On the wikipedia page for Black Body radiation, it states that:


where $I(\nu,T)$ is the energy per unit time (or the power) radiated per unit area of emitting surface in the normal direction per unit solid angle per unit frequency by a black body at temperature T.

I wanted to use this formula to find (roughly) the amount of U.V. light that the Earth receives from the sun. To do this, I supposed that I would first integrate against $\nu$ for all frequencies that lie in the U.V. range. Next, I would restrict that number to just the amount of U.V. light that points in the direction of Earth.

However, there seems to be two contradictory ways of limiting my answer to only the U.V. light that is incident upon Earth.

enter image description here

Method 1

First, find the total power outputted by the sun in the U.V. range, by integrating over the U.V. spectrum, and then multiplying the result by the area of emitting surface $(4\pi R^2)$ and the solid angles subtended by the surface $(4\pi)$. After that, we can limit the result to the amount of U.V. light incident upon earth by multiplying by the proportion of the area of the Earth to the area of the sphere at a distance d from the sun. $(\frac{A}{4\pi d^2})$ where $A$ is the surface area of the Earth facing the sun. We see that the factor that we have to include to account for the amount of U.V. light incident upon the earth is:

$$4 \pi A \frac{R^2}{d^2} $$

Method 2

Again, we will first integrate the expression over all the U.V. frequencies, but this time we will only consider the power emitted by the small area of the sun that is facing the earth (colored in red on the diagram). So now, the area that is producing the U.V. light that strikes the earth is given by $A \frac{R}{d}$, and the solid angle that this area subtends onto the sun's center is equal to the solid angle the earth subtends onto the Sun's center, which is given by $\frac{A}{d^2}$. We see that the factor we have to include to account for the amount of U.V. light falling onto Earth only using method 2 is

$$ A^2 \dfrac{R}{d^3}$$

Both methods give results which are dimensionally correct, though only one of them can be right. I'm pretty sure I've just messed up in regards to the 'per unit solid angle', this is the first time that I've encountered the term so that's probably what's wrong, though I'm not sure.

  • $\begingroup$ I can't think about it right now, but I find it very unlikely that you should be multiplying by area and solid angle. Usually it's one or the other. $\endgroup$
    – Javier
    Commented Oct 30, 2015 at 1:01
  • 1
    $\begingroup$ both methods are wrong $\endgroup$
    – user83548
    Commented Oct 30, 2015 at 1:05
  • $\begingroup$ To ask a thorny question, you want for the earth, as in the upper atmosphere, or what reaches the ground? The atmosphere absorbs at lot. $\endgroup$
    – MaxW
    Commented Oct 30, 2015 at 1:14
  • $\begingroup$ That model is for a point source $\endgroup$ Commented Oct 30, 2015 at 3:26

1 Answer 1


I don't think either is right.

You can find by integrating over solid angle that the spectral power per unit area from the surface is $\pi I_{\nu}$ in Watts per square metre per Hertz.

The luminosity spectrum (assuming isotropy) is this times $4\pi R^2$. The flux at Earth is this divided by $4\pi d^2$ and then you integrate over whatever spectral range you wish.

$$ f_{\nu} = \pi I_{\nu} \frac{R^2}{d^2}$$

The appropriate area to multiply by for the power intercepted by Earth is $\pi R_{E}^{2}$.

  • $\begingroup$ Shouldn't integrating over solid angle for the surface result in $4\pi I_\nu$, because there are $4\pi$ solid angles in a sphere? $\endgroup$ Commented Oct 30, 2015 at 8:40
  • 1
    $\begingroup$ @JoshuaLin No. The power per unit area is $\pi I_{\nu}$. You integrate $I_{\nu} \cos \theta\ \sin\theta d\theta\ d\phi$ over a hemisphere. $\endgroup$
    – ProfRob
    Commented Oct 30, 2015 at 13:17

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