The material time derivative of Jacobian of the deformation gradient

Question

The key step in the derivation of Reynolds transport theorem is time derivative of $J$, the determinant of deformation gradient $F$. Its result says

$$\dot{J}=\frac{\partial J(\xi,t)}{\partial t}=\frac{DJ}{Dt}=J\nabla\cdot\mathbf{v},$$ where $\mathbf{v}=\partial\mathbf{x}(\boldsymbol{\xi},t)/\partial t$; $\mathbf{x}=\mathbf{x}(\xi,t)$, is spatial coordinates while $\xi$ is material coordinates; The deformation gradient is expressed by the formula $${F}_{iJ}=\frac{\partial{x}_i}{\partial{\xi}_J},\;\;i, J=1,2,3$$ I tried to derive by myself: $$ \begin{split} \frac{\partial J(\boldsymbol\xi,t)}{\partial t}&=\sum_{i,J}\frac{\partial J}{\partial F_{iJ}}\frac{\partial F_{iJ}}{\partial t} =\sum_{i,J}\frac{\partial J}{\partial F_{iJ}}\frac{d}{dt}\bigg(\frac{\partial x_i}{\partial \xi_J}\bigg) =\sum_{i,J}\frac{\partial J}{\partial F_{iJ}}\frac{\partial}{\partial\xi_J}\bigg(\frac{\partial x_i}{\partial t}\bigg)\\ &=\sum_{i,J}\frac{\partial J}{\partial F_{iJ}}\frac{\partial v_i}{\partial\xi_J}\\ &=\sum_{i,J}\frac{\partial J}{\partial F_{iJ}}\sum_k\biggl(\frac{\partial v_i}{\partial x_k}\frac{\partial x_k}{\partial \xi_J}\biggr) =\sum_{i,J}\frac{\partial J}{\partial F_{iJ}}\sum_k\biggl(\frac{\partial v_i}{\partial x_k}F_{kJ}\biggr) \end{split}$$ Now we introduce velocity gradient $L\equiv(\nabla\mathbf{v})^T$, where $\mathbf{v}=\mathbf{v}(\mathbf{x},t)$. So $$\dot{J}=\sum_{i,J}\frac{\partial J}{\partial F_{iJ}}\sum_{k}L_{ik}F_{kJ}.$$ But how it actually turns to tensor notation? Hope someone can demonstrate how to prove it.

Answer 1

Let Understand Step By Step:

Deformation Gradient F:

The deformation gradient $F$ describes the change in the position of a material point from the reference configuration $\Omega_0$ to the current configuration $\Omega(t)$. It is defined as a matrix with components $$F_{ij} = \frac{\partial x_i}{\partial\xi_j},$$ where $i$ and $j$ range from 1 to 3.

Jacobian of the Deformation Gradient J:

The Jacobian of the deformation gradient, i.e., $J$, is the determinant of the deformation gradient matrix $F$. Mathematically, $J=\det(F)$.

Total Derivative of $J$ with Respect to Time:

We want to find how the Jacobian $J$ changes with respect to time. Using the chain rule, we can express this change as follows:

$$\frac{dJ}{dt} = \frac{\partial J}{\partial t}+\nabla J\cdot\mathbf{v},$$

where $\partial J/\partial t$ is the partial derivative of $J$ with respect to time, $\nabla J$ is the gradient of $J$, and $\mathbf{v}$ is the velocity field.

Expression for $\partial J/\partial t$: The expression for $\partial J/\partial t$ can be found from the given equation:

$$\frac{D}{Dt}\int_{\Omega(t)} f(\mathbf{x},t)d\Omega =\int_{\Omega_0} \bigg[f\frac{\partial J(\xi,t)}{\partial t}+\frac{\partial f(\xi,t)}{\partial t}J\bigg] dΩ_0.$$

Here, the left-hand side represents the material derivative of the integral of $f$ over $Ω(t)$, and the right-hand side involves the change of $f$ with respect to time in the material coordinates.

By comparing this equation with the total derivative of $J$ with respect to time, you can deduce that $∂J(ξ,t)/∂t = J\nabla\cdot\mathbf{v}$.

I hope that you can understand.

Answer 2

Jacobian may be expressed on terms of material coordinates $X_p$ by using Levi-Chivita tensor. $$J=e_{ijk}\frac{\partial x_1}{\partial X_i}\frac{\partial x_2}{\partial X_j}\frac{\partial x_3}{\partial X_k}.$$ $$\frac{DJ}{Dt}=e_{ijk}\left(\frac{\partial v_1}{\partial x_1}\frac{\partial x_1}{\partial X_i}\frac{\partial x_2}{\partial X_j}\frac{\partial x_3}{\partial X_k}+\frac{\partial x_1}{\partial X_i}\frac{\partial v_2}{\partial x_2}\frac{\partial x_2}{\partial X_j}\frac{\partial x_3}{\partial X_k}+\frac{\partial x_1}{\partial X_i}\frac{\partial x_2}{\partial X_j}\frac{\partial v_3}{\partial x_3}\frac{\partial x_3}{\partial X_k}\right)$$ $$\frac{DJ}{Dt}= \left(e_{ijk}\frac{\partial x_1}{\partial X_i}\frac{\partial x_2}{\partial X_j}\frac{\partial x_3}{\partial X_k}\right)\frac{\partial v_p}{\partial x_p}.$$ $$\frac{DJ}{Dt}=J\nabla\cdot\textbf{v}.$$ In obtaining this equation, I have used the definitión of velocity field, the chain rule and inditial notation.

Answer 3

$$ J=\Psi(x_i(t),\xi_j), $$ $$ \textrm{d}J=\frac{\partial\Psi}{\partial x_i}dx_i+\frac{\partial\Psi}{\partial\xi_i}\textrm{d}\xi_i. $$ Dividing by dt and since $\xi$ is independent of t, $\frac{\textrm{d}\xi}{\textrm{d}t}=0,$ so that $$ \dot{J} = \frac{\textrm{d} J}{\textrm{d}t}=\frac{\partial \Psi}{\partial x_i} \frac{\textrm{d} x_i}{\textrm{d}t} = \frac{\partial J}{\partial x_i} v_i = \boldsymbol{\nabla} J \cdot \mathbf{v}. $$