//---------------------------------------------------------------------------
// $Id$
//
-// Copyright (C) 2005, 2006, 2008, 2009, 2010, 2011 by the deal.II authors
+// Copyright (C) 2005, 2006, 2008, 2009, 2010, 2011, 2012 by the deal.II authors
//
// This file is subject to QPL and may not be distributed
// without copyright and license information. Please refer
/**
* Return the tensor of rank 4 that, when multiplied by a symmetric rank 2
- * tensor <tt>t</tt> returns the deviator <tt>dev t</tt>. It is the operator
+ * tensor <tt>t</tt> returns the deviator $\textrm{dev}\ t$. It is the operator
* representation of the linear deviator operator.
*
* For every tensor <tt>t</tt>, there holds the identity
/**
* Returns the fourth-order symmetric identity tensor which maps symmetric
- * second-order tensors to themselves.
+ * second-order tensors to themselves.
*
* Note that this tensor, even though it is the identity, has a somewhat funny
* form, and in particular does not only consist of zeros and ones. For
* except for <tt>id[0][0][0][0]=id[1][1][1][1]=1</tt> and
* <tt>id[0][1][0][1]=id[0][1][1][0]=id[1][0][0][1]=id[1][0][1][0]=1/2</tt>. To
* see why this factor of 1/2 is necessary, consider computing <tt>A=Id
- * . B</tt>. For the element <tt>a_01</tt> we have <tt>a_01=id_0100 b_00 +
+ * : B</tt>. For the element <tt>a_01</tt> we have <tt>a_01=id_0100 b_00 +
* id_0111 b_11 + id_0101 b_01 + id_0110 b_10</tt>. On the other hand, we need
* to have <tt>a_01=b_01</tt>, and symmetry implies <tt>b_01=b_10</tt>,
* leading to <tt>a_01=(id_0101+id_0110) b_01</tt>, or, again by symmetry,