# Does curvature tensor of a cylindrical surface depend on azimuthal and axial (linear) deformations?

I am interested in the curvature tensor of a cylinder undergoing small deformation.

Let us denote by $\mathbf{a} = R \, \mathbf{e}_r + z \, \mathbf{e}_z$ the position vector of the points located at the undisplaced cylindrical surface, with $R$ being the undeformed radius. Here $r$, $\theta$ and $z$ are used to refer to the radial, azimuthal and vertical coordinates respectively. After deformation, the vector position reads $$\mathbf{r} = (R + u_r)\, \mathbf{e}_r + u_\theta \, \mathbf{e}_\theta + (z + u_z) \, \mathbf{e}_z \, ,$$ where $\mathbf{u} (\theta,z)$ denotes the displacement vector field.

The cylindrical membrane can be defined by the covariant base vectors $\mathbf{g}_1 := \mathbf{r}_{,\theta}$ and $\mathbf{g}_2 := \mathbf{r}_{,z}$, where commas in indices stands for a partial derivative. The unit normal vector $\mathbf{n}$ is defined in the usual way as $$\mathbf{n} := \frac{\mathbf{g}_1 \times \mathbf{g}_2}{|\mathbf{g}_1 \times \mathbf{g}_2|} \, .$$

Hence, the covariant base vectors reads $$\mathbf{g}_1 = (u_{r,\theta} - u_\theta) \, \mathbf{e}_r + (R + u_r + u_{\theta, \theta}) \, \mathbf{e}_\theta + u_{z,\theta} \, \mathbf{e}_z \, , \\ \mathbf{g}_2 = u_{r,z} \, \mathbf{e}_r + u_{\theta, z} \, \mathbf{e}_\theta + (1 + u_{z,z} )\, \mathbf{e}_z \, ,$$ and the unit normal vector at leading order in deformation reads $$\mathbf{n} = \mathbf{e}_r + \frac{u_\theta - u_{r, \theta}}{R} \, \mathbf{e}_\theta - u_{r,z} \, \mathbf{e}_z \, .$$

The covariant components of the metric tensor are defined by the scalar product $g_{\alpha\beta} = \mathbf{g}_{\alpha} \cdot \mathbf{g}_{\beta}$. The contravariant tensor $g^{\alpha\beta}$ is the inverse of the metric tensor. In a linearized form, we obtain $$g_{\alpha\beta} = \left( \begin{array}{cc} R^2 + 2R (u_r + u_{\theta, \theta}) & u_{z,\theta} + R u_{\theta, z} \\ u_{z,\theta} + R u_{\theta, z} & 1+2u_{z,z} \end{array} \right) \, , \quad g^{\alpha\beta} = \left( \begin{array}{cc} \frac{1}{R^2} - 2\frac{u_r + u_{\theta, \theta}}{R^3} & -\frac{u_{z,\theta} + R u_{\theta, z}}{R^2} \\ -\frac{u_{z,\theta} + R u_{\theta, z}}{R^2} & 1-2u_{z,z} \end{array} \right) \, . \label{cocontravariantTensor}$$

The curvature tensor is defined in the usual way as $$b_{\alpha\beta} = {\mathbf{g}_\alpha}_{,\beta} \cdot \mathbf{n} = -\mathbf{g}_\alpha \cdot {\mathbf{n}}_{,\beta} \, .$$

In this way, for linear deformation (i.e. by neglecting the products of two displacements), we obtain after calculation $$b_{\alpha\beta} = \left( \begin{array}{cc} -(R + u_r + 2 u_{\theta, \theta}) +u_{r,\theta\theta} & -u_{\theta, z}+u_{r, \theta z} \\ -u_{\theta, z}+u_{r, \theta z} & u_{r,zz} \end{array} \right) \, .$$

Next, the mixed version of the curvature tensor $b_\alpha^\beta$ defined as $b_\alpha^\beta = b_{\alpha\delta} g^{\delta \beta}$ reads $$b_\alpha^\beta = \left( \begin{array}{cc} -\frac{1}{R}+\frac{u_r}{R^2}+\frac{u_{r,\theta\theta}}{R^2} & u_{r,z\theta}+\frac{u_{z,\theta}}{R} \\ \frac{u_{r,z\theta}}{R^2}-\frac{u_{\theta,z}}{R^2} & u_{r,zz} \end{array} \right) \, .$$

It can be seen that $b_\alpha^\beta$ involves derivatives of azimuthal deformation $u_\theta$ and axial deformation $u_z$. However, $u_\theta$ and $u_z$ do not change surface topology meaning that the cylindrical shape is maintained after deformation. I would expect that $b_\alpha^\beta$ is function only of $u_r$ and its derivatives.

I was wondering whether someone here could be of help to clarify this point.

Any help would be highly appreciated and rated

Thank you,

Federiko

• Aren't you missing the $\theta$ parameter in your unperturbed cylinder? I also don't understand your formula for the second fundamental form $b_{\alpha\beta}$. You should have all second-order partials. (I haven't gotten past that point.) – Ted Shifrin Jan 31 '17 at 21:17
• @TedShifrin Thank you for your feedback. For the unperturbed cylinder I simply write the vector position in cylindrical coordinates, is that wrong? Yes, you are wright, I have now corrected the formula for $b_{\alpha\beta}$. I would be happy if you could please provide with some hints, thank you – Daddy Feb 1 '17 at 7:07
• @TedShifrin I have just started a bounty for this question, it would be cool if you could provide with feedback. Thanks – Daddy Feb 2 '17 at 17:54

(This is now too long to fit in a comment, so I'm starting a preliminary answer.) OK, so lots of issues here. I assume you're thinking of the perturbation $(u_r,u_\theta,u_z)$ as functions of $(\theta,z)$. What you purport to be computing is the second fundamental form (up to sign) of the deformed cylinder; this is not yet the curvature, as you have to multiply by the inverse of the matrix of the first fundamental form. Next, it's not just the second derivatives of the position vector that need to be computed; more seriously, $\mathbf e_r$ is just the zeroth order term of the unit normal $\mathbf n$, but the cross product $\mathbf r_\theta\times\mathbf r_z$ is an awful mess. I actually cannot figure out how you got the matrix you did. I think you need to start by explaining what you have done. (Once again, I remind you that writing $\partial^2\mathbf r/\partial\theta\partial z$ is a misrepresentation, as well.)
EDIT: In brief, there's no reason to expect the $u_\theta$ and $u_z$ not to appear in the matrix. But if we set $u_r=0$, we should still see that $K=0$ (we maintain flatness of the surface). So I suspect that your computation is still not quite right.
• I'm not convinced by your approximation scheme. If the deformation terms are "small," their derivatives needn't be. So why, for example, is the $u_\theta u_{r,z}$ term missing from $g_{12}$? – Ted Shifrin Feb 5 '17 at 17:51
• Well, I stand by my comment that when $u_r=0$, we should maintain $K=0$, so your computation needs to be consistent with that! – Ted Shifrin Feb 5 '17 at 19:13
• You've computed wrong. As it stands, there's $u_{z,\theta}u_{\theta,z}/R^3$ in there. – Ted Shifrin Feb 5 '17 at 19:15
• Well, it shouldn't be a matter of first-order or second-order approximation. $K$ should remain $0$ on the nose if you do the computation correctly, I claim. – Ted Shifrin Feb 5 '17 at 19:20
• Sorry, @Federiko, I've been out most of the day. No, $u_r=0$ will give you the same cylindrical shape, but with some weird parametrization when $u_\theta$ and $u_z$ are nonzero. That will not affect $K=0$. That's worth your time to understand. – Ted Shifrin Feb 6 '17 at 2:48