\mathbf{X}, t)}{\partial t}\right] = \frac{\partial}{\partial \mathbf{X}}\left[\mathbf{V}(\mathbf{X}, t)\right] where \mathbf{V} is the (material) velocity. The derivative on the right hand side represents a
material velocity gradient. It is common to convert that into a spatial gradient by applying the chain rule for derivatives, i.e., \dot{\mathbf{F}} = \frac{\partial}{\partial \mathbf{X}}\left[\mathbf{V}(\mathbf{X}, t)\right] = \frac{\partial}{\partial \mathbf{X}}\left[\mathbf{v}(\mathbf{x}(\mathbf{X}, t),t)\right] = \left.\frac{\partial}{\partial \mathbf{x}}\left[\mathbf{v}(\mathbf{x},t)\right]\right|_{\mathbf{x} = \mathbf{x}(\mathbf{X}, t)} \cdot \frac{\partial \mathbf{x}(\mathbf{X}, t)}{\partial \mathbf{X}} = \boldsymbol{l}\cdot\mathbf{F} where \boldsymbol{l} = (\nabla_{\mathbf{x}} \mathbf{v})^T is the
spatial velocity gradient and where \mathbf{v}(\mathbf{x},t) = \mathbf{V}(\mathbf{X},t) is the spatial (Eulerian) velocity at \mathbf{x} = \mathbf{x}(\mathbf{X}, t). If the spatial velocity gradient is constant in time, the above equation can be solved exactly to give \mathbf{F} = e^{\boldsymbol{l}\, t} assuming \mathbf{F} = \mathbf{1} at t = 0. There are several methods of computing the
exponential above. Related quantities often used in continuum mechanics are the
rate of deformation tensor and the
spin tensor defined, respectively, as: \boldsymbol{d} = \tfrac{1}{2} \left(\boldsymbol{l} + \boldsymbol{l}^T\right) \,,~~ \boldsymbol{w} = \tfrac{1}{2} \left(\boldsymbol{l} - \boldsymbol{l}^T\right) \,. The rate of deformation tensor gives the rate of stretching of line elements while the
spin tensor indicates the rate of rotation or
vorticity of the motion. The material time derivative of the inverse of the deformation gradient (keeping the reference configuration fixed) is often required in analyses that involve finite strains. This derivative is \frac{\partial}{\partial t} \left(\mathbf{F}^{-1}\right) = - \mathbf{F}^{-1} \cdot \dot{\mathbf{F}} \cdot \mathbf{F}^{-1} \,. The above relation can be verified by taking the material time derivative of \mathbf{F}^{-1} \cdot d\mathbf{x} = d\mathbf{X} and noting that \dot{\mathbf{X}} = 0.
Polar decomposition of the deformation gradient tensor The deformation gradient \mathbf{F}, like any invertible second-order tensor, can be decomposed, using the
polar decomposition theorem, into a product of two second-order tensors (Truesdell and Noll, 1965): an orthogonal tensor and a positive definite symmetric tensor, i.e., \mathbf{F} = \mathbf{R} \mathbf{U} = \mathbf{V} \mathbf{R} where the tensor \mathbf{R} is a
proper orthogonal tensor, i.e., \mathbf R^{-1} = \mathbf R^T and \det \mathbf R = +1\,\!, representing a rotation; the tensor \mathbf{U} is the
right stretch tensor; and \mathbf{V} the
left stretch tensor. The terms
right and
left means that they are to the right and left of the rotation tensor \mathbf{R}\,\!, respectively. \mathbf{U} and \mathbf{V} are both
positive definite, i.e. \mathbf x \cdot \mathbf U \cdot \mathbf x > 0 and \mathbf x\cdot\mathbf V \cdot \mathbf x > 0 for all non-zero \mathbf x \in \R^3, and
symmetric tensors, i.e. \mathbf U = \mathbf U^T and \mathbf V = \mathbf V^T\,\!, of second order. This decomposition implies that the deformation of a line element d\mathbf X in the undeformed configuration onto d\mathbf x in the deformed configuration, i.e., d\mathbf x = \mathbf F \,d\mathbf X\,\!, may be obtained either by first stretching the element by \mathbf U\,\!, i.e. d\mathbf x' = \mathbf U \,d\mathbf X\,\!, followed by a rotation \mathbf R\,\!, i.e., d\mathbf x = \mathbf R \,d\mathbf x'\,\!; or equivalently, by applying a rigid rotation \mathbf R first, i.e., d\mathbf x' = \mathbf R \, d\mathbf X\,\!, followed later by a stretching \mathbf V\,\!, i.e., d\mathbf x = \mathbf V \, d\mathbf x' (See Figure 3). Due to the orthogonality of \mathbf R \mathbf V = \mathbf R \cdot \mathbf U \cdot \mathbf R^T so that \mathbf U and \mathbf V have the same
eigenvalues or
principal stretches, but different
eigenvectors or
principal directions \mathbf{N}_i and \mathbf{n}_i\,\!, respectively. The principal directions are related by \mathbf{n}_i = \mathbf{R} \mathbf{N}_i. This polar decomposition, which is unique as \mathbf F is invertible with a positive determinant, is a corollary of the
singular-value decomposition.
Transformation of a surface and volume element To transform quantities that are defined with respect to areas in a deformed configuration to those relative to areas in a reference configuration, and vice versa, we use '''Nanson's relation''', expressed as da~\mathbf{n} = J~dA ~\mathbf{F}^{-T} \cdot \mathbf{N} where da is an area of a region in the deformed configuration, dA is the same area in the reference configuration, and \mathbf{n} is the outward normal to the area element in the current configuration while \mathbf{N} is the outward normal in the reference configuration, \mathbf{F} is the
deformation gradient, and J = \det\mathbf{F}\,\!. The corresponding formula for the transformation of the
volume element is dv = J~dV {{math proof d\mathbf{A} = dA~\mathbf{N} ~;~~ d\mathbf{a} = da~\mathbf{n} The reference and current volumes of an element are dV = d\mathbf{A}^{T}\cdot d\mathbf{L} ~;~~ dv = d\mathbf{a}^{T} \cdot d\mathbf{l} where d\mathbf{l} = \mathbf{F}\cdot d\mathbf{L}\,\!. Therefore, d\mathbf{a}^{T} \cdot d\mathbf{l}= dv = J~dV = J~d\mathbf{A}^{T}\cdot d\mathbf{L} or, d\mathbf{a}^{T} \cdot \mathbf{F}\cdot d\mathbf{L} = dv = J~dV = J~d\mathbf{A}^{T} \cdot d\mathbf{L} so, d\mathbf{a}^{T} \cdot \mathbf{F} = J~d\mathbf{A}^{T} So we get d\mathbf{a} = J~\mathbf{F}^{-T} \cdot d\mathbf{A} or, da~\mathbf{n} = J~dA~\mathbf{F}^{-T} \cdot \mathbf{N}
Q.E.D. }} ==Fundamental strain tensors==