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Physics
Classical Physics
Optics
Rearranging the equation for the cutoff condition in optical fibers
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[QUOTE="EmilyRuck, post: 6322817, member: 416886"] The first expression, which is correct, is written using formula (A4) of the [URL='https://apps.dtic.mil/dtic/tr/fulltext/u2/674600.pdf']linked document[/URL]: $$\frac{J'_{\nu} (u)}{u J_{\nu}(u)} = \frac{J_{\nu - 1} (u)}{u J_{\nu}(u)} - \frac{\nu}{u^2} = \xi_1(u) - \frac{\nu}{u^2}$$ 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, $$\xi_1^2(u) - \xi_1(u) \left[ \frac{k_1^2 + k_2^2}{k_1^2} \cdot \xi_2(w) + \nu \left( \frac{2}{u^2} + \frac{k_1^2 + k_2^2}{k_1^2} \cdot \frac{1}{w^2} \right) \right] + \left[ \frac{k_2^2}{k_1^2} \cdot \xi_2^2(w) + \xi_2(w) \nu \left( \frac{k_1^2 + k_2^2}{k_1^2} \cdot \frac{1}{u^2} + 2 \frac{k_2^2}{k_1^2} \cdot \frac{1}{w^2} \right) \right] = 0$$ [/QUOTE]
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Classical Physics
Optics
Rearranging the equation for the cutoff condition in optical fibers
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