Derive the Momentum Equation from a Lagrangian (Fluid Dynamics)

I'm trying to derive the fully compressible Euler-Momentum equation for the given Lagrangian.

We wish to derive $$\rho\frac{D\boldsymbol{u}}{Dt} + \nabla P + \rho\nabla\phi = 0$$ from the Lagrangian $$\mathcal{L} = \rho \left( \frac{|\boldsymbol{u}|^{2}}{2} -e(\rho,s) - \phi \right)$$

I've gotten as far as

$$\frac{\partial}{\partial t}(\rho\boldsymbol{u}) + \nabla\cdot(\rho\boldsymbol{u}\boldsymbol{u}) + \rho\nabla\frac{|\boldsymbol{u}|^{2}}{2} -\rho\nabla \left(\frac{|\boldsymbol{u}|^{2}}{2}-e-\frac{P}{\rho}-\phi \right) - \rho T\nabla s = 0$$ which has been verified as correct. I've recognized the following relationships, $$\frac{\partial}{\partial t}(\rho\boldsymbol{u}) + \nabla\cdot(\rho\boldsymbol{u}\boldsymbol{u}) = -\nabla P$$ $$\frac{D\boldsymbol{u}}{Dt} = -\nabla \frac{P}{\rho}$$ $$\rho T\nabla s = \rho\nabla e - \frac{P}{\rho}\nabla\rho$$ which are the Euler-Momentum equation in conservative form, the Euler-Momentum equation, and the 2nd Law of Thermodynamics, respectively.

but after implementing these I arrive at:

$$\frac{\partial}{\partial t}(\rho\boldsymbol{u}) + \nabla\cdot(\rho\boldsymbol{u}\boldsymbol{u}) + \rho\nabla\frac{|\boldsymbol{u}|^{2}}{2} - \rho\nabla\frac{|\boldsymbol{u}|^{2}}{2} + \rho\nabla e + \rho\nabla\frac{P}{\rho} + \rho\nabla\phi - \rho\nabla e + \frac{P}{\rho}\nabla\rho = 0$$ $$\Rightarrow -\nabla P - \rho \frac{D\boldsymbol{u}}{Dt} + \rho\nabla\phi + \frac{P}{\rho}\nabla\rho = 0$$ which is "close" but incorrect. Where am I going wrong here?

• Welcome to Physics SE! Please do not post formulae as images, but use MathJax instead. – Nihar Karve Feb 22 at 2:43
• @NiharKarve Thanks! Roger that. – adamsthirdeye Feb 22 at 3:12
• Did you derive $\frac{D\boldsymbol{u}}{Dt} = -\nabla \frac{P}{\rho}$ from the equation above it? If so, it is wrong and should be $\frac{D\boldsymbol{u}}{Dt} = -\frac{1}{\rho}\nabla P$ unless the flow is incompressible. Given the parts which are deviating i suspect this will clear up the incorrect terms – nluigi Feb 22 at 21:12
• @nluigi I did not, got that from the definition of the nonlinear Eularian momentum eqn. – adamsthirdeye Feb 23 at 1:18

It works out just fine

Starting from your equation: $$\frac{\partial}{\partial t}(\rho\boldsymbol{u})+\nabla\cdot(\rho\boldsymbol{u}\boldsymbol{u})+\rho\nabla\frac{|\boldsymbol{u}|^{2}}{2}-\rho\nabla\left(\frac{|\boldsymbol{u}|^{2}}{2}-e-\frac{P}{\rho}-\phi\right)-\rho T\nabla s=0$$ it can be simplified: $$\frac{\partial}{\partial t}(\rho\boldsymbol{u})+\nabla\cdot(\rho\boldsymbol{u}\boldsymbol{u})+\rho\nabla\left(e+\frac{P}{\rho}+\phi\right)-\rho T\nabla s=0$$ by canceling kinetic energy terms.

Then applying the 2nd law of thermodynamics: $$\rho T\nabla s=\rho\nabla e-\frac{P}{\rho}\nabla\rho$$ substituting, cancelling and rearranging we get: $$\frac{\partial}{\partial t}(\rho\boldsymbol{u})+\nabla\cdot(\rho\boldsymbol{u}\boldsymbol{u})+\rho\nabla\frac{P}{\rho}+\frac{P}{\rho}\nabla\rho+\rho\nabla\phi=0$$

The pressure terms can be combined using the product rule: $$\nabla P = \nabla \left[\rho\frac{P}{\rho}\right]=\rho\nabla\frac{P}{\rho}+\frac{P}{\rho}\nabla\rho$$ to yield: $$\frac{\partial}{\partial t}(\rho\boldsymbol{u})+\nabla\cdot(\rho\boldsymbol{u}\boldsymbol{u})+\nabla P+\rho\nabla\phi=0$$

Then using the continuity equation: $$\frac{\partial\rho}{\partial t}+\nabla\cdot(\rho\boldsymbol{u})=0$$ we simplify the first two terms on the left: $$\frac{\partial}{\partial t}(\rho\boldsymbol{u})+\nabla\cdot(\rho\boldsymbol{u}\boldsymbol{u})=\rho\left[\frac{\partial\boldsymbol{u}}{\partial t}+\boldsymbol{u}\nabla\cdot\boldsymbol{u}\right]+\left[\frac{\partial\rho}{\partial t}+\nabla\cdot(\rho\boldsymbol{u})\right]\boldsymbol{u}=\rho\frac{D\boldsymbol{u}}{Dt}$$

Finally we get to what you wished to derive: $$\rho\frac{D\boldsymbol{u}}{Dt}+\nabla P+\rho\nabla\phi=0$$

• Oh perfect, thank you! – adamsthirdeye Feb 24 at 22:53