Sig Turbomachinery MRF Library

1 Continuity Equation

1.1 Mathematical Relations

The product rule is also valid with the Nabla-Operator and arbitrary products $\oplus$ and $\boxdot$. The arrows above indicates the appliance of the Nabla-Operator. $\nabla \oplus ( \overset{\downarrow}{ \Phi } \boxdot \overset{\downarrow}{ \Psi } ) = \nabla \oplus ( \overset{\downarrow}{ \Phi } \boxdot \Psi ) + \nabla \oplus ( \Phi \boxdot \overset{\downarrow}{ \Psi } ) \qquad (30)$

The total time derivation of any variable $\Phi$ can be split off into a local term and a convective term. In the absolute frame of reference it gives with the absolute velocity $\vec{c}$ $\frac{d \Phi}{d t} = \frac{\partial \Phi}{\partial t} + \vec{c} \bullet \nabla \Phi \qquad (3)$

And in the relative frame of reference it gives with the relative velocity $\vec{w}$ $\frac{d^\prime \Phi}{d t} = \frac{\partial^\prime \Phi}{\partial t} + \vec{w} \bullet \nabla^\prime \Phi \qquad (4)$

As for a scalar variable it is valid $\frac{d \alpha}{d t} = \frac{d^\prime \alpha}{d t} \qquad (1.4)$

The first time derivation of any vector $\vec{b}$ formulated in the absolute and relative frame of reference, according to , where $\vec{\omega}$ is the angular frequency between these two coordinate systems. Any quantity with an ' is related to the relative frame of reference. $\frac{d \vec{b}}{d t} = \frac{d^\prime \vec{b}}{d t}+\vec{\omega} \times \vec{b} \qquad (1)$

If $\nabla$ is a vector operator, then it is valid $\vec{\nabla} \equiv \vec{\nabla}^\prime \qquad (2)$

Consider, that the origin of the absolute and relative frame of reference is coincident. Then it is valid to say, that the position vector is equal in both systems $\vec{r} =\vec{r}^\prime$. But, be aware, the components of these vectors are not equal, because they are defined in different base systems. If we now use $\vec{b} =\vec{r}^\prime$ in eq. 1, it gives us the well known relation between the absolute and relative velocity. $\begin{array}{rcl} \dfrac{d \vec{r}^\prime}{d t} = \dfrac{d \vec{r}}{d t} & = & \dfrac{d^\prime \vec{r}^\prime}{d t} + \vec{\omega} \times \vec{r}^\prime \\ = \vec{c} & = & \vec{w} + \left( \vec{\omega} \times \vec{r}^\prime \right) = \vec{w} + \vec{u} \end{array}\qquad (14)$

If one replaces the left and right hand side of eq. 1.4 with the expressions from eq. 3 and eq. 4 one yields for a scalar variable $\frac{\partial \alpha}{\partial t} = \frac{\partial^\prime \alpha}{\partial t} - \left( \vec{c} - \vec{w} \right) \bullet \nabla \alpha$

If we insert eq. 2-4 into the 1, one can write for a vector variable $\frac{\partial\vec{b}}{\partial t} + \vec{c} \bullet \nabla \vec{b} = \frac{\partial^\prime\vec{b}}{\partial t} + \vec{w} \bullet \nabla \vec{b} + \vec{\omega} \times \vec{b} \qquad (5)$

With the relationship of eq. 14 this yields to $\frac{\partial\vec{b}}{\partial t} = \frac{\partial^\prime\vec{b}}{\partial t} - \vec{u} \bullet \nabla \vec{b} + \vec{\omega} \times \vec{b}$

The scalar product of two vectors is commutative : $\vec{a} \bullet \vec{b} = \vec{b} \bullet \vec{a} \qquad (1.6)$

furthermore the following is valid ${\nabla} \bullet (\vec{a} \times \vec{b})=\vec{b} \bullet ({\nabla} \times \vec{a})-\vec{a} \bullet ({\nabla} \times \vec{b}) \qquad (8)$

1.2 Transformation in a Relative Frame of Reference

The continuity equation in an inertial frame of reference is written as $\frac{\partial \varrho}{\partial t} + \nabla \bullet \left( \varrho \vec{c} \right) = 0$

If one now applies Eq. 1.4 and Eq. 14 one gets $\underbrace{ \frac{\partial^\prime \varrho}{\partial t} - \left( \vec{c} - \vec{w} \right) \bullet \nabla \varrho }_{\frac{\partial \varrho}{\partial t}} + \nabla \bullet [ \varrho \underbrace{ \left( \vec{w} + \vec{\omega} \times \vec{r}^\prime \right) }_\vec{c} ] = 0$

If one now uses the general product rule from Eq. 30, one gets $\frac{\partial^\prime \varrho}{\partial t} - \left( \vec{\omega} \times \vec{r}^\prime \right) \bullet \nabla \varrho + \nabla \bullet \left( \varrho \vec{w} \right) + \nabla \bullet [ \overset{\downarrow}{ \varrho } ( \vec{\omega} \overset{\downarrow}{ \times } \vec{r}^\prime ) ] = 0 \qquad (1.7)$

The last term on the left hand of Eq. 1.7 gets now expanded with Eq. 30, and the second term on the left hand side is changed according to Eq. 1.6 $\frac{\partial^\prime \varrho}{\partial t} + \nabla \bullet \left( \varrho \vec{w} \right) - \nabla \varrho \bullet \left( \vec{\omega} \times \vec{r}^\prime \right) + \nabla \varrho \bullet ( \vec{\omega} \times \vec{r}^\prime ) + \varrho \nabla \bullet ( \vec{\omega} \times \vec{r}^\prime ) = 0 \qquad (1.5)$

If we now apply eq. 8 to the last term on the left hand side of eq. 1.5, we get $\nabla \bullet \left( \vec{\omega} \times \vec{r}^\prime \right) = \vec{r}^\prime \bullet \underbrace{({\nabla} \times \vec{\omega})}_{0} - \vec{\omega} \bullet \underbrace{({\nabla} \times \vec{r}^\prime)}_{0} = 0 \qquad (16)$

If one consider for example $\vec{\omega} = (0, 0, \omega)^T$ and $\vec{r}^\prime = (x^\prime, y^\prime, z^\prime)^T$ in cartesian coordinates and apply the cross product component-by-component, then the result from eq. 16 maybe become a little bit more clearer.

The two terms in the mid of the left hand side vanish and according to Eq. 16 the last term on the left hand side is zero. Thus, the continuity equation in the relative frame of reference can be written as $\frac{\partial^\prime \varrho}{\partial t} + \nabla \bullet \left( \varrho \vec{w} \right) = 0 \qquad (1.18)$

2 Momentum Equation

2.1 Mathematical Relations

The following identities are valid with any vector $\vec{a}$ and a scalar $\alpha$ $(\alpha \vec{a}) \otimes \vec{b} = \vec{a} \otimes (\alpha \vec{b}) = \alpha (\vec{a} \otimes \vec{b}) \qquad (6)$

and ${\nabla} \bullet \left( {\vec{a} \otimes \vec{b}}\right) = \left( {\nabla}\bullet{\vec{a}} \right) \vec{b} + \vec{a} \bullet {\nabla}{\vec{b}} \qquad (7)$ $\left( \vec{a} \otimes \vec{b} \right) \,{\scriptscriptstyle \stackrel{\bullet}{\bullet}}\, \mathbb{G} = \vec{a} \bullet \vec{b} = \mathbb{G} \,{\scriptscriptstyle \stackrel{\bullet}{\bullet}}\, \left( \vec{a} \otimes \vec{b} \right) \qquad (2.3)$

2.2 Transformation in a Relative Frame of Reference

We are now deriving the momentum equation in the relative coordinate system but formulated with the absolute velocity.

If we use eq. 6, then it can be easily shown, that the following statement is valid $\left( \varrho \vec{c} \right) \otimes \vec{c} = \vec{c} \otimes \left( \varrho \vec{c} \right) \qquad (9)$

If we use the solution from eq. 9, and apply them to eq. 7, the following can be shown $\begin{array}{rcl} \nabla \bullet \left[ \left( \varrho \vec{c} \right) \otimes \vec{c} \right] & = & \left[\nabla \bullet \left( \varrho \vec{c} \right)\right] \vec{c} + \left( \varrho \vec{c} \right) \bullet \nabla \vec{c} \qquad (10)\\ = \nabla \bullet \left[ \vec{c} \otimes \left( \varrho \vec{c} \right) \right] & = & \left[\nabla \bullet \vec{c} \right] \left( \varrho \vec{c} \right) + \vec{c} \bullet \nabla \left( \varrho \vec{c} \right) \qquad (11) \end{array}$

Furthermore, if we apply $\vec{b} = \varrho \vec{c}$ in eq. 5, this gives us $\dfrac{\partial \left( \varrho \vec{c} \right)}{\partial t} + \vec{c} \bullet \nabla \left( \varrho \vec{c} \right) = \frac{\partial^\prime\left( \varrho \vec{c} \right)}{\partial t} + \vec{w} \bullet \nabla \left( \varrho \vec{c} \right) + \vec{\omega} \times \left( \varrho \vec{c} \right) \qquad (12)$

Now we add the first term on the right hand side of eq. 11 $\left[\nabla \bullet \vec{c} \right] \left( \varrho \vec{c} \right)$ to eq. 12 and apply furthermore eq. 11. Then it follows $\begin{array}{rcl} \dfrac{\partial \left( \varrho \vec{c} \right)}{\partial t} + \left[\nabla \bullet \vec{c} \right] \left( \varrho \vec{c} \right) + \vec{c} \bullet \nabla \left( \varrho \vec{c} \right) & = & \dfrac{\partial^\prime\left( \varrho \vec{c} \right)}{\partial t} + \vec{w} \bullet \nabla \left( \varrho \vec{c} \right) + \vec{\omega} \times \left( \varrho \vec{c} \right) + \left[\nabla \bullet \vec{c} \right] \left( \varrho \vec{c} \right) \\ = \dfrac{\partial \left( \varrho \vec{c} \right)}{\partial t} + \nabla \bullet \left[ \left( \varrho \vec{c} \right) \otimes \vec{c} \right] & = & \dfrac{\partial^\prime\left( \varrho \vec{c} \right)}{\partial t} + \vec{\omega} \times \left( \varrho \vec{c} \right) + \underbrace{\vec{w} \bullet \nabla \left( \varrho \vec{c} \right) + \left[\nabla \bullet \vec{c} \right] \left( \varrho \vec{c} \right)}_{A1} \\ \end{array} \qquad (13)$

If we now use the solution from eq. 14 and apply this to the first term on the right hand side of eq. 11, the following is valid $\begin{array}{rcl} \left[\nabla \bullet \vec{c} \right] \left( \varrho \vec{c} \right) & = & \left[\nabla \bullet \left( \vec{w}+\vec{u} \right) \right] \left( \varrho \vec{c} \right) \\ & = & \left[\nabla \bullet \vec{w} \right] \left( \varrho \vec{c} \right) + \underbrace{\left[\nabla \bullet \left( \vec{\omega} \times \vec{r}^\prime \right)\right]}_{A2} \left( \varrho \vec{c} \right) \end{array}\qquad (15)$

With eq. 16, eq. 15 becomes $\left[\nabla \bullet \vec{c} \right] \left( \varrho \vec{c} \right) = \left[\nabla \bullet \vec{w} \right] \left( \varrho \vec{c} \right) \qquad (17)$

Now with eq. 17 and eq. 7 term A1 from eq. 13 writes to $\begin{array}{rcl} \left[\nabla \bullet \vec{c} \right] \left( \varrho \vec{c} \right) + \vec{w} \bullet \nabla \left( \varrho \vec{c} \right) & = & \left[\nabla \bullet \vec{w} \right] \left( \varrho \vec{c} \right) + \vec{w} \bullet \nabla \left( \varrho \vec{c} \right) \\ & = & \nabla \bullet \left[ \vec{w} \otimes \left( \varrho \vec{c} \right) \right] \end{array}\qquad (18)$

If we now insert eq. 18 into eq. 13, we get $\dfrac{\partial \left( \varrho \vec{c} \right)}{\partial t} + \nabla \bullet \left[ \left( \varrho \vec{c} \right) \otimes \vec{c} \right] = \dfrac{\partial^\prime\left( \varrho \vec{c} \right)}{\partial t} + \nabla \bullet \left[ \vec{w} \otimes \left( \varrho \vec{c} \right) \right] + \vec{\omega} \times \left( \varrho \vec{c} \right)\qquad (19)$

The momentum equation of an inertial coordinate system is equal to $\dfrac{\partial \varrho \vec{c}}{\partial t} + \nabla \bullet \left( \varrho \vec{c} \otimes \vec{c}\right) = -\nabla p + \nabla \bullet \tau + \varrho \vec{k}\qquad (20)$

If we now apply eq. 19 to eq. 20 we get the momentum equation in a relative coordinate system, see for example $\dfrac{\partial^\prime \left( \varrho \vec{c}\right)}{\partial t} + \nabla \bullet \left( \varrho \vec{w} \otimes \vec{c} \right) + \vec{\omega} \times \left( \varrho \vec{c} \right) = -\nabla p + \nabla \bullet \tau + \varrho \vec{k}\qquad (21)$

Furthermore it can be shown that the tensor 2. rank from the dyadic product of $\vec{c} \otimes \vec{w}$ is not symmetric, if you write it down component-by-component in cartesian coordinates. So that it must be valid to write $\vec{c} \otimes \vec{w} \boldsymbol{\neq} \vec{w} \otimes \vec{c}$

3 Energy Equation

The energy equation formulated with the internal energy $e = c_v T \qquad (3.0)$

writes as $\varrho \dfrac{d e}{d t} = \sigma \,{\scriptscriptstyle \stackrel{\bullet}{\bullet}}\, \left( \nabla \vec{c} \right) + \varrho \dot{q}_s - \nabla \bullet \vec{q} \qquad (3.1)$

With Eq. 1.4 and Eq. 4, and if one and inserts Eq. 14 into the right hand side, it follows: $\varrho \dfrac{d e}{d t} = \varrho \dfrac{d^\prime e}{d t} = \varrho \dfrac{\partial^\prime e}{\partial t} + \varrho \vec{w} \bullet \nabla e = \sigma \,{\scriptscriptstyle \stackrel{\bullet}{\bullet}}\, \left( \nabla \left[ \vec{w} + \vec{u} \right] \right) + \varrho \dot{q}_s - \nabla \bullet \vec{q} \qquad (3.2)$ $\varrho \left( \dfrac{\partial^\prime e}{\partial t} + \vec{w} \bullet \nabla e \right) = \sigma \,{\scriptscriptstyle \stackrel{\bullet}{\bullet}}\, \left( \nabla \vec{w} \right) + \underbrace{ \sigma \,{\scriptscriptstyle \stackrel{\bullet}{\bullet}}\, \left( \nabla \vec{u} \right) }_{B1} + \varrho \dot{q}_s - \nabla \bullet \vec{q} \qquad (3.3)$

In cartesian coordinates, the two variables of term B1 are written as: $\sigma = \left( \begin{array}{ccc} \sigma_{xx} & \sigma_{xy} & \sigma_{xz} \\ \sigma_{xy} & \sigma_{yy} & \sigma_{yz} \\ \sigma_{xz} & \sigma_{yz} & \sigma_{zz} \end{array} \right) \qquad (3.4)$ $\nabla \vec{u} = \left( \begin{array}{ccc} 0 & -\omega_z^\prime & \omega_y^\prime \\ \omega_z^\prime & 0 & -\omega_x^\prime\\ -\omega_y^\prime & \omega_x^\prime & 0 \end{array} \right) \qquad (3.5)$

As the stress tensor is symmetric $\sigma = \sigma^T$, and thus $\sigma_{xy} = \sigma_{yx}$; $\sigma_{xz} = \sigma_{zx}$ and $\sigma_{zy} = \sigma_{yz}$. An the gradient of the rotational speed is skew-symmetric $\nabla \vec{u} = - \left( \nabla \vec{u} \right)^T$

Thus the double scalar product from term B1 is in cartesian coordinates: \begin{alignat}{2} \sigma \,{\scriptscriptstyle \stackrel{\bullet}{\bullet}}\, \left( \nabla \vec{u} \right) & = \left( \begin{array}{ccc} \sigma_{xx} & \sigma_{xy} & \sigma_{xz} \\ \sigma_{xy} & \sigma_{yy} & \sigma_{yz} \\ \sigma_{xz} & \sigma_{yz} & \sigma_{zz} \end{array} \right) \,{\scriptscriptstyle \stackrel{\bullet}{\bullet}}\, \left( \begin{array}{ccc} 0 & -\omega_z^\prime & \omega_y^\prime \\ \omega_z^\prime & 0 & -\omega_x^\prime \\ -\omega_y^\prime & \omega_x^\prime & 0 \end{array} \right) \\ & = 0 - \sigma_{xy} \omega_z^\prime + \sigma_{xz} \omega_y^\prime \\ & + \sigma_{xy} \omega_z^\prime + 0 - \sigma_{yz} \omega_x^\prime \\ & - \sigma_{xz} \omega_y^\prime + \sigma_{yz} \omega_x^\prime + 0 \\ & = 0 \qquad \qquad \qquad \qquad \qquad (3.6) \end{alignat}

Inserting Eq. 3.6 and the continuity equation Eq. 1.18 into to left hand side, results in $\varrho \left( \dfrac{\partial^\prime e}{\partial t} + \vec{w} \bullet \nabla e \right) + e \underbrace{ \left( \dfrac{\partial^\prime \varrho}{\partial t} + \nabla \bullet \left( \varrho \vec{w} \right) \right) }_{\text{= 0; s. } Eq. 1.18} = \sigma \,{\scriptscriptstyle \stackrel{\bullet}{\bullet}}\, \left( \nabla \vec{w} \right) + \varrho \dot{q}_s - \nabla \bullet \vec{q} \qquad (3.7)$ $\varrho \dfrac{\partial^\prime e}{\partial t} + \left( \varrho\vec{w} \right) \bullet \nabla e + e \dfrac{\partial^\prime \varrho}{\partial t} + e \left( \nabla \bullet \left( \varrho \vec{w} \right) \right) = \sigma \,{\scriptscriptstyle \stackrel{\bullet}{\bullet}}\, \left( \nabla \vec{w} \right) + \varrho \dot{q}_s - \nabla \bullet \vec{q} \qquad (3.8)$ $\dfrac{\partial^\prime \varrho e}{\partial t} + \nabla \bullet \left( \varrho \vec{w} e \right) = \sigma \,{\scriptscriptstyle \stackrel{\bullet}{\bullet}}\, \left( \nabla \vec{w} \right) + \varrho \dot{q}_s - \nabla \bullet \vec{q} \qquad (3.8)$

The stress tensor can be split up into two parts, the thermodynamic pressure and the shear stress tensor $\sigma = -p \mathbb{G} + \tau \qquad (3.9)$

Inserting Eq. 3.9 into Eq. 3.8 and reducing afterward with Eq. 2.3 yields to \begin{alignat}{2} \dfrac{\partial^\prime \varrho e}{\partial t} + \nabla \bullet \left( \varrho \vec{w} e \right) & = \left( -p \mathbb{G} + \tau \right) \,{\scriptscriptstyle \stackrel{\bullet}{\bullet}}\, \left( \nabla \vec{w} \right) + \varrho \dot{q}_s - \nabla \bullet \vec{q} \\ & = -p \mathbb{G} \,{\scriptscriptstyle \stackrel{\bullet}{\bullet}}\, \left( \nabla \vec{w} \right) + \tau \,{\scriptscriptstyle \stackrel{\bullet}{\bullet}}\, \left( \nabla \vec{w} \right) + \varrho \dot{q}_s - \nabla \bullet \vec{q} \\ & = -p \left( \nabla \bullet \vec{w} \right) + \tau \,{\scriptscriptstyle \stackrel{\bullet}{\bullet}}\, \left( \nabla \vec{w} \right) + \varrho \dot{q}_s - \nabla \bullet \vec{q} \qquad (3.10) \end{alignat}

The heat flux can be expressed with the Fourier's law, where $k$ is the conductivity. The temperature can be expressed with the internal energy with Eq. 3.0 $\vec{q} = - k \nabla T = - k \nabla \left( \frac{e}{c_v} \right) \qquad (3.11)$

Inserting Eq. 3.11 into Eq. 3.10, yields to an energy equation for the internal energy in a relative reference frame, formulated with the relative velocity. $\dfrac{\partial^\prime \varrho e}{\partial t} + \nabla \bullet \left( \varrho \vec{w} e \right) = -p \left( \nabla \bullet \vec{w} \right) + \tau \,{\scriptscriptstyle \stackrel{\bullet}{\bullet}}\, \left( \nabla \vec{w} \right) + \varrho \dot{q}_s + \nabla \bullet \left( k \nabla \frac{e}{c_v} \right) \qquad (3.12)$

4 Shear Stress Tensor

4.1 Mathematical Relations

For a cross product of two vectors it is valid $\vec{b} \times \vec{c} = -\vec{c} \times \vec{b} \qquad (31)$

In general one writes for a tensor $\mathbb{T} = \vec{b} \otimes \vec{c}$ : $\left( \vec{b} \otimes \vec{c}\right)^T = \vec{c} \otimes \vec{b} \qquad (32)$

And for a tensor $\mathbb{S}$ one can also write $\left( \mathbb{S}^T \right)^T = \mathbb{S} \qquad (33)$

Furthermore for a tensor $\mathbb{T} = \vec{b} \otimes \vec{c}$ it is valid : $\left( \vec{b} \otimes \vec{c}\right) \times \vec{a} = \vec{b} \otimes \left( \vec{c} \times \vec{a}\right) \qquad (34)$

and also $\vec{a} \times \left( \vec{b} \otimes \vec{c}\right) = \left( \vec{a} \times \vec{b} \right) \otimes \vec{c} \qquad (35)$

Therefore the following indentities are valid, starting with eq. 34: \begin{alignat}{2} \left( \vec{b} \otimes \vec{c}\right) \times \vec{a} & = & \vec{b} \otimes \left( \vec{c} \times \vec{a}\right) \qquad \qquad \\ \text{with eq. 31 follows: } & = & - \vec{b} \otimes \left( \vec{a} \times \vec{c}\right) \qquad (36) \\ \text{with eq. 32 follows: } & = & - \left[ \left( \vec{a} \times \vec{c}\right) \otimes \vec{b} \right]^T \qquad (37)\\ \text{with eq. 35 follows: } & = & - \left[ \vec{a} \times \left( \vec{c} \otimes \vec{b} \right) \right]^T \qquad (38)\\ \text{with eq. 32 follows: } & = & - \left[ \vec{a} \times \left( \vec{b} \otimes \vec{c} \right)^T \right]^T \qquad (39) \end{alignat}

With eq. 39 for the tensor $\mathbb{T} = \vec{b} \otimes \vec{c}$ the following is valid: $\mathbb{T} \times \vec{a} = - \left[ \vec{a} \times \mathbb{T}^T \right]^T \qquad (40)$

If eq. 40 gets transposed and then eq. 33 is applied to the result, one yields: $\left( \mathbb{T} \times \vec{a} \right)^T = - \left( \left[ \vec{a} \times \mathbb{T}^T \right]^T \right)^T = - \left( \vec{a} \times \mathbb{T}^T \right) \qquad (41)$

4.2 Transformation in a Relative Frame of Reference

For a newtonian fluid the shear stress tensor $\tau$ can be expressed in the inertial frame of reference with $\lambda = \mu^\prime - \frac{2}{n_D} \mu$ as $\tau = \mu \left[ \nabla \vec{c} + \left( \nabla \vec{c} \right)^T \right] + \lambda \nabla \bullet \vec{c} \qquad (50)$

Where in three-dimensional the number of dimensions $n_D = 3$. Furthermore the volume viscosity $\mu^\prime$ vanishes after Stokes' hypotheses.

As already expressed in eq. 16, the last term on the right hand side is equal to $\nabla \bullet \vec{c} = \nabla \bullet \vec{w} \qquad (51)$

The same procedure is also applied to the following expression $\nabla \vec{c} + \left( \nabla \vec{c} \right)^T = \nabla \left\lbrace \vec{w} + \vec{u} \right\rbrace + \left( \nabla \left\lbrace \vec{w} + \vec{u} \right\rbrace \right)^T = \nabla \vec{w} + \left( \nabla \vec{w} \right)^T + \nabla \vec{u} + \left( \nabla \vec{u} \right)^T \qquad (52)$

For the last two terms on the right hand side, one can write: $\nabla \vec{u} + \left( \nabla \vec{u} \right)^T = \nabla \left( \vec{\omega} \times \vec{r}^\prime \right) + \left( \nabla \left( \vec{\omega} \times \vec{r}^\prime \right) \right)^T \qquad (53)$

If one now applies the product rule from eq. 30, one yields the following: \begin{alignat}{2} \nabla \left( \vec{\omega} \times \vec{r}^\prime \right) + \left( \nabla \left( \vec{\omega} \times \vec{r}^\prime \right) \right)^T & = \underbrace{\nabla \overset{\downarrow}{\vec{\omega}} }_0 \times \vec{r}^\prime + \nabla ( \vec{\omega} \times \overset{\downarrow}{ \vec{r}^\prime } ) + ( \underbrace{\nabla \overset{\downarrow}{\vec{\omega}} }_0 \times \vec{r}^\prime )^T + ( \nabla ( \vec{\omega} \times \overset{\downarrow}{ \vec{r}^\prime } ) )^T \\ & = \nabla ( \vec{\omega} \times \overset{\downarrow}{ \vec{r}^\prime } ) + ( \nabla ( \vec{\omega} \times \overset{\downarrow}{ \vec{r}^\prime } ) )^T \qquad (54) \end{alignat}

As $\vec{\omega}$ is a single vector and not a vector field, the gradient of a single vector is always zero. If one now apply eq. 36 and eq. 34 at each term of the right hand side of eq. 54, one yields with the metric tensor $\mathbb{G}$: $\nabla \vec{u} + \left( \nabla \vec{u} \right)^T = - ( \underbrace{ \nabla \vec{r}^\prime }_{\mathbb{G}} ) \times \vec{\omega} - ( ( \underbrace{ \nabla \vec{r}^\prime }_{\mathbb{G}} ) \times \vec{\omega} )^T \qquad (55)$

With the help of eq. 41 it follows: $\nabla \vec{u} + \left( \nabla \vec{u} \right)^T = - \mathbb{G} \times \vec{\omega} + \vec{\omega} \times \mathbb{G}^T \qquad (56)$

As the metric tensor $\mathbb{G}$ is symmetric, one can write $\mathbb{G} = \mathbb{G}^T$. Thus from eq. 56 it follows $\nabla \vec{u} + \left( \nabla \vec{u} \right)^T = - \mathbb{G} \times \vec{\omega} + \vec{\omega} \times \mathbb{G} = 0 \qquad (57)$

With the help of eq. 57 it follows for eq. 52: $\nabla \vec{c} + \left( \nabla \vec{c} \right)^T = \nabla \vec{w} + \left( \nabla \vec{w} \right)^T \qquad (58)$

Finally with eq. 51 and eq. 58 the shear stress tensor for a newtonian fluid in the relative frame of reference can be expressed as $\tau = \mu \left[ \nabla \vec{w} + \left( \nabla \vec{w} \right)^T \right] + \lambda \nabla \bullet \vec{w} \qquad (59)$