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Can i exploit johnson noise to violate the second law?

  1. Jul 5, 2007 #1
    Thermal noise known as “Johnson noise” or “Nyquist noise” is well known and characterized. This noise is a result of thermal agitation of charge carriers inside a conductor. Its power spectral density is given by:

    v^2=4kTR


    where kB is Boltzmann's constant in joules per kelvin, T is the resistor's absolute temperature in kelvins, and R is the resistor value in ohms.

    Imagine a resistor in an adiabatic enclosure (one incapable of heat transfer) connected to a lossless diode. The wires leave the enclosure through sealed ports to power a circuit. Shouldn’t the rectification of this random fluctuation in the voltage across the resistor create a net positive voltage and hence allow us to violate the second law of thermodynamics (creating free energy)?

    I once asked a physics professor this question and he excitedly gave me a rushed explanation of why this wouldn’t work. It involved Brownian motion and he quickly digressed into a tangent about proving the continuity between classical (Rayleigh Jeans) and quantum blackbody theory. It involved some esoteric thought experiment where he defiled my adiabatic enclosure with a big parabolic dish. I wasn’t impressed.

    edit: I just realised the parallels between this hypothetical and Maxwell's demon, which seems to be still up for debate.
     
    Last edited: Jul 5, 2007
  2. jcsd
  3. Jul 5, 2007 #2

    Cthugha

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    Science Advisor

    It is not that easy. What you present here is a version of Smoluchowski's Trapdoor, I think.

    The problem is your diode. A diode is not a perfect binary switch. It is subject to thermal noise as well. Therefore you will only notice some net electron movement, if there is a temperature gradient between the diode and your conductor. This thing will work similar to a thermocouple.
    Have a look at the Seebeck-effect. It is very similar.
     
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