Consider the system of equations formulated in a previous question (optical fibers continuity of tangent field components across the core-cladding interface):

$$ \left\{ \begin{array}{c} e_z^{(1)}(r,\phi)|_{r = a} = e_z^{(2)}(r,\phi)|_{r = a}\\ e_{\phi}^{(1)}(r,\phi)|_{r = a} = e_{\phi}^{(2)}(r,\phi)|_{r = a}\\ h_z^{(1)}(r,\phi)|_{r = a} = h_z^{(2)}(r,\phi)|_{r = a}\\ h_{\phi}^{(1)}(r,\phi)|_{r = a} = h_{\phi}^{(2)}(r,\phi)|_{r = a} \end{array} \right. $$

Its determinant should be $0$, to obtain a non-trivial solution for the unknown amplitudes $C_1$, $C_2$, $D_1$, $D_2$. According to Maxwell's equations, $e_{\phi}$ and $h_{\phi}$ depend on the derivative of $e_z$ and $h_z$ with respect to $r$ and $\phi$. Books use some tricks to avoid having a function of $\phi$ in the computation of these derivatives. See for example John A. Buck, Fundamentals of Optical Fibers, which assumes (Table 3.2):

$$A(\phi) = \cos (\nu \phi), \ \mathrm{for} \ e_{\phi}\\ A(\phi) = \sin (\nu \phi), \ \mathrm{for} \ h_{\phi}$$

The resulting eigenvalue equation is:

$$\left[ \frac{J'_{\nu}(k_{c_1} a)}{k_{c_1}a J_{\nu}(k_{c_1} a)} + \frac{K'_{\nu}(|k_{c_2}| a)}{|k_{c_2}|a K_{\nu}(|k_{c_2}| a)} \right] \left[ \frac{n_1^2}{n_2^2}\frac{J'_{\nu}(k_{c_1} a)}{k_{c_1}a J_{\nu}(k_{c_1} a)} + \frac{K'_{\nu}(|k_{c_2}| a)}{|k_{c_2}|a K_{\nu}(|k_{c_2}| a)} \right] = \nu^2 \left( \frac{1}{k_{c_1}^2 a^2} + \frac{1}{|k_{c_2}|^2 a^2} \right)\left( \frac{n_1^2}{n_2^2} \frac{1}{k_{c_1}^2 a^2} + \frac{1}{|k_{c_2}|^2 a^2}\right)$$

Consider instead (respectively for the Electric and Magnetic fields)

$$A(\phi) = C_3 \sin(\theta) + C_4 \cos(\theta) = C' \cos (\nu \phi + \alpha)\\ A(\phi) = D_3 \sin(\theta) + D_4 \cos(\theta) = D' \cos (\nu \phi + \beta)$$

(in fact, according to this answer, $A(\phi) = A \sin(\theta) + B \cos(\theta)$ can be re-written as $A'\sin(\theta+\rho)$ or $A'\cos(\theta+\alpha)$). In the most general case, $\alpha \neq \beta$.

1) How would the eigenvalue equation be in this more general case?

2) Is there any textbook dealing with this computation?


2 Answers 2


It would be too long to copy the whole procedure here. If I did not make mistakes, the result should be:

$$\left[ \displaystyle \frac{n_1^2}{n_2^2} \frac{1}{a k_{c_1}} \frac{J'_{\nu} (a k_{c_1})}{J_{\nu} (a k_{c_1})} + \frac{1}{a |k_{c_2}|} \frac{K'_{\nu} (a |k_{c_2}|)}{K_{\nu} (a |k_{c_2}|)} \right] \cdot \left[ \displaystyle \frac{1}{a k_{c_1}} \frac{J'_{\nu} (a k_{c_1})}{J_{\nu} (a k_{c_1})} + \frac{1}{a |k_{c_2}|} \frac{K'_{\nu} (a |k_{c_2}|)}{K_{\nu} (a |k_{c_2}|)} \right] = \\ = - \tan (\nu \phi + \alpha) \tan (\nu \phi + \beta) \left( \displaystyle \frac{\nu \beta}{k_{c_1}^2} \right)^2 \left( \displaystyle \frac{1}{a^2 |k_{c_2}|^2} \right)^2 \displaystyle \frac{\omega^2 \mu_0 (n_1^2 - n_2^2)}{n_2^2}$$

I did not find any book dealing with this case. Maybe the most relevant difference with John A. Buck eigenvalue equation is the $\phi$ dependence. There is not a single eigenvalue equation for any $\phi$.


The $\phi$ dependence is given by the solution of the Helmholtz equation in cylindrical coordinates. It comes out to be the harmonic function $\exp(i m\phi)$ where $m$ is an integer equal to the order of the Bessel function. Due to the boundary conditions, these harmonic functions cancel out, so that the only remnants of the $\phi$-dependencies are factors of $m$. In the end, the determinant produces an equation: $$ \left(\frac{J'_m(Ua)}{U J_m(Ua)} + \frac{K'_m(Wa)}{W K_m(Wa)}\right) \left(\frac{n_1^2 J'_m(Ua)}{U J_m(Ua)} + \frac{n_2^2 K'_m(Wa)}{W K_m(Wa)}\right) = \left(\frac{m\beta}{a k}\right)^2 \left(\frac{1}{U^2} + \frac{1}{W^2}\right)^2 ,$$ where $U=\sqrt{n_1^2 k^2 - \beta^2}$ and $W=\sqrt{\beta^2 - n_2^2 k^2}$.

  • $\begingroup$ $e^{im\phi}$ is the solution assumed by D. Marcuse, Light Transmission Optics. What you state is substantially the procedure presented there and I agree. But if I correctly understood the Helmholtz equation, this is only one form for the representation of the solution. Note also that it is a complex quantity. Another form is the real quantity $a \cos(m \phi) + b \sin (m \phi)$, which is the one I assumed here. I can't state if they are equivalent. $\endgroup$
    – BowPark
    Jul 5, 2019 at 11:15
  • $\begingroup$ Yes one can use $\sin(m\phi)$ and $\cos(m\phi)$ instead of $\exp(i m\phi)$. Note however, the two trig functions represent different solutions replacing $\exp(i m\phi)$ and $\exp(-i m\phi)$. They would then cancel out in the boundary conditions. $\endgroup$ Jul 6, 2019 at 3:42

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