# What’s the relationship between thermal radiation and Johnson thermal noise?

All objects above absolute zero emit radiation due to random collisions between the atoms they are made of. The spectrum of radiation emitted varies according to the temperature of the object, I believe because of the random speeds at which atoms hit each other, producing photons of random energies/wavelengths that follow a probability distribution. The energy and frequencies increase as the temperature increases. This is called thermal radiation, related to black-body radiation.

All electrical components create small noise currents due to the random thermal motion of electrons they are made of. The current/voltage level increases as the temperature increases, with a constant white spectrum. This is called thermal noise or Johnson noise.

What's the relationship between these? Are they created by the same process? Why are their spectra different?

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If you short-circuit a resistor, the power dissipated in itself by its own thermal noise is ${v_n}^2/r = 4 k_\mathrm{B} T \Delta f$, which depends only on temperature and measurement bandwidth? But the amount of radiation emitted from the resistor due to its temperature will be the same whether shorted or open? – endolith May 14 '14 at 17:46

You can read a discussion of how to think about Johnson noise in terms of blackbody radiation in this classic 1946 paper by Robert Dicke. The physical point made there is that an antenna receiving blackbody radiation at temperature $T$ and a resistor at temperature $T$ experiencing Johnson noise must have equal power. The difference in the forms of the power spectra is apparently due to the frequency dependence of the antenna's detection pattern.