Consider Maxwell's equations in a general medium without free charges or currents:

$$\nabla\cdot\vec{D} = 0 \tag{1}$$ $$\nabla\cdot\vec{B} = 0 \tag{2}$$ $$\nabla\times\vec{E} = -\frac{\partial \vec{B}}{\partial t} \tag{3}$$ $$\nabla\times\vec{H} = \frac{\partial \vec{D}}{\partial t} \tag{4}$$

The constitutive relations are $\vec{D} = \varepsilon_0\vec{E} + \vec{P}$ and $\vec{B} = \mu_0(\vec{H} + \vec{M})$.

Suppose for simplicity that the medium is nonmagnetic, so that $\vec{M} = \vec{0}$.

Additionally, suppose the medium is isotropic, homogeneous and nondispersive, so that we can write the nonlinear polarization density as $\vec{P} = \epsilon_0\chi(|\vec{E}|)\vec{E}$, where $\chi$ is (in general) a nonconstant scalar function of $|\vec{E}|$.

Taking the curl of equation $(3)$ and using $\vec{B} = \mu_0\vec{H}$ we get $$\nabla \times \nabla \times \vec{E} = -\mu_0\frac{\partial}{\partial t} (\nabla \times \vec{H}) = -\mu_0\frac{\partial^2 \vec{D}}{\partial t^2} \tag{5}$$

Now we can use the vector identity $\nabla \times \nabla \times \vec{E} = \nabla(\nabla \cdot \vec{E}) - \nabla^2\vec{E}$ and use the constitutive equation for $\vec{D}$ to obtain $$\nabla^2\vec{E} - \mu_0\varepsilon_0\frac{\partial^2 \vec{E}}{\partial t^2} = \mu_0\frac{\partial^2 \vec{P}}{\partial t^2} + \nabla(\nabla \cdot \vec{E}) \tag{6}$$

Now every source I have seen apparently assumes that $\nabla(\nabla\cdot \vec{E}) = \vec{0}$, after which they obtain the standard nonlinear optics wave equation with polarization source term: $$\nabla^2\vec{E} - \mu_0\varepsilon_0\frac{\partial^2 \vec{E}}{\partial t^2} = \mu_0\frac{\partial^2 \vec{P}}{\partial t^2} \tag{7}$$

Unfortunately, I am unable to see why this assumption is justified. The usual argument is that since $\vec{D} = \varepsilon_0(1 + \chi)\vec{E}$ then since $\nabla \cdot \vec{D} = 0$ we also have $\nabla\cdot\vec{E} = 0$.

But this argument is flawed in the nonlinear case since $\chi$ is in fact a function of $|\vec{E}|$, so that we need to use the product rule when computing the divergence operator: $$0 = \nabla\cdot\vec{D} = \varepsilon_0\nabla\chi(|\vec{E}|)\cdot\vec{E} + \varepsilon_0(1 + \chi(|\vec{E}|))\nabla\cdot\vec{E} \tag{8}$$

We therefore obtain $$\nabla\cdot\vec{E} = -\frac{\nabla\chi(|\vec{E}|)\cdot\vec{E}}{1 + \chi(|\vec{E}|)} \tag{9}$$

and it is far from obvious to me that this equals $0$ (or that the gradient of this expression equals $\vec{0}$) (indeed, $|\vec{E}|$ is generally spatially dependent so that the spatial derivatives are nonzero).

So is this assumption flawed and the most basic equation of nonlinear optics plainly wrong, or is there a correct way to derive the nonlinear optics wave equation (7), perhaps with additional assumptions on $\vec{E}$?

  • $\begingroup$ Your Ansatz for the nonlinear polarization, $\vec{P} = \epsilon_0\chi(|\vec{E}|) \vec{E}$, isn't very appropriate. In the frequency domain this can indeed have a relationship roughly of that form, but in general the response function will be non-local in time (i.e. the non-linearity has memory) and it is given as an integral of the electric field at all past times multiplied by some response kernel. $\endgroup$ Oct 13, 2018 at 17:33
  • $\begingroup$ For simplicity I assumed that the medium was nondispersive. I think this should imply that the dependence of $\vec{P}$ on $\vec{E}$ is immediate so that there is no memory effect. I agree that in principle there should be this effect taken into account. $\endgroup$
    – Tob Ernack
    Oct 13, 2018 at 19:04

1 Answer 1


You are correct, in general the term $\nabla \nabla\cdot \mathbf E$ should be kept in the equation.

The reason it is customary to drop it is that it simplifies the equation. The best justification I can think of is that this term should be negligible in isotropics if the susceptibility $\chi$ depends on the field only weakly.

Consider the case where the susceptibility does not vary at all, so $\chi(E)$ can be replaced by a constant independent of coordinates $\chi_0$. Since charge density is given by divergence of electric field and also by negative divergence of polarization, we have

$$ \nabla \cdot \mathbf E = -\nabla \cdot (\chi_0 \mathbf E) $$ and the only way this could hold everywhere in the isotropic medium$^*$ is if $\nabla \cdot \mathbf E = 0$ everywhere in the medium.

Purely transversal plane waves obey the condition $\nabla \cdot \mathbf E =0.$

In those cases, the right-hand side of the wave equation does have two terms

$$ -\mu_0\frac{\partial^2\mathbf P}{\partial t^2} - \nabla \nabla\cdot \mathbf E $$ but the second one is zero, so we can simplify the equation and work with RHS $$ -\mu_0\frac{\partial^2\mathbf P}{\partial t^2}. $$

Now, the effects of nonlinear optics are usually very small and thus dependence of $\chi$ on field or position is weak. Thus the second term $\nabla \nabla \cdot \mathbf E = 0$ , although non-zero, should be small with respect to the first term

$$-\mu_0\frac{\partial^2\mathbf P}{\partial t^2}.$$

We should still be able to neglect the second term and use the same equation as in the linear case. The nonlinearity will be preserved - but the variability of $\chi$ will enter the equation only through the first term, where polarization is nonlinear function of electric field strength.

In other words: since the terms in the equation are continuous functions of field and susceptibility, the small change in susceptibility should cause only small change in divergence of $\mathbf E$. The value should therefore be still negligible, at least in some range of field strengths that are not too high.

$*$ In crystals, the situation is different: because susceptibility is a tensor with differing components, zero divergence of $\mathbf D$ prevents zero divergence of $\mathbf E$ in all but special cases, such as when only one component of $\chi$ is realized (when polarization of the wave agrees with one of the principal axes of the crystal).

  • $\begingroup$ Is there an easy way to get an estimate on the magnitude of $\mu_0\frac{\partial^2 \vec{P}}{\partial t^2}$ in order to compare it to the magnitude of $\nabla(\nabla\cdot\vec{E})$? If we move the linear part of $\vec{P}$ into the LHS of the wave equation and use $\mu_0\varepsilon_0(1+\chi_1)=\frac{n_1^2}{c_0^2}$, we get $\nabla^2\vec{E}-\frac{n_1}{c_0^2}\frac{\partial^2\vec{E}}{\partial t^2}=\mu_0\frac{\partial^2\vec{P_{NL}}}{\partial t^2}+\nabla(\nabla\cdot\vec{E})$ where the LHS is a wave equation with the linear part of the refractive index $n_1$, and the RHS is the nonlinear term. $\endgroup$
    – Tob Ernack
    Oct 13, 2018 at 19:27
  • 1
    $\begingroup$ So we want to argue that $\left|\nabla(\nabla\cdot\vec{E})\right| \ll \left|\mu_0\frac{\partial^2\vec{P_{NL}}}{\partial t^2}\right|$. It is somewhat tricky to get estimates for these because the exact expressions are complicated. I might be fussy, but I like to be confident in the approximations I use :) $\endgroup$
    – Tob Ernack
    Oct 13, 2018 at 19:27
  • $\begingroup$ (And also we would expect both of these terms to be small since the nonlinearity is weak, and the waves are going to be close to plane waves, so it is not immediately clear how the terms compare) $\endgroup$
    – Tob Ernack
    Oct 13, 2018 at 19:34
  • $\begingroup$ You are right, the divergence term should be compared to the nonlinear polarization term. Perhaps it is possible to make such comparison when both $E$ and $P_{NL}$ are expressed as Taylor series in space coordinates and time. I.e. $ E_k(\mathbf x,t) = E_{0,k} + g_{l}x_l + g_4t + ...$, put that into $P_{NL,l}(\mathbf E) = \chi_{lki}E_kE_i + ...$ to get $P_{NL}(\mathbf x,t)$, calculate the derivatives and try to find what condition on the coefficients must be satisfied in order for the above inequality to hold. $\endgroup$ Oct 13, 2018 at 21:38
  • $\begingroup$ That may be easier to work out if some specific form of $\mathbf E$ is assumed, such as, for example, $\mathbf E(\mathbf x,t) = (X_0\mathbf e_x + Z_0\mathbf e_z)M(z,t)\cos(\Omega t - kz)$, where $M(z,t)$ is unknown modulation function but expected to vary with much smaller k number than $k$ of the main wave. $\endgroup$ Oct 13, 2018 at 21:57

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