The thermal noise density can be written as:

$$nd_V=\sqrt{k_BTR}$$ or $$nd_I=\sqrt{\frac{k_BT}{R}}$$ The units begin V/sqrt(Hz) or A/sqrt(Hz). For the second expression, does this imply infinite current noise density for an ideal wire? This seems strange! I understand that the final noise power does not depend on resistance, but still infinite noise density seems absurd.

  • 1
    \$\begingroup\$ I think this might get better answers on the physics stack exchange, but don't have the rep to vote to migrate it. \$\endgroup\$ – Jack B Aug 6 '17 at 18:37
  • 1
    \$\begingroup\$ I have edited your question to fix the second equation. Presumably that was just a typo, but in case it wasn't, you get the noise current by dividing the noise voltage by R, and then you're left with a root R, not an R, on the bottom. \$\endgroup\$ – Jack B Aug 6 '17 at 18:49

This looks a bit ugly, but maybe if we think a bit more about what a zero resistance wire is, we can work out why we won't get something physically unrealistic.


One way to get a zero resistance would be to use superconductors. These are very weird materials - they have huge quantum effects, but the Johnson-Nyquist noise theory you're using in your question is semi-classical, so we might reasonably expect it to not work when lots of quantum things are going on.

In fact, in a superconductor, there are two conducting 'fluids' sharing the same space. One, the normal fluid, is made of electrons, and acts like a electrons in a normal material. This will have thermal fluctuations just like the ones which cause Johnson Nyquist noise. The other, called the superfluid, is made of cooper pairs, and has zero resistance. So it will short out any external current or voltage (which is what makes superconductors into perfect conductors). But it will also short out the noise voltage from the normal fluid. Every thermal fluctuation in the material will be immediately and completely cancelled by a movement in the superfluid, so there will be no Johnson-Nyquist noise. There may well be other noise, but that's a whole other topic.

Not superconductors

That leaves us making a zero resistance wire from normal materials, which is of course impossible. So the problem isn't that the current is infinite, it's that it tends to infinity as we reduce the resistance. To see if that makes sense, we have to think about what reducing the resistance to zero really means.

The resistance of a block of material is a material-dependent constant times the length divided by the cross sectional area. The two ways of getting zero resistance are then:

  1. To increase the area to infinity. Having infinite noise current in an infinite area seems reasonable, the current density is the same as it would be for a finite block of material.

  2. To reduce the length to zero. This one is a bit trickier, and I'm not sure my solution is correct. But I think this boils down to a geometry thing. If the loop circumference tends to zero, then the thickness of the wire must also tend to zero, or it isn't a loop of wire anymore. This means there is a minimum resistance, where you can reasonably apply Johnson-Nyquist theorem. Beyond that you have a plate of copper with a hole in, and you'd have to analyse that differently. There is a whole sub-field of physics called fluctuational electrodynamics and you'd probably find the detailed answer somewhere in there.


This seems strange! I understand that the final noise power does not depend on resistance, but still infinite noise density seems absurd.

No, it's neither strange nor absurd because you are dividing 0 by 0:

You get power from current by squaring and multiplying by R, so you get one R in the numerator and one in the denominator and both cancel out:

\$\require{cancel}P = \Delta f(nd_I)^2R = \Delta f \sqrt{\frac{k_B T}{R}}^2 R = \Delta f\frac{k_B T}{\cancel{R}} \cancel{R} = \Delta fk_BT\$
wich is independent of R.

So even if current noise density is infinite, noise power is not.


still infinite noise density seems absurd.

You're assuming \$R=0\$, which is just as absurd. But yeah, if you have the slightest voltage in a system without resistance, you get infinite current. Ohm.

However, the thermal noise formula is actually derived through the voltage case (ie. you get energy level fluctuation of charges (electrons), and these are observable as voltage fluctuation). So, in a superconductor, that way of looking at thermal noise breaks down.

  • 1
    \$\begingroup\$ $$R=0$$ is possible in superconductors so it might be less absurd, I would argue. \$\endgroup\$ – student1 Aug 6 '17 at 17:46
  • \$\begingroup\$ Both very short wire loops and superconductors exist, but without huge or infinite noise currents respectively so there might be something deeper going on. \$\endgroup\$ – Jack B Aug 6 '17 at 17:46
  • \$\begingroup\$ @JackB well, the truth is that above formula isn't true to the phenomenon that causes thermal noise in superconductors \$\endgroup\$ – Marcus Müller Aug 6 '17 at 17:49
  • 1
    \$\begingroup\$ @JackB :D trick answer to a trick question: the power of thermal noise voltage times thermal noise current will obviously be converted to heat. Does that mean that any washer will heat up, because it's always subject to thermal noise for any non-zero temperature? (if it does, can we both apply for a patent?) \$\endgroup\$ – Marcus Müller Aug 6 '17 at 18:02
  • 4
    \$\begingroup\$ For a M1 copper washer, it'a about 6nA/sqrt(Hz). And the noise current can't dissipate heat because it is the heat. So no perpetual motion machines today :-p. I'm still not sure how to resolve the divergence though, but I might have a stab at it. \$\endgroup\$ – Jack B Aug 6 '17 at 19:06

It does seem strange. Indeed it's wrong! Jack B made the crucial point: Johnson-Nyqvist noise is a semiclassical model, i.e. it's a simplified approximation that works well in the limit of large-scale systems (i.e., anything more than a couple hundred of atoms) at high temperature (which in solid-state physics means roughly, not cooled with liquid helium). It is at these conditions that the measurable behaviour “looks classical”, because thermal fluctuations destroy the phase coherence that would be necessary for the macroscopic quantum phenomena like superconductivity or quantum hall effect to show up. It so happens that in electronics, we basically always work in this classical regime for obvious practical reasons.

But the same thermal fluctuations (phonon collisions) inevitably also cause some nonzero resistivity. So you can only take the limit \$R=\tfrac{\rho\cdot\ell}{A}\to 0\$ by either making the cross-section \$A\$ infinitely large (in which case, as Jack said, it's completely reasonable the the currents become infinite too, as does the mass and everything else) or by reducing the length \$\ell\$ to virtually-zero, in which case you don't have the large-scale system that's necessary for the semiclassical description.

Read up on the ultraviolet catastrophe, which is a mostly analogous paradox in terms of radiation energy and was in fact one of the incentives for developing the theory quantum mechanics in the first place, seeing as classical physics evidently gave bogus results.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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