Factorising Harder Quadratics Revision Notes. (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°.) &\left.-\left(\left(\frac{ka}{q_0}\right)^2\mathrm{Ai}(-\xi_a)\mathrm{Bi}(-\xi_a)+\mathrm{Ai}'(-\xi_a)\mathrm{Bi}'(-\xi_a)\right)(\mathrm{Ai}(-\xi_0)\mathrm{Bi}'(-\xi_0)+\mathrm{Ai}'(-\xi_0)\mathrm{Bi}(-\xi_0)) \right], This is one more example that the raw result of measurement, without estimated uncertainty ("error bounds") is next to worthless. Thanks for contributing an answer to Physics Stack Exchange! For example, a blue light filter appears blue because it absorbs red and green wavelengths. It may seem surprising that t can be greater than unity. It is possible to have a reflection coefficient greater than 1 without violating any passivity conditions. Well, if true than you just have invented a perpetuum mobile, or, in other words, energy from nothing. site design / logo © 2020 Stack Exchange Inc; user contributions licensed under cc by-sa. What is the cost of health care in the US? If they travel less than a unit wavelength, then they are damped, evanescent waves. The time-independant Schrödinger equation for the wavefunction on each section reads, $$\frac{d^2\psi_I}{dx^2}+k^2\psi_I=0$$ Consider the finite 1D wedge-shaped potential well given by, $$V(x)=V_0\left(\frac{|x|}{a}-1\right) \hspace{10pt}\mathrm{for}\hspace{3pt} |x|a.$$. Try to calculate the attenuation due to dispersion for those wavelengths in the medium for which you are getting transmission coefficient greater than 1. Calculate the reflection probability instead Change $\psi_{IV}$ to $G e^{-i k x}$ If a body explodes then Coefficient of restitution can be greater than 1. If your transmission coefficient is the ratio of amplitudes then, Institute of Physics of the Polish Academy of Sciences. When they are nearly equal, then the apparent result may easily exceed 1. e^{-ika} & e^{ika} & -\mathrm{Ai}(-\xi_a) & -\mathrm{Bi}(-\xi_a) & 0 & 0 \\ $$\psi_{IV}(x)=Ge^{ikx},$$. Join ResearchGate to find the people and research you need to help your work. 0 \\ Using the fact that both $\psi(x)$ and $\psi'(x)$ are continuous, this gives the linear system: $$\left(\begin{matrix} \end{equation*}. More seriously: your transmission coefficient is a ratio of two uncertain (noisy) quantities. The penetration depth of the wave is not determined by the amplitude alone. A \\ a laser or maser). Axiomatic development of electrodynamics. Die Theorie zum Elektromagnetismus wird als Elektrodynamik bezeichnet. If your transmission coefficient is the ratio of power in to power out then Marek's answer is correct, unless you have a medium or boundary with gain (e.g. It may be calculated from the above system using Cramer's rule. \end{matrix} \right)=\left( \begin{matrix} Ge^{ika} \\ E \\ Why are Stratolaunch's engines so far forward? $$\frac{d^2\psi_{III}}{dx^2}+(\frac{x}{a}+\epsilon^2-1)k_0^2\psi_{III}=0$$ The reflection coefficient c may be positive or negative so the transmission coefficient t may be greater than unity. where $\mathrm{Ai}(z)$ and $\mathrm{Bi}(z)$ are the standard Airy functions of the first and second kind. What does it mean physically? a laser or maser). What would it even mean for the coefficient to be higher than 1? 0 & 0 & 0 & 0 & \mathrm{Ai}(-\xi_a) & \mathrm{Bi}(-\xi_a) \\ This does not violate any physical laws. A&=\frac{iq_0\pi^2e^{2ika}}{ka} \left[\left(\left(\frac{ka}{q_0}\right)^2\mathrm{Ai}^2(-\xi_a)+\mathrm{Ai}'^2(-\xi_a)\right)\mathrm{Bi}(-\xi_0)\mathrm{Bi}'(-\xi_0)\right.\\ &+\left(\left(\frac{ka}{q_0}\right)^2\mathrm{Bi}^2(-\xi_a)+\mathrm{Bi}'^2(-\xi_a)\right)\mathrm{Ai}(-\xi_0)\mathrm{Ai}'(-\xi_0)\\ reply from a potential PhD advisor? It follows that $T=\frac{J_{tr}}{J_{in}}=\frac{1}{|A|^2}$ and $R=1-\frac{1}{|A|^2}$. Has it something to do with the change of variables discussed in akhmeteli's answer and comments? Sorry, i pressed enter and the page sent my brocken comment. By using our site, you acknowledge that you have read and understand our Cookie Policy, Privacy Policy, and our Terms of Service. Where should small utility programs store their preferences? In case of few heterostructure problem, the transmission vs fermi energy curve shows more than 1. what does it signify? See this, National Institute of Science Education and Research. I know quantum espresso can calculate transport. Can the President of the United States pardon proactively? Now, I've gone over my calculations a dozen times, and I still can't spot a mistake, so I want to know wheather it's possible to have a transmission coefficient greater than one for certain values of energy, like in this scenario. It only takes a minute to sign up. When are they same when are they different? Why does Chrome need access to Bluetooth? 0 & 0 & \mathrm{Ai}(-\xi_0) & \mathrm{Bi}(-\xi_0) & -\mathrm{Ai}(-\xi_0) & -\mathrm{Bi}(-\xi_0) \\ $$J_{IV}=\frac{\hbar}{m}\Im\left(\psi_{IV}^*\frac{d\psi_{IV}}{dx}\right)=\frac{\hbar}{m}|G|^2=J_{tr}$$. I've found this as a related question, but I'm not really satisfied with the answers. Why did you set (1,3), (1,4), (3,5) and (3,6) with -1 multiplier? Cutting out most sink cabinet back panel to access utilities. In other words, I didn't forget about setting that amplitude to 0, but thank you for pointing that out. The classic case of having activity coefficient greater than 1 is the solution of very hydrophobic organic solvents in water. For some wavelengths, the transmission coefficient is going above 1. It's nothing to be alarmed about. 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. In order to do that, I've divided the domain into 4 parts (like in the picture), solved the Schrödinger equation on each one, got a linear system for the coefficients of each wavefunction and solved it. The disease will stay alive and stable, but there won’t be an outbreak or an epidemic.

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