When is the Curl of Electric Displacement Zero?

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The curl of electric displacement (D) is zero under specific conditions, primarily in electrostatics when the electric field (E) is static. This occurs when there is no polarization or uniform polarization, both of which ensure that the curl of polarization (P) is also zero. Additionally, if there are no free charges present, the electric displacement D can be zero, leading to a zero curl. The relationship between zero curl of D and symmetry is highlighted in cases of spherical, cylindrical, or plane symmetry, where D can be derived using Gauss's law, confirming that curl of P is automatically zero. Overall, understanding the conditions for zero curl is essential for accurately applying electrostatic principles.
fricke
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In what condition(s) curl of electric displacement is zero?

Is it okay to say curl of electric displacement is zero in:
1) in electrostatics (curl of E is zero)
then followed by the following conditions:
2) when there is no polarization (curl of P is zero)
3) in uniform polarization (which makes curl of P is zero)
4) when there is no free charge presents (directly obtain electric displacement D is zero and thus curl of D is zero)

and one more, how does zero curl of electric displacement related with the symmetry?
because in Griffith 4th edition book says:
"If the problem exhibits spherical, cylindrical, or plane symmetry, then you can get D directly from equation by the usual Gauss's law methods. (Evidently in such cases curl of P is automatically zero)"
 
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We have ##\mathbf{D} = \epsilon_0 \mathbf{E} + \mathbf{P}##. Applying curl on both sides and requiring the curl of electric field displacement to be zero we obtain ##\epsilon_0 \nabla \times\mathbf{E} = -\nabla \times \mathbf{P}##. This is the general requirement for both electrostatic and electrodynamics for curl of D to be zero. So, point 2) and 3) are strictly for electrostatic case. In addition to point 2) and 3), consider also the case where ##\mathbf{P} = \nabla f## where ##f## is some scalar function. This last condition might be too rare to find in reality as compared to P being zero or uniform, but it still satisfy ##\nabla \times \mathbf{P} = 0##. As for point 4), what you mention there is only a special case for which there is no free charge. The point to remember is that in the Gauss equation ##\nabla \cdot \mathbf{D}(\mathbf{r}) = \rho_f (\mathbf{r})##, ##\rho_f (\mathbf{r})## is the charge density at a position ##\mathbf{r}##. Even if there is no free charge at this point, there might be another charge at other positions causing the D to be non zero. More generally, Gauss law only tells us the divergence of D, in order to completely characterize D we also need to know its curl. In other words, we can't really say anything about the curl from the divergence or vice versa, for there are certain vector fields whose divergence is zero but the curl is not.
fricke said:
"If the problem exhibits spherical, cylindrical, or plane symmetry, then you can get D directly from equation by the usual Gauss's law methods. (Evidently in such cases curl of P is automatically zero)"
For example, if the system exhibits spherical symmetry, the D field vector can only be radial, it has no angular components. Furthermore, it must be only a function of ##r##. So ##\mathbf{D} = \hat{r}D(r)##. Now compute the curl of P which is equal to the curl of D in electrostatic and you should see curl of P is zero.
 
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