The issue is that with shapes other than ellipsoids, the polarizability shall be non-uniform. (only with ellipsoids is the additional field created by the surface charges of the polarized body is uniformity preserved). Polarization then is not only a function of the applied field but also depends on the coordinates within the particle. The polarization tensor is then not universally defineable for the entire particle but itself a function of spatial coordinates.

To prove your point you would first have to define a quantity that represents the particle as a whole (and show that it makes sense).

Btw., the same applies to magnetic susceptibility and the demagnetization tensor.

Hi Kai Fauth,

You are absolutely right. For an arbitrary geometry, the charge density distribution would be sophisticated, and the near-field as well. Still, I am interested in the far-field scattering. If the particle is uncharged, the far field has predominantly dipolar component, defined by the total dipolar moment of the particle (integral rho*r dr). This quantity is a single vector, and it depends on the applied external field. If I look at the Taylor expansion of that dependency, zero order term is zero (I consider particles that have no dipolar moment without applied field), and the first-order term is linear in the applied field. It it this tensor coefficient, a complex matrix 3x3, which I am thinking about. For an ellipsoidal particle, it has been calculated from first principles. I want to claim that even for an arbitrary particle it is always possible to find orientation(s), for which the induced dipolar moment is parallel to the applied field.

So, you are completely right that near-field distribution is complex. But I am struggling to find a general principle that governs the symmetry of total polarizability of the particle.

Thank you for valuable discussion!

OK, I understand your goal now and have to admit that while it sounds pretty logical, I cannot come up with a valid strategy to demonstrate the point you want to make.

Some brainstorming (no guarantee of usefulness): it might be useful to start with some homogeneous material and assume linear response. For every point in the volume of the particle there will be a local polarization given by polarizability times (1 minus (local) depolarization tensor) times external field. I could imagine that a transformation to e.g. the center of mass could be done, resulting in some kind of multipole expansion for each term. Maybe the total dipolar moment is then the integral over just the dipole term(s) of that expansion? (Higher multipoles should die off towards the far field more quickly than does the dipolar term). So if you could construct from this an effective depolarization/polarization tensor, then the problem should be diagonizeable and you'd have made the point, wouldn't you.

The above was a pretty much constructive kind of approach. Maybe it's easier to disprove the opposite assumption by some kind of argument that the response (total dipole moment) must be continous under infinitesimal changes of the external field direction. So if in some sense there is a kind of "positive" and "negative" contribution (suitably defined) to the deviation of collinearity between external field and induced dipole moment for two different particle orientations, then there must be an intermediate position with zero deviation (easy in 1D, needs translation to 2D of course). Maybe the fact that the trace of the depolarization tensor is zero for every point in the particle volume (is it in the general case ?!) could be the starting point for such an analysis.

Good luck!!

I just got an idea. If the total dipolar moment is not parallel to E, then the particle will experience a torque. If we allow it to rotate freely with friction, it will sooner or later find an equilibrium, when p||E. So, there must be at least one axis with real eigenvalue. Still to prove that two other axes can be found.

There is an error in what I wrote above. The trace of the depolarisation tensor is not zero but one! However,  it can be decomposed into 1/3 times the unity tensor plus a term with zero trace. The structure of the ansatz thus is still as described above.

The remaining term is (I guess)  a symmetrical tensor (still depending on coordinates). And yes, the problem seems equivalent to an eigenvalue problem. How about the eigenvalues of symmetrical tensors?! The symmetry property shall not change as a function of spatial coordinates, therefore I assume that you can use this property to derive that also the integration over the particle volume leaves this property untouched.

With all this I am sure you will be able to find that the eigenvalue problem has three real solutions.

Note added: the corresponding eigenvectors are linearly i dependent and hence can be chosen to be orthogonal.  Just as with ellipsoids.

I'll say that for an arbitrary shape, the tensor is in general not symmetric so not diagonalizable.

I'll come back when i get a proof ! For now, this guess stands on the basis of a dumb numerical simulation of the polarization of a screw-like particle, that gives a clearly non-symmetric response.

Yes one real eigenvector always exists for a 3x3 non-degenerate matrix, but the other two can still be complex, unless some special properties of the polarization tensor.

I don't know anything about "polarizability tensor" but if you have a 2-tensor, and a metric (I assume there is one), there is an isomorphism between this 2-tensor and a vector space operator. Whether such operators are diagonalizable depends on whether you want to diagonalise over the complex numbers, or the reals, and also whether the allowable transformations are just orthogonal, or more general linear basis changes. If it is symmetric (with respect to the metric), it clear it is diagonalizable using orthogonal matrices. If you allow more general basis changes, then if you have all real and distinct eigenvalues, then it is diagonizable. If you allowing for complex basis changes, then distinct eigenvalues is sufficent for diagonalizability (so 1 real and two complex conjugate eigenvalues implies diagonalizablity over the complex numbers). This can be seen by considering the Jordan canonical form. The case when two (or more) is equal, is a bit more hairy, but Jordan canonical form is the key.

Giorgio, if you have screw-like symmetry (i.e. handedness is inherent to the material, e.g. something magnetized) then the "natural" states of polarization to use would not be x,y,and z, but L,R and z  (or, +,-,z) i.e. left & right circular (propagation along axis defining handedness and linear along this axis). My guess would be that by transforming to this basis, real Eigenvalues will be obtained, while when working in x,y,z basis, one real eigenvalue and a complex pair will emerge. Does this mean that the tensor is antisymmetric?