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I want to integrate by part the following integral in cylindrical coordinates $$\int \vec{r} \times (\nabla \cdot \overline{T}) ~d^3\vec{r} $$ where $\overline{T}$ is a second order symmetric tensor and $\times$ is the vectorial product.

Once I integrated by part I want to use the divergence theorem to obtain a surface integral. It works very well in cartesian coordinates but I cannot manage to do it properly in cylindrical.

In cylindrical coordinates $\hat{\boldsymbol{\rho}},\hat{\boldsymbol{\theta}},\hat{\mathbf{z}}$,

\begin{align}&\nabla \cdot \overline{T}= \left[ \frac{1}{\rho} \frac{\partial}{\partial \rho}(\rho T_{\rho\rho}) + \frac{1}{\rho} \frac{\partial T_{\theta \rho}}{\partial \theta} + \frac{\partial T_{z \rho }}{\partial z} - \frac{T_{\theta \theta}}{\rho} \right] \hat{\boldsymbol{\rho}} + \\ &\qquad\qquad\left[ \frac{1}{\rho} \frac{\partial}{\partial \rho}(\rho T_{\rho\theta}) + \frac{1}{\rho} \frac{\partial T_{\theta\theta}}{\partial \theta} + \frac{\partial T_{z\theta}}{\partial z} + \frac{T_{\theta \rho }}{\rho} \right] \hat{\boldsymbol{\theta}} + \\ &\qquad\qquad\left[ \frac{1}{\rho} \frac{\partial}{\partial \rho}(\rho T_{\rho z}) + \frac{1}{\rho} \frac{\partial T_{\theta z}}{\partial \theta} + \frac{\partial T_{z z}}{\partial z} \right] \hat{\mathbf{z}} \end{align}

I tried a lot of different ways (5 days I am trying to do it) of doing the integration by part, but by checking numerically I know that my results is wrong. The value I obtain numerically seems to be more like ($\overline{T}$ is symmetric) \begin{align} \int \vec{r} \times (\nabla \cdot \overline{T}) ~d^3\vec{r}&= \oint \vec{r} \times \left( \hat{\boldsymbol{\rho}} \cdot \overline{T} \right) ~dS + \\ &\left( \int T_{\rho z} ~d^3\vec{r} - \int_0^{\rho_\text{max}} \left[ T_{\theta z}(2\pi) - T_{\theta z}(0) \right] ~\rho d\rho \right) \hat{\boldsymbol{\theta}} + \\ &\int_0^{\rho_\text{max}} \left[ T_{\theta\theta}(2 \pi) - T_{\theta\theta}(0) \right] \rho d\rho \hat{\mathbf{z}} \end{align} First, I am not sure of the result and second, I would like to know how to derive this analytically. Someone can help me? thank you very much

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I should stress that the expression $\mathbf{r}\times\nabla\cdot\mathbf{T}$ is independent of the coordinate system, while the symmetry of $\mathbf{T}$ is easily expressed with respect to Cartesian coordinates, then \begin{align} \mathbf{r}\times\nabla\cdot\mathbf{T} &=\mathbf{e}_i\varepsilon_{ijk}x_j\partial_lT_{kl}\\ &=\mathbf{e}_i\varepsilon_{ijk}\partial_l(x_jT_{kl})-\mathbf{e}_i\varepsilon_{ijk}(\partial_lx_j)T_{kl}\\ &=\mathbf{e}_i\varepsilon_{ijk}\partial_l(x_jT_{kl})-\mathbf{e}_i\varepsilon_{ijk}\delta_{jl}T_{kl}\\ &=\mathbf{e}_i\partial_l(\varepsilon_{ijk}x_jT_{kl})-\mathbf{e}_i\varepsilon_{ijk}T_{kj}\\ &=\mathbf{e}_i\partial_l(\varepsilon_{ijk}x_jT_{kl})\\ &=\nabla\cdot(\mathbf{r}\times\mathbf{T}). \end{align} The result obtained is independent from the coordinate system used, so $$ \int_B\mathbf{r}\times\nabla\cdot\mathbf{T}\,dV=\int_B\nabla\cdot(\mathbf{r}\times\mathbf{T})\,dV $$ then applying the divergence theorem, this becomes $$ \int_{\partial B}\mathbf{n}\cdot(\mathbf{r}\times\mathbf{T})\,dS $$ now you can evaluate this flux in whatever coordinate system you prefer.

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  • $\begingroup$ I am sorry but I do not think the first equality is correct. If you do the maths and explicitly calculate the equation, you will see that the left hand side has no radial component because of the cross product with $\mathbf{r}=\begin{pmatrix} r \\0 \\0 \end{pmatrix}$. However, if you look at the expression for the divergence of a tensor in my question above, you will see that the right hand side can have a radial component if the component $(\mathbf{r}\times\mathbf{T})_{\theta\theta}$ of the tensor $\mathbf{r}\times\mathbf{T}$ is different from zero. So I think your answer is not correct $\endgroup$ – lambertmular Jun 1 '15 at 21:00
  • $\begingroup$ @lambertmular: see my editing. $\endgroup$ – enzotib Jun 2 '15 at 7:31
  • $\begingroup$ @ enzotib: I have developed the first expression you gave, and I have noted that in cylindrical coordinates it lacks all the terms without derivative (see the expression of $\nabla \cdot \mathbf{T}$ in my question above). So I think that your definition of $\nabla \cdot \mathbf{T}$ you use in the first expression is only valid in cartesian coordinates. You have to take the derivatives of the unit vectors into account when you express the divergence: $\nabla \cdot \mathbf{T}=\left( \hat{\vec{e}}_k \frac{\partial}{\partial x_k} \right) \cdot \left( T_{ij} \hat{\vec{e}}_i\hat{\vec{e}}_j \right)$ $\endgroup$ – lambertmular Jun 2 '15 at 9:30

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