0

In a normal conductor, the current density is directly proportional to the electric field strength: $$ \vec{J} = \sigma \vec{E} $$ In a perfect conductor, $\vec{E} =0$, even if there is a current.

Inside a perfect conductor, if $\vec{E}=0$, then what determines the magnitude and direction of the current density $\vec{J}$ in the first place?

DanielSank
  • 24,439
A Slow Learner
  • 326

2 Answers2

1

what determines the magnitude and direction of the current density $\vec{J}$ in the first place?

Something else does, depending on what you connect your perfect conductor up to.

It could be the current limitation of the source that's driving the current. It could be the inductance of the circuit and the time that voltage has been applied. It could be a resistor or other circuit element that the perfect wire is connected to.

A Slow Learner
  • 326
The Photon
  • 27,393
  • I am not sure I understand the big picture here: Since resistivity is $0$, even a tiny amount of voltage applied would cause charges to have an infinite acceleration. Thus, how come do we still have a finite current inside a perfect conductor? – A Slow Learner May 07 '18 at 23:21
  • 1
    @ASlowLearner, resistance is zero, but inductance is non-zero. So you cannot ramp current infinitely fast. https://physics.stackexchange.com/questions/179374/is-current-in-superconductors-infinite-if-they-have-0-resistance-then-i-v-r-s – BowlOfRed May 07 '18 at 23:43
  • Also: Zero DC resistance does not mean zero AC resistance (and I mean actual resistance not reactive impedance). Even a superconductor or perfect metal do not even theoretically have zero AC resistance. – Sebastian Riese May 08 '18 at 00:08
  • @ASlowLearner, to expand on BowlOfRed's comment, consider an inductor made from turns of ideal conductor. There can be no net electric field inside the conductor but the total electric field is the superposition of the conservative electric field from separated charge (at say, the ends of the inductor) and the non-conservative electric field associated with a changing magnetic flux. Remember, a rapidly changing electric current through the ideal conductor is accompanied by a rapidly changing magnetic field surrounding the conductor. – Alfred Centauri May 08 '18 at 00:57
  • If inductance and non-zero AC resistance play a role in preventing the current from being infinite, why are they not taken into account in Ohm's law? i.e. when calculating the current in a normal conductor, we usually just apply Ohm's law without mentioning the other effects. – A Slow Learner May 08 '18 at 01:58
  • 1
    @ASlowLearner Because we normally are interested in the steady state (which is reached very fast for most circuits). – Sebastian Riese May 08 '18 at 11:18
1

The current density is defined by the history of $E(t)$. For simplicity we will concentrate on the case of a "perfect metal" and not a superconductor (where the underlying physics is more intricate but also real).

Zero (DC) resistance does not mean infinite acceleration. Technically, (DC) conductivity is defined as the linear response in a steady state that is achieved when the perturbation is turned on adiabatically from $t=-\infty$. So zero resistance (or infinite conductivity), tells us the current will be infinite after the non-zero electric field has been applied for eternity!

If we consider the real problem, we have to consider the actual time dependence: The electric field has been turned on at some time $t_0$, or more generally the field is $E(t)$. Now, the charge carriers have an effective mass and are accelerated according to $eE = F = m_\text{eff} a$, so due to this effect the current is increasing linearly with time (since $I(t) = env(t)$ and $v(t) = -eEt/m_\text{eff}$). There are additional complications due Bloch oscillations in crystals and due to the non-conservation of the quasi-particle number in superconductors, which we put aside for now.

An additional effect preventing instantaneous infinite current is the inductance of the wire, which will limit the current increase with time (since the change in current induces an electric field opposing the accelerating field).

Note that this discussion is not hypothetical: superconductors come very close to allowing dissipation free currents and there the actual current density depends on the history of the applied field (superconducting MRI or accelerator magnets to not need to be connected to a voltage source constantly, they can be short-circuited and the current keeps flowing). Also note, that a perfect metal (with free electrons, not impurities and no phonons) does indeed have non-zero AC conductivity due to the inertia of the electrons. The characteristic quantity describing the scaling of the AC conductivity is the plasma frequency (above which the metal becomes non-conductive).

Sebastian Riese
  • 11,682
  • "Technically, (DC) conductivity is defined as the linear response in a steady state that is achieved when the perturbation is turned on adiabatically from t=−∞." Are you referring to Ohm's law here? – A Slow Learner May 08 '18 at 02:24
  • 1
    Ohm's law is either the phenomenological statement that current is proportional to the voltage in the steady state (note that the steady state is reached within at most a few milliseconds for typical circuits). This version of Ohm's law is a very crude approximation of reality. Or the relation $j = \sigma E$ which is no "law" but a general form if we allow $\sigma$ to depend on the field strength (yes this is cheating) and $\sigma$ is usually constant for small $E$. This equation gets especially useful, when we consider it in Fourier space, then it can also handle transients. – Sebastian Riese May 08 '18 at 11:48