# 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?

• 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? Commented May 14, 2014 at 17:46
• Blackbody radiation and Johnson noise are both related to the fluctuation-dissipation theorem. Commented Aug 29, 2018 at 18:55

Blackbody radiation is an idealized description of thermal radiation of a substance which is in thermal equilibrium with the photon field. Your description in the question, which equates thermal radiation and blackbody radiation is therefore not quite accurate. Indeed, blackbody radiation is quite simple compared to thermal radiation in general -- blackbody radiation doesn't depend at all on material properties, all it depends on is the temperature and fundamental constants of the electromagnetic field. Why is this? Philosophically, when two things are in thermal equilibrium, we can "set an equals sign" between many of their properties. Therefore, when an object is in thermal equilibrium with the field of photons, to understand the radiation it emits, we only need properties arising from the statistical mechanics and spectrum of photons in three dimensions.

Johnson noise also is independent of the properties of the material, and as it also has to do with electromagnetic fields, one might expect it to be related to blackbody radiation inside the conductor. This is indeed the case, but now we must concern ourselves with photons in one dimension, since the typical setting for Johnson noise is in a wire, rather than free space! This explains the difference in the formulas. Derivations of Johnson noise in this context can be found here and here (just from a google search of "Johnson noise" and "blackbody").

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 titled "The Measurement of Thermal Radiation at Microwave Frequencies". 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.

• I second j.c.'s reference to Dicke's justifiably classic paper. It's highly readable. Also let me put in a plug for Harry Nyquist's 1928 Phys Rev paper "thermal agitation of electric charge in conductors." Google the complete title and you'll get a PDF. In my view this paper ranks up there with Carnot's on heat engines in that by considering a very specific (and simple) model, Nyquist is able to derive a deep, universal result which also explains Johnson's noise measurement. After Einstein's Brownian motion paper this is the second (to my knowledge) fluctuation-dissipation theorem
– user27777
Commented Aug 13, 2013 at 19:10
• Unfortunately, Dicke's paper link is dead. I assume the paper was "Atmospheric Absorption Measurements with a Microwave Radiometer" for anyone interested. Commented Aug 25, 2018 at 12:43
• @user54826 Thanks for letting me know. However, the paper was a different one. I've edited my answer to link to an internet archive of the scan I originally referenced, but here's the DOI link doi.org/10.1063/1.1770483
– j.c.
Commented Aug 29, 2018 at 18:52