I'm studying variational principles applied to continuum mechanics. In particular I am stuck on the derivation of the variational derivative of the jacobian.
So, given a smooth path that maps the reference configuration $\mathbf{X}$ of a body $\mathcal{B}$ with a boundary $\partial \mathcal{B}$, to the position vector $\mathbf{x}$ of the same body in a configuration at time $t$ ($\mathcal{B}_t, \partial\mathcal{B}_t$):
$$ \mathbf{x}(t) = \boldsymbol{\chi}(\mathbf{X}, t)$$
the deformation tensor is defined as: $$ \mathbf{F} = \frac{\partial \boldsymbol{\chi}}{\partial \mathbf{X}}$$
The jacobian of the trasformation is $$ J = \det(\mathbf{F}) $$
In the calculus of variations, an admissible family of paths1, parametrized in $\epsilon$ is defined as (for $\boldsymbol{\eta}(\mathbf{X}, t_1) = \boldsymbol{\eta}(\mathbf{X}, t_2) = \mathbf{0}$): $$ \boldsymbol{\chi}^*(\mathbf{X}, t) = \boldsymbol{\chi}(\mathbf{X}, t) + \epsilon\boldsymbol{\eta}(\mathbf{X}, t) $$
such that we hence have: $$ \mathbf{F}^* = \frac{\partial \boldsymbol{\chi}^*}{\partial \mathbf{X}} = \frac{\partial \boldsymbol{\chi}}{\partial \mathbf{X}} + \epsilon \frac{\partial \boldsymbol{\eta}}{\partial \mathbf{X}} = \mathbf{F} + \epsilon \frac{\partial \boldsymbol{\eta}}{\partial \mathbf{X}} $$
What I would like to check is the variational derivative of the jacobian: $$\delta J = \lim_{\epsilon \rightarrow 0}\frac{\text{d}}{\text{d}\epsilon}\left(\det(\mathbf{F}^*)\right) = J \; \boldsymbol{\nabla} \cdot \boldsymbol{\eta}$$
What I did so far is applying the Jacobi relation for the derivative of the determinant: $$\frac{\text{d}}{\text{d}\epsilon}\left(\det(\mathbf{F}^*)\right) = \text{tr}\left(\text{adj}\left(\mathbf{F}^*\right)\frac{\text{d} \mathbf{F}^*}{\text{d}\epsilon}\right) = \text{tr}\left(\text{adj}\left(\mathbf{F}^*\right)\frac{\partial \boldsymbol{\eta}}{\partial \mathbf{X}}\right)$$
but now I don't realize how to proceed. I see that $\text{tr}\left(\frac{\partial \boldsymbol{\eta}}{\partial \mathbf{X}}\right) = \boldsymbol{\nabla} \cdot \boldsymbol{\eta}$ but I don't see how I can now extract the additional term needed $\det(\mathbf{F}) = J$. And moreover how to get them indipendently out of the trace. Am I missing some properties of the $\text{adj}$ operator?
1: The "comparison motion" given by $\boldsymbol{\chi}^*$ must also satisfy boundary conditions on $\partial \mathcal{B}$
PS: I have actually an engineering background, so feel free to suggest improvements on the notation or in the descriptions if they are not clear or not correct.
I think I managed to get an answer to my own question. There were some blurry concepts I did not clarify, first of all to me, well.
When we write the map $$ \mathbf{x} = \boldsymbol{\chi}(\mathbf{X}, t) $$ we also say that there exist the inverse mapping: $$ \mathbf{X} = \boldsymbol{\chi}^{-1}(\mathbf{x}(t))$$
So when we write a vector field as $\boldsymbol{\eta}$, we can write it as dependent on the reference configuration particle position $\mathbf{X}$, i.e. $\hat{\boldsymbol{\eta}}(\mathbf{X}, t)$, but also as $\boldsymbol{\eta}(\boldsymbol{\chi}^{-1}(\mathbf{x}(t)) = \boldsymbol{\eta}(\mathbf{x}, t)$. So with the $\hat{\left(\cdot\right)}$ notation we indicate, for clarity, that the underlying function is written in reference variables1.
Moreover $\det(\mathbf{F}^*)$ as an indirect dependency on $\epsilon$, i.e.
$$ \det(\mathbf{F}^*) = \det(\mathbf{F}^*(\epsilon)) $$
In addition to that, let's interpret the determinant as a function of the tensor components:
$$\det(\mathbf{F}^*) = f(F^*_{ij}(\epsilon))$$
So when we try to do the derivative of the determinant, we need to use the chain rule:
$$ \frac{\text{d}}{\text{d}\epsilon}\left(\det(\mathbf{F}^*)\right) = \sum_i\sum_j\frac{\partial\det(\mathbf{F}^*)}{F^*_{ij}}\frac{\text{d}F^*_{ij}}{\text{d}\epsilon } \tag{+}$$
The second term in the multiplication by definition is: $$\frac{\text{d}F^*_{ij}}{\text{d}\epsilon } = \frac{\text{d}}{\text{d}\epsilon }\left(\frac{\partial\chi^*_i}{\partial{X_j}}\right) = \frac{\partial}{\partial X_j}\left(\frac{\text{d}\chi_i^*}{\text{d}\epsilon}\right) = \frac{\partial \hat{\eta}_i}{\partial X_j}$$
But if we want to use the variables in the configuration at time $t$, i.e. using $\boldsymbol{\eta}(\mathbf{x}, t)$ (and not $\hat{\boldsymbol{\eta}}(\mathbf{X}, t)$), again with the chain rule: $$\frac{\partial \hat{\eta}_i}{\partial X_j} = \sum_k\frac{\partial \eta_i}{\partial x_k}\frac{\partial x_k}{\partial X_i} =\sum_k \frac{\partial \eta_i}{\partial x_k}\frac{\partial \chi_k}{\partial X_i}=\sum_k\frac{\partial \eta_i}{\partial x_k}F_{ki} \tag{1}$$
Not how the $F_ki$ does not have the asterisk. That is subtle, but when we will go to $\epsilon \rightarrow 0;\; \mathbf{F}^* \rightarrow \mathbf{F}$.
Let's now focus on the derivative of the determinant. Using the Laplace formula ($\ell$ can be chosen arbitrarily from $1...n$, $n$ number of rows):
$$ \det(\mathbf{F}) = \sum_m F_{\ell m}\text{cof}\left(\mathbf{F}\right)_{\ell m}\tag{2}$$
and the derivative is hence (derivative of a product):
$$\frac{\partial\det(\mathbf{F}^*)}{F^*_{ij}} = \sum_m \frac{\partial F^*_{\ell m} }{\partial F^*_{ij}}\text{cof}\left(\mathbf{F}^*\right)_{\ell m} + F^*_{\ell m}\frac{\partial \text{cof}\left(\mathbf{F}^*\right)_{\ell m}}{\partial F^*_{ij}}$$
Since $\ell$ can be chosen arbitrarily, we set $\ell = i$. In this case the term:
$$\frac{\partial \text{cof}\left(\mathbf{F}^*\right)_{i m}}{\partial F^*_{ij}} = 0$$ because the cofactors do not depend on the elements of the same row ($i$). The term $$\frac{\partial F^*_{i m} }{\partial F^*_{ij}} = \delta_{jm}$$
leading to:
$$\frac{\partial\det(\mathbf{F}^*)}{F^*_{ij}} = \sum_m \frac{\partial F^*_{i m} }{\partial F^*_{ij}}\text{cof}\left(\mathbf{F^*}\right)_{i m} = \sum_m\delta_{jm}\text{cof}\left(\mathbf{F}^*\right)_{i m} = \text{cof}\left(\mathbf{F^*}\right)_{i j}$$
And if we use the Laplace formula (2), we can also extract that: $$\det(\mathbf{F}^*)\left(F_{ij}^*\right)^{-1}=\text{cof}\left(\mathbf{F^*}\right)_{i j} \tag{3}$$
Finally injecting (1), (2), and (3) in our target (+) equation, and noting that for $\epsilon\rightarrow 0;\; \mathbf{F}^* \rightarrow \mathbf{F}$:
\begin{split} \lim_{\epsilon \rightarrow 0}\frac{\text{d}}{\text{d}\epsilon}\left(\det(\mathbf{F}^*)\right) &=\sum_i\sum_j\sum_k \det(F) F^{-1}_{ij}\frac{\partial \eta_i}{\partial x_k}F_{kj} \\ &= \det(F)\sum_i\sum_j\sum_k F_{ij}^{-1}F_{kj}\frac{\partial \eta_i}{\partial x_k}\\ \end{split}
noting that:
$$ \sum_k F_{ij}^{-1}F_{kj} = \delta_{ki}$$ \begin{split} \lim_{\epsilon \rightarrow 0}\frac{\text{d}}{\text{d}\epsilon}\left(\det(\mathbf{F}^*)\right) &=\det(F)\sum_i\sum_j\delta_{ki}\frac{\partial \eta_i}{\partial x_k} = \det(F)\sum_i\frac{\partial \eta_i}{\partial x_i} \\ &= \det(F)\boldsymbol{\nabla} \cdot \boldsymbol{\eta} = J\; \boldsymbol{\nabla} \cdot \boldsymbol{\eta} \end{split}
Hence we obtained what we were looking for: $$ \boxed{\delta J = J \boldsymbol{\nabla} \cdot \boldsymbol{\eta}} $$
1Lagrangian?