Here is a machine which seems to violate the second law of thermodynamics:

- $A$ and $B$ are point black bodies of the
**same**temperature (*initially*). - everything is rotationally symmetric around the axis $AB$
- $e$ and $f$ are ellipsoids with foci $A$ and $B$, made of a reflective material
- $CD$ and $EF$ are sections of a reflective annulus
- there is no air

The stable state of the machine is $A$ having **higher** temperature than $B$ because

- The heat radiated by $B$ is all absorbed by $A$ (via paths $B\rightarrow K\rightarrow A$ and $B\rightarrow J\rightarrow A$).
- The heat radiated by $A$ is either absorbed by $B$ (via paths $A\rightarrow K\rightarrow B$ and $A\rightarrow J\rightarrow B$)
*OR*by $A$ (via paths $A\rightarrow G\rightarrow I\rightarrow A$)

This seems to violate the 2nd law.

So, where is the hole here?

PS. While point bodies and perfect mirrors do not exits, note that we have quite a lot of margin here: a huge left ellipsoid and a tiny right ellipsoid will lead to almost 50% of all radiation from $A$ reflecting back to $A$. So, "small" bodies and 90%-efficient mirrors should be fine.

## Best Answer

Ingenious. A and B are small, but they cannot be points.The image of B is magnified at A. Therefore if A and B are the same size, some of the light from B will miss A.