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Star Zeta Puppis lies at a distance $460$ pc from Earth. Its apparent visual magnitude $m_V$ is $2.25$, its absolute bolometric magnitude $M_{\mathrm{bol}}$ is $-9.9$, and its angular diameter is $4.3 \times 10^{−4}$ arcseconds.

(You may take for the Sun $M_{\mathrm{bol}} = +4.8$)

a) Calculate the absolute visual magnitude $M_V$ of Zeta Puppis.

b) Calculate the luminosity of Zeta Puppis in solar units $L_{\bigodot}$

For part a) I find that $M_V=m_V-5\log_{10} d +5=2.25-5\log_{10} 460 + 5\approx-6.06$

This is the correct answer.


For part b) I make use of the relations

$$F=\frac{L}{4 \pi^2 d^2}\tag{1}$$

$$m_{\zeta}-m_{\bigodot}=-2.5\log_{10}\left(\frac{F_{\zeta}}{F_{\bigodot}}\right)\tag{2}$$

Where $L$ is the luminosity, $d$ is the distance from the star to earth in parsecs, $F_{\zeta}$, $F_{\bigodot}$ are the fluxes received at Earth from Zeta Puppis and the Sun respectively. I already know that the apparent magnitude of the Sun, $m_{\bigodot}=-26.7$ (as seen from the Earth) from an earlier part of the problem sheet.

From $(1)$ and $(2)$, then

$$2.25--26.7=-2.5\log_{10}\left[\frac{L_{\zeta}}{L_{\bigodot}}\left(\frac{d_{\bigodot}}{d_{\zeta}}\right)^2\right]\tag{3}$$

Since the distances are in parsecs $$d_{\bigodot}=\frac{1.5\times 10^{11}\mathrm{m}\,\mathrm{pc}}{3.1\times 10^{16}\mathrm{m}}\approx 4.84\times 10^{-6}\mathrm{pc}$$

So substituting given values into $(3)$ and simplifying

$$28.95=-2.5\log_{10}\left[\frac{L_{\zeta}}{L_{\bigodot}}\left(\frac{4.84\times 10^{-6}}{460}\right)^2\right]$$

$$\implies 10^{-\dfrac{28.95}{2.5}}=\left[\frac{L_{\zeta}}{L_{\bigodot}}\left(\frac{4.84\times 10^{-6}}{460}\right)^2\right]$$

$$\implies \left(\frac{460}{4.84\times 10^{-6}}\right)^2\times10^{-11.58}=\frac{L_{\zeta}}{L_{\bigodot}}$$

$$\implies L_{\zeta}= \left(\frac{460}{4.84\times 10^{-6}}\right)^2\times10^{-11.58}L_{\bigodot}\approx \color{red}{2.38 \times 10^4 L_{\bigodot}}$$


The problem is that the correct answer is $7.6 \times 10^5 L_{\bigodot}$

The author's solution states that

To calculate the luminosity, we need to use the bolometric absolute magnitude as only bolometric quantities account for the power emitted over the whole wavelength range. As $M = −2.5 \log L+c$, where $c$ is a constant, we can relate $\zeta$ Pup’s luminosity $L_{\zeta}$ to the solar luminosity $L_{\bigodot}$ $$M_{\zeta}-M_{\bigodot}=-2.5\log_{10}\left(\frac{L_{\zeta}}{L_{\bigodot}}\right)$$ and thus $$\frac{L_{\zeta}}{L_{\bigodot}}=10^{-4 \left(M_{\zeta}-M_{\bigodot}\right)}=10^{0.4(4.8+9.9)}=7.6\times 10^5$$

Okay, so I understand why the author's solution is correct. But I don't understand why my solution is incorrect. This is because from my lecture notes I have that

Bolometric Magnitude

The total magnitude, $\color{green}{\text{apparent or absolute}}$, of a star represents the flux of the star summed over all wavelengths. This is termed the bolometric magnitude, $m_{\mathrm{bol}}$ or $M_{\mathrm{bol}}$ for $\color{green}{\text{apparent or absolute}}$. The difference between the bolometric magnitude of a star and its magnitude in a given passband, for example $V$, is called the bolometric correction, $BC$. For a given stellar type and luminosity class you can go from the magnitude measured in a given passband, say $V$, to the bolometric magnitude by adding the bolometric correction. Thus $$m_{\mathrm{bol}} = m_V + BC$$

So from the above, it clearly states that apparent magnitudes can be used also (which is what I used in my solution) but the author uses absolute magnitudes.


I don't really understand what I am missing here (probably something straightforward). So put simply, why is my solution wrong?

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Your first method didn't work because you appear to have just used the apparent visual magnitudes. This would work for two stars with the same bolometric corrections, but the Sun and Zeta Pup have different spectral types and very different bolometric corrections.

The values of $F$ in your equation (2) are the fluxes in the V-band only. You can only rewrite the ratio of these fluxes as a ratio of luminosities if the fraction of the luminosity appearing in the V-band is the same for both stars. Accounting for this is the point of the bolometric correction.

The bolometric correction of the Sun is close to zero, whereas according to your first calculation, the BC of Zeta Pup is -3.84.

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