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Radiation Pressure - Piston

  1. Jun 15, 2011 #1
    Hi All,

    First off, everyone on this forum is amazing. Period.

    Second, I had a thought experiment the other day that was interesting and was hoping you could comment on it:

    The System:

    Imagine an infinitely long hollow cylinder in one direction, made of a metal (perhaps aluminum). At the beginning of the cylinder is a stationary black-body radiator that has a high emissivity (say, carbon black). Somewhere in the middle of the cylinder is a movable mirror that may move up or down the cylinder, acting as a piston.

    Suppose the entire system is at thermal equilibrium. The mirror-piston and metallic cylindrical container have a low emissivity, so that the carbon black emitter is emitting more thermal radiation that either the mirror or the cylinder.

    In addition, at any time a mirror may be slid in front of the carbon black emitter, effectively reflecting radiation that would otherwise have traveled down the cylinder back onto the emitter.

    Finally, whenever the emitter is uncovered, thermal radiation travels down the cylinder from the emitter to the mirror-piston, producing a small radiation pressure on the mirror-piston.

    My Question:

    Imagine this engine cycle:
    1. The mirror-piston travels down the cylinder, due to the radiation pressure
    2. The emitter cover is slid over the emitter
    3. The mirror-piston is compressed up the cylinder
    4. The emitter cover is retracted, uncovering the emitter

    Since compressing the piston requires less work than expanding it (less radiation pressure), the temperature of the system would spontaneously decrease.

    My only explanation would be that the work required to slide the emitter cover into place over the emitter might be the culprit - However, since it is a finite quantity of work, it could always be compensated for by allowing the mirror-piston to travel further down the cylinder.

    Thoughts?
     
  2. jcsd
  3. Jun 15, 2011 #2

    Drakkith

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    What causes the mirror to move back down? Something has to apply a force.
    Also, is this piston a vacuum or is it filled with something?
     
  4. Jun 15, 2011 #3
    Yes, I'd assume the piston is a vacuum. Also, to keep it simple, let's assume the cylinder is on Earth and vertical, so that gravity will pull the mirror back down.
     
  5. Jun 15, 2011 #4

    Drakkith

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    Then the work pulling the piston back down is provided by gravity. This is just like picking up a rock and dropping it to the ground. No temperature decrease would happen.
     
  6. Jun 15, 2011 #5

    jambaugh

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    Consider this mirror you are placing between BB emitter and piston.

    There are two extremes of possibilities.

    1.) The back, facing the piston is of low emissivity (i.e. you have a two-sided mirror). Then the light which passes the mirror, and bounces off walls and piston will bounce off the back of the mirror and the back to the piston again. You'll have a photon gas at equilibrium temperature and the pressure is unchanged by the presence of that mirror.

    2.) The back, facing the piston is of high emissivity and the mirror will have reached equilibrium temperature equal to the BB radiator. Then the mirror's back emits just as much radiation as does the BB radiator and again the photon gas is at equilibrium temperature with the pressure unchanged.

    Something to remember about emissivity of surfaces in these thermal experiments. They are two way streets. Use a diffusing gas as an analogy. The analogue of a low emissivity surface for the photons will be a non-porous surface for the gas. Likewise the analogue of a high emissivity surface is a porous surface. The gas can enter and leave the surface.

    Try making your engine work with conventional gas with these analogues. See what goes wrong. Note that the emitter cover is no different from a valve between upper and lower half cylinders (and the black body acts as a gas reservoir). Imagine trying to get net work out of a piston and cylinder merely by switching a valve to a tank open and closed at various parts of a cycle.

    Back to the photon gas "engine" note that even if your completely close off the cylinder with the intervening mirror, you are still compressing the photon gas behind the mirror. In step 3 you'll find that it is in fact even harder to compress the cylinder than it was for it to expand since you're compressing a smaller volume...(although it isn't quite the same as compressing a gas) the temperature of the upper half will rise above the initial temperature. This means you will, through a cycle, be doing net work on the system. At step 4 you will have the higher temperature photon gas flooding into the lower half of the cylinder warming up your black body radiator.
     
  7. Jun 16, 2011 #6
    jambaugh -

    I didn't realize that by covering the emitter we were trapping the photon gas in the cylinder. I can see now that compressing the piston will require more work than the amount that would be extracted from the expansion, so the temperature would rise. Thanks for your comment.
     
  8. Jun 16, 2011 #7

    Drakkith

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    I'm still confused on this temperature thing involving the piston.
     
  9. Jun 16, 2011 #8

    jambaugh

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    The thermal energy of the walls, piston, and especially the black body emitter of this example pumps energy into the interior electro-magnetic field (equiv photon gas). Even with no air the "empty" cylinder has a temperature and energy and entropy in the form of a photon gas.

    If the walls and piston are 100% reflective (0 emissivity) then as the piston reduces the volume of the cylinder it will (a) do work on this photon gas since it has a pressure, and (b.) increase the energy density since at 0 emissivity no heat transfer can occur between photon gas and walls of cylinder.

    It is quite similar to heating a conventional gas by compressing it.

    But in a typical situation, even if you have 99.999% reflective surfaces the interior photon gas will couple to the walls quickly enough to be in equilibrium with the temperature of the walls.

    Now if piston, walls, and BB emitter are not all at the same temperature one will have a flux between hotter and cooler (just as with e.g. our sun and our planet and the vast icy void of dark intergalactic space.)
     
  10. Jun 16, 2011 #9

    Drakkith

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    Ah ok, thanks.
     
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