# How to prove $\operatorname{div} \mathbf{A}=\operatorname{Div} \mathbf{A} \mathbf{F}^{-\mathrm{T}}$?

I recently focus on solid mechanics and I am reading Nonlinear Solid Mechanics A Continuum Approach for Engineering by Gerhard A. Holzapfel. However, I was confused by a mathematical formula eq(2.49), it is as follows, $$\begin{equation} \operatorname{div} \mathbf{A}=\operatorname{Div} \mathbf{A} \mathbf{F}^{-\mathrm{T}} \tag{2.49} \end{equation}$$ where $$\mathbf{A}$$ is a smooth second-order tensor field, $$\mathbf{F}^{-1}$$ is the inverse of the deformation gradient. I tried to prove this formula but failed, my current progress is as follows,

$$\begin{equation} \operatorname{div} \mathbf{A}= \frac{\partial A_{i j}}{\partial x_{j}} \mathbf{e}_{i} \\ \operatorname{Div} \mathbf{A} \mathbf{F}^{-\mathrm{T}} = \frac{\partial A_{i j}}{\partial X_{j}} \mathbf{e}_{i} \frac{\partial X_{k}}{\partial x_{l}} \mathbf{e}_{l} \otimes \mathbf{e}_{k}=\frac{\partial A_{i j}}{\partial X_{j}}\frac{\partial X_{k}}{\partial x_{l}}\delta_{il}\mathbf{e}_{k} = \frac{\partial A_{i j}}{\partial X_{j}}\frac{\partial X_{k}}{\partial x_{i}}\mathbf{e}_{k} \end{equation}$$

I stopped here and can't reach that equation(2.49) in the book. Can someone help me?

• What is the difference between a div and a Div? – G. Smith Jun 22 at 4:02
• And what is $F^{-T}$ – Eli Jun 22 at 6:20
• $\operatorname{Div}$ is the divergence with respect to material coordinates while $\operatorname{div}$ is the divergence with respect to spatial coordinates. $F^{-T}$ is the transpose of $F^{-1}$. – John Lionel Jun 22 at 18:12