# Difference between Eulerian and Lagrangian formulation of Fluid Dynamics

Difference between Eulerian and Lagrangian formulation of Fluid Dynamics.

I am completely new to fluid mechanics. Until now definition $F = ma$ was sufficient for me to solve any rigid body problems in classical mechanics. With problems involving change of mass (for example, rocket propulsion) more basic definition comes to play, i.e. $F = \frac{\partial (mv)}{\partial t}$.

I tried to apply this definition to a fixed volume of sides $\Delta x, \Delta y, \Delta z$ respectively and proceeded as follows (in Eulerian frame): \begin{eqnarray} \mathbf{F} &=& \frac{\partial (m \mathbf{v})}{\partial t} \\ &=& \Delta x \Delta y \Delta z \frac{\partial (\rho \mathbf{v})}{\partial t} \\ &=& \Delta x \Delta y \Delta z \left[ \rho\frac{\partial ( \mathbf{v})}{\partial t} + \mathbf{v}\frac{\partial (\rho)}{\partial t} \right] \\ &=& \Delta x \Delta y \Delta z\left[ \rho\frac{\partial ( \mathbf{v})}{\partial t} - \mathbf{v}\left (\nabla (\rho \mathbf{v}) \right) \right] \hspace{1cm} continuity \hspace{.5cm} equation \\ \end{eqnarray}

The textbook proceeds as follows: Since $m\mathbf{v}$ is function of both velocity and space it is differentiated with respect to all variables to get total change, i.e.

\begin{eqnarray} \mathbf{F} &=& \frac{\partial (m \mathbf{v})}{\partial t} + \frac{\partial (m \mathbf{v})}{\partial x} \frac{\partial x}{\partial t} +\frac{\partial (m \mathbf{v})}{\partial y} \frac{\partial y}{\partial t} +\frac{\partial (m \mathbf{v})}{\partial z} \frac{\partial z}{\partial t}\\ &=& \Delta x \Delta y \Delta z \left[\frac{\partial (\rho \mathbf{v})}{\partial t} + \frac{\partial (\rho \mathbf{v})}{\partial x} v_x+\frac{\partial (\rho \mathbf{v})}{\partial y} v_y +\frac{\partial (\rho \mathbf{v})}{\partial z} v_z \right] \\ \end{eqnarray}

Clearly the two results are not identical. What am I missing here?

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Force is not $\frac{\partial (m v)}{\partial t}$, it is $\frac{d(mv)}{d t}$. Expand the derivative in terms of partial derivatives and you will get the same equation in both cases. – SMeznaric Jul 12 '13 at 13:47
– SMeznaric Jul 12 '13 at 14:05
To clarify SMeznaric, you want to distinguish very carefully between two things. How $mv$ evaluated at particular point in space changes with time, and how $mv$ for a particular packet of particles changes with time. These are different because the packets are moving. – BebopButUnsteady Jul 12 '13 at 14:06
You derived your formula for a fixed volume, i.e. $\Delta x, \Delta y, \Delta z$ do not change with time. But in that textbook derivation you provided, they have not considered them fixed. That's why those $v_x,v_y,v_z$ terms. Put them as $0$ and it turns into the same equation you got. – udiboy1209 Jul 12 '13 at 14:13