From: gusenet@gmx.net   
      
   Am Dienstag, 22. April 2014 22:12:40 UTC+2 schrieb gghe...@gmail.com:   
   >    
   > > So when one beam can get from a certain point P at one blackbody to a   
   point Q to the other blackbody, there is always a way back, from Q to P - and   
   vice  versa. The prove focuses on another fact, that the intensity differ.   
   >   
   > Hmm, how does that happen?  (different intensities.)   
   >    
   > I mean if the two bodies are at the same temperature.   
      
   This is the main point of the idea. Firstly, the sheet with the lower   
   refractive index acts like a filter for certain angles (but the same for both   
   directions). Secondly, this filter is not parallel to one of the black bodies.   
   The combination of both    
   facts will generate this effect. But nevertheless the optical propagation here   
   is reciprocal.   
      
      
   > > The simplest statement for "Reciprocity" regarding optics states that 'if   
   I can see you, then you can see me'.  This remains true for this setup. The   
   equality of the amount is proved by using the second law of thermodynamics,   
   which is however only an    
   empirically validated postulate.   
   >    
   > Well as far as I'm concerned everything we know in science is empirically   
   validated.  You do the experiment to test your theory.  (You can't fool me   
   with those big words :^)   
      
   Many laws are mathematical proved, like the first law, and of course this fact   
   is empirically validated.    
   The second law about "heat can flow spontaneously from a colder to the hotter   
   body" is only a validated postulate, since in daily life you don't cannot see   
   such an effect. However it may exists like the Maxwell's demon h   
   tp://en.wikipedia.org/wiki/   
   Maxwell%27s_demon    
   I wrote a simple simulation program (for another purpose) using the Monte   
   Carlo method, starting beams randomly from both blackbodies and calculate the   
   reflections and absorption. When I simulate that all sheets are parallel, no   
   net flux appear. But    
   using this setup I described (including the sheet with the lower refractive   
   index) the net flux will appear.   
      
    Gerhard   
      
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