# Calculate radiance of Lambertian emitters: Is the solid angle $\pi$ or $2 \pi$?

Given a Lambertian emitter (or reflector) of area $$A$$ that is emitting a total power (resp. flux) of $$\Phi$$ (units $$W$$). To calculate the radiance $$L$$ (units $$\frac{W}{m^2 sr}$$) the solid angle $$\Omega$$ is needed:

$$L = \frac{\Phi}{A \cdot \Omega}$$

Question: Is the solid angle $$\Omega = \pi$$ or $$\Omega = 2 \pi$$?

Approach 1: My initial feeling was $$2 \pi$$, because $$L$$ is constant in all directions of the hemisphere for Lambertian emitters, and the hemisphere has the solid angle $$2 \pi$$.

Approach 2: The radiance of a Lambertian emitter is $$L = \frac{I(\theta)}{A \cdot cos(\theta)}$$ (which is constant as $$I(\theta) = I_{max} \cdot cos(\theta)$$ for Lambertian emitters). Thus, $$I(\theta) = L \cdot A \cdot cos(\theta)$$ (units $$\frac{W}{sr}$$), which gives the total flux or power $$\Phi$$ (units $$W$$) when integrated over the hemisphere: $$\Phi =\int_{0}^{2\pi} \int_{0}^{\frac{\pi}{2}} I(\theta) \cdot sin(\theta) d\theta d\varphi$$ $$= \int_{0}^{2\pi} \int_{0}^{\frac{\pi}{2}} L \cdot A \cdot cos(\theta) \cdot sin(\theta) d\theta d\varphi$$ $$= 2\pi \cdot L \cdot A \int_{0}^{\frac{\pi}{2}} cos(\theta) \cdot sin(\theta) d\theta$$ $$= 2\pi \cdot L \cdot A \cdot \frac{1}{2} \int_{0}^{\frac{\pi}{2}} sin(2\theta) d\theta$$ $$= \pi \cdot L \cdot A$$

And thus $$L = \frac{\Phi}{\pi \cdot A}$$ and $$\Omega = \pi$$, which is contradiction to the initial approach of $$2\pi$$.

So which one is correct and why? (Calculation says $$\pi$$, but $$2 \pi$$ is more intuitive...).

• Thanks, I understand now that the projected solid angle of the hemisphere is $\pi$. But is there a way to see why a projected angle is needed at all? Maybe it's naive, but as there is a solid angle in the definition of the radiance I would have just used the real solid angle, not a projected one. Jun 19 '21 at 8:29
The projected solid angle of the hemisphere is $$\pi$$ from its definition $$\Omega=A/r^2$$, where $$A$$ is the corresponding surface area.