From ae1eba6b816da108b5f2cb836f34f078310d9f88 Mon Sep 17 00:00:00 2001 From: Jean-Paul Pelteret Date: Fri, 12 Aug 2022 02:30:40 +0200 Subject: [PATCH] Consistently update notation in SymmetricTensor docs --- include/deal.II/base/symmetric_tensor.h | 24 ++++++++++++------------ 1 file changed, 12 insertions(+), 12 deletions(-) diff --git a/include/deal.II/base/symmetric_tensor.h b/include/deal.II/base/symmetric_tensor.h index 6396f05d58..462ef41b70 100644 --- a/include/deal.II/base/symmetric_tensor.h +++ b/include/deal.II/base/symmetric_tensor.h @@ -54,7 +54,7 @@ DEAL_II_CONSTEXPR inline DEAL_II_ALWAYS_INLINE SymmetricTensor<2, dim, Number> * operator representation of the linear deviator operator $\mathbb P$, also * known as the volumetric projection tensor, calculated as: * \f{align*}{ - * \mathbb{P} &=\mathbb{I} -\frac{1}{\text{dim}} \mathbf I \otimes \mathbf I + * \mathbb{P} &=\mathbb{S} -\frac{1}{\text{dim}} \mathbf I \otimes \mathbf I * \\ * \mathcal{P}_{ijkl} &= \frac 12 \left(\delta_{ik} \delta_{jl} + * \delta_{il} \delta_{jk} \right) @@ -81,10 +81,10 @@ DEAL_II_CONSTEXPR inline DEAL_II_ALWAYS_INLINE SymmetricTensor<4, dim, Number> deviator_tensor(); /** - * Return the fourth-order symmetric identity tensor $\mathbb I$ which maps + * Return the fourth-order symmetric identity tensor $\mathbb S$ which maps * symmetric second-order tensors, such as $\mathbf A$, to themselves. * \f[ - * \mathbb I : \mathbf A = \mathbf A + * \mathbb S : \mathbf A = \mathbf A * \f] * * Note that this tensor, even though it is the identity, has a somewhat funny @@ -92,25 +92,25 @@ DEAL_II_CONSTEXPR inline DEAL_II_ALWAYS_INLINE SymmetricTensor<4, dim, Number> * example, for dim=2, the identity tensor has all zero entries * except for * \f[ - * \mathcal{I}_{0000} = \mathcal{I}_{1111} = 1 + * \mathcal{S}_{0000} = \mathcal{S}_{1111} = 1 * \f] * \f[ - * \mathcal{I}_{0101} = \mathcal{I}_{0110} = \mathcal{I}_{1001} - * = \mathcal{I}_{1010} = \frac 12. + * \mathcal{S}_{0101} = \mathcal{S}_{0110} = \mathcal{S}_{1001} + * = \mathcal{S}_{1010} = \frac 12. * \f] * In index notation, we can write the general form * \f[ - * \mathcal{I}_{ijkl} = \frac 12 \left( \delta_{ik} \delta_{jl} + + * \mathcal{S}_{ijkl} = \frac 12 \left( \delta_{ik} \delta_{jl} + * \delta_{il} \delta_{jk} \right). * \f] * To see why this factor of $1 / 2$ is necessary, consider computing * $\mathbf A= \mathbb I : \mathbf B$. - * For the element $A_{01}$ we have $A_{01} = \mathcal{I}_{0100} B_{00} + - * \mathcal{I}_{0111} B_{11} + \mathcal{I}_{0101} B_{01} + - * \mathcal{I}_{0110} B_{10}$. On the other hand, we need + * For the element $A_{01}$ we have $A_{01} = \mathcal{S}_{0100} B_{00} + + * \mathcal{S}_{0111} B_{11} + \mathcal{S}_{0101} B_{01} + + * \mathcal{S}_{0110} B_{10}$. On the other hand, we need * to have $A_{01} = B_{01}$, and symmetry implies $B_{01}=B_{10}$, - * leading to $A_{01} = (\mathcal{I}_{0101} + \mathcal{I}_{0110}) B_{01}$, or, - * again by symmetry, $\mathcal{I}_{0101} = \mathcal{I}_{0110} = \frac 12$. + * leading to $A_{01} = (\mathcal{S}_{0101} + \mathcal{S}_{0110}) B_{01}$, or, + * again by symmetry, $\mathcal{S}_{0101} = \mathcal{S}_{0110} = \frac 12$. * Similar considerations hold for the three-dimensional case. * * This issue is also explained in the introduction to step-44. -- 2.39.5