5
$\begingroup$

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] \qquad \text{ continuity 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?

$\endgroup$
  • 5
    $\begingroup$ 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. $\endgroup$ – SMeznaric Jul 12 '13 at 13:47
  • 2
    $\begingroup$ See also: en.wikipedia.org/wiki/Material_derivative $\endgroup$ – SMeznaric Jul 12 '13 at 14:05
  • 4
    $\begingroup$ 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. $\endgroup$ – BebopButUnsteady Jul 12 '13 at 14:06
  • 2
    $\begingroup$ 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. $\endgroup$ – udiboy1209 Jul 12 '13 at 14:13
1
$\begingroup$

I thought I would post an answer to this, to piece together and summarize some of the helpful information in the comments:

I think the key misconception is that, in the Eulerian case, momentum is considered as a continuous field that is a function of both time and space, i.e. $m\mathbf{v}=m\mathbf{v}(\mathbf{x},t)$, whereas in the Lagrangian case, we are considering an individual parcel of fluid, that is moving with the flow. Therefore, momentum is a function only of time ($m\mathbf{v}=m\mathbf{v}(t)$), but the parcel also has a position, which is also a function of time ($\mathbf{x}=\mathbf{x}(t)$).

So, looking at the Eulerian case, as SMeznaric's comment states:

$$\mathbf{F}=\frac{d(m\mathbf{v})}{d t}$$

not

$$\mathbf{F}=\frac{\partial(m\mathbf{v})}{\partial t}$$

Force is the total derivative of momentum, from Newton's Second Law. Because momentum is a function of both time and space ($m\mathbf{v}=m\mathbf{v}(\mathbf{x},t)$), the partial derivatives with respect to position and time have to be expanded, as shown in the textbook. This then results in:

$$\mathbf{F}=\Delta x\Delta y\Delta z\Bigl[\frac{\partial(\rho\mathbf{v})}{\partial t}+\mathbf{v}\cdot\nabla(\rho\mathbf{v})\Bigr]$$

which, on the right-hand side, gives the transient and advective terms of the Navier-Stokes momentum equation. The $\mathbf{F}$ on the left-hand-side is made up of contributions to the force on the volume from the pressure gradient, gravity and viscous forces.

In the Lagrangian case, Newton's Second Law still holds, i.e.:

$$\mathbf{F}=\frac{d(m\mathbf{v})}{d t}$$

But the key difference is that now we are considering a parcel of fluid moving with the flow; therefore, as explained above, momentum is now a function only of time. Mass of the parcel is constant, so that can be brought out of the derivative:

$$\mathbf{F}=m\frac{d(\mathbf{v})}{d t}$$

So, now those same forces from the pressure gradient, gravity and viscosity on the left-hand-side are imposed by the wider flow field on the parcel and are equal to the mass times the total derivative of the parcel velocity. So here, if we know the forces being imposed, then we can simply integrate to find the velocity of the parcel and track its path through the flow field.

I hope this helps.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.