Relativistic cyclotron frequency

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Discussion Overview

The discussion revolves around the concept of relativistic cyclotron frequency, particularly focusing on the role of transverse mass and momentum in the context of charged particles moving in a magnetic field. Participants explore the relationships between acceleration, velocity, and forces in a relativistic framework.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests that since acceleration is transverse to velocity, the transverse mass should be considered in the formula for centripetal force, proposing a relationship involving gamma, velocity, radius, charge, and magnetic field.
  • Another participant inquires about expressing centripetal force in terms of momentum and seeks clarification on the definition of relativistic momentum.
  • A later reply confirms the initial suggestion about using transverse mass, defining it as gamma times mass.
  • Another participant presents an alternative expression for the centripetal force in terms of momentum, indicating a direct relationship between momentum, charge, magnetic field, and radius.

Areas of Agreement / Disagreement

Participants express differing views on the best way to approach the problem, particularly regarding the use of transverse mass and the formulation of centripetal force. The discussion remains unresolved with multiple competing perspectives presented.

Contextual Notes

Participants do not fully resolve the definitions or implications of transverse mass and relativistic momentum, leaving some assumptions and mathematical steps unaddressed.

jdstokes
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Since the acceleration is transverse to the velocity, should we consider the transverse mass in the formula [itex]mv^2/r[/itex] ie

[itex]\gamma \frac{mv^2}{r} = qvB \implies \frac{v}{\sqrt{1-(v/c)^2}} = \frac{qBr}{m}[/itex]?
 
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How can you express the centripetal force in terms of momentum (with other terms)?
How is the relativistic momentum?
 
jdstokes said:
Since the acceleration is transverse to the velocity, should we consider the transverse mass in the formula [itex]mv^2/r[/itex] ie

[itex]\gamma \frac{mv^2}{r} = qvB \implies \frac{v}{\sqrt{1-(v/c)^2}} = \frac{qBr}{m}[/itex]?
Yes. But just to be clear, using your symbols, the transverse mass = gamma*m.

Pete
 
It's clearer in terms of momentum pv/r=qvB-->p=qBr.
 

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