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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?

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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.)

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