4.1.1 Two ways of deriving the equation

As mentioned earlier, the basis vectors $\boldsymbol{e}_{1}$ , $\boldsymbol{e}_{2}$ , $\boldsymbol{e}_{3}$ are not unit vectors, so that, to introduce the ‘physical’ components of velocity $\boldsymbol{v}$ , we need to normalize these vectors:

(4.1) $$\begin{eqnarray}\boldsymbol{v}=v^{1}\boldsymbol{e}_{1}+v^{2}\boldsymbol{e}_{2}+v^{3}\boldsymbol{e}_{3}=v^{1}|\boldsymbol{e}_{1}|\frac{\boldsymbol{e}_{1}}{|\boldsymbol{e}_{1}|}+v^{2}|\boldsymbol{e}_{2}|\frac{\boldsymbol{e}_{2}}{|\boldsymbol{e}_{2}|}+v^{3}|\boldsymbol{e}_{3}|\frac{\boldsymbol{e}_{3}}{|\boldsymbol{e}_{3}|}=u_{\unicode[STIX]{x1D709}}\hat{\boldsymbol{e}}_{1}+u_{\unicode[STIX]{x1D702}}\hat{\boldsymbol{e}}_{2}+u_{\unicode[STIX]{x1D703}}\hat{\boldsymbol{e}}_{3},\end{eqnarray}$$

where

(4.2a-c ) $$\begin{eqnarray}u_{\unicode[STIX]{x1D709}}=v^{1}|\boldsymbol{e}_{1}|=v^{1}\sqrt{\unicode[STIX]{x1D628}_{11}},\quad u_{\unicode[STIX]{x1D702}}=v^{2}|\boldsymbol{e}_{2}|=v^{2},\quad u_{\unicode[STIX]{x1D703}}=v^{3}|\boldsymbol{e}_{3}|=v^{3}\unicode[STIX]{x1D702}\end{eqnarray}$$

are the physical components of $\boldsymbol{v}$ and $\hat{\boldsymbol{e}}_{1}=\boldsymbol{e}_{1}/|\boldsymbol{e}_{1}|$ , $\hat{\boldsymbol{e}}_{2}=\boldsymbol{e}_{2}/|\boldsymbol{e}_{2}|$ , $\hat{\boldsymbol{e}}_{3}=\boldsymbol{e}_{3}/|\boldsymbol{e}_{3}|$ are now unit vectors.

The continuity equation $\unicode[STIX]{x1D735}\boldsymbol{\cdot }\boldsymbol{v}=0$ , i.e.  $\unicode[STIX]{x1D735}_{i}v^{i}=0,$ written down explicitly has the form

(4.3) $$\begin{eqnarray}\frac{\unicode[STIX]{x2202}v^{i}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D709}^{i}}+\unicode[STIX]{x1D6E4}_{ik}^{i}v^{k}=0,\end{eqnarray}$$

that is

(4.4) $$\begin{eqnarray}\displaystyle & & \displaystyle \frac{\unicode[STIX]{x2202}v^{1}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D709}^{1}}+\unicode[STIX]{x1D6E4}_{11}^{1}v^{1}+\unicode[STIX]{x1D6E4}_{12}^{1}v^{2}+\unicode[STIX]{x1D6E4}_{13}^{1}v^{3}+\frac{\unicode[STIX]{x2202}v^{2}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D709}^{2}}+\unicode[STIX]{x1D6E4}_{21}^{2}v^{1}+\unicode[STIX]{x1D6E4}_{22}^{2}v^{2}+\unicode[STIX]{x1D6E4}_{23}^{2}v^{3}\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\frac{\unicode[STIX]{x2202}v^{3}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D709}^{3}}+\unicode[STIX]{x1D6E4}_{31}^{3}v^{1}+\unicode[STIX]{x1D6E4}_{32}^{3}v^{2}+\unicode[STIX]{x1D6E4}_{33}^{3}v^{3}=0,\end{eqnarray}$$

or

(4.5) $$\begin{eqnarray}\displaystyle & & \displaystyle \frac{\unicode[STIX]{x2202}v^{1}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D709}^{1}}+\frac{\unicode[STIX]{x2202}v^{2}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D709}^{2}}+\frac{\unicode[STIX]{x2202}v^{3}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D709}^{3}}+(\unicode[STIX]{x1D6E4}_{11}^{1}+\unicode[STIX]{x1D6E4}_{21}^{2}+\unicode[STIX]{x1D6E4}_{31}^{3})v^{1}+(\unicode[STIX]{x1D6E4}_{12}^{1}+\unicode[STIX]{x1D6E4}_{22}^{2}+\unicode[STIX]{x1D6E4}_{32}^{3})v^{2}\nonumber\\ \displaystyle & & \displaystyle \qquad +\,(\unicode[STIX]{x1D6E4}_{13}^{1}+\unicode[STIX]{x1D6E4}_{23}^{2}+\unicode[STIX]{x1D6E4}_{33}^{3})v^{3}=0.\end{eqnarray}$$

Then, using (4.2), which express the velocity components $v^{1}$ , $v^{2}$ , $v^{3}$ in terms of the physical components $u_{\unicode[STIX]{x1D709}}$ , $u_{\unicode[STIX]{x1D702}}$ , $u_{\unicode[STIX]{x1D703}}$ , and the expressions for the Christoffel symbols $\unicode[STIX]{x1D6E4}_{jk}^{i}$ , $(i,j,k=1,2,3)$ calculated in appendix A with the results summarized in § 3.3, we finally have the continuity equation in the following form:

(4.6) $$\begin{eqnarray}\displaystyle & & \displaystyle \frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D709}}\left(\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}\right)+\frac{\unicode[STIX]{x2202}u_{\unicode[STIX]{x1D702}}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D702}}+\frac{1}{\unicode[STIX]{x1D702}}\frac{\unicode[STIX]{x2202}u_{\unicode[STIX]{x1D703}}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}-\frac{\unicode[STIX]{x1D702}\cos \unicode[STIX]{x1D703}}{1-\unicode[STIX]{x1D702}\unicode[STIX]{x1D705}_{1}\cos \unicode[STIX]{x1D703}}\frac{\text{d}\unicode[STIX]{x1D705}_{1}}{\text{d}\unicode[STIX]{x1D709}}\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}\nonumber\\ \displaystyle & & \displaystyle \qquad +\left(1-\frac{\unicode[STIX]{x1D702}\unicode[STIX]{x1D705}_{1}\cos \unicode[STIX]{x1D703}}{1-\unicode[STIX]{x1D702}\unicode[STIX]{x1D705}_{1}\cos \unicode[STIX]{x1D703}}\right)\frac{u_{\unicode[STIX]{x1D702}}}{\unicode[STIX]{x1D702}}+\frac{\unicode[STIX]{x1D705}_{1}\sin \unicode[STIX]{x1D703}}{1-\unicode[STIX]{x1D702}\unicode[STIX]{x1D705}_{1}\cos \unicode[STIX]{x1D703}}u_{\unicode[STIX]{x1D703}}=0.\end{eqnarray}$$

The component $\unicode[STIX]{x1D628}_{11}$ of the metric tensor, which accounts for the torsion of the baseline, is given in (3.15), and the curvature $\unicode[STIX]{x1D705}_{1}$ of the baseline is expressed in terms of $X(\unicode[STIX]{x1D709})$ , $Y(\unicode[STIX]{x1D709})$ , $Z(\unicode[STIX]{x1D709})$ in (3.6). Note that in (4.6), besides differential terms, there are also algebraic terms proportional to each of the three components of velocity.

An alternative and much shorter way of deriving the continuity equation is by using the well-known formula for the divergence (Sedov Reference Sedov1997)

(4.7) $$\begin{eqnarray}\unicode[STIX]{x1D735}_{i}v^{i}=\frac{1}{\sqrt{\unicode[STIX]{x1D6E5}}}\frac{\unicode[STIX]{x2202}(v^{i}\sqrt{\unicode[STIX]{x1D6E5}})}{\unicode[STIX]{x2202}\unicode[STIX]{x1D709}^{i}},\end{eqnarray}$$

where $\unicode[STIX]{x1D6E5}=\text{det}(\unicode[STIX]{x1D628}_{ij})$ was calculated earlier (3.16) and the expressions of $v^{i}$ $(i=1,2,3)$ in terms of the physical components of velocity are given by (4.2). Once applied, this formula verifies (4.6) and hence implicitly the expressions for the Christoffel symbols calculated in appendix A and summarized in a compact form in § 3.3.

4.1.2 Typical mistakes

As one can see from the derivation above, there is again no room for making a mistake as all that the derivation of (4.6) requires is the definition of a divergence as $\unicode[STIX]{x1D735}_{i}v^{i}$ , the definition of the covariant derivative leading to (4.3) and the definition of the Christoffel symbols (3.18). The latter, of course, rely on the correct definition of the basis vectors (3.10). The formula (4.7) requires even less, namely the expression for $\unicode[STIX]{x1D6E5}$ (3.16), which, again, relies on the correct definition of the basis vectors, and the definition of the physical components of velocity resulting in (4.2).

In the published work on the topic, there is invariably a gap between the introduction of the local curvilinear coordinate system $(\unicode[STIX]{x1D709},\unicode[STIX]{x1D702},\unicode[STIX]{x1D703})$ with the supposedly corresponding basis vectors and the continuity equation. The details of the derivation are covered by the phrases such as ‘we obtain the conservation of mass  $\ldots$ ’ (Wallwork et al. Reference Wallwork, Decent, King and Schulkes2002), ‘using standard methods, we obtain the continuity equation  $\ldots$ ’ (Părău et al. Reference Părău, Decent, Simmons, Wong and King2007) or simply ‘we obtain the continuity equations as  $\ldots$ ’ (Decent et al. Reference Decent, King and Wallwork2002), so that the continuity equation becomes the first litmus test as to how these ‘standard methods’ have been applied.

By comparing equation (4.6) with the versions featuring, for example, in Decent et al. (Reference Decent, King and Wallwork2002), Wallwork et al. (Reference Wallwork, Decent, King and Schulkes2002), Părău et al. (Reference Părău, Decent, King, Simmons and Wong2006, Reference Părău, Decent, Simmons, Wong and King2007), Decent et al. (Reference Decent, King, Simmons, Părău, Wallwork, Gurney and Uddin2009) one can see immediately that the latter are erroneous in all that is related to the jet being curved and, notably, irrespective of whether the coordinate system is orthogonal ( $\unicode[STIX]{x1D705}_{2}=0$ ) or not ( $\unicode[STIX]{x1D705}_{2}\neq 0$ ). In other words, even where by chance the local coordinate system ‘introduced’ as orthogonal happens to be orthogonal (Wallwork et al. Reference Wallwork, Decent, King and Schulkes2002) and hence the technique of ‘scaling factors’ should work, the resulting continuity equation is still erroneous. The term proportional to $u_{\unicode[STIX]{x1D709}}$ and the derivative of the baseline’s curvature with respect to $\unicode[STIX]{x1D709}$ is simply missing whilst the factors in front of other algebraic terms involve the (non-normalized) component of the binormal, in our notation $X^{\prime }Y^{\prime \prime }-Y^{\prime }X^{\prime \prime }$ , but not the curvature of the baseline given by (3.6) as they should.

Ribe (Reference Ribe2004), one of the few who recognized that the local coordinate system for a coiling liquid jet is non-orthogonal, arrived at the continuity equation as the first invariant of the rate-of-strain tensor taking the latter from Green & Zerna (Reference Green and Zerna1992). The result was again the absence of the term proportional to the variation of the baseline’s curvature along the baseline.

4.2.1 Velocity and acceleration in Frenet’s basis

One way of deriving the expression for the acceleration term on the left-hand side of the equation of motion (2.1) in projection on the Frenet basis is simply by writing it down as $\unicode[STIX]{x2202}\boldsymbol{v}/\unicode[STIX]{x2202}t+\boldsymbol{v}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\boldsymbol{v}$ , doing the required covariant differentiation using the Christoffel symbols calculated in appendix A and then projecting the result on the Frenet basis using (3.11)–(3.13). This procedure is rather lengthy and cumbersome, so it would have to be hidden behind ‘after lengthy calculations, we obtain  $\ldots$ ’. To remain verifiable, we will take an alternative route where calculating the Christoffel symbols and projecting the result onto the Frenet basis is built into the very procedure. This route is used in textbooks when the equations of motion in curvilinear coordinates are introduced. The subtlety in our case is that the velocity components we are dealing with correspond to our curvilinear non-orthogonal coordinate system whilst what we need is these components in the Frenet basis. To find the latter, we will, first, consider the velocity of a ‘fluid particle’, i.e. a particle with fixed Lagrangian coordinates moving with respect to the Eulerian coordinate frame.

If we take a fluid particle whose position vector at moment $t$ in our (Eulerian) coordinate system $(\unicode[STIX]{x1D709},\unicode[STIX]{x1D702},\unicode[STIX]{x1D703})$ is given by $\boldsymbol{r}(t)=\boldsymbol{r}(\unicode[STIX]{x1D709}(t),\unicode[STIX]{x1D702}(t),\unicode[STIX]{x1D703}(t))$ , then the particle’s velocity is

(4.8) $$\begin{eqnarray}\displaystyle \boldsymbol{v}=\frac{\text{d}\boldsymbol{r}}{\text{d}t}=\frac{\unicode[STIX]{x2202}\boldsymbol{r}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D709}}\frac{\text{d}\unicode[STIX]{x1D709}}{\text{d}t}+\frac{\unicode[STIX]{x2202}\boldsymbol{r}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D702}}\frac{\text{d}\unicode[STIX]{x1D702}}{\text{d}t}+\frac{\unicode[STIX]{x2202}\boldsymbol{r}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}\frac{\text{d}\unicode[STIX]{x1D703}}{\text{d}t}=\frac{\text{d}\unicode[STIX]{x1D709}}{\text{d}t}\boldsymbol{e}_{1}+\frac{\text{d}\unicode[STIX]{x1D702}}{\text{d}t}\boldsymbol{e}_{2}+\frac{\text{d}\unicode[STIX]{x1D703}}{\text{d}t}\boldsymbol{e}_{3}. & & \displaystyle\end{eqnarray}$$

Comparing this with (4.1) we see that

(4.9a-c ) $$\begin{eqnarray}\frac{\text{d}\unicode[STIX]{x1D709}}{\text{d}t}=v^{1},\quad \frac{\text{d}\unicode[STIX]{x1D702}}{\text{d}t}=v^{2},\quad \frac{\text{d}\unicode[STIX]{x1D703}}{\text{d}t}=v^{3}\end{eqnarray}$$

are components of the particle’s velocity in the local basis $\boldsymbol{e}_{1}$ , $\boldsymbol{e}_{2}$ , $\boldsymbol{e}_{3}$ .

Now, given that for a fluid particle, according to (3.9),

(4.10) $$\begin{eqnarray}\displaystyle \boldsymbol{r}(\unicode[STIX]{x1D709}(t),\unicode[STIX]{x1D702}(t),\unicode[STIX]{x1D703}(t))=\boldsymbol{R}(\unicode[STIX]{x1D709}(t))+\unicode[STIX]{x1D702}(t)\cos \unicode[STIX]{x1D703}(t)\,\boldsymbol{n}(\unicode[STIX]{x1D709}(t))+\unicode[STIX]{x1D702}(t)\sin \unicode[STIX]{x1D703}(t)\,\boldsymbol{b}(\unicode[STIX]{x1D709}(t)), & & \displaystyle\end{eqnarray}$$

all that we need to do is differentiate $\boldsymbol{r}$ with respect to $t$ and use the Frenet formulae (3.5) together with the definitions (4.9) and (4.2):

(4.11) $$\begin{eqnarray}\displaystyle \boldsymbol{v} & = & \displaystyle \frac{\text{d}\boldsymbol{R}}{\text{d}\unicode[STIX]{x1D709}}\frac{\text{d}\unicode[STIX]{x1D709}}{\text{d}t}+\left(\frac{\text{d}\unicode[STIX]{x1D702}}{\text{d}t}\cos \unicode[STIX]{x1D703}-\unicode[STIX]{x1D702}\sin \unicode[STIX]{x1D703}\frac{\text{d}\unicode[STIX]{x1D703}}{\text{d}t}\right)\boldsymbol{n}+\unicode[STIX]{x1D702}\cos \unicode[STIX]{x1D703}\frac{\text{d}\boldsymbol{n}}{\text{d}\unicode[STIX]{x1D709}}\frac{\text{d}\unicode[STIX]{x1D709}}{\text{d}t}+\left(\frac{\text{d}\unicode[STIX]{x1D702}}{\text{d}t}\sin \unicode[STIX]{x1D703}+\unicode[STIX]{x1D702}\cos \unicode[STIX]{x1D703}\frac{\text{d}\unicode[STIX]{x1D703}}{\text{d}t}\right)\boldsymbol{b}\nonumber\\ \displaystyle & & \displaystyle +\,\unicode[STIX]{x1D702}\sin \unicode[STIX]{x1D703}\frac{\text{d}\boldsymbol{b}}{\text{d}\unicode[STIX]{x1D709}}\frac{\text{d}\unicode[STIX]{x1D709}}{\text{d}t}=\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}\unicode[STIX]{x1D749}+(u_{\unicode[STIX]{x1D702}}\cos \unicode[STIX]{x1D703}-u_{\unicode[STIX]{x1D703}}\sin \unicode[STIX]{x1D703})\boldsymbol{n}+\unicode[STIX]{x1D702}\cos \unicode[STIX]{x1D703}\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}(-\unicode[STIX]{x1D705}_{1}\unicode[STIX]{x1D749}+\unicode[STIX]{x1D705}_{2}\boldsymbol{b})\nonumber\\ \displaystyle & & \displaystyle +\,(u_{\unicode[STIX]{x1D702}}\sin \unicode[STIX]{x1D703}+u_{\unicode[STIX]{x1D703}}\cos \unicode[STIX]{x1D703})\boldsymbol{b}-\unicode[STIX]{x1D702}\unicode[STIX]{x1D705}_{2}\sin \unicode[STIX]{x1D703}\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}\boldsymbol{n}.\end{eqnarray}$$

After rearrangement, we finally have:

(4.12) $$\begin{eqnarray}\displaystyle \boldsymbol{v} & = & \displaystyle (1-\unicode[STIX]{x1D705}_{1}\unicode[STIX]{x1D702}\cos \unicode[STIX]{x1D703})\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}\unicode[STIX]{x1D749}+\left(-\unicode[STIX]{x1D705}_{2}\unicode[STIX]{x1D702}\sin \unicode[STIX]{x1D703}\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}+u_{\unicode[STIX]{x1D702}}\cos \unicode[STIX]{x1D703}-u_{\unicode[STIX]{x1D703}}\sin \unicode[STIX]{x1D703}\right)\boldsymbol{n}\nonumber\\ \displaystyle & & \displaystyle +\left(\unicode[STIX]{x1D705}_{2}\unicode[STIX]{x1D702}\cos \unicode[STIX]{x1D703}\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}+u_{\unicode[STIX]{x1D702}}\sin \unicode[STIX]{x1D703}+u_{\unicode[STIX]{x1D703}}\cos \unicode[STIX]{x1D703}\right)\boldsymbol{b}.\end{eqnarray}$$

The coefficients in front of $\unicode[STIX]{x1D749}$ , $\boldsymbol{n}$ , $\boldsymbol{b}$ , i.e.

(4.13a,b ) $$\begin{eqnarray}v_{\unicode[STIX]{x1D70F}}=(1-\unicode[STIX]{x1D705}_{1}\unicode[STIX]{x1D702}\cos \unicode[STIX]{x1D703})\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}},\quad v_{n}=-\unicode[STIX]{x1D705}_{2}\unicode[STIX]{x1D702}\sin \unicode[STIX]{x1D703}\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}+u_{\unicode[STIX]{x1D702}}\cos \unicode[STIX]{x1D703}-u_{\unicode[STIX]{x1D703}}\sin \unicode[STIX]{x1D703},\end{eqnarray}$$

(4.14) $$\begin{eqnarray}v_{b}=\unicode[STIX]{x1D705}_{2}\unicode[STIX]{x1D702}\cos \unicode[STIX]{x1D703}\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}+u_{\unicode[STIX]{x1D702}}\sin \unicode[STIX]{x1D703}+u_{\unicode[STIX]{x1D703}}\cos \unicode[STIX]{x1D703},\end{eqnarray}$$

are components of velocity in the Frenet basis $\unicode[STIX]{x1D749}$ , $\boldsymbol{n}$ , $\boldsymbol{b}$ expressed in terms of the physical components of velocity $u_{\unicode[STIX]{x1D709}}$ , $u_{\unicode[STIX]{x1D702}}$ , $u_{\unicode[STIX]{x1D703}}$ in our curvilinear coordinate system $(\unicode[STIX]{x1D709},\unicode[STIX]{x1D702},\unicode[STIX]{x1D703})$ whose (non-orthogonal) basis is $\boldsymbol{e}_{1}$ , $\boldsymbol{e}_{2}$ , $\boldsymbol{e}_{3}$ .

Now, we can consider the velocity field of a moving continuum and treat our velocity components in (4.12) as functions of $(t,\unicode[STIX]{x1D709},\unicode[STIX]{x1D702},\unicode[STIX]{x1D703})$ . To calculate the acceleration term in the second equation (2.1), we need to differentiate (4.12) with respect to $t$ , keeping in mind that (i) the derivatives of $\unicode[STIX]{x1D709}$ , $\unicode[STIX]{x1D702}$ , $\unicode[STIX]{x1D703}$ with respect to $t$ are, according to definitions (4.9), components $v^{1}$ , $v^{2}$ , $v^{3}$ , which, following definitions (4.2), can be expressed in terms of the physical components $u_{\unicode[STIX]{x1D709}}$ , $u_{\unicode[STIX]{x1D702}}$ , $u_{\unicode[STIX]{x1D703}}$ and (ii) the Frenet basis $\unicode[STIX]{x1D749}$ , $\boldsymbol{n}$ , $\boldsymbol{b}$ varies with $\unicode[STIX]{x1D709}$ and the derivatives can be expressed back in terms of $\unicode[STIX]{x1D749}$ , $\boldsymbol{n}$ , $\boldsymbol{b}$ using Frenet’s formulae (3.5). This is basically the textbook derivation of the acceleration in a curvilinear coordinate system with the only difference that $v_{\unicode[STIX]{x1D70F}}$ , $v_{n}$ , $v_{b}$ in (4.13), (4.14) are expressed in terms of $u_{\unicode[STIX]{x1D709}}$ , $u_{\unicode[STIX]{x1D702}}$ , $u_{\unicode[STIX]{x1D703}}$ . Thus, differentiating (4.12) we obtain

(4.15) $$\begin{eqnarray}\displaystyle \frac{\text{d}\boldsymbol{v}}{\text{d}t} & = & \displaystyle \left\{(1-\unicode[STIX]{x1D705}_{1}\unicode[STIX]{x1D702}\cos \unicode[STIX]{x1D703})\left[\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}t}\left(\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}\right)+\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D709}}\left(\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}\right)\right.\right.+u_{\unicode[STIX]{x1D702}}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D702}}\left(\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}\right)\nonumber\\ \displaystyle & & \displaystyle \left.+\,\frac{u_{\unicode[STIX]{x1D703}}}{\unicode[STIX]{x1D702}}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}\left(\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}\right)\right]-\unicode[STIX]{x1D702}\cos \unicode[STIX]{x1D703}\frac{\text{d}\unicode[STIX]{x1D705}_{1}}{\text{d}\unicode[STIX]{x1D709}}\frac{u_{\unicode[STIX]{x1D709}}^{2}}{\unicode[STIX]{x1D628}_{11}}-\unicode[STIX]{x1D705}_{1}\cos \unicode[STIX]{x1D703}\frac{u_{\unicode[STIX]{x1D709}}u_{\unicode[STIX]{x1D702}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}+\unicode[STIX]{x1D705}_{1}\sin \unicode[STIX]{x1D703}\frac{u_{\unicode[STIX]{x1D709}}u_{\unicode[STIX]{x1D703}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}\nonumber\\ \displaystyle & & \displaystyle \left.-\,\unicode[STIX]{x1D705}_{1}\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}\left(-\unicode[STIX]{x1D705}_{2}\unicode[STIX]{x1D702}\sin \unicode[STIX]{x1D703}\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}+u_{\unicode[STIX]{x1D702}}\cos \unicode[STIX]{x1D703}-u_{\unicode[STIX]{x1D703}}\sin \unicode[STIX]{x1D703}\right)\right\}\unicode[STIX]{x1D749}\nonumber\\ \displaystyle & & \displaystyle +\left\{-\unicode[STIX]{x1D705}_{2}\unicode[STIX]{x1D702}\sin \unicode[STIX]{x1D703}\left[\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}t}\left(\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}\right)+\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D709}}\left(\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}\right)\right.\right.\nonumber\\ \displaystyle & & \displaystyle \left.+\,u_{\unicode[STIX]{x1D702}}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D702}}\left(\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}\right)+\frac{u_{\unicode[STIX]{x1D703}}}{\unicode[STIX]{x1D702}}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}\left(\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}\right)\right]\nonumber\\ \displaystyle & & \displaystyle -\,\unicode[STIX]{x1D702}\sin \unicode[STIX]{x1D703}\frac{\text{d}\unicode[STIX]{x1D705}_{2}}{\text{d}\unicode[STIX]{x1D709}}\frac{u_{\unicode[STIX]{x1D709}}^{2}}{\unicode[STIX]{x1D628}_{11}}-\unicode[STIX]{x1D705}_{2}\sin \unicode[STIX]{x1D703}\frac{u_{\unicode[STIX]{x1D709}}u_{\unicode[STIX]{x1D702}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}-\unicode[STIX]{x1D705}_{2}\cos \unicode[STIX]{x1D703}\frac{u_{\unicode[STIX]{x1D709}}u_{\unicode[STIX]{x1D703}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}\nonumber\\ \displaystyle & & \displaystyle +\,\cos \unicode[STIX]{x1D703}\left(\frac{\unicode[STIX]{x2202}u_{\unicode[STIX]{x1D702}}}{\unicode[STIX]{x2202}t}+\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}\frac{\unicode[STIX]{x2202}u_{\unicode[STIX]{x1D702}}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D709}}+u_{\unicode[STIX]{x1D702}}\frac{\unicode[STIX]{x2202}u_{\unicode[STIX]{x1D702}}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D702}}+\frac{u_{\unicode[STIX]{x1D703}}}{\unicode[STIX]{x1D702}}\frac{\unicode[STIX]{x2202}u_{\unicode[STIX]{x1D702}}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}\right)\nonumber\\ \displaystyle & & \displaystyle -\,\sin \unicode[STIX]{x1D703}\left(\frac{\unicode[STIX]{x2202}u_{\unicode[STIX]{x1D703}}}{\unicode[STIX]{x2202}t}+\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}\frac{\unicode[STIX]{x2202}u_{\unicode[STIX]{x1D703}}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D709}}+u_{\unicode[STIX]{x1D702}}\frac{\unicode[STIX]{x2202}u_{\unicode[STIX]{x1D703}}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D702}}+\frac{u_{\unicode[STIX]{x1D703}}}{\unicode[STIX]{x1D702}}\frac{\unicode[STIX]{x2202}u_{\unicode[STIX]{x1D703}}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}\right)\nonumber\\ \displaystyle & & \displaystyle -\,\frac{u_{\unicode[STIX]{x1D703}}}{\unicode[STIX]{x1D702}}(u_{\unicode[STIX]{x1D702}}\sin \unicode[STIX]{x1D703}+u_{\unicode[STIX]{x1D703}}\cos \unicode[STIX]{x1D703})+\unicode[STIX]{x1D705}_{1}(1-\unicode[STIX]{x1D705}_{1}\unicode[STIX]{x1D702}\cos \unicode[STIX]{x1D703})\frac{u_{\unicode[STIX]{x1D709}}^{2}}{\unicode[STIX]{x1D628}_{11}}\nonumber\\ \displaystyle & & \displaystyle \left.-\,\unicode[STIX]{x1D705}_{2}\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}\left(\unicode[STIX]{x1D705}_{2}\unicode[STIX]{x1D702}\cos \unicode[STIX]{x1D703}\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}+u_{\unicode[STIX]{x1D702}}\sin \unicode[STIX]{x1D703}+u_{\unicode[STIX]{x1D703}}\cos \unicode[STIX]{x1D703}\right)\right\}\boldsymbol{n}\nonumber\\ \displaystyle & & \displaystyle +\left\{\unicode[STIX]{x1D705}_{2}\unicode[STIX]{x1D702}\cos \unicode[STIX]{x1D703}\left[\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}t}\left(\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}\right)+\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D709}}\left(\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}\right)\right.\right.\nonumber\\ \displaystyle & & \displaystyle \left.+\,u_{\unicode[STIX]{x1D702}}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D702}}\left(\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}\right)+\frac{u_{\unicode[STIX]{x1D703}}}{\unicode[STIX]{x1D702}}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}\left(\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}\right)\right]\nonumber\\ \displaystyle & & \displaystyle +\,\unicode[STIX]{x1D705}_{2}\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}\left(u_{\unicode[STIX]{x1D702}}\cos \unicode[STIX]{x1D703}-u_{\unicode[STIX]{x1D703}}\sin \unicode[STIX]{x1D703}\right)+\unicode[STIX]{x1D702}\cos \unicode[STIX]{x1D703}\frac{\text{d}\unicode[STIX]{x1D705}_{2}}{\text{d}\unicode[STIX]{x1D709}}\frac{u_{\unicode[STIX]{x1D709}}^{2}}{\unicode[STIX]{x1D628}_{11}}\nonumber\\ \displaystyle & & \displaystyle +\,\sin \unicode[STIX]{x1D703}\left(\frac{\unicode[STIX]{x2202}u_{\unicode[STIX]{x1D702}}}{\unicode[STIX]{x2202}t}+\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}\frac{\unicode[STIX]{x2202}u_{\unicode[STIX]{x1D702}}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D709}}+u_{\unicode[STIX]{x1D702}}\frac{\unicode[STIX]{x2202}u_{\unicode[STIX]{x1D702}}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D702}}+\frac{u_{\unicode[STIX]{x1D703}}}{\unicode[STIX]{x1D702}}\frac{\unicode[STIX]{x2202}u_{\unicode[STIX]{x1D702}}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}\right)\nonumber\\ \displaystyle & & \displaystyle +\,\cos \unicode[STIX]{x1D703}\left(\frac{\unicode[STIX]{x2202}u_{\unicode[STIX]{x1D703}}}{\unicode[STIX]{x2202}t}+\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}\frac{\unicode[STIX]{x2202}u_{\unicode[STIX]{x1D703}}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D709}}+u_{\unicode[STIX]{x1D702}}\frac{\unicode[STIX]{x2202}u_{\unicode[STIX]{x1D703}}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D702}}+\frac{u_{\unicode[STIX]{x1D703}}}{\unicode[STIX]{x1D702}}\frac{\unicode[STIX]{x2202}u_{\unicode[STIX]{x1D703}}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}\right)\nonumber\\ \displaystyle & & \displaystyle +\,\frac{u_{\unicode[STIX]{x1D703}}}{\unicode[STIX]{x1D702}}\left(u_{\unicode[STIX]{x1D702}}\cos \unicode[STIX]{x1D703}-u_{\unicode[STIX]{x1D703}}\sin \unicode[STIX]{x1D703}\right)+\unicode[STIX]{x1D705}_{2}\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}\nonumber\\ \displaystyle & & \displaystyle \left.\times \left(-\unicode[STIX]{x1D705}_{2}\unicode[STIX]{x1D702}\sin \unicode[STIX]{x1D703}\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}+u_{\unicode[STIX]{x1D702}}\cos \unicode[STIX]{x1D703}-u_{\unicode[STIX]{x1D703}}\sin \unicode[STIX]{x1D703}\right)\right\}\boldsymbol{b}.\end{eqnarray}$$

After rearrangement, we finally have

(4.16) $$\begin{eqnarray}\displaystyle \frac{\text{d}\boldsymbol{v}}{\text{d}t} & = & \displaystyle \left\{\vphantom{\frac{u_{\unicode[STIX]{x1D709}}^{2}}{\unicode[STIX]{x1D628}_{11}}}(1-\unicode[STIX]{x1D705}_{1}\unicode[STIX]{x1D702}\cos \unicode[STIX]{x1D703})\left[\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}t}\left(\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}\right)+\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D709}}\left(\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}\right)\right.\right.\nonumber\\ \displaystyle & & \displaystyle \left.+\,u_{\unicode[STIX]{x1D702}}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D702}}\left(\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}\right)+\frac{u_{\unicode[STIX]{x1D703}}}{\unicode[STIX]{x1D702}}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}\left(\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}\right)\right]\nonumber\\ \displaystyle & & \displaystyle \left.+\,\unicode[STIX]{x1D702}\frac{u_{\unicode[STIX]{x1D709}}^{2}}{\unicode[STIX]{x1D628}_{11}}\left(\unicode[STIX]{x1D705}_{1}\unicode[STIX]{x1D705}_{2}\sin \unicode[STIX]{x1D703}-\cos \unicode[STIX]{x1D703}\frac{\text{d}\unicode[STIX]{x1D705}_{1}}{\text{d}\unicode[STIX]{x1D709}}\right)-2\unicode[STIX]{x1D705}_{1}\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}(u_{\unicode[STIX]{x1D702}}\cos \unicode[STIX]{x1D703}-u_{\unicode[STIX]{x1D703}}\sin \unicode[STIX]{x1D703})\right\}\unicode[STIX]{x1D749}\nonumber\\ \displaystyle & & \displaystyle +\left\{-\unicode[STIX]{x1D705}_{2}\unicode[STIX]{x1D702}\sin \unicode[STIX]{x1D703}\left[\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}t}\left(\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}\right)+\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D709}}\left(\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}\right)\right.\right.\nonumber\\ \displaystyle & & \displaystyle \left.+\,u_{\unicode[STIX]{x1D702}}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D702}}\left(\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}\right)+\frac{u_{\unicode[STIX]{x1D703}}}{\unicode[STIX]{x1D702}}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}\left(\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}\right)\right]\nonumber\\ \displaystyle & & \displaystyle +\,\frac{u_{\unicode[STIX]{x1D709}}^{2}}{\unicode[STIX]{x1D628}_{11}}\left(\unicode[STIX]{x1D705}_{1}-\unicode[STIX]{x1D702}\cos \unicode[STIX]{x1D703}(\unicode[STIX]{x1D705}_{1}^{2}+\unicode[STIX]{x1D705}_{2}^{2})-\unicode[STIX]{x1D702}\sin \unicode[STIX]{x1D703}\frac{\text{d}\unicode[STIX]{x1D705}_{2}}{\text{d}\unicode[STIX]{x1D709}}\right)\nonumber\\ \displaystyle & & \displaystyle -\left(2\unicode[STIX]{x1D705}_{2}\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}+\frac{u_{\unicode[STIX]{x1D703}}}{\unicode[STIX]{x1D702}}\right)(u_{\unicode[STIX]{x1D702}}\sin \unicode[STIX]{x1D703}+u_{\unicode[STIX]{x1D703}}\cos \unicode[STIX]{x1D703})\nonumber\\ \displaystyle & & \displaystyle +\,\cos \unicode[STIX]{x1D703}\left(\frac{\unicode[STIX]{x2202}u_{\unicode[STIX]{x1D702}}}{\unicode[STIX]{x2202}t}+\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}\frac{\unicode[STIX]{x2202}u_{\unicode[STIX]{x1D702}}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D709}}+u_{\unicode[STIX]{x1D702}}\frac{\unicode[STIX]{x2202}u_{\unicode[STIX]{x1D702}}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D702}}+\frac{u_{\unicode[STIX]{x1D703}}}{\unicode[STIX]{x1D702}}\frac{\unicode[STIX]{x2202}u_{\unicode[STIX]{x1D702}}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}\right)\nonumber\\ \displaystyle & & \displaystyle \left.-\,\sin \unicode[STIX]{x1D703}\left(\frac{\unicode[STIX]{x2202}u_{\unicode[STIX]{x1D703}}}{\unicode[STIX]{x2202}t}+\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}\frac{\unicode[STIX]{x2202}u_{\unicode[STIX]{x1D703}}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D709}}+u_{\unicode[STIX]{x1D702}}\frac{\unicode[STIX]{x2202}u_{\unicode[STIX]{x1D703}}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D702}}+\frac{u_{\unicode[STIX]{x1D703}}}{\unicode[STIX]{x1D702}}\frac{\unicode[STIX]{x2202}u_{\unicode[STIX]{x1D703}}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}\right)\right\}\boldsymbol{n}\nonumber\\ \displaystyle & & \displaystyle +\left\{\unicode[STIX]{x1D705}_{2}\unicode[STIX]{x1D702}\cos \unicode[STIX]{x1D703}\left[\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}t}\left(\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}\right)+\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D709}}\left(\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}\right)\right.\right.\nonumber\\ \displaystyle & & \displaystyle \left.+\,u_{\unicode[STIX]{x1D702}}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D702}}\left(\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}\right)+\frac{u_{\unicode[STIX]{x1D703}}}{\unicode[STIX]{x1D702}}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}\left(\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}\right)\right]\nonumber\\ \displaystyle & & \displaystyle +\left(2\unicode[STIX]{x1D705}_{2}\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}+\frac{u_{\unicode[STIX]{x1D703}}}{\unicode[STIX]{x1D702}}\right)(u_{\unicode[STIX]{x1D702}}\cos \unicode[STIX]{x1D703}-u_{\unicode[STIX]{x1D703}}\sin \unicode[STIX]{x1D703})+\unicode[STIX]{x1D702}\frac{u_{\unicode[STIX]{x1D709}}^{2}}{\unicode[STIX]{x1D628}_{11}}\left(\cos \unicode[STIX]{x1D703}\frac{\text{d}\unicode[STIX]{x1D705}_{2}}{\text{d}\unicode[STIX]{x1D709}}-\unicode[STIX]{x1D705}_{2}^{2}\sin \unicode[STIX]{x1D703}\right)\nonumber\\ \displaystyle & & \displaystyle +\,\sin \unicode[STIX]{x1D703}\left(\frac{\unicode[STIX]{x2202}u_{\unicode[STIX]{x1D702}}}{\unicode[STIX]{x2202}t}+\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}\frac{\unicode[STIX]{x2202}u_{\unicode[STIX]{x1D702}}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D709}}+u_{\unicode[STIX]{x1D702}}\frac{\unicode[STIX]{x2202}u_{\unicode[STIX]{x1D702}}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D702}}+\frac{u_{\unicode[STIX]{x1D703}}}{\unicode[STIX]{x1D702}}\frac{\unicode[STIX]{x2202}u_{\unicode[STIX]{x1D702}}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}\right)\nonumber\\ \displaystyle & & \displaystyle \left.+\,\cos \unicode[STIX]{x1D703}\left(\frac{\unicode[STIX]{x2202}u_{\unicode[STIX]{x1D703}}}{\unicode[STIX]{x2202}t}+\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}\frac{\unicode[STIX]{x2202}u_{\unicode[STIX]{x1D703}}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D709}}+u_{\unicode[STIX]{x1D702}}\frac{\unicode[STIX]{x2202}u_{\unicode[STIX]{x1D703}}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D702}}+\frac{u_{\unicode[STIX]{x1D703}}}{\unicode[STIX]{x1D702}}\frac{\unicode[STIX]{x2202}u_{\unicode[STIX]{x1D703}}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}\right)\right\}\boldsymbol{b}.\end{eqnarray}$$

An alternative way of calculating the acceleration is to substitute into

(4.17) $$\begin{eqnarray}\frac{\text{d}\boldsymbol{v}}{\text{d}t}=\left(\frac{\unicode[STIX]{x2202}v^{i}}{\unicode[STIX]{x2202}t}+v^{j}\unicode[STIX]{x1D735}_{j}v^{i}\right)\boldsymbol{e}_{i}=\left(\frac{\unicode[STIX]{x2202}v^{i}}{\unicode[STIX]{x2202}t}+v^{j}\frac{\unicode[STIX]{x2202}v^{i}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D709}^{j}}+v^{j}v^{k}\unicode[STIX]{x1D6E4}_{jk}^{i}\right)\boldsymbol{e}_{i}\end{eqnarray}$$

the expressions for $\unicode[STIX]{x1D6E4}_{jk}^{i}$ $(i,j,k=1,2,3)$ from § 3.3 and the expressions (4.2) of $v^{i}$ $(i=1,2,3)$ in terms of the physical components $u_{\unicode[STIX]{x1D709}}$ , $u_{\unicode[STIX]{x1D702}}$ , $u_{\unicode[STIX]{x1D703}}$ and project the result on the Frenet basis by expressing $\boldsymbol{e}_{i}$ $(i=1,2,3)$ in terms of $\unicode[STIX]{x1D749}$ , $\boldsymbol{n}$ , $\boldsymbol{b}$ using (3.11)–(3.13). We used this (slightly more cumbersome) way for an independent verification of (4.16).

4.2.2 Pressure gradient in Frenet’s basis

To calculate the pressure gradient, we need only its definition and expressions (3.11)–(3.13) relating the local basis with Frenet’s basis:

(4.18) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D735}p & = & \displaystyle \boldsymbol{e}_{i}\unicode[STIX]{x1D735}^{i}p=\unicode[STIX]{x1D628}^{ij}\frac{\unicode[STIX]{x2202}p}{\unicode[STIX]{x2202}\unicode[STIX]{x1D709}^{j}}~\boldsymbol{e}_{i}\nonumber\\ \displaystyle & = & \displaystyle \left(\unicode[STIX]{x1D628}^{11}\frac{\unicode[STIX]{x2202}p}{\unicode[STIX]{x2202}\unicode[STIX]{x1D709}}+\unicode[STIX]{x1D628}^{13}\frac{\unicode[STIX]{x2202}p}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}\right)[(1-\unicode[STIX]{x1D702}\unicode[STIX]{x1D705}_{1}\cos \unicode[STIX]{x1D703})\unicode[STIX]{x1D749}-\unicode[STIX]{x1D702}\unicode[STIX]{x1D705}_{2}\sin \unicode[STIX]{x1D703}\,\boldsymbol{n}+\unicode[STIX]{x1D702}\unicode[STIX]{x1D705}_{2}\cos \unicode[STIX]{x1D703}\,\boldsymbol{b}]\nonumber\\ \displaystyle & & \displaystyle +\,\unicode[STIX]{x1D628}^{22}\frac{\unicode[STIX]{x2202}p}{\unicode[STIX]{x2202}\unicode[STIX]{x1D702}}(\cos \unicode[STIX]{x1D703}\,\boldsymbol{n}+\sin \unicode[STIX]{x1D703}\,\boldsymbol{b})+\left(\unicode[STIX]{x1D628}^{31}\frac{\unicode[STIX]{x2202}p}{\unicode[STIX]{x2202}\unicode[STIX]{x1D709}}+\unicode[STIX]{x1D628}^{33}\frac{\unicode[STIX]{x2202}p}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}\right)(-\unicode[STIX]{x1D702}\sin \unicode[STIX]{x1D703}\,\boldsymbol{n}+\unicode[STIX]{x1D702}\cos \unicode[STIX]{x1D703}\,\boldsymbol{b}).\nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

Using expressions for $\unicode[STIX]{x1D628}^{ij}$ , $(i,j=1,2,3)$ from (3.17), we finally have

(4.19) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D735}p & = & \displaystyle \frac{1}{1-\unicode[STIX]{x1D702}\unicode[STIX]{x1D705}_{1}\cos \unicode[STIX]{x1D703}}\left(\frac{\unicode[STIX]{x2202}p}{\unicode[STIX]{x2202}\unicode[STIX]{x1D709}}-\unicode[STIX]{x1D705}_{2}\frac{\unicode[STIX]{x2202}p}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}\right)\unicode[STIX]{x1D749}+\left(\cos \unicode[STIX]{x1D703}\frac{\unicode[STIX]{x2202}p}{\unicode[STIX]{x2202}\unicode[STIX]{x1D702}}-\frac{\sin \unicode[STIX]{x1D703}}{\unicode[STIX]{x1D702}}\frac{\unicode[STIX]{x2202}p}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}\right)\boldsymbol{n}\nonumber\\ \displaystyle & & \displaystyle +\left(\sin \unicode[STIX]{x1D703}\frac{\unicode[STIX]{x2202}p}{\unicode[STIX]{x2202}\unicode[STIX]{x1D702}}+\frac{\cos \unicode[STIX]{x1D703}}{\unicode[STIX]{x1D702}}\frac{\unicode[STIX]{x2202}p}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}\right)\boldsymbol{b}.\end{eqnarray}$$

Note that for non-zero torsion of the baseline, $\unicode[STIX]{x1D705}_{2}\neq 0$ , the $\unicode[STIX]{x1D749}$ -projection of the gradient includes an azimuthal derivative of the pressure.

4.2.3 Centrifugal, Coriolis and gravitational forces in Frenet’s basis

Centrifugal force. Using the parametrization (3.9), for the centrifugal force we have

(4.20) $$\begin{eqnarray}\displaystyle -\unicode[STIX]{x1D734}\times (\unicode[STIX]{x1D734}\times \boldsymbol{r})=-\unicode[STIX]{x1D734}\times (\unicode[STIX]{x1D734}\times \boldsymbol{R})-\unicode[STIX]{x1D702}\cos \unicode[STIX]{x1D703}\,\unicode[STIX]{x1D734}\times (\unicode[STIX]{x1D734}\times \boldsymbol{n})-\unicode[STIX]{x1D702}\sin \unicode[STIX]{x1D703}\,\unicode[STIX]{x1D734}\times (\unicode[STIX]{x1D734}\times \boldsymbol{b}). & & \displaystyle \nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

Given that $\unicode[STIX]{x1D734}=\unicode[STIX]{x1D6FA}\hat{\boldsymbol{z}}$ , $\boldsymbol{R}=X\hat{\boldsymbol{x}}+Y\hat{\boldsymbol{y}}+Z\hat{\boldsymbol{z}}$ and using the notation $\unicode[STIX]{x1D749}=\unicode[STIX]{x1D70F}_{x}\hat{\boldsymbol{x}}+\unicode[STIX]{x1D70F}_{y}\hat{\boldsymbol{y}}+\unicode[STIX]{x1D70F}_{z}\hat{\boldsymbol{z}}$ , $\boldsymbol{n}=n_{x}\hat{\boldsymbol{x}}+n_{y}\hat{\boldsymbol{y}}+n_{z}\hat{\boldsymbol{z}}$ , $\boldsymbol{b}=b_{x}\hat{\boldsymbol{x}}+b_{y}\hat{\boldsymbol{y}}+b_{z}\hat{\boldsymbol{z}}$ , we obtain

(4.21) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D734}\times (\unicode[STIX]{x1D734}\times \boldsymbol{R})=\unicode[STIX]{x1D6FA}^{2}\hat{\boldsymbol{z}}\times (-Y\hat{\boldsymbol{x}}+X\hat{\boldsymbol{y}})=-\unicode[STIX]{x1D6FA}^{2}(X\hat{\boldsymbol{x}}+Y\hat{\boldsymbol{y}}) & & \displaystyle\end{eqnarray}$$

and similarly

(4.22a,b ) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D734}\times (\unicode[STIX]{x1D734}\times \boldsymbol{n})=-\unicode[STIX]{x1D6FA}^{2}(n_{x}\hat{\boldsymbol{x}}+n_{y}\hat{\boldsymbol{y}}),\quad \unicode[STIX]{x1D734}\times (\unicode[STIX]{x1D734}\times \boldsymbol{b})=-\unicode[STIX]{x1D6FA}^{2}(b_{x}\hat{\boldsymbol{x}}+b_{y}\hat{\boldsymbol{y}}). & & \displaystyle\end{eqnarray}$$

Then the centrifugal force takes the form

(4.23) $$\begin{eqnarray}-\unicode[STIX]{x1D734}\times (\unicode[STIX]{x1D734}\times \boldsymbol{r})=\unicode[STIX]{x1D6FA}^{2}(X+\unicode[STIX]{x1D702}n_{x}\cos \unicode[STIX]{x1D703}+\unicode[STIX]{x1D702}b_{x}\sin \unicode[STIX]{x1D703})\hat{\boldsymbol{x}}+\unicode[STIX]{x1D6FA}^{2}(Y+\unicode[STIX]{x1D702}n_{y}\cos \unicode[STIX]{x1D703}+\unicode[STIX]{x1D702}b_{y}\sin \unicode[STIX]{x1D703})\hat{\boldsymbol{y}},\end{eqnarray}$$

or, in projection on the Frenet basis,

(4.24) $$\begin{eqnarray}\displaystyle -\unicode[STIX]{x1D734}\times (\unicode[STIX]{x1D734}\times \boldsymbol{r}) & = & \displaystyle \unicode[STIX]{x1D6FA}^{2}[(X+\unicode[STIX]{x1D702}n_{x}\cos \unicode[STIX]{x1D703}+\unicode[STIX]{x1D702}b_{x}\sin \unicode[STIX]{x1D703})\unicode[STIX]{x1D70F}_{x}+(Y+\unicode[STIX]{x1D702}n_{y}\cos \unicode[STIX]{x1D703}+\unicode[STIX]{x1D702}b_{y}\sin \unicode[STIX]{x1D703})\unicode[STIX]{x1D70F}_{y}]\unicode[STIX]{x1D749}\nonumber\\ \displaystyle & & \displaystyle +\,\unicode[STIX]{x1D6FA}^{2}\left[(X+\unicode[STIX]{x1D702}n_{x}\cos \unicode[STIX]{x1D703}+\unicode[STIX]{x1D702}b_{x}\sin \unicode[STIX]{x1D703})n_{x}\right.\nonumber\\ \displaystyle & & \displaystyle +\,(Y+\unicode[STIX]{x1D702}n_{y}\cos \unicode[STIX]{x1D703}+\unicode[STIX]{x1D702}b_{y}\sin \unicode[STIX]{x1D703})n_{y}\left.\!\right]\boldsymbol{n}\nonumber\\ \displaystyle & & \displaystyle +\,\unicode[STIX]{x1D6FA}^{2}\left[(X+\unicode[STIX]{x1D702}n_{x}\cos \unicode[STIX]{x1D703}+\unicode[STIX]{x1D702}b_{x}\sin \unicode[STIX]{x1D703})b_{x}\right.\nonumber\\ \displaystyle & & \displaystyle +\,(Y+\unicode[STIX]{x1D702}n_{y}\cos \unicode[STIX]{x1D703}+\unicode[STIX]{x1D702}b_{y}\sin \unicode[STIX]{x1D703})b_{y}\left.\!\right]\boldsymbol{b}.\end{eqnarray}$$

Coriolis’ force. Using the representation of $\boldsymbol{v}$ in the Frenet basis (4.12), the Coriolis force can be written down as

(4.25) $$\begin{eqnarray}\displaystyle -2\unicode[STIX]{x1D734}\times \boldsymbol{v} & = & \displaystyle -2(1-\unicode[STIX]{x1D705}_{1}\unicode[STIX]{x1D702}\cos \unicode[STIX]{x1D703})\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}\unicode[STIX]{x1D734}\times \unicode[STIX]{x1D749}\nonumber\\ \displaystyle & & \displaystyle -\,2\left(-\unicode[STIX]{x1D705}_{2}\unicode[STIX]{x1D702}\sin \unicode[STIX]{x1D703}\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}+u_{\unicode[STIX]{x1D702}}\cos \unicode[STIX]{x1D703}-u_{\unicode[STIX]{x1D703}}\sin \unicode[STIX]{x1D703}\right)\unicode[STIX]{x1D734}\times \boldsymbol{n}\nonumber\\ \displaystyle & & \displaystyle -\,2\left(\unicode[STIX]{x1D705}_{2}\unicode[STIX]{x1D702}\cos \unicode[STIX]{x1D703}\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}+u_{\unicode[STIX]{x1D702}}\sin \unicode[STIX]{x1D703}+u_{\unicode[STIX]{x1D703}}\cos \unicode[STIX]{x1D703}\right)\unicode[STIX]{x1D734}\times \boldsymbol{b},\end{eqnarray}$$

and for its projections onto the Frenet basis one has

(4.26) $$\begin{eqnarray}\displaystyle -(2\unicode[STIX]{x1D734}\times \boldsymbol{v})\boldsymbol{\cdot }\unicode[STIX]{x1D749} & = & \displaystyle -2\left(-\unicode[STIX]{x1D705}_{2}\unicode[STIX]{x1D702}\sin \unicode[STIX]{x1D703}\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}+u_{\unicode[STIX]{x1D702}}\cos \unicode[STIX]{x1D703}-u_{\unicode[STIX]{x1D703}}\sin \unicode[STIX]{x1D703}\right)\left(\unicode[STIX]{x1D734}\times \boldsymbol{n}\right)\boldsymbol{\cdot }\unicode[STIX]{x1D749}\nonumber\\ \displaystyle & & \displaystyle -\,2\left(\unicode[STIX]{x1D705}_{2}\unicode[STIX]{x1D702}\cos \unicode[STIX]{x1D703}\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}+u_{\unicode[STIX]{x1D702}}\sin \unicode[STIX]{x1D703}+u_{\unicode[STIX]{x1D703}}\cos \unicode[STIX]{x1D703}\right)(\unicode[STIX]{x1D734}\times \boldsymbol{b})\boldsymbol{\cdot }\unicode[STIX]{x1D749},\qquad\end{eqnarray}$$

(4.27) $$\begin{eqnarray}\displaystyle -(2\unicode[STIX]{x1D734}\times \boldsymbol{v})\boldsymbol{\cdot }\boldsymbol{n} & = & \displaystyle -2(1-\unicode[STIX]{x1D705}_{1}\unicode[STIX]{x1D702}\cos \unicode[STIX]{x1D703})\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}\left(\unicode[STIX]{x1D734}\times \unicode[STIX]{x1D749}\right)\boldsymbol{\cdot }\boldsymbol{n}\nonumber\\ \displaystyle & & \displaystyle -\,2\left(\unicode[STIX]{x1D705}_{2}\unicode[STIX]{x1D702}\cos \unicode[STIX]{x1D703}\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}+u_{\unicode[STIX]{x1D702}}\sin \unicode[STIX]{x1D703}+u_{\unicode[STIX]{x1D703}}\cos \unicode[STIX]{x1D703}\right)\left(\unicode[STIX]{x1D734}\times \boldsymbol{b}\right)\boldsymbol{\cdot }\boldsymbol{n},\qquad\end{eqnarray}$$

(4.28) $$\begin{eqnarray}\displaystyle -(2\unicode[STIX]{x1D734}\times \boldsymbol{v})\boldsymbol{\cdot }\boldsymbol{b} & = & \displaystyle -2(1-\unicode[STIX]{x1D705}_{1}\unicode[STIX]{x1D702}\cos \unicode[STIX]{x1D703})\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}\left(\unicode[STIX]{x1D734}\times \unicode[STIX]{x1D749}\right)\boldsymbol{\cdot }\boldsymbol{b}\nonumber\\ \displaystyle & & \displaystyle -\,2\left(-\unicode[STIX]{x1D705}_{2}\unicode[STIX]{x1D702}\sin \unicode[STIX]{x1D703}\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}+u_{\unicode[STIX]{x1D702}}\cos \unicode[STIX]{x1D703}-u_{\unicode[STIX]{x1D703}}\sin \unicode[STIX]{x1D703}\right)\left(\unicode[STIX]{x1D734}\times \boldsymbol{n}\right)\boldsymbol{\cdot }\boldsymbol{b}.\nonumber\\ \displaystyle \qquad & & \displaystyle\end{eqnarray}$$

For $\unicode[STIX]{x1D734}=\unicode[STIX]{x1D6FA}\hat{\boldsymbol{z}}$ , we obviously have

(4.29a-c ) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D734}\times \unicode[STIX]{x1D749}=\unicode[STIX]{x1D6FA}(-\unicode[STIX]{x1D70F}_{y}\hat{\boldsymbol{x}}+\unicode[STIX]{x1D70F}_{x}\hat{\boldsymbol{y}}),\quad \unicode[STIX]{x1D734}\times \boldsymbol{n}=\unicode[STIX]{x1D6FA}(-n_{y}\hat{\boldsymbol{x}}+n_{x}\hat{\boldsymbol{y}}),\quad \unicode[STIX]{x1D734}\times \boldsymbol{b}=\unicode[STIX]{x1D6FA}(-b_{y}\hat{\boldsymbol{x}}+b_{x}\hat{\boldsymbol{y}}), & & \displaystyle \nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

and the scalar products on the right-hand side of (4.26)–(4.28) are given by

(4.30a,b ) $$\begin{eqnarray}\displaystyle (\unicode[STIX]{x1D734}\times \unicode[STIX]{x1D749})\boldsymbol{\cdot }\boldsymbol{n}=\unicode[STIX]{x1D6FA}(-\unicode[STIX]{x1D70F}_{y}n_{x}+\unicode[STIX]{x1D70F}_{x}n_{y}),\quad (\unicode[STIX]{x1D734}\times \unicode[STIX]{x1D749})\boldsymbol{\cdot }\boldsymbol{b}=\unicode[STIX]{x1D6FA}(-\unicode[STIX]{x1D70F}_{y}b_{x}+\unicode[STIX]{x1D70F}_{x}b_{y}), & & \displaystyle\end{eqnarray}$$

(4.31a,b ) $$\begin{eqnarray}\displaystyle (\unicode[STIX]{x1D734}\times \boldsymbol{n})\boldsymbol{\cdot }\unicode[STIX]{x1D749}=\unicode[STIX]{x1D6FA}(-n_{y}\unicode[STIX]{x1D70F}_{x}+n_{x}\unicode[STIX]{x1D70F}_{y}),\quad (\unicode[STIX]{x1D734}\times \boldsymbol{n})\boldsymbol{\cdot }\boldsymbol{b}=\unicode[STIX]{x1D6FA}(-n_{y}b_{x}+n_{x}b_{y}), & & \displaystyle\end{eqnarray}$$

(4.32a,b ) $$\begin{eqnarray}\displaystyle (\unicode[STIX]{x1D734}\times \boldsymbol{b})\boldsymbol{\cdot }\unicode[STIX]{x1D749}=\unicode[STIX]{x1D6FA}(-b_{y}\unicode[STIX]{x1D70F}_{x}+b_{x}\unicode[STIX]{x1D70F}_{y}),\quad (\unicode[STIX]{x1D734}\times \boldsymbol{b})\boldsymbol{\cdot }\boldsymbol{n}=\unicode[STIX]{x1D6FA}(-b_{y}n_{x}+b_{x}n_{y}). & & \displaystyle\end{eqnarray}$$

Thus, finally projections of the Coriolis force onto the Frenet basis are:

(4.33) $$\begin{eqnarray}\displaystyle -(2\unicode[STIX]{x1D734}\times \boldsymbol{v})\boldsymbol{\cdot }\unicode[STIX]{x1D749} & = & \displaystyle -2\left(-\unicode[STIX]{x1D705}_{2}\unicode[STIX]{x1D702}\sin \unicode[STIX]{x1D703}\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}+u_{\unicode[STIX]{x1D702}}\cos \unicode[STIX]{x1D703}-u_{\unicode[STIX]{x1D703}}\sin \unicode[STIX]{x1D703}\right)\unicode[STIX]{x1D6FA}(-n_{y}\unicode[STIX]{x1D70F}_{x}+n_{x}\unicode[STIX]{x1D70F}_{y})\nonumber\\ \displaystyle & & \displaystyle -\,2\left(\unicode[STIX]{x1D705}_{2}\unicode[STIX]{x1D702}\cos \unicode[STIX]{x1D703}\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}+u_{\unicode[STIX]{x1D702}}\sin \unicode[STIX]{x1D703}+u_{\unicode[STIX]{x1D703}}\cos \unicode[STIX]{x1D703}\right)\unicode[STIX]{x1D6FA}(-b_{y}\unicode[STIX]{x1D70F}_{x}+b_{x}\unicode[STIX]{x1D70F}_{y}),\nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

(4.34) $$\begin{eqnarray}\displaystyle -(2\unicode[STIX]{x1D734}\times \boldsymbol{v})\boldsymbol{\cdot }\boldsymbol{n} & = & \displaystyle -2(1-\unicode[STIX]{x1D705}_{1}\unicode[STIX]{x1D702}\cos \unicode[STIX]{x1D703})\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}\unicode[STIX]{x1D6FA}(-\unicode[STIX]{x1D70F}_{y}n_{x}+\unicode[STIX]{x1D70F}_{x}n_{y})\nonumber\\ \displaystyle & & \displaystyle -\,2\left(\unicode[STIX]{x1D705}_{2}\unicode[STIX]{x1D702}\cos \unicode[STIX]{x1D703}\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}+u_{\unicode[STIX]{x1D702}}\sin \unicode[STIX]{x1D703}+u_{\unicode[STIX]{x1D703}}\cos \unicode[STIX]{x1D703}\right)\unicode[STIX]{x1D6FA}(-b_{y}n_{x}+b_{x}n_{y}),\nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

(4.35) $$\begin{eqnarray}\displaystyle -(2\unicode[STIX]{x1D734}\times \boldsymbol{v})\boldsymbol{\cdot }\boldsymbol{b} & = & \displaystyle -2(1-\unicode[STIX]{x1D705}_{1}\unicode[STIX]{x1D702}\cos \unicode[STIX]{x1D703})\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}\unicode[STIX]{x1D6FA}(-\unicode[STIX]{x1D70F}_{y}b_{x}+\unicode[STIX]{x1D70F}_{x}b_{y})\nonumber\\ \displaystyle & & \displaystyle -\,2\left(-\unicode[STIX]{x1D705}_{2}\unicode[STIX]{x1D702}\sin \unicode[STIX]{x1D703}\frac{u_{\unicode[STIX]{x1D709}}}{\sqrt{\unicode[STIX]{x1D628}_{11}}}+u_{\unicode[STIX]{x1D702}}\cos \unicode[STIX]{x1D703}-u_{\unicode[STIX]{x1D703}}\sin \unicode[STIX]{x1D703}\right)\unicode[STIX]{x1D6FA}(-n_{y}b_{x}+n_{x}b_{y}).\nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

Gravity. The gravity force $\boldsymbol{g}=-g\hat{\boldsymbol{z}}$ in projection onto the Frenet basis is simply $\boldsymbol{g}=-g\unicode[STIX]{x1D70F}_{z}\unicode[STIX]{x1D749}-gn_{z}\boldsymbol{n}-gb_{z}\boldsymbol{b}$ .

5.3.1 Case 1: $\mathit{We}=O(1)$ as $\unicode[STIX]{x1D716}\rightarrow 0$

For $\mathit{We}=O(1)$ , as we can see from equations (5.10)–(5.12), the term with the pressure gradient is of the same order, i.e. of $O(1)$ , as other terms in the projection of equations of motion on $\unicode[STIX]{x1D749}$ (5.10) whilst in normal and binormal projections (5.11), (5.12) it is dominant (of $O(\unicode[STIX]{x1D716}^{-1})$ ) so that the next-order approximation and hence the first-order corrections to $p$ and then to $h$ will have to be brought in.

On substitution of expansions (5.18) into equations (5.11), (5.12), to leading order (i.e. to order $\unicode[STIX]{x1D716}^{-1}$ ), these equations become

(5.24a,b ) $$\begin{eqnarray}\cos \unicode[STIX]{x1D703}\frac{\unicode[STIX]{x2202}p_{0}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D702}}-\frac{\sin \unicode[STIX]{x1D703}}{\unicode[STIX]{x1D702}}\frac{\unicode[STIX]{x2202}p_{0}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}=0,\quad \sin \unicode[STIX]{x1D703}\frac{\unicode[STIX]{x2202}p_{0}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D702}}+\frac{\cos \unicode[STIX]{x1D703}}{\unicode[STIX]{x1D702}}\frac{\unicode[STIX]{x2202}p_{0}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}=0,\end{eqnarray}$$

giving that $p_{0}=p_{0}(\unicode[STIX]{x1D709})$ . Then, from the dynamic boundary condition for the zeroth-order terms (5.15) subject to periodicity conditions

(5.25a,b ) $$\begin{eqnarray}h_{0}(\unicode[STIX]{x1D709},0)=h_{0}(\unicode[STIX]{x1D709},2\unicode[STIX]{x03C0}),\quad \frac{\unicode[STIX]{x2202}h_{0}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}(\unicode[STIX]{x1D709},0)=\frac{\unicode[STIX]{x2202}h_{0}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}(\unicode[STIX]{x1D709},2\unicode[STIX]{x03C0})\end{eqnarray}$$

we have that the jet’s cross-section is a circle

(5.26) $$\begin{eqnarray}h_{0}^{2}+\unicode[STIX]{x1D702}_{c}^{2}-2h_{0}\unicode[STIX]{x1D702}_{c}\cos (\unicode[STIX]{x1D703}-\unicode[STIX]{x1D703}_{c})=1/p_{0}^{2},\end{eqnarray}$$

where $\unicode[STIX]{x1D702}_{c}$ and $\unicode[STIX]{x1D703}_{c}$ are coordinates of the circle’s centre. It can be easily shown that this is the general solution: $h_{0}$ in (5.15) can be rescaled to ‘absorb’ $p_{0}$ since the numerator and the denominator of (5.15) are homogeneous with respect to $h_{0}$ and for the resulting parameter-free equation $h_{0}$ given by (5.26) (with 1 on the right-hand side) will be a solution depending on two arbitrary constants, i.e. the general solution.

Now, since the ‘centre’ of the cross-section appears to be well defined as the centre of the circle we can, without loss of generality, choose our ‘baseline’ to be the ‘centreline’, i.e. make $\unicode[STIX]{x1D702}_{c}=0$ , so that $h_{0}$ satisfying (5.15), (5.25) now becomes $h_{0}=h_{0}(\unicode[STIX]{x1D709})=1/p_{0}(\unicode[STIX]{x1D709})$ , and hence from (5.23) one has $Q_{1}=\text{const.}$ , whilst the pressure is obviously given by

(5.27) $$\begin{eqnarray}p_{0}(\unicode[STIX]{x1D709})=\frac{1}{h_{0}(\unicode[STIX]{x1D709})}.\end{eqnarray}$$

With $Q_{1}=\text{const.}$ and $h_{0}=h_{0}(\unicode[STIX]{x1D709})$ , we can now see that (5.23) is essentially a conservation law stating that the volumetric flux, $\unicode[STIX]{x03C0}Q_{1}$ , remains constant along the jet.

Now, substituting (5.27) into the $\unicode[STIX]{x1D70F}$ -projection of the equations of motion (5.10), expressing $h_{0}$ in terms of $u_{\unicode[STIX]{x1D709},0}$ from (5.23) and using (3.2) to write down explicitly the Cartesian projections of $\unicode[STIX]{x1D749}$ , we obtain

(5.28) $$\begin{eqnarray}\displaystyle u_{\unicode[STIX]{x1D709},0}\frac{\text{d}u_{\unicode[STIX]{x1D709},0}}{\text{d}\unicode[STIX]{x1D709}}=-\frac{1}{\mathit{We}Q_{1}^{1/2}}\frac{\text{d}u_{\unicode[STIX]{x1D709},0}^{1/2}}{\text{d}\unicode[STIX]{x1D709}}-\frac{1}{\mathit{Fr}^{2}}Z^{\prime }+\frac{1}{\mathit{Rb}^{2}}(XX^{\prime }+YY^{\prime }). & & \displaystyle\end{eqnarray}$$

After integrating this equation, we arrive at a depressed quartic in $u_{\unicode[STIX]{x1D709},0}^{1/2}$ :

(5.29) $$\begin{eqnarray}u_{\unicode[STIX]{x1D709},0}^{2}+\frac{2}{\mathit{We}Q_{1}^{1/2}}u_{\unicode[STIX]{x1D709},0}^{1/2}+\frac{2}{\mathit{Fr}^{2}}Z-\frac{1}{\mathit{Rb}^{2}}(X^{2}+Y^{2})+Q_{2}=0,\end{eqnarray}$$

where $Q_{2}$ is the constant of integration. This equation specifies $u_{\unicode[STIX]{x1D709},0}$ as a function of $X$ , $Y$ , $Z$ . Although it can be solved analytically in a standard way (Korn & Korn Reference Korn and Korn1968), in practice, for its application below it is more convenient to solve it numerically.

Now, considering equations of motion (5.11), (5.12) to $O(1)$ as $\unicode[STIX]{x1D716}\rightarrow 0$ , we have

(5.30a,b ) $$\begin{eqnarray}\cos \unicode[STIX]{x1D703}\frac{\unicode[STIX]{x2202}p_{1}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D702}}-\frac{\sin \unicode[STIX]{x1D703}}{\unicode[STIX]{x1D702}}\frac{\unicode[STIX]{x2202}p_{1}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}=\mathit{We}A(\unicode[STIX]{x1D709},u_{\unicode[STIX]{x1D709},0}),\quad \sin \unicode[STIX]{x1D703}\frac{\unicode[STIX]{x2202}p_{1}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D702}}+\frac{\cos \unicode[STIX]{x1D703}}{\unicode[STIX]{x1D702}}\frac{\unicode[STIX]{x2202}p_{1}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}=\mathit{We}B(\unicode[STIX]{x1D709},u_{\unicode[STIX]{x1D709},0}),\end{eqnarray}$$

where for brevity we introduced the notation

(5.31) $$\begin{eqnarray}\displaystyle & \displaystyle A(\unicode[STIX]{x1D709},u_{\unicode[STIX]{x1D709},0})=-u_{\unicode[STIX]{x1D709},0}^{2}\unicode[STIX]{x1D705}_{1}-\frac{1}{\mathit{Fr}^{2}}n_{z}-\frac{2}{\mathit{Rb}}u_{\unicode[STIX]{x1D709},0}(\unicode[STIX]{x1D70F}_{x}n_{y}-\unicode[STIX]{x1D70F}_{y}n_{x})+\frac{1}{\mathit{Rb}^{2}}(Xn_{x}+Yn_{y}), & \displaystyle\end{eqnarray}$$

(5.32) $$\begin{eqnarray}\displaystyle & \displaystyle B(\unicode[STIX]{x1D709},u_{\unicode[STIX]{x1D709},0})=-\frac{1}{\mathit{Fr}^{2}}b_{z}-\frac{2}{\mathit{Rb}}u_{\unicode[STIX]{x1D709},0}(\unicode[STIX]{x1D70F}_{x}b_{y}-\unicode[STIX]{x1D70F}_{y}b_{x})+\frac{1}{\mathit{Rb}^{2}}(Xb_{x}+Yb_{y}). & \displaystyle\end{eqnarray}$$

In writing down the arguments of $A$ and $B$ we emphasize that there is a ‘direct’ dependence on $\unicode[STIX]{x1D709}$ via $X$ , $Y$ , $Z$ (and projections of $\unicode[STIX]{x1D749}$ , $\boldsymbol{n}$ , $\boldsymbol{b}$ expressible in terms of $X$ , $Y$ , $Z$ via (3.2)–(3.4)) and also the ‘indirect’ dependence on $\unicode[STIX]{x1D709}$ via $u_{\unicode[STIX]{x1D709},0}$ , which in this case is given by (5.29) and also depends only on $\unicode[STIX]{x1D709}$ .

We can take instead of (5.30) their linear combinations

(5.33a,b ) $$\begin{eqnarray}\frac{\unicode[STIX]{x2202}p_{1}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D702}}=\mathit{We}(A\cos \unicode[STIX]{x1D703}+B\sin \unicode[STIX]{x1D703}),\quad \frac{\unicode[STIX]{x2202}p_{1}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}=\mathit{We}\unicode[STIX]{x1D702}(-A\sin \unicode[STIX]{x1D703}+B\cos \unicode[STIX]{x1D703}).\end{eqnarray}$$

After integrating the first of these equations, we obtain $p_{1}=\mathit{We}\unicode[STIX]{x1D702}(A\cos \unicode[STIX]{x1D703}+B\sin \unicode[STIX]{x1D703})+C(\unicode[STIX]{x1D709},\unicode[STIX]{x1D703}),$ where $C(\unicode[STIX]{x1D709},\unicode[STIX]{x1D703})$ is the ‘constant’ of integration. Substituting this expression into the second equation (5.33) gives $\unicode[STIX]{x2202}C/\unicode[STIX]{x2202}\unicode[STIX]{x1D703}=0,$ i.e.  $C=C(\unicode[STIX]{x1D709})$ . Thus, we now have

(5.34) $$\begin{eqnarray}p_{1}=\mathit{We}\,\unicode[STIX]{x1D702}(A\cos \unicode[STIX]{x1D703}+B\sin \unicode[STIX]{x1D703})+C(\unicode[STIX]{x1D709}),\end{eqnarray}$$

which we need to substitute into the dynamic boundary condition for the first-order terms (5.16) where for $h_{0}$ independent of $\unicode[STIX]{x1D703}$ the right-hand side considerably simplifies:

(5.35) $$\begin{eqnarray}\displaystyle \mathit{We}\,h_{0}(A\cos \unicode[STIX]{x1D703}+B\sin \unicode[STIX]{x1D703})+C=-\unicode[STIX]{x1D705}_{1}\cos \unicode[STIX]{x1D703}-\frac{1}{h_{0}^{2}}h_{1}-\frac{1}{h_{0}^{2}}\frac{\unicode[STIX]{x2202}^{2}h_{1}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}^{2}}, & & \displaystyle\end{eqnarray}$$

or, to write it in the standard form,

(5.36) $$\begin{eqnarray}\frac{\unicode[STIX]{x2202}^{2}h_{1}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}^{2}}+h_{1}=-h_{0}^{2}(\mathit{We}\,h_{0}A+\unicode[STIX]{x1D705}_{1})\cos \unicode[STIX]{x1D703}-\mathit{We}\,h_{0}^{3}B\sin \unicode[STIX]{x1D703}-Ch_{0}^{2}.\end{eqnarray}$$

This is a linear ordinary differential equation with $\unicode[STIX]{x1D703}$ as the independent variable and $\unicode[STIX]{x1D709}$ as a parameter. Since both $\sin \unicode[STIX]{x1D703}$ and $\cos \unicode[STIX]{x1D703}$ satisfy the corresponding homogeneous equation, (5.36) has $2\unicode[STIX]{x03C0}$ -periodic solutions for $h_{1}$ only if the coefficients in front of $\sin \unicode[STIX]{x1D703}$ and $\cos \unicode[STIX]{x1D703}$ on the right-hand side are both zero, i.e. if

(5.37a,b ) $$\begin{eqnarray}\mathit{We}\,h_{0}A+\unicode[STIX]{x1D705}_{1}=0,\quad B=0.\end{eqnarray}$$

These equations, together with the second equation in (3.2), namely $X^{\prime 2}+Y^{\prime 2}+Z^{\prime 2}=1$ , form a closed set of ordinary differential equations (ODEs) describing the jet’s trajectory, i.e. the functions $X(\unicode[STIX]{x1D709})$ , $Y(\unicode[STIX]{x1D709})$ , $Z(\unicode[STIX]{x1D709})$ . To write down this set of equations explicitly, we recall expressions (3.2)–(3.4) for the Cartesian components of $\unicode[STIX]{x1D749}$ , $\boldsymbol{n}$ , $\boldsymbol{b}$ , expression (3.6) for $\unicode[STIX]{x1D705}_{1}$ , use equation (5.23) to eliminate $h_{0}$ and, to get rid of the square roots, multiply (5.37) by $-\unicode[STIX]{x1D705}_{1}$ . Then, the set of ODEs for the jet’s centreline takes the form

(5.38) $$\begin{eqnarray}\displaystyle & & \displaystyle \left(u_{\unicode[STIX]{x1D709},0}^{2}-\frac{u_{\unicode[STIX]{x1D709},0}^{1/2}}{\mathit{We}\,Q_{1}^{1/2}}\right)(X^{\prime \prime 2}+Y^{\prime \prime 2}+Z^{\prime \prime 2})+\frac{1}{\mathit{Fr}^{2}}Z^{\prime \prime }\nonumber\\ \displaystyle & & \displaystyle \quad +\,\frac{2}{\mathit{Rb}}u_{\unicode[STIX]{x1D709},0}(X^{\prime }Y^{\prime \prime }-Y^{\prime }X^{\prime \prime })-\frac{1}{\mathit{Rb}^{2}}(XX^{\prime \prime }+YY^{\prime \prime })=0,\end{eqnarray}$$

(5.39) $$\begin{eqnarray}\displaystyle & & \displaystyle \frac{1}{\mathit{Fr}^{2}}(X^{\prime }Y^{\prime \prime }-Y^{\prime }X^{\prime \prime })+\frac{2}{\mathit{Rb}}u_{\unicode[STIX]{x1D709},0}[X^{\prime }(Z^{\prime }X^{\prime \prime }-X^{\prime }Z^{\prime \prime })-Y^{\prime }(Y^{\prime }Z^{\prime \prime }-Z^{\prime }Y^{\prime \prime })]\nonumber\\ \displaystyle & & \displaystyle \quad -\,\frac{1}{\mathit{Rb}^{2}}[X(Y^{\prime }Z^{\prime \prime }-Z^{\prime }Y^{\prime \prime })+Y(Z^{\prime }X^{\prime \prime }-X^{\prime }Z^{\prime \prime })]=0,\end{eqnarray}$$

(5.40) $$\begin{eqnarray}\displaystyle & X^{\prime 2}+Y^{\prime 2}+Z^{\prime 2}=1, & \displaystyle\end{eqnarray}$$

where $u_{\unicode[STIX]{x1D709},0}(X,Y,Z)$ is specified by (5.29). To compute the centreline, one needs to specify its initial point and slope, i.e. prescribe $X(0)$ , $Y(0)$ , $Z(0)$ and $X^{\prime }(0)$ , $Y^{\prime }(0)$ , $Z^{\prime }(0)$ , of which, obviously, only two can be specified independently, and, knowing $u_{\unicode[STIX]{x1D709},0}(0)$ and $h_{0}(0)$ at this point, specify $Q_{1}$ via (5.23) and, using $u_{\unicode[STIX]{x1D709},0}(0)$ together with $X(0)$ , $Y(0)$ , $Z(0)$ , $Q_{2}$ via (5.29).

Note that the analogy with the motion of a material point in the Frenet basis that we pointed out in § 5.2 here becomes even more transparent: as in the case of a material point (Butenin et al. Reference Butenin, Lunz and Merkin1979), the $\unicode[STIX]{x1D70F}$ -projection of the equation of motion determines the velocity (5.29) whilst the fact that it is a free jet dictates that the ‘reaction’ from the pressure in terms of its gradient across the jet must be zero. The latter gives the two solvability conditions (5.37), i.e. (5.38)–(5.39), which, together with the normalization condition (5.40), determine the jet’s trajectory. In other words, the prescribed trajectory of a material point (e.g. in the design of a roller-coaster track) determines the normal and binormal components of the reaction forces (from the track) whilst for a free jet it is the other way round: the equality to zero of the normal and binormal ‘reactions’ from the pressure gradient give the conditions that determine the jet’s trajectory.

It is worth noting here that numerical analysis of (5.29), (5.38)–(5.40) shows that for the trajectory to be, as one expects, an outgoing spiral, i.e. the process to be dominated by inertia rather than capillarity, the expression in the first bracket of (5.38) must be positive.

5.3.2 Case 2: $\unicode[STIX]{x1D716}\mathit{We}=O(1)$ as $\unicode[STIX]{x1D716}\rightarrow 0$

For $\mathit{We}=O(\unicode[STIX]{x1D716}^{-1})$ we will be dealing only with the zeroth-order terms of the expansions (5.18) and, having in mind the results of the previous section, look for a solution where the cross-section of the jet is circular, so that the baseline is again the centreline, $h_{0}=h_{0}(\unicode[STIX]{x1D709})$ (and hence $Q_{1}=\text{const.}$ ), to see whether such a solution exists.

As one can see from equations (5.10)–(5.12), for $\mathit{We}=O(\unicode[STIX]{x1D716}^{-1})$ as $\unicode[STIX]{x1D716}\rightarrow 0$ , the terms with the pressure gradient are of the same $O(1)$ as other terms in the normal and binormal projections of the equations of motion (5.11), (5.12), whilst, to leading order, the corresponding term drops out from the tangential projection (5.10), so that, instead of (5.10), one now has

(5.41) $$\begin{eqnarray}\frac{\unicode[STIX]{x2202}u_{\unicode[STIX]{x1D709}}}{\unicode[STIX]{x2202}t}+u_{\unicode[STIX]{x1D709}}\frac{\unicode[STIX]{x2202}u_{\unicode[STIX]{x1D709}}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D709}}+u_{\unicode[STIX]{x1D702}}\frac{\unicode[STIX]{x2202}u_{\unicode[STIX]{x1D709}}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D702}}+\frac{u_{\unicode[STIX]{x1D703}}}{\unicode[STIX]{x1D702}}\frac{\unicode[STIX]{x2202}u_{\unicode[STIX]{x1D709}}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}=-\frac{1}{\mathit{Fr}^{2}}\unicode[STIX]{x1D70F}_{z}+\frac{1}{\mathit{Rb}^{2}}(X\unicode[STIX]{x1D70F}_{x}+Y\unicode[STIX]{x1D70F}_{y}).\end{eqnarray}$$

As before, using the definition of $\unicode[STIX]{x1D749}$ (3.2) and our assumptions (5.20), we can write down this equation for $u_{\unicode[STIX]{x1D709},0}$ as

(5.42) $$\begin{eqnarray}\displaystyle u_{\unicode[STIX]{x1D709},0}\frac{\text{d}u_{\unicode[STIX]{x1D709},0}}{\text{d}\unicode[STIX]{x1D709}}=-\frac{1}{\mathit{Fr}^{2}}Z^{\prime }+\frac{1}{\mathit{Rb}^{2}}(XX^{\prime }+YY^{\prime }), & & \displaystyle\end{eqnarray}$$

which, after integration, gives

(5.43) $$\begin{eqnarray}u_{\unicode[STIX]{x1D709},0}=\left[-\frac{2}{\mathit{Fr}^{2}}Z+\frac{1}{\mathit{Rb}^{2}}(X^{2}+Y^{2})+Q_{2}\right]^{1/2},\end{eqnarray}$$

where $Q_{2}$ is the constant of integration.

Considering equations (5.11), (5.12), which now have the form

(5.44a,b ) $$\begin{eqnarray}\cos \unicode[STIX]{x1D703}\frac{\unicode[STIX]{x2202}p_{0}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D702}}-\frac{\sin \unicode[STIX]{x1D703}}{\unicode[STIX]{x1D702}}\frac{\unicode[STIX]{x2202}p_{0}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}=\unicode[STIX]{x1D716}\mathit{We}\,A(\unicode[STIX]{x1D709},u_{\unicode[STIX]{x1D709},0}),\quad \sin \unicode[STIX]{x1D703}\frac{\unicode[STIX]{x2202}p_{0}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D702}}+\frac{\cos \unicode[STIX]{x1D703}}{\unicode[STIX]{x1D702}}\frac{\unicode[STIX]{x2202}p_{0}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}=\unicode[STIX]{x1D716}\mathit{We}\,B(\unicode[STIX]{x1D709},u_{\unicode[STIX]{x1D709},0}),\end{eqnarray}$$

where $A$ and $B$ are defined by (5.31), (5.32), we can repeat the steps that led from (5.30) to (5.34) arriving at

(5.45) $$\begin{eqnarray}p_{0}=\unicode[STIX]{x1D716}\mathit{We}\,\unicode[STIX]{x1D702}(A\cos \unicode[STIX]{x1D703}+B\sin \unicode[STIX]{x1D703})+C(\unicode[STIX]{x1D709}).\end{eqnarray}$$

Substituting this expression into the dynamic boundary condition (5.15), which for a circular cross-section has the form $p_{0}=1/h_{0}$ at $\unicode[STIX]{x1D702}=h_{0},$ we obtain

(5.46) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D716}\mathit{We}\,h_{0}(\unicode[STIX]{x1D709})[A(\unicode[STIX]{x1D709},u_{\unicode[STIX]{x1D709},0}(\unicode[STIX]{x1D709}))\cos \unicode[STIX]{x1D703}+B(\unicode[STIX]{x1D709},u_{\unicode[STIX]{x1D709},0}(\unicode[STIX]{x1D709}))\sin \unicode[STIX]{x1D703}]+C(\unicode[STIX]{x1D709})=\frac{1}{h_{0}(\unicode[STIX]{x1D709})}. & & \displaystyle\end{eqnarray}$$

Unlike (5.36), this is now an algebraic equation, and it can be satisfied only if

(5.47a,b ) $$\begin{eqnarray}A(\unicode[STIX]{x1D709},u_{\unicode[STIX]{x1D709},0})=0,\quad B(\unicode[STIX]{x1D709},u_{\unicode[STIX]{x1D709},0})=0,\end{eqnarray}$$

and $C(\unicode[STIX]{x1D709})=1/h_{0}(\unicode[STIX]{x1D709})$ . Thus, we have that the pressure is uniform across the jet,

(5.48) $$\begin{eqnarray}p_{0}=\frac{1}{h_{0}(\unicode[STIX]{x1D709})},\end{eqnarray}$$

and equations (5.47) together with the second equation (3.2), namely $X^{\prime 2}+Y^{\prime 2}+Z^{\prime 2}=1$ , form a system of three ODEs which, together with the appropriate boundary conditions, determine $X(\unicode[STIX]{x1D709})$ , $Y(\unicode[STIX]{x1D709})$ , $Z(\unicode[STIX]{x1D709})$ , i.e. the shape of the centreline. To write these equations down explicitly, we again recall the expressions in terms of $X$ , $Y$ , $Z$ for the components of $\unicode[STIX]{x1D749}$ , $\boldsymbol{n}$ , $\boldsymbol{b}$ given by (3.2)–(3.4), for the curvature $\unicode[STIX]{x1D705}_{1}$ in (3.6) and expression (5.43) obtained earlier for $u_{\unicode[STIX]{x1D709},0}$ , substitute all these expressions into the definitions (5.31), (5.32) for $A$ and $B$ and the obtained expressions for $A$ and $B$ into equations (5.47). Multiplying the result by $-\unicode[STIX]{x1D705}_{1}$ to get rid of square roots, we finally arrive at the following set of equations describing the shape of the centreline, i.e. the trajectory of the jet:

(5.49) $$\begin{eqnarray}u_{\unicode[STIX]{x1D709},0}^{2}(X^{\prime \prime 2}+Y^{\prime \prime 2}+Z^{\prime \prime 2})+\frac{1}{\mathit{Fr}^{2}}Z^{\prime \prime }+\frac{2}{\mathit{Rb}}u_{\unicode[STIX]{x1D709},0}(X^{\prime }Y^{\prime \prime }-Y^{\prime }X^{\prime \prime })-\frac{1}{\mathit{Rb}^{2}}(XX^{\prime \prime }+YY^{\prime \prime })=0,\end{eqnarray}$$

(5.50) $$\begin{eqnarray}\displaystyle & & \displaystyle \frac{1}{\mathit{Fr}^{2}}(X^{\prime }Y^{\prime \prime }-Y^{\prime }X^{\prime \prime })+\frac{2}{\mathit{Rb}}u_{\unicode[STIX]{x1D709},0}[X^{\prime }(Z^{\prime }X^{\prime \prime }-X^{\prime }Z^{\prime \prime })-Y^{\prime }(Y^{\prime }Z^{\prime \prime }-Z^{\prime }Y^{\prime \prime })]\nonumber\\ \displaystyle & & \displaystyle \qquad -\,\frac{1}{\mathit{Rb}^{2}}[X(Y^{\prime }Z^{\prime \prime }-Z^{\prime }Y^{\prime \prime })+Y(Z^{\prime }X^{\prime \prime }-X^{\prime }Z^{\prime \prime })]=0,\end{eqnarray}$$

(5.51) $$\begin{eqnarray}X^{\prime 2}+Y^{\prime 2}+Z^{\prime 2}=1,\end{eqnarray}$$

where

(5.52) $$\begin{eqnarray}u_{\unicode[STIX]{x1D709},0}=\left[-\frac{2}{\mathit{Fr}^{2}}Z+\frac{1}{\mathit{Rb}^{2}}(X^{2}+Y^{2})+Q_{2}\right]^{1/2}.\end{eqnarray}$$

This set of equations differs from equations (5.38)–(5.40) of § 5.3.1 in two ways: (i) the first term in (5.38) includes the influence of capillarity on the jet’s trajectory whilst in (5.49) there is no such influence, and, more importantly, (ii) $u_{\unicode[STIX]{x1D709},0}$ is calculated differently, namely using (5.29) if $\mathit{We}=O(1)$ and using (5.52) if $\unicode[STIX]{x1D716}\mathit{We}=O(1)$ as $\unicode[STIX]{x1D716}\rightarrow 0$ . It is noteworthy, however, that although the two cases involved different asymptotic procedures, with the first one requiring two-term expansion of $p$ and $h$ , the end results exhibit reassuring continuity: if we formally take the limit $\mathit{We}\rightarrow \infty$ in (5.38) and (5.29), the system (5.38)–(5.40), (5.29) will be reduced to (5.49)–(5.52).

To compute the jet’s trajectory using (5.49)–(5.51), one, again, needs to specify initial conditions, i.e. $X(0)$ , $Y(0)$ , $Z(0)$ and two of $X^{\prime }(0)$ , $Y^{\prime }(0)$ , $Z^{\prime }(0)$ , and, with the known value of $u_{\unicode[STIX]{x1D709},0}(0)$ together with $X(0)$ , $Y(0)$ , $Z(0)$ , specify $Q_{2}$ using (5.51).

Note that by looking for a solution with a circular cross-section we simply took a shortcut and, by reducing the problem to a closed set of ODEs (5.49)–(5.52) from which the flow parameters can be calculated via (5.43), (5.23) and (5.48), demonstrated that such a solution exists. An alternative, less insightful and hence more cumbersome way would be, as in § 5.3.1, not to make an assumption that the cross-section is circular, substitute (5.34) into the dynamic boundary condition (5.15) and show that $h_{0}$ as a $2\unicode[STIX]{x03C0}$ -periodic function of $\unicode[STIX]{x1D703}$ exists only if conditions (5.47) are satisfied and that the cross-section is a circle. This is similar, for example, to the situation with the equilibrium shapes of a liquid drop sitting on a solid substrate in the gravity field (Shikhmurzaev Reference Shikhmurzaev1997): the linear (due to gravity) pressure distribution (5.34) results in a drop touching the substrate at one point (i.e. with the cross-section being a circle) only if (5.47) hold, i.e. if $p$ is a constant across the drop; otherwise, the drop has a finite base on which it sits, i.e. a $2\unicode[STIX]{x03C0}$ -periodic free-surface profile does not exist.

The analogy with the motion of a material point mentioned earlier is even more obvious here than in the previous section, as the fluid’s motion along the trajectory, as can be seen from (5.52), is determined entirely by the body forces with no influence of the pressure gradient. The two solvability conditions (5.49), (5.50), as before, come from the fact that it is a free jet and hence there is no ‘reaction’ from the pressure in terms of its gradient in the normal and binormal directions to the trajectory.

5.3.3 Typical mistakes

Setting aside the errors in the derivation of the governing equations and boundary conditions mentioned earlier, here we will point out some mistakes of principle occurring in the way by which equations for the jet’s trajectory are derived in the slender-jet approximation.

As shown above, in either of the two cases specifying the behaviour of $\mathit{We}$ as $\unicode[STIX]{x1D716}\rightarrow 0$ , the closed set of three ODEs for $X$ , $Y$ , $Z$ is formed by (i) two solvability conditions, which essentially state that we are dealing with a free jet and hence the pressure gradient across it must be zero, and (ii) the normalization equation $X^{\prime 2}+Y^{\prime 2}+Z^{\prime 2}=1$ stating that the independent variable $\unicode[STIX]{x1D709}$ is the arclength. This set of equations remains the same regardless whether gravity is included into consideration or not as the Froude number $\mathit{Fr}$ is just a parameter in two of these equations. It is also worth noting that even if gravity is neglected, the jet’s trajectory can still have all three of its Cartesian coordinates varying if the conditions at the outlet from which this jet is produced specify that the jet will not lie in the horizontal coordinate plane – and hence one will still need all three equations for $X$ , $Y$ , $Z$ . This is invariably the case in all applications as no set-up generating spiralling jets is symmetric with respect to any horizontal plane.

In Wallwork et al. (Reference Wallwork, Decent, King and Schulkes2002), where gravity is neglected, and its twin paper (Decent et al. Reference Decent, King and Wallwork2002), which differs only by the addition of gravity, the situation is different. In Wallwork et al. (Reference Wallwork, Decent, King and Schulkes2002), the vertical coordinate of the jet’s trajectory is assumed to be zero, so that the authors formulate only equations for the two horizontal coordinates of the trajectory (which somehow appear to be inseparably linked to an equation for the variation of the radius of the jet’s circular cross-section along the trajectory). As a result of this ad hoc approach, when in Decent et al. (Reference Decent, King and Wallwork2002) gravity is taken into consideration and the jet’s trajectory can no longer be assumed to remain in the horizontal plane, the authors, as they admit, find themselves short of one equation. The way out appears to be, to put it mildly, more than ad hoc: Decent et al. (Reference Decent, King and Wallwork2002) take a linear combination of two equations of motion (normal and azimuthal) and declare $p_{1}$ of the form of our (5.34) to be its solution from which one missing equation for the trajectory can be obtained as the solvability condition. What should be noted in this regard is that this ‘solution’ satisfies neither of the equations of motion (33) and (35) of Decent et al. (Reference Decent, King and Wallwork2002), as one can easily see just from the dependence of the terms on the azimuthal angle, nor does it satisfy even their linear combination (44) should this linear combination be written down correctly. This is not to mention that two equations cannot be reduced to one (their linear combination) as the number of equations should stay the same, and hence the solution should satisfy both of them. Setting aside this departure from mathematics and the passing of a non-solution for a solution, it is worth noting that, if $p_{1}$ of the form (5.34) is required to satisfy the (correct version of the) normal and azimuthal projections of the equation of motion, this produces, as in the present paper, two solvability conditions whereas in Decent et al. (Reference Decent, King and Wallwork2002) only one ‘extra’ equation was required as otherwise the system becomes overdetermined. This fact alone exposes a flaw of principle in the whole procedure.

5.4.1 Case 1: $\mathit{We}=O(1)$ as $\unicode[STIX]{x1D716}\rightarrow 0$

If $\mathit{We}=O(1)$ and $\unicode[STIX]{x1D716}\unicode[STIX]{x1D6FF}^{-2}=O(1)$ as $\unicode[STIX]{x1D716}\rightarrow 0$ , then, from (5.56), (5.57) one has that the leading, i.e.  $O(1)$ , terms are those with the pressure gradient, so that one immediately has $p_{0}=p_{0}(\unicode[STIX]{x1D709},t)$ . Used in the dynamic boundary condition (5.15), this means that the curvature of the jet’s free surface in the cross-sectional plane is a constant and hence the cross-section of the jet is a circle. Assuming that its centre remains on the baseline, we have that $h_{0}=h_{0}(\unicode[STIX]{x1D709},t)$ and hence

(5.61) $$\begin{eqnarray}\displaystyle p_{0}=\frac{1}{h_{0}}. & & \displaystyle\end{eqnarray}$$

Using this in the leading-order equation obtained from (5.55), we arrive at

(5.62) $$\begin{eqnarray}\frac{\unicode[STIX]{x2202}u_{\unicode[STIX]{x1D709},0}}{\unicode[STIX]{x2202}\bar{t}}+u_{\unicode[STIX]{x1D709},0}\frac{\unicode[STIX]{x2202}u_{\unicode[STIX]{x1D709},0}}{\unicode[STIX]{x2202}\bar{\unicode[STIX]{x1D709}}}+\frac{1}{\mathit{We}}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\bar{\unicode[STIX]{x1D709}}}\left(\frac{1}{h_{0}}\right)=0.\end{eqnarray}$$

Equations (5.60), (5.62) form a closed system describing peristaltic waves in the slender-jet approximation in the case $\mathit{We}=O(1)$ as $\unicode[STIX]{x1D716}\rightarrow 0$ . These equations are exactly the ones derived by Ting & Keller (Reference Ting and Keller1990) for a straight axisymmetric jet. This is what one should expect as in the case where the wavelength is much shorter than the radius of curvature of the jet’s trajectory the latter will not feature at leading order in the slender-jet approximation as, to put it simply, for such waves the jet is locally straight. The body forces do not appear at this order also since their influence determines the jet’s trajectory and on a much shorter scale they, to leading order, play no role.

In the case of a straight axisymmetric jet, corrections to (5.60), (5.62) corresponding to shorter wavelengths have been considered by Markova & Shkadov (Reference Markova and Shkadov1972) and by Papageorgiou & Orellana (Reference Papageorgiou and Orellana1998) who included higher-order derivatives of $h_{0}$ with respect to $\unicode[STIX]{x1D709}$ to account for the steeper slope of the free surface. Our equations (5.54)–(5.57) together with the corresponding boundary conditions allow one to perform a similar analysis for a spiralling jet which would bring in the curvature of the jet’s trajectory. Another way of analysing (5.54)–(5.57) could be via the radial coordinate expansions similar to the analysis presented by Garcia & Castellanos (Reference Garcia and Castellanos1994) for a variety of models for axisymmetric viscous jets.

5.4.2 Case 2: $\unicode[STIX]{x1D716}\mathit{We}=O(1)$ as $\unicode[STIX]{x1D716}\rightarrow 0$

If $\unicode[STIX]{x1D716}\mathit{We}=O(1)$ and $\unicode[STIX]{x1D716}\unicode[STIX]{x1D6FF}^{-2}=O(1)$ as $\unicode[STIX]{x1D716}\rightarrow 0$ , at leading order, the first term on the right-hand side of (5.55) drops out, so that we immediately have

(5.63) $$\begin{eqnarray}\frac{\unicode[STIX]{x2202}u_{\unicode[STIX]{x1D709},0}}{\unicode[STIX]{x2202}\bar{t}}+u_{\unicode[STIX]{x1D709},0}\frac{\unicode[STIX]{x2202}u_{\unicode[STIX]{x1D709},0}}{\unicode[STIX]{x2202}\bar{\unicode[STIX]{x1D709}}}=0,\end{eqnarray}$$

and the two equations (5.60), (5.63) form a closed system. Note that, as with trajectories, by formally taking the limit $\mathit{We}\rightarrow \infty$ we turn equation (5.62) into (5.63). As with the trajectory, the peristaltic disturbances in the case $\unicode[STIX]{x1D716}\mathit{We}=O(1)$ as $\unicode[STIX]{x1D716}\rightarrow 0$ do not involve capillarity, essentially corresponding to the regime which could be labelled as ‘free flow’.

Thus, from the above analysis we can see that the non-existence of an unsteady solution to (5.9)–(5.12) means that (i) the jet’s trajectory can be unsteady only due to the conditions specifying how it is produced, e.g. at the orifice where it comes from should its direction there vary with time and (ii) the body forces which determine the trajectory should not feature in the equations describing the unsteady motion.