In this example we find that the polarization in a rotating cylinder is given by an iterative method (see section 2.2 in the link) using an effective $E$-field of $\vec E_0=\omega B \vec r/c$. Why are these iterations to find the final polarization necessary since (by definition of $\chi_e$) the final state of the polarization is: $$\vec P=\chi_E \vec E_0$$ Surly if you just applied an electric field to a dielectric sphere the above equation would hold, why doesn't it under the current situation?
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asked Nov 4, 2015 at 13:32
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I think that your confusion is due to a subtlety associated with the solution approach of going to a rotating frame of reference. If you're just trying to find out the polarization of a stationary rectangular, block-shaped slab of dielectric material due to a uniform electric field, then you could simply write down an electric polarization equation similar to the one you wrote above and be finished.
However, when you go to a rotating frame of reference to solve the current problem, you are considering the response of the electrons in the dielectric rotating cylinder not to a real E-field, but the response due to what you yourself acknowledged to be an "effective" or "pseudo" E-field of of $\vec E_0=ωB\vec r /c$. So you can't just stick this pseudo E-field of $E_o$ into your polarization equation and say that you're done because you forgot something: The real E-field. Initially, there is no real E-field. But when the cylinder starts rotating, the electrons start moving in response to the $\vec v \times\vec B$ "pseudo" E-field. But when they start moving and therefore start polarizing the dielectric medium of the cylinder then what happens? The polarization that the electrons produce generates an E-field of its own! So now you have both the "pseudo" E-field and a "real" E-field, so you have to re-consider and re-calculate the polarization of the medium in response to the sum of BOTH E-fields. But, wait, you're not done yet. When you re-calculate the polarization and find that it is different from your first solution, then that means that you have to go back and re-calculate the real E-field again. Do you see where this is leading? You have to iteratively re-calculate the polarization and the (real) E-field over and over again because they depend on each other. Again, this complication arises because you have a contribution to the polarization from a "pseudo" E-field that was produced by going to a rotating frame of reference. As your linked example says, it's a "chicken-and-egg" problem.
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$\begingroup$ Hi, thanks for your answer, but for the case of a stationary block you will still get due to the polarization a new $E$ field which means for the stationary block you should also do this iteration but you don't? $\endgroup$ Commented Nov 4, 2015 at 16:28
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$\begingroup$ @Joseph - Actually, you're right that this sort of iteration is also needed for just pure E-fields and stationary dielectrics, but that effect has already been taken into account and calculated for many standard shapes so you don't have to give any thought to it. For example, the polarization P of a dielectric sphere in an E-field E is not the same as the polarization P' of a dielectric thin slab placed in the same E-field. For a sphere you have to use a factor of 1/3 (I recall) in the relationship between the E-field and the polarization P and other factors for other shapes. $\endgroup$– user93237Commented Nov 4, 2015 at 16:41
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$\begingroup$ Interesting... are then saying that $\chi_E$ depends on the shape of the object or do these factors have to be put in by hand? $\endgroup$ Commented Nov 4, 2015 at 18:09
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$\begingroup$ @Joseph - χE is an intrinsic property of the dielectric material which doesn't depend on its shape or size. The "shape factor" is put in separately. It's called the "demagnetization factor" for the analogous case of relating magnetization M to the H-field in magnetism, but I forget what it's called for the case of polarization and electric fields. Again, for spheres the factor is 1/3. See, for example: physicspages.com/2012/10/21/… $\endgroup$– user93237Commented Nov 4, 2015 at 18:28