# Perfect fluid stress tensor

In Thorne and Balndford's new book, they approach the subject of classical physics and tensors from the geometric viewpoint (as in relativity) that is independent from coordinates, instead from a coordinates-based viewpoints. This is quite a powerful approach and it deepens ones understanding.

At one point they mentioned that a perfect fluid has only isotropic (but no shear) pressures, and so its tensor was simply $$\boldsymbol{T}=P\boldsymbol{g}$$, where $$g_{ij}=\delta_{ij}$$ is the Euclidean metric tensor. Which is clear to me, because the pressure $$P$$ is only exerted normal to the faces of an element of volume now. Then they introduce the reader to the conservation of momentum and how the stress tensor is actually the flux of momentum, and the differential form of the conservation law $$\partial \boldsymbol{G}/\partial t+\nabla\cdot \boldsymbol{T}=0$$, where $$\boldsymbol{G}$$ is momentum density.

But then they ask the reader in a later exercise to prove that, for a perfect fluid with density $$\rho$$, momentum density $$\boldsymbol{G}=\rho \boldsymbol{v}$$, pressure $$P$$ and velocity $$\boldsymbol{v}$$ that vary in space and time, we will have a momentum flux (stress tensor) equal to

$$\boldsymbol{T}=P\boldsymbol{g}+\rho \boldsymbol{v}\otimes\boldsymbol{v},$$ where $$\otimes$$ indicates tensor product. We can also express this relation in index format as $$T_{ij}=Pg_{ij}+\rho v_{i}v_{j}$$

This is were I couldn't follow them. The first term is clear to me from the above isotropic pressure discussion, but how does the second term naturally arise? I would really appreciate a simple and clear geometric explanation in terms of these vectors (independent of coordinates) if possible, similar to their treatment. Most explanations I noticed in common literature are coordinates-based and obscure the physics a bit.

• Are you familiar with the 3D version of this in the derivation of the Navier Stokes equation? Mar 9, 2020 at 11:54
• @chetmiller No. Mar 9, 2020 at 19:15
• What about the Euler equation? Mar 9, 2020 at 20:07

In non-relativistic fluid mechanics, the 3D acceleration vector is given by: $$\mathbf{a}=\frac{\partial \mathbf{v}}{\partial t}+\mathbf{v}\centerdot \nabla \mathbf{v}$$where $$\mathbf{v}$$ is the fluid velocity vector.

And Newton's 2nd law (per unit volume) of the flowing inviscid fluid is given by Euler's equation: $$\rho \mathbf{a}=\rho \left(\frac{\partial \mathbf{v}}{\partial t}+\mathbf{v}\centerdot \nabla \mathbf{v}\right)=-\nabla p\tag{1}$$where $$\rho$$ is the fluid density, p is the fluid pressure, and $$-\nabla p$$ is the net force per unit volume driving the flow. In addition to this, the equation for conservation of mass is given by $$\frac{\partial \rho}{\partial t}+\nabla \centerdot (\rho \mathbf{v})=0\tag{2}$$ If we multiply Eqn.2 by $$\mathbf{v}$$ and add the result to Eqn. 1, we obtain: $$\frac{\partial (\rho \mathbf{v})}{\partial t}+(\rho \mathbf{v})\centerdot \nabla \mathbf{v}+\mathbf{v}\nabla \centerdot (\rho \mathbf{v})=-\nabla p\tag{3}$$It follows from the mathematical identity $$[(\rho \mathbf{v})\centerdot \nabla \mathbf{v}+\mathbf{v}\nabla \centerdot (\rho \mathbf{v})]=\nabla \centerdot (\rho \mathbf{v} \otimes \mathbf{v})$$that Eqn. 3 becomes:$$\frac{\partial (\rho \mathbf{v})}{\partial t}+\nabla \centerdot (\rho \mathbf{v} \otimes \mathbf{v})=-\nabla p\tag{4}$$where $$\rho \mathbf{v} \otimes \mathbf{v}$$ is the momentum flux (dyadic) tensor. Writing Eqn. 4 in a slightly different form, we then have:$$\frac{\partial (\rho \mathbf{v})}{\partial t}+\nabla \centerdot \mathbf{T}=0\tag{5}$$where $$\mathbf{T}=p\mathbf{I}+\rho \mathbf{v} \otimes \mathbf{v}$$, with $$\mathbf{I}$$ representing the identity (metric) tensor.

The 2nd term is the momentum flux due to advection (i.e. momentum being physically transported by the fluid velocity)

Using your notation, the equation of conservation of momentum in an ideal fluid can be written as $$\partial_t G_i + \nabla_j T_{ji} = 0$$ .

Here, $$G_i$$ is the density of the $$i$$th component of momentum, $$T_{ji}$$ is the $$j$$th component of flux of $$i$$th component of momentum. Note that the two indices are formally not the same (at this point anyway)

At a fluid element that is at rest, there is still an isotropic Pressure felt: For this fluid element at rest, $$T_{ji} = P \delta_{ji}$$.

What if the element is in motion with velocity $$v$$? We can see that now, there must be an additional contribution to the momentum flux (ie, momentum crossing over across surfaces) from the fact that the fluid element, carrying some momentum, is also physically crossing some surfaces (as it is in motion with velocity $$v$$), and hence, carrying its momentum with it. This mechanism of momentum (replace by any other conserved quantity) transfer is called "advection". We can see that the contribution to $$T_{ji}$$ should be $$v_j G_i$$. Altogether we have $$T_{ji} = P \delta_{ji} + v_j G_i$$

To proceed further, note that momentum $$G_i$$ itself can be given as $$\rho v_i$$, where $$\rho$$ is the mass density. This leads to the final form of the stress tensor: $$T_{ji} = P \delta_{ji} + \rho v_j v_i$$

I think you are confusing the stress tensor $$\sigma_{ij}$$ and the momentum flux tensor $$\Pi_{ij}$$ - they are two different entities: The stress tensor is only part of the entire momentum flux.

The stress tensor $$\underline{\sigma}$$ is a result of the material law and describes the correlation of a deformations $$\vec d$$ and/or deformation speeds $$\vec u$$ and the corresponding induced stress and is formally similar for solids as well as fluids. Solids generally can be assumed to behave like (linear) springs, proportional to the deformation, while fluids like huge (linear) dampeners, proportional to the deformation speeds. As rigid-body transformations do not result in a stress they have to be excluded by only taking the symmetric part of the corresponding deformation measures, the displacement $$\underline{\epsilon}$$ and the strain-rate tensor $$\underline{S}$$

$$\underline{\epsilon} := \frac{1}{2} \left( \vec \nabla \otimes \vec d + (\vec \nabla \otimes \vec d)^T \right) \phantom{abcdefg} \underline{S} := \frac{1}{2} \left( \vec \nabla \otimes \vec u + (\vec \nabla \otimes \vec u)^T \right)$$

$$\epsilon_{ij} = \frac{1}{2} \left( \frac{\partial d_i}{\partial x_j} + \frac{\partial d_j}{\partial x_i} \right) \phantom{abcdefg} S_{ij} = \frac{1}{2} \left( \frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i} \right)$$

The stress tensor for any continuum can be any (non-linear) function of the two

$$\underline{\sigma} = f( \underline{\epsilon}, \underline{S} ).$$

For fluids it is generally assumed that the dissipation (internal friction) dominates, is linearly proportional to the strain, the material is isotropic and the viscous stress tensor traceless and symmetric. This combination is generally referred to as Newtonian fluid and results in a stress tensor

$$\sigma_{ij} = - p \delta_{ij} + \underbrace{2 \mu S_{ij} - \frac{2}{3} \mu \sum\limits_{k \in \mathcal{D}} S_{kk} \delta_{ij}}_{\tau_{ij}}.$$

The stress tensor in this case consists of an isotropic resting stress $$- p \delta_{ij}$$ (Pascal's law) and the viscous stresses $$\tau_{ij}$$ arising from friction between the fluid particles themselves as well as with their surroundings. A simple derivation for this as well as the conservation equations of fluid mechanics, the Navier-Stokes equations, - assuming Cartesian coordinates for simplicity - can be found here.

The momentum flux $$\underline{\Pi}$$ is a measure for the specific momentum felt by a certain volume, consisting of the stress tensor and additionally the advected momentum (second line corresponds to Cartesian coordinates)

$$\underline{\Pi} := \rho \vec u \otimes \vec u - \underline{\sigma} = p \underline{\delta} + \rho \vec u \otimes \vec u - \underline{\tau}.$$

$$\Pi_{ij} = \rho u_i u_j - \sigma_{ij} = p \delta_{ij} + \rho u_i u_j - \tau_{ij}.$$

Clearly when sticking your hand out a moving car the "pressure" you feel is higher than the static pressure, it is enhanced by the moving fluid (stagnation pressure).

The momentum flux allows you to rewrite the momentum equation

$$\frac{\partial (\rho \vec u )}{\partial t} + \vec \nabla \cdot (\rho \vec u \otimes \vec u ) = \vec \nabla \cdot \underline{\sigma} + \rho \vec g$$

$$\frac{\partial (\rho u_i )}{\partial t} + \sum\limits_{j \in \mathcal{D}}\frac{\partial (\rho u_i u_j )}{\partial x_j} = \sum\limits_{j \in \mathcal{D}} \frac{\partial \sigma_{ij}}{\partial x_j } + \rho g_i$$

in a pretty convenient way

$$\frac{\partial (\rho \vec u )}{\partial t} + \vec \nabla \cdot \underline{\Pi} = \rho \vec g$$

$$\frac{\partial (\rho u_i )}{\partial t} + \sum\limits_{j \in \mathcal{D}}\frac{\partial \Pi_{ij}}{\partial x_j} = \rho g_i.$$

In a perfect fluid there is an absence of friction, the viscous stress part is neglected and you are left with the expression given in your book

$$\underline{\Pi} \approx p \underline{\delta} + \rho \vec u \otimes \vec u$$

$$\Pi_{ij} \approx p \delta_{ij} + \rho u_i u_j.$$

This is a viable assumption far away from walls. Near walls the flow is dominated by dissipation that results in the formation of a boundary layer and potentially more intricate flow patterns such as eddies and flow separation.