Consistently update notation in SymmetricTensor docs

diff --git a/include/deal.II/base/symmetric_tensor.h b/include/deal.II/base/symmetric_tensor.h
index 6396f05d58fc95c132e3c6cef8634202f564d79b..462ef41b70afc9ed04be44f65db5d0cd97480e4e 100644 (file)
--- 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 <tt>dim=2</tt>, 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.