# total noise power of a resistor (all frequencies)

Let's calculate the power generated by Johnson-Nyquist noise (and then immediately dissipated as heat) in a short-circuited resistor. I mean the total power at all frequencies, zero to infinity...

$$(\text{Noise power at frequency }f) = \frac{V_{rms}^2}{R} = \frac{4hf}{e^{hf/k_BT}-1}df$$ $$(\text{Total noise power}) = \int_0^\infty \frac{4hf}{e^{hf/k_BT}-1}df$$ $$=\frac{4(k_BT)^2}{h}\int_0^\infty \frac{\frac{hf}{k_BT}}{e^{hf/k_BT}-1}d(\frac{hf}{k_BT})$$ $$=\frac{4(k_BT)^2}{h}\int_0^\infty \frac{x}{e^x-1}dx=\frac{4(k_BT)^2}{h}\frac{\pi^2}{6}$$ $$=\frac{\pi k_B^2}{3\hbar}T^2$$ i.e. temperature squared times a certain constant, 1.893E-12 W/K2.

Is there a name for this constant? Or any literature discussing its significance or meaning? Is there any intuitive way to understand why total blackbody radiation goes as temperature to the fourth power, but total Johnson noise goes only as temperature squared?

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The thermal radiation from the resistor will be the same whether shorted or not, right? It will only depend on temperature? So does shorting it into a loop actually change anything? (Wolfram Alpha says it's 4 * Stefan-Boltzmann constant in 1 dimension) – endolith May 14 '14 at 23:24
@endolith -- Yes, I just said it was shorted because I wanted my question to be very concrete and specific. If you have a transmission line, it has a series of modes (standing waves), and in thermal equilibrium each mode has kT of energy (or less at high frequency). These modes exchange energy with a resistor: They give energy via joule heating, and get energy via johnson noise. This quantity 1.893E-12W/K2 is related to how fast the energy is exchanging. But, depending on what exactly you're calculating, you may need to take into account impedance matching etc. – Steve B May 28 '14 at 17:11