Anonymous asked in Science & MathematicsPhysics · 1 decade ago

How can we defy the second law of thermodynamics?

A property of ellipsoids is that if the inside surface is a perfect reflector, a ray of light originating at one focus will reflect off of the ellipsoid and pass through the other focus.

In the figure above FC is a portion of ellipse with foci A and B. DE is a portion of a larger ellipse also having foci A and B. Arcs CD and EF are part of the circle with center B. Points A, C, and D lie on the same line and A, E, and F are also collinear.

Think of a perfect reflector made by revolving this shape about line AB to form an axisymmetric shape. Any photon originating at A must strike one of the ellipsoid surfaces and will be reflected toward B. For photons orginating at B, some strike the ellipsoids and reflect toward A, but some also strike the spherical portion of the reflector and come right back to B.

Take two small identical black body objects of the same temperature and place one at A and the other at B inside the reflector. Also the rest of the space inside the reflector is vacuum. Objects emit thermal radiation based on temperature in the form of photons. All the radiation from A strikes B. Only a fraction of that emitted from B arrives at A. So with time B will become hotter than A. In theory, this temperature difference can be used to generate power.

3 Answers

  • 1 decade ago
    Favorite Answer

    Maxwell's demon at work.

    As an aside, Scythian and I have previously had e-mail exchanges about vacuums and thermo which is very apropos to this matter. Very interesting stuff.

    I had to be absolutely sure that the geometry was right. I imagined one photon originating at A with points A and B being perfect, diffuse reflectors. Our photon would hit B about 12 % more often than A. So, indeed, it would defy the second law.

    But the escape has to be that there is no such thing as a perfect reflector. Disorder happens. Light diverges. There is also nothing with an albedo of 1 throughout the entire EM spectrum. I suspect that you would need a container at absolute zero to get the perfect reflections which means that you would need to defy the second law just to make your reflector. lol.


    The answer to your problem is:

    1. There is no such thing as a point source. Everything (except maybe a fundamental particle) has a radius. This means that a portion of the EM will not go from A to B or B to B, or B to A, but instead be bounced around your urn. This is because it is only from the foci that you can guarantee that you will bounce to a foci. Anytime that you have a diffuse ray from a point off of the foci, it can bounce around for a long time.

    2. A big wave can go around a small object without being absorbed. The bigger radiowave type EM can pass around a foci. (A beam could go from A, flow around B, then be reabsorbed by A).

    3. The uncertainty principle applies to light. No matter what you do, a portion of the light will miss the target. This beam will end up bouncing around the container.

    The bottom line is that most of the EM will end up bouncing around the container eventually running into either A or B. I realize that is more ad hoc than a complete solution, but the temperature difference between A and B will be less than the geometry would predict.

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  • 1 decade ago

    You can generalize this even for gases instead of radiation, if we could control the scattering function of the surfaces that the gases come into contact with. This would be an example of how the 2nd Law of Thermodynamics can break down at nanoscales. Ordinarily, a surface will scatter incoming molecules isotropically, which preserves the law and maintains a pressure equilibrium. But with the right surface characteristics and chamber geometry (at nanoscales), one can actually create a pressure differential and hence gas flow.

    Modelling statistical behavior of gases as radiation is an approximate way to analyze this problem. Imagine if the walls of your chamber of ellipsoids was not mirror reflective, but scattered light uniformly like white paper, the "radiative lfux differences" insdie the chamber that you speak of would vanish. This would be like how gases ordinarily behave in a typical (very small) chamber, upholding the 2nd law.

    There's a classic problem where it asks if there could be an enclosed chamber with mirror walls where light from a source inside would fail to reach some places inside. This has been solved, proving that there can exist such examples.

    The reason why the example you've given won't work is because that for every ray reaching from A to B, there's a counter ray. Nevertheless, it's not an useless exercise to think about more general versions of this.

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  • Pneuma
    Lv 5
    1 decade ago

    I was rethinking the problem and I came up with this new interpretation.

    Black bodies are not geometrical points. Even the black body cavity is not a perfect black body but just the best approximation, and it radiates only through the cavity hole.

    So our black bodies are spatial entities. Let S be the surface of A and S' the surface of B. Now

    a) S1 is the part of S which radiation is reflected by the FC ellipse and received by the S'1 portion of the surface S' of B.

    b) S2 is the part of S which radiation is reflected by the DE ellipse and received by the S'2 portion of the surface S' of B.

    A and B being identical, S'1< S1 and S'2 > S2 (we can suppose them to be spherical for the shake of simplicity).

    Now we can think object A as composed by two surfaces S1 and S2, and object B as composed by two surfaces S'1 and S'2 (we can ignore S'3, the one that strikes the spherical portions of the reflector). In 1859 Gustav Kirchoff formulated its law of thermal radiation, that states that at thermal equilibrium a body radiates as much energy as it receives, with independence of its nature and surface. So S'1 radiates back as much energy as it receives from S1, and it goes back to S1 of A, and the same happens for S'2/S2.

    If we wonder why is it possible that at a same temperature T different surfaces radiate same energy quantity, we can think the thermal equilibrium as being rather dynamical than static. As soon as temperature raises by a minimum quantity, the radiation augments dramatically, for it depends on the 4th power of T (Stefan-Boltzmann's law).

    So both bodies will keep its original temperature in a thermal equilibrium, according with the second law of thermodynamics. One of the consequences of this law is that bodies in an isolated system will finish with the same temperature (maximum entropy).


    Here down is my old interpretation. Now I think it is wrong.

    Right, but just for a while.

    Suppose the fraction of the radiation from B to A is f, with f<1, a constant depending only on the geometry of the frame. Black body A at T temperature emits R(T) energy per unit of time, and all that energy reaches B, being absorbed by it due to its black body nature. On the other hand, black body B at temperature T' (initially T=T') emits R(T') energy/time unit, but only R(T') * f reaches A. So, as soon as we have

    R(T) = R(T') * f

    the process will reach an state of equilibrium, so no "Perpetuum Mobile" will be possible. Of course T' will be greater than T, but both of them will remain constant, with the entropy having reached its maximum value.

    As the emission-radiation processes are not symmetrical we end up with a non symmetrical status.

    A black body is a thermal source of radiation (Planck's law). If you do not give back to it the same energy that it emits, then it will loss energy content. If we use some of the energy content of B to generate power, a new state of equilibrium will be obtained, with colder temperatures and less total energy content. If we insist, we will exhaust the energy content of both A and B bodies, but I do not see a contradiction to the second law.

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