# Why is calculating the resulting field from polarization not recursive?

In a parallel-plate capacitor filled with a dielectric, if the plates are kept at a potential $$V$$, and the dielectric has some (non-linear) polarization $$P$$ that is dependent on the magnitude of $$E$$ (so $$P$$ is a vector of which the magnitude is some function of $$E$$, for simplicity sake we can say it is in the direction of $$E$$), then the electric field applied to the space within the plates is $$E = V/d$$.

Then, from this the E-field inside the dielectric can be calculated, which opposes this applied $$E$$ field.

However, what I don't understand is, why is this not done recursively? Why is the resulting $$E$$ field not the one that again determines the polarization, which then again determines the dipole moments and thus how the E-field inside the dielectric behaves? Like finding an equilibrium? But in all exercises on this it is just done naively, once. I don't understand this physically.

• Well, one way to look at it is that you can perform the infinite sum and set the total difference to be the polarisation. Commented May 9, 2023 at 5:21

This is a good question. It is not "recursive" because in the equation
$$\mathbf D= \epsilon_0 \mathbf E + \mathbf P \tag{1}\label{1}$$ and in the constitutive equation $$\mathbf P = \mathbf P(\mathbf E)\tag{2}\label{2}$$ the field $$\mathbf E$$ is defined to be the end-result of the implicit recursion you are alluding to. In other words, it implies an electrostatic & thermodynamic equilibrium that has been established after all the dipoles will have aligned properly so that the macroscopic fields $$\mathbf {\{D, E, P\}}$$ so that $$\eqref {1}$$ and $$\eqref{2}$$ are satisfied along with all the macroscopic boundary conditions. Note that the microscopic field of which there is only one, say $$\mathfrak E$$, with $$\mathbf E$$ being the macroscopic average of $$\mathfrak E$$, satisfying the Maxwell equations $$\nabla \times \mathfrak E = 0$$ and $$\nabla \cdot \mathfrak E = \rho/\epsilon_0$$ everywhere, does not have to satisfy the macroscopic boundary conditions on the dielectric, such as the continuity of $$\mathfrak E_t$$ or $$\mathfrak E_n$$.

When calculating the energy/work relationship in electrostatics it is "derived" that spatial energy density is $$\mathbf E \cdot d \mathbf D$$. In fact, this is not the internal energy but rather the (Helmholtz) free energy at some temperature where the temperature dependence is expressed in the temperature dependent constitutive relationship $$\mathbf P = \mathbf P(\mathbf E, T)$$ because it is related to the infinitesimal isothermal work needed to create and maintain the electrostatic configuration. Actually it is better to consider the full volume integral $$\delta w =\int_{\infty} dV \mathbf E \cdot \delta \mathbf D$$ as being representative of that work as the amount it takes to change the displacement vector field by $$\delta \mathbf D$$. Interestingly and significantly, using some standard vector analytical identities this integral taken over all space can be transformed to another one $$\delta w = \delta \frac{1}{2}\int_{\infty} dV \mathbf E_0^2 - \int_{\mathcal V}dV\mathbf P \cdot \delta \mathbf E_0 \tag{3}\label{3}$$ where $$\mathcal V$$ is the space occupied by the dielectric and $$\mathbf E_0$$ is the vacuum field without the dielectric, and in this form $$\delta w$$ is the electrostatic work it takes to maintain the sources (free charges) of $$\mathbf E_0$$ to be the same amount at the same places as it was before the dielectric was inserted.

• Thanks, this answer elucidates the matter, and surprises me. I have used a wrong interpretation where the E in P(E) is the vacuum field. Commented May 9, 2023 at 12:12
• If you have a vacuum energy like E = V/d, because two plates are put at potential diff V at distance d, and I know the formula for polarization P(E), how then do I have to calculate the E field? Commented May 9, 2023 at 12:45
• Many years ago I made the same mistake as the one you are alluding to, specifically, in the magnetic case ( H v. E and B v. D). While nowadays this issue is better discussed in most books it is rare to find a fully detailed analysis of it. Probably the first ones are in Stratton or Guggenheim or Heine but few read those, see similar discussion here and here Commented May 9, 2023 at 12:45
• If it is a "simple" material then there is a local macroscopic relationship between the field $E$ and the induced polarization $P$. The relationship can be linear (e.g., glass) and then there is a susceptibility coefficient $\chi = \chi (E)$, or relative permittivity $\epsilon_r = 1+ \chi$, that gives you P (or D). So the question is how to get $E$ from $E_0$? That is quite complicated problem unless the geometry is simple. The parallel plate capacitor with negligible fringing field has a simple geometry where you can make the assumption that $E\approx E_0 = V/d$ and then $P = \chi V/d$. Commented May 9, 2023 at 13:17
• So it is a nonlinear dielectric $P=\chi(E)E$ with $\chi = \chi_0 + \chi_2 E^2+....$. As long as the geometry is simple, such as a parallel plate condenser with negligible fringing fields, you can still write $E \approx E_0 = V/d$. Commented May 9, 2023 at 13:51

When puzzling over problems like this, you should always consider how you would do the experiment. Here, the dielectric constant is a parameter you measure by making a capacitor and determining the polarization by measuring the capacitance. So, the experiment finds the equilibrium you seek.

• I'll update the question because the polarization is a formula and I'm looking for a theoretical solution, not an experimental one. Commented May 8, 2023 at 17:50
• @buddhabrot Physics is not mathematics. Every physical question is fundamentally about the results of experiments and observations. Polarization is a theoretical model crafted to match experimental results. Commented May 8, 2023 at 17:55