3
$\begingroup$

In Photonic Devices by Jia-Ming-Liu, it is mentioned that magnetization at optical frequencies is zero. I know that electrons behave like an oscillator in EM fields and the amplitude of oscillation as : $\frac{F_0}{m((\omega_0-\omega_d)^2+\gamma^2\omega_d^2)^{1/2}}$ and decreases as the frequency of forcing function increases, so is the case ? i.e the natural frequency of oscillation is low for the electron in materials compared to optical frequencies?

At optical frequencies, the magnetization vanishes, M = 0. Consequently, for optical fields, the following relation is always true: B(r, t) = μ0H(r, t).

$\endgroup$

3 Answers 3

3
$\begingroup$

As Pieter said, there are two main reasons for vanishing magnetization at optical frequencies. I will elaborate on this a little more.

Amplitude: Without an EM field, the net magnetization of a magnetic material will be zero because magnetic domains will form. All these domains will have different orientations which results in zero net magnetization. Net magnetization is obtained when all domains are aligned in the same direction. The field required for this alignment is the coercive field. As Pieter said, the magnetic field inside the optical EM wave is typically not as large as the coercive field. Hence, it cannot align the magnetization.

Frequency: The resonance frequencies of ferromagnetic materials are in the GHz-range. At much higher frequencies, the magnetization can simply not follow the field anymore. Hence, at optical frequencies, the dynamic response of the magnetization is very small.

$\endgroup$
2
$\begingroup$

For sizeable magnetization one needs ferromagnetism: domains of ordered moments. At optical frequencies, fields are smaller than coercive fields. And domain walls could not move that fast anyway.

But there are ferrites for radio frequencies.

$\endgroup$
0
1
$\begingroup$

The magnetic permeability approaches unity at high frequencies because magnetic moments cannot follow the EM field at such frequencies.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.