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Acceleration of a fluid particle

  1. May 29, 2010 #1
    1. The velocity vector which describes the motion of a particle (point) in a fluid is [tex] \vec {u} = \vec {u}(\vec{x},t), [/tex] so that the particle follows the path on which [tex] \frac {dx}{dt}=\vec{U}(t)=\vec{u}[\vec{x}(t),t]. [/tex]
    Write [tex] \vec{x}\equiv (x,y,z) [/tex] and [tex] \vec{u} \equiv (u,v,w) [/tex] (in rectangular Cartesian coordinates), and hence show that the acceleration of the particle is [tex] \frac {d\vec{U}}{dt} = \frac {\partial \vec{u}}{\partial t} + (\vec {u} . \nabla) \vec {u} \equiv \frac {D\vec {u}}{Dt} [/tex], the material derivative




    2. Relevant equations



    Acceleration of particle = [tex] \frac{d^2 x}{dt^2} = \frac {dU}{dt}=\frac{du}{dt}[\frac {\partial x}{\partial t}, \frac {\partial y}{\partial t}, \frac {\partial z}{\partial t}, 1] [/tex]

    [tex](\vec{u}.\nabla)\vec{u} = uu_x\vec{i} + uv_x\vec{j} + uw_z\vec{k} + vu_y\vec{i} + vv_y \vec{j} + vw_y \vec {k} + wu_z \vec{i} + wv_z \vec{j} + ww_z \vec {k} [/tex]
     
  2. jcsd
  3. May 29, 2010 #2
  4. May 30, 2010 #3
    Got it! Thanks..
     
  5. Jun 16, 2010 #4
    Wait! The total derivative. How does that help ?

    [tex] \frac {dU}{dt} = \frac {\partial U}{\partial t} + \frac {\partial U}{\partial x} \frac {dx}{dt} + \frac {\partial U}{\partial t} \frac {dt}{dt} [/tex]
     
  6. Jun 16, 2010 #5
    This isn't the total derivative, this is:

    [tex]\frac{dU}{dt}=\frac{\partial u}{\partial t}+\frac{dx}{dt}\frac{\partial u}{\partial x}+\frac{dy}{dt}\frac{\partial u}{\partial y}+\frac{dz}{dt}\frac{\partial u}{\partial z}[/tex]

    Note that the last three terms look like:

    [tex]\frac{dx}{dt}\frac{\partial }{\partial x}+\frac{dy}{dt}\frac{\partial }{\partial y}+\frac{dz}{dt}\frac{\partial }{\partial z}=u_x\frac{\partial}{\partial x}+u_y\frac{\partial}{\partial y}+u_z\frac{\partial}{\partial z}=\mathbf{u}\cdot\nabla[/tex]

    Which is what you're looking for:

    [tex]\frac{dU}{dt}=\frac{\partial u}{\partial t}+\left(\mathbf{u}\cdot\nabla\right)u[/tex]

    This last formula is also called the convective derivative or the material derivative.
     
  7. Jun 16, 2010 #6
    Very helpful thank you. Just one quick question. Why does U become u in your total derivative expressions. Thanks again.

     
  8. Jun 16, 2010 #7
    Because you have [itex]U(t)[/itex] defined as [itex]U(t)=u\left[x(t),t\right][/itex] in the problem. If [itex]U(t)[/itex] is only a function of [itex]t[/itex], then [itex]\left(\mathbf{u}\cdot\nabla\right)u=0[/itex] and [itex]dU/dt=\partial U/\partial t[/itex], which is not what you want for this problem.


    Glad I can help.
     
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