# Euler's Equation applied to Perfect Fluid

Consider this form of Euler's Equation: $$\rho \vec{a}=\nabla \cdot T+\rho \vec{f}$$ Where: $$\rho$$ is the density, $$\vec{a}$$ is the acceleration, $$T$$ is Cauchy's Stress Tensor and $$\vec{f}$$ is the force density (or if you prefer we can say that $$\vec{f}$$ is the acceleration per unit mass).
(This equation in practice is the generalization of Newton's Second Law to the continuum, where $$\nabla \cdot T$$ represents the surface forces and $$\rho \vec{f}$$ represents the "volume forces", like gravity).

Ok, now we want to apply this equation to a perfect fluid, which by definition has surface forces acting only in the perpendicular direction of the surface (no shear stresses). My book states that the Euler's Equation becomes: $$\rho \vec{a}=-\nabla P+\rho \vec{f}$$ where $$P$$ is the pressure. How can we prove that this is true?

Furthermore in component notation this seems strange: $$\nabla \cdot T\equiv\partial _i T^{ij}$$ $$-\nabla P\equiv -\partial _j P$$ $$\nabla \cdot T = -\nabla P \ \ \Rightarrow \ \ \partial _i T^{ij}=-\partial _j P$$ the components don't add up, in fact we have one covariant vector on the right side and one contravariant vector on the other side.

The stress tensor is: $$T_{ij} = p\delta_{ij} + \sigma_{ij}$$
Assuming stresses are indeed negligible, $$\sigma_{ij} = 0$$: $$T_{ij} = p\delta_{ij}$$
Taking the divergence of the stress tensor: $$\partial_j T_{ij} = \partial_j p \delta_{ij} = \partial_i p$$
• These are Cartesian tensors. There is no distiction between covariant and contravariant becuse we are restricting ourselves to metric $g_{ij}= \delta_{ij}$. – mike stone Jul 25 '20 at 18:26