Electrons of conductors Free? [duplicate]

Are electrons of any conductor really free ? I mean are they always already moving or do they move only when electrostatic field of some sort is applied across them. I suppose if they were always moving each and every conductor must have some varying magnetic field around it at all times which though negligible musy be detected. Is that really so ? Or are the so called free electrons not really free ?

Also is there some other experiment which i do not know about by which we have already proven that they are always moving ? does it take into account the earths and other electric fields which may be causing that movement ?

Addendum : the question is not duplicate of Are free electrons in a metal really free as that post talks about ease of electrons to interact with external fields, my question is about the nature of motion of electrons in absense of external field interference as in are electrons in free motion even when they are not under the influence of any electromagnetic field.

marked as duplicate by John Rennie, Emilio Pisanty, Qmechanic♦Oct 25 '13 at 10:49

• possible duplicate of Are free electrons in a metal really free – John Rennie Oct 24 '13 at 8:46
• How do you define "free"? The second sentence seems to imply you equate free with "zero velocity". But even if you neglect all the non-electron particles in the metal completely, at finite temperature, the velocity distribution function (consider the Maxwell–Boltzmann distribution) takes finite values for any possible speed. – Nikolaj-K Oct 24 '13 at 9:07
• @John Rennie : not a duplicate, that post is about the ability of electrons to inreract freely with electromagnetic fields while I am asking about the original state of said free electrons without the influence of any external fields. – Rijul Gupta Oct 24 '13 at 9:22
• Well there appears to be a number of different schools of thought about "free" electrons. To me, a free electron is not associated with the structure of any atom or molecule. That does not mean free of motion ( in what frame of reference), nor does it mean free of electric or magnetic fields. The EM field is infinite range, so no point in the universe is free of EM fields. – user26165 Oct 26 '13 at 6:06
• @ George E. Smith : do you imply that the elctrons do not move in absence of electromagnetic interference ? – Rijul Gupta Oct 26 '13 at 8:19

In conductor physics, mean or drift electron velocity $v_\text{dr}$ determined by the anisotropic part of the distribution function $\tfrac{\partial f}{\partial v}$, which indeed is zero for vanishing electrical field $E=0$. The mediating quantity is the electron mobility $v_\text{dr}=\mu(E)\cdot E$.
Then the current $j=q\ n_e v_\text{dr}$ and hence magnetic field via $\nabla \times B\propto j$ goes down to zero as well.
• Regarding magnetic field $B$: If you have an isotropic distribution of velocities, the expectation value $\langle {\bf v} \rangle=\int_{\mathbb{R}^3} {\bf v}\cdot f({\bf v})\ \mathrm d^3v=4\pi\int_0^\infty |{\bf v}|\cdot f(|{\bf v}|)\ \mathrm d^3|{\bf v}|=0$ is zero and hence $B$ is too. – Nikolaj-K Oct 24 '13 at 9:45
• The way you'd be able to measure a magnetic field is via acceleration of nearby charged particles with velocity $w$. The force they feel is given by $F=w\times B$. By Newton, the forces from all the electrons have to be added up: $\sum_i F_i$ and since $w$ in your experiemtn is fixed, the magnetic field the solid generates would be $\sum_i B_i$, i.e. the effect is linear and no net velocity results in no net magnetic field. In any case, microscopically speaking you can't measure any properties of the solid without applying an electrical field to it: If you shine light, you have an E field. – Nikolaj-K Oct 24 '13 at 9:59