# Is the inverse of the deformation gradient simply the deformation gradient of the inverse transformation?

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.

• Would Mathematics be a better home for this question? – Qmechanic Feb 25 '17 at 10:08
• @Qmechanic yeah, I waffled over this, but ultimately chose Physics since it seemed to have more continuum-mechanics related questions. – nnn Feb 25 '17 at 13:18

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