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I am having trouble understanding the bolometric magnitude and it relation to this question:

Show if there were a white dwarf instead of the sun in our Solar System, it would have an apparent visual magnitude of $-18.6 $mag. Assume hat the White Dwarf has a surface temperature of three times that of the Sun (which is $5780K$) and a radius of $0.01$ times that of the Sun. The Sun has an absolute bolometric magnitude of $4.75$mag and the visual magnitude of the White Dwarf will be three magnitudes fainter than its bolometric magnitude.

There are a couple of thing that puzzle me here but first I will show working to were I am at.

What been given in the question as follows: $$T_{white\:Dwarf}=3T_{sun}$$

$$R_{White\:Dwarf}=0.01R_{sun}$$

I know of the equation which relates the absolute bolometric magnitudes is as follows:

$$M_{bol\:White\:Dwarf}-M_{bol\:sun}=-2.5log(\frac{L_{wWhit\:Dwarf}}{L_{Sun}}) \:[1]$$

So using the luminosity radius relation I can form the Equation for $\frac{L_{White\:Dwarf}}{L_{Sun}}$ like so:

$$\frac{L_{White\:Dwarf}}{L_{Sun}}=(\frac{R_{White\:Dwarf}}{R_{sun}})^2\cdot(\frac{T_{White\:Dwaf}}{T_{sun}})^4 \:[2]$$

So I would just sub the given value in a get a numerical ans

Now my next step is to use the absolute metric equation ans rearrange for the absolute bolometric of the white dwarf:

$$M_{bol\:White\:Dwarf}=M_{bol\:sun}-2.5log(\frac{L_{wWhit\:Dwarf}}{L_{Sun}})$$

Once again subbing the value that I obtain for [2] and the given absolute bolometric for the sun, I will obtain a numerical ans for $M_{White\:Dwarf}$

But it the last part of the question that troubles me which is this:

"the visual magnitude of the White Dwarf will be three magnitudes fainter than its bolometric magnitude."

Now what I am not sure of is what are they defining as the bolometric, is it magnitudes is the "apparent bolometric" which is designated $m_{bol}$ or is it the "absolute bolometric" $M_{bol}$?

I am inclined to think that it is $m_bol$ as they are referring to the apparent visual magnitude. So in that case the visual magnitudes is a third of the bolometric like so:

$$m_{white\:Dwarf}=\frac{1}{3}\cdot M_{white\:Dwarf}\:[4]$$

But this dose not seem correct, well I know it not because my value is neither negative or even close to the value of what is wanted in the question. What troubles me is that the question is asking for "visual apparent magnitudes" but they state is as "visual magnitude" what is the difference are they the same thing? I have googled but nothing seem to come up which is confusing me more and more. As I read further there ref to bolometric constant, which as I understand will only be in relation when viewing from earth, but I don’t have the necessary data in the question for that.

I have read through a couple of book but nothing really explain the relations for what bolometric magnitudes is, how it relates to visual magntided.

Could someone please maybe expand upon some of the troubles I am having, I personally cant see how bolometric and visual tie together nor how it applies to this question.

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The bolometric magnitude is a representation of the flux at all wavelengths. The visual magnitude is that flux that is in the "visual band". For the Sun, most of the flux it emits (or at least a large fraction) is in the visual band. If your white dwarf is three times hotter than the Sun, then Wien's law dictates that the peak of its spectral energy distribution will be at a wavelength only 1/3 of that where the solar spectral energy distribution peaks.

That means that most of the bolometric flux of the white dwarf falls outside the visual magnitude, so I can easily believe that the bolometric correction (the difference between bolometric and visual magnitudes) is 3 magnitudes for the white dwarf (whereas it is roughly zero for the Sun).

NB Your equation (4) is incorrect. The question is telling you that $M_V-M_{\rm bol} = 3$ mag. (or indeed $m_V - m_{\rm bol} = 3$ mag. if we want to work in apparent magnitudes)

Also not that the zeropoints of the bolometric and visual magnitude systems are not the same (hence why the difference is $\sim 0$ for the Sun).

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  • $\begingroup$ Ah thank you, that clear some stuff up, much appreciated. $\endgroup$ – james2018 Apr 26 '18 at 18:37

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