UV energy output confusion

[VictordeHolland]

The Sun is a G2V type star in the OBAFGKM spectral system.
It outputs 1 Solar luminosity which corresponds to 3.828×10^26 Watt

The spectrum of the Sun closely resembles that of of a black body with a temperature of about 5,800 Kelvin. Sunlight in space at the top of Earth's atmosphere is composed of about 50% infrared light, 40% visible light, and 10% ultraviolet light" So most of the energy of the Sun is in the visible(V) and IR part of the spectrum and only a little in the UV part. Most (77%) of the dangerous UV is blocked by our atmosphere (almost all the UV-C).

More massive, hotter stars have higher luminosities and I would expect them to produce their peak EM radiation output in the UV, as that corresponds to a higher black-body temperature. Yet if I look at a graph of a A0V star ( https://en.wikipedia.org/wiki/File:A...comparison.png ). It looks as if the peak is still in the visible part of the spectrum (~420nm). But the total spectral power corresponds to a 9500K black-body object. The visible part of the spectrum looks more like 15000K.

If we take for instance the star Vega (Spectral type A0 Va)

U−B color index     0.00
B−V color index     0.00
Absolute magnitude (MV)     +0.582
Mass     2.135 ± 0.074 M☉
Radius     2.362 × 2.818 R☉
Luminosity     40.12 ± 0.45 L☉

So Vega outputs about 40 times the energy of the Sun across the entire EM spectrum.
3.828×10^26 x 40.12 = 1.54 ×10^28 W

The Sun has absolute Magnitude of 4.83 in the V band.
So Vega would appear 4.83-0.582 = 4.248
10^(0.4*4.248) = 50.02 times brighter than the Sun in the V band. Right?

To measure the index, one observes the magnitude of an object successively through two different filters, such as U and B, or B and V, where U is sensitive to ultraviolet rays, B is sensitive to blue light, and V is sensitive to visible (green-yellow) light. The set of passbands or filters is called a photometric system. The difference in magnitudes found with these filters is called the U−B or B−V color index respectively.

So I would think the luminosity in UV would be equal to the luminosity in visible light (since U-B and B-V are 0). But this hardly seems to be the case from SPD diagram earlier?

So are the O B A classes not outputting as much UV as you would expect based on their effective temperature? Am I missing something big?

[kladner]

I don't know, but I am very interested in the possible answers to this intriguing question.

[xilman]

Originally Posted by VictordeHolland

Am I missing something big?

TL;DR: Yes.

To go into somewhat more detail, the fundamental thing you are missing is that stars are not black bodies. They have all sorts of lines in their spectrum, some in emission and some in absorption. Further, UBVRIJ (etc) photometry is really low resolution spectroscopy.

A0V stars show strong atomic hydrogen absorption lines, G2V stars do as well but not as strongly and also have many lines from metals such as Fe. The main reason for your confusion is that the Balmer discontinuity occurs at 364nm which is within the U band pass. (The Balmer discontinuity lies at the energy required to ionize a hydrogen atom whose electron is in the n=2 state.) The population of hydrogen atoms with n=2 depends greatly on temperature. O and B stars are too hot to have a significant population; K stars are too cool.

There are other factors, arising from the H- ion, from metal and He lines, from the fact that light from a star comes from a substantial range of depths within the stellar photosphere, and more but the principal cause is the effect of the Balmer discontinuity.

Photometry in other bands, uvby for instance, is much easier to interpret in terms of stellar temperature because the filters do not straddle the Balmer discontinuity

[xilman]

More on this: B-V is a much better measure of stellar temperature because those pass bands aren't as affected by spectral features as is the U band.

U-B is much more useful for determining spectral class without having to go to the trouble of taking spectra.

[fivemack]

Cool, thank you! I knew that absorption bands existed, but I had them associated only with molecules in colder stars; I really didn't realise the effect was so dramatic for hydrogen in an A-class star.

[fivemack]

http://iopscience.iop.org/article/10.1086/420715/pdf is interesting - very carefully calibrated Hubble spectroscopy of Vega, where you can see the intensity increasing again into the UV after a minimum at the Balmer discontinuity.

[VictordeHolland]

Thanks for the explanation and the link to the Vega paper. I've got some reading to do on Balmer lines/series .

https://en.wikipedia.org/wiki/Balmer_series
https://en.wikipedia.org/wiki/Hydrogen_spectral_series


So In hindsight the 'missing' UV due to the Hydrogen Balmer series absorption lines is actually used to define the A class . And the lines become fainter as Hydrogen get ionized (loses the electron that causes the absorption) in the hotter B and O types. In the cooler stars F G K M the more Hydrogen electrons can go back to the ground state (n=1), leaving less in the 'excited' n=2 state needed for absorption of the Balmer lines. Brilliant!