Physics Asked on September 2, 2021
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:
$$nabla cdot Tequivpartial _i T^{ij}$$
$$-nabla Pequiv -partial _j P$$
$$nabla cdot T = -nabla P Rightarrow partial _i T^{ij}=-partial _j P$$
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: $$T_{ij} = pdelta_{ij} + sigma_{ij}$$
Assuming stresses are indeed negligible, $sigma_{ij} = 0$: $$T_{ij} = pdelta_{ij}$$
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.
Answered by nluigi on September 2, 2021
Get help from others!
Recent Questions
Recent Answers
© 2024 TransWikia.com. All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP