I'm trying to figure this out for the enter image description here case...

enter image description here

Supposedly the average luminosity will be equal to enter image description here but how :S

I know enter image description here is the number density of galaxies per unit volume and luminosity but that isn't helping me find an average luminosity!


1 Answer 1


So first of all, $\Phi(L)dL$ is, as you say, the number density (per unit volume) of galaxies with a luminosity between $L$ and $L+dL$.

The constant $\Phi_0$ is a normalisation constant, and defines the overall number of galaxies per unit volume.

To calculate the average luminosity we simply add up the luminosity from each galaxy and divide by the number of galaxies - but since the luminosity function is continuous we will need to turn this summation into integration.

The total luminosity (per unit volume) from all galaxies with a luminosity between $L+dL$ is $\Phi(L)LdL$, and the total number of galaxies per unit volume is $\Phi_0$, so our average luminosity can be calculated:

$\bar{L}=\frac{\int_0^\infty\Phi(L)LdL}{\Phi_0}\\ =\frac{1}{\Phi_0} \int^\infty_0 \Phi_0\left(\frac{L}{L_*}\right)^{-1}e^{-\frac{L}{L_*}}\frac{L}{L_*}dL\\ =\int_0^\infty e^{-\frac{L}{L_*}}dL\\ =\left[-L_*e^{-\frac{L}{L_*}}\right]_0^\infty\\ =(-L_*e^{-\frac{\infty}{L_*}})-(-L_*e^{-\frac{0}{L_*}})\\ =0+L_*\\ =L_* $

So we find after some maths that the average luminosity if $L_*$ when $\alpha=-1$.

Good luck with the studies!

  • $\begingroup$ Is $\bar L$ the same thing as $\langle L\rangle$? If so, do you think this method would be useful for this question where I asked about finding $\langle V/V_{max}\rangle$? $\endgroup$
    – John Doe
    Commented Apr 30, 2018 at 18:58

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