**B. Looking for the demonstration of the Eq. (9)**

$$\frac{d}{dt}\delta V = \frac{d}{dt}[\delta \mathbf{x}(\delta \mathbf{y} \delta \mathbf{z})] \tag{\text{B1}}$$

$$\frac{d}{dt}\delta V = \delta \mathfrak{x} \frac{d}{dt}(\delta y \delta \mathfrak{z}) + \delta \mathfrak{x} \frac{d}{dt}(\delta y \delta \mathfrak{z}) \tag{\text{B2}}$$

$$\frac{d}{dt}\delta V = \delta \mathbf{x} \frac{d}{dt}(\delta y \delta \mathbf{z}) + (\delta y \delta \mathbf{z}) \frac{d}{dt} \delta \mathbf{x} \tag{\text{B3}}$$

$$\frac{d}{dt}\delta V = \delta \mathbf{x} \left[ \delta \mathbf{y} \frac{d}{dt}(\delta \mathbf{z}) + \delta \mathbf{z} \frac{d}{dt}(\delta \mathbf{y}) \right] + (\delta \mathbf{y} \delta \mathbf{z}) \frac{d}{dt} \delta \mathbf{x} \tag{\text{B4}}$$

$$\frac{d}{dt}\delta V = \delta \eta \delta \mathbf{z} \frac{d}{dt}(\delta \mathbf{x}) + \delta \mathbf{x} \delta \mathbf{z} \frac{d}{dt}(\delta \mathbf{y}) + \delta \mathbf{x} \delta \mathbf{y} \frac{d}{dt}(\delta \mathbf{z}) \tag{B5}$$

$$\frac{d}{dt}\delta V = \delta \mathbf{y} \delta \mathbf{z} \,\, \boldsymbol{v}\_{\mathbf{x}} + \delta \mathbf{x} \delta \mathbf{z} \,\, \boldsymbol{v}\_{\mathbf{y}} + \delta \mathbf{x} \delta \mathbf{y} \,\, \boldsymbol{v}\_{\mathbf{z}} \tag{\text{B6}}$$

$$\frac{d}{dt}\delta V = \delta y \delta \mathbf{z} \frac{d}{d\mathbf{x}} v\_x \delta \mathbf{x} + \delta \mathbf{x} \delta \mathbf{z} \frac{d}{d\mathbf{y}} v\_\mathbf{y} \delta \mathbf{y} + \delta \mathbf{x} \delta \mathbf{y} \frac{d}{d\mathbf{z}} v\_\mathbf{z} \delta \mathbf{z} \tag{\text{B7}}$$

$$\frac{d}{dt}\delta V = \delta \mathbf{x} \delta \mathbf{y} \delta \mathbf{z} \frac{d}{d\mathbf{x}} v\_{\mathbf{x}} + \delta \mathbf{x} \delta \mathbf{y} \delta \mathbf{z} \frac{d}{d\mathbf{y}} v\_{\mathbf{y}} + \delta \mathbf{x} \delta \mathbf{y} \delta \mathbf{z} \frac{d}{d\mathbf{z}} v\_{\mathbf{z}} \tag{\text{B8}}$$

Hence

$$\frac{d}{dt}\delta V = \delta V \,\,(\nabla.\nu)\tag{\text{B9}}$$
