Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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?

share|improve this question
    
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 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 at 17:11
add comment

1 Answer 1

up vote 5 down vote accepted

I think you've just derived the Stefan-Boltzman law for a one-dimensional system. The T^4 comes from three dimensions. The more dimensions the quanta can populate the higher power of T you get.

share|improve this answer
    
Thanks, that's what I was looking for. Maybe there is no universal official name for the constant, but "one-dimensional analogue of the Stefan-Boltzmann constant" is pretty good. Since the oscillations in a wire or resistor are stuck in one dimension, there are many fewer high-energy modes, so extra temperature does not as dramatically grow the total power. That makes sense. –  Steve B Aug 2 '12 at 23:08
    
This related question has a nice discussion. The answer cites a "classic" 1946 paper by Robert Dicke (free link , official link). The discrepancy between 3D mode density and 1D mode density helps explain some facts in antenna theory, particularly the fact that antennas larger than 1 square wavelength have to have a narrow acceptance angle. –  Steve B Aug 2 '12 at 23:10
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.