Timeline for Perpetuum Mobile
Current License: CC BY-SA 2.5
4 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 6, 2010 at 6:14 | comment | added | Yuri Bakhtin | You can mentally split the red one in two parts. One is in equilibrium with its own gas of photons. The other one is in equilibrium with the blue one. These two red parts are also in equilibrium with each other. How about that? | |
| Jan 6, 2010 at 6:11 | comment | added | Yuri Bakhtin | I do not know the actual physics of this, but I have convinced myself that writing detailed balance equations should produce equal temperatures for the two bodies. At least I see no contradiction. the balance equations should include energy exchanges: blue body<-->fotons between blue and red bodies<-->red body<-->bath of fotons that do not reach the blue one. (in fact, one should separate the fotons traveling from red to blue and vice versa). | |
| Jan 6, 2010 at 5:21 | comment | added | Anton Petrunin | Assume it is equilibrium then temperature of red and blue bodies should be the same --- it does not matter how mush energy flying in between. You can rewrite it as equality of some integrals. BUT here we have funny thing: if size is zero those integrals are NOT equal, if little more --- you expect them to be equal --- why? It should be simple math... | |
| Jan 6, 2010 at 4:42 | history | answered | Yuri Bakhtin | CC BY-SA 2.5 |