Total luminosity of a galactic disk

[quarky2001]

This is basically problem 2.8 from Sparke & Gallagher "Galaxies in the universe".

There's only one area I'm having trouble with.

I've solved the surface density for stars in the galactic disk (# stars per unit area) as

$$\Sigma(R)=2\exp(\frac{-R}{h_R})h_R$$

Now, with L_0 being the luminosity of one star, and knowing that the surface brightness (luminosity per unit area) is

$$I(R)=L_0 \Sigma(R)$$

I'm supposed to find the luminosity of the entire disk. Likely by integrating, and the answer that's supposed to come out is

$$L_D=2\pi I(R=0) {h_R}^2$$

I keep getting factors showing up that don't belong in the answer though.

 

[mark726]

Hello, thank you for sharing your progress and question with us. It seems like you have made good progress in solving the surface density for stars in the galactic disk. However, you are correct in thinking that you need to integrate to find the total luminosity of the disk. Let me walk you through the steps to get to the correct answer.

First, let's define some variables for clarity:
- \Sigma(R) = surface density of stars in the galactic disk (in units of # stars per unit area)
- L_0 = luminosity of one star
- I(R) = surface brightness (luminosity per unit area)
- L_D = total luminosity of the galactic disk

To find the total luminosity of the disk, we need to integrate the surface brightness over the entire disk. This can be done using the following equation:

L_D = \int_{0}^{\infty} I(R) 2\pi R dR

Using the equation you provided for the surface brightness, we can substitute in I(R) and solve the integral:

L_D = \int_{0}^{\infty} L_0 \Sigma(R) 2\pi R dR
L_D = L_0 2\pi \int_{0}^{\infty} \Sigma(R) R dR

Next, we can substitute in the equation for surface density that you have already solved:

L_D = L_0 2\pi \int_{0}^{\infty} 2\exp(\frac{-R}{h_R})h_R R dR

To solve this integral, we can use integration by parts. Let u = 2\exp(\frac{-R}{h_R}) and dv = h_R R dR. This gives us du = -\frac{2}{h_R}\exp(\frac{-R}{h_R}) dR and v = \frac{1}{2}h_R R^2. Substituting these into the integral, we get:

L_D = L_0 2\pi \left[ -\frac{1}{2}h_R R^2 \exp(\frac{-R}{h_R}) \right]_{0}^{\infty} + \frac{1}{2}h_R \int_{0}^{\infty} R^2 \frac{2}{h_R}\exp