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.
