Help With Negative Phases in a Numerical Term

Click For Summary
The discussion focuses on resolving issues related to negative phases in a numerical term. The original poster has made progress with the numerator but is struggling to integrate the negative phases effectively. They suspect an error in their calculations, particularly near the value of 4.14. A key suggestion is to multiply "terms" by its complex conjugate when evaluating |terms|^2. Clarifying these points is essential for achieving a correct final result.
Blanchdog
Messages
56
Reaction score
22
Homework Statement
Verify that ##T^{tot}=\frac{n_2~cos \theta_2}{n_0~cos \theta_0}\frac{|t^{0\rightarrow1}|^2|t^{1\rightarrow2}|^2}{|e^{-ikd~cos\theta_1} - r^{0\leftarrow1}r^{1\rightarrow2}e^{ikd~cos\theta_1}|^2}## simplifies to ##T^{tot}=\frac{T^{max}}{1 + F~sin^2 \frac{\Phi}{2}}## assuming all angles are real. Assume that all light is s polarized as the equations are precisely the same for p polarized light in terms of Fresnel coefficients.
Relevant Equations
Included in image below because I didn't want to have to LaTeX them all.
Finesse attempt pt 1.png

Finesse attempt pt2.png

I think I've got the numerator part figured out, but I'm really stuck on what to do with those negative phases in the last term and how to get this to all come together in the end. I feel like I must have made a mistake somewhere, but can't find it. Thanks in advance for the help!
 
Physics news on Phys.org
One error that I think I spotted is right by 4.14: In evaluating ## |terms|^2 ##, you need to multiply "terms" by its complex conjugate.
 
At first, I derived that: $$\nabla \frac 1{\mu}=-\frac 1{{\mu}^3}\left((1-\beta^2)+\frac{\dot{\vec\beta}\cdot\vec R}c\right)\vec R$$ (dot means differentiation with respect to ##t'##). I assume this result is true because it gives valid result for magnetic field. To find electric field one should also derive partial derivative of ##\vec A## with respect to ##t##. I've used chain rule, substituted ##\vec A## and used derivative of product formula. $$\frac {\partial \vec A}{\partial t}=\frac...