# Landau & Lifshitz - Euler's equation for one-dimensional flow

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 $t=t_0$.

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 $\rho_0 = \rho(a)$ the given initial density distribution):

$$\rho\,\mathrm{d}x = \rho_0 \mathrm{d}a$$

or alternatively:

$$\rho\left(\frac{\partial x}{\partial a}\right)_t = \rho_0$$

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

$$\left(\frac{\partial v}{\partial t}\right)_a = -\frac{1}{\rho_0} \left(\frac{\partial p}{\partial a}\right)_t$$

How are the authors arriving at that equation?

In particular, when looking at Euler's equation: $$\frac{\partial\mathbb{v}}{\partial t} + \left( \mathbf{v} \cdot \textbf{grad} \right) \mathbf{v} = - \frac{1}{\rho} \textbf{grad}\, p$$ what happens with the second term on the LHS $\left( \mathbf{v} \cdot \textbf{grad} \right) \mathbf{v}$? Why does it not appear in the authors' solution?

So it turns out that the answer to my question can be found explained in great detail in the following answer to another question: https://physics.stackexchange.com/a/109661/8254

(Taken directly from the linked answer above:) The main message is that in the Euler equation we are considering Langrangian (Material) derivatives of tensor fields $S(t,x)$ in the Eulerian picture of the form: $$\frac{\mathrm{D}S}{\mathrm{D}t} := \frac{\partial S}{\partial t} + v(t,x)\cdot \nabla_x S(t,x)\text{,}$$ where it turns out that

$$\left.\frac{\mathrm{D}S}{\mathrm{D}t}(t,x)\right|_{x=x(t,y)}= \frac{\partial}{\partial t} S_L(t,y)\text{,}$$ which simply is the same time derivative but now in the Lagrangian picture.

This is just what I was looking for. (Have a look at the very good derivation in the link above.)