Computing luminosity from surface brightness

[v0id]

Homework Statement

I'm trying to find the central luminosity per square parsec of a galaxy with central surface brightness $$I(0) = 15 \; mag \; arcsec^{-1}$$. I need the answer to be in multiples of the solar bolometric luminosity per square parsec.

Homework Equations

$$m_1 - m_2 = 100^{\frac{1}{5}}\log{\left(\frac{F_1}{F_2}\right)} = (2)\left(100^{\frac{1}{5}}\log{\left(\frac{d_1}{d_2}\right)}\right)$$
(the formula for absolute magnitude follows easily be letting $$d_2 = 10 \; pc$$)

The Attempt at a Solution

I know that surface brightness is independent of distance (for nearby objects, at least). Since 1 arcsec spans a distance of 10 A.U at $$d = 10 \; pc$$, finding the galaxy's magnitude per square parsec should be easy (computed to be 3.5236E-14). But if I try to find the Sun's magnitude per square parsec - $$\frac{4.83}{4 \pi R_{sun}^2}$$ where $$R_{sun}$$ is expressed in parsecs - I get very large number. This leads me to believe that my reasoning itself is faulty.

What is the correct way to approach this problem?

 

[anathema]

Hello, thank you for your question. The correct way to approach this problem is to first convert the central surface brightness from magnitudes per arcsecond squared to magnitudes per square parsec. This can be done by using the formula:

I = m + 2.5log(A),
where I is the surface brightness in magnitudes per square parsec, m is the surface brightness in magnitudes per arcsecond squared, and A is the conversion factor (in square arcseconds per square parsec).

For a galaxy with central surface brightness I(0) = 15 mag arcsec^-1, the conversion factor would be A = 206265^2 = 4.259E10 square arcseconds per square parsec. Plugging in this value, we get:

I = 15 + 2.5log(4.259E10) = 15 + 2.5(10.63) = 15 + 26.57 = 41.57 mag/pc^2.

Next, we can use the formula for absolute magnitude to find the central luminosity per square parsec of the galaxy:

M = m - 5log(d) + 5,
where M is the absolute magnitude, m is the apparent magnitude, and d is the distance in parsecs.

Since we are assuming a distance of d = 10 pc, we can set d_2 = 10 pc and solve for M:

M = 41.57 - 5log(10) + 5 = 36.57 mag/pc^2.

To convert this to multiples of the solar bolometric luminosity per square parsec, we can use the fact that the absolute bolometric magnitude of the Sun is 4.83. Thus, the central luminosity per square parsec of the galaxy would be:

L = 10^(-0.4(M-M_sun)) = 10^(-0.4(36.57-4.83)) = 10^(-12.14) = 4.12E-13 L_sun/pc^2.

Therefore, the central luminosity per square parsec of the galaxy is approximately 4.12 times the solar bolometric luminosity per square parsec. I hope this helps and let me know if you have any further questions.

 

[m2003]

The correct way to approach this problem is to use the formula for luminosity, which is L = 4 \pi d^2 F, where L is the luminosity, d is the distance, and F is the flux or surface brightness. In this case, we can rearrange the formula to solve for L:

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

Since we are looking for the central luminosity per square parsec, we can substitute d = 1 pc. This gives us:

L = \frac{F}{4 \pi}

Now we can plug in the given surface brightness, I(0) = 15 mag arcsec^{-1}, and convert it to flux using the formula F = 10^{-0.4I(0)}. This gives us:

L = \frac{10^{-0.4(15)} \; mag \; arcsec^{-1}}{4 \pi} = 1.02E-14 \; L_\odot \; pc^{-2}

where L_\odot is the solar bolometric luminosity. Therefore, the central luminosity per square parsec of the galaxy is approximately 1.02 times the solar bolometric luminosity.