1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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. :-)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook