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The stress tensor can be written as: $$\sigma_{ij}=-p \delta_{ij}+\sigma'_{ij}\label{1}\tag{1}$$ where $\sigma_{ij}'$ is called the extra stress tensor.

From what I understand pressure is pressure is force per unit area, so that the force acting on the surface $d\vec A$ is given by: $$\vec F=-p d\vec A\label{2}\tag{2}$$ (This has the effect that the force on an infinitesimally thin surface due to a continuous pressure is zero). However, $\sigma_{ij}$ represent the force in the $i$ direction on a surface with normal in the $\hat e_j$ direction. So the force acting on a surface $d \vec A$ must be: $$F_i=\sigma_{ij}dA_j\label{3}\tag{3}$$ Equations (2) and (3) clearly do not agree (even if we look soley at the force in the $d\vec A$ direction they don't agree). So my question is what is the exact meaning of pressure in expression (1) and why do we not have: $$\sigma_{ii}=-p$$

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    $\begingroup$ Is $\sigma'$ traceless? See en.wikipedia.org/wiki/Cauchy_stress_tensor at Stress deviator tensor $\endgroup$ – Valter Moretti Oct 15 '16 at 9:42
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    $\begingroup$ See en.wikipedia.org/wiki/Cauchy_stress_tensor item Stress deviator tensor $\endgroup$ – Valter Moretti Oct 15 '16 at 9:47
  • $\begingroup$ Observe that, in the static definition of pressure, one should only consider the normal force to the surface. Any tangential force contributes to the shear terms in the stress tensor. Hence the right identity is $(\mathbf F,\hat{\mathbf n}) = A\sigma(\mathbf n,\mathbf n)$. If the $p$ in your expression is the actual pressure, then $\sigma'$ is a symmetric bilinear form with vanishing diagonal (hence zero-trace; the symmetry condition follows from the requirement that angular momentum should be conserved). $\endgroup$ – Phoenix87 Oct 15 '16 at 10:01
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In fluid mechanics, stress tensor $\sigma_{ij}$ is the primary quantity. Pressure is defined by means of equation (1) mentioned in your question. Clearly, $-p\equiv \frac{1}{3}\sigma_{ii}$ ($\sigma'_{ij}$ is traceless by definition). In other words pressure is defined as the average of normal stresses on three orthogonal planes passing through the point where stress tensor is calculated. Pressure defined in this way is known as mechanical pressure. When interpreted this way equation (2) mentioned in your question becomes invalid in general. Equation (3) mentioned in your question is the correct formula to calculate net force on fluid element in direction $i$.

Strictly speaking, equation (2) mentioned in your question is applicable only in fluid statics, because then $\sigma'_{ij}=0$, and so $\sigma_{ij}=-p\delta_{ij}$, which means that normal stress is the same in all directions and thus equation (2) becomes unambiguous. However people equate this average value to thermodynamic pressure when applying relations from equilibrium thermodynamics to flows. If this approximation is made, then within that approximation equation (2) may be applied to flows. For example, if a balloon is being inflated, flow inside the balloon will be complicated. However if temperature inside the balloon is uniform, and if ideal gas equation is applicable, you may calculate (thermodynamic) pressure at inside wall of the balloon, and pretend as if this is also the pressure that appears in Navier-Stokes equation (which is the pressure defined by equation (1)).

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Within the usual framework of continuum mechanics, it is assumed that there are two types of forces: body forces and surface forces. The latter can be shown to be representable by a tensor $\textbf{T}$, the Cauchy stress tensor. This tensor yields the local stress on a surface by: $$ \vec F= \textbf{T} d\vec A $$ We can then decompose the Cauchy stress tensor into the isotropic pressure contribution and the stress contribution (which we need constitutive equations for): $$ \textbf{T} = \tau -p \textbf{1} $$ So your equation (3) is correct. Your equation (2) gives you a contribution to the force on the surface but not the total force.

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