(Eulerian) Velocity of an elementary vector

[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.

 

[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}.

 

[trabo]

Thanks for the reply.

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​

that is adimitting that

\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.

 

[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}.

 

[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