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CP-volation -> violation of 2nd law of thermodynamics?

  1. Apr 2, 2010 #1
    The amount of black body radiation is proportional to the albedo. Microscopically, this is a result of the fact that the probability of process ‘wall absorbs a photon’ is equal to the probability of the reverse process: ‘wall emits a photon’

    Lets assume that it is not true, and probability of absorption is greater than the probability of emission for some material. We make a sphere and make one segment of it from that weird material. We heat the sphere, and when it is filled with Black Body radiation, the wall made of such material overheats, and thermal equilibrium is broken.

    However, we know (from the decay of neutral kaons) that CP symmetry is broken. Hence, based on CPT theorem, T-symmetry is broken. Hence, there are processes, where the probability of the process and the reverse process are (slightly) different. Hence, there is a loophole in 2nd law?
     
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  3. Apr 3, 2010 #2
    CP symmetry is broken in the weak force yes, so if you consider "heat" for weak processes then there would be no 2nd law of thermodynamics for those reactions...
     
  4. Apr 4, 2010 #3
    Hm, I was actually expecting someone to tell me where I am wrong, not to agree with me :)
    So, if 2nd law is violated in some processes, then it is simply violated.

    Unless it is possible to prove, that the increase of entropy in "normal" processes surrounding "exotic" processes always masks the effect.
     
  5. Apr 4, 2010 #4
    I have a feeling that CPT invariance is enough to protect such a case from violating the second law of thermodynamics. But, I don't have an argument, offhand, to demonstrate this.
     
  6. Apr 4, 2010 #5

    Haelfix

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    You can find it in Weinberg vol 1 page 51, where he proves the H theorem. CPT invariance and unitarity constraints suffice.
     
  7. Apr 5, 2010 #6
    For the global wavefuncton of the Universe - yes, I also have that feeling
    However, it is not automatically valid in any given basic (branch in MWI). For example, it might be not valid in our (branch) world, where CP symmetry is broken is a specific way and there is more matter than antimatter
     
  8. Apr 5, 2010 #7

    Demystifier

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    I don't get it. I've heard arguments that T-violation might explain the 2nd law (which are not very convincing arguments, but do make some sense), but I've never heard an argument for the opposite - that T-violation might be in contradiction with the 2nd law. I mean, the 2nd law itself is a kind of violation of T-symmetry on the macroscopic level, so how can it be in contradiction with particle-physics T-violation at the microscopic level?
     
  9. Apr 5, 2010 #8
    Demystifier, I tried to explain it it the very first post.
    So I was not clear, but could you reread it - what step exactly is not clear?

    Do you agree that if you have some weird material (say, unobtanium :) ) which has broken symmetry between photon absobtion and emission, then making a perpetual mobile of the second kind using that material is trivial?
     
  10. Apr 5, 2010 #9
    There are systems which allow transmission only in one direction:
    http://en.wikipedia.org/wiki/Optical_isolator
    So it is not that simple.

    Plus I have seen sunglasses with logos made on the the glass such that you can see the logo from the outside, but not when wearing them: Using carefully chosen thin films, they can make the transmission different in the two directions.

    Why don't these violate the second law of thermodynamics? I've actually never really understood. It seems like if the transmission in each direction is different, then you could put the device in an equilibrium thermal bath and create a thermal gradient ... and then get energy from that.

    There is of course solutions for why this doesn't violate the 2nd law (if someone has very simple reasons, I'd love to hear them). The point for this discussion is just that violating the 2nd law seems to require more.
     
  11. Apr 5, 2010 #10

    Demystifier

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    I don't agree. Instead of argument (which I will present in the next post as a reply to Haelfix), here let me write a famous quote by Sir Arthur Stanley Eddington:
    "If someone points out to you that your pet theory of the universe is in disagreement with Maxwell's equations—then so much the worse for Maxwell's equations. If it is found to be contradicted by observation—well these experimentalists do bungle things sometimes. But if your theory is found to be against the second law of thermodynamics I can give you no hope; there is nothing for it but to collapse in deepest humiliation."
     
  12. Apr 5, 2010 #11
    Thank you for the poetic argument :)
    But what step in my logic is flawed?
     
  13. Apr 5, 2010 #12

    Demystifier

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    Any proof of the H-theorem contains an assumption that breaks T-symmetry by hand. The Weinberg's proof is not an exception. In the Weinberg case it is Eq. (3.6.19) at page 150 (the proof ends at page 151, not 51 as you said) that explicitly breaks T-symmetry without really explaining it. Indeed, logically, on the left hand side of (3.6.19) t could be replaced by some other evolution parameter, such as -t, or perhaps something that depends on t in a much more complicated way, or does not depend on t at all. The choice of t in (3.6.19) corresponds to our intuitive understanding of the concept of the "flow of time", rather than to something that can be derived mathematically.

    Yet, as a reply to Dmitry, the Weinberg's proof shows that the dynamical violation of T-symmetry, such as that occuring in QFT due to CP violation, cannot affect the validity of the H-theorem, and thus of the 2nd law. Essentially, the 2nd law is the principle that the system always evolves to a more probable state (which is valid only statistically for sufficiently large ensembles), and it does not depend on details of the microscopic laws that describe fundamental forces.
     
  14. Apr 5, 2010 #13

    Demystifier

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    See #12.
     
  15. Apr 5, 2010 #14

    Demystifier

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    The point is that is that the entropy of the WHOLE system (including the environment and everything) increases, even if the entropy of a subsystem decreases. Is that simple enough?

    The best example is a fridge. Locally, it decreases the amount or thermal energy, but globally, it increases it. You cannot lower the temperature in your kitchen by opening the door of your fridge (before turning the fridge on).
     
    Last edited: Apr 5, 2010
  16. Apr 5, 2010 #15
    Yes, I am not that stupid.
    Of course, we know that perpetuum mobile does not work. But there is also some value in the explanation why that particulare model of PM does not work.

    I am interested not in general proof (which is believe is valid) but what is wrong with my toy example. So I have unobtanium from Pandora, which absorbs more photons then it emits. So it is always hotter than the environment. Even my unobtanium is not stable (K0-mesons) and I use pions instead of photons, still the question is valid.
     
  17. Apr 5, 2010 #16
    Exactly. If the heat exhaust were directed outside the room, it would be a different story. But that's not the case will residential frigs, of course.
    And if the frigs' compressor apparatus were placed INSIDE the frig, not only would there be no net-cooling, the frig would get hot!
     
  18. Apr 5, 2010 #17
    All you did is restate the second law. So your answer to my question of "if someone has very simple reasons, I'd love to hear them", is you saying "well, the second law is correct, is that good enough?". I'm not denying the second law is correct. But I want to understand how it is satisfied in such cases.

    It looks like I can place that apparatus into an equilibrium heat bath and create a thermal gradient. Once we have a thermal gradient we can of course use that to convert some heat energy to useful power. If we can indeed use this device to create a thermal gradient, then this will never end and we have perpetual motion. So there should be some reason this DOESN'T allow us to create a thermal gradient. Your answer does not address this.

    Basically:
    Imagine I can make a material in which the transmission coefficient is different in different directions. What conditions must be satisfied so that this does not violate the second law of thermodynamics? There clearly exist such material, and the second law is correct, so how do we reconcile this?
     
  19. Apr 6, 2010 #18

    Demystifier

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    I think the point is that one should distinguish statistical entropy from thermodynamic entropy. In "normal" systems these are the same, but your example above is not normal. The H-theorem is a theorem on statistical entropy, not necessarily on thermodynamic entropy. So in your example the system will move towards the statistical equilibrium, but not towards the thermodynamic equilibrium.
     
  20. Apr 6, 2010 #19

    Demystifier

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    I believe my post #18 answers this as well. Of course, in a finite system, such an extraction of work cannot last forever, so in this sense there is no perpetual motion.
     
    Last edited: Apr 6, 2010
  21. Apr 6, 2010 #20

    Demystifier

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    Another (clearer) example of violation of naive 2nd law occurs when one replaces thermodynamic equilibrium with uniformity equilibrium. Intuitively, the entropy is the largest when the space distribution of matter is uniform (homogeneous). Thus, naively, the system should evolve towards a uniform distribution of matter. Yet, gravity (or any other attractive force) violates this because it causes tendency for matter to form localized lumps of matter. Still, the statistical entropy in such systems increases. The lumped non-uniform distributions of matter are more probable then the uniform ones.
     
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