# I Understanding Luminosity,Brightness,Intensity,Flux

#### Arman777

Gold Member
I am taking astrophysics course and we are constantly talking about Luminosity,Brightness,Intensity,Flux etc. I thought that I understand the idea properly but I see now that I dont.

Intensity is the light passing through a steradian per second

$$I=E/dtd\Omega$$

Luminosity is intensity per area

$$L=E/dtdA\Omega=I/dA$$

And Flux is the Luminosity per area ?

$$F= L/4\pi r^2$$

And brightness = Flux

Are these statements true ?

I am confusing these definitions and their meanings constantly. Can someone help me to understand it better.

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#### jambaugh

Gold Member
Luminosity, according to wikipedia is the total power output of a star. That's energy per unit time.

Flux is the power per unit area so given its luminosity is $L$ then the light flux through the a sphere $r$ units away from the center of the star is $\frac{1}{4\pi r^2}L$ because that is the total Luminosity divided by the surface area at that radius. All the light is flowing outward across that spherical surface.

Luminous Intensity is the power emitted per unit solid angle from a point source or spherically symmetric source. It is basically the flux (power per unit area) through the unit sphere (1 steradians = 1 unit area) centered at the source. Thus Intensity is simply rescaled Luminosity $I=\frac{1}{4\pi}L$ assuming a spherically symmetric source. An object could have directionally dependent luminosity (as an extreme example consider a pulsar).

#### hutchphd

Two usual points of confusion:
Radiometry is what Maxwell and energy flux is about. Usual physical definitions.
Photometry (illuminence) convolves everything with the sensitivity of our eyeball and is what photographers have historically used. For instance a lumen is a photometric measure of light output and lux is lumen/sq.m.
Here is a good explanation

I suggest printing off a comprehensive conversion table and sticking it to your wall!

#### Arman777

Gold Member
I understand it I think. I looked a optic textbook, in there it says the unit of radiant flux is (W). And the notation etc confused me a lot.

#### Arman777

Gold Member
So then the Energy transfer to earth from sun per second would be,

$$dE=dt\frac{L_{sun}A_{earth}}{4\pi r^2}$$ for $r=1AU$ ?

#### hutchphd

Yes (where A is the area of the flat earth disc)

#### jambaugh

Gold Member
The energy transfer per second would be: $dE/dt$, what you wrote is the amount of energy transferred in$dt$ seconds. Note that as hutchphd mentioned you want the area of Earth's shadow which is going to be $\pi R^2$. So your formula would simplify to:
$$\frac{dE}{dt} = L_{sun} \frac{R^2}{4r^2}$$
Basically what hits the Earth is what proportion of the area of the sphere 1AU from the sun is covered by the Earth's shadow times what the sun puts out, L.

#### Arman777

Gold Member
Thank you all

"Understanding Luminosity,Brightness,Intensity,Flux"

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