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AB magnitudes

The definition of the AB magnitude is:

\begin{displaymath}
m_{AB}(\nu) = -2.5 \cdot (log(F_{\nu}) + 19.436 )
\end{displaymath}


\begin{displaymath}
\hspace{13.2mm}
= -2.5 \cdot log \Big( \frac{F_{\lambda} \cdot \lambda^2}{c} \Big) - 48.59
\end{displaymath} (5)

where c is the speed of light. From equation (11) (see Appendix B) and after some algebra, it can be shown that:

\begin{displaymath}
m_{AB}(\nu) = -2.5 \cdot log(C_{\lambda})
+ 2.5 \cdot log \...
...a} }
{S_{\lambda} \cdot \lambda^2} \Big) - k_{\lambda} \cdot X
\end{displaymath}


\begin{displaymath}
\hspace{16.6mm}
- 48.59
\end{displaymath} (6)

where $S_{\lambda}$ is the ``counts-to-energy'' conversion factor and $W_{\lambda}$ is equal to $\int T_{\lambda} d\lambda$, i.e. the equivalent width of the filter.

We can rewrite eq 6 as follows:

\begin{displaymath}
m_{AB}(\nu) = m_{inst}(\lambda)
- k_{\lambda} \cdot X + Z_{\lambda}
\end{displaymath} (7)

where $Z_{\lambda}$ is the zero point defined as
\begin{displaymath}
Z_{\lambda} = 2.5 \cdot log \Big( \frac{c \cdot W_{\lambda} }
{S_{\lambda} \cdot \lambda^2} \Big) - 48.59
\end{displaymath} (8)

In this way, the AB magnitudes can be determined from eq. 7 where X is the airmass, $k_{\lambda}$ the extinction coefficient and $Z_{\lambda}$ the zero point reported in Table 1.


next up previous
Next: Field scattering effect: a Up: The intermediate-band calibration Previous: The intermediate-band calibration
Juan Alcala
2002-02-05