Rheology, rheometers, and matching models to experiments

The dynamic (or absolute) viscosity µ is defined typically as the ratio of stress σ and strain rate:

$$\begin{equation} \mu\frac{\partial v}{\partial x}=\sigma. \end{equation} \tag{ 2.3 }$$

Stress is defined as the force per unit area across an infinitesimal surface. Force has units $ML/T^{\,2}$, so stress has units $M/(LT^{\,2})$. Therefore the units in (2.3) are consistent. But v and x are vector quantities, so in general µ would have to be a tensor of order (or arity) 4 in general (Nunan and Keller 1984). For Newtonian fluids, this tensor reduces to a scalar times the identity tensor.

The units of µ are mass divided by length×time, and the units of mass density ρ are mass divided by length cubed. Thus the units of $\mu/\rho$ are

$$\begin{equation*} \frac{M/(LT)}{M/L^{\,3}}=\frac{L^{\,2}}{T} \end{equation*}$$

which are the units for diffusion.

The kinematic viscosity ν is simply $\mu/\rho$. Table 1 gives the kinematic viscosity for various fluids under various conditions.

Apparent viscosity is typically defined as (2.3) without significant elaboration. In Nijenhuis et al (2007), more complicated notions are explored. But there are various ways to generate v and the resulting σ will often have several components. So a precise experimental or computational framework is necessary to make a clear definition. Here we will give two such examples. We will see that the ultimate relationship is tensorial, so it is not so clear how to define a scalar value for 'viscosity.' Indeed, we seem to be lacking a good term to explain the principal effect of rheology beyond Newtonian viscosity. One way is simply to define the apparent viscosity ν_a as

$$\begin{equation} \nu_a(\mathbf{v})=\frac{\Vert\, \mathbf{T} \,\Vert}{\Vert\, \nabla\mathbf{v} \,\Vert} \end{equation} \tag{ 2.4 }$$

for some norm on matrices, where T is the observed stress related to the flow velocity v. The same fluid could produce different values of apparent viscosity in different flow geometries, however.

If we use the definition (2.4) for apparent viscosity, then we can say that the fluid is thinning if

$$\begin{equation*} \nu_a(t\mathbf{v})<\nu_a(\mathbf{v}) \end{equation*}$$

when t > 1. When this holds for the expression (2.4) involving a norm, then any other notion thinning would also apply. But (2.4) may be too strong to catch subtle behaviors.

Thickening would involve the reverse inequality:

$$\begin{equation*} \nu_a(t\mathbf{v})>\nu_a(\mathbf{v}) \end{equation*}$$

when t > 1. Often, the terminology suggests a particular flow regime such as shear, which we will study. But we will also consider a flow problem involving pure extension, in which $\nabla\mathbf{v}$ has quite different structure.

For a given constant speed U, extensional flow $\mathbf{u}(x,y) = U\,(x,-y)^t$, depicted in figure 1, exhibits a constraint for the Oldroyd models. This flow is evidently incompressible, and we will show that there is a corresponding solution with constant T, so that ${{\nabla\cdot} \,}\mathbf{T} = \mathbf{0}$. We have

$$\begin{equation} \mathbf{E}=\nabla\mathbf{u}=U\begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix},\qquad \mathbf{R}=\mathbf{0}. \end{equation} \tag{ 3.8 }$$

Thus

$$\begin{equation*} \mathbf{u}\cdot\nabla\mathbf{u}=\mathbf{E}\mathbf{u}=U^{\,2}\begin{pmatrix} x \\ y\end{pmatrix} =\frac{U^{\,2}}{2}\nabla\left(x^{\,2}+y^{\,2}\right), \end{equation*}$$

so we can take

$$\begin{equation} p(x,y)=\frac{U^{\,2}}{2}\left(x^{\,2}+y^{\,2}\right) \end{equation} \tag{ 3.9 }$$

and solve

$$\begin{equation} -\beta\Delta\mathbf{u}+\mathbf{u}\cdot\nabla\mathbf{u}+\nabla p =\mathbf{0} \;\hbox{in}\,\Omega, \end{equation} \tag{ 3.10 }$$

for any β.

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If we take

$$\begin{equation*} \mathbf{u}(x,y)=U\,(x,-y)^t, \end{equation*}$$

then (3.6) (for constant T) simplifies to

$$\begin{equation} \mathbf{T}-\mu_1 (\mathbf{E}\kern-.15em\circ\kern-.15em\mathbf{T}+\mathbf{T}\kern-.15em\circ\kern-.15em\mathbf{E}) =2\nu\mathbf{E} . \end{equation} \tag{ 3.11 }$$

Suppose that

$$\begin{equation*} \mathbf{T}=\begin{pmatrix} a & b\\ b & c \end{pmatrix}, \end{equation*}$$

for unknown constants a, b, c. Then

$$\begin{equation*} \mathbf{E}\kern-.15em\circ\kern-.15em\mathbf{T}+\mathbf{T}\kern-.15em\circ\kern-.15em\mathbf{E}= U\begin{pmatrix} a & b \\ -b & - c \end{pmatrix} + U\begin{pmatrix} a & -b \\ b & - c \end{pmatrix} = 2U\begin{pmatrix} a & 0 \\ 0 & - c \end{pmatrix} . \end{equation*}$$

From (3.8) and (3.11), we conclude that

$$\begin{equation*} \begin{pmatrix} a & b\\ b & c \end{pmatrix}-2\mu_1 U \begin{pmatrix} a & 0 \\ 0 & - c \end{pmatrix} = 2\nu U \begin{pmatrix} 1 & 0 \\ 0 & - 1 \end{pmatrix} . \end{equation*}$$

Thus b = 0 and

$$\begin{equation} a=\frac{2\nu U }{1-2\mu_1 U }, \qquad c=-\frac{2\nu U }{1+2\mu_1 U }. \end{equation} \tag{ 3.12 }$$

We can simplify this using the expressions

$$\begin{equation} \frac{1}{1-x}=1+\frac{x}{1-x},\qquad \frac{1}{1+x}=1-\frac{x}{1+x}, \end{equation} \tag{ 3.13 }$$

with $x = 2\mu_1 U$. Thus there is a solution with constant T given by

$$\begin{align} \mathbf{T}&=2\nu U \begin{pmatrix} \frac{1 }{1-2\mu_1 U} & 0 \\ 0 & -\frac{1}{1+2\mu_1 U} \end{pmatrix}\nonumber\\ &=2\nu U \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} +4\nu \mu_1 U^{\,2} \begin{pmatrix} \frac{1}{1-2\mu_1 U } & 0 \\ 0 & \frac{1}{1+2\mu_1 U } \end{pmatrix} \end{align} \tag{ 3.14 }$$

for extensional flow $\mathbf{u}(x,y) = U\,(x,-y)^t$. The first part of the stress T is the Newtonian stress

$$\begin{equation*} \mathbf{T}_\mathrm{N} =2\nu U \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \end{equation*}$$

and the second part we can think of as the Oldroydian part:

$$\begin{equation*} \mathbf{T}_\mathrm{O} =4\nu \mu_1 U^{\,2} \begin{pmatrix} \frac{1}{1-2\mu_1 U } & 0 \\ 0 & \frac{1}{1+2\mu_1 U } \end{pmatrix}. \end{equation*}$$

There is a well known singularity (Owens and Phillips 2002) for flow at speed

$$\begin{equation} U=1/2|\mu_1|. \end{equation} \tag{ 3.15 }$$

For $\mu_1\gt0$, the coefficient a becomes singular for this value of U, and if µ₁ were negative, the coefficient c in (3.12) would be the one to indicate a singularity, instead of a. The limitation U on µ₁ can be viewed as a defect, but it can also be viewed as just a feature of the model. That is, it says that there is a limit on the size of $|\mu_1|$ related to the range of appropriate flow speeds being modeled.

By considering extensional flow for the Oldroyd model, we have learned that

• the parameter µ₁ can be very small and yet have a big effect, and
• the Oldroyd stress in extension does not depend directly on λ₁.

We can imagine an extensional rheometer where we measure the normal stress on the outlet of the channel

$$\begin{equation} \int_0^1 \mathbf{n}^t\mathbf{T}\mathbf{n}(L,y)-p(L,y)\,\mathrm{d}y, \end{equation} \tag{ 3.16 }$$

where $\mathbf{n} = (1,0)^t$ and L indicates the end of the channel. Physically, we could put a membrane over the outlet and measure its deformation, at least for small deformations. The integral of the normal stress gives the force on the membrane.

If we are mainly interested in how (3.16) differs from Newtonian flow, this simplifies to be

$$\begin{equation} \frac{4\nu U^{\,2} \mu_1}{1-2\mu_1 U}, \end{equation} \tag{ 3.17 }$$

since the pressure is the same for Newtonian flow. Thus the Oldroyd fluid can be shear-thickening or shear-thinning depending on the sign of µ₁.

The force F measured in the extensional flow rheometer, that is the Newtonian component together with the non-Newtonian force in (3.17), is

$$\begin{equation} F(U)=2\nu U+ U^{\,2} \left(\frac{4\nu \mu_1}{1-2\mu_1 U}-c_p\right), \end{equation} \tag{ 3.18 }$$

where (3.9) implies that

$$\begin{equation} c_p=U^{-2}\int_0^1 p(L,y)\,\mathrm{d}y =\int_0^1 (L^{\,2}+y^{\,2})\,\mathrm{d}y. \end{equation} \tag{ 3.19 }$$

We claim that plotting (3.18) as a function of U allows identification of ν and µ₁, although it would give no information on λ₁. We will demonstrate this as follows.

We can measure the slope $s(U) = F^{\,^{\prime}}(U)$ from the plot of (3.18) as a function of U to get

$$\begin{equation} s(U)=2\nu + 4U \left(\frac{2\nu \mu_1}{1-2\mu_1 U}-c_p\right) + U^{\,2} \left(\frac{8\nu \mu_1^{\,2}}{(1-2\mu_1 U)^{\,2}}\right). \end{equation} \tag{ 3.20 }$$

We have $2\nu = \lim_{U\to 0} s(U) = s(0)$. In particular, this shows that the extensional rheometer can measure the viscosity. Moreover,

$$\begin{equation} \begin{split} s(U)/s(0)&=1 + \frac{4\mu_1 U}{1-2\mu_1 U}-4U c_p/s(0) + \frac{(2 \mu_1 U)^{\,2}}{(1-2\mu_1 U)^{\,2}} \\ &=1 +2f(2\mu_1 U) -4U c_p/s(0)+ f(2\mu_1 U)^{\,2}, \end{split} \end{equation} \tag{ 3.21 }$$

where

$$\begin{equation} f(x)=x/(1-x). \end{equation} \tag{ 3.22 }$$

Differentiating (3.21) with respect to U, we find

$$\begin{equation*} s^{\,^{\prime}}(U)/s(0)=-4c_p/s(0)+4 \mu_1 f^{\,^{\prime}}(2\mu_1 U)\big(1+ f(2\mu_1 U)\big). \end{equation*}$$

Thus

$$\begin{equation} s^{\,^{\prime}}(0)/s(0)=-4c_p/s(0)+4 \mu_1 f^{\,^{\prime}}(0)=-4c_p/s(0)+4 \mu_1, \end{equation} \tag{ 3.23 }$$

so that $c_p/s(0) = \mu_1-{\textstyle{\frac{1}{4}}} s^{\,^{\prime}}(0)/s(0)$. Therefore

$$\begin{align} s(U)/s(0) &=1 -2U\big(\mu_1-{\textstyle{\frac{1}{4}}} s^{\,^{\prime}}(0)/s(0)\big) +2f(2\mu_1 U)+ f(2\mu_1 U)^{\,2}, \nonumber \\ s^{\,^{\prime}}(U)/s(0) - s^{\,^{\prime}}(0)/s(0) &=4\mu_1 \Big(-1+ f^{\,^{\prime}}(x)\big(1+ f(x)\big)\Big), x=2\mu_1 U. \end{align} \tag{ 3.24 }$$

But

$$\begin{align} g(x)&=f^{\,^{\prime}}(x)\big(1+ f(x)\big)-1=\frac{1}{(1-x)^{\,2}}\frac{1}{1-x}-1\nonumber\\ &=\frac{1}{(1-x)^{\,3}}-1 =\frac{3x-3x^{\,2}+x^{\,3}}{(1-x)^{\,3}}. \end{align} \tag{ 3.25 }$$

Thus

$$\begin{equation} s^{\,^{\prime}}(U)/s(0) - s^{\,^{\prime}}(0)/s(0) =4\mu_1 g(2\mu_1 U) =4\mu_1 g(x),\quad x=2\mu_1 U. \end{equation} \tag{ 3.26 }$$

Moreover,

$$\begin{equation} g^{\,^{\prime}}(x)=\big((1-x)^{-3}-1\big)^{\prime} =3(1-x)^{-4}>0,\quad 0\leqslant x< 1, \end{equation} \tag{ 3.27 }$$

so g is monotonically increasing. From (3.26)

$$\begin{equation} \frac{U\big(s^{\,^{\prime}}(U)- s^{\,^{\prime}}(0)\big)}{s(0)} =4\mu_1 U g(2\mu_1 U) = 2 x g(x),\quad x=2\mu_1 U. \end{equation} \tag{ 3.28 }$$

Since g is monotonically increasing, the function φ defined by

$$\begin{equation*} \phi(x)= 2 x g(x) \end{equation*}$$

is also monotonically increasing for $0\leqslant x\lt1$. Thus we can write

$$\begin{equation} \mu_1=\frac{1}{2U} \phi^{-1}\bigg( \frac{U\big(s^{\,^{\prime}}(U)- s^{\,^{\prime}}(0)\big)}{s(0)} \bigg). \end{equation} \tag{ 3.29 }$$

Note that solving $\phi(x) = d$ can easily be done via Newton's method. Both the slope and its derivative can be obtained reliably from the data via the Savitsky–Golay algorithm (Scott and Scott 1989). The value of µ₁ can thus be reliably estimated by averaging (3.29) over a suitable range of U values. A plot of ${\textstyle{\frac{1}{2}}}\phi$ is in figure 2.

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To get a more precise understanding of the behavior of µ₁ on the data, we can also write

$$\begin{equation*} x g(x) =x\frac{3x-3x^{\,2}+x^{\,3}}{(1-x)^{\,3}} = x^{\,2} q(x),\quad \hbox{where}\; q(x) =\frac{3-3x+x^{\,2}}{(1-x)^{\,3}} . \end{equation*}$$

Then

$$\begin{equation} \begin{split} q^{\,^{\prime}}(x)&=\frac{(-3+2x)(1-x)^{\,3} +3 \big(3-3x+x^{\,2}\big)(1-x)^{\,2}}{(1-x)^6} \\ &=\frac{(-3+2x)(1-x) +3\big(3-3x+x^{\,2}\big)}{(1-x)^4} \\ &=\frac{(-3+5x-2x^{\,2})+ \big(9-9x+3x^{\,2}\big)}{(1-x)^4} \\ &=\frac{6- 4x +x^{\,2}}{(1-x)^4} =\frac{2+(x-2)^{\,2}}{(1-x)^4} \geqslant 3(1-x)^{-4}, \;0\leqslant x < 1. \end{split} \end{equation} \tag{ 3.30 }$$

Therefore, q is strictly increasing for $0\leqslant x\lt1$. For U small, we have

$$\begin{equation} f(x)=xg(x)=x^{\,2} q(x)=\frac{U\big(s^{\,^{\prime}}(U)- s^{\,^{\prime}}(0)\big)}{s(0)} \approx U^{\,2} \frac{s^{\,^{\prime\prime}}(0)}{s(0)}, \end{equation} \tag{ 3.31 }$$

so that

$$\begin{equation*} q(0)x^{\,2} = 3 x^{\,2} \approx U^{\,2} \frac{s^{\,^{\prime\prime}}(0)}{s(0)}, \end{equation*}$$

and hence

$$\begin{equation} x \approx U \sqrt{\frac{s^{\,^{\prime\prime}}(0)}{3s(0)}}\implies \mu_1\approx \sqrt{\frac{s^{\,^{\prime\prime}}(0)}{12 s(0)}}. \end{equation} \tag{ 3.32 }$$

We do not need to differentiate the slope of the data twice to compute µ₁ via the formula (3.29). Inverting the function φ can be done numerically to high accuracy via Newton's method, without relying on any smoothness of the data.

We consider two-dimensional flow in a channel ${\Omega} = [-L,L]\times [0,1]$. Couette, or shear, flow, has $\mathbf{u} = U(y,0)^t$ and p constant if T is constant as shown in figure 3. Now we consider (3.7). We have

$$\begin{align} \nabla\mathbf{u} &=\begin{pmatrix} 0 & U \\ 0 & 0 \end{pmatrix},\quad \mathbf{E}=\frac12\begin{pmatrix} 0 & U \\ U & 0 \end{pmatrix},\quad \mathbf{R}=\frac12\begin{pmatrix} 0 & -U\\ U & 0 \end{pmatrix},\quad\nonumber\\ \mathbf{R}^t &=\frac12\begin{pmatrix} 0 & U\\ -U & 0 \end{pmatrix}, \end{align} \tag{ 3.33 }$$

and suppose that

$$\begin{equation*} \mathbf{T}=\begin{pmatrix} a & b\\ b & c \end{pmatrix}. \end{equation*}$$

Then

$$\begin{equation*} \mathbf{E}\kern-.15em\circ\kern-.15em\mathbf{T}+\mathbf{T}\kern-.15em\circ\kern-.15em\mathbf{E}= \frac{U}{2}\begin{pmatrix} b & c \\ a & b \end{pmatrix} + \frac{U}{2}\begin{pmatrix} b & a \\ c & b \end{pmatrix} = U\begin{pmatrix} b & {\textstyle{\frac{1}{2}}}(a+c) \\ {\textstyle{\frac{1}{2}}}(a+c) & b \end{pmatrix} . \end{equation*}$$

Similarly,

$$\begin{equation*} \mathbf{R}\kern-.15em\circ\kern-.15em\mathbf{T} + \mathbf{T}\kern-.15em\circ\kern-.15em\mathbf{R}^t = \frac{U}{2} \begin{pmatrix} -b & -c\\ a & b \end{pmatrix}+ \frac{U}{2} \begin{pmatrix} -b & a\\-c & b \end{pmatrix}= U \begin{pmatrix} -b & {\textstyle{\frac{1}{2}}}(a-c)\\ {\textstyle{\frac{1}{2}}}(a-c) & b \end{pmatrix}. \end{equation*}$$

Therefore

$$\begin{align} &\mathbf{T}+\lambda_1\big(\mathbf{R}\kern-.15em\circ\kern-.15em\mathbf{T}+\mathbf{T}\kern-.15em\circ\kern-.15em\mathbf{R}^t\big) -\mu_1 (\mathbf{E}\kern-.15em\circ\kern-.15em\mathbf{T}+\mathbf{T}\kern-.15em\circ\kern-.15em\mathbf{E})\nonumber \\ &\quad=\begin{pmatrix} a-U(\lambda_1+\mu_1)b & b+{\textstyle{\frac{1}{2}}} U\big(\lambda_1(a-c)-\mu_1(a+c)\big) \\ b+{\textstyle{\frac{1}{2}}} U\big(\lambda_1(a-c)-\mu_1(a+c)\big) & c+U(\lambda_1-\mu_1)b \end{pmatrix}. \nonumber \\ \end{align} \tag{ 3.34 }$$

Therefore the equation

$$\begin{equation*} \mathbf{T}-\lambda_1\big((\nabla\mathbf{u})\kern-.15em\circ\kern-.15em\mathbf{T}+\mathbf{T}\kern-.15em\circ\kern-.15em(\nabla\mathbf{u})^t\big) =2\nu\mathbf{E}= \nu \begin{pmatrix} 0 & U \\ U & 0 \end{pmatrix} \end{equation*}$$

reduces to

$$\begin{equation*} a=(\lambda_1+\mu_1) b U , \quad c=-(\lambda_1-\mu_1) b U , \quad a-c=2\lambda_1 b U , \quad a+c=2\mu_1 b U , \end{equation*}$$

so that

$$\begin{equation} \begin{split} \nu U &=b+{\textstyle{\frac{1}{2}}} \lambda_1 (a-c) U -{\textstyle{\frac{1}{2}}} \mu_1 (a+c)U \\ &=b\big(1+(U )^{\,2} (\lambda_1^{\,2} - \mu_1^{\,2} )\big) .\\ \end{split} \end{equation} \tag{ 3.35 }$$

Therefore

$$\begin{equation*} b=\frac{\nu U }{1+(U )^{\,2} (\lambda_1^{\,2} - \mu_1^{\,2} )}. \end{equation*}$$

Thus

$$\begin{equation*} a=(\lambda_1+\mu_1)b U =\frac{\nu(\lambda_1+\mu_1) U^{\,2}}{1+U^{\,2} (\lambda_1^{\,2} - \mu_1^{\,2} )},\quad c=\frac{-\nu(\lambda_1-\mu_1)U^{\,2}}{1+U^{\,2} (\lambda_1^{\,2} - \mu_1^{\,2} )}, \end{equation*}$$

and so

$$\begin{equation} \mathbf{T}= \frac{\nu}{1+U^{\,2} (\lambda_1^{\,2} - \mu_1^{\,2} )} \begin{pmatrix} (\lambda_1+\mu_1)U^{\,2} & U \\ U & -(\lambda_1-\mu_1)U^{\,2} \end{pmatrix}. \end{equation} \tag{ 3.36 }$$

Let us define the stress $\mathbf{T}_\mathrm{N}$ for Newtonian shear flow by

$$\begin{equation} \mathbf{T}_\mathrm{N}= \nu U \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \end{equation} \tag{ 3.37 }$$

and we can denote the Oldroyd stress (3.36) by $\mathbf{T}_\mathrm{O}$. Then

$$\begin{equation} \mathbf{T}_\mathrm{O}= \frac{1}{1+U^{\,2} (\lambda_1^{\,2} - \mu_1^{\,2} )}\mathbf{T}_\mathrm{N} +\frac{\nu U^{\,2}}{1+U^{\,2} (\lambda_1^{\,2} - \mu_1^{\,2} )} \begin{pmatrix} \lambda_1+\mu_1 & 0 \\ 0 & -\lambda_1+\mu_1 \end{pmatrix}. \end{equation} \tag{ 3.38 }$$

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We can imagine a rheometer based on shear flow, as follows. We have a rotating belt at the top of the channel enforcing the velocity $U(1,0)$. If we did not constrain the bottom of the channel, the whole apparatus would move to the right. So we measure the force required to keep it in place. This force must be balanced by the fluid shear stress

$$\begin{equation} \int_{-L}^L \mathbf{n}^t\mathbf{T}_\mathrm{O}\mathbf{t}(x,0)\,\mathrm{d}x. \end{equation} \tag{ 3.39 }$$

Here, $\mathbf{n} = (0,-1)$ is the outward normal to the bottom of the channel, and $\mathbf{t} = (1,0)$ is tangent to the bottom. But

$$\begin{equation} \begin{pmatrix} \lambda_1+\mu_1 & 0 \\ 0 & -\lambda_1+\mu_1 \end{pmatrix} \begin{pmatrix} 1\\0\end{pmatrix} = \begin{pmatrix} \lambda_1+\mu_1\\0\end{pmatrix} \end{equation} \tag{ 3.40 }$$

so that

$$\begin{equation} \mathbf{n}^t\begin{pmatrix}\lambda_1+\mu_1 & 0\\ 0 & -\lambda_1+\mu_1 \end{pmatrix} \mathbf{t}=0 . \end{equation} \tag{ 3.41 }$$

Thus the force measured for an Oldroyd fluid is the same as that measured for a Newtonian fluid, multiplied by

$$\begin{equation*} \frac{1}{1+U^{\,2} (\lambda_1^{\,2} - \mu_1^{\,2} )}. \end{equation*}$$

Plotting

$$\begin{equation*} f(U)=\frac{2\nu U}{1+U^{\,2} (\lambda_1^{\,2} - \mu_1^{\,2} )} \end{equation*}$$

as a function of U would allow identification of both ν and $\lambda_1^{\,2}-\mu_1^{\,2}$, but it would not determine λ₁ and µ₁ separately. But if we combine this with the results of the extensional rheometer, we can recover λ₁, although the sign would be ambiguous. In particular,

$$\begin{equation*} f(U)+U^{\,2} \lambda_1^{\,2} f(U)=U^{\,2}\mu_1^{\,2} f(U)+2\nu U. \end{equation*}$$

Thus

$$\begin{equation*} \lambda_1^{\,2}=\frac{U^{\,2}\mu_1^{\,2} f(U)+2\nu U-f(U)}{U^{\,2} f(U)} =\mu_1^{\,2} + \frac{2\nu U-f(U)}{U^{\,2} f(U)}. \end{equation*}$$

Note that for small U,

$$\begin{equation*} f(U)\approx 2\nu U\Big(1-U^{\,2}(\lambda_1^{\,2} -\mu_1^{\,2}) +\big(U^{\,2}(\lambda_1^{\,2}-\mu_1^{\,2})\big)^{\,2}+{{\cal O}(U^6)}\Big), \end{equation*}$$

so that

$$\begin{equation} \begin{split} \frac{2\nu U-f(U)}{U^{\,2} f(U)} &\approx \frac{2\nu U^{\,3}(\lambda_1^{\,2}-\mu_1^{\,2})+2\nu U^5(\lambda_1^{\,2}-\mu_1^{\,2})^{\,2}+{{\cal O}(U^7)}}{2\nu U^{\,3} +{{\cal O}(U^5)}} \\ &\approx \lambda_1^{\,2}-\mu_1^{\,2}+ \frac{U^{\,2}(\lambda_1^{\,2}-\mu_1^{\,2})^{\,2}+{{\cal O}(U^4)}}{1 +{{\cal O}(U^{\,2})}} \\ \end{split} \end{equation} \tag{ 3.42 }$$

remains bounded as U → 0.

We have shown that a shear-flow rheometer can be used to determine both ν and the combination $\lambda_1^{\,2}-\mu_1^{\,2}$, whereas the extensional-flow rheometer can be used to determine both ν and µ₁, but not λ₁. Combining measurements from these two rheometers allows the determination of all three coefficients, and it includes a cross check on the viscosity parameter ν.

We saw in section 3.2 that extensional flow, given by

$$\begin{equation*} \mathbf{u}(x,y)=U (x,-y)^t\quad\hbox{and}\quad p(x,y)=\frac{U^{\,2}}{2}\big(x^{\,2}+y^{\,2}\big), \end{equation*}$$

is a solution of the Navier–Stokes equations (3.10) in the domain depicted in figure 1. Thus we also have a solution of the grade-two model in that domain provided that

$$\begin{equation} \alpha_1\big(\mathbf{u}\cdot\nabla\mathbf{A}+\mathbf{A}\kern-.15em\circ\kern-.15em(\nabla\mathbf{u})+(\nabla\mathbf{u})^t\kern-.15em\circ\kern-.15em\mathbf{A}\big) + \alpha_2 \mathbf{A}\kern-.15em\circ\kern-.15em\mathbf{A} \end{equation} \tag{ 4.47 }$$

is constant. Since $\nabla\mathbf{u}$ is symmetric in this case, we have

$$\begin{equation*} \mathbf{A}=2\nabla\mathbf{u}=2U\begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}, \end{equation*}$$

and thus $\mathbf{u}\cdot\nabla\mathbf{A} = \mathbf{0}$ and

$$\begin{equation*} 2\mathbf{A}\nabla\mathbf{u} =2\big(\nabla\mathbf{u}^t\mathbf{A}^t\big)^t =2\nabla\mathbf{u}^t\mathbf{A} =\mathbf{A}^{\,2} =4U^{\,2}\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix}=4U^{\,2}{\mathcal I}, \end{equation*}$$

where ${\mathcal I}$ is the identity matrix. Thus the expression in (4.47) is constant. Then

$$\begin{equation} \begin{split} \mathbf{T}&=\nu \mathbf{A} + \alpha_1\big(\mathbf{A}\kern-.15em\circ\kern-.15em(\nabla\mathbf{u})+(\nabla\mathbf{u})^t\kern-.15em\circ\kern-.15em\mathbf{A}\big) + \alpha_2 \mathbf{A}\kern-.15em\circ\kern-.15em\mathbf{A}\\ &=2\nu U \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} +4(\alpha_1 +\alpha_2) U^{\,2} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}. \end{split} \end{equation} \tag{ 4.48 }$$

The extensional flow rheometer would report a difference from Newtonian flow for the normal stress at the outlet given by

$$\begin{equation*} \mathbf{T}_\mathrm{G}=4(\alpha_1 +\alpha_2) U^{\,2}. \end{equation*}$$

Thus the grade-two fluid can be shear-thickening or shear-thinning depending on the sign of $\alpha_1 +\alpha_2$.

The force measured will be a combination of ν and $\alpha_1 +\alpha_2$, namely

$$\begin{equation*} 2\nu U+ U^{\,2}\big(4(\alpha_1 +\alpha_2)-c_p\big), \end{equation*}$$

where c_p is defined in (3.19). By plotting the measured force against U, it is possible to determine both ν and $\alpha_1 +\alpha_2$.

We saw in section 3.5 that $\mathbf{u} = U(y,0)^t$ and p is constant if T is constant. We have

$$\begin{align*}\begin{split} \nabla\mathbf{u}&=\begin{pmatrix} 0 & U \\ 0 & 0 \end{pmatrix},\; \mathbf{A}=\begin{pmatrix} 0 & U \\ U & 0 \end{pmatrix},\; \mathbf{A}\kern-.15em\circ\kern-.15em\mathbf{A}=\begin{pmatrix} U^{\,2} & 0\\ 0&U^{\,2} \end{pmatrix},\;\\ \mathbf{A}\kern-.15em\circ\kern-.15em\nabla\mathbf{u} &= \nabla\mathbf{u}^t\kern-.15em\circ\kern-.15em\mathbf{A}= \begin{pmatrix} 0 & 0 \\ 0 & U^{\,2} \end{pmatrix}.\end{split} \end{align*}$$

Thus (4.44) implies that

$$\begin{equation} \mathbf{T}=\nu U \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} +2\alpha_1 U^{\,2} \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} +\alpha_2 U^{\,2} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}.\end{equation} \tag{ 4.49 }$$

Here we have $\mathbf{T} = \mathbf{T}_\mathrm{N}+\mathbf{T}_\mathrm{G}$ where

$$\begin{equation*} \mathbf{T}_\mathrm{N}= \nu U \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix},\qquad \mathbf{T}_\mathrm{G}= 2\alpha_1 U^{\,2} \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} +\alpha_2 U^{\,2} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}.\end{equation*}$$

The shear-flow rheometer described in section 3.6 for the grade-two model will measure the same quantity as in (3.39) but with $\mathbf{T}_{{\text{O}}}$ replaced by $\mathbf{T}_\mathrm{G} = \mathbf{T}-\mathbf{T}_\mathrm{N}$. And like the Oldroyd model, $\mathbf{n}\mathbf{T}_\mathrm{G}\mathbf{t} = 0$, so the simple shear rheometer will give the same result as for a Newtonian fluid, independent of the parameters α_i . Thus a pure shear-flow rheometer will report ν and a pure extensional flow rheometer will report a combination of ν and $\alpha_1+\alpha_2$. Combining the two rheometers, we can determine $\alpha_1+\alpha_2$, but it does not seem possible to determine α₁ and α₂ separately with these two geometries alone.

We have seen that the simple shear-flow rheometer described here does not distinguish the coefficients α₁ and α₂ in the grade-two model, despite the fact that the induced stress difference $\mathbf{T}_\mathrm{G}$ is quite complex in shear. If we take the definition (2.4) for apparent shear, then the grade-two model is shear thinning when α₂ is sufficiently negative, and otherwise it is shear thickening (assuming $\alpha_1\gt0$ in both cases).

For extensional flow (section 3.1), the Newtonian stress $\mathbf{T}_\mathrm{N}$ and Oldroyd stress $\mathbf{T}^\mathrm{O}$ are given in (3.14):

$$\begin{equation} \mathbf{T}_\mathrm{N} =2\nu U \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}, \qquad \mathbf{T}^\mathrm{O}=\mathbf{T}_\mathrm{N}+ 4\nu \mu_1 U^{\,2} \begin{pmatrix} \frac{1}{1-2\mu_1 U } & 0 \\ 0 & \frac{1}{1+2\mu_1 U } \end{pmatrix}. \end{equation} \tag{ 5.52 }$$

Thus $\mathbf{T}^\mathrm{O}-\mathbf{T}_\mathrm{N} = {\cal O}(\mu_1)$. By contrast, the grade-two stress $\mathbf{T}^\mathrm{G}$ is given in (4.48) as

$$\begin{equation} \mathbf{T}^\mathrm{G}=\mathbf{T}_\mathrm{N}+4(\alpha_1 +\alpha_2) U^{\,2} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} =\mathbf{T}_\mathrm{N}+4\nu\mu_1 U^{\,2} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \end{equation} \tag{ 5.53 }$$

where we have invoked (5.50). Thus we see that $\mathbf{T}^\mathrm{G}-\mathbf{T}_\mathrm{N} = {{\cal O}(\mu_1)}$. But more importantly, we can use (3.13) to show that

$$\begin{equation} \begin{split} \mathbf{T}^\mathrm{O}-\mathbf{T}^G&=4\nu \mu_1 U^{\,2} \Bigg( \begin{pmatrix} \frac{1}{1-2\mu_1 U} & 0 \\ 0 & \frac{1}{1+2\mu_1 U} \end{pmatrix} - \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\Bigg) \\ &=8\nu \mu_1^{\,2} U^{\,3} \begin{pmatrix} \frac{1}{1-2\mu_1 U} & 0 \\ 0 & \frac{-1}{1+2\mu_1 U} \end{pmatrix} ={{\cal O}(\mu_1^{\,2})}, \end{split} \end{equation} \tag{ 5.54 }$$

in agreement with Tanner duality. On the other hand, $\mathbf{T}^\mathrm{O}$ and $\mathbf{T}^\mathrm{G}$ diverge rapidly as $U\to 1/2|\mu_1|$ in accordance with (3.15). Note that $\mathbf{T}^\mathrm{G}$ displays no singularity in this limit.

For shear flow (section 3.5), the Newtonian stress $\mathbf{T}_\mathrm{N}$ and Oldroyd stress T^O are given in (3.38):

$$\begin{equation} \begin{split} \mathbf{T}_\mathrm{N}&= \nu U \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix},\\ \mathbf{T}^O&= \frac{1}{1+U^{\,2} (\lambda_1^{\,2} - \mu_1^{\,2} )}\mathbf{T}_\mathrm{N} +\frac{\nu U^{\,2}}{1+U^{\,2} (\lambda_1^{\,2} - \mu_1^{\,2} )} \begin{pmatrix} \lambda_1+\mu_1 & 0 \\ 0 & -\lambda_1+\mu_1 \end{pmatrix}. \end{split} \end{equation} \tag{ 5.55 }$$

Therefore

$$\begin{equation*} \big(1+U^{\,2} (\lambda_1^{\,2} - \mu_1^{\,2} )\big)\mathbf{T}^\mathrm{O}=\mathbf{T}_\mathrm{N}+ {\nu U^{\,2}} \lambda_1 \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}+ {\nu U^{\,2}} \mu_1\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}. \end{equation*}$$

By contrast, the grade-two stress $\mathbf{T}^\mathrm{G}$ is given in (4.49) as

$$\begin{equation} \begin{split} \mathbf{T}^\mathrm{G} &= \mathbf{T}_\mathrm{N} +2\alpha_1 U^{\,2} \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} + \alpha_2 U^{\,2} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\\ &= \mathbf{T}_\mathrm{N} +U^{\,2} \begin{pmatrix}\alpha_2 & 0\\ 0 & 2\alpha_1+ \alpha_2 \end{pmatrix} \\ &= \mathbf{T}_\mathrm{N} -\alpha_1 U^{\,2} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} + (\alpha_1+\alpha_2) U^{\,2} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}. \end{split} \end{equation} \tag{ 5.56 }$$

Therefore invoking (5.50), we see that

$$\begin{align} \big(1+U^{\,2} (\lambda_1^{\,2} - \mu_1^{\,2} )\big)\mathbf{T}^\mathrm{O}=\mathbf{T}^\mathrm{G}.\end{align} \tag{ 5.57 }$$

Thus

$$\begin{equation} \mathbf{T}^\mathrm{O}-\mathbf{T}^\mathrm{G}=U^{\,2} (\lambda_1^{\,2} - \mu_1^{\,2} )\mathbf{T}^\mathrm{O}, \end{equation} \tag{ 5.58 }$$

again confirming Tanner duality.

Tanner duality allows us to pick models appropriately that match certain experiments. Let us re-examine shear flow and postulate a rheometer that measures the normal force on the top and bottom of the domain. Such a theoretical rheometer shares some features with the Lodge rheometer (Lodge et al 1991).

Let us simplify to the case where $\mu_1 = \alpha_1+\alpha_2 = 0$. The Newtonian stress contributes nothing to the normal force, and the pressure is constant in shear flow, so the normal force is proportional to λ₁ for the Oldroyd model and $-\alpha_1$ for the grade-two model, where the constants of proportionality have the same sign. We can imagine two complementary materials A and B, one that pushes out on the channel walls and one that pulls in, as the fluids are sheared. Both types of materials are known in solid mechanics, so one cannot say a priori either is unrealistic. Then picking the Oldroyd model for material A might require $\lambda_1\lt0$, which has several drawbacks from a modeling perspective. Correspondingly, picking the grade-two model for material B would require $\alpha_1\lt0$, with its own set of drawbacks. However, picking the Oldroyd model for material B would have $\lambda_1\gt0$ due to Tanner duality, so it would be the preferred model for simulation. Similarly, picking the grade-two model for material A would have $\alpha_1\gt0$ due to Tanner duality, so it would be the preferred model for simulation.

A rheometer that emphasizes extensional flow is based on a contraction nozzle (Nyström et al 2017). Fluid is forced through the contraction either by pressure or a rod. The flow domain is defined by

$$\begin{equation*} \Omega=\left\lbrace (x,y,z) \; \big| \; b\leqslant z\leqslant t,\; x^{\,2}+y^{\,2}\leqslant f\left(x^{\,2}+y^{\,2}\right) \right\rbrace \end{equation*}$$

for a given function f.

Since the two theoretical rheometers fail to detect α₁ for the grade-two fluid, a natural question is whether or not a contraction rheometer can do so. Since the extensional flow rheometer detects $\alpha_1+\alpha_2$, a natural approach is the consider the special case $\alpha_1+\alpha_2 = 0$, for which the grade-two model simplifies (Girault and Scott 1999, 2001). Similarly, it makes sense to consider a two-dimensional contraction as a first step, again simplifying the grade-two model. Such a geometry is considered in Pollock and Scott (2022b).

One common type of rheometer is based on counter-rotating concentric cylinders, essentially the original experiment of Couette (Gallot et al 2013). What is measured is the torque on one cylinder induced by the other cylinder. For example, the outer cylinder is rotated at a fixed speed and the torque on the inner cylinder required to keep it stationary is recorded. This measures quantities similar to our pure shear rheometer, so we do not pursue this further here.

There are two different, but related rheometers that involve a plate below and either (a) a cone above or (b) a parallel plate above. For example, the lower plate could be fixed and the upper structure rotated. Thus one obtains the cone-and-plate rheometer (Markovitz et al 1955, Ellenberger and Fortuin 1985) and the parallel-plate rheometer (Yamamoto 1958, Ellenberger and Fortuin 1985). In both cases, the geometry and (steady) flow have radial symmetry.

Lodge (1996) constructed a device to measure normal stress differences. The flow in a two-dimensional version of this device has been proposed as a test problem (Lodge et al 1991). Such approaches have been used effectively in rheological measurements in food technology (Padmanabhan and Bhattacharya 1993).

In Robert et al (1985), two-dimensional computational simulations of journal bearings were used to evaluate six different rheological models.

We first need to resolve the matrix on the right in (3.14). Using (7.60), we have

$$\begin{equation} \begin{split} \begin{pmatrix} \frac{1}{1-2\mu_1 U} &0\\ 0&\frac{1}{1+2\mu_1 U} \end{pmatrix}&= \frac12\bigg(\frac{1}{1-2\mu_1 U} + \frac{1}{1+2\mu_1 U}\bigg)\mathcal{I} \\ &\quad+ \frac12\bigg(\frac{1}{1-2\mu_1 U} - \frac{1}{1+2\mu_1 U}\bigg)\mathcal{J}\\ &=\frac{1}{1-(2\mu_1 U)^{\,2}} \big(\mathcal{I}+ 2\mu_1 U \mathcal{J}\big). \end{split} \end{equation} \tag{ 7.62 }$$

Therefore

$$\begin{equation*} \mathbf{T}_\mathrm{N}=2\nu U \mathcal{J},\qquad \mathbf{T}=\mathbf{T}_\mathrm{N}+\frac{4\nu\mu_1 U^{\,2}}{1-(2\mu_1 U)^{\,2}} \big(\mathcal{I}+ 2\mu_1 U \mathcal{J}\big). \end{equation*}$$

Thus

$$\begin{equation} \begin{split} \Vert\, \mathbf{T} \,\Vert_F^{\,2}&=\bigg(2\nu U+ \frac{8\nu\mu_1^{\,2} U^{\,3}}{1-(2\mu_1 U)^{\,2}} \bigg)^{\,2} +\bigg(\frac{4\nu\mu_1 U^{\,2}}{1-(2\mu_1 U)^{\,2}} \bigg)^{\,2}\\ &=(2\nu U)^{\,2}\bigg(1+ \frac{4\mu_1^{\,2} U^{\,2}}{1-(2\mu_1 U)^{\,2}} \bigg)^{\,2} +(2\nu U)^{\,2} \bigg(\frac{2\mu_1 U}{1-(2\mu_1 U)^{\,2}} \bigg)^{\,2}\\ &=(2\nu U)^{\,2}\bigg(\frac{1}{1-(2\mu_1 U)^{\,2}} \bigg)^{\,2} +(2\nu U)^{\,2} \bigg(\frac{2\mu_1 U}{1-(2\mu_1 U)^{\,2}} \bigg)^{\,2}\\ &=\frac{(2\nu U)^{\,2}}{\big(1-(2\mu_1 U)^{\,2}\big)^{\,2}} \big(1+(2\mu_1 U)^{\,2}\big). \end{split} \end{equation} \tag{ 7.63 }$$

The strain (3.8) for extensional flow is proportional to U, so the apparent viscosity (2.4) tends to infinity as $U\to 1/2|\mu_1|$, and this would be described as extensional thickening.

From (3.38), we find

$$\begin{equation*} \mathbf{T}_\mathrm{N}=\nu U \mathcal{K},\ \ \mathbf{T}=\frac{1}{1+U^{\,2}(\lambda_1^{\,2}-\mu_1^{\,2})}\Big( \mathbf{T}_\mathrm{N} +\nu U^{\,2}\big(\lambda_1 \mathcal{J} + \mu_1\mathcal{I}\big)\Big). \end{equation*}$$

If $\lambda_1 = \mu_1$, this simplifies to

$$\begin{equation*} \mathbf{T}= \mathbf{T}_\mathrm{N} +\nu U^{\,2}\lambda_1 \big(\mathcal{J} + \mathcal{I}\big). \end{equation*}$$

In this special case,

$$\begin{equation*} \Vert\, \mathbf{T} \,\Vert_F^{\,2}=(\nu U)^{\,2}\big(1+2U^{\,2}\lambda_1^{\,2}\big), \end{equation*}$$

and asymptotically as $U\to\infty$,

$$\begin{equation*} \Vert\, \mathbf{T} \,\Vert_F\approx \nu U(1+\sqrt{2}U\lambda_1). \end{equation*}$$

In the general case,

$$\begin{equation*} \Vert\, \mathbf{T} \,\Vert_F^{\,2}=(\nu U)^{\,2} \frac{1+U^{\,2}\big(\lambda_1^{\,2}+\mu_1^{\,2})}{\big(1+U^{\,2}(\lambda_1^{\,2}-\mu_1^{\,2})\big)^{\,2}} = \nu^{\,2}\frac{U^{-2}+\big(\lambda_1^{\,2}+\mu_1^{\,2})}{\big(U^{-2}+(\lambda_1^{\,2}-\mu_1^{\,2})\big)^{\,2}} . \end{equation*}$$

If $\lambda_1\neq \pm\mu_1$, then

$$\begin{equation*} \Vert\, \mathbf{T} \,\Vert_F\to \nu\frac{\sqrt{\lambda_1^{\,2}+\mu_1^{\,2}}}{|\lambda_1^{\,2}-\mu_1^{\,2}|} . \end{equation*}$$

as $U\to\infty$. This would qualify as shear thinning, since (3.33) implies that the Frobenius norm of the strain is proportional to U.

From (4.48), we find

$$\begin{equation*} \mathbf{T}_\mathrm{N}=2\nu U \mathcal{J},\qquad \mathbf{T}=\mathbf{T}_\mathrm{N} + 4(\alpha_1+\alpha_2)U^{\,2}\mathcal{I}. \end{equation*}$$

Thus

$$\begin{equation*} \Vert\, \mathbf{T} \,\Vert_F^{\,2}=(2\nu U)^{\,2} + 16(\alpha_1+\alpha_2)^{\,2}U^4. \end{equation*}$$

Since the strain (3.8) for extensional flow is proportional to U, the apparent viscosity (2.4) grows monotonically as U increases, and this would be described as extensional thickening.

From (4.49), we find

$$\begin{equation*} \mathbf{T}_\mathrm{N}=\nu U \mathcal{K},\ \ \, \mathbf{T}=\mathbf{T}_\mathrm{N} + U^{\,2}\big(\alpha_1(\mathcal{I}-\mathcal{J})+\alpha_2\mathcal{I}\big) =\mathbf{T}_\mathrm{N} + U^{\,2}\big((\alpha_1+\alpha_2)\mathcal{I}-\alpha_1\mathcal{J}\big). \end{equation*}$$

Thus

$$\begin{equation*} \Vert\, \mathbf{T} \,\Vert_F^{\,2}=(\nu U)^{\,2} + (\alpha_1+\alpha_2)^{\,2} U^4 +\alpha_1^{\,2} U^4. \end{equation*}$$

Again, the strain (3.8) for shear flow is proportional to U, so the apparent viscosity (2.4) grows monotonically as U increases, and this would be described as shear thickening.

A norm is blunt instrument, but we saw in section 7.2 that it can identify shear thinning. But there may be a finer tool to analyze the impact of stress. The expression $\boldsymbol{\sigma}\mathbf{n}$ is the force on the plane perpendicular to the unit vector n due to the stress σ . Then $\mathbf{n}^t\boldsymbol{\sigma}\mathbf{n}$ is the force due to σ in the direction n, that is, against the surface represented by the plane. It t is another direction, then $\mathbf{t}^t\boldsymbol{\sigma}\mathbf{n}$ is the force due to σ in the direction t. One direction of interest would be a direction tangent to the plane.

In the two-dimensional case, we can use the decomposition (7.60) for stresses and take t to be orthogonal to n and consider the force magnitudes

$$\begin{equation*} \mathbf{n}^t\boldsymbol{\sigma}\mathbf{n},\qquad \mathbf{t}^t\boldsymbol{\sigma}\mathbf{n}. \end{equation*}$$

These represent the 'observables' for σ . Note that for symmetric σ , $\mathbf{t}^t\boldsymbol{\sigma}\mathbf{n} = \mathbf{n}^t\boldsymbol{\sigma}\mathbf{t}$.

To quantify this, let us assume that

$$\begin{equation*} \mathbf{n}=(\cos\theta,\sin\theta),\qquad \mathbf{t}=(-\sin\theta,\cos\theta). \end{equation*}$$

Then we have

$$\begin{equation*} \mathbf{n}^t\mathcal{I}\mathbf{n}=1,\qquad \mathbf{t}^t\mathcal{I}\mathbf{n}=0, \end{equation*}$$

$$\begin{equation*} \mathbf{n}^t\mathcal{J}\mathbf{n}=\cos^{\,2}\theta-\sin^{\,2}\theta=\cos 2\theta,\qquad \mathbf{t}^t\mathcal{J}\mathbf{n}=-2\cos\theta \sin\theta=-\sin 2\theta. \end{equation*}$$

$$\begin{equation*} \mathbf{n}^t\mathcal{K}\mathbf{n}=2\cos\theta \sin\theta=\sin 2\theta,\qquad \mathbf{t}^t\mathcal{K}\mathbf{n}=\cos^{\,2}\theta-\sin^{\,2}\theta=\cos 2\theta . \end{equation*}$$

Therefore

$$\begin{equation} \begin{pmatrix} \mathbf{n}^t(a\mathcal{I}+b\mathcal{J}+c\mathcal{K})\mathbf{n} \\ \mathbf{t}^t(a\mathcal{I}+b\mathcal{J}+c\mathcal{K})\mathbf{n} \end{pmatrix} = \begin{pmatrix} a \\ 0 \end{pmatrix}+ \begin{pmatrix} \cos 2\theta & \sin 2\theta \\ -\sin 2\theta & \cos 2\theta \end{pmatrix} \begin{pmatrix} b \\ c \end{pmatrix}. \end{equation} \tag{ 7.64 }$$

We recognize the matrix in (7.64) as a rotation by $-2\theta$.

As a first application, let us apply this methodology to Newtonian fluids. For shear flow, $\mathbf{T}_\mathrm{N} = \nu U\mathcal{K}$, and for extensional flow, $\mathbf{T}_\mathrm{N} = 2\nu U\mathcal{J}$. Thus

$$\begin{equation} \begin{split} \begin{pmatrix} \mathbf{n}^t\mathbf{T}_\mathrm{N}\mathbf{n} \\ \mathbf{t}^t\mathbf{T}_\mathrm{N}\mathbf{n} \end{pmatrix} &=\begin{pmatrix} \cos 2\theta & \sin 2\theta \\ -\sin 2\theta & \cos 2\theta \end{pmatrix} \begin{pmatrix} 2 \nu U \\ 0\end{pmatrix} \quad\hbox{(extensional flow)} \\ \begin{pmatrix} \mathbf{n}^t\mathbf{T}_\mathrm{N}\mathbf{n} \\ \mathbf{t}^t\mathbf{T}_\mathrm{N}\mathbf{n} \end{pmatrix} &=\begin{pmatrix} \cos 2\theta & \sin 2\theta \\ -\sin 2\theta & \cos 2\theta \end{pmatrix} \begin{pmatrix} 0 \\ \nu U \end{pmatrix} \quad\hbox{(shear flow).} \\ \end{split} \end{equation} \tag{ 7.65 }$$

For example, if we take θ = 0 (meaning the plane is perpendicular to the x-axis), then the normal force on this plane would be proportional to $2\nu U$ in extensional flow (with the shear stress zero), whereas the tangential force on this plane would be proportional to $\nu U$ in shear flow (with the normal stress zero).

As another application, we apply this methodology to grade-two shear flow. Then

$$\begin{equation} \begin{pmatrix} \mathbf{n}^t\mathbf{T}\mathbf{n} \\ \mathbf{t}^t\mathbf{T}\mathbf{n} \end{pmatrix} = \begin{pmatrix} U^{\,2}(\alpha_1+\alpha_2) \\ 0 \end{pmatrix}+ \begin{pmatrix} \cos 2\theta & \sin 2\theta \\ -\sin 2\theta & \cos 2\theta \end{pmatrix} \begin{pmatrix} -U^{\,2} \alpha_1 \\ \nu U \end{pmatrix}.\end{equation} \tag{ 7.66 }$$

For θ = 0, the normal force on the plane perpendicular to the x-axis is proportional to $U^{\,2} \alpha_2$. This could be measured with a net held initially in this plane, the deformation of the net being proportional to the force. Together with the measurement of $\alpha_1+\alpha_2$ via the extensional flow rheometer, α₁ can then be determined.