Curl of the Polarization (Electrostatics)

[ManuJulian]

I've been reading Griffith's "Introduction to Electrodynamics" and I've got to this part where it says:

"When you are asked to compute the electric displacement, first look for symmetry. If the problem exhibits spherical, cylindrical, or plane symmetry, then you can get $\vec{D}$directly from Gauss's equation (for the displacement) in integral form. (Evidently in such cases ∇x$\vec{P}$ is automatically zero, but since symmetry alone dictates the answer you're not really obliged to worry about the curl.)"

Now, why is it that the curl of the polarization is always zero in those cases where there is symmetry?

Is it just because in every case that exhibits symmetry the polarization is perpendicular to the boundary?

 

[Andy Resnick]

Are you considering electrostatics or electrodynamics?
In what sense do you mean 'symmetry': a symmetric homogenous body, or some sort of crystal symmetry?

 

[ManuJulian]

Electrostatics. Symmetry meaning the first one, a symmetric homogeneous body.

 

[Andy Resnick]

Since ∇x E = -∂B/∂t = 0 for statics and D = E + P, the result immediately follows.

 

[sardar]

I can explain the concept of curl of polarization in electrostatics. The curl of the polarization is a mathematical concept that represents the circulation or rotation of the polarization vector in a given region. In electrostatics, the polarization vector represents the distribution of electric dipole moments in a material.

Now, when a problem exhibits spherical, cylindrical, or plane symmetry, it means that the electric field and polarization are symmetrically distributed in the region. This symmetry also implies that the electric displacement vector is constant and perpendicular to the boundary in that region.

In such cases, the curl of the polarization, which is given by the cross product of the gradient of the polarization vector and the polarization vector itself, becomes zero. This is because the gradient of the polarization vector is also constant and perpendicular to the boundary, and when multiplied by the polarization vector, the result is always perpendicular to both vectors, resulting in a zero curl.

In other words, the symmetry of the problem dictates that the curl of the polarization must be zero, and this is why we do not need to worry about it while calculating the electric displacement using Gauss's law. It is not just because the polarization is perpendicular to the boundary, but rather because of the overall symmetry of the problem.

This concept is important in understanding the behavior of electric fields in materials with different symmetries, and it allows us to simplify calculations and focus on the essential components of the problem.