What is the right way to calculating the acoustic reflection/transmission coefficients? Now, I've gone over my calculations a dozen times, and I still can't spot a mistake, so I want to know wheather it's possible to have a transmission coefficient greater than one for certain values of energy, like in this scenario. You have to specify how you arrive at this T>1 result and give some pertinent formulas. What is the cost of health care in the US? Also did you mention that for one of $\psi_{II}$ and $\psi_{III}$ : $\frac{d}{dx} = \frac{d}{d\xi}$ and for the other $\frac{d}{dx} = -\frac{d}{d\xi}$, Meaning of transmission coefficient greater than one in a potential well problem, mathworld.wolfram.com/AiryDifferentialEquation.html, “Question closed” notifications experiment results and graduation, MAINTENANCE WARNING: Possible downtime early morning Dec 2/4/9 UTC (8:30PM…. The disease will stay alive and stable, but there won’t be an outbreak or an epidemic. Now, I've gone over my calculations a dozen times, and I still can't spot a mistake, so I want to know wheather it's possible to have a transmission coefficient greater than one for certain values of energy, like in this scenario. When are they same when are they different? Plotting $\psi$ for finite square well potential. Why did you set (1,3), (1,4), (3,5) and (3,6) with -1 multiplier? I followed the method (see file CAL.png) but the results seems incorrect. Asking for help, clarification, or responding to other answers. See, how far such wavelengths can propagate. $\begingroup$ A laser amplifier has a transmission coefficient greater than 1 (aka gain), for some range of wavelengths. 0 & 0 & \mathrm{Ai}(-\xi_0) & \mathrm{Bi}(-\xi_0) & -\mathrm{Ai}(-\xi_0) & -\mathrm{Bi}(-\xi_0) \\ © 2008-2020 ResearchGate GmbH. I'm trying to find the reflection and transmission coefficients for a stream of electrons coming from $x=-\infty$ (so $E>0$, no bound states involved). Also see this article. Does anybody know how can I order figures exactly in the position we call in Latex template? The classic case of having activity coefficient greater than 1 is the solution of very hydrophobic organic solvents in water. This does not violate any physical laws. The reflection coefficient of any device (S11), can it be greater than one? $\endgroup$ – Jon Custer Jul 25 '16 at 12:55 where use was made of the Wronskian for Airy functions: $$\mathcal{W}[\mathrm{Ai}(z),\mathrm{Bi}(z)]=\pi^{-1}.$$. I've tried $V_0=0$, but the algorhitm crashes because somewhere a division with zero occurs (since $A\propto V_0$). I am working on a metal-dielectric geometry. If some light is absorbed by the substance, then the transmitted light will be a combination of the wavelengths of the light that was transmitted and not absorbed. It probably might also be because of the numerical approximations assumed by the software package you use for calculating the coefficients (due to the medium properties and existing external properties involved in a study, it generally is not be possible to to get exact numerical solution). Consider the finite 1D wedge-shaped potential well given by, $$V(x)=V_0\left(\frac{|x|}{a}-1\right) \hspace{10pt}\mathrm{for}\hspace{3pt} |x|a.$$. incident wave . I haven't studied your solution in detail, but, for what it's worth, the formulas for the solutions of Airy equation differ depending on the sign of the coefficient. Does a metal have One, Zero or Infinite Relative Permittivity? Well, if true than you just have invented a perpetuum mobile, or, in other words, energy from nothing. What does it mean physically? Indian Institute of Technology (ISM) Dhanbad. Using of the rocket propellant for engine cooling. All was well untill I plugged the analytic solution for the transmission coefficient into Python to render a graph and got this. @Sobanoodles: That's not what I mean. In such cases, how do you define transmission coefficient? Join ResearchGate to find the people and research you need to help your work. F This does not violate any physical laws. When they are nearly equal, then the apparent result may easily exceed 1. I'm aware that this breaks conservation of energy (red alert! then find Why Is an Inhomogenous Magnetic Field Used in the Stern Gerlach Experiment? @ Dipantil, The reflection coefficient may be positive or negative so the transmission coefficient may be greater than unity. rev 2020.11.24.38066, The best answers are voted up and rise to the top, Physics Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us. We are not able to install and run the softare. This type of work was a popular research topic in the 50-70s. Has it something to do with the change of variables discussed in akhmeteli's answer and comments? The currents are given by, $$J_I=\frac{\hbar}{m}\Im\left(\psi_I^*\frac{d\psi_I}{dx}\right)=\frac{\hbar}{m}(|A|^2-|B|^2)=J_{in}-J_{ref}$$ where $\mathrm{Ai}(z)$ and $\mathrm{Bi}(z)$ are the standard Airy functions of the first and second kind. You may wish to check if your solutions give correct results for $V_0=0$. Where should small utility programs store their preferences? Reflection Coefficients … Now, I've gone over my calculations a dozen times, and I still can't spot a mistake, so I want to know wheather it's possible to have a transmission coefficient greater than one for certain values of energy, like in this scenario. (For example, try light incident from a medium of n 1 =1.5 upon a medium of n 2 =1.0 with an angle of incidence of 30°.) Since $T+R=1$, it follows that $|A|^2-|B|^2=|G|^2$. What would it even mean for the coefficient to be higher than 1? Since I've got the Airy equation in the form $\frac{d^2\psi}{d\xi^2}+\xi\psi=0$, I've simply declared that the solutions are of the form $\psi(\xi)=K\mathrm{Ai}(-\xi)+J\mathrm{Bi}(-\xi),$ which is justified as a change of variables $\xi=-z$ in the conventional Airy equation $\frac{d^2\psi}{dz^2}-z\psi=0$.