In deriving continuity equations using Lagrangian.
We consider the element of fluid which occupied a rectangular parallelopiped having its centre at the point $(a,b,c)$ and its edges $\delta a$ , $\delta b$ ,$\delta c $ parallel to the axes . At the time $t$ the same element for an oblique parallelepiped . The centre now has for its co-ordinates $x$ , $y$ , $z \ ;$ and the projections of the edges on the co-ordinate axes are respectively
$$ \frac{\partial x}{\partial a} \delta a \ , \ \frac{\partial y}{\partial a} \delta a \ , \ \frac{\partial z}{\partial c} \delta a$$
$$\frac{\partial x}{\partial b} \delta b \ , \ \frac{\partial y}{\partial b} \delta b \ , \ \frac{\partial z}{\partial b} \delta b$$
$$\frac{\partial x}{\partial c} \delta c \ , \ \frac{\partial y}{\partial c} \delta c \ , \ \frac{\partial z}{\partial c} \delta c$$
How can i get these projections ?
The volume of the parallelepiped is therefore
$$\begin{vmatrix}
\frac{\partial x}{\partial a} & \frac{\partial y}{\partial a} & \frac{\partial z}{\partial a} \\
\frac{\partial x}{\partial b} & \frac{\partial y}{\partial b} & \frac{\partial z}{\partial b} \\
\frac{\partial x}{\partial c} & \frac{\partial y}{\partial c} & \frac{\partial z}{\partial c}
\end{vmatrix} \delta a \delta b \delta c$$
or as its often written $$\frac{D(x,y,z)}{D(a,b,c)} \delta a \delta b \delta c$$
since the fluid mass is unchanged and the fluid is incompressble we have
$$\frac{D(x,y,z)}{D(a,b,c)} =1$$
Is there a way to prove that $$\frac{D(a,b,c)}{D(x,y,z)}= 1$$
without expanding the determinant?
Best Answer
In the Lagrangian flow picture ${\bf a}\equiv(a,b,c)$ typically denote continuous labels of a fluid parcel distributed such that $$d(\text{mass})~=~da~db~dc,\tag{2.1}$$ cf. e.g. Ref. 1.
On the other hand ${\bf x}\equiv(x,y,z)$ typically denote the position coordinates of a fluid parcel. Therefore the mass density becomes $$ \rho~=~ |\det\frac{\partial{\bf a}}{\partial{\bf x}} |.\tag{2.2}$$
The flow velocity is defined as $${\bf u}~\equiv~ \frac{d{\bf x}}{dt}.\tag{2.4}$$ The mass continuum equation follows in the Lagrangian flow picture from $$\begin{align} -\frac{d\ln\rho}{dt}~=~&\frac{d}{dt}\ln |\det \frac{\partial{\bf x}}{\partial{\bf a}}|\cr ~=~&{\rm tr}\left( \frac{\partial{\bf a}}{\partial{\bf x}}\frac{d}{dt}\frac{\partial{\bf x}}{\partial{\bf a}}\right)\cr ~=~&{\rm tr}\left( \frac{\partial}{\partial{\bf x}}\frac{d{\bf x}}{dt}\right)\cr ~=~&{\bf \nabla}\cdot {\bf u}.\end{align} \tag{2.3}$$ For an incompressible flow, the density $\rho$ is constant along the fluid flow.
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