spectro-aras.com

Hi Etienne,

This is because of the Balmer jump where all the Balmer lines merge together and most of the light at the blue end is used up in the ionisation of Hydrogen.
The problem is stars are not really black bodies and Teff is not really the temperature of the photosphere. It is the temperature of a black body which would emit the same amount of radiation. That is why you do not get the "right" answer when fitting a black body curve. See this diagram for example
https://en.wikipedia.org/wiki/Color_tem ... arison.png
You cannot measure the temperature of stars using this method. (interstellar extinction also changes the result) The power of spectroscopy is in the lines. You can tell the temperature of a star by looking at which lines are present, not by looking at the shape of the continuum
https://www.astronomynotes.com/starprop/linestr.gif

Cheers
Robin

Olivier GARDE wrote:Etienne,

Tu trouveras la réponse dans l'article suivant :
https://www.shelyak.com/mesurer-les-tem ... s-etoiles/

Olivier,

This question is often coming up. You should not claim you can measure the temperature of stars by fitting a planck curve to the continuum. No astronomer actually does this. Spectroscopic classification uses the strength of different lines, not the continuum shape.

The method just does not work. For example the claim is that the a 7700K Planck curve is a good fit to Altair (A7v) is but if we look at the fit is is not good at all.
We can fit a Planck curve to the part of the spectrum which is a black body but that gives a temperature of 10000K. The figure 7750 K Kaler gives is Teff which is the temperature that a black body would have to produce the same total amount of radiation. It is theoretical and does not represent the actual temperature of the star photosphere.

Robin