# Solar Absolute Magnitudes

1. Jul 5, 2012

### StarDweller

I'm trying to calculate the luminosity of the sun in the infrared (J, H and K filters). According to http://mips.as.arizona.edu/~cnaw/sun.html, the absolute magnitudes of the Sun in each of these filters are

J: 3.64
H: 3.32
K: 3.28.

The absolute bolometric magnitude of the Sun is about 4.72. What I can't quite wrap my head around is this: how is it possible that the absolute magnitude in specific filters is smaller (as in brighter) than the bolometric absolute magnitude? Does this not mean that the luminosity of the Sun in each of these filters is greater than the bolometric luminosity? That is what the equation M2-M1 = -2.5 log (L2/L1) would seem to imply (though I'm not entirely sure if it's legal to have M1 as M_bol and M2 as M_J for instance; I'm fairly certain the filters have to be the same).

I'm sure it's just an interpretation error on my part. I understand that as a blackbody, to find the total luminosity the emission must be integrated over all wavelengths, but it doesn't make sense to me that the Sun would be brighter in one filter than in ALL filters. (Looking at the V magnitude as well this implies that the Sun is brighter in the IR than the visible but the peak of the blackbody distribution of the Sun is clearly in the visible.)

Any help is appreciated!

2. Jul 6, 2012

### Staff: Mentor

I believe they are comparing the brightness of the Sun to Vega. I'm not sure, but I think that Vega, being a higher temperature than the Sun, puts out more of it's energy in the higher energy wavelengths, like the UBV bands, than the lower energy bands, like the JHK. So when you compare the two perhaps the lower difference in the JHK bands is due to the different shape of the spectrums.