# Rearranging the equation for the cutoff condition in optical fibers

• EmilyRuck

#### 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.

Emily

P. S.

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

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##.

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,

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