(Eulerian) Velocity of an elementary vector

1. Jun 20, 2014

trabo

Hello everyone,

In order to define the eulerian rate deformation tensor, one should first express $\dfrac{d}{dt}(\underline{dx})$ in gradient velocity terms (denoted $\underline{\nabla v}$ with $v$ equal to the partial time derivative of the geometrical mapping that relates the inital configuration to the current one).
In an article, we claim that
$\dfrac{d}{dt}(\underline{dx})=v(\underline{x}+\underline{dx},t)-v(\underline{x},t)$ (1)​
and thus
$\dfrac{d}{dt}(\underline{dx})$$=\underline{\nabla v} . \underline{dx}$ ​
I'm not sure about (1). It says actually that
$\dfrac{d}{dt}(\underline{x} +\underline{dx}-\underline{x})$ $=\dfrac{d}{dt}(\underline{x} +\underline{dx})-\dfrac{d}{dt}(\underline{x} )$​
I can't see that clearly though. Is there any physical explanation ? May be an approximation since we're dealing with elementary vectors....

Regards.

2. Jun 20, 2014

vanhees71

The material time derivative of an arbitrari vector field is derived as follows: In a little time increment $\mathrm{d}t$ on the one hand the vector field changes due to its own time dependence, $\mathrm{d} t \partial_t \vec{A}$. On the other hand, the fluid element that was at position $\vec{x}$ at time $t$, will have moved by $\mathrm{d} \vec{x}=\mathrm{d}t \vec{v}$. So the material time derivative is given by
$$\mathrm{D}_t \vec{A}=\partial_t \vec{A}+(\vec{v}\cdot \nabla)\vec{A}.$$

3. Jun 23, 2014

trabo

The expression that I know for the material derivative of a vector (and in general a tensor of any order) $\vec{A}$ is
$D_t \vec{A} = \dfrac{\partial \vec{A}}{ \partial t} + \nabla \vec{A} .\underline{v}$​
Anyhow, a way of seing the equality is :
$\dfrac{d}{dt} (dx)=dv$​
$\dfrac{d}{dt dv} (dx)=\dfrac{d^2 x}{dt dv}=\dfrac{d}{dv}\Big ( \dfrac{dx}{dt}\Big) =1$​
but I can't tell why is it peculiar to the eulerian description.

4. Jun 23, 2014

vanhees71

I'm not so sure what your notation in the 2nd term of the material time derivative should mean. As derived in my previous posting, the correct expression in usual nabla-calculus notation is
$$\mathrm{D}_t \vec{A}=\partial_t \vec{A} + (\vec{v} \cdot \vec{\nabla}) \vec{A}.$$
In component-Ricci calculus notation, including Einstein-summation notation and strictly distinguishing co- and contravariant components (which is very useful also for Cartesian components as here!) what's meant is
$$(\mathrm{D}_t A^{j})=\partial_t A^j + v^k \partial_k A^j.$$
Note that
$$\partial_k = \frac{\partial}{\partial x^k}.$$

5. Jun 23, 2014

trabo

Yes this is what I meant too, it was just a matter of notation. But as you notice, there is no velocity gradient in the material derivative of dx, whereas it is stated that $\dfrac{d}{dt} (dx)=\nabla v . dx$

Last edited: Jun 23, 2014