How does intensity differ from apparent luminosity in the context of photometry?

Context: photometry in astronomy.

Background: The total luminosity $L$ of a star is the energy that radiates in all spatial directions in all wavelengths and is given by the following formula, where $F$ is the flux, $\lambda$ the wavelength and $R$ the radius of the star:

\begin{align*} L = 4\pi R^2 \int_0^\infty F(\lambda) \mathbb{d} \lambda \end{align*}

If we "divide" that total luminosity on a sphere with radius $d$ where $d$ is the distance between us, the observers, and the star, then we get the apparent luminosity, that is:

\begin{align*} \ell = \frac{L}{4\pi d^2} \end{align*}

and is measured in $W/m^2$.

Question: How does the apparent luminosity, $\ell$, differ from the intensity $I$ ?

• Well, what's your equation for $I$ ? And BTW, luminosity (per Wikipedia) is watts, not per area, so I suspect you're mixing units. – Carl Witthoft Jun 17 '14 at 14:04

The terms used in photometry and radiometry have specific meanings that may not match the meanings that the words have in other context. (The same is true of the words "heat" and "work", which mean different things outside of a physics context.) Furthermore, not everyone uses the same words. The astronomy community may have their own standard usage that differs from the optics community. (I hope not, but I'm not familiar with what astronomers do.)

The first thing to straighten out is that your question concerns radiometry, not photometry. Photometric quantities are weighted by the spectral response of the eye. It does not appear that your question involves what a human sees with his eyes.

You can get the radiant intensity of your star by dividing $L$ by $4\pi$, the total solid angle subtended by a sphere. (Assuming the star radiates isotropically.) $$I=\frac{L}{4\pi}$$
The quantity $F(\lambda)$ might be better called spectral radiant power or spectral power (the word "radiant" being redundant, but makes it clear that we are not talking about photometry.) Calling $L$ "flux" is ok, but the word "flux" is one of those words that different people give different meanings. To avoid that problem, I usually call $L$ radiant power, although if I'm writing in a context where radiant flux has already been established as the definition, I'll use "radiant flux".
What you call apparent luminosity I would call irradiance, the total power incident on a unit area, which in this case, is a unit area on earth. ("Apparent luminosity might be a term astronomers use. I don't know.) Watts per square meter. It is obtained by multiplying the intensity by the solid angle at the star subtended by one square meter on earth: $$\ell = I\Omega = I \frac{1}{d^2} = \frac{L}{4\pi d^2}$$