I've got a problem in some geometry of flow.
For the sake of completeness I will give the complete derivation of the equation of interest, but I will seperate it into derivation part and question part:
So for the derivation part I've got a domain $\Omega$, ratational symmetric w.r.t. the $z$-axis and tip-growing in $z$-direction with time, see e.g. $\Gamma\subset\partial\Omega$ in the following picture
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Now, I can calculate the increase of the cell surface $\Gamma$ by integrating the flow ($F$ insides $\Omega$) along $\Gamma$. Using Gauss’s formula for the first variation of area, this leads to
\begin{align}\frac{\partial ||\Gamma||}{\partial t} = \int_{\Gamma} F = - \int_\Gamma H(\hat{n}\cdot v)dS + \oint_\Gamma (\hat{m}\cdot v)dl \ \end{align}
where $\hat{n}$ is the outpointing normal an $\partial \Omega$, $\hat{m}$ is the outward pointing normal to $\partial\Gamma$ tangent to $\partial\Omega$, $v$ is the velocity field and $H$ the mean curvature.
As it is shown in the figure, the origin of the coordinate system is locatet at the tip of the domain. By using a reference frame moving along with the tip $\partial\Omega$ (and so $\partial\Gamma$) appears stationary while growing in $z$-direction. See e.g.
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In this reference frame Gauss’s formula for the first variation of area became $\int_{\Gamma} F = - \int_\Gamma H(\hat{n}\cdot (v- \hat{e}_z))dS + \oint_\Gamma (\hat{m}\cdot (v- \hat{e}_z))dl \ $
where $\hat{e}_z$ is the unit vector in $z$-direction.
And now, the question part is: The observation that the surface $\Gamma$ will be stationary in the described moving frame should lead to the implikation, that $v-\hat{e}_z$ must lie tangent to $\partial\Omega$ (and so tangent to $\Gamma$). At this point I see no way to prove that statement. In addition I've got no idea what "stationary" means to surfaces, beside time-independence.
So I would be glad for every response.