Is it possible to derive Ohm's law (perhaps in some appropriate limit) from Maxwell's Equations?

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    $\begingroup$ For a simple theory behind Ohm's law, see e.g. the Drude model at Wikipedia or this Phys.SE post. $\endgroup$
    – Qmechanic
    Oct 16, 2012 at 0:09

5 Answers 5


Ohm's law $\vec\jmath=\sigma\vec{E}$ can be derived in the limit of small electric fields using linear response theory. This leads to Kubo's formula for the electric conductivity, which relates $\sigma$ to the zero frequency limit of the retarded current-current correlation function.

$$ \sigma^{\alpha\beta}(q)=\lim_{\omega\to0}\frac{1}{-i\omega}\left\{\frac{ne^2}{m}\delta^{\alpha\beta} - i\langle[j^\alpha(\omega,q),j^\beta(-\omega,-q)]\rangle \right\} $$

(This derivation, of course, involves more than just Maxwell's equation. This is properly derived in the context of non-equilibrium field theory.) The Drude model is a model for the spectral function of the current-current correlation function in terms of a single ``collision time''. This model can be derived within kinetic theory, which is applicable when interactions are weak and the correlation function can be computed in terms of quasi-particles.

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    $\begingroup$ But one needs to keep in mind that this linear response theory assumes linearity, which is a basic feature of Ohm's law. So, Kubo's approach is an incomplete derivation. $\endgroup$ Jun 23, 2018 at 20:00
  • $\begingroup$ @JasonArthurTaylor Obviously, the response is non-linear at large field. What we are talking about here are small fields. If linear response were to break down in this regime, you would see that from IR divergences in the current-current correlation functions. I'm not aware of any "real" system in which this happens in the thermodynamic limit. (The usual exceptions for linear response are weird systems, like the Navier-Stokes law in con starch, or Ohm's law in ballistic channels.) $\endgroup$
    – Thomas
    Jun 24, 2018 at 14:12
  • $\begingroup$ Yes, but what's "large"? I don't think Kubo made any attempt to estimate what the next order term/contribution is at high fields. If true, the regime of linearity, a key element of Ohm's Law, is not supplied there. (This means it is linear, but no idea where.) Incidentally, I can think of, for example, a sandpile, that might be difficult to model using this approach. Also, if you would be so kind and have a moment, can you take a quick look on page 2 of the reference I cited in my answer and comment on it? Do you think the Hamiltonian was mis-modeled? $\endgroup$ Jun 25, 2018 at 0:13
  • $\begingroup$ I think you should probably regard Ohm's law as saying that the resistance is the derivative of the current with respect to the voltage at zero voltage. In this case, linear response is fine. $\endgroup$ Mar 25 at 3:58

No, not in the way you are probably thinking. You can do a lot with Maxwell's equations, but you have to step outside of them to derive Ohm's law. There is a trivial relationship going from points to macroscopic objects (e.g., multiplying by lengths and cross sectional areas), but this is just giving different forms of what is still referred to as Ohm's law.

As I pointed out in a comment on Thomas' currently-accepted answer, I think that Kubo's solution implicitly assumes (i.e., does not derive from scratch) a linear relationship between the current and field. It is already is going way beyond Maxwell's laws.

A full answer requires going even beyond that. See, e.g., Riess (2004). So that's why I'm saying no is correct the answer to your actual question.

Importantly, I don't think Kubo's original paper on this attempts to compute any actual values of $\sigma.$ So, neither aspect of Ohm's law was really derived by Kubo. Rather, Kubo's formalism allows computation of $\sigma$ assuming a linear relationship should exist.

For these reasons, I would object to Thomas' use of the phrase "derived rigorously" in describing Kubo's contribution as described. This is also partly why I think my own answer is worth submitting. (I am somewhat bothered by the use of the phrase in this context, especially if also saying that the problematic Drude model also gives it, like it is a trivial equation to derive or something.)


No, it is an approximation and not derived from first principles. It is based on empirical observations.

  • $\begingroup$ Are you talking about Ohm's law of the Maxwell equations? :D $\endgroup$
    – Nikolaj-K
    Oct 16, 2012 at 7:36
  • $\begingroup$ You say "is ... not derived," present tense. But, IMO, one can actually compute it. Here's a paper where the conductivity or resistivity was numerically computed for aluminum: arxiv.org/abs/1310.4013 . $\endgroup$ Jun 23, 2018 at 21:27

I have added this answer because some comments on this and other similar questions (marked as duplicate) have requested additional details of the Quantum Mechanical derivation of Ohm's law.

The derivation given here is appropriate for single particle QM and seeks to derive Ohm's law in the form: $$ j_i = \sigma_{ij}E_j\;, $$ where $j_i$ is the ith component of the current (density) and E_j is the jth component of the electric field and $\sigma_{ij}$ is the conductivity (tensor).

By definition we have the current operator: $$ \hat {j_i} = q \hat v_i = \frac{q}{m}\hat p_i\;, $$ where $\hat p$ is the momentum operator, q is the charge, m is the mass, and the current (not the current operator) is: $$ j_i(t) = \langle\Psi(t)|\hat{j_i}|\Psi(t)\rangle\;, $$ where $\Psi$ is the QM state.

We now solve for the state in the interaction picture (as opposed to the Heisenberg picture or the Schrodinger picture). The hamiltonian is decomposed as $H=H_0+V(t)$ where $V$ is the interaction Hamiltonian and $H_0$ is the unperturbed Hamiltonian whose ground state we call $|0\rangle$. Then, to first order in the interaction, the interaction picture state is: $$ |\Psi^I(t)\rangle = |0\rangle - i\int^t dt' V^I(t')|0\rangle\;, $$ where $$ V^I(t) = e^{iH_0t}V(t)e^{-iH_0t}\;. $$

Now, let's specialize to a interaction that is linear in the spacially constant field $\vec E$: $$ V = -qf(t)\vec E\cdot \vec r\;. $$ N.b., remember that $-\nabla\phi = E$. The time dependent part $f(t)$ can be something like $\sin(\omega_0 t)$ or whatever other pure time dependence we like.

So, keeping enough terms in the expansion of the state to solve for j to first order in E, we have: $$ j_i =\left(\langle0|-iqE_j\int^t dt'f(t')\langle 0| \hat r^I_j(t')\right) \frac{q}{m}\hat p_i \left(|0\rangle+iqE_j\int^t dt'f(t')\hat r^I_j(t')|0\rangle\right) $$ Or: $$ j_i = \sigma_{ij}E_j\;, $$ where $$ \sigma_{ij} = \frac{-iq^2}{m}\int^t dt'f(t')\langle0|[r^I_j(t'),p_i]|0\rangle\;. $$

This specific relationship between j and E also is based on the assumption that there is no current in the unperturbed ground state $|0\rangle$.


First we have the following equation taken from the constitutive relationships:


Furthermore, we will assume that sigma is constant throughout all the medium and it is not temporarily dispersive. Therefore the convolution operator is equivalent to multiplication.


We also know that the electrostatic potential is (only if $\epsilon$ is constant in all the medium):


So we can write the electric field as a function of the potential difference or commonly known as voltage:

$$W = \int ^2_{1}\vec{F}\cdot\vec{dl}= \int ^2_{1}q\vec{E}\cdot\vec{dl} = q\int ^2_{1}\vec{E}\cdot\vec{dl}=-q\int ^2_{1}\nabla\phi\cdot\vec{dl}=-q\int ^2_{1}d\phi=q\cdot(\phi_1 - \phi_2)$$

$$V=(\phi_1 - \phi_2)=\int ^2_{1}\vec{E}\cdot\vec{dl} ==> V=E \cdot l$$

The previous relation ( $ V=E \cdot l$ ) is only valid if the electric field is constant along the l curve. Therefore, we will apply that approximation that is fulfilled in materials of few losses, that is to say permittivity $\epsilon$ is not temporarily dispersive and constant throughout the medium $\epsilon(\vec{r},t)\cong\epsilon$.

Finally we can derive the Omh's law as:

$$\int\vec{J}\cdot\vec{dS} = \int\sigma\cdot\vec{E}\cdot\vec{dS} ==> I = \int\sigma\cdot\frac{V}{l}\cdot dS = \sigma \cdot \frac{V}{l}\cdot S$$

$$V=I\cdot R$$ $$R= \frac{1}{\sigma} \cdot \frac{l}{S}$$ $$Resistivity=\rho= \frac{1}{\sigma}$$

Reminder: The medium permitivity $\epsilon$ must be constant and not temporarily dispersive. This condition is fullfilled in the vast majority of conductors materials.

  • $\begingroup$ how do we prove that E = V/L = del(V)/del(L) (constant) all across the length of R ? $\endgroup$ Mar 8, 2021 at 10:36
  • $\begingroup$ @SilverMoon If the length of the resistance is small enough in comparisson to the wavelength of the electric field, we can ignore the propagation effect. Remember that the solution of the wave equation is E(t)=|E|⋅cos(w⋅t+(2pi/lambda)⋅R). You can check it out really easily. $\endgroup$ Mar 9, 2021 at 11:22
  • $\begingroup$ @SilverMoon And also the medium can not be temporarily dispersive, because if it is it will make the electric field vary. $\endgroup$ Mar 25, 2021 at 0:12

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