electromagnetism - Optical fibers field analysis, $phi$ dependence not considered

In optical fibers, the core (with $\varepsilon = \varepsilon_1$) is usually represented by a cylinder of radius $a$, whose axis coincides with the $z$-axis in cylindrical coordinates. The cladding is the external region, with $\varepsilon = \varepsilon_2$. The longitudinal Electric field is assumed to be:

$$e_z(r,\phi) = B(r)A(\phi)$$

the well-known dependence $e^{- j \beta z}$ on $z$ has been omitted. $B(r)$ is the solution of a Bessel (in the core) or Modified Bessel (outside the core) Differential Equation; $A(\phi)$ is the solution of a Helmholtz equation (the same, for both the core and the outside).

In the core:

$$B(r) = C_1 J_{\nu}(k_{c_1} r)$$

Outside the core:

$$B(r) = C_2 K_{\nu}(|k_{c_2}| r)$$

In both regions:

$$A(\phi) = C_3 \sin (\nu \phi) + C_4 \cos (\nu \phi)$$

where $k_{c_1}^2 = k_1^2 - \beta$ and $|k_{c_2}|^2 = \beta - k_2^2$.

The Magnetic field $h_z(r,\phi) = B(r)A(\phi)$ has the same form, but it is featured by different amplitudes $D_1$, $D_2$, $D_3$, $D_4$.

The field components which are tangent to the interface $r=a$ between the core and the cladding must be continuous across it:

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

In textbooks, only the quantities $C_1$, $C_2$, $D_1$, $D_2$ are presented as the unknowns of this system.

$C_3$, $C_4$, $D_3$, $D_4$ are uncorrelated to them (the equation for $A(\phi)$ is different from the equation for $B(r)$), and they actually represent a further degree of freedom. But why are they never considered?

Answer

Because $A(\phi)$ is the same in both the core and cladding, you can't use the core cladding boundary conditions determine $C_3$, $C_4$, $D_3$, and $D_4$.

Instead, these will be determined by normalization, $C_3^2+C_4^2=1$. And because of cylindrical symmetry, we can arbitrarily choose $C_3=0$, $C_4=1$ (I chose this so the $\nu=0$ case doesn't have to be taken as a special case). Any other choice that respects the normalization is just a rotation of our coordinate system that doesn't change the behavior of the modes, because $A\sin(\theta)+B\cos(\theta)$ can be re-written as $A'\sin(\theta+\rho)$.