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Euler's equation for one-dimensional flow (Landau Lifshitz)

  1. Jun 27, 2014 #1
    One page 5 in Landau & Lifshitz Fluid Mechanics (2nd edition), the authors pose the following problem:
    The authors then go on to give their solutions and assumptions. Here are the important parts:

    For the condition of mass conversation the authors arrive at (where [itex]ρ_0=ρ(a)[/itex] is the given initial density distribution):
    [tex]
    ρ\mathrm{d}x=ρ_0 \mathrm{d}a
    [/tex]

    or alternatively:
    [tex]
    ρ\left(\frac{∂x}{∂a}\right)_t=ρ_0
    [/tex]

    Now the authors go on to write out Euler's equation, where I start to miss something. With the velocity of the fluid particle [itex]v=\left(\frac{∂x}{∂t}\right)_a[/itex] and [itex]\left(\frac{∂v}{∂t}\right)_a[/itex] the rate of change of the velocity of the particle during its motion, they write for Euler's equation:
    How are the authors arriving at that equation?

    In particular, when looking at the full Euler's equation:
    [tex]
    \frac{∂v}{∂t}+(\mathbf{v}⋅\textbf{grad})\mathbf{v}=−1 ρ\, \textbf{grad}\, p
    [/tex]

    what happens with the second term on the LHS, [itex](\mathbf{v}⋅\textbf{grad})\mathbf{v}[/itex]? Why does it not appear in the authors' solution?
     
  2. jcsd
  3. Jun 27, 2014 #2

    jedishrfu

    Staff: Mentor

    Could it be that (v.grad)v is zero meaning the v is perpendicular to grad v?

    In the wikipedia article on Euler flow:

    http://en.wikipedia.org/wiki/Euler_equations_(fluid_dynamics)

    They mention the characteristics of Euler flow that the divergence of the flow velocity is zero and I think thats why it is zero.
     
    Last edited: Jun 27, 2014
  4. Jun 27, 2014 #3

    boneh3ad

    User Avatar
    Science Advisor
    Gold Member

    This. It's a 1-D flow, so the gradient and velocity must be parallel or anti-parallel to one another.
     
  5. Jun 27, 2014 #4

    maajdl

    User Avatar
    Gold Member

    That's precisely the point of this exercise.
    The derivation is simple: consider a bunch of particles in a range [a,a+da],
    and apply Newtons law on this bunch of particles.
    You will immediately get the result.
    The point is that

    [tex]\left(\frac{dv}{dt}\right) = \left(\frac{∂v}{∂t}\right)_a[/tex]

    Note that by definition:

    [tex]\left(\frac{∂v}{∂t}\right) = \left(\frac{∂v}{∂t}\right)_x[/tex]

    The gradient term (convective term) is precisely the variation of velocity that comes from "following the fluid".
    The Euler view, where a is taken constant, follows the fluid.
    Therefore the "Euler derivative" incorporates the convective term: it is the change of speed when following the fluid.
     
  6. Jun 27, 2014 #5

    Maybe I am being dense, but isn't this a contradiction to what the author above said (namely that gradient and vector are perpendicular)?
     
  7. Jun 27, 2014 #6

    maajdl

    User Avatar
    Gold Member

    Landau never said that "gradient and vector are perpendicular".
    Here is what he said in the solution to this exercise:

    The message of this exercice is that the "Euler derivative" (derivative when following the fluid) is the same thing as the convective derivative

    [tex]\left(\frac{∂v}{∂t}\right)_a = \frac{dv}{dt} = \frac{∂v}{∂t}+(\mathbf{v}⋅\textbf{grad})\mathbf{v}[/tex]

    where by definition

    [tex]\left(\frac{∂v}{∂t}\right) = \left(\frac{∂v}{∂t}\right)_x[/tex]
     
  8. Jun 27, 2014 #7

    Thank you! Your explanation pointed me in the right direction. :-)
     
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