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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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up vote 4 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.

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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
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There is a significant typo in the original question. In the case of Fermi-Dirac statistics the constant term in the denominator is +1 rather than -1 as stated.

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But photons are bosons... –  Benjamin Hodgson Aug 20 '12 at 10:56
    
Indeed, I made a silly mistake :) Please ignore my comment. –  Chris Aug 20 '12 at 14:20
    
And why do photons considered? I don't get what particles should be considered to find voltage fluctuations, is there a quick explanation suitable for a comment? –  Yrogirg Aug 21 '12 at 7:21
    
Rough-and-ready explanation: the OP's calculation was carried out in analogy to Planck's black-body calculation and Debye's phonon calculation. It makes sense for those theories (and this one!) to use bosons because you can have lots of energy at a single frequency, which is the same as having lots of particles in a single mode. The proper proof that photons and phonons are bosons is deep. Hope that explains things a little more. –  Benjamin Hodgson Aug 22 '12 at 14:54
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