Rearranging the equation for the cutoff condition in optical fibers

[EmilyRuck]

Hello!

In Optical fibers, let $k_1$ and $k_2$ be respectively the propagation constants in core and cladding, $\beta$ the propagation costant of a mode along the direction $z$, $a$ the radius of the fiber. Using the normalized quantities $u=a \sqrt{k_1^2 − \beta^2}$ and $w=a \sqrt{\beta^2 − k_2^2}$, the characteristic equation for modes is:

$$\left[ \displaystyle \frac{J'_{\nu} (u)}{u J_{\nu}(u)} + \frac{K'_{\nu} (w)}{w K_{\nu}(w)} \right] \left[ \displaystyle \frac{k_1^2 J'_{\nu} (u)}{u J_{\nu}(u)} + \frac{k_2^2 K'_{\nu} (w)}{w K_{\nu}(w)} \right] = \nu^2 \beta^2 \left( \displaystyle \frac{1}{u^2} + \frac{1}{w^2} \right)^2$$

where $n_1$ and $n_2$ are the refractive indices of core and cladding, $\nu$ is the order of Bessel functions.

As specified in this document, page 15, the cutoff condition for modes is obtained taking the limit of the above expression for $w \to 0$. But, before this, I can't obtain the same expression as in the document. Consider:

$$\frac{J'_{\nu} (u)}{u J_{\nu}(u)} = \frac{J_{\nu - 1} (u)}{u J_{\nu}(u)} - \frac{\nu}{u^2} = \xi_1(u) - \frac{\nu}{u^2}\\
\frac{K'_{\nu} (w)}{w K_{\nu}(u)} = \frac{K_{\nu - 1} (w)}{w K_{\nu}(w)} + \frac{\nu}{w^2} = \xi_2(w) + \frac{\nu}{w^2}$$

Substituting in the characteristic equation, dividing both sides by $k_1^2$ and rearranging the terms, I obtain:

$$\xi_1^2(u) + \xi_1(u) \left[ \frac{k_1^2 + k_2^2}{k_1^2} \cdot \xi_2(w) - \nu \left( \frac{2}{u^2} - \frac{k_1^2 + k_2^2}{k_1^2} \cdot \frac{1}{w^2} \right) \right] + \left[ \frac{k_2^2}{k_1^2} \cdot \xi_2^2(w) - \xi_2(w) \nu \left( \frac{k_1^2 + k_2^2}{k_1^2} \cdot \frac{1}{u^2} - 2 \frac{k_2^2}{k_1^2} \cdot \frac{1}{w^2} \right) \right] + \left( \frac{\nu}{u^2} \right)^2 - \frac{\nu^2}{u^2 w^2} - \frac{k_2^2}{k_1^2} \cdot \frac{\nu^2}{u^2 w^2} + \frac{k_2^2}{k_1^2} \left( \frac{\nu}{w^2} \right)^2 - \frac{1}{k_1^2} \left( \frac{\nu^2 \beta^2}{u^4} + 2 \frac{\nu^2 \beta^2}{u^2 w^2} + \frac{\nu^2 \beta^2}{w^4} \right) = 0$$

I double checked this result, but the document shows instead:

$$\xi_1^2(u) - \xi_1(u) \left[ \frac{k_1^2 + k_2^2}{k_1^2} \cdot \xi_2(w) + \nu \left( \frac{2}{u^2} + \frac{k_1^2 + k_2^2}{k_1^2} \cdot \frac{1}{w^2} \right) \right] + \left[ \frac{k_2^2}{k_1^2} \cdot \xi_2^2(w) + \xi_2(w) \nu \left( \frac{k_1^2 + k_2^2}{k_1^2} \cdot \frac{1}{u^2} + 2 \frac{k_2^2}{k_1^2} \cdot \frac{1}{w^2} \right) \right] = 0$$

First, terms with $\frac{k_1^2 + k_2^2}{k_1^2}$ have swapped signs; moreover, I can't figure out how it could be

$$\left( \frac{\nu}{u^2} \right)^2 - \frac{\nu^2}{u^2 w^2} - \frac{k_2^2}{k_1^2} \cdot \frac{\nu^2}{u^2 w^2} + \frac{k_2^2}{k_1^2} \left( \frac{\nu}{w^2} \right)^2 - \frac{1}{k_1^2} \left( \frac{\nu^2 \beta^2}{u^4} + 2 \frac{\nu^2 \beta^2}{u^2 w^2} + \frac{\nu^2 \beta^2}{w^4} \right) = 0$$

Is the result in the document correct? Or am I doing something wrong? As an alternative, I'm also looking for a textbook which deals with the procedure to obtain the cutoff condition.EmilyP. S.
The linked document should be similar, or equal, to this article:

Cylindrical Dielectric Waveguide Modes, E. Snitzer, Journal of the Optical Society of America Vol. 51, Issue 5, pp. 491-498 (1961)

which I can't find for free.

 

[EmilyRuck]

Temporarily putting aside the $\frac{k_1^2 + k_2^2}{k_1^2}$ terms signs, consider the part which should be $0$. The first 4 terms come from the espansion of the LHS (which involves $\xi_1$, $\xi_2$) of the original characteristic equation. The last 3 terms directly come from the RHS of the characteristic equation, divided by $k_1^2$. Considering the homologous terms:

$$\frac{\nu^2}{u^4} \left( 1 - \frac{\beta^2}{k_1^2} \right) - \frac{\nu^2}{u^2 w^2} \left( \frac{k_1^2 + k_2^2}{k_1^2} + 2 \frac{\beta^2}{k_1^2} \right) + \frac{\nu^2}{w^4} \left( \frac{k_2^2}{k_1^2} - \frac{\beta^2}{k_1^2} \right) = \\
= \frac{\nu^2}{u^4} \left( \frac{k_1^2 - \beta^2}{k_1^2} \right) - \frac{\nu^2}{u^2 w^2} \left( \frac{k_1^2 + k_2^2 + 2\beta^2}{k_1^2} \right) + \frac{\nu^2}{w^4} \left( \frac{k_2^2 - \beta^2}{k_1^2} \right)$$

Now, $u = a k_{t_1}$ and $w = a |k_{t_2}|$. $k_{t_1}$ is, in the core, the wavevector component orthogonal to the direction of propagation: the transverse wavevector; $k_{t_2}$ is its homologous in the cladding, but for a confined mode it is pure imaginary, therefore $k_{t_2}^2 = - |k_{t_2}|^2$. Then,

$$k_1^2 - \beta^2 = k_{t_1}^2 = \frac{u^2}{a^2}\\
k_2^2 - \beta^2 = - |k_{t_2}|^2 = - \frac{w^2}{a^2}\\
k_{t_1}^2 - |k_{t_2}|^2 = k_1^2 + k_2^2 - 2 \beta^2 \Rightarrow k_1^2 + k_2^2 - 2 \beta^2 = \frac{u^2 - w^2}{a^2}$$

Substituting in the above equation:

$$\frac{\nu^2}{a^2 u^2} - \frac{\nu^2}{a^2 w^2} + \frac{\nu^2}{a^2 u^2} - \frac{\nu^2}{u^2 w^2} 4 \beta^2 - \frac{\nu^2}{a^2 w^2} = \\
= 2 \frac{\nu^2}{a^2 u^2} - 2\frac{\nu^2}{a^2 w^2} - 4 \frac{\nu^2}{u^2 w^2} \beta^2 = 2 \frac{\nu^2}{a^2} \cdot \frac{w^2 - u^2}{u^2 w^2} - 4 \frac{\nu^2}{u^2 w^2} \beta^2 = \\
= - 2 \nu^2 \cdot \frac{k_1^2 + k_2^2 - 2 \beta^2}{u^2 w^2} - 4 \frac{\nu^2}{u^2 w^2} \beta^2 = - 2 \nu^2 \cdot \frac{k_1^2 + k_2^2}{u^2 w^2}$$

I hope I did not make errors. But, still, it doesn't seem that this expression can be $0$.

 

[EmilyRuck]

The first expression, which is correct, is written using formula (A4) of the linked document:

$$\frac{J'_{\nu} (u)}{u J_{\nu}(u)} = \frac{J_{\nu - 1} (u)}{u J_{\nu}(u)} - \frac{\nu}{u^2} = \xi_1(u) - \frac{\nu}{u^2}$$

Formula (A6), used for the second expression, is wrong. It should be:

$$\frac{K'_{\nu} (w)}{w K_{\nu}(u)} = - \frac{K_{\nu - 1} (w)}{w K_{\nu}(w)} - \frac{\nu}{w^2}$$

And therefore the second expression should be:

$$\frac{K'_{\nu} (w)}{w K_{\nu}(u)} = - \frac{K_{\nu - 1} (w)}{w K_{\nu}(w)} - \frac{\nu}{w^2} = - \xi_2(w) - \frac{\nu}{w^2}$$

Substituting in the characteristic equation, the mentioned issues are no more present. In fact, signs are correct and the extra-term becomes:

$$\left( \frac{\nu}{u^2} \right)^2 + \frac{\nu^2}{u^2 w^2} + \frac{k_2^2}{k_1^2} \cdot \frac{\nu^2}{u^2 w^2} + \frac{k_2^2}{k_1^2} \left( \frac{\nu}{w^2} \right)^2 - \frac{1}{k_1^2} \left( \frac{\nu^2 \beta^2}{u^4} + 2 \frac{\nu^2 \beta^2}{u^2 w^2} + \frac{\nu^2 \beta^2}{w^4} \right)$$

Consider the homologous terms, it becomes:

$$\frac{\nu^2}{u^4} \left( 1 - \frac{\beta^2}{k_1^2} \right) - \frac{\nu^2}{u^2 w^2} \left( - \frac{\left( k_1^2 + k_2^2 \right)}{k_1^2} + 2 \frac{\beta^2}{k_1^2} \right) + \frac{\nu^2}{w^4} \left( \frac{k_2^2}{k_1^2} - \frac{\beta^2}{k_1^2} \right)$$

Substituting

$$ - \left( k_1^2 + k_2^2 \right) + 2 \beta^2 = \frac{w^2 - u^2}{a^2}\\
k_1^2 - \beta^2 = \frac{u^2}{a^2}\\
k_2^2 - \beta^2 = - \frac{w^2}{a^2}$$

The whole term vanishes.

The final result is, as expected,

$$\xi_1^2(u) - \xi_1(u) \left[ \frac{k_1^2 + k_2^2}{k_1^2} \cdot \xi_2(w) + \nu \left( \frac{2}{u^2} + \frac{k_1^2 + k_2^2}{k_1^2} \cdot \frac{1}{w^2} \right) \right] + \left[ \frac{k_2^2}{k_1^2} \cdot \xi_2^2(w) + \xi_2(w) \nu \left( \frac{k_1^2 + k_2^2}{k_1^2} \cdot \frac{1}{u^2} + 2 \frac{k_2^2}{k_1^2} \cdot \frac{1}{w^2} \right) \right] = 0$$