# Optics thought experiment and 2nd law of thermodynamics

1. Aug 12, 2010

### theDoc

Imagine a closed isolated system consisting of 2 rooms separated by a window. The window is built with one layer of material that has refractive index 1.5 and one layer of material that has refractive index 2, for a wide range of wavelenghts (say 1 $$\mu$$ m - 50 $$\mu$$ m), and low absorption. The side with lower refractive index is in room A, the side with higher refractive index is in room B. Alice is standing in room A and Bob is standing in room B.
Alice and Bob radiate electromagnetic waves with power $$\sigma A T^4$$ (Stefan–Boltzmann law) with wavelengths almost entirely in the range 1 $$\mu$$ m - 50 $$\mu$$ m (Planck's law). An easy calculation using Fresnel equations shows that more radiation is transmitted through the window from room A to room B than from room B to room A. So Alice should get colder and Bob warmer. Where is the mistake?

2. Aug 12, 2010

### Andy Resnick

3. Aug 12, 2010

### theDoc

Andy: It's been calculated by Lord Rayleigh in 1880
http://www.nature.com/nphoton/journal/v1/n3/full/nphoton.2007.26.html
and the calculation is really straightforward. The formulas are in http://en.wikipedia.org/wiki/Fresnel_equations
It is well known that a gradient of increasing refractive indices lets more light through than
going directly to a high refractive index.
One-way mirrors have no built-in asymmetry, and that is the reason why they are not real one-way mirrors.
Real one-way mirrors might be impossible to build because they would contradict the 2nd law of thermodynamics (which has never been done). For the same reason: either 1) my experiment doesn't work or 2) the 2nd law of thermodynamics is not always true.
I was hoping for an answer other than it's number 1 because it's not number 2.

Last edited by a moderator: May 4, 2017
4. Aug 12, 2010

Forgive me, Im a little slow...

How does this violate the 2nd law?

5. Aug 12, 2010

### mgb_phys

Because it generates order - and you could put a stirling engine between the two rooms and generate free energy for ever.

6. Aug 12, 2010

Free energy? It requires the Alice and Bob to radiate though. I could use that radiation directly and do work.

Also, it generates order? So the entropy of the two rooms, alice and bob after radiating is lower than the entropy of the two rooms, alice and bob before radiating? Intuitively I would think the opposite...

Last edited: Aug 12, 2010
7. Aug 12, 2010

### Mapes

For those who are shaky on what's being described: Fresnel's Equations describe the transmission (or reflection) of EM radiation between materials of different refractive indices, as a function of angle and polarization. For example, the transmission of a beam perpendicular to the interface is predicted to be

$$\frac{2n_1}{n_1+n_2}$$

when moving from material 1 to 2 and

$$\frac{2n_2}{n_1+n_2}$$

when moving from material 2 to 1. I believe theDoc is asking essentially why a temperature difference wouldn't develop between two transparent materials (with difference refractive indices) that were originally at the same temperature. The difference would be sustained by this imbalance in energy transfer.

I am also interested in hearing the resolution. Perhaps a junction is created, analogous to the PN junction in electronics, where thermal conduction balances the asymmetry? Or perhaps one of the simplifications involved in deriving Fresnel's Laws is leading to an apparent paradox? I don't know.

8. Aug 12, 2010

### Andy Resnick

Ah- I see.

The question can be resolved by considering, not the transmission of the field amplitude, but the transmission of *energy* (intensity or power), this is given by (for normal incidence):

T = t^2 *(n_i/n_t)

http://www.inyourfacefotos.com/fresnel.htm

The power transmission is symmetric: there is no imbalance in energy transfer.

9. Aug 12, 2010

### JaWiB

I don't follow this. T is the power transmission? And t the field? The link you posted seems to say that T = t^2, which is what I would expect...

10. Aug 12, 2010

### Andy Resnick

T is the transmission of the intensity, 't' the transmission of the field.

That's not what the link says. The same formula is written on the wiki page (halfway down) as well.

Edit: on the link I posted, the text just above the equation is correct. The equation itself appears to be misprinted.

11. Aug 12, 2010

Im still confused...

I dont see the discrepancy here. Could somebody walk me through this. So two people emit EM radiation, and one room gets hotter than the other. So what? Thats not order being created, that doesnt violate the second law. If I put two light bulbs in two rooms and turn them on, one room gets hotter than the other - but entropy has still increased.

12. Aug 12, 2010

### Andy Resnick

The OP invented a scenario where it appears that a cold room could radiate into a hot room but not vice-versa, thus making the cold room colder. This would violate the second law of thermodynamics.

I showed that the transfer of energy is not unidirectional: the cold room and hot room radiate into each other equally, which warms the cold room in accordance with the second law.

13. Aug 12, 2010

### Redbelly98

Staff Emeritus
None of these links seem to talk about the issue at hand: the net transmission of light going in one direction vs. in the reverse direction. I could only see the abstract for the Nature article, but it discusses the graded index case, i.e. a continuous change in n. But we are talking about discrete changes in the index, from 1 to 1.5 to 2 to 1.

Calculating the transmission for multiple layers is rather tricky, as you need to consider multiple reflections within the two materials even if you are neglecting interference effects. To really settle the matter, somebody will have to do the calculation for the OP's example:
n=1 | n=1.5 | n=2 | n=1​
If I can find the time this weekend, I may try to produce actual numbers for this example. But others should feel free to jump in and do it, as I am not making any promises here.

Last edited by a moderator: May 4, 2017
14. Aug 13, 2010

### cesiumfrog

According to wikipedia, the fraction of power that is transmitted doesn't depend on which side the radiation is coming from:
$$1-\left(\frac{n_1-n_2}{n_1+n_2}\right) ^2$$
(Would the OP detail exactly how they obtain equations which say otherwise? Strictly we should be considering extra details of the situation, like integrating over all directions and polarisations, and accounting for the temperature of the window, but I won't bother before I see that there's any paradox to resolve in the first place.. I see now, this is also what Andy was referring to.)

Last edited: Aug 13, 2010
15. Aug 13, 2010

### Mapes

Makes sense, thanks!

16. Aug 14, 2010

### theDoc

you guys are right, my mistake: the transmission is symmetric.
It's not so simple to build something with asymmetric transmission, but maybe possible:
http://arxiv.org/abs/1005.1970