Second Law of Thermodynamics and heating a blackbody with another blackbody Given a large blackbody with surface area $A_1$ and temperature $T_1$, let's assume I can use some mirror and lens system to capture all the emitted radiation and transfer this energy to a smaller blackbody of area $A_2$ such that $A_2<A_1$. Let the temperature of the second body be $T_2$. 
The second body receives $Q_{in}=\sigma T_1^4 A_1$. It emits $Q_{out}=\sigma T_2^4 A_2$. By assumption $A_2< A_1$, so in a steady state, we must have $T_2>T_1$. However, this violates the second law of thermodynamics (heat transfer from cold to hot body without doing any work). Where is the argument going wrong? 
 A: Mirrors and lenses cannot do what you ask.
Light has a specific intensity, meaning a power per unit area per unit solid angle per unit wavelength.  Mirrors and lenses can never increase the specific intensity.
As an example, consider using a lens to focus sunlight onto a target.  The specific intensity of the sunlight is the same with or without the lens.  What the lens does is increase the solid angle of sunlight that the target receives.  But, there is a maximum: it is not possible for a flat surface to receive a solid angle of light greater than $2\pi$ steradians.  This is why the target can never get hotter than the source, regardless of the optical system used.  
Consequently, the mirrors and lenses cannot deliver $Q_{in}=\sigma T_1^4 A_1$ to body 2 with $A_2<A_1$.  The most that it can deliver is  $Q_{in}=\sigma T_1^4 A_2$
A: Your argument is wrong in that the second body receives Qin=σT41A1.  You can direct radiant energy at it - does not mean it will receive it.  Heat does not flow from cold to hot.  Temperature wins every time.  Double the heat at the same temperature is still the same temperature.
