\begin{table}%t7
\caption{{\bf b)} Line-strength indices used in this work as a function of the logarithm
of age and metallicity using the models by V06. The last two  columns express
the required variation in metallicity with velocity dispersion to explain the
slopes of the index--$\sigma$ relations, assuming that the slope is due to an
exclusive variation of this parameter with the velocity
dispersion.\label{parametrization}}
%\centering
\par
\small
\begin{tabular} {@{}l r c c@{}}
\hline\hline
                & &\multicolumn{1}{c}{(LDEG)}& \multicolumn{1}{c@{}}{(HDEG)}\\
  Index         & &$\displaystyle\frac{\partial [{\rm M/H}]}{\partial\log\sigma}$&$\displaystyle\frac{\partial [{\rm M/H}]}{\partial\log\sigma}$\\
\hline
  D4000$'$       & $-2.7672 + 0.3817\;$[M/H]$+0.3724\log$(age)&   0.440  &  0.532 \\
  H$\delta_A'$   & $+1.0737 - 0.1206\;$[M/H]$-0.1155\log$(age)&   0.403  &  0.507\\
  H$\delta_F'$   & $+1.0202 - 0.0878\;$[M/H]$-0.1029\log$(age)&   0.593  &  0.380\\
  CN$_2$         & $-0.5655 + 0.1210\;$[M/H]$+0.0640\log$(age)&   1.470  &  1.464\\
  Ca4227$'$      & $-0.6957 + 0.0857\;$[M/H]$+0.0845\log$(age)&   0.313  &  0.304\\
  G4300$'$       & $-0.4797 + 0.0651\;$[M/H]$+0.0669\log$(age)&   0.267  &  0.504\\
  H$\gamma_A'$   & $+0.9424 - 0.0993\;$[M/H]$-0.1092\log$(age)&   0.565  &  0.554\\
  H$\gamma_F'$   & $+1.3029 - 0.1080\;$[M/H]$-0.1388\log$(age)&   0.900  &  0.738 \\
  Fe4383$'$      & $-0.4544 + 0.0907\;$[M/H]$+0.0583\log$(age)&   0.290  &  0.363\\
  Ca4455$'$      & $-0.1952 + 0.0509\;$[M/H]$+0.0282\log$(age)&   0.479  &  0.350\\
  Fe4531$'$      & $-0.1611 + 0.0374\;$[M/H]$+0.0252\log$(age)&   0.406  &  0.524\\
  C4668$'$       & $-0.1022 + 0.0734\;$[M/H]$+0.0171\log$(age)&   0.742  &  0.911\\
  H$\beta'$      & $+0.5974 - 0.0240\;$[M/H]$-0.0538\log$(age)&   1.271  &  0.487\\
  Fe5015$'$      & $-0.0133 + 0.0374\;$[M/H]$+0.0097\log$(age)&   0.342  &  0.551\\
  Mgb$'$         & $-0.4036 + 0.0775\;$[M/H]$+0.0545\log$(age)&   1.289  &  1.174\\
  Fe5270$'$      & $-0.1415 + 0.0414\;$[M/H]$+0.0229\log$(age)&   0.391  &  0.579\\
  Fe5335$'$      & $-0.1561 + 0.0472\;$[M/H]$+0.0236\log$(age)&   0.432  &  0.739\\
  \hline
\end{tabular}
\end{table}