Deformation gradient is defined as $$F_{iJ}=\frac{\partial x_i}{\partial X_J},\;\mathbf{F}=\frac{\partial\mathbf{x}}{\partial\mathbf{X}},$$ where $\mathbf{x}$ is spatial coordinates; $\mathbf{X}$ is material coordinates. I wonder whether $\nabla\mathbf{F}$ has physical meaning. Here is my derivation: $$\nabla\mathbf{F}=\frac{\partial\mathbf{F}}{\partial\mathbf{x}}=\frac{\partial}{\partial\mathbf{x}}\frac{\partial\mathbf{x}}{\partial\mathbf{X}},$$ where $\nabla$ is the del operator with respect to spatial coordinates. Then we interchange the differential order on the right-hand side, $$\nabla\mathbf{F}=\frac{\partial}{\partial\mathbf{X}}\frac{\partial\mathbf{x}}{\partial\mathbf{x}}=\frac{\partial\mathbf{I}}{\partial\mathbf{X}},$$ where $\mathbf{I}$ is a second-order unit tensor. Now we have $$\nabla\mathbf{F}=\nabla_0\mathbf{I},$$ where $\nabla_0$ is the del operator with respect to material coordinates. The answer seems to be zero if there is no mistake in my derivation, but what is its physical meaning?


1 Answer 1



The deformation is the application $$ \begin{array}{rl} \boldsymbol y:\Omega_0 &\longmapsto\Omega\\ \boldsymbol X &\longmapsto \boldsymbol y(\boldsymbol X) \end{array} $$ where $\Omega_0$ and $\Omega$ designate the initial (material) and current (spatial) configurations, respectively.

The deformation gradient has components $$ F_{i\alpha}=(\mathrm{grad}\,\boldsymbol y)_{i\alpha}=\frac{\partial y_i}{\partial X_\alpha} $$ and is also a function of the material coordinates $\boldsymbol F(\boldsymbol X)$.

The gradient of the deformation gradient naturally reads $$ (\mathrm{grad}\,\boldsymbol F)_{i\alpha\beta}=\frac{\partial^2 y_i}{\partial X_\alpha\partial X_\beta}. $$ The deformation mapping is not a function of the spatial coordinates, hence they have no reason to appear at any step.


  • Since the deformation is a function of the material coordinates, its derivatives are unequivocally defined with respect to the material coordinates.
  • It is preferable not to mix up the spacial variable $\boldsymbol x$ (an independent variable in $\Omega$) with the deformation $\boldsymbol y$ (a mapping from $\Omega_0$ to $\Omega$). In OP's suggestion, the identity tensor emerges precisely because of this confusion.

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