This must be an absolute naive question, but I’m going to ask it anyway. Aside from the obvious thermodynamic law violation, if I short cut a resistor’s terminals, why doesn’t thermal Nyquist noise in the resistor produce heat and raise its temperature until it burns up?
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\$\begingroup\$ It is going out as fast as it is going in. But is you did connect a resistor at room temperature to a resistor at -200C, the latter would be slowly heated up by the thermal noise power from the former. \$\endgroup\$– Henry CrunAug 16, 2018 at 2:09
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Because of the mechanism of noise production, shorting the resistor (or leaving it open) does not heat the resistor. This is analogous to asking why reflecting the black-body radiation back to a hot sphere does not cause it to heat up and melt.
Connecting it to a noise-free resistive load draws power from the resistor and cools it (optimally transfer is between equal value resistors). If the load is resistive and exhibits Johnson-Nyquist noise the two resistors will equalize in temperature. This is like black body radiation equalizing the temperature between two surfaces.
The roots of this behavior are deep and you can refer to the fluctuation dissipation theorem for more information. For example, a damped mechanical system will exhibit the same kind of temperature-dependent noise.
answered Aug 16, 2018 at 4:01