User:Kamil Kielczewski/sandbox

Main equation
The Cauchy momentum equation is a vector partial differential equation put forth by Cauchy that describes the non-relativistic momentum transport in any continuum. In convective (or Lagrangian) form it is written:


 * $$ \frac{D \mathbf{u}}{D t} = \frac 1 \rho \nabla \cdot \boldsymbol{\sigma} + \mathbf{f}$$

where \frac{\partial \sigma_{11}}{\partial x_1} + \frac{\partial \sigma_{21}}{\partial x_2} + \frac{\partial \sigma_{31}}{\partial x_3} \\ \frac{\partial \sigma_{12}}{\partial x_1} + \frac{\partial \sigma_{22}}{\partial x_2} + \frac{\partial \sigma_{32}}{\partial x_3} \\ \frac{\partial \sigma_{13}}{\partial x_1} + \frac{\partial \sigma_{23}}{\partial x_2} + \frac{\partial \sigma_{33}}{\partial x_3} \\ \end{bmatrix}[\mathrm{Pa/m=kg/m^2\cdot s^2 }] $$ is divergence of stress tensor.
 * $$\mathbf{u}\ [\mathrm{m/s}]$$ is the flow velocity vector field, which depends on time and space.
 * $$t\ [\mathrm{s}]$$ is the time,
 * $$\frac{D \mathbf{u}}{D t}\ [\mathrm{m/s^2}]$$ is material derivative equal to $$\partial_t\mathbf{u} + \mathbf{u}\cdot \nabla\mathbf{u}$$,
 * $$\rho\ [\mathrm{kg/m^3}]$$ is the density at the point considered in the continuum (for which the continuity equation holds),
 * $$\boldsymbol{\sigma}\ [\mathrm{Pa=N/m^2 = kg/m\cdot s^2 }]$$ is the stress tensor,
 * $$\mathbf{f}=\begin{bmatrix}f_x\\ f_y\\ f_z\end{bmatrix}\ [\mathrm{m/s^2 }]$$ contains all of the body forces acceleration (sometimes simply gravitational acceleration) (,
 * $$\nabla\cdot\boldsymbol{\sigma}= \begin{bmatrix}

Note that above we use column vectors only for clarity - however equation is written using physical components (which are neither covariant(bottom index - row vectors) nor contravariant(top index - column vectors) ). However if we chose non-orthogonal curvilinear coordinate system the we should calculate and write equation in covariant(row vectors) or contravariant(column vectors) form.

After an appropriate change of variables, it can also be written in conservation form:


 * $$ \frac {\partial \mathbf j }{\partial t}+ \nabla \cdot \mathbf F  = \mathbf s $$

where $j$ is the momentum density at a given space-time point, $F$ is the flux associated to the momentum density, and $s$ contains all of the body forces per unit volume.

Differential derivation
Let’s start from generalized momentum conservation principle which can be write as follows: "The change in the momentum of the system is proportional to the resultant force acting on this system." it is expressed by formula :


 * $$\vec p(t+\Delta t)-\vec p(t)=\Delta t \vec\bar F$$

where $$\vec p(t)$$ is momentum in time t, $$\vec\bar F$$ is force averaged in $$\Delta t$$. After dividing by $$\Delta t$$ and passing to the limit $$\Delta t\rightarrow 0$$ we get (derivative):


 * $$\frac{d\vec p}{dt}=\vec F$$

Lets analyse each side of above equation

Right side


We divide forces to body force and surface force
 * $$\vec F=\vec F_p + \vec F_m$$

Surface forces works on walls of cubic fluid element. The X component of this forces, for each wall, was placed in the picture with a cubic element (in the form of a product of stress and surface area e.g. $$-\sigma_{xx}dydz\ [Pa\cdot m\cdot m=\frac{N}{m^2}\cdot m\cdot m=N]$$).


 * {| class="toccolours collapsible collapsed" width="60%" style="text-align:left"

!Explanation of the value of forces (approximations and minus signs) on the walls of the cube It requires some explanation why the stress on the walls lying on the coordinate axes has a minus sign (e.g. on left wall we have $$-\sigma_{xx}$$). For simplicity, let's focus on the left wall with tension $$-\sigma_{xx}$$. The minus sign is caused by the fact that normal vector to this wall $$\vec n =[-1,0,0]=-\vec e_x$$ is a negative unit vector. Then, when calculating stress vector by definition $$\vec s=\vec n\cdot \boldsymbol\sigma=[-\sigma_{xx},-\sigma_{xy},-\sigma_{xy}]$$ therefore the x component of this vector is $$s_x=-\sigma_{xx}$$ (we use similar reasoning for stresses on the bottom and back walls, i.e.: $$-\sigma_{yx},-\sigma_{zx}$$).

The second element requiring explanation is the approximation of the stress values on the walls opposite the walls lying on the axes. Let's focus on the right wall where stress is an approximation of stress $$\sigma_{xx}$$ from the left wall at points with coordinates $$x+dx$$ and it is equal $$\sigma_{xx}+\frac{\partial\sigma_{xx}}{\partial x}dx$$. This approximation came from the application Taylor's formula for function approximate i.e.


 * $$\sigma_{xx}(x+dx)=\sigma_{xx}(x) + dx\frac{\partial\sigma_{xx}(x)}{\partial x} +\frac{dx^2}{2}\frac{\partial^2\sigma_{xx}(x)}{\partial x^2}+... $$

Because value $$dx^2$$ is infinitely less than value $$dx$$ so all components with $$dx$$ in powers greater than one we can abandon considering it insignificant. In this way we obtained the sought tension approximation on the opposite wall. More intuitive representation of value approximation $$\sigma_{xx}$$ in poiont $$x+dx$$ is shown on picture below cube. We carry out similar reasoning for stress approximations $$\sigma_{yx},\sigma_{zx}$$ Adding forces (their X components) acting on each of the cube walls, we get
 * }
 * $$F_p^x=(\sigma_{xx}+\frac{\partial\sigma_{xx}}{\partial x}dx)dydz-\sigma_{xx}dydz+

(\sigma_{yx}+\frac{\partial\sigma_{yx}}{\partial y}dy)dxdz-\sigma_{yx}dxdz+ (\sigma_{zx}+\frac{\partial\sigma_{zx}}{\partial z}dz)dxdy-\sigma_{zx}dxdy $$ After ordering $$F_p^x$$ and after performing similar reasoning for components $$F_p^y, F_p^z$$ (they are not in the drawing - they will be parallel vectors to the Y and Z axes, respectively) we get:


 * $$F_p^x=\frac{\partial\sigma_{xx}}{\partial x}dxdydz

+\frac{\partial\sigma_{yx}}{\partial y}dydxdz +\frac{\partial\sigma_{zx}}{\partial z}dzdxdy $$
 * $$F_p^y=\frac{\partial\sigma_{xy}}{\partial x}dxdydz

+\frac{\partial\sigma_{yy}}{\partial y}dydxdz +\frac{\partial\sigma_{zy}}{\partial z}dzdxdy $$
 * $$F_p^z=\frac{\partial\sigma_{xz}}{\partial x}dxdydz

+\frac{\partial\sigma_{yz}}{\partial y}dydxdz +\frac{\partial\sigma_{zz}}{\partial z}dzdxdy $$

We can then write it in the symbolic operational form:


 * $$\vec F_p=(\nabla\cdot\boldsymbol\sigma)dxdydz$$

Mass forces act on the inside of the control volume - we can write them using the acceleration field $$\mathbf{f}$$ (e.g. gravitational acceleration):


 * $$\vec F_m= \mathbf f \rho dxdydz$$

Left side
Lets calculate momentum of the cube:
 * $$\vec p= \mathbf u m=\mathbf u \rho dxdydz$$

Because we assume that that tested mass (cube) $$m=\rho dxdydz$$ is constant in time, so


 * $$\frac{d\vec p}{dt}=\frac{d\mathbf u}{dt}\rho dxdydz$$

Left and Right side comparison
We have


 * $$\frac{d\vec p}{dt}=\vec F$$

then


 * $$\frac{d\vec p}{dt}=\vec F_p + \vec F_m$$

then
 * $$\frac{d\mathbf u}{dt}\rho dxdydz = (\nabla\cdot\boldsymbol\sigma)dxdydz + \mathbf f\rho dxdydz$$

Divide both sides by $$\rho dxdydz$$, and because $$\frac{d\mathbf u}{dt}=\frac{D\mathbf u}{Dt}$$ we get:
 * $$\frac{D\mathbf u}{Dt} = \frac{1}{\rho}\nabla\cdot\boldsymbol\sigma + \mathbf f$$

which finish derivation.

Integral derivation
Applying Newton's second law ($i$th component) to a control volume in the continuum being modeled gives:


 * $$m a_i = F_i$$

Then, based on the Reynolds transport theorem and using material derivative notation, one can write


 * $$\begin{align}

\int_{\Omega} \rho \frac{D u_i}{D t} \, dV &= \int_{\Omega} \nabla_j\sigma_i^j \, dV + \int_{\Omega} \rho f_i \, dV \\ \int_{\Omega} \left(\rho \frac{D u_i}{D t} - \nabla_j\sigma_i^j - \rho f_i \right)\, dV &= 0 \\ \rho \frac{D u_i}{D t}- \nabla_j\sigma_i^j - \rho f_i &= 0 \\ \frac{D u_i}{D t}- \frac {\nabla_j\sigma_i^j}{\rho} - f_i &= 0 \end{align}$$

where $Ω$ represents the control volume. Since this equation must hold for any control volume, it must be true that the integrand is zero, from this the Cauchy momentum equation follows. The main step (not done above) in deriving this equation is establishing that the derivative of the stress tensor is one of the forces that constitutes $F_{i}$.

Conservation form
Cauchy equations can also be put in the following form:

simply by defining:


 * $$ \begin{align} {\mathbf j}&= \rho \mathbf u \\

{\mathbf F}&=\rho \mathbf u \otimes \mathbf u - \boldsymbol \sigma \\ {\mathbf s}&= \rho \mathbf f \end{align}$$

where $j$ is the momentum density at the point considered in the continuum (for which the continuity equation holds), $F$ is the flux associated to the momentum density, and $s$ contains all of the body forces per unit volume. $u ⊗ u$ is the dyad of the velocity.

Here $j$ and $s$ have same number of dimensions $N$ as the flow speed and the body acceleration, while $F$, being a tensor, has $N^{2}$.

In the Eulerian forms it is apparent that the assumption of no deviatoric stress brings Cauchy equations to the Euler equations.

Convective acceleration


A significant feature of the Navier–Stokes equations is the presence of convective acceleration: the effect of time-independent acceleration of a flow with respect to space. While individual continuum particles indeed experience time dependent acceleration, the convective acceleration of the flow field is a spatial effect, one example being fluid speeding up in a nozzle.

Regardless of what kind of continuum is being dealt with, convective acceleration is a nonlinear effect. Convective acceleration is present in most flows (exceptions include one-dimensional incompressible flow), but its dynamic effect is disregarded in creeping flow (also called Stokes flow). Convective acceleration is represented by the nonlinear quantity $j$, which may be interpreted either as $σ$ or as $F$, with $F$ the tensor derivative of the velocity vector $F$. Both interpretations give the same result.

Advection operator vs tensor derivative
The convection term $$D\mathbf{u}/Dt$$ can be written as $σ$, where $u · ∇u$ is the advection operator. This representation can be contrasted to the one in terms of the tensor derivative. The tensor derivative $(u · ∇)u$ is the component-by-component derivative of the velocity vector, defined by $u · (∇u)$, so that
 * $$\left[\mathbf{u}\cdot\left(\nabla \mathbf{u}\right)\right]_i=\sum_m v_m \partial_m v_i=\left[(\mathbf{u}\cdot\nabla)\mathbf{u}\right]_i\,.$$

Lamb form
The vector calculus identity of the cross product of a curl holds:


 * $$ \mathbf{v} \times \left( \nabla \times \mathbf{a} \right) = \nabla_a \left( \mathbf{v} \cdot \mathbf{a} \right) - \mathbf{v} \cdot \nabla \mathbf{a} \, $$

where the Feynman subscript notation $∇u$ is used, which means the subscripted gradient operates only on the factor $a$.

Lamb in his famous classical book Hydrodynamics (1895), still in print, used this identity to change the convective term of the flow velocity in rotational form, i.e. without a tensor derivative:


 * $$\mathbf{u} \cdot \nabla \mathbf{u} = \nabla \left( \frac{\|\mathbf{u}\|^2}{2} \right) + \left( \nabla \times \mathbf{u} \right) \times \mathbf{u}$$

where the vector $$\mathbf l = \left( \nabla \times \mathbf{u} \right) \times \mathbf{u}$$ is called the Lamb vector. The Cauchy momentum equation becomes:


 * $$\frac{\partial \mathbf{u}}{\partial t} + \frac{1}{2} \nabla \left(u^2\right) + (\nabla \times \mathbf u) \times \mathbf u = \frac 1 \rho \nabla \cdot \boldsymbol \sigma + \mathbf{f}$$

Using the identity:


 * $$\nabla \cdot \left( \frac {\boldsymbol \sigma}{\rho} \right) = \frac 1 \rho \nabla \cdot \boldsymbol \sigma - \frac{1}{\rho^2} \boldsymbol \sigma \cdot \nabla \rho$$

the Cauchy equation becomes:


 * $$\nabla \cdot \left(\frac{1}{2} u^2 + \frac {\boldsymbol \sigma} \rho \right) - \mathbf f = \frac{1}{\rho^2} \boldsymbol \sigma \cdot \nabla \rho + \mathbf u \times (\nabla \times \mathbf u) - \frac{\partial \mathbf u}{\partial t}$$

In fact, in case of an external conservative field, by defining its potential $φ$:


 * $$\nabla \cdot \left( \frac{1}{2} u^2 + \phi + \frac {\boldsymbol \sigma} \rho \right) = \frac{1}{\rho^2} \boldsymbol \sigma \cdot \nabla \rho + \mathbf u \times (\nabla \times \mathbf u) - \frac{\partial \mathbf u}{\partial t}$$

In case of a steady flow the time derivative of the flow velocity disappears, so the momentum equation becomes:


 * $$\nabla \cdot \left( \frac{1}{2} u^2 + \phi + \frac {\boldsymbol \sigma} \rho \right) = \frac{1}{\rho^2} \boldsymbol \sigma \cdot \nabla \rho + \mathbf u \times (\nabla \times \mathbf u)$$

And by projecting the momentum equation on the flow direction, i.e. along a streamline, the cross product disappears due to a vector calculus identity of the triple scalar product:


 * $$\mathbf u \cdot \nabla \cdot \left( \frac{1}{2} u^2 + \phi + \frac {\boldsymbol \sigma} \rho \right) = \frac{1}{\rho^2} \mathbf u \cdot (\boldsymbol \sigma \cdot \nabla \rho)$$

If the stress tensor is isotropic, then only the pressure enters, and the Euler momentum equation in the steady incompressible case becomes:


 * $$\mathbf u \cdot \nabla \cdot \left( \frac{1}{2} u^2 + \phi + \frac p \rho \right) = \frac{p}{\rho^2} \mathbf u \cdot \nabla \rho$$

In the steady incompressible case the mass equation is simply:


 * $$\mathbf u \cdot \nabla \rho = 0\,,$$

that is, the mass conservation for a steady incompressible flow states that the density along a streamline is constant. This leads to a considerable simplification of the Euler momentum equation:


 * $$\mathbf u \cdot \nabla \left( \frac{1}{2} u^2 + \phi + \frac p \rho \right) = 0$$

The convenience of defining the total head for an inviscid liquid flow is now apparent:


 * $$b_l \equiv \frac{1}{2} u^2 + \phi + \frac p \rho\,,$$

in fact, the above equation can be simply written as:


 * $$\mathbf u \cdot \nabla b_l = 0$$

That is, the momentum balance for a steady inviscid and incompressible flow in an external conservative field states that the total head along a streamline is constant.

Irrotational flows
The Lamb form is also useful in irrotational flow, where the curl of the velocity (called vorticity) $u$ is equal to zero. In that case, the convection term $$D\mathbf{u}/Dt$$ reduces to


 * $$\mathbf{u} \cdot \nabla \mathbf{u} = \nabla \left( \frac{\|\mathbf{u}\|^2}{2} \right).$$

Stresses
The effect of stress in the continuum flow is represented by the $(u · ∇)u$ and $u · ∇$ terms; these are gradients of surface forces, analogous to stresses in a solid. Here $∇u$ is the pressure gradient and arises from the isotropic part of the Cauchy stress tensor. This part is given by the normal stresses that occur in almost all situations. The anisotropic part of the stress tensor gives rise to $[∇u]_{mi} = ∂_{m} v_{i}$, which usually describes viscous forces; for incompressible flow, this is only a shear effect. Thus, $∇_{a}$ is the deviatoric stress tensor, and the stress tensor is equal to:


 * $$\boldsymbol \sigma = - p \mathbf I + \boldsymbol \tau$$

where $ω = ∇ × u$ is the identity matrix in the space considered and $∇p$ the shear tensor.

All non-relativistic momentum conservation equations, such as the Navier–Stokes equation, can be derived by beginning with the Cauchy momentum equation and specifying the stress tensor through a constitutive relation. By expressing the shear tensor in terms of viscosity and fluid velocity, and assuming constant density and viscosity, the Cauchy momentum equation will lead to the Navier–Stokes equations. By assuming inviscid flow, the Navier–Stokes equations can further simplify to the Euler equations.

The divergence of the stress tensor can be written as


 * $$\nabla \cdot \boldsymbol{\sigma} = -\nabla p + \nabla \cdot \boldsymbol{\tau}.$$

The effect of the pressure gradient on the flow is to accelerate the flow in the direction from high pressure to low pressure.

As written in the Cauchy momentum equation, the stress terms $p$ and $∇ · τ$ are yet unknown, so this equation alone cannot be used to solve problems. Besides the equations of motion—Newton's second law—a force model is needed relating the stresses to the flow motion. For this reason, assumptions based on natural observations are often applied to specify the stresses in terms of the other flow variables, such as velocity and density.

External forces
The vector field $∇p$ represents body forces per unit mass. Typically, these consist of only gravity acceleration, but may include others, such as electromagnetic forces. In non-inertial coordinate frames, other "inertial accelerations" associated with rotating coordinates may arise.

Often, these forces may be represented as the gradient of some scalar quantity $χ$, with $∇ · τ$ in which case they are called conservative forces. Gravity in the $z$ direction, for example, is the gradient of $τ$. Because pressure from such gravitation arises only as a gradient, we may include it in the pressure term as a body force $I$. The pressure and force terms on the right-hand side of the Navier–Stokes equation become


 * $$-\nabla p + \mathbf{f} = -\nabla p + \nabla \chi = -\nabla \left( p - \chi \right) = -\nabla h.$$

Nondimensionalisation
In order to make the equations dimensionless, a characteristic length $τ$ and a characteristic velocity $τ$ need to be defined. These should be chosen such that the dimensionless variables are all of order one. The following dimensionless variables are thus obtained:


 * $$\begin{align}

\rho^* &\equiv \frac \rho {\rho_0} & u^* &\equiv \frac u {u_0} & r^* &\equiv \frac r {r_0} & t^*&\equiv \frac {u_0}{r_0} t \\[6pt] \nabla^* &\equiv r_0 \nabla & \mathbf f^* &\equiv \frac {\mathbf f} {f_0} & p^* &\equiv \frac p {p_0} & \boldsymbol \tau^* &\equiv \frac {\boldsymbol \tau} {\tau_0} \end{align}$$

Substitution of these inverted relations in the Euler momentum equations yields:


 * $$\frac {\rho_0 u_0^2}{r_0}\frac{\partial \rho^* \mathbf u^*}{\partial t^*}+ \frac {\nabla^*}{r_0} \cdot \left( \rho_0 u_0^2 \rho^* \mathbf u^* \otimes \mathbf u^* + p_0 p^* \right)= - \frac {\tau_0}{r_0} \nabla^* \cdot \boldsymbol \tau^* + f_0 \mathbf f^*$$

and by dividing for the first coefficient:


 * $$\frac{\partial \mathbf \rho^* u^*}{\partial t^*}+ \nabla^* \cdot \left(\rho^* \mathbf u^* \otimes  \mathbf u^* + \frac {p_0}{\rho_0 u_0^2} p^* \right)= - \frac {\tau_0}{\rho_0 u_0^2} \nabla^* \cdot \boldsymbol \tau^* + \frac { f_0 r_0}{u_0^2} \mathbf f^*$$

Now defining the Froude number:


 * $$\mathrm{Fr}=\frac{u_0^2}{f_0 r_0},$$

the Euler number:


 * $$\mathrm{Eu}=\frac{p_0}{\rho_0 u_0^2},$$

and the coefficient of skin-friction or the one usually referred as 'drag' co-efficient in the field of aerodynamics:


 * $$C_\mathrm{f}=\frac{2 \tau_0}{\rho_0 u_0^2},$$

by passing respectively to the conservative variables, i.e. the momentum density and the force density:


 * $$\begin{align} \mathbf j &= \rho \mathbf u \\ \mathbf g &= \rho \mathbf f \end{align}$$

the equations are finally expressed (now omitting the indexes):

Cauchy equations in the Froude limit $f$ (corresponding to negligible external field) are named free Cauchy equations:

and can be eventually conservation equations. The limit of high Froude numbers (low external field) is thus notable for such equations and is studied with perturbation theory.

Finally in convective form the equations are:

Cartesian 3D coordinates
In general for asymmetric stress tensor, equations have following form


 * $$\begin{align}

x&: & \frac{\partial u_x}{\partial t} + u_x \frac{\partial u_x}{\partial x} + u_y \frac{\partial u_x}{\partial y} + u_z \frac{\partial u_x}{\partial z}       &= \frac 1 \rho \left( \frac{\partial \sigma_{xx}}{\partial x} + \frac{\partial \sigma_{yx}}{\partial y} + \frac{\partial \sigma_{zx}}{\partial z} \right) + f_x \\[8pt] y&: & \frac{\partial u_y}{\partial t} + u_x \frac{\partial u_y}{\partial x} + u_y \frac{\partial u_y}{\partial y} + u_z \frac{\partial u_y}{\partial z}       &= \frac 1 \rho \left( \frac{\partial \sigma_{xy}}{\partial x} + \frac{\partial \sigma_{yy}}{\partial y} + \frac{\partial \sigma_{zy}}{\partial z} \right) + f_y \\[8pt] z&: & \frac{\partial u_z}{\partial t} + u_x \frac{\partial u_z}{\partial x} + u_y \frac{\partial u_z}{\partial y} + u_z \frac{\partial u_z}{\partial z}       &= \frac 1 \rho \left( \frac{\partial \sigma_{xz}}{\partial x} + \frac{\partial \sigma_{yz}}{\partial y} + \frac{\partial \sigma_{zz}}{\partial z} \right) + f_z\end{align}$$

Cylindrical 3D coordinates
Below we write main equation in pressure-tau form assuming that stress tensor is symmetric ($$\sigma_{ij}=\sigma_{ji}\quad \Longrightarrow \quad \tau_{ij}=\tau_{ji}$$):


 * $$\begin{align}

r&: &\frac{\partial u_r}{\partial t} + u_r \frac{\partial u_r}{\partial r} + \frac{u_\phi}{r} \frac{\partial u_r}{\partial\phi} + u_z \frac{\partial u_r}{\partial z} - \frac{u_\phi^2}{r} &= -\frac{1}{\rho} \frac{\partial P}{\partial r} + \frac{1}{r\rho}\frac{\partial\left(r\tau_{rr}\right)}{\partial r} + \frac{1}{r\rho} \frac{\partial\tau_{\phi r}}{\partial\phi} + \frac{1}{\rho} \frac{\partial\tau_{zr}}{\partial z} - \frac{\tau_{\phi\phi}}{r\rho} + f_r \\[8pt] \phi&: &\frac{\partial u_\phi}{\partial t} + u_r \frac{\partial u_\phi}{\partial r} + \frac{u_\phi}{r} \frac{\partial u_\phi}{\partial\phi} + u_z \frac{\partial u_\phi}{\partial z} + \frac{u_r u_\phi}{r} &= -\frac{1}{r\rho} \frac{\partial P}{\partial\phi} + \frac{1}{r\rho}\frac{\partial\tau_{\phi \phi}}{\partial\phi} + \frac{1}{r^2 \rho} \frac{\partial\left(r^2 \tau_{r\phi}\right)}{\partial r} + \frac{1}{\rho} \frac{\partial\tau_{z\phi}}{\partial z} + f_\phi \\[8pt] z&: &\frac{\partial u_z}{\partial t} + u_r \frac{\partial u_z}{\partial r} + \frac{u_\phi}{r} \frac{\partial u_z}{\partial\phi} + u_z \frac{\partial u_z}{\partial z}        &= -\frac{1}{\rho} \frac{\partial P}{\partial z} + \frac{1}{\rho} \frac{\partial\tau_{zz}}{\partial z} + \frac{1}{r\rho}\frac{\partial\tau_{\phi z}}{\partial\phi} + \frac{1}{r\rho}\frac{\partial\left(r\tau_{rz}\right)}{\partial r} + f_z \end{align}$$