according to a crude (classical) model of the drift velocity as given in my high school textbook, using an example of cylindrical conductor, we derived ohm's law as

$j=\sigma E$ where $\sigma $

is the conductivity and is given as

$\sigma=\frac{ne^2\tau}{m}$ now we treat $\sigma $ as infinity.


(a) why do we treat conductivity as infinity?

(b) if we treat conductivity as infinity, then $E$ will be 0 but $j$ will be non zero. how can it be possible that charges are ccelarated without any external electric field. why then is this approximation accepted.

  • $\begingroup$ relaxation time in conductors is NOT very high , if it were ,would you get your light on after you press the switch? in fact it is quite low also for your information if something is high, how can you treat it as infinity? like is 10 to power 15 be treated as infinity when there are many small quantities present in formula like charge of e. $\endgroup$
    – Curious
    Mar 24 at 11:23
  • $\begingroup$ Related, might answer your question: London theory, an electromagnetic description? $\endgroup$ Mar 24 at 17:33

1 Answer 1


The conductivity in usual metals is high but not infinite. Thus they have a very short (nonzero) dielectric relaxation time, which is the time that any local free electron accumulation in the metal would disappear due its own electric field. The assumption of infinite conductivity (and zero electric field) is used as a good approximation in certain boundary value problems (e.g., in electrostatics or reflection of electromagnetic waves).

Note 1: Mathematically, the current density can stay finite when the conductivity goes to infinity and the electric field goes to zero.

Note 2: The tau in your formula for the conductivity is not the dielectric relaxation time, it is the momentum relaxation time of the electrons in the metal due to scattering with the lattice (see Drude model).


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