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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?

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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 $Q_{in}=\sigma T_1^4 A_2$

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  • $\begingroup$ I can't quite see why $Q_{in} \sim A_2$ though even though I agree that a flat surface can only receive radiation from half the solid angle. That restricts us to receiving radiation from area $A_{1}/2$ but $A_{1}$ is arbitrary. Could you expand a little but further and prove that $Q_{in}\leq \sigma T_{1}^4 A_{2}$? $\endgroup$ – user1936752 Aug 15 '15 at 7:46
  • $\begingroup$ The radiation emitted per unit area from the surface of body 1 is $\sigma T_1^4$ and it is emitted into $2\pi$ steradians. If, on the surface of body 2, the mirrors and lenses produce a $2\pi$ steradian image of body 1, then the best intensity per unit area received on the surface of body 2 will be $\sigma T_1^4$. If we integrated that over the area of body 2, we get $\sigma T_1^4 A_2$. $\endgroup$ – John1024 Aug 15 '15 at 7:56
  • $\begingroup$ I disagree with your example. Using a parabolic mirror you can get almost all 4$\pi$ light onto the second surface. $\endgroup$ – Rol Aug 15 '15 at 9:33
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    $\begingroup$ @Rol Any point on the surface of either object is exposed to at most 2π steradians of illumination on the outside, with the other 2π being the interior of the object. It's only possible to transfer all the radiation onto the second surface if it's as big as the image of the first surface. (That might be the basis of a nice alternative argument.) Doing so would require an elliptical mirror… a parabola loses light through its open end. $\endgroup$ – Blackbody Blacklight Aug 15 '15 at 10:40
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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.

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  • $\begingroup$ Down vote care to comment? $\endgroup$ – paparazzo Aug 15 '15 at 8:10
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    $\begingroup$ I'm not the downvoter, but you've only begged the question. "Why isn't heat flowing from cold to hot here?" "Because it doesn't." Also, it's short, poorly formatted, and a bit condescending in tone. The definition of an ideal blackbody is that it does absorb all incident radiant energy, and only re-emits it in a pure blackbody spectrum. So before begging the question, you denied the premise… $\endgroup$ – Blackbody Blacklight Aug 15 '15 at 8:54
  • $\begingroup$ @BlackbodyBlacklight I used the term second body. A black body is idealized. Just because the idealization breaks the 2nd law of thermodynamics does not mean it will happen. $\endgroup$ – paparazzo Aug 15 '15 at 9:11
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    $\begingroup$ Ideal blackbodies can be approximated well enough. The idealization does not break the law: See the accepted answer. $\endgroup$ – Blackbody Blacklight Aug 15 '15 at 10:33
  • $\begingroup$ @BlackbodyBlacklight So, yes, I see the accepted answer. My statement "Your argument is wrong in that the second body receives Qin=σT41A1" is correct and it is because you can't violate the 2nd law. The model is able to honor 2nd law is not the same. The causality is temperature wins every time. $\endgroup$ – paparazzo Aug 15 '15 at 19:13

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