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: $$\partial_j T_{ij} = \partial_j p \delta_{ij} = \partial_i p$$

The components seem to add up just fine.

  • $\begingroup$ Should the stress tensor not have both index contravariant? $\endgroup$ – Noumeno Jul 25 '20 at 15:49
  • $\begingroup$ These are Cartesian tensors. There is no distiction between covariant and contravariant becuse we are restricting ourselves to metric $g_{ij}= \delta_{ij}$. $\endgroup$ – mike stone Jul 25 '20 at 18:26

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