A resonator is a device that can store wave energy. The waves have an amplitude and frequency. If you try to input a wave of arbitrary frequency, then it will be almost totally reflected. If part of the wave is not reflected, its amplitude will quickly be attenuated to zero. The waves can be electromagnetic with frequency all the way from radio frequency to x-ray frequency. Or they can be acoustic (sound or vibrational) waves of any frequency. Here we will discuss waves like those of visible light which are usually considered to have an index of refraction. Resonators for visible light are part of a laser device. An example would be a solid state laser diode which is a staple of our modern technology. Here we have aa transparent material (called a slab since this is a one dimension resonator) partially filling the resonator. Our variables will be the material's index of refraction and its thickness.
Here we have a dielectric slab of thickness, `t_s`, of material centered between perfectly reflecting mirrors at both ends of a linear 1D resonator of width `L` (see Canvas 0).The Maxwell differential equation for the electric field, `vecE` is: (See Appendix):
`-1/n^2(del^2E(x))/(delx^2)=(k_r)^2E(x)`
where `n` is the index of refraction of the slab and `k_r` is the effective wave vector of the resonator. Obviously `E(x)`, when doubly differentiated, is a self repeating function. The only example of a function like this is the exponential. And since `n` is real, the real versions of the exponential are the sine and cosine. The cosine will be symmetric about the center of the resonator and the sine will be antisymmetric.
From the above discussion we may conclude the following:
The expressions for `E(x)` in the resonator are:
Inside the slab:
`E_s(x)=Scos(k_sx)\ \ \ symmetric`
`E_s(x)=Ssin(k_sx)\ \ \ antisymmetric`
`E_v(x)=Vcos(k_vx) \ \ \ symmetric`
`E_v(x)=Vsin(k_vx) \ \ \ antisymmetric`
`Vcos(k_vL/2)=0 \ \ \ symmetric `
`Vsin(k_vL/2)=0 \ \ \ antisymmetric`
`-(d^2E_s)/(dx^2)=k_s^2E_s` `-(d^2E_v)/(dx^2)=k_v^2E_v`
Repeating the wave equation we have:`-1/n^2(d^2E(x))/(dx^2)=k_r^2E(x)`
Then`k_s^2/n^2-k_r^2=0` `k_v^-k_r^2=0`
which just results in the following.
`k_s=nk_r`
`k_v=k_r`
`(n_s)*k_rt_s+n_vk_r(L-t_s)=(2m+1)pi \ \ \ symmetric`
`(n_s)*k_rt_s+n_vk_r(L-t_s)=(2m)pi \ \ \ antisymmetric`
`kr=(2m+1)pi/((n_s-1)t_s+L) \ \ \ symmetric`
`kr=(2m)pi/((n_s-1)t_s+L) \ \ \ antisymmetric`
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`grad xx vec H=(delD)/(delt)\ \ \ \ (1)`
Also, if there are no magnetic materials in proximity:`grad xx vec E=-(delB)/(delt) \ \ \ (2)`
where `vec B=mu_0vecH` and `mu_0` is the magnetic permeability of vacuum. Takine the curl of both sides of equation 2 we get:`grad xx grad xx vec E=-mu_0(del)/(delt)(gradxxvecH) \ \ \ (3)`
Using equation 1 in equation 3 we have:`grad xx grad xx vec E=-mu_0(del^2)/(delt^2)(D)=-mu_0epsilon(del^2)/(delt^2)(vecE) \ \ \ (4)`
The only vector variable in equation 4 is `vec E`. Equation 4 is called the wave equation for the `vecE`.An identity for `grad xx grad xx vec E` is:
`grad xx grad xx vec E=-grad * gradE\ \ \ \ (5)`
Then:`grad * gradE=mu_0epsilon(del^2)/(delt^2)(vecE) \ \ \ (6)`
Since `epsilon` may be a function of `x` we will multiply both sides of the equation by `epsilon_0/epsilon` and get:`epsilon_0/epsilon grad * gradE=mu_0epsilon_0(del^2)/(delt^2)(vecE) \ \ \ (7)`
A common name for `epsilon/epsilon_0` is the square of the index of refraction, `n^2`, and we already gave the name for `1/(mu_0epsilon_0)` as `c^2` so we use these symbols in equation 7:`1/n^2 grad * gradE=1/c^2(del^2)/(delt^2)(vecE) \ \ \ (8)`
Also, since we are working with waves we can expect that `vecE` will have periodic time variation at frequency `omega` for example `vec E(t)=E_0cos omegat` so a further simplification of equation 8 becomes:`1/n^2 grad * gradE=-(omega^2)/c^2(vecE) \ \ \ (9)`
In a resonator with perfect nodes at each end, only certain values of `omega` called the eigenfrequencies will be present. Since `omega/c` has units of a wave vector we will call this quantity `k_r=omega/c` so equation 9 becomes:`1/(n(x)^2) grad * gradE=-k_r^2(vecE) \ \ \ (10)`
To find the modes of a resonator, we will solve equation 10 for `k_r^2` and `E(x)`.Here we start with the finite difference equation as well a valid eigenvalue of that equation and iteratively solve for the eigenvector associated with that eigenvalue.
`a_i=m[i][i"+/"-1]` is an off diagonal element and `d_i=m[i][i]` where `m` is the tridiagonal finite difference equation matrix.
`((d_1-lambda,a_1,0,0),(a_2,d_2-lambda,a_2,0),(0,a_3,d_3-lambda,a_3),(0,0,a_4,d_4-lambda))((v_1),(v_2),(v_3),(v_4))=((0),(0),(0),(0))`
`(((d_1-lambda)v_1+a_1v_2),(a_2v_1+(d_2-lambda)v_2+a_2v_3), (a_3v_2+(d_3-lambda)v_3+a_3v_4),(a_4v_3+(d_4-lambda)v_4))=((0),(0),(0),(0))`
Let `v_1=b` where `b` is the left hand bouldary value. In order to have non-zero eigenvector elements, b cannot be zero but it can be made very small and the rest of the elements will be scaled to it. Then: First row:`v_2=-((d_1-lambda)v_1)/a_1`
Second row:`v_3=-(a_(2)v_2+(d_(i-1)-lambda)v_1)/a_2`
For all but the first and last rows:`v_i=-(a_(i-1)v_(i-2)+(d_(i-1)-lambda)v_(i-1))/a_(i-1)`
Last or Nth row:`v_N=-(d_N-lambda)*v_(N-1)/a_(N-1)`
So the algorithm is as follows:
`v_1=1;`
`v_2=-((d_1-lambda)v_1)/a_1;`
`"for"(i=3;i"<="N;i"++"")`
`v_i=-(a_(i-1)*v_(i-2)+d_(i-1)*v_(i-1))/a_(i-1)`
`v[N]=-(a_N-1)v_(N-1)")/a_N;`
`((d_11-lambda,a_12,0,0),(a_21,d_22-lambda,a_23,0),(0,a_32,d_33-lambda,a_34),(0,0,a_43,d_4-lambda))((v_1),(v_2),(v_3),(v_4))=((0),(0),(0),(0))`
`(((d_11-lambda)v_1+a_12v_2),(a_21v_1+(d_22-lambda)v_2+a_23v_3), (a_32v_2+(d_33-lambda)v_3+a_34v_4),(a41v_3+(d_44-lambda)v_4))=((0),(0),(0),(0))`