Waves in composites and metamaterials/Anisotropic mass and generalization

The content of these notes is based on the lectures by Prof. Graeme W. Milton (University of Utah) given in a course on metamaterials in Spring 2007.

Rigid Bar with Frequency Dependent Mass
Recall the a rigid bar with $$n$$ cavities, each containing a spring-mass system with mass $$m$$ and complex spring constant $$K$$ (see Figure 1).

The momentum of the bar ($$\widehat{P}$$) is related to its velocity ($$\widehat{V}$$) by the relation

\widehat{P} = M(\omega)~\widehat{V} $$ where $$\omega$$ is the frequency and $$M(\omega)$$ is the  effective mass which is given by

M(\omega) = M_0 + \cfrac{2~K~n~m}{2~K - m~\omega^2} $$ where $$M_0$$ is the mass of the rigid bar~.

Let us now consider a specific model for the springs. A simple model is the one-dimensional Maxwell model shown in Figure 2. In this case instead of a spring with a complex spring constant we have an elastic spring (with real spring constant $$k$$) and a dashpot (with a real viscosity $$\eta$$) in series. Let $$u_1$$ be the displacement of the elastic spring and let $$u_2$$ be that of the dashpot under the action of a force $$f$$.

For the elastic spring, we have
 * $$ \text{(1)} \qquad

u_1(t) = \cfrac{f(t)}{k} ~. $$ For the dashpot, we have
 * $$ \text{(2)} \qquad

\cfrac{d u_2}{d t} = \cfrac{f(t)}{\eta} ~. $$ Once again, assuming that the functions $$f(t)$$, $$u_1(t)$$, and $$u_2(t)$$ can be expressed as harmonic functions, we have
 * $$ \text{(3)} \qquad

u_1(t) = \widehat{u}_1~e^{-i\omega t} ~; u_2(t) = \widehat{u}_2~e^{-i\omega t} ~; f(t) = \widehat{f}~e^{-i\omega t} ~. $$ Plugging equations (3) into equations (1) and (2), we get
 * $$ \text{(4)} \qquad

\widehat{u}_1 = \cfrac{\widehat{f}}{k} \qquad \text{and} \qquad -i\omega~\widehat{u}_2 = \cfrac{\widehat{f}}{\eta} ~. $$ Recall that the displacement $$u(t)$$ of the sphere of mass $$m$$ inside the cavity is related to the applied force $$f(t)$$ by the relation

u(t) = \cfrac{f(t)}{K} \qquad \implies \qquad \widehat{u} = \cfrac{\widehat{f}}{K} ~. $$ Now, $$\widehat{u} = \widehat{u}_1 + \widehat{u}_2$$. Hence,

\cfrac{\widehat{f}}{K} = \cfrac{\widehat{f}}{k} - \cfrac{1}{i\omega}~\cfrac{\widehat{f}}{\eta} $$ or,

{   \cfrac{1}{K} = \cfrac{1}{k} + \cfrac{i}{\omega\eta} ~. } $$ So, with the Maxwell model, the effective mass is

M(\omega) = M_0 + \cfrac{2~n~m}{2 - \cfrac{1}{K}~m~\omega^2} = M_0 + \cfrac{2~n~m}{2 - \cfrac{1}{k}~m~\omega^2 - \cfrac{i}{\omega\eta}~m~\omega^2} $$ or,
 * $$ \text{(5)} \qquad

{ M(\omega) = M_0 + \cfrac{2~n~k}{\cfrac{2~k}{m} - \omega^2 - \cfrac{i\omega~k}{\eta}} ~. } $$ This model is remarkably similar to a simple model for the frequency dependent dielectric constant $$\epsilon(\omega)$$.

Comparison with a simple model for $$\epsilon(\omega)$$
Consider the following simple model of an electron bound to an atom by a harmonic force under the influence of a slowly varying electric field (a detailed description can in found in Jackson75, Sec. 7.5). A schematic of the situation is shown in Figure 3.

Let the electric field be $$E(x,t)$$ and assume that it varies slowly in space over the distance that the electron moves, i.e., $$E(x,t) \approx E(t)$$. Let $$m$$ be the mass of the electron and let $$-e$$ be its charge. Let $$\omega_0$$ be the frequency of the harmonic force binding the electron to the atom and let $$\gamma$$ be a damping coefficient due to interactions with obstacles.

Then, the equation of motion of the electron is given by
 * $$ \text{(6)} \qquad

m \left(\cfrac{d^2 x}{d t^2} + \gamma\cfrac{d x}{d t} + \omega_0^2~x\right) = -e~E(t)~. $$ Let
 * $$ \text{(7)} \qquad

x(t) = \widehat{x}~e^{-i\omega t} \text{and} E(t) = \widehat{E}~e^{-i\omega t} ~. $$ Plugging equations (7) into equation (6), we get

m (-\omega^2 - i\omega\gamma + \omega_0^2) \widehat{x} = - e~\widehat{E} $$ or

\widehat{x} = -\cfrac{e}{m} \left(\omega_0^2 - \omega^2 - i\omega\gamma\right)^{-1} ~\widehat{E} ~. $$ Let $$q(x,t)$$ be the charge density of the electron-atom system, i.e.,

q(x,t) = -\delta[x(t)-x_0]~e + \delta(x_0-x_0)~e $$ where $$\delta(x)$$ is the Dirac delta function and $$x_0$$ is the position of the atom. Then, the dipole moment (the first moment of the charge) contributed by the electron-atom system is given by

p(t) = \int q(x,t)~x(t)~\text{d}x = -e~\left[\int \delta[x(t)-x_0]~x(t)~\text{d}x + \int \delta(0)~x(t)~\text{d}x\right] ~. $$ or,

p(t) = -e~x(t) \qquad \implies \qquad \widehat{p} = -e~\widehat{x} = \cfrac{e^2}{m} \left(\omega_0^2 - \omega^2 - i\omega\gamma\right)^{-1} ~\widehat{E} ~. $$ Suppose now that instead of a binding frequency $$\omega_0$$ for all the atoms and all the electrons in a volume, there are $$N_i$$ atoms per unit volume with binding frequencies $$\omega_i$$ and damping constants $$\gamma_i$$ and. Then, we can write the polarization as
 * $$ \text{(8)} \qquad

\widehat{P} = \cfrac{e^2}{m} \left[\sum_i \cfrac{N_i}{ \omega_0^2 - \omega^2 - i\omega\gamma}\right]~\widehat{E} = \left[\sum_i \cfrac{a_i}{\omega_0^2 - \omega^2 - i\omega\gamma}\right] ~\widehat{E}~. $$ Recall that the electric displacement is related to the electric field and the polarization by
 * $$ \text{(9)} \qquad

\widehat{D} = \epsilon_0~\widehat{E} + \widehat{P} = \epsilon~\widehat{E} ~. $$ Therefore, from equations (8) and (9) we get
 * $$ \text{(10)} \qquad

{ \epsilon = \epsilon_0 + \sum_i \cfrac{a_i}{\omega_0^2 - \omega^2 - i\omega\gamma} ~. } $$ Comparing equation (5) with equation (10) we observe that they have the same form which implies that the effective permittivity is analogous to the effective mass. This also implies that the electric field is analogous to the velocity. Similarly, the polarization is analogous to the momentum.

More on frequency dependence of the mass
More generally, we get a frequency dependent density if all the constituents do not move in lock step. Lock step motion almost never occurs because there are thermal vibrations at the microscale. There are also many macroscopic situations in which lock step motion does not occur. Consider the example of a porous rock containing some water (see Figure 4). Both the rock grains and the water are connected, though this is not obvious from the figure

In this case, the water will move with a different frequency that the rock and the density of the composite will be dependent on the frequency.

At a molecular level, we can have a crystal with lead atoms attached by single bonds to the structure (see Figure 5). Presumably, the resonant frequency of such molecules is very high. so we will see the frequency dependence of the mass only at very high frequencies.

In fact, Sheng et al. Sheng03 have shown in experiments that materials indeed have frequency dependent masses. An example of such a material is shown in Figure 6.

Generalizations of the Rigid Bar Model
Consider the rigid body containing cavities shown in Figure~7. This is just a two-dimensional extension of the model shown in Figure~1. Here $$K$$ and $$L$$ are complex spring constants in the $$x$$ and $$y$$ directions.

In this case, the effective mass along the $$x$$-direction is given by
 * $$ \text{(11)} \qquad

M_x(\omega) = M_0 + \cfrac{2~K~n~m}{2~K - m~\omega^2} $$ while that along the $$y$$-direction is given by
 * $$ \text{(12)} \qquad

M_y(\omega) = M_0 + \cfrac{2~L~n~m}{2~L - m~\omega^2}~. $$ In matrix form, we then have

{ \left[\mathsf{M}\right](\omega) = M_0~\left[\mathsf{I}\right] + n~m~ \begin{bmatrix} \cfrac{2~K}{2~K - m~\omega^2} & 0 \\ 0 & \cfrac{2~L}{2~L - m~\omega^2} \end{bmatrix} ~. } $$ Hence, the effective mass is clearly  anisotropic. Note however that from a macroscopic perspective it is not the average velocity in the matrix which is important. In fact, such a quantity does not even make sense because the velocity is not defined in the void phase. Rather it is the velocity of the matrix that is the relevant quantity in this model.

One can generalize the model one step further by having the springs be oriented at different angles to each other as shown in Figure 8. Also let the springs in each cavity have different spring constants and the masses in each cavity are different.

In this case, let $$\boldsymbol{R}$$ be the rotation that is needed to orient each set of springs with the $$x$$ and $$y$$ axes. Then, if $$\left[\mathsf{R}\right]_j$$ is the rotation matrix for cavity $$j$$ containing a mass $$m_j$$, and if $$K_j$$ and $$L_j$$ are the complex spring constants for that cavity, the effective mass can be written as

{ \left[\mathsf{M}\right](\omega) = M_0~\left[\mathsf{I}\right] + \sum_{j=1}^n \left[\mathsf{R}\right]_j^T~ \begin{bmatrix} \cfrac{2~K_j~m_j}{2~K_j - m_j~\omega^2} & 0 \\ 0 & \cfrac{2~L_j~m_j}{2~L_j - m_j~\omega^2} \end{bmatrix} \left[\mathsf{R}\right]_j ~. } $$ The eigenvalues of $$\boldsymbol{M}(\omega)$$ can therefore depend on $$\omega$$.

General form of $$\boldsymbol{M}(\omega)$$
Take any body with a rigid matrix. Suppose, instead of applying harmonically varying velocities, we apply a time varying velocity $$\mathbf{V}(t)$$ and observe what the momentum $$P(t)$$ is. There will be some linear constitutive relation
 * $$ \text{(13)} \qquad

\mathbf{P}(t) = \int_{-\infty}^{\infty} \boldsymbol{H}(t - \tau)\cdot\mathbf{V}(\tau)~\text{d}\tau ~. $$ The kernel $$\boldsymbol{H}(t-\tau)$$ is second-order tensor valued and may possibly be singular. Also, since both $$\mathbf{P}$$ and $$\mathbf{V}$$ are physical and real, $$\boldsymbol{H}$$ must be real. Causality implies that $$\boldsymbol{H}(s) = \boldsymbol{\mathit{0}}$$ when $$s = t - \tau < 0$$ (or $$t < \tau$$) since the inertial force cannot depend on velocities in the future.

Taking the Fourier transform of equation (13) and using the convolution theorem, we get
 * $$ \text{(14)} \qquad

\widehat{\mathbf{P}}(\omega) = \boldsymbol{M}(\omega)\cdot\widehat{\mathbf{V}}(\omega) $$ where
 * $$ \text{(15)} \qquad

\boldsymbol{M}(\omega) := \widehat{\boldsymbol{H}}(\omega) = \int_{-\infty}^{\infty} \boldsymbol{H}(s)~e^{i\omega s}~\text{d}s ~. $$ The quantity $$\boldsymbol{M}(\omega)$$ can be shown to satisfy the Cauchy-Riemann equations only if $$\text{Im}(\omega) > 0$$. This is a consequence of causality $$s \ge 0$$ and the fact that the integral in equation (15) only converges in the upper half of the complex $$\omega$$ plane. \footnote{ Note the positive sign in the power of $$i\omega s$$ in equation (15). This occurs because we have chosen the inverse Fourier transform to be of the form

g(s) = \cfrac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \widehat{g}(\omega)~e^{-i\omega s}    ~\text{d}\omega ~. $$ The effect of such a choice is that the imaginary part of $$\omega$$ is positive instead of negative.} Hence, $$\boldsymbol{M}(\omega)$$ is analytic in $$\omega$$ when $$\text{Im}(\omega) > 0$$. The quantity $$\boldsymbol{H}(s)$$ is real.

Now, the complex conjugate of $$\boldsymbol{M}(\omega)$$ is given by
 * $$ \text{(16)} \qquad

\boldsymbol{M}^{*}(\omega) = \int_{-\infty}^{\infty} \boldsymbol{H}(s)~e^{-i\omega^{*} s}~\text{d}s = \boldsymbol{M}(-\omega^{*}) $$ where $$z^{*}$$ denotes the complex conjugate of $$z$$ (for any complex $$z$$).

Also assume that, for large enough frequencies, the dynamic mass tends toward the static mass, i.e.,
 * $$ \text{(17)} \qquad

\lim_{\omega\rightarrow\infty} \boldsymbol{M}(\omega) = M_0~\boldsymbol{\mathit{1}} ~. $$

Equation (15) can be used to establish the Kramers-Kronig equations for the material. To do that, recall Cauchy's formula for a function $$f$$ which is analytic on a domain that is enclosed in a piecewise smooth curve $$C$$:

f(z) = \cfrac{1}{2\pi i} \oint_C \cfrac{f(\zeta)}{\zeta - z}~\text{d}\zeta ~. $$ Since the function $$\boldsymbol{M}(\omega)$$ is analytic on the upper-half $$\omega$$ plane, for any point $$z$$ in a closed contour $$C$$ in the upper-half $$\omega$$ plane, we have
 * $$ \text{(18)} \qquad

\boldsymbol{M}(z) - M_0~\boldsymbol{\mathit{1}} = \cfrac{1}{2\pi i}       \oint_C \cfrac{\boldsymbol{M}(\omega') - M_0~\boldsymbol{\mathit{1}}}{\omega' - z}~\text{d}\omega' ~. $$ Let us now choose the contour $$C$$ such that it consists of the real $$\omega$$ axis and a great semicircle at infinity in the upper half plane (see Figure~9).

Also, from equation (17) we observe that

\boldsymbol{M}(\omega) - M_0~\boldsymbol{\mathit{1}} = \boldsymbol{0} \qquad \text{as} \qquad \omega \rightarrow \infty~. $$ Hence, there is no contribution to the integral in equation (18) due to the semicircular part of the contour and we just have to perform an integration only over the real line:
 * $$ \text{(19)} \qquad

\boldsymbol{M}(z) = M_0~\boldsymbol{\mathit{1}} + \cfrac{1}{2\pi i}       \int_{-\infty}^{\infty} \cfrac{\boldsymbol{M}(\omega') - M_0~\boldsymbol{\mathit{1}}}{\omega' - z}~\text{d}\omega' ~. $$

Now, let us consider the integral

\boldsymbol{I} = \int_{-\infty}^{\infty} \cfrac{\boldsymbol{M}(\omega') - M_0~\boldsymbol{\mathit{1}}}{\omega' - z}~\text{d}\omega' ~. $$ There is a pole at the point $$\omega' = z$$ (in the figure, the contour is shown as a semicircle of radius $$\varepsilon$$ centered at the pole). In the limit $$\varepsilon \rightarrow 0$$, this integral may be interpreted to mean the Cauchy principal value (not the same as principal values in complex analysis)

\boldsymbol{I} = P \int_{-\infty}^{\infty} \cfrac{\boldsymbol{M}(\omega') - M_0~\boldsymbol{\mathit{1}}}{\omega' - z}~\text{d}\omega' = \int_{-\infty}^{z-\varepsilon} \cfrac{\boldsymbol{M}(\omega') - M_0~\boldsymbol{\mathit{1}}}{\omega' - z}~\text{d}\omega' + \int_{z-\varepsilon}^{\infty} \cfrac{\boldsymbol{M}(\omega') - M_0~\boldsymbol{\mathit{1}}}{\omega' - z}~\text{d}\omega' ~. $$ From Figure~9 we observe that this integral can be broken up into the integral over the path $$A$$ minus the sum of the integrals over the paths $$B$$ and $$D$$. From the Cauchy-Goursat theorem, the integral over the closed path $$A$$ is zero. We have also seen that since $$\boldsymbol{M}(\omega) - M_0\boldsymbol{\mathit{1}} \rightarrow 0$$ as $$\omega \rightarrow \infty$$, the integral over $$B$$ is zero. The integral over the path $$D$$ around the pole is obtained from the Residue Theorem (where the value is divided by two because the integral is over a semicircle), i.e.,

\oint_{D} \cfrac{\boldsymbol{M}(\omega') - M_0~\boldsymbol{\mathit{1}}}{\omega' - z}~\text{d}\omega' = -\pi i~[\boldsymbol{M}(z) - M_0\boldsymbol{\mathit{1}}] ~. $$ The negative sign arises because the curve is traversed in the counter-clockwise direction.

Therefore,

\boldsymbol{I} = P \int_{-\infty}^{\infty} \cfrac{\boldsymbol{M}(\omega') - M_0~\boldsymbol{\mathit{1}}}{\omega' - z}~\text{d}\omega' = \pi i~[\boldsymbol{M}(z) - M_0\boldsymbol{\mathit{1}}] ~. $$ or,

\boldsymbol{M}(z) = M_0~\boldsymbol{\mathit{1}} + \cfrac{1}{\pi i} ~ P \int_{-\infty}^{\infty} \cfrac{\boldsymbol{M}(\omega') - M_0~\boldsymbol{\mathit{1}}}{\omega' - z}~\text{d}\omega' ~. $$ Letting $$z = \omega + i\varepsilon$$ and taking the limit as $$\varepsilon \rightarrow 0$$, we get
 * $$ \text{(20)} \qquad

{ \boldsymbol{M}(\omega) = M_0~\boldsymbol{\mathit{1}} + \cfrac{1}{\pi i}~ P \int_{-\infty}^{\infty} \cfrac{\boldsymbol{M}(\omega') - M_0~\boldsymbol{\mathit{1}}}{\omega' - \omega}~\text{d}\omega' ~. } $$ Note that, in the above equation, both $$\omega$$ and $$\omega'$$ are real. Expanding equation (20) into real and imaginary parts and collecting terms, we get the first form of the Kramers-Kronig relations
 * $$ \text{(21)} \qquad

{ \begin{align} \text{Re}[\boldsymbol{M}(\omega)] & = M_0~\boldsymbol{\mathit{1}} + \cfrac{1}{\pi}~ P \int_{-\infty}^{\infty} \cfrac{\text{Im}[\boldsymbol{M}(\omega')]}{\omega' - \omega}~\text{d}\omega' \\ \text{Im}[\boldsymbol{M}(\omega)] & = - \cfrac{1}{\pi}~ P \int_{-\infty}^{\infty} \cfrac{\text{Re}[\boldsymbol{M}(\omega')] - M_0~\boldsymbol{\mathit{1}}} {\omega' - \omega}~\text{d}\omega' ~. \end{align} } $$ Therefore, the real part of the frequency dependent mass can be determined if we know the imaginary part and vice versa.

We can also eliminate the negative frequencies from equations (21). Recall from equation (16) that

\boldsymbol{M}^{*}(\omega) = \boldsymbol{M}(-\omega^{*}) ~. $$ Since, in equations (21), $$\omega$$ and $$\omega'$$ are real, we have

\boldsymbol{M}^{*}(\omega) = \boldsymbol{M}(-\omega) ~. $$ This implies that

\text{Re}[\boldsymbol{M}(\omega)] = \text{Re}[\boldsymbol{M}(-\omega)] \qquad \text{and} \qquad \text{Im}[\boldsymbol{M}(\omega)] = -\text{Im}[\boldsymbol{M}(-\omega)] ~. $$ Consider the first of equations (21). We can write this relation as

\begin{align} \text{Re}[\boldsymbol{M}(\omega)] & = M_0~\boldsymbol{\mathit{1}} + \cfrac{1}{\pi}~\left\{ P \int_{-\infty}^{0} \cfrac{\text{Im}[\boldsymbol{M}(\omega')]}{\omega' - \omega}~\text{d}\omega' + P \int_0^{\infty} \cfrac{\text{Im}[\boldsymbol{M}(\omega')]}{\omega' - \omega}~\text{d}\omega' \right\}\\ & = M_0~\boldsymbol{\mathit{1}} + \cfrac{1}{\pi}~\left\{ P \int_0^{\infty} \cfrac{\text{Im}[\boldsymbol{M}(-\omega')]}{-\omega' - \omega}~\text{d}\omega' + P \int_0^{\infty} \cfrac{\text{Im}[\boldsymbol{M}(\omega')]}{\omega' - \omega}~\text{d}\omega' \right\}\\ & = M_0~\boldsymbol{\mathit{1}} + \cfrac{1}{\pi}~\left\{ -P \int_0^{\infty} \cfrac{-\text{Im}[\boldsymbol{M}(\omega')]}{\omega' + \omega}~\text{d}\omega' + P \int_0^{\infty} \cfrac{\text{Im}[\boldsymbol{M}(\omega')]}{\omega' - \omega}~\text{d}\omega' \right\}\\ & = M_0~\boldsymbol{\mathit{1}} + \cfrac{1}{\pi}~ P \int_0^{\infty} \text{Im}[\boldsymbol{M}(\omega')]\left[\cfrac{1}{\omega' + \omega}+ \cfrac{1}{\omega'-\omega}\right]~\text{d}\omega' \\ & = M_0~\boldsymbol{\mathit{1}} + \cfrac{2}{\pi}~ P \int_0^{\infty} \cfrac{\omega'~\text{Im}[\boldsymbol{M}(\omega')]}{\omega'^2 - \omega^2} ~\text{d}\omega' ~. \end{align} $$ Similarly, the second of equations (21) may be written as

\begin{align} \text{Im}[\boldsymbol{M}(\omega)] & = - \cfrac{1}{\pi}~\left\{ P \int_{-\infty}^{0} \cfrac{\text{Re}[\boldsymbol{M}(\omega')] - M_0~\boldsymbol{\mathit{1}}} {\omega' - \omega}~\text{d}\omega' + P \int_0^{\infty} \cfrac{\text{Re}[\boldsymbol{M}(\omega')] - M_0~\boldsymbol{\mathit{1}}} {\omega' - \omega}~\text{d}\omega' \right\} \\ & = - \cfrac{1}{\pi}~\left\{ -P \int_0^{\infty} \cfrac{\text{Re}[\boldsymbol{M}(-\omega')] - M_0~\boldsymbol{\mathit{1}}} {\omega' + \omega}~\text{d}\omega' + P \int_0^{\infty} \cfrac{\text{Re}[\boldsymbol{M}(\omega')] - M_0~\boldsymbol{\mathit{1}}} {\omega' - \omega}~\text{d}\omega' \right\} \\ & = - \cfrac{1}{\pi}~\left\{ P \int_0^{\infty} (\text{Re}[\boldsymbol{M}(-\omega')] - M_0~\boldsymbol{\mathit{1}})\left[ -\cfrac{1}{\omega' + \omega} + \cfrac{1} {\omega' - \omega}\right]~\text{d}\omega' \right\} \\ & = - \cfrac{2~\omega}{\pi}~\left\{ P \int_0^{\infty} \cfrac{\text{Re}[\boldsymbol{M}(-\omega')] - M_0~\boldsymbol{\mathit{1}}} {\omega'^2 - \omega^2}~\text{d}\omega' \right\} ~. \end{align} $$ Therefore, the alternative form of the Kramers-Kronig relations is
 * $$ \text{(22)} \qquad

{ \begin{align} \text{Re}[\boldsymbol{M}(\omega)] & = M_0~\boldsymbol{\mathit{1}} + \cfrac{2}{\pi}~ P \int_0^{\infty} \cfrac{\omega'~\text{Im}[\boldsymbol{M}(\omega')]}{\omega'^2 - \omega^2} ~\text{d}\omega' \\ \text{Im}[\boldsymbol{M}(\omega)] & = - \cfrac{2~\omega}{\pi}~\left\{ P \int_0^{\infty} \cfrac{\text{Re}[\boldsymbol{M}(-\omega')] - M_0~\boldsymbol{\mathit{1}}} {\omega'^2 - \omega^2}~\text{d}\omega' \right\} ~. \end{align} } $$

Significance of $$\text{Im}[\boldsymbol{M}(\omega)]$$
Consider harmonically varying $$\mathbf{F}(t)$$ and $$\mathbf{V}(t)$$ given by
 * $$ \text{(23)} \qquad

\mathbf{F}(t) = \text{Re}(\widehat{\mathbf{F}}~ e^{-i\omega t}) \quad \text{and} \quad \mathbf{V}(t) = \text{Re}(\widehat{\mathbf{V}}~ e^{-i\omega t})~. $$ Alternatively, we may write these as

\mathbf{F}(t) = \text{Re}(\widehat{\mathbf{F}})\cos(\omega t) + \text{Im}(\widehat{\mathbf{F}})\sin(\omega t) ~; \mathbf{V}(t) = \text{Re}(\widehat{\mathbf{V}})\cos(\omega t) + \text{Im}(\widehat{\mathbf{V}})\sin(\omega t) ~. $$ Then,

\widehat{\mathbf{F}} = -i\omega\boldsymbol{M}(\omega)\cdot\widehat{\mathbf{V}} $$ which implies that

\begin{align} \text{Re}(\widehat{\mathbf{F}}) & = \omega~[\text{Re}(\boldsymbol{M})\cdot\text{Im}(\widehat{\mathbf{V}}) + \text{Im}(\boldsymbol{M})\cdot\text{Re}(\widehat{\mathbf{V}})] \\ \text{Im}(\widehat{\mathbf{F}}) & = \omega~[\text{Im}(\boldsymbol{M})\cdot\text{Im}(\widehat{\mathbf{V}}) - \text{Re}(\boldsymbol{M})\cdot\text{Re}(\widehat{\mathbf{V}})] ~. \end{align} $$ The average rate of work done on the system in a cycle of oscillation will be

\begin{align} W & = \cfrac{\omega}{2~\pi} \int_{0}^{2\pi/\omega}~\mathbf{F}(t)\cdot\mathbf{V}(t)~\text{d}t \\ & = \cfrac{\text{Re}(\widehat{\mathbf{F}})\cdot\text{Re}(\widehat{\mathbf{V}}) + \text{Im}(\widehat{\mathbf{F}})\cdot\text{Im}(\widehat{\mathbf{V}})}{2} \\ & = \omega~[\text{Re}(\widehat{\mathbf{V}})\cdot\text{Im}[\boldsymbol{M}(\omega)]\cdot\text{Re}(\widehat{\mathbf{V}}) + \text{Im}(\widehat{\mathbf{V}})\cdot\text{Im}[\boldsymbol{M}(\omega)]\cdot\text{Im}(\widehat{\mathbf{V}})]. \end{align} $$ This quadratic form will be non-negative for all choices of $$\widehat{\mathbf{V}}$$ if and only if $$\text{Im}(\boldsymbol{M}(\omega))$$ is positive semidefinite for all real $$\omega>0$$. Note that the quadratic form does not contain $$\text{Re}{\boldsymbol{M}}$$. Since the work done in a cycle should be zero in the absence of dissipation, this implies that the imaginary part of the mass is connected to the energy dissipation (for instance, into heat).