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(Eulerian) Velocity of an elementary vector

  1. Jun 20, 2014 #1
    Hello everyone,

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

  2. jcsd
  3. Jun 20, 2014 #2


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    The material time derivative of an arbitrari vector field is derived as follows: In a little time increment [itex]\mathrm{d}t[/itex] on the one hand the vector field changes due to its own time dependence, [itex]\mathrm{d} t \partial_t \vec{A}[/itex]. On the other hand, the fluid element that was at position [itex]\vec{x}[/itex] at time [itex]t[/itex], will have moved by [itex]\mathrm{d} \vec{x}=\mathrm{d}t \vec{v}[/itex]. So the material time derivative is given by
    [tex]\mathrm{D}_t \vec{A}=\partial_t \vec{A}+(\vec{v}\cdot \nabla)\vec{A}.[/tex]
  4. Jun 23, 2014 #3
    Thanks for the reply.

    The expression that I know for the material derivative of a vector (and in general a tensor of any order) [itex]\vec{A}[/itex] is
    [itex]D_t \vec{A} = \dfrac{\partial \vec{A}}{ \partial t} + \nabla \vec{A} .\underline{v} [/itex]​
    Anyhow, a way of seing the equality is :
    [itex]\dfrac{d}{dt} (dx)=dv [/itex]​
    that is adimitting that
    [itex]\dfrac{d}{dt dv} (dx)=\dfrac{d^2 x}{dt dv}=\dfrac{d}{dv}\Big ( \dfrac{dx}{dt}\Big) =1 [/itex]​
    but I can't tell why is it peculiar to the eulerian description.
  5. Jun 23, 2014 #4


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    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
    [tex]\mathrm{D}_t \vec{A}=\partial_t \vec{A} + (\vec{v} \cdot \vec{\nabla}) \vec{A}.[/tex]
    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
    [tex](\mathrm{D}_t A^{j})=\partial_t A^j + v^k \partial_k A^j.[/tex]
    Note that
    [tex]\partial_k = \frac{\partial}{\partial x^k}.[/tex]
  6. Jun 23, 2014 #5
    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 [itex]\dfrac{d}{dt} (dx)=\nabla v . dx [/itex]
    Last edited: Jun 23, 2014
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