Suppose that the radius of a star increases by some factor, how does this affect the absolute magntiude of the star?

I know that $M_1 - M_2 = \Delta M = 2.5 \log \frac{L_1}{L_2}$, so if I knew the luminosities, I could make the calculation. I know that luminosity is given by the formula $$L = 4 \pi \sigma R^2 T^4.$$ However, this shows a dependence on not just on the radius but also on the temperature and I'm not sure how to account for a change of temperature on the luminosity.


1 Answer 1


An answer cannot be unambiguously given beceause there is insufficient information in your question about what causes the inflation.

If the inflation takes place at constant luminosity then of course the absolute (bolometric) magnitude will stay the same.

If you are talking about a magnitude through a given filter then that will change because if the temperature changes then so does the spectrum of the star and a different proportion of the luminosity will emerge in a particular wavelength band and you would have to calculate an appropriate change in the bolometric correction.

If the luminosity changes purely because of a change in radius (at fixed temperature), then for a small perturbation you can argue that since $L \propto R^2 T^4$ then $$ \frac{\Delta L}{L} \simeq 2\frac{\Delta R}{R}$$ and hence $$\Delta M \simeq -2.5 \log_{10} \frac{L + \Delta L}{L}\ .$$ If the temperature changes too, then $$\frac{\Delta L}{L} \simeq 2\frac{\Delta R}{R} + 4\frac{\Delta T}{T}$$ and you will have to take account of any temperature change on the stellar spectrum if $M$ is not the bolometric absolute magnitude.


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