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Consider a black hole in vacuum at Temperature T. The setup that I am interested is a one that collects thermal photons from a black hole by enclosing in a spherical photographic plate and then lowering it back into the black hole.

If one assumes that photographic plates are massless.(which they are not). Then the very fact that I can lower the photographic plates(after the photons are collected) back into the black hole allows for a possibility of for the creation of a perpetual motion machine, from the potential energy generated by lowering the collected photons back into the black hole.

The mass of the photons emitted is equal to the mass of the black hole that has evaporated away. When I drop the photons back in the black hole, The mass of the black hole is equal to the original mass that we started off with. Which is why i thought it could be perpetual. This can continue for ever.

Of course I know something is wrong with such a setup. I think It can be traced to my assumption about photographic being massless.

My understanding of black hole thermodynamics is pretty elementary. Perhaps some one could comment on what is going on?

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The Hawking radiation emitted from the Black Hole lowers its mass and reduces the energy in it's gravitational field. If you now drop the photographic plate back into the Black Hole you will get energy back. However if you take that energy away, by using it to generate electricity or whatever, you leave the Black Hole with a lower mass. That's where the energy has come from.

Suppose you take a small enough Black Hole that it radiates a lot of energy and put it in the boiler of a steam engine. You can now use the heat to generate electricity, but of course the Black Hole eventually evaporates away to nothing. Take the same Black Hole and put it inside a perfect mirror, that reflects back all the emitted radiation, and the Black Hole remains unchanged but of course now you can't generate any electricity from it.

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The mass of the emitted photons is equal to the mass of the black hole that has evaporated away. After I dropping the photons back in the black hole, The mass of the black hole is equal to the original mass that we started off with. Which is why i thought it could be perpetual. –  Prathyush Nov 3 '12 at 13:02
    
@Prathyush: the photons redshift on their way from the black hole to the plate. The energy lost to the redshifting equals the energy gained from the lowering of the plate. Also, photons have no mass--their energy equals their momentum. –  Jerry Schirmer Nov 3 '12 at 13:36
    
@Jerry: I am using mass as Energy in a synonymous manner. If you drop an object mass m into a black hole, and you wait for it to evaporate away. The Total Mass(energy) of the photons you collect at far off from the horizon will be equal to the mass of the object you threw in Is it not? So redshift should not count. –  Prathyush Nov 3 '12 at 14:16
    
@Prathyush: nope--the redshift is precisely a loss of energy, analogous to the slowing down of a massive particle as it rises in a gravitational field. There is no conserved mass of a photon. –  Jerry Schirmer Nov 3 '12 at 14:30
    
@Jerry: I know what redshift is. If a photon is absorbed by a photographic plate there will be an increase in the mass of the photographic plate. The Increase in the mass of the photographic plate is equal to energy of the photon absorbed.(In this sense I use energy and mass as equivalent). When the black hole radiates away its mass as photons, the energy radiated through photons(as seen from a distant observer) will be equal to change in the mass of the black hole. Which is why I say redshift of the photons will not explain this at all. –  Prathyush Nov 3 '12 at 16:06
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protected by Qmechanic Dec 19 '12 at 12:54

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