# Apparent 2nd law violation? (thermodynamics)

1. Jul 25, 2010

### Curl

This might be easy but apparently I haven't thought about long enough:

Suppose a collection of particles with significant thermal energy are vibrating such that they emit electromagnetic radiation. Because of this, they cool down. Cooling means that their entropy decreases (classical entropy). But, you might say, the photons are scattered randomly, which represents an increase in entropy.

I point out that this collection of particles (a metal ball i.e.) is at the focal point of a polished parabolic mirror. The E&M waves are reflected parallel, and can be collected and used to do work (create a current in a wire, charge a capacitor, whatever).

Apparently this is a heat engine with no cold reservoir. Whats the problem here? I haven't been able to figure it out.

2. Jul 25, 2010

### mikelepore

When they say that a heat engine requires a cold reservoir, they are talking about making a device operate repeatedly, performing a cycle over and over.

3. Jul 25, 2010

### Curl

My example is repeatable, the process can continue until the thermal energy of the ball is exhausted (cooled to the point where electromagnetic radiation is insignificant).

Or it can be placed in thermal contact with a reservoir of higher temperature and the process repeated.

4. Jul 25, 2010

### K^2

Curl, radiation has a temperature, related to its spectrum. It can only be used to do work if the collector for that radiation is colder than the source of the radiation. A solar battery at 300K, for example, cannot convert to energy radiation from a 200K source. Any other device you might come up with will face similar difficulties.

5. Jul 25, 2010

### AJ Bentley

Repeatable means that it returns to the same state after each cycle. Once your ball is cold, that's the end of it.
Going to a new reservoir means bringing in something new from outside - that's not what is meant by a 'cycle'. Eventually you will run out of reservoirs.

6. Jul 25, 2010

### Curl

Yes but it doesn't matter. If the wire is superconducting with ~0 resistance the "solar cell" will not increase in temperature, therefore the cold reservoir is never exhausted. This still violates the 2nd law, since entropy is not increased in the cold reservoir but entropy is decreased in the hot reservoir.

7. Jul 25, 2010

### Mapes

Even at perfect efficiency (superconducting wire, the whole works), the solar cell will heat up when it absorbs incoming photons (and the entropy increase will equal, at the very least, the entropy carried by the radiation). There's no Second Law violation here.

8. Jul 25, 2010

### Curl

Okay explain how it will "warm up" the wire. The waves coming by the wire will create an electric field in the wire which gives rise to a current. The potential difference between ends of wires can draw charge from a capacitor, storing useful energy. The only way any of this will rise in temperature is if there is resistance in the wire or in the capacitor plates, so I don't see what you're saying.

9. Jul 25, 2010

### Mapes

If your wire is absorbing radiative energy, its entropy will increase unless this entropy is transferred away by the refrigeration unit keeping the wire at constant temperature. It has nothing to do with Joule, or resistive, heating.

10. Jul 25, 2010

### Curl

???

the wire is not absorbing energy, the capacitor is. All the wire is doing is conducting electrical charges from one capacitor plate to another, with the help of the electric field supplied by the bypassing EM waves.

11. Jul 25, 2010

### Mapes

You're just playing a word game. In your thought experiment, radiation impinges upon a wire, and some of the radiative energy is converted into electrical energy. This is universally described as absorption of radiation.

Do you know what I meant when I said that heating an object with light does not depend on the object having a non-zero resistance? You are effectively claiming that superconductors do not heat up when you shine light on them; do you have a reference for this claim?

12. Jul 25, 2010

### K^2

Yes it will. Solar cells have limited efficiency. The limits actually come from thermodynamics.

13. Jul 26, 2010

### Curl

omg why is this tough to understand?

The wire doesn't need to be touched by anything, I can make a loop and shoot the EM beam through it and induce a current. The wire will not warm up since its resistance is ~0. All energy absorbed from the light beam will be put onto the capacitor.

Doesn't even matter, there is still an apparent violation because the particles are cooling off without the expense of a cold reservoir. How is this possible? The entropy of the photons does not count because they can easily be ordered without doing work ( by a mirror, for example).

14. Jul 26, 2010

### Mapes

I ask again, could you provide a reference for your claim that shining light (i.e., any EM wave) at a superconductor does not heat it up? This forum is for discussing consensus physics. It sounds like you're less interested in finding out where your mistake is and more interested in arguing that you've found some exception to the Second Law.

15. Jul 27, 2010

### Curl

It is obvious you chose to ignore my previous post. If you are interested in super conductors read here; there are many references on that page:
http://en.wikipedia.org/wiki/Superconductivity

If you go back to read my post, I have said that it is not necessary to shine a beam AT the conductor itself, just inside a loop (can be 10 meters across) to induce an electric field within the superconductor.

But even if you aren't familiar with superconductivity and Faraday's law, then as I said above, there is still an apparent violation because you can DECREASE entropy of an object without doing work or exhausting a cold reservoir.

A parabolic mirror does no work on light as it reflects it. Its free and it causes order in photons (parallel). You cannot say the photons have carry the particles' entropy because they can be captured in a mirror box if you wll. That's where I'm having the difficulty.

It's okay if you don't know, we'll just wait for someone who can figure this out.

16. Jul 27, 2010

### Mapes

You are veering away from legitimate physics here.

17. Jul 27, 2010

### Curl

Okay then explain it, that's what I've been searching for in the past few days. All you have done is shout "physics" but failed to explain anything.

18. Jul 27, 2010

### pallidin

I'm having a hard time understanding the question(s) here.
With questions properly put forth without vagueness or unnecessary convolution, you will eventually get legitimate answers here at PF.
But that's up to you. You MUST frame your questions correctly AND be open to new understanding and corrected perspective.

19. Jul 27, 2010

### cesiumfrog

The cold reservoir is the surrounding 0K darkness into which you assume the radiation heads. The entropy and thermal energy of the surrounds increases no less than the entropy and thermal energy of your object decreases, and eventually the system will approach an equilibrium (picture the ball absorbing as much radiation, even from the cosmic background, as what it is emitting).

It might help to explain if you would explicitly specify what the parabolic mirror is pointing at? If it is a solar generator, what type? Or is the radiation simply harnessed to heat up a different object? Whatever you choose, you'll find the process only works efficiently if your original ball is much hotter than the receiving system, but as the process proceeds it approaches a point where the reverse process (your generator shining heat back at the ball) occurs at the same rate (or in other words, energy ceases to be harnassed for useful work).

20. Jul 27, 2010

### Andy Resnick

This appears to be an interesting paradox, but it is ultimately flawed. Here's why:

First, let's list your assumptions (in no particular order):

"I point out that this collection of particles (a metal ball i.e.) is at the focal point of a polished parabolic mirror. The E&M waves are reflected parallel, and can be collected and used to do work"

Assumption 1: the distribution of thermal radiant energy has been converted from isotropic to collimated.

Assumption 2: no work has been performed by the radiation during the process of reflection.

Assumption 1 is false; light emitted from an extended object cannot be perfectly collimated.

Even so, this is a weak assumption- one could make it stronger by only requiring that the entropy associated with the radiance decreases. IIRC, the more narrow the radiant intensity, the lower the entropy associated with the radiation field- but I can't quickly find a reference. Then, assumption 1.1 may be true.

Unfortunately, assumption 2 is false due to radiation pressure. The light exerts pressure on the mirror, and if it were free to move, it would get pushed backwards. So work is performed when the light is reflected- anchoring the mirror won't save you because the radiation pressure is then stored as elastic energy (elastic deformation of the support).

21. Jul 27, 2010

### cesiumfrog

Andy,

2. If the mirror is arbitrarily heavily anchored, you know no work will be performed on it. (Elastic sink is just silliness, implying that constant pressures produce constantly increasing deformation perpetually?)

1. Isn't it enough that as the collimation is improved the beamwidth grows? (A pitfall here may stem from imagining the entire finite extent of the ball object to be at the singular point of a parabolla's focus.)

Last edited: Jul 27, 2010
22. Jul 27, 2010

### Curl

1. cesiumfrog is right. Its foolish to point out the "work" done by the radiation pressure. I can have the system in a box, which cannot move by conservation of momentum.
Also energy/momentum ratio is extremely large for photons: that's why nobody uses flashlights for rocket engines.

2. Doesn't matter if the beams are "perfectly" parallel, although if we assume the radiation is coming normal to the sphere's surface, the focal point will be the center of the sphere which is a point of zero size. Therefore they WOULD mathematically be perfectly parallel by the definition of a parabola.

However, it doesn't matter if they are parallel. I can make a loop of radius 10 meters and I would still capture all the rays which would have very small deviation.

I already thought of these debunks, but none of them work. Normally I can figure these type of things out on my own, but this one is giving me trouble. I KNOW it cannot work I just don't know WHY. I need someone intelligent to notice something that I"m missing.

23. Jul 27, 2010

### jajknight

24. Jul 27, 2010

### cesiumfrog

Please, would you simply (and concisely) state the entire complete closed cycle system? What each component is? What your problem is (that is, how you expect each component to act)?

25. Jul 28, 2010

Not true.