The relation for total energy (E ) and momentum (p) for a relativistic particle is E 2 = c2 p2 + m2c4, where m is the rest mass and c is the velocity of light. Using the relativistic relations E = ω and p = k, where ω is the angular frequency and k is the wave number, show that the product of group velocity (vg) and the phase velocity (vp) is equal to c2, that is vpvg = c2
The relation for total energy (E ) and momentum (p) for a relativistic particle is E 2 = c2 p2 + m2c4, where m is the rest mass and c is the velocity of light. Using the relativistic relations E = ω and p = k, where ω is the angular frequency and k is the wave number, show that the product of group velocity (vg) and the phase velocity (vp) is equal to c2, that is vpvg = c2
Related questions
Question
The relation for total energy (E ) and momentum (p) for a relativistic particle
is E 2 = c2 p2 + m2c4, where m is the rest mass and c is the velocity of light.
Using the relativistic relations E = ω and p = k, where ω is the angular
frequency and k is the wave number, show that the product of group velocity
(vg) and the phase velocity (vp) is equal to c2, that is vpvg = c2
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 2 steps
