--- Robin van Spaandonk <[EMAIL PROTECTED]> wrote: > In reply to Paul's message of Mon, 13 Nov 2006 > 11:27:11 -0800 > (PST): > Hi, > [snip] > >Lets simply. Neither experiment A or B have a power > >source except thermal noise. Experiment B radiates > >more power. It is a very simple circuit. Over time, > >more energy is leaving experiment B than experiment > A. > >Therefore experiment B will be colder than experiment A. > Please forgive my ignorance, but I thought that > thermal noise occurred as a consequence of a current > passing through a resistor. A current driven by an external > voltage. I didn't think that resistors actually generated > anything by themselves. Hence, I wouldn't expect either > experiment to radiate anything at all (through the antenna). > (The only radiation I would expect would be normal > thermal radiation, from the body of the resistor, as a > consequence of their being at a specific temperature.) > > Regards, > > Robin van Spaandonk
Hi Robin, Thermal noise is caused by vibrating thermal charges in matter. It's not related to any applied current. The rms voltage of thermal noise is : Vn = (4 K T R dF) ^ 0.5 K is Boltzmann constant, 1.3806503E-023 T is temperature in Kelvin R is resistance dF is bandwidth The total noise power in a matched load circuit is Pt = 4 K T dF Power across the matched load is Pt = 2 K T dF Of course that's not much power, but size is not a factor. So the object could be a nanometer. When you consider trillions of such objects with say 1 THz bandwidth then you are in well in the kilowatt region. 2 * 1.38E-023 * 295 K * 1.00e+12 = 8.142E-009 W 1 trillion such objects = 8.14 KW With just 1 GHz bandwidth, 8.14 W Regards, Paul Lowrance ____________________________________________________________________________________ Sponsored Link Mortgage rates near 39yr lows. $420k for $1,399/mo. Calculate new payment! http://www.LowerMyBills.com/lre
