Proving the Relationship Between \omega, \lambda and c

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SUMMARY

The discussion focuses on proving the relationship between the variables \(\omega\), \(\lambda\), and \(c\) through the equation \(\frac{dc}{d\omega}=\frac{1}{k}\left(1-\frac{c}{v_{g}}\right)\). The user initially struggled with the algebraic manipulation required to derive the relationship but ultimately resolved the issue by correcting a mistake in their calculations. The key takeaway is the interdependence of \(\omega\), \(\lambda\), and \(c\) as functions of each other, emphasizing the importance of accurate algebraic handling in physics problems.

PREREQUISITES
  • Understanding of calculus, specifically differentiation
  • Familiarity with wave mechanics concepts, including phase velocity and group velocity
  • Knowledge of algebraic manipulation techniques
  • Basic comprehension of functions and their interdependencies
NEXT STEPS
  • Study the derivation of wave equations in physics
  • Learn about the relationship between phase velocity and group velocity in wave propagation
  • Explore advanced calculus techniques for solving differential equations
  • Investigate the implications of variable interdependencies in physical systems
USEFUL FOR

Students and educators in physics, particularly those studying wave mechanics and calculus, as well as anyone interested in the mathematical relationships between physical quantities.

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Homework Statement



Show that:

\frac{dc}{d\omega}=\frac{1}{k}\left(1-\frac{c}{v_{g}}\right)

The Attempt at a Solution



\frac{dc}{d\omega}=\frac{dc}{d\omega}\frac{d\omega}{d\omega}=\frac{dc}{d\omega}\frac{dck}{d\omega}=\frac{dc}{d\omega}k+\frac{dk}{d\omega}c=\frac{dc}{d\omega}k+\frac{c}{v_{g}}=\left\frac{1}{1-k}\right\frac{c}{v_{g}}

...Yeah, I have no clue. \omega, \lambda and c are all functions of each other, and I'm confused :(.
 
Last edited:
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Nevermind, solved it. I was really close, I just made a stupid mistake with my algebra.
 

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