Group velocity of relativistic wave packet

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SUMMARY

The discussion centers on the calculation of group and phase velocities for a relativistic wave packet. The user derives the phase velocity as v_phase = c * sqrt(1 + (rest mass * c / p)^2) and the group velocity as v_group = c / sqrt(1 + (rest mass * c / p)^2), confirming that the product of phase and group velocities equals c^2. The results align with established physics principles, where the phase velocity can exceed the speed of light while the group velocity remains below it, consistent with the behavior of particles described by wave packets.

PREREQUISITES
  • Understanding of relativistic energy-momentum relationship
  • Familiarity with Planck-Einstein relations
  • Knowledge of wave packet theory
  • Basic concepts of phase and group velocities
NEXT STEPS
  • Study the derivation of the energy-momentum relationship in special relativity
  • Learn about the implications of phase and group velocities in quantum mechanics
  • Explore the concept of tachyons and their theoretical implications
  • Review modern physics textbooks that cover wave-particle duality
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Physicists, students of modern physics, and anyone interested in the behavior of relativistic particles and wave packets.

L-adara7
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Hi, I'm trying to figure out what the group and phase velocities of a wave packet describing a particle are in the relativistic case.

I started with the relationship between energy and impulse : E squared = p squared X c squared + rest mass squared X c to the fourth. In this, I input the Planck-Einstein relationships : E = h bar X omega and p = h bar X k.

I divide each side by h bar squared, take the square root to get omega, then divide by k to get the phase velocity or take the derivative relative to k to get the group velocity.

Here is what I get : phase velocity = c X square root of (one + (rest mass X c/p)squared)

group velocity = c / square root of (one + (rest mass X c/p)squared)

So that v phase X v group = c squared

I could not find this result anywhere so I have a doubt : can anyone tell me if I'm doing this correctly?

If it's correct, it's pretty interesting : we have a group velocity that is always less than c (as it should) and a phase velocity that is always greater than c, as if the particle was governed by waves that can't travel at less than the speed of light, just like tachyons...

Anyway, I'd be happy just knowing if I have the correct result for this, thanks for helping.
 
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Yes, that result is correct. Many introductory modern physics textbooks derive

[tex]v_{phase} = \frac {c^2} {v_{particle}}[/tex]

[tex]v_{group} = v_{particle}[/tex]

which are equivalent to your result.

By the way, you can make it easier for others to read your equations by studying this thread:

https://www.physicsforums.com/showthread.php?t=8997
 
L-adara7 said:
If it's correct, it's pretty interesting : we have a group velocity that is always less than c (as it should) and a phase velocity that is always greater than c, as if the particle was governed by waves that can't travel at less than the speed of light, just like tachyons...
QUOTE]

The phase velocity>c has nothing to do with the velocity of the particle, which is given by the group velocity<c, unlike tachyons.
 

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