If we have a continuum where the initial positions are denoted $X$ and the positions after some deformation are denoted $x$, the deformation gradient is defined:

$$ F = \frac{\partial x}{\partial X} $$

If we instead start at $x$ and undergo a deformation to $X$, we might write:

$$ F_* = \frac{\partial X}{\partial x} $$

Is $F_*$ just equal to $F^{-1}$? It seems like it should be true intuitively, but I don't know how to formally demonstrate it.

  • $\begingroup$ Would Mathematics be a better home for this question? $\endgroup$
    – Qmechanic
    Feb 25, 2017 at 10:08
  • $\begingroup$ @Qmechanic yeah, I waffled over this, but ultimately chose Physics since it seemed to have more continuum-mechanics related questions. $\endgroup$
    – nnn
    Feb 25, 2017 at 13:18

1 Answer 1


$$(FF_*)_{ij} = \sum_k \frac{\partial x_i}{\partial X_k} \frac{\partial X_k}{\partial x_j} = \frac{\partial x_i}{\partial x_j} = \delta_{ij}\:.$$ In other words $$FF_* =I$$ so that, since the matrices are square ones, $$F_*F =I\:.$$ That is equivalent to saying $F_* = F^{-1}.$


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