# Challenge: How can this meta-material be in thermal equilibrium?

The following simple ideal device comprises just two objects radiating at each other with a radiant concentrator in between. As far as I can tell, this device can't be in thermal equilibrium AND be at the same temperature. Either I am missing something, or it challenges some interpretations of the 2nd Law of Thermodynamics. Here is the description: The device comprises just two objects radiating at each other, with another device in between them. The device in between is a compound parabolic concentrator (CPC), which is a radiant concentrator - similar to a lens, but non-imaging. The CPC is filled with, or comprises a transparent material. (See the drawing above, which shows a 2D cross-section.) Let's look at the ideal case. So, assume that the materials are perfect. This means the absorptive (black) surfaces are blackbodies, and the others (apart from the CPC) are perfectly reflective - white bodies. The CPC would be 100% transmissible, with a 1.5 index of refraction. To name the parts, the top object is the emissive object 7, and the black surface on the bottom of it is the emissive surface 6. The concentrator is the CPC 5, and the absorber at the bottom is the receiver 2. The top of the CPC is the aperture 9.

Assume these objects are in a vacuum, so there is no conduction. Since the very top and bottom surfaces are 100% reflective, there would be no energy transfer with whatever enclosure surrounds it.

Rays of radiant energy emitted by the emissive surface 6 reach the aperture 9 coming at angles from a 180 degrees range with respect to a y-axis, which points downward on the page. Due to the index of refraction, the range of angles lowers to about 84 degrees (with a 42 degree half angle) upon entering the CPC. (Both glass and potassium bromide have indexes of refraction about 1.5, and that gives an approximately 42-degree critical angle, so these are reasonable numbers.)

Now there are two methods to handle the sides of the CPC. One method is that the CPC has reflective sides, and in this ideal case - 100% reflective. In this, case, there is no emittance. Also in this case, all rays that enter the CPC reach the bottom provided the half acceptance angle of the CPC is less than or equal to the critical angle, and all radiant rays coming from the emissive surface are 100% absorbed by the receiver, as is the definition of a blackbody. In this ideal case, there is more radiation that will be absorbed at the receiver than the receiver emits back at the emissive object. This is true for a range of temperature differences that end when the emissive object is somewhat cooler than bottom selective surface.

(The other method to handle the sides of the CPC is that the CPC is completely comprised of the transmissible material. This method relies on internal reflection. If it does, and all of the vectors that enter internally reflect, then the angle at the bottom of the sides of the CPC can’t exceed the critical angle for all rays to reach the bottom. I’m not going into detail here, but there can be some concentration, which again leads to a selective surface in the ideal case.)

Now let’s assume an initial condition where the temperature of the emissive object on top is the same as the selective surface on the bottom object. At this temperature, both objects emit the same radiation flux. In this case, the EMF energy radiated per area is the same. But the total area of the emitting surface of the top emissive object is much larger, so it emits more total radiation energy at the receiver than the receiver emits back at it. This is due to the CPC concentrator. In this ideal example, all the radiation emitted from the top emissive surface reaches and is absorbed by the receiver. The flux is concentrated and increased by the concentrator. The concentrator is similar to a concentrator in one direction, and a diffuser in the other.

The 2nd law is generally interpreted as stating that this system will be in thermal equilibrium if no net energy is transferred between the two objects, AND the two objects are at the same temperature (which is the initial condition here). However, for this to be true of this device, either radiant energy has to be destroyed at the receiver, as its absorption of a higher flux of radiant energy would have to match the emittance, which occurs at a lower relative flux. Or energy has to be created at the emissive surface.

In other words, there are three choices here. First, this device invalidates the first law of thermodynamics for this condition. Second, it invalidates the second law. Or third, it is merely that many have made an assumption about the second law that is not valid in all circumstances- particularly for two bodies radiating at each other with a radiant concentrator between them, which do not comprise a thermodynamic cycle.

So, what have I missed here? Or does this ideal device force a rethink of some previously held beliefs?

Regarding real world potential, it should be noted that real materials exist that have greater than 95% of the ideal radiant values used in the example. For example, Germanium with an AR coating transmits over 95%, and has a really high index of refraction. A 3-D CPC can be made with a concentration ratio of 16. For an ideal Germanium device, I calculated that it could initially move 402 Watts/m^2. This is a description of a heat pump with an undefined COP which does NOT comprise a thermodynamic cycle. And yes, I understand that it is generally accepted that this is not possible - which is why I have posed this question. Enjoy! ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯

Notes: (In response to some suggestions posted.)

The property of Etendue can be used to find the maximum concentration ratio of an optic. The formula is: C(max) = n²/sin²α, which is for a 3D optic. The symbol α represents the input half-angle, which is 90 degrees for this optic - as it accepts a full 180 degrees of vectors. Thus, the denominator equals 1 for this optic. For an n (index of refraction) of 4 (Ge) the property of Etendue thus yields a maximum concentration of 16 (4^2 - for a 3D object). This equals the maximum this CPC concentrator can deliver (3D object). Thus, etendue confirms the increase in radiant flux at the receiver, and that the concentrator described here in not an impossible device. (BTW, in the 2nd answer below, the author assumed his result to find his result. I disagree with that.)

Regarding irreversibilities, I don’t see any vectors that aren’t reversable. The non-proportionality comes from Stephan-Boltzmann that defines what the receiver can emit, whereas the receiver can absorb more radiation energy. (If it couldn’t, then radiation energy couldn’t be used to heat an object up.) As long as this device does not comprise top and bottom objects of infinite mass, it is not a perpetual motion machine.

This device in no way challenges any statements about thermodynamic cycles or heat engines, as this device does not operate on a cycle, and there is no work input or output.

• Please don't edit meta-commentary (like "This question is still unanswered") into your question - everyone can see whether you have accepted an answer or not even without that. Mar 8, 2021 at 19:14

Interesting question! I always enjoying trying to see where the second law is hiding.

Rather than considering the details of the CPCs, I'm going to steel man the concept and consider a perfect optical element which conserves optical etendue.

Energy balance from the perspective of plate 1

At best, etendue ($$\epsilon$$) of heat fluxes will be conserved between the plates,

$$\epsilon_1 = \epsilon_2 \\ A_1 n_1^2 \sin(\theta_1) = A_2 n_2^2 \sin(\theta_2)$$

Note, for the actual configuration the optical elements are not perfect, $$\epsilon_1 > \epsilon_2$$. So actual system will not reach this limit.

Top plate has subscript 1 and bottom subscript 2.

The optical energy balance between the plates is the etendue multiplied by the power emitted by the blackbody (Stefan–Boltzmann law).

For plate 1,

$$\epsilon_1 \sigma T_1^4$$

For plate 2, $$\epsilon_2 \sigma T_2^4$$

At equilibrium, the rates of energy exchanged between the plates must be equal to each other, because otherwise temperatures will carry on increasing forever — you have a perpetual motion machine.

As you already said the emission angles are $$\theta_1 = \theta_2 = \pi/2$$, which leads to the condition that $$\sin(\theta_1) = \sin(\theta_2) = 1$$.

From the perspective of plate 1 the heat is exchanged in vacuum so $$n_1 = n_2 = 1$$, and by definition this exchange occurs over the surface area of plate 1, therefore, the energy balance is,

$$0 = -A_1 \sigma T_1^4 + A_1 \sigma T_2^4$$

The first term is the energy emitted by plate 1 and the seconds term is the energy absorbed by plate 1.

$$T_1 = T_2$$

So if etendue is conserved (the ideal case) the plates will have equal temperatures.

In practice any optical element introduces loses so,

$$A_1 n_1^2 \sin(\theta_1) > A_2 n_2^2 \sin(\theta_2)$$

$$T_1 > T_2$$

Energy balance from the perspective of plate 2

Now let's apply energy balance from the perspective of plate 2.

Plate 2 absorbs the heat flux from plate 1,

$$A_2 n_2^2 \sin^2(\theta_c) \sigma T_1^4$$

The critical angles appears here for the reasons shown below: rays hitting the optical element over all external angles get transformed to rays with internal angles which cover the acceptance/escape cone. This is just the application of Snell's Law. The heat flux emitted by plate 2 which can escape and be absorbed by plate 1 must be restricted to angles inside the escape cone,

$$A_2 n_2^2 \sin^2(\theta_c) \sigma T_2^4$$

Light that is emitted outside the escape cone is totally internally reflected and reabsorbed by plate 2.

I think this is where a lot of people get struck. They don't realise that their idea has introduced an irreversibility in the direction of travel of light: there are no optical diodes.

Simplifying the above using,

$$\sin^2(\theta_c) = \frac{1}{n_2^2}$$

The energy balance from the perspective of plate 2 is, therefore,

$$0 = A_2 \sigma T_1^4 - A_2 \sigma T_2^4$$

Again, this leads to the same results as before,

$$T_1 = T_2$$

for the ideal case and

$$T_1 > T_2$$

when etendue is not conserved and losses are introduced.

• Thanks, but I am confused. The emission angle is π, and not π/2. The emission from the top plate and the receiver emit a full 180 degrees. Also, A₁ is not equal to A₂, and n₁ is not equal to n₂ unless you are using the aperture as the emissive source - which has no emissivity in this ideal example. The radiation coming through the CPC from the receiver is transmitted, so the Stephan Boltzmann equation can't be used there. Since the areas at the place where the emittance happens are unequal, the temperatures have to be different to satisfy the energy balance. Feb 22, 2021 at 4:36
• The energy balance from the perspective of plate 2 simply can't be: 0 = A₂σT⁴₁ − A₂σT⁴₂. The amount of radiant flux arriving at A₂ from the top plate can't be known just from the temperature of the top plate alone - with a concentrator in between the two plates. You have to use A₁, which is a different value from A₂. Or I suppose one could use A₂C,σT⁴₁ where "C" is the concentration factor of the CPC. Feb 23, 2021 at 0:47
• This answer show you how to apply the principle of etendue to your configuration. It's definitely correct because at equilibrium, and, without losses, both plates reach the same temperature - how could it be any other way? Feb 23, 2021 at 18:49
• So let's remove the top plate and use the sun instead. From your energy balance equation 0 = A₂σT⁴₁ − A₂σT⁴₂, T₂ then has to equal T₁, which is 5778 K. By Stephan-Boltzmann, the flux arriving at A₂ would be 63200617 W/m². Both obviously wrong. The solar flux drops by distance from the sun to 1367 W/m². Point is, you can't take the flux at the source and use it as the value at the receiver in your energy balance equation. How the temperatures of the plates are different at equlibrium in this ideal device is because the concentrator concentrates all radiation of the top plate to the receiver. Feb 23, 2021 at 21:48
• @WillTemple You actually can do that, unless energy's leaking out the sides. You can view the plates as planes, not points, so the light is emitted according to the locus of planes (planes), not the locus of points (circles). Mar 6, 2021 at 21:57

In other words, there are three choices here.

There is a fourth choice; your analysis is incorrect. We know this because it violates the Second Law, which has extraordinarily credibility—far higher than you or I.

With enough effort, one can analyze any arbitrary specific reflection–transmission arrangement in detail to identify why a perpetual-motion-type outcome is impossible. Browne addresses several such systems in Browne, "Focused radiation, the second law of thermodynamics and temperature measurements" Journal of Physics D: Applied Physics (1993), for example. Also see Yoder and Adkins, "Resolution of the ellipsoid paradox in thermodynamics", American Journal of Physics (2011).

The most common errors are to assume that a finite object is actually a geometric point, or to declare that rays will travel in one direction but not the other for some reason, or to assume a physically unrealizable object such as a perfect one-way mirror, or to state that some component will not heat up itself and emit radiation, or to rely on a real material not conducting heat, or some other idealization or wishful thinking.

This analysis is often laborious, as one can obtain from the above articles, and it's probably always easier to invoke the Second Law.

This answer will probably not satisfy you. I encourage you to read the papers I linked so that you can be familiar with their methods. Have you drawn raytracing diagrams and simulated a large number of rays? Have you eliminated all idealizations? Have you studied the application of physical laws such as conservation of etendue? Such study to try to debunk one's own unconventional scientific claims is a mark of a non-crackpot.