A Lambertian surface has a constant radiance $R$, regardless of the angle from which it is observed, where $A = A_0\cos(\theta)$ is the apparent surface area of an observer at angle $\theta$ normal to the surface.
$$
R = \frac{I}{A} = \frac{I}{A_0\cos(\theta)} = \text{const.}
$$
For that to be true the radiant intensity $I$ must be proportional to the $\cos(\theta)$ equal like
$$
I = I_0 \cos(\theta)
$$
so the $\cos(\theta)$ cancel
$$
R = \frac{I_0 \cos(\theta)}{A_0\cos(\theta)} = \frac{I_0}{A_0} = \text{const.}
$$
Now for an observer moving with speed $v$ close to the speed of light $c$ the apparent surface gets length contracted like $A = A_0/\gamma(v)$ and then radiance would be dependent on the speed like $$ R = \gamma\frac{I_0}{A_0} = \frac{I_0}{A_0 \sqrt{1-(v/c)^2} } \neq const. $$ Does this make sense? Is it true that for observers moving at very high speeds the radiance of a surface depends on the speed of the observer?
