From continuity equation and Ohm's law it it's possible to say that: $$ \nabla \cdot (\sigma \mathbf{E}) = -\frac{\partial \rho}{\partial t} \Rightarrow -\frac{\partial \rho}{\partial t} \frac{1}{\sigma} = \nabla \cdot \mathbf{E}$$ and using Gauss's law: $$ -\frac{\partial \rho}{\partial t} \frac{1}{\sigma} = \frac{\rho}{\epsilon_0}$$ This is a first-order ordinary differential equation, so its solution is given by: $$ \rho (\mathbf{r},t) = \rho(\mathbf{r},0) e^{-\frac{\sigma t}{\epsilon_0}} $$ From that expression we can say that $\rho \rightarrow 0$ as $ t $ increases, but also will its derivative, so going back to the continuity equation: $$ \nabla \cdot \mathbf{J}= -\frac{\partial \rho}{\partial t} \rightarrow 0 $$ For good conductors, this seems to mean that in practically no time $$ \nabla \cdot \mathbf{J} \approx 0 $$ This is one of the assumptions for the quasistatic approximation, but as far as I know, this is not always true, for instance while working at very high frequencies. But all the steps made in the derivation were independent from the frequency of $ E $, so why this doesn't apply for all frequencies? My intuition dictates me that the problem would be Ohm's law, that may not be valid for very high frequencies as the conductivity starts to behave differently (imaginary part), nevertheless, I've been told that for almost every practical application the conductivity matches its DC value, so Ohm's law can be applied, isn't this true? and if it is, when is not possible to say that $ \nabla \cdot \mathbf{J} \approx 0 $ ?

  • 1
    $\begingroup$ The classical Drude model (en.wikipedia.org/wiki/Drude_model ) shows the connection between the electric field and current to have a complex part which which frequency dependent. (Kudos to DocScience and DanielSank for sending me to my old notes.) (I see that Thomas has answered more completely than my comment). $\endgroup$ – ZeroTheHero Mar 21 '17 at 20:51

Yes, Ohm's law is valid at finite frequency $$ j(\omega)=\sigma(\omega)E(\omega). $$ This is simply a linear response relation, so the only assumption is that of a sufficiently weak field. A typical approximation for $\sigma(\omega)$ is the Drude form $$ \sigma(\omega) = \frac{\sigma_0}{1+i\omega\tau} $$ which is not exact, but valid in kinetic theory. Indeed, in many typical applications you can neglect the frequency dependence, because the time scale is set by a microscopic collision time $\tau\sim\tau_{coll}$. Note that in real time, the frequency dependent response becomes a convolution integral $$ j(t) = \int dt'\, G(t-t')E(t'). $$ where $G$ is the retarded (or response) function, the one-sided Fourier transform of $\sigma(\omega)$.

With regard to your manipulations: I think that $\nabla \cdot j=0$ is indeed a good approximation, and that the main correction arises from non-localities in the response that are important for high frequencies and short wavelength (compared to collision frequencies and mean free paths). A typical problem you find discussed in texts book on EM is the skin effect. Indeed, the skin depth $\delta=\sqrt{2/(\mu\omega\sigma)}$ is governed (approximately) by the DC conductivity $\sigma$ ($\mu$ is the permeability).

  • $\begingroup$ where should I introduce the $ \partial B/ \partial t $ term? I'm just plugging in ohm's relation into the continuity equation in order to find the charge density and relate it to the divergence of the current. My goal is to understand how, at very high frequencies, where we have a propagation effect (i.e. the E fields is not the same for every part of the circuit, and neither is the current density) the volume charge density also goes to 0. where can i find the charge density resulting from the difference of the currents density flux? $\endgroup$ – diegobatt Mar 22 '17 at 3:58
  • $\begingroup$ my mistake, I misunderstood what you were trying to do. $\endgroup$ – Thomas Mar 22 '17 at 4:23
  • $\begingroup$ I understand that the title I've chosen for the question can be very misleading. I'm sorry for that. Regarding my question, I have been pointed out that the charge density I am looking for may be a surface charge density rather than a volume one, is this reasoning correct ? $\endgroup$ – diegobatt Mar 22 '17 at 4:29
  • 1
    $\begingroup$ I think the main correction does indeed arise from Ohm's law becoming gnon-local in space and time. This is not a surface charge density, but it becomes important in the high frequency regime where the current does not penetrate far (because of the skin effect). In order to estimate these effects you have to go beyond the Maxwell equations, and study the current in a more microscopic theory, like kinetic theory. $\endgroup$ – Thomas Mar 22 '17 at 13:19

If you assume that the only property of the material is that of conductivity, then yes. Ohms law is valid at all frequencies.

But we know real materials have properties of capacitance and inductance as well as conductance and which also depend on the geometry as well as the material itself. So in this case, no, you need to include additional physical models that address these properties.

In general impedance, which encompasses capacitance, inductance as well as conductance will change according to frequency.

  • $\begingroup$ @ZeroTheHero Capacitance depends on $\epsilon$ and geometry of the conductor. $\endgroup$ – DanielSank Mar 21 '17 at 1:39
  • $\begingroup$ So is it correct to say that for a material that only has conductivity as property $ \nabla \cdot J \approx 0 $ for all frequencies? if it is, when would the Kirchhoff's current law (KCL) doesn't apply ? I thought that the hypothesis of it was the null divergence of the current density $\endgroup$ – diegobatt Mar 21 '17 at 2:25
  • $\begingroup$ @diegobatt without capacitance, inductance, a hypothetical material indeed would allow electrons to travel with unlimited velocity; well not really, no faster than $c$. Do the Maxwell field equations impose that limit in this case? Or do we need Einstein's reformulation? Also without capacitance, inductance no energy 'storing' component in the material. KCL should always apply according to conservation laws. $\endgroup$ – docscience Mar 21 '17 at 17:57
  • $\begingroup$ I think you are confusing a few things: Conductivity is a local relation between current density and electric field. It is needed (i.e. an input) to solve the Maxwell equations in a material. Capacitance, inductance etc. are global properties of some arrangement of conductors. They are determined by the solution of the Maxwell equations (i.e. an output). $\endgroup$ – Thomas Mar 21 '17 at 20:38

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.