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

[mSSM]

One page 5 in Landau & Lifshitz Fluid Mechanics (2nd edition), the authors pose the following problem:

Write down the equations for one-dimensional motion of an ideal fluid in terms of the variables a, t, where a (called a Lagrangian variable) is the x coordinate of a fluid particle at some instant t=t0.

The authors then go on to give their solutions and assumptions. Here are the important parts:

The coordinate x of a fluid particle at an instant t is regarded as a function of t and its coordinate a at the initial instant: x=x(a,t).

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

or alternatively:
<br /> ρ\left(\frac{∂x}{∂a}\right)_t=ρ_0<br />

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

<br /> \left(\frac{∂v}{∂t}\right)_a=−1ρ_0 \left(\frac{∂p}{∂a}\right)_t<br />

How are the authors arriving at that equation?

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

what happens with the second term on the LHS, (\mathbf{v}⋅\textbf{grad})\mathbf{v}? Why does it not appear in the authors' solution?

 

[jedishrfu]

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 that's why it is zero.

 

[boneh3ad]

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

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

 

[maajdl]

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

\left(\frac{dv}{dt}\right) = \left(\frac{∂v}{∂t}\right)_a

Note that by definition:

\left(\frac{∂v}{∂t}\right) = \left(\frac{∂v}{∂t}\right)_x

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.

 

[mSSM]

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

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

 

[maajdl]

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

SOLUTION. In these variables the co-ordinate x of any fluid particle at any instant is regarded
as a function of t and its co-ordinate a at the initial instant: x = x(a, t). The condition
of conservation of mass during the motion of a fluid element (the equation of continuity)
is accordingly written ... , or
where ... is a given initial density distribution. The velocity of a fluid particle is, by
definition, ... , and the derivative ... gives the rate of change of the velocity
of the particle during its motion. Euler's equation becomes
and the adiabatic equation is ... .

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

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

where by definition

\left(\frac{∂v}{∂t}\right) = \left(\frac{∂v}{∂t}\right)_x

 

[mSSM]

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

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

where by definition

\left(\frac{∂v}{∂t}\right) = \left(\frac{∂v}{∂t}\right)_x

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