Linear force operator

Using Eqs. (36) and (37) to eliminate $p_1$ and $\mathbf{B}_1$ from Eq. (35), we obtain

$$- \omega^2 \rho_0 \ensuremath{\boldsymbol{\xi}}=\mathbf{F} (\ensuremath{\boldsymbol{\xi}}),$$ (47)

where $\mathbf{F} (\ensuremath{\boldsymbol{\xi}})$, the linear force operator, is given by

$$\mathbf{F} (\ensuremath{\boldsymbol{\xi}}) = \nabla (\ensuremath{... ...tmu}_0 \times \nabla \times (\ensuremath{\boldsymbol{\xi}} \times \mathbf{B}_0)$$ (48)

It can be proved that the linear force operator $\mathbf{F} (\ensuremath{\boldsymbol{\xi}})$ is self-adjoint (or Hermitian) (I have not proven this), i.e., for any two functions $\ensuremath{\boldsymbol{\eta}}_1$ and $\ensuremath{\boldsymbol{\eta}}_2$ that satisfy the same boundary condition, we have

$$\int \ensuremath{\boldsymbol{\eta}}_2 \cdot \mathbf{F} (\ensurema... ...\boldsymbol{\eta}}_1 \cdot \mathbf{F} (\ensuremath{\boldsymbol{\eta}}_2) d^3 x.$$ (49)

As a consequence of the self-adjointness, the eigenvalue, $\omega^2$, must be a real number, which implies that $\omega$ itself is either purely real or purely imaginary. [Proof: Taking the complex conjugate of Eq. (47), we obtain

$$- \rho_0 (\omega^2 \ensuremath{\boldsymbol{\xi}})^{\star} = [\mathbf{F}(\ensuremath{\boldsymbol{\xi}})]^{\star} .$$ (50)

Note that the expression of $\mathbf{F} (\ensuremath{\boldsymbol{\xi}})$ given in Eq. (48) have the property $[\mathbf{F}(\ensuremath{\boldsymbol{\xi}})]^{\star} =\mathbf{F} (\ensuremath{\boldsymbol{\xi}}^{\star})$, Using this, equation (50) is written

$$- \omega^{2 \star} \rho_0 \ensuremath{\boldsymbol{\xi}}^{\star} =\mathbf{F} (\ensuremath{\boldsymbol{\xi}}^{\star}),$$ (51)

Taking the scalar product of both sides of the above equation with $\ensuremath{\boldsymbol{\xi}}$ and integrating over the entire volume of the system, gives

$$- \omega^{2 \star} \int \rho_0 \ensuremath{\boldsymbol{\xi}}^{\st... ...ensuremath{\boldsymbol{\xi}}^{\star}) \cdot \ensuremath{\boldsymbol{\xi}}d^3 x,$$ (52)

Using the self-ajointness of $\mathbf{F}$, the above equation is written

$$- \omega^{2 \star} \int \rho_0 \ensuremath{\boldsymbol{\xi}}^{\st... ...nsuremath{\boldsymbol{\xi}}) \cdot \ensuremath{\boldsymbol{\xi}}^{\star} d^3 x,$$ (53)

$$\Rightarrow - \omega^{2 \star} \int \rho_0 \ensuremath{\boldsymbo... ...ensuremath{\boldsymbol{\xi}} \cdot \ensuremath{\boldsymbol{\xi}}^{\star} d^3 x,$$ (54)

$$\Rightarrow (\omega^2 - \omega^{2 \star}) \int \rho_0 \ensuremath{\boldsymbol{\xi}} \cdot \ensuremath{\boldsymbol{\xi}}^{\star} d^3 x = 0$$ (55)

Since $\int \rho_0 \ensuremath{\boldsymbol{\xi}} \cdot \ensuremath{\boldsymbol{\xi}}^{\star} d^3 x$ is non-zero for any non-trivial eigenfunction, it follows from Eq. (55) that $\omega^2 = \omega^{2 \star}$, i.e., $\omega^2$ must be a real number, which implies that $\omega$ is either purely real or purely imaginary.] It can also be proved that two eigenfunctions with different $\omega^2$ are orthogonal to each other. [Proof:

$$- \omega^{2 \star}_m \rho_0 \ensuremath{\boldsymbol{\xi}}^{\star}_m =\mathbf{F} (\ensuremath{\boldsymbol{\xi}}^{\star}_m),$$ (56)

$$- \omega^2_n \rho_0 \ensuremath{\boldsymbol{\xi}}_n =\mathbf{F} (\ensuremath{\boldsymbol{\xi}}_n),$$ (57)

Taking the scalar product of both sides of Eq. (56) with $\ensuremath{\boldsymbol{\xi}}_n$ and integrating over the entire volume of the system, gives

$$\Rightarrow - \omega^{2 \star}_m \int \rho_0 \ensuremath{\boldsym... ...emath{\boldsymbol{\xi}}^{\star}_m) \cdot \ensuremath{\boldsymbol{\xi}}_n d^3 x,$$ (58)

Taking the scalar product of both sides of Eq. (57) with $\ensuremath{\boldsymbol{\xi}}_m^{\star}$ and integrating over the entire volume of the system, gives

$$\Rightarrow - \omega^2_n \int \rho_0 \ensuremath{\boldsymbol{\xi}... ...emath{\boldsymbol{\xi}}_n) \cdot \ensuremath{\boldsymbol{\xi}}^{\star}_m d^3 x,$$ (59)

Combining the above two equations, we obtain

$$(\omega^{2 \star}_m - \omega^2_n) \int \rho_0 \ensuremath{\boldsymbol{\xi}}^{\star}_m \cdot \ensuremath{\boldsymbol{\xi}}_n d^3 x = \int [\mathbf{F}(\ensuremath{\boldsymbol{\xi}}_n) \cdot \ensurema... ...emath{\boldsymbol{\xi}}^{\star}_m) \cdot \ensuremath{\boldsymbol{\xi}}_n] d^3 x$$ (60)

Using the self-ajointness of $\mathbf{F}$, we know the right-hand side of Eq. (60) is zero. Thus Eq. (60) is written

$$(\omega^{2 \star}_m - \omega^2_n) \int \rho_0 \ensuremath{\boldsymbol{\xi}}^{\star}_m \cdot \ensuremath{\boldsymbol{\xi}}_n d^3 x = 0.$$ (61)

Using $\omega^{2 \star}_m = \omega_m^2$, the above equation is written

$$(\omega^2_m - \omega^2_n) \int \rho_0 \ensuremath{\boldsymbol{\xi}}^{\star}_m \cdot \ensuremath{\boldsymbol{\xi}}_n d^3 x = 0.$$ (62)

Since we assume $\omega_m^2 \neq \omega_n^2$, the above equation reduces to

$$\int \rho_0 \ensuremath{\boldsymbol{\xi}}^{\star}_m \cdot \ensuremath{\boldsymbol{\xi}}_n d^3 x = 0,$$ (63)

i.e., $\ensuremath{\boldsymbol{\xi}}_m^{\star}$ and $\ensuremath{\boldsymbol{\xi}}_n$ are orthogonal to each other.]