I'm trying to understand how to calculate radiance ($L_{e,\nu}$), so I am using the sun/earth as a comparison as I have seen it stated that the radiance of sunlight is equal to:

$$ L_{e,\Omega} \approx 1.54\times10^7 \frac{W}{sr \cdot m^2} $$

My understanding of radiance is:

  1. Radiance is the radiant flux ($\Phi_e$) emitted or received by a surface, per unit solid angle per unit projected area
  2. Radiance is constant along a ray


\begin{align} \Phi_e &= 3.7889e26 \ W \\ R_{sun} &= 6.957\times10^8 \ m\\ R_{earth} &= 6.3781\times 10^6 \ m\\ d &= 1.4776\times10^11 \ m \end{align}

Then the irradiance at Earth and the radiant exitance at the Sun are, respectively: $$ E_e = 1.381\times10^3 \ \frac{W}{m^2}\\ M_e = 6.2296\times10^7 \ \frac{W}{m^2} $$

Using the equation to calculate the solid angle of distant spherical object:

$$ \Omega = 2\pi\left(1 - \frac{\sqrt{d^2 - R^2}}{d}\right) $$

we can find that the solid angle subtended by the Earth from the perspective of the Sun, and the solid angle subtended by the Sun from the perspective of the Earth are, respectively: $$ \Omega_{earth} = 5.8635\times10^-9 \ sr \\ \Omega_{sun} = 6.9644\times10^-5 \ sr $$

So far, all of these values seems to match up with what I've found online.

On wikipedia it is stated that when calculating the radiance emitted by a source, $\Omega$ refers to the solid angle into which the light is emitted, while when calculating radiance received, $\Omega$ refers to the solid angle subtended by the source as viewed from that detector. This leads me to believe we could calculate radiance one of two ways.

The first way would be to divide the irradiance received at earth by the solid angle subtended by the sun: \begin{align} L_{e,\Omega} &= E_e/\Omega_{sun}\\ &= \frac{1.381\times10^3 \ \frac{W}{m^2}}{6.9644\times10^-5 \ sr} \\ &= 1.9829\times10^7 \frac{W}{sr\cdot m^2} \end{align}

This answer appears to be correct, as while it differs slightly from the amount I've seen quoted, the source also states the solar irradiance would be $\approx1050\frac{W}{m^2}$. Swapping this out for my calculated irradiance matches exactly.

The second way to calculate radiance therefore would be to take the radiant exitance and divide it by the angular size of earth. Doing this however, gives the following result: \begin{align} L_{e,\Omega} &= M_e / \Omega_{earth}\\ &= \frac{6.2296\times10^7 \frac{W}{m^2}}{5.8635\times10^-9 \ sr} \\ &= 1.0642\times10^{16} \frac{W}{sr \cdot m^2} \end{align}

This doesn't match at all, and my expectation is that the radiance, whether calculated from the sun's perspective heading towards the Earth or from the Earth's perspective received from the sun, should be the same.

Exploring this a bit further, I attempted to see if I could rearrange terms to use my (assuming to be) correct $L_{e,\nu}$ and $M_e$ to determine what $\Omega_{earth}$ should be. Doing this yielded: \begin{align} \Omega_{earth} &= \frac{M_e}{L_{e,\Omega}}\\ &= \frac{6.2296\times10^7 \ \frac{W}{m^2}}{1.9829\times10^7 \frac{W}{sr\cdot m^2}}\\ &= 3.1416 \end{align}

Assuming that that truly is exactly $\pi$, and not a coincidence, that would imply that the radiance can be calculated by dividing the radiant exitance by $\pi$. I cannot wrap my head around why this would be the case.

Any guidance on what I am missing here would be greatly appreciated!


1 Answer 1


After re-reading some of my notes, I think I have misinterpreted the radiance/radiant exitance relationship. It is not related to the angle subtended by the observer, but rather the sum of all possible directions the light is emitted from. This seems obvious in hindsight.

The answer to my confusion comes from looking at the relevant relations between quantities. Namely the following:

\begin{align} M_e &= \frac{\partial \Phi_e}{\partial A} \\ L_{e,\Omega} &= \frac{\partial^2 \Phi_e}{\partial\Omega \partial (A\cos{\theta})}\\ \end{align}

We can see through substitution that we can express $L_{e,\Omega}$ in terms of $M_e$:

\begin{align} L_{e,\Omega} &= \frac{\partial M_e}{\partial \Omega} \frac{1}{\cos{\theta}}\\ \int_SL_{e,\Omega} \cos{\theta} \ d\Omega &= M_e \end{align}

If we assume that the sun is an isotropic emitter (i.e. a Lambertian emitter) than we can integrate over the entire hemisphere of a point on the sun, giving:

\begin{align} M_e &= \int_0^{2\pi}\int_0^{\pi/2} L_{e,\Omega}\cos{\theta}\sin{\theta} \ d\theta \ d\phi = L_{e,\Omega} \pi \end{align}


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